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SENIOR, SECONDARY SCHOOL, MATHEMATICS, FOR CLASS 11, , [In accordance with the latest CBSE syllabus], , R S Aggarwal, MSc, PhD
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Preface, It gives me great pleasure in presenting the new edition of this book. In this, edition, the modifications have been dictated by the changes in the CBSE, syllabus. The structure and the methods used in the previous editions, which, have been appreciated by teachers using the book in classroom conditions,, remain unchanged., In this edition certain topics that are no longer a part of the CBSE syllabus, have been retained so that students do not face difficulty in competitive, examinations, etc. These topics include conditional identities involving the, angles of a triangle, infinite geometric series and exponential series., The main consideration in writing the book was to present the considerable, requirements of the syllabus in as simple a manner as possible. Special, attention has been paid to the gradation of problems. This will help students, gain confidence in problem-solving., One problem faced by students is the lack of a comprehensive and, carefully selected set of solved problems in textbooks of this kind. I have given, due weightage to this aspect. Each set of solved examples is followed by a, comprehensive exercise section in which students will get enough questions, for practice. Hints have been given to the more difficult questions. Students, should take their help as a last resort., I have received many suggestions and letters of appreciation from teachers, all over the country. I thank them all for contributing in the improvement of, the book and for their encouragement. I hope they will like this edition as well., And as always, I would like to hear their views on the book., R S Aggarwal, , (iii)
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Mathematics Syllabus, For Class 11, , UNIT I. Sets and Functions, 1. Sets, 20 Periods, Sets and their representations. Empty set. Finite and Infinite sets., Equal sets. Subsets. Subsets of a set of real numbers especially intervals, (with notation). Power set. Universal set. Venn diagrams. Union and, Intersection of sets. Difference of sets. Complement of a set. Properties, of complement., 2. Relations and Functions, 20 Periods, Ordered pairs, Cartesian product of sets. Number of elements in the, cartesian product of two finite sets. Cartesian product of the set of reals, with itself (up to R × R × R). Definition of relation, pictorial diagrams,, domain, co-domain and range of a relation. Function as a special type of, relation. Pictorial representation of a function, domain, co-domain and, range of a function. Real valued functions, domain and range of these, functions, constant, identity, polynomial, rational, modulus, signum;, exponential, logarithmic and greatest integer functions with their, graphs. Sum, difference, product and quotient of functions., 3. Trigonometric Functions, 20 Periods, Positive and negative angles. Measuring angles in radians and in, degrees and conversion from one measure to another. Definition of, trigonometric functions with the help of unit circle. Truth of the identity, sin 2 x cos 2 x 1, for all x. Signs of trigonometric functions. Domain and, range of trigonometric functions and their graphs. Expressing sin (x ! y), and cos (x ! y) in terms of sin x, sin y, cos x, cos y, and their simple, applications. Deducing identities like the following:, tan x ! tan y, cot x cot y " 1, , cot (x ! y) , ,, cot y ! cot x, 1 " tan x tan y, 1, 1, sin ! sin 2 sin ( ! ) cos ( " ),, 2, 2, 1, 1, cos cos 2 cos ( ) cos ( ),, 2, 2, 1 , 1, , , , cos cos , 2 sin ( ) sin ( ) ., 2, 2, Identities related to sin 2x, cos 2x, tan 2x, sin 3x, cos 3x and tan 3x., General solution of trigonometric equations of the type sin y sin a,, cos y cos a and tan y tan a., tan (x ! y) , , (v)
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UNIT II. Algebra, 1. Principle of Mathematical Induction, 10 Periods, Process of the proof by induction, motivating the application of the, method by looking at natural numbers as the least inductive subset, of real numbers. The principle of mathematical induction and simple, applications., 2. Complex Numbers and Quadratic Equations, 15 Periods, Need for complex numbers, especially 1 , to be motivated by inability, to solve some of the quadratic equations. Algebraic properties of, complex numbers. Argand plane and polar representation of complex, numbers. Statement of Fundamental Theorem of Algebra, solution, of quadratic equations (with real coefficients) in the complex number, system. Square root of a complex number., 3. Linear Inequalities, 15 Periods, Linear inequalities. Algebraic solutions of linear inequalities in one, variable and their representation on the number line. Graphical, representation of linear inequalities in two variables. Graphical method, of finding a solution of system of linear inequalities in two variables., 4. Permutations and Combinations, 10 Periods, Fundamental principle of counting. Factorial n(n !) . Permutations, and combinations, derivation of formulae for nPr and nCr and their, connections, simple applications., 5. Binomial Theorem, 10 Periods, History, statement and proof of the binomial theorem for positive, integral indices. Pascal’s triangle, general and middle term in binomial, expansion, simple applications., 6. Sequence and Series, 10 Periods, Sequence and Series. Arithmetic progression (AP), Arithmetic mean, (AM), Geometric progression (GP), general terms of a GP, sum of first, n terms of a GP, infinite GP and its sum, geometric mean (GM), relation, between AM and GM. Formulae for the following special sums, n, , n, , 2, , n, , 3, , / k, / k and / k ., , k1, , k1, , k1, , UNIT III. Coordinate Geometry, 1. Straight Lines, 10 Periods, Brief recall of two dimensional geometry from earlier classes. Shifting, of origin. Slope of a line and angle between two lines. Various forms of, equations of a line: parallel to axis, point–slope form, slope–intercept, form, two-point form, intercept form and normal form. General, (vi)
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equation of a line. Equation of family of lines passing through the point, of intersection of two lines. Distance of a point from a line., 2. Conic Sections, 20 Periods, Sections of a cone: circle, ellipse, parabola, hyperbola, a point, a straight, line and a pair of intersecting lines as a degenerated case of a conic, section. Standard equations and simple properties of parabola, ellipse, and hyperbola. Standard equation of a circle., 3. Introduction to Three-dimensional Geometry, 10 Periods, Coordinate axes and coordinate planes in three dimensions. Coordinates, of a point. Distance between two points and section formula., , UNIT IV. Calculus, 1. Limits and Derivatives, 30 Periods, Derivative introduced as rate of change both as that of distance function, and geometrically., Intutive idea of limit. Limits of polynomials and rational functions;, trigonometric, exponential and logarithmic functions. Definition of, derivative, relate it to slope of tangent of the curve, derivative of sum,, difference, product and quotient of functions. Derivatives of polynomial, and trigonometric functions., , UNIT V. Mathematical Reasoning, 1. Mathematical Reasoning, 10 Periods, Mathematically acceptable statements. Connecting words/phrases—, consolidating the understanding of “if and only if (necessary and, sufficient) condition”, “implies”, “and/or”, “implied by”, “and”, “or”,, “there exists” and their use through variety of examples related to real, life and mathematics. Validating the statements involving the connecting, words, difference between contradiction, converse and contrapositive., , UNIT VI. Statistics and Probability, 1. Statistics, 15 Periods, Measures of dispersion: Range, mean deviation, variance and standard, deviation of ungrouped/grouped data. Analysis of frequency, distributions with equal means but different variances., 2. Probability, 15 Periods, Random experiments: outcomes, sample spaces (set representation)., Events: occurrence of events, ‘not’, ‘and’ and ‘or’ events, exhaustive, events, mutually exclusive events, Axiomatic (set theoretic) probability,, connections with other theories studied in earlier classes. Probability of, an event, probability of ‘not’, ‘and’ and ‘or’ events., (vii)
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Contents, Set and Functions, 1. Sets, , 1, , 2. Relations, , 50, , 3. Functions, , 78, , Algebra, 4. Principle of Mathematical Induction, , 131, , 5. Complex Numbers and Quadratic Equations, , 150, , 6. Linear Inequations (In one variable), , 217, , 7. Linear Inequations (In two variables), , 245, , 8. Permutations, , 264, , 9. Combinations, , 322, , 10. Binomial Theorem, , 350, , 11. Arithmetic Progression, , 383, , 12. Geometrical Progression, , 433, , 13. Some Special Series, , 495, , Trigonometry, 14. Measurement of Angles, , 509, , 15. Trigonometric, or Circular, Functions, , 519, , 16. Conditional Identities Involving the, , Angles of a Triangle, , 587, , 17. Trigonometric Equations, , 596, , 18. Solution of Triangles, , 614, , 19. Graphs of Trigonometric Functions, , 634, , Coordinate Geometry, 20. Straight Lines, , 641, , 21. Circle, , 719, , 22. Parabola, , 733, , 23. Ellipse, , 743, (ix)
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24. Hyperbola, , 760, , 25. Applications of Conic Sections, , 775, , 26. Three-Dimensional Geometry, , 783, , Calculus, 27. Limits, , 795, , 28. Differentiation, , 838, , Mathematical Reasoning, 29. Mathematical Reasoning, , 894, , Statistics and Probability, 30. Statistics, , 918, , 31. Probability, , 946, , Logarithm, , 980, , Appendix Mathematical Modelling, , 985, , Objective Questions, Complex Numbers, , 988, , Permutations and Combinations, , 1001, , Binomial Theorem, , 1013, , Sequences and Series (AP, GP, HP), , 1026, , Trigonometry, , 1053, , Coordinate Geometry, , 1112, , , , (x)
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Sets, , 1, , 1, , Sets, INTRODUCTION, , In our mathematical language, everything in this universe, whether living or nonliving, is called an object., A given collection of objects is said to be well defined, if we can definitely say, whether a given particular object belongs to the collection or not., A well-defined collection of objects is called a set., The objects in a set are called its members or elements or points., We denote sets by capital letters A, B, C, X, Y, Z, etc., If a is an element of a set A, we write, ad A, which means that a belongs to A, or that a is an element of A., If a does not belong to A, we write, az A., SET, , ILLUSTRATIONS, , (i) The collection of all vowels in the English alphabet contains five, elements, namely a, e, i, o, u., So, this collection is well defined and therefore, it is a set., (ii) The collection of all odd natural numbers less than 10 contains the, numbers 1, 3, 5, 7, 9., So, this collection is well defined and therefore, it is a set., (iii) The collection of all prime numbers less than 20 contains the numbers, 2, 3, 5, 7, 11, 13, 17, 19., So, this collection is well defined and therefore, it is a set., (iv) All possible roots of the quadratic equation x 2 x 6 0 are –2 and 3., So, the collection of all possible roots of x 2 x 6 0 is well defined, and therefore, it is a set., (v) The collection of all rivers of India, is clearly well defined and, therefore, it is a set., Clearly, river Ganga belongs to this set while river Nile does not, belong to it., (vi) The collection of five most talented writers of India is not a set, since, no rule has been given for deciding whether a given writer is talented, or not., 1
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2, , Senior Secondary School Mathematics for Class 11, , (vii) The collection of most dangerous animals of the world is not a set,, since no rule has been given for deciding whether a given animal is, dangerous or not., (viii) The collection of five most renowned mathematicians of the world, is not a set, since there is no criterion for deciding whether a, mathematician is renowned or not., (ix) The collection of all beautiful girls of India is not a set, since the term, ‘beautiful‘ is vague and it is not well defined., Similarly, ‘rich persons‘, ‘honest persons‘, ‘good players‘, ‘old people‘,, ‘young men‘, etc., do not form sets., However, ‘blind persons‘, ‘dumb persons‘, ‘illiterate persons‘, ‘retired, persons‘, etc., form sets., HOW TO DESCRIBE OR SPECIFY A SET?, , There are two methods of describing a set., Under this method, we list all the members, of the set within braces { } and separate them by commas., Note that the order in which the elements are listed, is immaterial., , I. ROSTER FORM, OR TABULATION METHOD, , EXAMPLES, , SOLUTION, , Write each of the following sets in the roster form:, (i) A = set of all factors of 24., (ii) B = set of all prime numbers between 50 and 70., 3, 11 $, (iii) C = set of all integers between and, 2, 2, (iv) D = set of all consonants in the English alphabet which precede k., (v) E = set of all letters in the word ‘TRIGONOMETRY‘., (vi) F = set of all months having 30 days., (i) All factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24., , A = {1, 2, 3, 4, 6, 8, 12, 24}., (ii) All prime numbers between 50 and 70 are 53, 59, 61, 67., , B = {53, 59, 61, 67}., 3, 11, (iii) All integers between and, are –1, 0, 1, 2, 3, 4, 5., 2, 2, , C = {–1, 0, 1, 2, 3, 4, 5}., (iv) All consonants preceding k are b, c, d, f, g, h, j., , D = {b, c, d, f, g, h, j}., (v) It may be noted here that the repeated letters are taken only, once each., , E = {T, R, I, G, O, N, M, E, Y}., (vi) We know that the months having 30 days are April, June,, September, November., , F = {April, June, September, November}.
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Sets, NOTE, , 3, , We denote the sets of all natural numbers, all integers, all rational numbers, and all real numbers by N, Z, Q and R respectively., , Under this method of describing a set, we list the property or, properties satisfied by all the elements of the set., We write, {x : x has properties P}., , II. SET-BUILDER FORM, , We read it as, ‘the set of all those x such that each x satisfies properties P’., ILLUSTRATIONS, EXAMPLE 1, SOLUTION, , EXAMPLE 2, SOLUTION, , EXAMPLE 3, SOLUTION, , EXAMPLE 4, SOLUTION, , EXAMPLE 5, SOLUTION, , EXAMPLE 6, SOLUTION, , EXAMPLE 7, SOLUTION, , Write the set A = {1, 2, 3, 4, 5, 6, 7} in the set-builder form., Clearly, A = set of all natural numbers less than 8., Thus, in the set-builder form, we write it as, A {x : xd N and x 8} ., Write the set B = {1, 2, 4, 7, 14, 28} in the set-builder form., Clearly, B = set of all factors of 28., Thus, in the set-builder form, we write it as, B {x : xd N and x is a factor of 28}., Write the set C = {2, 4, 8, 16, 32} in the set-builder form., Clearly, C {21, 2 2, 2 3, 2 4, 2 5} ., Thus, in the set-builder form, we write it as, C {x : x 2 n, where nd N and 1 # n # 5} ., Write the set D = {–6, –4, –2, 0, 2, 4, 6} in the set-builder form., Clearly, D = set of even integers from –6 to 6., Thus, in the set-builder form, we write it as, D {x : x 2n, where ndZ and 3 # n # 3} ., Write the set E = {3, 6, 9, 12, 15, 18} in the set-builder form., Clearly, E {3 #1, 3 # 2, 3 # 3, 3 # 4, 3 # 5, 3 # 6} ., Thus, in the set-builder form, we write it as, E {x : x 3n, where nd N and 1 # n # 6} ., 1 2 3 4 5 6 7 8, Write the set F & , , , , , , , 0 in the set-builder form., 2 3 4 5 6 7 8 9, Clearly, we have, n ,, F &x : x , where nd N and 1 # n # 80 ·, (n 1), Write the set G = {1, 3, 5, 7, 9, 11, …} in the set-builder form., Clearly, G = set of all odd natural numbers., Thus, in the set-builder form, we write it as, G {x : xd N and x is odd} .
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4, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 8, SOLUTION, , EXAMPLE 9, , SOLUTION, , Write the set H = {1, 4, 9, 16, 25, 36, …} in the set-builder form., Clearly, H is the set of the squares of all natural numbers., So, in the set-builder form, we write it as, H {x : x n 2, where nd N} ., Match each of the sets on the left in the roster form with the same set on, the right given in set-builder form:, (i) {23, 29}, (a) {x : x 3 n, nd N and 1 # n # 5}, (ii) {B, E, T, R}, (b) {x : x n 3, nd N and 2 # n # 6}, (iii) {3, 9, 27, 81, 243}, (c) {x : x is prime, 20 x 30}, (iv) {8, 27, 64, 125, 216}, (d) {x : x is a letter of the word ‘BETTER‘}, (i) {23, 29} = set of prime numbers between 20 and 30, = {x : x is prime, 20 < x < 30}., , (i) (c)., (ii) {B, E, T, R} = set of letters in the word ‘BETTER‘}, = {x : x is a letter in the word ‘BETTER‘}., , (ii) (d)., (iii) {3, 9, 27, 81, 243} {31, 3 2, 3 3, 3 4, 3 5}, {x : x 3 n, nd N and 1 # n # 5} ., , (iii) (a)., (iv) {8, 27, 64, 125, 216} {2 3, 3 3, 4 3, 5 3, 6 3}, {x : x n 3, nd N, 2 # n # 6} ., , (iv) (b)., , EXERCISE 1A, 1. Which of the following are sets? Justify your answer., , (i) The collection of all whole numbers less than 10., (ii) The collection of good hockey players in India., (iii) The collection of all questions in this chapter., (iv) The collection of all difficult chapters in this book., (v) A collection of Hindi novels written by Munshi Prem Chand., (vi) A team of 11 best cricket players of India., (vii) The collection of all the months of the year whose names begin with the, letter M., (viii) The collection of all interesting books., (ix) The collection of all short boys of your class., (x) The collection of all those students of your class whose ages exceed, 15 years., (xi) The collection of all rich persons of Kolkata.
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Sets, , 5, , (xii) The collection of all persons of Kolkata whose assessed annual, incomes exceed (say) ` 20 lakh in the financial year 2016–17., (xiii) The collection of all interesting dramas written by Shakespeare., 2. Let A be the set of all even whole numbers less than 10., , (a) Write A in roster form., (b) Fill in the blanks with the approximate symbol or :, (i) 0 ..... A, (ii) 10 ..... A, (iii) 3 ..... A, (iv) 6 ..... A, 3. Write the following sets in roster form:, , (i), (ii), (iii), (iv), (v), (vi), (vii), (viii), (ix), (x), , A = {x : x is a natural number, 30 # x 36} ., B = {x : x is an integer and 4 x 6} ., C = {x : x is a two-digit number such that the sum of its digits is 9}., D = {x : x is an integer, x 2 # 9} ., E = {x : x is a prime number, which is a divisor of 42}., F = {x : x is a letter in the word ‘MATHEMATICS’}., G = {x : x is a prime number and 80 < x < 100}., H = {x : x is a perfect square and x < 50}., J {x : xd R and x 2 x 12 0} ., K {x : xd N, x is a multiple of 5 and x 2 400} ., , 4. List all the elements of each of the sets given below:, , (i) A {x : x 2n, nd N and n # 5} ., (ii) B {x : x 2n 1, ndW and n 5} ., 1, (iii) C &x : x n , nd N and n 60 ·, (iv) D {x : x n 2, nd N and 2 # n # 5} ., (v) E {x : xdZ and x 2 x} ., 13, 1, (vi) F &x : xdZ and x 0 ·, 2, 2, 1, ,, (vii) G 'x : x , nd N and 1 # n # 51 ·, (2n 1), (viii) H {x : xdZ,|x|# 2} ., 5. Write each of the sets given below in set-builder form:, , (i) A &1,, , 1, 1, 1 , 1 , 1 , 1, 0, 4 9 16 25 36 49, (iii) C {53, 59, 61, 67, 71, 73, 79}, (v) E {14, 21, 28, 35, 42, …, 98}, , 1 2 3 4 5 , 6 , 7, 0, (ii) B & , , , ,, 2 5 10 17 26 37 50, (iv) D {1, 1}, , 6. Match each of the sets on the left described in roster form with the same set, , on the right described in the set-builder form:, (i) {–5, 5}, (a) {x : xdZ and x 2 16}, (ii) {1, 2, 3, 6, 9, 18}, (b) {x : xd N and x 2 x}
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6, , Senior Secondary School Mathematics for Class 11, , (iii) {–3, –2, –1, 0, 1, 2, 3}, (iv) {P, R, I, N, C, A, L}, (v) {1}, , (c) {x : xdZ and x 2 25}, (d) {x : xd N and x is a factor of 18}, (e) {x : x is a letter in the word ‘PRINCIPAL‘}, ANSWERS (EXERCISE 1A), , 1. (i), (iii), (v), (vii), (x), (xii), 2. (a) A {0, 2, 4, 6, 8}, , (i), (iii), (iv), (vi), (viii), (x), 4. (i), , 3., , (iii), (v), (vii), 5., , (i), (ii), (iii), , (b) (i) (ii) (iii) (iv) , A {30, 31, 32, 33, 34, 35} (ii) B {3, 2, 1, 0, 1, 2, 3, 4, 5}, C {18, 81, 27, 72, 36, 63, 45, 54, 90}, (v) E {2, 3, 7}, D {3, 2, 1, 0, 1, 2, 3}, F {M, A, T, H, E, I, C, S} (vii) G {83, 89, 97}, H {1, 4, 9, 16, 25, 36, 49} (ix) J { 4, 3}, K {5, 10, 15}, (ii) B {1, 3, 5, 7, 9}, A {2, 4, 6, 8, 10}, 1, 1, 1, 1, (iv) D {4, 9, 16, 25}, C &1, , , , 0, 2 3 4 5, (vi) F {0, 1, 2, 3, 4, 5, 6}, E {0, 1}, 1, 1, 1, 1, (viii) H {2, 1, 0, 1, 2}, G &1, , , , 0, 3 5 7 9, 1, A {x : x 2 , nd N and 1 # n # 7}, n, n ,, B 'x : x 2, nd N and 1 # n # 71, (n 1), C {x : x is prime, 50 x 80}, , (iv) D {x : xdZ, x 2 1}, (v) E {x : x 7n, nd N, 2 # n # 14}, 6. (i) (c), (ii) (d), (iii) (a), (iv) (e), (v) (b), HINTS TO SOME SELECTED QUESTIONS, 1. (i) Clearly, all whole numbers less than 10 are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9., (ii) Clearly, there is no specific criterion to decide whether a given hockey player of, India, is good or not., So, the given collection is not a set., (iii) The collection of all questions in this chapter is a set, because, if a question is given, then we can easily decide whether it is a question of the chapter or not., (iv) Clearly, the term ‘difficult‘ is vague. So, the given collection is not well defined and, therefore, it is not a set., (v) Suppose we are given a collection of Hindi novels written by Munshi Prem, Chand. Now, if we take any Hindi novel then we can clearly decide whether it, belongs to our collection or not., So, the given collection is a set., (vi) The term ‘best‘ is vague. So, the given collection is not a set.
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Sets, , 7, , (vii) The given collection has definite members, namely March and May. So, this, collection is a set., (viii) The term ‘interesting‘ is vague. So, the given collection is not a set., (ix) The term ‘short’ is vague. So, the given collection is not a set., (x) Clearly, it contains definite members. So, it is a set., (xi) The term ‘rich‘ is vague. So, the given collection is not a set., (xii) The given collection is clearly well defined. So, it is a set., (xiii) The term ‘interesting‘ is vague. So, the given collection is not a set., , SOME TERMS RELATED TO SETS, EMPTY SET A set containing no element at all is called the empty set, or the null set,, or the void set, denoted by , or { }., A set which has at least one element is called a nonempty set., , Examples of empty sets, , (i) {x : xd N and 2 x 3} , since there is no natural number lying, between 2 and 3., (ii) {x : x is a number, x ! x} , since there is no number which is not, equal to itself., (iii) {x : xd N, x 5 and x 7} , since there is no natural number which, is less than 5 and greater than 7., (iv) {x : xd R and x 2 1} , since there is no real number whose square, is –1., (v) {x : x is rational and x 2 2 0} , since there is no rational number, whose square is 2., (vi) {x : x is an even prime number greater than 2} = , since there is no, prime number which is even and greater than 2., (vii) {x : x is a point common to two parallel lines} = , since there is no, point common to two parallel lines., SINGLETON SET, , A set containing exactly one element is called a singleton set., , Examples of singleton sets, , (i) {0} is a singleton set whose only element is 0., (ii) {15} is a singleton set whose only element is 15., (iii) {–8} is a singleton set whose only element is –8., (iv) {x : xd N and x 2 4} {2}, since 2 is the only natural number whose, square is 4., However, {x : xdZ and x 2 4} { 2, 2}, which is not a singleton set., (v) Consider the set {x : xd R and x 3 1 0} ., Now, x 3 1 0 & (x 1) (x 2 x 1) 0, & x 1 or x , , 1 ! 1 4 1 ! 3, , ·, 2, 2
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8, , Senior Secondary School Mathematics for Class 11, , Thus, the given equation has one real root, namely x 1., , {x : xd R and x 3 1 0} {1}, which is a singleton set., FINITE AND INFINITE SETS, , An empty set or a nonempty set in which the process of counting of elements surely, comes to an end is called a finite set. A set which is not finite is called an infinite set., The number of distinct elements contained in a finite set A is denoted, by n(A) ., Examples of finite sets, , (i) Let A {2, 4, 6, 8, 10, 12} ., Then, A is clearly a finite set and n(A) 6., (ii) Let B set of all letters in the English alphabet., Then, n(B) 26 and therefore, B is finite., (iii) Let C {x : xdZ and x 2 36 0} ., Then, C {6, 6}, which is clearly a finite set and n(C) 2., (iv) The set of all persons on earth is a finite set., (v) The set of all animals on earth is a finite set., Examples of infinite sets, , (i), (ii), (iii), (iv), , The set of all points on the arc of a circle is an infinite set., The set of all points on a line segment is an infinite set., The set of all circles passing through a given point is an infinite set., The set of all straight lines parallel to a given line, say the x-axis, is an, infinite set., (v) The set of all positive integral multiples of 5 is an infinite set., Let Z be the set of all positive integers., Then, {5x : xdZ } {5, 10, 15, 20, 25, …} is an infinite set., (vi) Each of the sets N, Z, Q and R is an infinite set., , NOTE, , All infinite sets cannot be described in roster form., For example, the set R of all real numbers cannot be described in this, form, since the elements of this set do not follow a particular pattern., , Two nonempty sets A and B are said to be equal, if they have exactly the, same elements and we write, A B., Otherwise, the sets are unequal and we write, A ! B., , EQUAL SETS, , REMARKS, , (i) The elements of a set may be listed in any order., Thus, {1, 2, 3} = {2, 1, 3} = {3, 2, 1}., (ii) The repetition of elements in a set has no meaning., Thus, {1, 1, 2, 2, 3} = {1, 2, 3}.
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Sets, , 9, , Some examples of equal sets, , SOLUTION, , Let A = set of letters in the word ‘follow’,, and B = set of letters in the word ‘wolf ‘., Show that A = B., Clearly, we have, A = {f, o, l, w} and B = {w, o, l, f}., Clearly, A and B have exactly same elements., , A B., , EXAMPLE 2, , Let A {p, q, r, s} and B {q, r, p, s} . Are A and B equal?, , SOLUTION, , Since A and B have exactly the same elements, so A B., , EXAMPLE 1, , EXAMPLE 3, , Show that , {0} and 0 are all different., , SOLUTION, , We know that is a set containing no element at all., And, {0} is a set containing one element, namely 0., Also, 0 is a number, not a set., Hence, , {0} and 0 are all different., , EXAMPLE 4, , Let A {x : xd N, x 2 9 0} and B {x : xdZ, x 2 9 0} ., Show that A ! B., x 2 9 0 & (x 3) (x 3) 0 & x 3 or x 3., , A {x : xd N, x 2 9 0} {3}, [a 3z N], 2 , , , , and B {x : xdZ, x 9 0} { 3, 3} ., Hence, A ! B., , SOLUTION, , Two finite sets A and B are said to be equivalent, if n(A) n(B) ., Equal sets are always equivalent. But, equivalent sets need not be equal., , EQUIVALENT SETS, , EXAMPLE 1, , Let A {1, 3, 5} and B {2, 4, 6} ., Then, n(A) n(B) 3., So, A and B are equivalent., Clearly, A ! B., Hence, A and B are equivalent sets but not equal., , EXAMPLE 2, , Show that {0} and are not equivalent sets., Let A {0} and B ., Then, clearly n(A) 1 and n(B) 0., , n(A) ! n(B) and hence A and B are not equivalent sets., , SOLUTION, , EXERCISE 1B, 1. Which of the following are examples of the null set?, (i) Set of odd natural numbers divisible by 2., (ii) Set of even prime numbers.
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10, , Senior Secondary School Mathematics for Class 11, , (iii) A {x : xd N, 1 x # 2}, (iv) B {x : xd N, 2x 3 4}, (v) C {x : x is prime, 90 x 96}, (vi) D {x : xd N, x 2 1 0}, (vii) E {x : xdW, x 3 # 3}, (viii) F {x : xdQ, 1 x 2}, (ix) G {0}, 2. Which of the following are examples of the singleton set?, (i) {x : xdZ, x 2 4}, (ii) {x : xdZ, x 5 0}, (iii) {x : xdZ,|x| 1}, (iv) {x : xd N, x 2 16}, (v) {x : x is an even prime number}, 3. Which of the following are pairs of equal sets?, (i) A = set of letters in the word, ‘ALLOY’, B = set of letters in the word, ‘LOYAL’, (ii) C = set of letters in the word, ‘CATARACT’, D = set of letters in the word ‘TRACT’, (iii) E {x : xdZ, x 2 # 4} and F {x : xdZ, x 2 4}, (iv) G {1, 1} and H {x : xdZ, x 2 1 0}, (v) J {2, 3} and K {x : xdZ, (x 2 5x 6) 0}, 4. Which of the following are pairs of equivalent sets?, (i) A {2, 1, 0} and B {1, 2, 3}, (ii) C {x : xd N, x 3} and D {x : xdW, x 3}, (iii) E {a, e, i, o, u} and F {p, q, r, s, t}, 5. State whether the given set is finite or infinite:, (i) A = set of all triangles in a plane, (ii) B = set of all points on the circumference of a circle, (iii) C = set of all lines parallel to the y-axis, (iv) D = set of all leaves on a tree, (v) E = set of all positive integers greater than 500, (vi) F {xd R: 0 x 1], (vii) G {xdZ: x 1}, (viii) H {xdZ: 15 x 15}, (ix) J {x : xd N and x is prime}, (x) K {x : xd N and x is odd}, (xi) L = set of all circles passing through the origin (0, 0), 6. Rewrite the following statements using set notation:, (i) a is an element of set A., (ii) b is not an element of A., (iii) A is an empty set and B is a nonempty set.
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Sets, , 11, , (iv) Number of elements in A is 6., (v) 0 is a whole number but not a natural number., ANSWERS (EXERCISE 1B), , 1., 3., 4., 5., , (i), (iv), (v), (vi), 2. (ii), (iv), (v), (i) A = B (ii) C = D (iv) G = H, (i) A and B are equivalent sets (iii) E and F are equivalent sets, (i) infinite, (ii) infinite, (iii) infinite, (iv) finite, (v) infinite, (vi) infinite (vii) infinite (viii) finite, (ix) infinite (x) infinite, (xi) infinite, 6. (i) ad A, (ii) bz A, (iii) A and B! , (iv) n(A) 6, (v) 0dW but 0z N, , SUBSETS, SUBSET A set A is said to be a subset of set B if every element of A is also an element, of B, and we write, A 3 B., SUPERSET, , If A 3 B, then B is called a superset of A, and we write, B 4 A., , PROPER SUBSET, , write, A 1 B., REMARK, , If A 3 B and A ! B then A is called a proper subset of B and we, , If there exists even a single element in A which is not in B, then A is not, Y B., a subset of B, and we write, A 1, , Examples of subsets, EXAMPLE 1, , Let A {2, 3, 5} and B {2, 3, 5, 7, 9} ., Then, every element of A is an element of B., , A 3 B but A ! B., Hence, A is a proper subset of set B, i.e., A 1 B., , EXAMPLE 2, , Let A {1, 2} and B {2, 3, 5} ., Then, 1d A but 1zB., , A M B., Again, 3dB but 3z A., , B M A., Thus, A M B and B M A., , EXAMPLE 3, , Clearly, N 1 W 1 Z 1 Q 1 R., But, 0dW and 0z N., , W M N.
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12, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 4, , SOLUTION, , Let A {1, {2, 3}, 4} ., Then, which of the following statements is true?, (i) {2, 3}d A, (ii) {2, 3} 1 A, Rectify the wrong statement., Clearly, A is a set containing three elements, namely 1, {2, 3} and 4., (i) {2, 3}d A is a true statement., (ii) {2, 3} 1 A is wrong., On rectifying this statement, we get {{2, 3}} 1 A as true, statement., , SOME RESULTS ON SUBSETS, , Every set is a subset of itself., Let A be any set., Then, each element of A is in A., , A 3 A., Hence, every set is a subset of itself., , THEOREM 1, PROOF, , The empty set is a subset of every set., Let A be any set and be the empty set., Since contains no element at all, so there is no element of which is, not contained in A., Hence, 1 A., , THEOREM 2, PROOF, , The total number of subsets of a set containing n elements is 2 n ., Let A be a finite set containing n elements. Then,, number of subsets of A each containing no element 1 n C0 ., number of subsets of A each containing 1 element n C1 ., number of subsets of A each containing 2 elements n C2 ., ………, ………, ………, ………, ………, ………, ………, ………, ………, ………, n, number of subsets of A each containing n elements Cn ., , total number of subsets of A, (n C0 n C1 n C2 … n Cn), ( 1 1) n 2 n, [using binomial theorem]., , THEOREM 3, PROOF, , If there are some sets under consideration then there happens to be a, set which is a superset of each one of the given sets. Such a set is known as the universal, set for those sets. We shall denote a universal set by U., , UNIVERSAL SET, , EXAMPLE 1, , Let A {1, 2, 3}, b {2, 3, 4, 5} and C {6, 7} ., If we consider the set U {1, 2, 3, 4, 5, 6, 7} then clearly, U is a, superset of each of the given sets., Hence, U is the universal set., , EXAMPLE 2, , When we discuss sets of lines, triangles or circles in twodimensional geometry, the plane in which these lines, triangles or, circles lie, is the universal set.
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Sets, , 13, , SUBSETS OF THE SET R OF ALL REAL NUMBERS, , (i) N {1, 2, 3, 4, 5, …} is the set of all natural numbers., (ii) W {0, 1, 2, 3, 4, 5, …} is the set of all whole numbers., (iii) Z {…, 4, 3, 2, 1, 0, 1, 2, 3, 4, …} is the set of all integers., Z {1, 2, 3, 4, 5, …} is the set of all positive integers., Z {1, 2, 3, 4, …} is the set of all negative integers., Sometimes, we denote the set of all integers by I., p, (iv) Q 'x : x q , where, p, qdZ and q ! 01 is the set of all rational, numbers., The set of all positive rational numbers is denoted by Q ., (v) T {x : xd R and xzQ} is the set of all irrational numbers., INTERVALS AS SUBSETS OF R, , Let a, bd R and a b. Then, we define:, (i) Closed Interval [a, b] {xd R : a # x # b} ., (ii) Open Interval (a, b) or ]a, b[ {xd R : a x b} ., (iii) Right Half Open Interval [a, b) or [a, b[ {xd R : a # x b} ., (iv) Left Half Open Interval (a, b] or ]a, b] {xd R : a x # b} ., On the real line, we represent these intervals as shown below:, a, , b, , a, , a, , b, [a, b), , LENGTH OF AN INTERVAL, , (a, b] is (b – a)., , b, (a, b), , [a, b], a, , b, (a, b], , The length of each of the intervals [a, b], (a, b) [a, b) and, , Examples on intervals, , (i) [2, 3] {xd R : 2 # x # 3}, (iii) [2, 3) {xd R : 2 # x 3}, , (ii) (2, 3) {xd R : 2 x 3}, (iv) (2, 3] {xd R : 2 x # 3}, , The set of all subsets of a given set A is called the power set of A, denoted, by P(A) ., If n(A) m then n[P(A)] 2 m ., , POWER SET, , SOLVED EXAMPLES ON SUBSETS, POWER SET AND INTERVALS, EXAMPLE 1, SOLUTION, , Write down all possible subsets of A {4} ., All possible subsets of A are , {4} ., , P(A) {, {4}} ., Here, n(A) 1 and n[P(A)] 2 21 .
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14, EXAMPLE 2, SOLUTION, , EXAMPLE 3, SOLUTION, , EXAMPLE 4, SOLUTION, , EXAMPLE 5, SOLUTION, , EXAMPLE 6, , SOLUTION, , EXAMPLE 7, SOLUTION, , Senior Secondary School Mathematics for Class 11, , Write down all possible subsets of A {2, 3} ., All possible subsets of A are, , {2}, {3}, {2, 3} ., , P(A) {, {2}, {3}, {2, 3}}, Thus, n(A) 2 and n{P(A)} 4 2 2 ., Write down all possible subsets of A {1, 0, 1} ., All possible subsets of A are, , {1}, {0}, {1}, {1, 0}, {0, 1}, {1, 1}, and {1, 0, 1} ., , P(A) {, {1}, {0}, {1}, {1, 0}, {0, 1}, {1, 1}, {1, 0, 1}} ., Thus, n(A) 3 and n{P(A)} 8 2 3 ., Write down all possible subsets of A {1, {2, 3}} ., Here, A contains two elements, namely 1 and {2, 3}., Let {2, 3} B, then A {1, B} ., , P(A) {, {1}, {B}, {1, B}}, &, P(A) {, {1}, {{2, 3}}, {1, {2, 3}}} ., Write down all possible subsets of ., has only one subset, namely ., , P() {} ., Write each of the following subsets of R as an interval:, (i) A {x : xd R, 3 x # 5}, (ii) B {x : xd R, 5 x 1}, (iii) C {x : xd R, 2 # x 0}, (iv) D {x : xd R, 1 # x # 4}, Find the length of each of the above intervals., We have, (i) A {x : xd R, 3 x # 5} (3, 5] . Length (A) 5 (3) 8., (ii) B {x : xd R, 5 x 1} (5, 1) . Length (B) 1 (5) 4., (iii) C {x : xd R, 2 # x 0} [2, 0) . Length (C) 0 (2) 2., (iv) D {x : xd R, 1 # x # 4} [1, 4] . Length (D) 4 (1) 5., Write each of the following intervals in the set-builder form:, (i) A (2, 5) (ii) B [ 4, 7] (iii) C [8, 0) (iv) D (5, 9], We have, (i) A (2, 5) {x : xd R, 2 x 5}., (ii) B [ 4, 7] {x : xd R, 4 # x # 7}., (iii) C [8, 0) {x : xd R, 8 # x 0}., (iv) D (5, 9] {x : xd R, 5 x # 9}., , EQUAL SETS Two sets A and B are said to be equal, if every element of A is in B and, every element of B is in A, and we write, A = B., REMARKS (i) The elements of a set may be listed in any order., Thus, {1, 2, 3} {3, 1, 2} {2, 3, 1}, etc., (ii) The repetition of elements in a set is meaningless., Thus, [2, 4, 6} {2, 2, 4, 6, 6} .
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Sets, , 15, , Let A and B be two sets. Then, prove that A B + A 3 B and B 3 A., Let A B., Then, by definition of equal sets, every element of A is in B and every, element of B is in A., , A 3 B and B 3 A., Thus, (A B) & (A 3 B and B 3 A) ., Again, let A 3 B and B 3 A., Then, by the definition of a subset, it follows that every element of A, is in B and every element of B is in A., Consequently, A B., Thus, (A 3 B and B 3 A) & A B., Hence, A B + (A 3 B and B 3 A) ., , THEOREM 4, PROOF, , EXAMPLE 8, , SOLUTION, , Let A {1, {2}, {3, 4}, 5} . Which of the following are incorrect statements?, Rectify each:, (ii) {2} 1 A, (iii) {1, 2} 1 A, (i) 2d A, (iv) {3, 4} 1 A, (v) {1, 5} 1 A, (vi) {} 1 A, (vii) 11 A, (viii) {1, 2, 3, 4} 1 A, Clearly, A contains four elements, namely 1, {2}, {3, 4} and 5., (i) 2d A is incorrect. The correct statement would be {2}d A., (ii) {2} 1 A is incorrect. The correct statement is {{2}} 1 A., (iii) Clearly, 2z A and therefore, {1, 2} 1 A is incorrect., The correct statement would be {1, {2}} 1 A., (iv) Clearly, {3, 4} is an element of A., So, {3, 4} 1 A is incorrect and {{3, 4}} 1 A is correct., (v) Since 1 and 5 are both elements of A, so {1, 5} 1 A is correct., (vi) Since z A, so {} 1 A is incorrect while 1 A is correct., (vii) Since 1d A, so 11 A is incorrect and therefore, {1} 1 A is, correct., (viii) Since 2z A and 3z A, so {1, 2, 3, 4} 1 A is incorrect., The correct statement would be {1, {2}, {3, 4}} 1 A., , EXERCISE 1C, Y B., 1. State in each case whether A 1 B or A 1, (i) A {0, 1, 2, 3}, B {1, 2, 3, 4, 5}, (ii) A , B {0}, (iii) A {1, 2, 3}, B {1, 2, 4}, (iv) A {x : xdZ, x 2 1}, B {x : xd N, x 2 1}, (v) A {x : x is an even natural number}, B {x : x is an integer}, (vi) A {x : x is an integer}, B {x : x is a rational number}
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16, , Senior Secondary School Mathematics for Class 11, , (vii) A {x : x is a real number}, B {x : x is a complex number}, (viii) A {x : x is an isosceles triangle in a plane},, B {x : x is an equilateral triangle in the same plane}, (ix) A {x : x is a square in a plane},, B {x : x is a rectangle in the same plane}, (x) A {x : x is a triangle in a plane},, B {x : x is a rectangle in the same plane}, (xi) A {x : x is an even natural number less than 8},, B {x : x is a natural number which divides 32}, 2. Examine whether the following statements are true or false:, Y {b, c, a}, (i) {a, b} 1, (ii) {a}d{a, b, c}, (iii) {} 1 {a, b, c}, (iv) {a, e} 1 {x : x is a vowel in the English alphabet}, (v) {x : xdW, x 5 5} , (vi) ad{{a}, b}, (vii) {a} 1 {{a}, b}, (viii) {b, c} 1 {a, {b, c}}, (ix) {a, a, b, b} {a, b}, (x) {a, b, a, b, a, b, …} is an infinite set., (xi) If A set of all circles of unit radius in a plane and B set of all circles, in the same plane then A 1 B., Y B., 3. If A {1} and B {{1}, 2} then show that A 1, Hint, , 1d A but 1z B., , 4. Write down all subsets of each of the following sets:, (ii) B {a, b}, (iii) C {2, 3}, (i) A {a}, (iv) D {1, 0, 1}, (v) E , (vi) F {2, {3}}, (vii) G {3, 4, {5, 6}}, 5. Express each of the following sets as an interval:, (i) A {x : xd R, 4 x 0} (ii) B {x : xd R, 0 # x 3}, (iii) C {x : xd R, 2 x # 6}, (iv) D {x : xd R, 5 # x # 2}, (v) E {x : xd R, 3 # x 2} (vi) F {x : xd R, 2 # x 0}, 6. Write each of the following intervals in the set-builder form:, (i) A (2, 3), (ii) B [4, 10], (iii) C [1, 8), (iv) D (4, 9], (v) E [ 10, 0), (vi) F (0, 5], 7. If A {3, {4, 5}, 6}, find which of the following statements are true., (i) {4, 5} 3 A, (ii) {4, 5}d A, (iii) {{4, 5}} 3 A, (iv) 4d A, (v) {3} 3 A, (vi) {} 3 A, (vii) 3 A, (viii) {3, 4, 5} 3 A, (ix) {3, 6} 3 A, 8. If A {a, b, c}, find P(A) and n{P(A)} ., 9. If A {1, {2, 3}}, find P(A) and n{P(A)} ., 10. If A then find n{P(A)} ., 11. If A {1, 3, 5}, B {2, 4, 6} and C {0, 2, 4, 6, 8} then find the universal set., 12. Prove that A 3 B, B 3 C and C 3 A & A C.
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Sets, , 17, , 13. For any set A, prove that A 3 + A ., 14. State whether the given statement is true or false:, (i) If A 1 B and xzB then xz A., (ii) If A3 then A ., (iii) If A, B and C are three sets such that AdB and B 1 C then A 1 C., Hint, , Let A {a}, B {{a}, b} and C {{a}, b, c} ., Y C., Then, {a}d B and B 1 C. But, {a} 1, , (iv) If A, B and C are three sets such that A 1 B and BdC then AdC., Hint, , Let A {a}, B {a, b} and C {{a, b}, c} ., Then, A 1 B and B d C. But, A z C., , Y B and B 1, Y C then A 1, Y C., (v) If A, B and C are three sets such that A 1, Hint, , Let A {a}, B {b, c} and C {a, c} ., Y B and B 1, Y C. But A 1 C., Then, A 1, , (vi) If A and B are sets such that xd A and AdB then xdB., Hint, , Let A {x}, B {{x}, y} ., Then, x d A and A d B. But, x z B., ANSWERS (EXERCISE 1C), , YB, YB, YB, 1. (i) A 1, (ii) AfB, (iii) A 1, (iv) A 1, (v) AfB, YB, YB, (vi) AfB, (vii) AfB, (viii) A 1, (ix) AfB, (x) A 1, YB, (xi) A 1, 2. (i) False, (ii) False, (iii) False, (iv) True, (v) False, (vi) False, (vii) False, (viii) False, (ix) True, (x) False, (xi) True, 4. (i) , {a}, (ii) , {a}, {b}, {a, b}, (iii) , {2}, {3}, {2, 3}, (iv) , {1}, {0}, {1}, {1, 0}, {0, 1}, {1, 1}, {1, 0, 1}, (v) , (vi) , {2}, {{3}}, {2, {3}}, (vii) , {3}, {4}, {{5, 6}}, {3, 4}, {3, {5, 6}}, {4, {5, 6}}, {3, 4, {5, 6}}, 5. (i) A ( 4, 0), (ii) B [0, 3), (iii) C (2, 6], (iv) D [5, 2], (v) E [3, 2), (vi) F [2, 0), 6. (i) A {x : xd R, 2 x 3}, (ii) B {x : xd R, 4 # x # 10}, (iii) C {x : xd R, 1 # x 8}, (iv) D {x : xd R, 4 x # 9}, (v) E {x : xd R, 10 # x 0}, (vi) F {x : xd R, 0 x # 5}, 7. (i) False, (ii) True, (iii) True, (iv) False, (v) True, (vi) False, (vii) True, (viii) False, (ix) True, 8. P(A) {, {a}, {b}, {c}, {a, b}, {b, c}, {a, c}, {a, b, c}} and n{P(A)} 2 3 8, 9. P(A) {, {1}, {{2, 3}}, {1, {2, 3}}} and n{P(A)} 2 2 4, 10. P(A) {} & n{P(A)} 1, 11. U {0, 1, 2, 3, 4, 5, 6, 8}, 14. (i) True, (vi) False, , (ii) True, , (iii) False, , (iv) False, , (v) False
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18, , Senior Secondary School Mathematics for Class 11, HINTS TO SOME SELECTED QUESTIONS, , Y B., 1. (iv) A {1, 1} and B {1} . So, A 1, (v) A {2, 4, 6, 8, …} and B {…, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, …] . So, A f B., (vi) Every integer is a rational number., (vii) We may write a real number x as x 0i., So, every real number is a complex number., (ix) Every square is a rectangle., Y B., (xi) A {2, 4, 6} and B {1, 2, 4, 8, 16, 32} . So, A 1, 2. (ii) ad{a, b, c} is true. So, {a}d{a, b, c} is false., (iii) 1 {a, b, c} is true and {} 1 {a, b, c} is false., (iv) {a, e} 1 {a, e, i, o, u} is true., (v) {x : xdW, x 5 5} {0} ., (vi) {{a}, b} has two elements, namely {a} and b., (vii) {a}d{{a}, b} . So, {a} 1 {{a}, b} is false., (viii) {b, c}d{a, {b, c}} . So, {b, c} 1 {a, {b, c}} is false., (ix) Repetition of elements in a set has no meaning., (x) Given set {a, b}, which is finite., 9. Let A {1, a} . Then, P(A) {, {1}, {a}, {1, a}}, where a {2, 3} ., , , P(A) {, {1}, {{2, 3}}, {1, {2, 3}}} ., , OPERATIONS ON SETS, UNION OF SETS The union of two sets A and B, denoted by A , B, is the set of all those, elements which are either in A or in B or in both A and B., Thus, A , B {x : xd A or xdB}, , , , xd A , B + xd A or xdB., , EXAMPLE 1, SOLUTION, EXAMPLE 2, SOLUTION, , EXAMPLE 3, SOLUTION, EXAMPLE 4, SOLUTION, , If A {3, 4, 5, 6} and B {4, 6, 8, 10}, find A , B., Clearly, A , B {3, 4, 5, 6, 8, 10} ., Let A {x : x is a prime number less than 10} and B {x : xd N, x is a, factor of 12}. Find A , B., We have, A {2, 3, 5, 7} and B {1, 2, 3, 4, 6, 12} ., , A , B {2, 3, 5, 7},{1, 2, 3, 4, 6, 12} {1, 2, 3, 4, 5, 6, 7, 12}., Let A {x : x is a positive integer} and let B {x : x is a negative integer}., Find A , B., Clearly, A , B {x : x is an integer and x ! 0} ., If A {x : x 2n 1, ndZ} and B {x : x 2n, ndZ} then find A , B., We have, A , B {x : x is an odd integer },{x : x is an even integer}, {x : x is an integer} = Z.
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Sets, REMARK, , 19, , The union of n sets A1, A2, A3, …, An is denoted by, , n, , (A1 , A2 , A3 ,…, An) , Ai ., i1, , The intersection of two sets A and B, denoted by A + B, is the, set of all elements which are common to both A and B., Thus, A + B {x : xd A and xdB} ., INTERSECTION OF SETS, , , , xd A + B + xd A and xdB;, xz A + B & xz A or xzB., , EXAMPLE 5, SOLUTION, , EXAMPLE 6, SOLUTION, , EXAMPLE 7, SOLUTION, , EXAMPLE 8, SOLUTION, , Let A {1, 3, 5, 7, 9} and B {2, 3, 5, 7, 11, 13} . Find A + B., We have, A + B {1, 3, 5, 7, 9}+{2, 3, 5, 7, 11, 13} {3, 5, 7} ., If A {x : xd N, x is a factor of 12} and B {x : xd N, x is a factor of 18},, find A + B., We have, A {x : xd N, x is a factor of 12} = {1, 2, 3, 4, 6, 12},, B {x : xd N, x is a factor of 18} = {1, 2, 3, 6, 9, 18}., , A + B {1, 2, 3, 4, 6, 12}+{1, 2, 3, 6, 9, 18} {1, 2, 3, 6} ., If A {x : x 3n, ndZ} and B {x : x 4n, ndZ} then find A + B., We have, A {x : xdZ and x is a multiple of 3}, and B {x : xdZ and x is a multiple of 4}., , A + B {x : xdZ and x is a multiple of both 3 and 4}, {x : xdZ and x is a multiple of 12}, {x : x 12n, ndZ} ., Hence, A + B {x : x 12n, ndZ} ., If A (2, 4) and B [3, 5), find A + B., We have, A (2, 4) {x : xd R, 2 x 4}, B [3, 5) {x : xd R, 3 # x 5}, x¢, , 0, , 1, , (, 2, , [, 3, , ), 4, , ), 5, , Clearly, A + B {x : xd R, 3 # x 4} [3, 4) ., REMARK, , The intersection of n sets A1, A2, A3, …, An is denoted by, n, , (A1 + A2 + A3 +…+ An) + Ai ., i1, , DISJOINT SETS, , Two sets A and B are said to be disjoint if A + B ., , INTERSECTING SETS, , Two sets A and B are said to be intersecting if A + B !., , x
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20, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 9, SOLUTION, , EXAMPLE 10, SOLUTION, , EXAMPLE 11, SOLUTION, , If A {1, 3, 5, 7, 9}, B {2, 4, 6, 8} and C {2, 3, 5, 7, 11}, find (A + B), and (A + C) . What do you conclude?, We have, A + B {1, 3, 5, 7, 9}+{2, 4, 6, 8} , and A + C {1, 3, 5, 7, 9}+{2, 3, 5, 7, 11} {3, 5, 7} !., Thus, A and B are disjoint sets while A and C are intersecting sets., Give examples of three sets A, B, C such that (A + B) ! , (B+ C) ! ,, (A + C) ! and (A + B+ C) ., Consider the sets A {1, 2}, B {2, 3, 4} and C {1, 3, 5} ., Then, (A + B) {1, 2}+{2, 3, 4} {2} !;, (B+ C) {2, 3, 4}+{1, 3, 5} {3} !;, (A + C) {1, 2}+{1, 3, 5} {1} !;, (A + B+ C) {1, 2}+{2, 3, 4}+{1, 3, 5} ., Thus, A + B ! , B+ C ! ; A + C ! and A + B+ C ., Give an example of three sets A, B, C such that A + B A + C but B ! C., Consider the sets A {1, 2, 3}, B {3, 4} and C {3, 5, 7} ., Then, A + B {1, 2, 3}+{3, 4} {3} ., And, A + C {1, 2, 3}+{3, 5, 7} {3} ., Thus, A + B A + C, and clearly, B ! C., , For any sets A and B, their difference (A B) is defined as, (A B) {x : xd A and xzB} ., Thus, xd(A B) & xd A and xzB., , DIFFERENCE OF SETS, , EXAMPLE 12, SOLUTION, , If A {x : xd N, x is a factor of 6} and B {xd N : x is a factor of 8} then, find (i) A , B, (ii) A + B, (iii) A B, (iv) B A., We have, A {x : xd N, x is a factor of 6} {1, 2, 3, 6}, and B {x : xd N, x is a factor of 8} {1, 2, 4, 8} ., (i) A , B {1, 2, 3, 6},{1, 2, 4, 8} {1, 2, 3, 4, 6, 8}., (ii) A + B {1, 2, 3, 6}+{1, 2, 4, 8} {1, 2} ., (iii) A B {1, 2, 3, 6} {1, 2, 4, 8} {3, 6} ., (iv) B A {1, 2, 4, 8} {1, 2, 3, 6} {4, 8} ., , SYMMETRIC DIFFERENCE OF TWO SETS, , The symmetric difference of two sets A and B, denoted by A T B, is defined as, A T B (A B),(B A) ., EXAMPLE 13, SOLUTION, , Let A {a, b, c, d} and B {b, d, f, g} . Find A T B., We have, (A B) {a, b, c, d} {b, d, f, g} {a, c} .
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Sets, , , , 21, , (B A) {b, d, f, g} {a, b, c, d} {f, g} ., A T B (A B),(B A) {a, c},{f, g} {a, c, f, g} ., , COMPLEMENT OF A SET, , Let U be the universal set and let AfU. Then, the complement of A, denoted by Al or, (U A), is definied as, Al {xdU : xz A} ., Clearly, xd Al + xz A., EXAMPLE 14, SOLUTION, , EXAMPLE 15, , SOLUTION, , If U {1, 2, 3, 4, 5, 6, 7, 8} and A {2, 4, 6, 8}, find (i) Al (ii) (Al)l ., We have, (i) Al U A, {1, 2, 3, 4, 5, 6, 7, 8} {2, 4, 6, 8} {1, 3, 5, 7} ., (ii) (Al)l U Al, {1, 2, 3, 4, 5, 6, 7, 8} {1, 3, 5, 7} {2, 4, 6, 8} A., Let N be the universal set., (i) If A {x : xd N and x is odd}, find Al., (ii) If B {x : xd N, x is divisible by 3 and 5}, find Bl., We have, (i) Al {x : xd N and x is not odd} {x : xd N and x is even}., (ii) Bl {x : xd N, x is not divisible by 3 or x is not divisible by 5}., , SOME RESULTS ON COMPLEMENTATION, EXAMPLE 16, , SOLUTION, , If A 1 U, prove that:, (i) Ul (ii) l U (iii) (Al)l A (iv) A , Al U (v) A + Al , We have, (i) Ul {xdU : xzU} ., (ii) l {xdU : xz} U., (iii) (Al)l {xdU : xz Al} [xdU : xd A} A., (iv) A , Al {xdU : xd A , Al}, {xdU : xd A or xd Al}, {xdU : xd A or xz A} U., (v) A + Al {xdU : xd A + Al}, {xdU : xd A and xd Al}, {xdU : xd A and xz A} ., , EXERCISE 1D, 1. If A {a, b, c, d, e, f }, B {c, e, g, h} and C {a, e, m, n}, find:, (i) A , B, (ii) B, C, (iii) A , C, (iv) B+ C, (v) C + A, (vi) A + B
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22, , Senior Secondary School Mathematics for Class 11, , 2. If A {1, 2, 3, 4, 5}, B {4, 5, 6, 7, 8}, C {7, 8, 9, 10, 11} and, D {10, 11, 12, 13, 14}, find:, (i) A , B, (ii) B, C, (iii) A , C, (iv) B, D, (v) (A , B), C, (vi) (A , B)+ C, (vii) (A + B), D, (viii) (A + B),(B+ C), (ix) (A , C)+(C , D), 3. If A {3, 5, 7, 9, 11}, B {7, 9, 11, 13}, C {11, 13, 15} and D {15, 17}, find:, (i) A + B, (ii) A + C, (iii) B+ C, (iv) B+ D, (v) B+(C , D), (vi) A +(B, C), 4. If A {x : xd N}, B {x : xd N and x is even}, C {x : xd N and x is odd} and, D {x : xd N and x is prime} then find:, (i) A + B, (ii) A + C, (iii) A + D, (iv) B+ C, (v) B+ D, (vi) C + D, 5. If A {2x : xd N and 1 # x 4}, B {(x 2) : xd N and 2 # x 5} and, C {x : xd N and 4 x 8}, find:, (i) A + B, (ii) A , B, (iii) (A , B)+ C, 6. If A {2, 4, 6, 8, 10, 12} and B {3, 4, 5, 6, 7, 8, 10}, find:, (i) (A B), (ii) (B A), (iii) (A B),(B A), 7. If A {a, b, c, d, e}, B {a, c, e, g} and C {b, e, f, g}, find:, (i) A +(B C), (ii) A (B, C), (iii) A (B+ C), 1, 1, 8. If A & x : xd N and x 80 and B & : xd N and x # 40 , find:, 2x, , (ii) A + B, (iii) A B, (iv) B A, (i) A , B, 9. If R is the set of all real numbers and Q is the set of all rational numbers, then what is the set (R Q) ?, 10. If A {2, 3, 5, 7, 11} and B , find:, (i) A , B, (ii) A + B, 11. If A and B are two sets such that A 3 B then find:, (i) A , B, (ii) A + B, (iii) A B, 12. Which of the following sets are pairs of disjoint sets? Justify your answer:, (i) A {3, 4, 5, 6} and B {2, 5, 7, 9}, (ii) C {1, 2, 3, 4, 5} and D {6, 7, 9, 11}, (iii) E {x : xd N, x is even and x 8}, F {x : x 3n, nd N and n 4}, (iv) G {x : xd N, x is even} and H {x : xd N, x is prime}, (v) J {x : xd N, x is even} and K {x : xd N, x is odd}, 13. If U {1, 2, 3, 4, 5, 6, 7, 8, 9}, A {1, 2, 3, 4}, B {2, 4, 6, 8} and C {1, 4, 5, 6},, find:, (i) Al, (ii) Bl, (iii) Cl, (iv) (Bl)l, (v) (A , B)l, (vi) (A + C)l, (vii) (B C)l
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Sets, , 23, , 14. If U {a, b, c, d, e}, A {a, b, c} and B {b, c, d, e} then verify that:, (i) (A , B)l (Al+ Bl), (ii) (A + B)l (Al, Bl), 15. If U is the universal set and AfU then fill in the blanks:, (ii) A + Al ……, (i) A , Al ……, (iii) l+ A ……, (iv) Ul+ A ……, ANSWERS (EXERCISE 1D), , 1., 2., , 3., 4., 5., 6., 7., 8., 9., 10., 12., 13., 15., , (i), (iv), (i), (iii), (v), (vii), (ix), (i), (v), (i), (v), (i), (i), (i), , {a, b, c, d, e, f, g, h} (ii) {a, c, e, g, h, m, n} (iii) {a, b, c, d, e, f, m, n}, {e}, (v) {a, e}, (vi) {c, e}, {1, 2, 3, 4, 5, 6, 7, 8}, (ii) {4, 5, 6, 7, 8, 9, 10, 11}, {1, 2, 3, 4, 5, 7, 8, 9, 10, 11}, (iv) {4, 5, 6, 7, 8, 10, 11, 12, 13, 14}, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}, (vi) {7, 8}, {4, 5, 10, 11, 12, 13, 14}, (viii) {4, 5, 7, 8}, {7, 8, 9, 10, 11}, {7, 9}, (ii) {11}, (iii) {11, 13}, (iv) , {11, 13}, (vi) {7, 9, 11}, B, (ii) C, (iii) D, (iv) , {2}, (vi) D {2}, {4, 6}, (ii) {2, 4, 5, 6}, (iii) {5, 6}, {2, 12}, (ii) {3, 5, 7}, (iii) {2, 3, 5, 7, 12}, {a, c}, (ii) {d}, (iii) {a, b, c, d}, , 1, 1 1 1, 1, 1, 1, 1, 1, 1, 1, 1 1 1, 0, (ii) & , , 0, (iii) &1, , , 0 (iv) & 0, 8, 3 5 7, 2 3 4 5 6 7 8, 2 4 6, (R Q) {x : xd R, x is irrational}, (i) 2, 3, 5, 7, 11} (ii) , 11. (i) B (ii) A (iii) , (ii) C and D, since C + D (v) J and K, since J + K , (i) {5, 6, 7, 8, 9}, (ii) {1, 3, 5, 7, 9}, (iii) {2, 3, 7, 8, 9} (iv) {2, 4, 6, 8}, (v) {5, 7, 9}, (vi) {2, 3, 5, 6, 7, 8, 9}, (vii) {1, 3, 4, 5, 6, 7, 9}, (i) U, (ii) , (iii) A, (iv) , (i) &1,, , LAWS OF OPERATIONS ON SETS, , For any set A, prove that:, (i) A , A A, (ii) A + A A, We have, (i) A , A {x : xd A or xd A} {x : xd A} A., (ii) A + A {x : xd A and xd A} {x : xd A} A., , THEOREM 1 (Idempotent laws), , PROOF, , For any set A, prove that:, (i) A , A, (ii) A +U A, where U is the universal set, , THEOREM 2 (Identity laws)
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24, PROOF, , NOTE, , Senior Secondary School Mathematics for Class 11, , We have, (i) A , {x : xd A or xd} {x : xd A} A, [a xz], , , , (ii) A +U {x : xd A and xdU} {x : xd A} A, [a AfU], and U are the identity elements for union and intersection of sets, respectively., For any two sets A and B, prove that:, I. A , B B, A, [Commutative law for union of sets], II. A + B B+ A, [Commutative law for intersection of sets], I. Let x be an arbitrary element of A , B. Then,, xd A , B & xd A or xdB, & xdB or xd A, & xdB, A., , … (i), A , B 3 B , A., Again, let y be an arbitrary element of B, A. Then,, y dB, A & y dB or y d A, & y d A or y dB, & y d A , B., , … (ii), B, A 3 A , B., From (i) and (ii), we get A , B B, A., , THEOREM 3 (Commutative laws), , PROOF, , II. Let x be an arbitrary element of A + B. Then,, xd A + B & xd A and xdB, & xdB and xd A, & x d B + A., , A + B 3 B + A., Again, let y be an arbitray element of B+ A. Then,, y dB+ A & y dB and y d A, & y d A and y dB, & y d A + B., B+ A 3 A + B., From (iii) and (iv), we get A + B B+ A., , … (iv), , For any sets A, B, C, prove that:, I. (A , B), C A ,(B, C) [Associative law for union of sets], II. (A + B)+ C A +(B+ C) [Associative law for intersection of sets], I. Let x be an arbitrary element of (A , B), C. Then,, xd(A , B), C & xd(A , B) or xdC, & (xd A or xdB) or xdC, & xd A or (xdB or xdC), & xd A or xd(B, C), & xd A ,(B, C) ., , THEOREM 4 (Associative laws), , PROOF, , … (iii)
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Sets, , 25, , (A , B), C 3 A ,(B, C) ., … (i), Again, let y be an arbitrary element of A ,(B, C) . Then,, y d A ,(B, C) & y d A or y d(B, C), & y d A or (y dB or y dC), & (y d A or y dB) or y dC, & y d(A , B) or y dC, & y d(A , B), C., A ,(B,C) 3 (A , B), C., … (ii), From (i) and (ii), we get (A , B), C A ,(B, C) ., II. Let x be an arbitrary element of (A + B)+ C. Then,, xd(A + B)+ C & xd(A + B) and xdC, & (xd A and xdB) and xdC, & xd A and (xdB and xdC), & xd A and xd(B+ C), & xd A +(B+ C)., (A + B)+ C 3 A +(B+ C) ., … (iii), Again, let y be an arbitrary element of A +(B+ C) . Then,, y d A +(B+ C) & y d A and y d(B+ C), & y d A and (y dB and y dC), & (y d A and y dB) and y dC, & y d(A + B) and y dC, & y d(A + B)+ C, A +(B+C) 3 (A + B)+ C., … (iv), From (iii) and (iv), we get (A + B)+ C A +(B+ C) ., For any three sets A, B, C prove that:, I. A ,(B+ C) (A , B)+(A , C), [Distributive law of union over intersection], II. A +(B, C) (A + B),(A + C), [Distributive law of intersection over union], I. Let x be an arbitrary element of A ,(B+ C) . Then,, xd A ,(B+ C) & xd A or xd(B+ C), & xd A or (xdB) and xdC, & (xd A or xdB) and (xd A or xdC)), [ ‘or‘ distributes ‘and‘], & xd(A , B) and xd(A , C)., & xd(A , B)+(A , C) ., A ,(B+ C) 3 (A , B)+(A , C) ., … (i), Again, let y be an arbitrary element of (A , B)+(A , C) . Then,, , THEOREM 5 (Distributive laws), , PROOF
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26, , Senior Secondary School Mathematics for Class 11, , y d(A , B)+(A , C) & y d(A , B) and y d(A , C), & (y d A or y dB) and (y d A or y dC), & y d A or (y dB and y dC), [ ‘or‘ distributes ‘and‘], & y d A or y d(B+ C), & y d A ,(B+ C)., (A , B)+(A , C) 3 A ,(B+ C) ., … (ii), From (i) and (ii), we get A ,(B+ C) (A , B)+(A , C) ., II. Let x be an arbitrary element of A +(B, C) . Then,, xd A +(B, C) & xd A and xd(B, C), & xd A and (xdB or xdC), & (xd A and xdB) or (xd A and xdC), [ ‘and‘ distributes ‘or‘], & xd(A + B) or xd(A + C), & xd(A + B),(A + C)., A +(B, C) 3 (A + B),(A + C) ., … (iii), Again, let y be an arbitrary element of (AkB)j(AkC) . Then,, y d(A + B),(A + C) & y d(A + B) or y d(A + C), & (y d A and y dB) or (y d A and y dC), & y d A and (y dB or y dC), [ ‘and‘ distributes ‘or‘], & y d A and y d(B, C), & y d A +(B, C)., (A + B),(A + C) 3 A +(B, C)., … (iv), From (iii) and (iv), we get A +(B, C) (A + B),(A + C) ., For any two sets A and B, prove that:, I. (A , B)l (Al+ Bl), II. (A + B)l (Al, Bl), I. Let x be an arbitrary element of (A , B)l . Then,, xd(A , B)l & xz A , B, [note this point], & xz A and xzB, & xd Al and xdBl, & xd(Al+ Bl)., (A , B)l3 (Al+ Bl) ., … (i), Again, let y be an arbitrary element of (Al+ Bl) . Then,, y d(Al+ Bl) & y d Al and y dBl, & y z A and y zB, [note this point], & y z(A , B), & y d(A , B)l., , THEOREM 6 (De Morgan’s laws), , PROOF
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Sets, , 27, , (Al+ Bl) 3 (A , B)l ., … (ii), From (i) and (ii), we get (A , B)l (Al+ Bl) ., II. Let x be an arbitrary element of (A + B)l . Then,, xd(A + B)l & xz(A + B), [note this point], & xz A or xzB, & xd Al or xdBl, & xd(Al, Bl), (A + B)l3 (Al, Bl) ., … (iii), Again, let y be an arbitrary element of (Al, Bl) . Then,, y d(Al, Bl) & y d Al or y dBl, & y z A or y zB, [note this point], & y z(A + B), & y d(A + B)l., (Al, Bl) 3 (A + B)l ., … (iv), From (iii) and (iv), we get (A + B)l (Al, Bl) ., SOME MORE RESULTS, THEOREM 7, , PROOF, , For any two sets A and B, prove that:, I. A 3 B & Bl3 Al, II. A B A + Bl, III. (A B)+ B , IV. (A B),(B A) (A , B) (A + B), V. (A B) A + A + B , I. Let A 3 B be given and let x be an arbitrary element of Bl. Then,, xdBl & xzB, & xz A, [a A 3 B], & xd Al., Bl3 Al., Hence, A 3 B & Bl3 Al., II. Let x be an arbitrary element of (A B) . Then,, xd(A B) & xd A and xzB, & xd A and xdBl, & xd A + Bl., (A B) 3 A + Bl., … (i), Again, let y be an arbitrary element of (A + Bl) . Then,, xd(A + Bl) & xd A and xdBl, & xd A and xzB, & xd(A B)., (A + Bl) 3 (A B) ., … (ii), , , Hence, from (i) and (ii), we get (A B) (A + Bl) ., III. If possible, let (A B)+ B ! and let xd(A B)+ B. Then,, xd(A B)+ B & xd(A B) and xdB
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28, , Senior Secondary School Mathematics for Class 11, , & (xd A and xzB) and xdB, & xd A and (xzB and xdB) ., But, xzB and xdB can never hold simultaneously., Thus, we arrive at a contradiction., Since the contradiction arises by assuming that (A B)+ B !, and hence (A B)+ B ., IV. Let x be an arbitrary element of (A B),(B A) . Then,, xd(A B),(B A) & xd(A B) or xd(B A), & (xd A and xzB) or (xdB and xz A), & (xd A or xdB) and (xz A or xzB), [note it], & xd(A , B) and xz(A + B), & xd{(A , B) (A + B)} ., (A B),(B A) 3 {(A , B) (A + B)} ., … (iii), Again, let y be an arbitrary element of (A , B) (A + B) . Then,, y d(A , B) (A + B) & y d(A , B) and y z(A + B), & (y d A or y dB) and (y z A or y zB), [note it], & (y d A and y zB) or (y dB and y z A), [note it], & y d(A B) or y d(B A), & y d(A B),(B A) ., ( A , B ) ( A + B) 3 ( A B) , ( B A ) ., … (iv), From (iii) and (iv), we get (A B),(B A) (A , B) (A + B) ., V. Let (A B) A be given and we have to show that A + B ., If possible, let A + B ! and let xd A + B. Then,, xd A + B & xd A and xdB, & xd(A B) and xdB [a A (A B) (given)], & (xd A and xzB) and xdB, & xd A and (xzB and xdB) ., But, xzB and xdB both can never hold simultaneously., Thus, we arrive at a contradiction., Since the contradiction arises by assuming that A + B !, so, A + B ., Thus, (A B) A & A + B ., Again, let (A + B) and we have to show that (A B) A., Now, xd(A B) & xd A and xzB, & xd A (surely)., ( A B) 3 A ., … (v)
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Sets, , 29, , [a A + B ], , Again, y d A & y zB, & y d A and y zB, & y d(A B) ., A 3 (A B) ., Thus, from (v) and (vi), we get (A B) A., A + B & (A B) A., Hence, (A B) A + (A + B) ., , … (vi), , If (A , B) (A + B) then prove that A B., Let (A , B) (A + B) be given., , THEOREM 8, PROOF, , Let x be an arbitrary element of A. Then,, xd A & xd A, B, & xd A+ B, & xd A and xdB, & xdB (surely)., , A 3 B., Again, let y dB. Then,, y dB & y d A , B, & yd A+ B, & y d A and y dB, & y d A (surely)., , B 3 A., Thus, from (i) and (ii), we get A B., THEOREM 9, PROOF, , … (i), [a B 3 A , B], [a A , B A + B], , … (ii), , If A 3 B then for any set C, prove that (C B) 3 (C A) ., , Let A 3 B be given., Let xd(C B) . Then,, xd(C B) & xdC and xzB, & xdC and xz A, & xd(C A) ., , ( C B) 3 ( C A ) ., Hence, A 3 B & (C B) 3 (C A) ., , THEOREM 10, , PROOF, , [a A 3 A , B], [a A , B A + B], , [a A 3 B], , For any sets A and B, prove that:, (i) A ,(A + B) A, (ii) A +(A , B) A, , (i) Since (A + B) 3 A, we have A ,(A + B) A [a X 3 Y & X ,Y Y] ., (ii) Since A 3 (A , B), we have A +(A , B) A [a X 3 Y & X +Y X] ., , THEOREM 11, , For any sets A and B, prove that:, (i) (A + B),(A B) A, (ii) A ,(B A) (A , B)
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30, PROOF, , Senior Secondary School Mathematics for Class 11, , (i) We have, (A + B),(A B) (A + B),(A + Bl), A +(B, Bl), A +U, A., Hence, (A + B),(A B) A., (ii) We have, A ,(B A) A ,(B+ Al), (A , B)+(A , Al), (A , B)+U, ( A , B) ., Hence, A ,(B A) (A , B) ., , [a (A B) (A + Bl)], [by distributive law], [a B, Bl U], , [a (B A) (B+ Al)], [by distributive law], [a A , Al U], , If A + Bl then prove that A A + B and hence show that A 3 B., Let A + Bl be given. Then,, A (A +U), where U is the universal set, A +(B, Bl), [a B, Bl U], (A + B),(A + Bl), (A + B),, [a A + Bl ], ( A + B) ., Hence, A (A + B) ., Further, let A A + B and let xd A. Then,, xd A & xd A+ B, [a A A + B], & xd A and xdB, & xdB (surely)., , A 3 B., , THEOREM 12, PROOF, , If A, B and C be the sets such that A , B A , C and A + B A + C, then prove that B C., Let A , B A , C and A + B A + C be given. Then,, A, B A,C, , (A , B)+ B (A , C)+ B and (A , B)+ C (A , C)+ C, , B (A + B),(C + B) and (A +C),(B+ C) C, [a B 3 (A , B) and C 3 (A , C)], , B (A + B),(B+ C) and (A + B),(B+ C) C [a A + C A + B], , B C., Hence, B C., , THEOREM 13, PROOF, , THEOREM 14, , For any sets A, B and C, prove that:, (i) A (B, C) (A B)+(A C), (ii) A (B+ C) (A B),(A C), (iii) (A , B) C (A C),(B C), (iv) (A + B) C (A C)+(B C)
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Sets, PROOF, , We have, (i) A (B, C) A +(B, C)l, A +(Bl+ Cl), (A + Bl)+(A + Cl), (A B)+(A C) ., A (B, C) (A B)+(A C) ., (ii) A (B+ C) A +(B+ C)l, A +(Bl, Cl), (A + Bl),(A + Cl), (A B),(A C), , A (B+ C) (A B),(A C) ., (iii) (A , B) C (A , B)+ Cl, (A +Cl),(B+ Cl), (A C),(B C) ., (A , B) C (A C),(B C) ., (iv) (A + B) C (A + B)+ Cl, (A +Cl)+(B+ Cl), (A C)+(B C) ., (A + B) C (A C)+(B C) ., , 31, , [a X Y X +Yl], [by De Morgan’s law], , [a X Y X +Yl], [by De Morgan‘s law], [by distributive law], [a X +Yl X Y]., [a X Y X +Yl], [by distributive law], [a X +Yl X Y], [a X Y X +Yl], [note], [a X +Yl X Y], , Let A and B be sets. If A + X B+ X and A , X B, X for some, set X, show that A B., [given], A, X B, X, , A + (A , X) A + (B , X), , [by distributive law], (A + A),(A + X) (A + B),(A + X), , , A , ( A + B) , , [a A + X ], , A ( A + B), , … (i) [a A + B A & A 3 B]., A3B, Again, A , X B, X, , B + ( A , X) B + ( B , X), , [by distributive law], ( B + A ) , ( B + X ) ( B + B ) , (B + X ), , ( A + B) , B , , [a B+ X and B+ A A + B], , A+ B B, , … (ii) [a A + B B & B 3 A]., B3 A, From (i) and (ii), we get A B., , THEOREM 15, PROOF, , Show that the following four conditions are equivalent:, (i) A 1 B, (ii) A B , (iii) A , B B, (iv) A + B A, In order to prove the required result, we will show that:, (i) (ii) (iii) (iv) (i)., , THEOREM 16, , PROOF
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32, , Senior Secondary School Mathematics for Class 11, , Now, (i) (ii):, Let A 1 B be given., Then, there is no element of A which is not in B., , A B {x : xd A and xzB} , [ there is no element of A which is not in B]., Hence, A 1 B & A B and therefore, (i) & (ii)., (ii) & (iii):, Let A B be given. Then,, A B & every element of A is in B, & A1B, & A , B B., Thus, A B & A , B B and therefore, (ii) & (iii)., (iii) & (iv):, Let A , B B be given. Then,, A , B B & A 1 B & A + B A., Thus, A , B B & A + B A and therefore, (iii) (iv)., (iv) (i):, , Let A + B A be given. Then,, xd A & xd A+ B, & xd A and xdB, & xdB (surely)., , A 3 B., Thus, A + B A & A 3 B and therefore, (iv) (i)., , (i) (ii) (iii) (iv) (i)., Hence, the given four conditions are equivalent., For any sets A and B, prove that, P ( A + B) P ( A ) + P (B) ., Let X d P(A + B) . Then,, X d P ( A + B) & X 3 A + B, & X 3 A and X 3 B, & X d P(A) and X d P(B), & X d P ( A ) + P (B) ., , P(A + B) 3 P(A)+ P(B)., Again, let Y d P(A)+ P(B) . Then,, Y d P(A)+ P(B) & Y d P(A) and Y d P(B), & Y 3 A and Y 3 B, & Y 3 A+ B, & Y d P ( A + B) ., , [a A A + B], , THEOREM 17, , PROOF, , … (i)
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Sets, , 33, , , P ( A ) + P (B) 3 P ( A + B) ., From (i) and (ii), we get P(A + B) P(A)+ P(B) ., , … (ii), , For any two sets A and B, prove that, P ( A ) , P (B) 1 P ( A , B) ., But, P(A , B) is not necessarily a subset of P(A), P(B) ., Let X be an arbitrary element of P(A), P(B) . Then,, X d P(A), P(B) & X d P(A) or X d P(B), & X 1 A or X 1 B, & X 1 ( A , B), & X d P ( A , B) ., , P ( A ) , P (B) 1 P ( A , B) ., However, P(A , B) 1 P(A), P(B) is not always true., For example, let A {1} and B {2} . Then, A , B {1, 2} ., , P(A) {, {1}}, P(B) {, {2}}, and P(A , B) {, {1}, {2}, {1, 2}} ., Also, P(A), P(B) {, {1}, {2}} ., Y P ( A ) , P (B) ., , P ( A , B) 1, Hence, in general, P(A , B) ! P(A), P(B) ., , THEOREM 18, , PROOF, , THEOREM 19, PROOF, , If P(A) P(B), prove that A B., , Let P(A) P(B) . Then,, A 3 A & Ad P ( A ), & A d P ( B), & A 3 B., Again, B 3 B & Bd P(B), & Bd P(A), & B 3 A., From (i) and (ii), we get A 3 B and B 3 A., Hence, A = B., , [a P (A) P(B)], … (i), [a P(B) P(A)], … (ii), , EXERCISE 1E, 1. If A {a, b, c, d, e}, B {a, c, e, g} and C {b, e, f, g}, verify that:, (i) A , B B, A, (ii) A , C C , A, (iii) B, C C , B, (iv) A + B B+ A, (v) B+ C C + B, (vi) A + C C + A, (vii) (A , B), C A ,(B, C), (viii) (A + B)+ C A +(B+ C), 2. If A {a, b, c, d, e}, B {a, c, e, g} and C {b, e, f, g}, verify that:, (i) A +(B C) (A + B) (A + C), (ii) A (B+ C) (A B),(A C), 3. If A {x : xd N, x # 7}, B {x : x is prime, x 8} and C {x : xd N, x is odd, and x 10}, verify that:, (i) A ,(B+ C) (A , B)+(A , C), (ii) A +(B, C) (A + B),(A + C)
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34, , Senior Secondary School Mathematics for Class 11, , 4. If U {1, 2, 3, 4, 5, 6, 7, 8, 9}, A {2, 4, 6, 8} and B {2, 3, 5, 7}, verify that:, (i) (A , B)l (Al+ Bl), (ii) (A + B)l (Al, Bl), 5. Let A {a, b, c}, B {b, c, d, e} and C {c, d, e, f } be subsets of, U {a, b, c, d, e, f } . Then, verify that:, (i) (Al)l A, (ii) (A , B)l (Al+ Bl) (iii) (A + B)l (Al, Bl), 6. Given an example of three sets A, B, C such that A + B ! , B+ C ! ,, A + C ! and A + B+ C ., 7. For any sets A and B, prove that:, (i) (A B)+ B , (ii) A ,(B A) A , B, (iii) (A B),(A + B) A, (iv) (A , B) B A B, (v) A (A + B) A B, 8. For any sets A and B, prove that:, (i) A + Bl & A 1 B, (ii) Al, B U & A 1 B, HINTS TO SOME SELECTED QUESTIONS, 6. Take A {1, 2}, B {2, 3} and C {1, 3} ., 7. (i) (A B)+ B (A + Bl)+ B A +(Bl+ B) A + ., (ii) A ,(B A) A ,(B+ Al) (A , B)+(A , Al) (A , B)+U (A , B) ., (iii) (A B),(A + B) (A + Bl),(A + B) A +(Bl, B) A +U A., (iv) (A , B) B (A , B)+ Bl (A + Bl),(B + Bl) (A + Bl), A + Bl A B., (v) A (A + B) A +(A + B)l A +(Al, Bl) (A + Al),(A + Bl) ,(A + Bl), (A + Bl) (A B) ., 8. (i) A + Bl & A B & A 1 B., (ii) Al, B U & A +(Al, B) A +U & (A + Al),(A + B) A, & ,(A + B) A & A + B A & A 1 B., , VENN DIAGRAMS, In order to express the relationship among sets in perspective, we represent, them pictorially by means of diagrams, called Venn diagrams., In these diagrams, the universal set is represented by a rectangular region, and its subsets by circles inside the rectangle. We represent disjoint sets by, disjoint circles and intersecting sets by intersecting circles., VENN DIAGRAMS IN DIFFERENT SITUATIONS, CASE 1, , When the universal set and its subset are given, Let U be the universal set and let A 3U., We draw a circle inside a rectangle., The rectangular region represents U and the circular region, represents A.
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Sets, EXAMPLE, , CASE 2, , EXAMPLE, , CASE 3, , EXAMPLE, , CASE 4, , Let U {1, 2, 3, 4, 5, 6, 7} and, A {1, 3, 5, 7} ., Then, we draw the Venn diagram,, as shown in the given figure., Clearly, Al {2, 4, 6} ., , 35, U, 2, 4, 6, , 5, , 1, 3, , 7, , A, , When two intersecting subsets of U are, U, given, For representing two intersecting, A – B A ÈB, subsets A and B of U, we draw two, intersecting circles within the rectangle., B–A, A, The common region of these circles, B, represents A + B., Excluding the region of B from that of A shows (A B) ., Excluding the region of A from that of B shows (B A) ., Let U {1, 2, 3, 4, 5, 6, 7, 8} be the U, universal set, and let, A {1, 3, 4, 5} and B {2, 4, 5, 6} be 7, its subsets., 8, Then, A + B {4, 5} ., We draw the Venn diagram, as, shown in the given figure., Clearly, (A B) {1, 3} and (B A) {2, 6} ., , 4, 5, , 1, 3, A, , 2, 6, B, , When two disjoint subsets of a set be given, In order to represent two disjoint subsets A and B of the universal set, U, we draw two disjoint circles within a rectangle., Let U {1, 2, 3, 4, 5, 6, 7} be the U, universal set, and let A {1, 3, 5}, and B {2, 4} be two of its disjoint, 1, 2, 3, subsets., 4, 5, Clearly, A + B ., So, we may draw the Venn diagram, 6, B, A, 7, as shown in the adjoining figure., Clearly, A + B , (A B) {1, 3, 5} A and (B A) {2, 4} B., Al {2, 4, 6, 7} and Bl {1, 3, 5, 6, 7} ., When B 3 A 3U, In this case, we draw two concentric circles within a rectangular, region., The inner circle represents B and the outer circle represents A.
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36, , Senior Secondary School Mathematics for Class 11, , EXAMPLE, , Let U {1, 2, 3, 4, 5, 6, 7, 8} be the U, universal set, and let A {1, 3, 5, 7}, 2, and B {3, 7} be its subsets., 4, Then, clearly B 3 A., 6, Now, we may draw the Venn 8, diagram, as shown in the given, figure., A + B B {3, 7}, A , B A {1, 3, 5, 7},, (A B) {1, 5}, (B A) ,, Al {2, 4, 6, 8} and Bl {1, 5, 2, 4, 6, 8} ., , 1, 5, 3, 7, B, A, , IMPORTANT RESULTS FROM VENN DIAGRAMS, , Let A and B be two intersecting subsets of U., In counting the elements of (A , B), the elements, of A + B are counted twice, once in counting the, elements of A and second time in counting the, elements of B., n ( A , B ) n ( A ) n (B ) n (A + B ) ., , A–B A, , È, , B, , A, , B–A, , B, , If A + B , then n(A + B) 0 and therefore, in this case, we have, n ( A , B) n ( A ) n ( B ) ., From the Venn diagram, it is also clear that, (i) n(A B) n(A + B) n(A), (ii) n(B A) n(A + B) n(B), (iii) n(A B) n(A + B) n(B A) n(A , B) ., , EXERCISE 1F, 1. Let A {a, b, c, e, f }, B {c, d, e, g} and C {b, c, f, g} be subsets of the set, U {a, b, c, d, e, f, g, h} ., Draw Venn diagrams to represent the following sets:, (i) A + B, (ii) A ,(B+ C), (iii) A B, (iv) B A, (v) A (B+ C), (vi) (B C),(C B), 2. Let A {2, 4, 6, 8, 10}, B {4, 8, 12, 16} and C {6, 12, 18, 24} ., Using Venn diagrams, verify that:, (i) (A , B), C A ,(B, C), (ii) (A + B)+ C A +(B+ C), 3. Let A {a, e, i, o, u}, B {a, d, e, o, v} and C {e, o, t, m} ., Using Venn diagrams, verify the following:, (i) A ,(B+ C) (A , B)+(A , C), (ii) A +(B, C) (A + B),(A + C) ., 4. Let A 1 B 1 U. Exhibit it in a Venn diagram., 5. Let A {2, 3, 5, 7, 11, 13} and B {5, 7, 9, 11, 15} be subsets of, U {2, 3, 5, 7, 9, 11, 13, 15} .
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Sets, , 37, , Using Venn diagrams, verify that:, (i) (A , B)l (Al+ Bl), (ii)(A + B)l (Al, Bl), 6. Using Venn diagrams, show that (A B), (A + B) and (B A) are disjoint, sets, taking A {2, 4, 6, 8, 10, 12} and B {3, 6, 9, 12, 15} ., , SOME RESULTS DERIVED FROM VENN DIAGRAMS, , When A and B are disjoint sets, then we have, n ( A , B) n ( A ) n ( B ) ., , RESULT 1, , ´´´, ´´, A, , For any sets A and B, prove that, n ( A , B ) n ( A ) n (B ) n (A + B ) ., , RESULT 2, , A–B, , PROOF, , B, , A, , È, , B, , B–A, , From the given Venn diagram, it is clear, that the sets (A B), (A + B) and (B A), are disjoint and their union is (A , B) ., A, B, , n ( A , B) n ( A B) n ( A + B) n ( B A ), n ( A B ) n ( A + B ) n ( B A ) n ( A + B ) n (A + B ), [adding and subtracting n(A + B)], n ( A ) n ( B ) n (A + B ), [a n(A B) n(A + B) n(A) and n(B A) n(A + B) n(B)], Hence, n(A , B) n(A) n(B) n(A + B) ., Prove that n(A B) n(A + B) n(A) ., Clearly, (A B) and (A + B) are disjoint sets and their union is A., , n ( A B) n ( A + B ) n ( A ) ., , Corollary 1, PROOF, , Corollary 2, , Prove that n(B A) n(A + B) n(B) ., , PROOF, , Clearly, (B A) and (A + B) are disjoint sets whose union is B., , n ( B A ) n (A + B ) n (B) ., , RESULT 3, , For any sets A, B, C prove that, n(A , B, C) [n(A) n(B) n(C) n(A + B+ C)], [n(A + B) n(B+ C) n(A + C)] ., We have, n(A , B, C) n[(A , B), C], n(A , B) n(C) n[(A , B)+ C], {n(A) n(B) n(A + B)} n(C) n[(A + C),(B+ C), {n(A) n(B) n(C) n(A + B)} {n(A + C) n(B+ C), n(A + C + B+ C)}, , PROOF
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38, , Senior Secondary School Mathematics for Class 11, , n(A) n(B) n(C) n(A + B) n(A + C) n(B+ C), n(A + B+ C), {n(A) n(B) n(C) n(A + B+ C)}, {n(A + B) n(B+ C) n(A + C)} ., Hence, the result follows., SUMMARY, , For any sets A, B, C we have:, (i) n(A , B) n(A) n(B) n(A + B) ., (ii) If A + B , then n(A , B) n(A) n(B) ., , A–B, , A, , È, , B, , B–A, , (iii) n(A B) n(A + B) n(A) ., (iv) n(B A) n(A + B) n(B) ., , A, , B, , (v) n(A , B, C) {n(A) n(B) n(C) n(A + B+ C)}, {n(A + B) n(B+ C) n(A + C)} ., SOLVED EXAMPLES, EXAMPLE 1, SOLUTION, , EXAMPLE 2, , SOLUTION, , CASE 1, , CASE 2, , EXAMPLE 3, SOLUTION, , If A and B are two sets such that n(A) 27, n(B) 35 and n(A , B) 50,, find n(A + B) ., We know that, n ( A , B ) n ( A ) n (B ) n (A + B ), , n(A + B) n(A) n(B) n(A , B) (27 35 50) 12., Hence, n(A + B) 12., If A and B are two sets containing 3 and 6 elements respectively, what can, be the maximum number of elements in A , B?, Find also the minimum number of elements in A , B., We know that, ... (i), n ( A , B ) n ( A ) n (B ) n (A + B ) ., From (i), it is clear that n(A , B) will be maximum when n(A + B) 0., In that case, n(A , B) n(A) n(B) (3 6) 9., , maximum number of elements in (A , B) 9., From (i), it is clear that n(A , B) will be minimum when n(A + B) is, maximum, i.e., when n(A + B) 3., In this case, n(A , B) n(A) n(B) n(A + B) (3 6 3) 6., , minimum number of elements in A , B 6., A survey shows that 73% of the Indians like apples, whereas 65% like, oranges. What percentage of Indians like both apples and oranges?, Let A set of Indians who like apples, and B set of Indians who like oranges.
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Sets, , 39, , Then, n(A) 73, n(B) 65 and n(A , B) 100., n(A + B) n(A) n(B) n(A , B) (73 65 100) 38., Hence, 38% of the Indians like both apples and oranges., EXAMPLE 4, , SOLUTION, , EXAMPLE 5, , SOLUTION, , EXAMPLE 6, , In a survey of 425 students in a school, it was found that 115 drink apple, juice, 160 drink orange juice and 80 drink both apple as well as orange, juice. How many drink neither apple juice nor orange juice?, Let U set of all students surveyed;, A set of all students who drink apple juice, and B set of all students who drink orange juice., Then, n(U) 425, n(A) 115, n(B) 160 and n (A + B) 80., , n(A , B) n(A) n(B) n(A + B) (115 160 80) 195., Set of students who drink neither apple juice nor orange juice, (Al+ Bl) (A , B)l, , n{(A , B)l} n(U) n(A , B) (425 195) 230., Hence, 230 students drink neither apple juice nor orange juice., In a group of 850 persons, 600 can speak Hindi and 340 can speak Tamil., Find (i) how many can speak both Hindi and Tamil,, (ii) how many can speak Hindi only,, (iii) how many can speak Tamil only., Let A set of persons who can speak Hindi, and B set of persons who can speak Tamil., , n(A) 600, n(B) 340 and n(A , B) 850., (i) Set of persons who can speak both Hindi and Tamil (A + B) ., Now, n(A + B) n(A) n(B) n(A , B), (600 340 850) 90., Thus, 90 persons can speak both, Hindi and Tamil., È, (ii) Set of persons who can speak Hindi, A–B A B, B–A, only (A B) ., Now, n(A B) n(A + B) n(A), , n ( A B) n ( A ) n ( A + B ), A, B, (600 90) 510., Thus, 510 persons can speak Hindi only., (iii) Set of persons who can speak Tamil only (B A) ., Now, n(B A) n(A + B) n(B), , n(B A) n(B) n(A + B) (340 90) 250., Hence, 250 persons can speak Tamil only., A market research group conducted a survey of 1000 consumers and, reported that 745 consumers like product A and 430 consumers like, product B. What is the least number that must have liked both products?
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40, , Senior Secondary School Mathematics for Class 11, , SOLUTION, , Let P and Q be the sets of consumers who like product A and, product B respectively., Then, n(P) 745 and n(Q) 430., Now, n(P ,Q) n(P) n(Q) n(P +Q), , n(P ,Q) 745 430 n(P +Q), , n(P ,Q) 1175 n(P +Q) ., Clearly, n(P +Q) is least when n(P ,Q) is maximum and therefore,, n(P ,Q) 1000., So, 1000 1175 n(P +Q) & n(P +Q) (1175 1000) 175., Hence, the least number of consumers liking both the products, is 175., , EXAMPLE 7, , Out of 600 car owners investigated, 500 owned car A; 200 owned car B, and 50 owned both A and B cars. Verify whether the given data is correct, or not., Let P and Q be the sets of those who own car A and car B, respectively. Then,, n(P) 500, n(Q) 200 and n(P +Q) 50., Now, n(P ,Q) n(P) n(Q) n(P +Q), , n(P ,Q) (500 200 50) 650., This is a contradiction, since the maximum value of n(P ,Q) is 600., Hence, the given data is incorrect., , SOLUTION, , EXAMPLE 8, , SOLUTION, , In a group of 52 persons, 16 drink tea but not coffee and 33 drink tea., Find (i) how many drink tea and coffee both;, (ii) how many drink coffee but not tea., Let A set of persons who drink tea, and B set of persons who drink coffee., È, A–B A B B–A, Then, (A B) set of persons who drink, tea but not coffee., And, (B A) set of persons who drink, A, B, coffee but not tea., Given: n(A , B) 52, n(A B) 16 and n(A) 33., (i) Set of persons who drink tea and coffee both (A + B) ., Now, n(A B) n(A + B) n(A), , n(A + B) n(A) n(A B) (33 16) 17., Thus, 17 persons drink tea and coffee both., (ii) n(A , B) n(A) n(B) n(A + B), , n(B) n(A , B) n(A + B) n(A) (52 17 33) 36., Now, n(B A) n(A + B) n(B) ., , n(B A) n(B) n(A + B) (36 17) 19., Thus, number of persons who drink coffee but not tea = 19.
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Sets, EXAMPLE 9, , SOLUTION, , 41, , A school awarded 58 medals in three sports, namely 38 in football; 15 in, basketball and 20 in cricket. If 3 students got medals in all the three sports,, how many received medals in exactly two sports?, Let A, B and C denote the sets of students who won medals in, football, basketball and cricket respectively., Then, n(A) 38, n(B) 15, n(C) 20, n(A + B+ C) 3, and n(A , B, C) 58., We know that, n(A , B, C) n(A) n(B) n(C) n(A + B) n(B+ C), n(A + C) n(A + B+ C), , n(A + B) n(B+ C) n(A + C), n(A) n(B) n(C) n(A + B+ C) n(A , B, C), {(38 15 20 3) 58} (76 58) 18., Let a, b, c and d denote respectively, the number of students who won A, B, medals in football and basketball, a, both; basketball and cricket both;, d, football and cricket both and all, b, c, the 3 sports., Then, n(A + B) n(B+ C) n(A + C), = 18, C, , (a d) (b d) (c d) 18, , (a b c) 3d 18, , (a b c) 3 # 3 18 & a b c 9, [a d 3]., Hence, 9 students received medals in exactly two sports., , AN EASY APPROACH, EXAMPLE 10, , SOLUTION, , In a survey it is found that 21 people like product A, 26 people like product, B and 29 like product C. If 14 people like products A and B; 15 people like, products B and C; 12 people like products C and A; and 8 people like all, the three products, find, (i) how many people are surveyed in all;, (ii) how many like product C only., Let A, B, C denote respectively the, sets of people who like product A, B, a, e, b, B, and C respectively, as shown in the A, given figure., c, Let us denote the number of elements, d, f, contained in bounded regions by a,, b, c, d, e, f, g as shown in the given, g, figure., C
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42, , Senior Secondary School Mathematics for Class 11, , Then, we have, a b c d 21,, b c e f 26,, c d f g 29,, b c 14, c f 15, c d 12, and c 8., On solving these equations, we get, c 8, d 4, f 7, b 6, g 10, e 5, a 3., (i) Total number of surveyed people (a b c d e f g) 43., (ii) Number of persons who like product C only g 10., EXAMPLE 11, , SOLUTION, , In a survey of 25 students, it was found that 12 have taken physics,, 11 have taken chemistry and 15 have taken mathematics; 4 have taken, physics and chemistry; 9 have taken physics and mathematics; 5 have, taken chemistry and mathematics while 3 have taken all the three subjects., Find the number of students who have taken, (i) physics only;, (ii) chemistry only;, (iii) mathematics only;, (iv) physics and chemistry but not mathematics;, (v) physics and mathematics but not chemistry;, (vi) only one of the subjects;, (vii) at least one of the three subjects;, (viii) none of the three subjects., Let P, C and M be the sets of students who have taken physics,, chemistry and mathematics respectively., Let a, b, c, d, e, f and g denote the number of students in the respective, regions, as shown in the adjoining Venn diagram., As per data given, we have, Z]a b c d 12,, ]], ]] b c e f 11,, P, C, ]], a, e, b, ]] c d f g 15,, ]], ], c, [], b c 4,, ]], d, f, ]], c d 9,, ]], ]], c f 5,, g, ]], ], c 3., M, \, From these equations, we get, c 3, f 2, d 6, b 1., Now, c d f g 15 & 3 6 2 g 15 & g 4;, b c e f 11 & 1 3 e 2 11 & e 5;
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Sets, , 43, , a b c d 12 & a 1 3 6 12 & a 2;, , a 2, b 1, c 3, d 6, e 5, f 2 and g 4., So, we have:, (i) Number of students who offered physics only a 2., (ii) Number of students who offered chemistry only e 5., (iii) Number of students who offered mathematics only g 4., (iv) Number of students who offered physics and chemistry but, not mathematics b 1., (v) Number of students who offered physics and mathematics but, not chemistry d 6., (vi) Number of students who offered only one of the given subjects, (a e g) (2 5 4) 11., (vii) Number of students who offered at least one of the given, subjects (a b c d e f g) (2 1 3 6 5 2 4) 23., (viii) Number of students who offered none of the three given, subjects (25 23) 2., , EXERCISE 1G, 1. If A and B are two sets such that n(A) 37, n(B) 26 and n(A , B) 51, find, n ( A + B) ., 2. If P and Q are two sets such that n(P ,Q) 75, n(P +Q) 17 and n(P) 49,, find n(Q) ., 3. If A and B are two sets such that n(A) 24, n(B) 22 and n(A + B) 8, find:, (i) n(A , B), (ii) n(A B), (iii) n(B A), 4. If A and B are two sets such that n(A B) 24, n(B A) 19 and, n(A + B) 11, find:, (i) n(A), (ii) n(B), (iii) n(A , B), 5. In a committee, 50 people speak Hindi, 20 speak English and 10 speak both, Hindi and English. How many speak at least one of these two languages?, 6. In a group of 50 persons, 30 like tea, 25 like coffee and 16 like both. How, many like, (i) either tea or coffee?, (ii) neither tea nor coffee?, 7. There are 200 individuals with a skin disorder, 120 had been exposed to the, chemical C1, 50 to chemical C 2, and 30 to both the chemicals C1 and C 2 . Find, the number of individuals exposed to, (i) chemical C1 but not chemical C 2, (ii) chemical C 2 but not chemical C1, (iii) chemical C1 or chemical C 2, 8. In a class of a certain school, 50 students offered mathematics, 42 offered, biology and 24 offered both the subjects. Find the number of students offering
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44, , 9., , 10., 11., , 12., , 13., , 14., , 15., , Senior Secondary School Mathematics for Class 11, , (i) mathematics only,, (ii) biology only,, (iii) any of the two subjects., In an examination, 56% of the candidates failed in English and 48% failed, in science. If 18% failed in both English and science, find the percentage of, those who passed in both the subjects., In a group of 65 people, 40 like cricket and 10 like both cricket and tennis., How many like tennis only and not cricket? How many like tennis?, A school awarded 42 medals in hockey, 18 in basketball and 23 in cricket., If these medals were bagged by a total of 65 students and only 4 students, got medals in all the three sports, how many students received medals in, exactly two of the three sports?, In a survey of 60 people, it was found that 25 people read newspaper H, 26, read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both, H and T, 8 read both T and I, and 3 read all the three newspapers. Find, (i) the number of people who read at least one of the newspapers,, (ii) the number of people who read exactly one newspaper., In a survey of 100 students, the number of students studying the various, languages is found as: English only 18; English but not Hindi 23; English, and Sanskrit 8; Sanskrit and Hindi 8; English 26; Sanskrit 48 and no, language 24. Find, (i) how many students are studying Hindi,, (ii) how many students are studying English and Hindi both., In a town of 10,000 families, it was found that 40% of the families buy, newspaper A, 20% buy newspaper B, 10% buy newspaper C, 5% buy A and, B; 3% buy B and C and 4% buy A and C. If 2% buy all the three newspapers,, find the number of families which buy, (i) A only,, (ii) B only,, (iii) none of A, B and C., A class has 175 students. The following description gives the number of, students studying one or more of the subjects in this class: mathematics, 100, physics 70, chemistry 46; mathematics and physics 30; mathematics, and chemistry 28; physics and chemistry 23; mathematics, physics and, chemistry 18. Find, (i) how many students are enrolled in mathematics alone, physics alone, and chemistry alone,, (ii) the number of students who have not offered any of these subjects., ANSWERS (EXERCISE 1G), , 1. 12, 5. 60, , 2. 43, 6. (i) 39 (ii) 11, , 3. (i) 38 (ii) 16 (iii) 14, 7. (i) 90 (ii) 20 (iii) 140, , 4. (i) 35 (ii) 30 (iii) 54, 8. (i) 26 (ii) 18 (iii) 68
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Sets, , 9. 14%, 13. (i) 18 (ii) 3, , 10. 25, 35, 11. 22, 14. (i) 3300 (ii) 1400 (iii) 4000, , 45, , 12. (i) 52 (ii) 30, 15. (i) 60, 35, 13 (ii) 32, , HINTS TO SOME SELECTED QUESTIONS, 1. n(A + B) n(A) n(B) n(A , B)., 2. n(P ,Q) n(P) n(Q) n(P +Q)., 3. (i) n(A , B) n(A) n(B) n(A + B) ., (ii) n(A B) n(A + B) n(A) ., (iii) n(B A) n(A + B) n(B) ., , 16, A–B, , A, , 8, È, , 14, B, , A, , 4. (i) n(A B) n (A + B) n(A), , n(A) n(A B) n(A + B), , n(A) (24 11) 35., , A–B, , (ii) n(B A) n(A + B) n(B), , n (B) (19 11) 30., , B, , A, , 24, , (iii) n(A , B) n(A) n(B) n(A + B) ., , B–A, , È, , B, , 11, , A, , B–A, 19, , B, , 5. n(A , B) n(A) n(B) n(A + B) (50 20 10) 60., 6. n(A) 30, n(B) 25 and n(A + B) 16., (i) n(A , B) n(A) n(B) n(A + B) 39., (ii) n(Al+ Bl) n{(A , B)l} n (U) n (A , B) (50 39) 11., 7. n(U) 200, n(C1) 120, n(C2) 50 and n(C1 + C2) 30., (i) n(C1) n(C1 C2) n(C1 + C2), , n(C1 C2) n(C1) n(C1 + C2) (120 30) 90., (ii) n(C2) n(C2 C1) n(C1 + C2), , n(C2 C1) n(C2) n(C1 + C2) (50 30) 20., (iii) n(C1 , C2) n(C1) n(C2) n(C1 + C2), (120 50 30) 140., 8. n(A) 50, n(B) 42 and n(A + B) 24., (i) n(A) n(A B) n(A + B), , n(A B) n(A) n(A + B) (50 24) 26., (ii) n(B) n(B A) n(A + B), , n(B A) n(B) n(A + B) (42 24) 18., (iii) n(A , B) n(A) n(B) n(A + B) (50 42 24) 68., 9. Failed in English only (56 18) 38., Failed in science only (48 18) 30., Failed in both English and science = 18., Failed in one or both of the subjects (38 30 18) 86., Passed in both the subjects (100 86) 14., , È, C1 – C2 C1 C 2 C2 – C1, , C2, , C1, , A–B, , A, , A, , È, , B, , B–A, , B
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46, , Senior Secondary School Mathematics for Class 11, , 10. Let A set of people who like cricket., B set of people who like tennis., n(A) 40, n(A + B) 10., 30, 10, 25, n(A B) (40 10) 30, , , , , , n(B A) 65 (30 10) 25., Number of people who like tennis only = 25., A, B, Number of people who like tennis (25 10) 35., 11. Let A, B, C denote the sets of students who bagged medals in hockey, basketball and, cricket respectively., Then, n(A) 42, n(B) 18, n(C) 23, n(A , B , C) 65 and n(A + B+ C) 4. Then,, n(A , B, C) {n(A) n(B) n(C) n(A + B+ C)} {n(A + B) n(B+ C) n(A + C)}, 65 (42 18 23 4) x + x (87 65) 22., H, T, 12. a b c d 25, b c e f 26, c d f g 26,, c d 9, b c 11, c f 8 and c 3., , f 5, b 8, d 6, c 3, g 12, e 10 and a 8., (i) Number of people who read at least one of the papers, (a b c d e f g) 52., (ii) Number of people who read exactly one newspaper, (a e g) (8 10 12) 30., , e, , b, , a, , c, f, , d, g, , I, , 13. We have, E, a 18, a d 23, c d 8, c f 8,, e, b, a b c d 26, c d f g 48,, a, a b c d e f g 100 24 76., c, a 18, d 5, c 3, f 5 and b 0., f, d, g 48 (3 5 5) 35, and e 76 (18 0 3 5 5 35) 76 66 10., g, (i) Number of students studying Hindi (b c e f ), S, (0 3 10 5) 18., (ii) Number of students studying English and Hindi both (b c) (0 3) 3., 40, 14. n(A) (40% of 10000) a, #10000k 4000,, 100, n(B) (20% of 10000) 2000, n (C) (10% of 10000) 1000,, n(A + B) (5% of 10000) 500, n(B+ C) (3% of 10000) 300,, n (A + C) (4% of 10000) 400, n(A + B+ C) (2% of 10000) 200,, and n(U) 10000., ]Z]a b c d 4000,, A, ]], ]]b c e f 2000,, a, ]], ]]c d f g 1000,, ]], [], b c 500,, ]], d, ]], c f 300,, ]], ]], c d 400,, ]], ], c 200., \, , H, , B, e, , b, c, f, g, C
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Sets, , 47, , On solving these equations, we get, c 200, d 200, f 100, b 300, g 500, e 1400, a 3300., (i) a 3300,, (ii) e 1400, (iii) 10000 (3300 300 200 200 1400 100 500) (10000 6000) 4000., , EXERCISE 1H, Very-Short-Answer Questions, 1. If a set A has n elements then find the number of elements in its power set, P (A) ., 2. If A then write P(A) ., 3. If n(A) 3 and n(B) 5, find:, (i) the maximum number of elements in A , B,, (ii) the minimum number of elements in A , B., 4. If A and B are two sets such that n(A) 8, n(B) 11 and n(A , B) 14 then, find n(A + B) ., 5. If A and B are two sets such that n(A) 23, n(B) 37 and n(A B) 8 then, find n(A , B) ., Hint, , n(A) n(A B) n(A + B) & n(A + B) (23 8) 15., , 6. If A and B are two sets such that n(A) 54, n(B) 39 and n(B A) 13 then, find n(A , B) ., Hint n(B) n(B A) n(A + B) & n(A + B) (39 13) 26., 7. If A 1 B, prove that Bl1 Al., 8. If A 1 B, show that (Bl Al) ., Hint A 1 B & Bl1 Al & Bl Al ., 9. Let A {x : x 6n, nd N} and B {x : x 9n, nd N}, find A + B., 10. If A {5, 6, 7}, find P(A) ., 11. If A {2, {2}}, find P(A) ., 12. Prove that A +(A , B)l ., 13. Find the symmetric difference A T B, when A {1, 2, 3} and B {3, 4, 5} ., 14. Prove that A B A + Bl., 15. If A {x : xd R, x 5} and B {x : xd R, x 4}, find A + B., Hint, , A (3, 5) and B (4, 3) . So, A + B (4, 5) ., ANSWERS (EXERCISE 1H), , 1., 9., 10., 11., , 2. {}, 3. (i) 8 (ii) 5, 4. 5, 5. 45, 6. 67, 2, A + B {x : x 18n, nd N}, P(A) {, {5}, {6}, {7}, {5, 6}, {5, 7}, {6, 7}, {5, 6, 7}}, 13. {1, 2, 4, 5}, 15. (4, 5), P(A) {, {2}, {{2}}, {2, {2}}}, n
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48, , Senior Secondary School Mathematics for Class 11, , SUMMARY OF KEY FACTS, 1., 2., 3., 4., 5., , 6., 7., 8., , 9., 10., 11., 12., 13., 14., , 15., , 16., , 17., , 18., 19., , A well-defined collection of objects is called a set., The objects in a set are called its elements, or members, or points., Usually, we denote sets by capital letters A, B, C, X, Y, Z, etc., A set is usually described in tabular form or set-builder form., In tabulation method, we make a list of all the objects of the set and put, them within braces { }. In set-builder form, we write {x : x satisfies properties P}, which means, the set of all those x such that each x satisfies properties P., A set having no element at all is called a null set, or a void set and it is, denoted by ., A set having a single element is called a singleton set. For example, {3}., A set having finite number of elements is called a finite set, otherwise it is, called an infinite set., The number of elements in a finite set A is denoted by n(A) ., Two sets A and B having exactly the same elements are known as equal sets, and we write, A B., A set A is called a subset of a set B, if every element of A is in B and we write,, A 3 B., If A is a subset of set B and A ! B then A is called a proper subset of set B and, we write, AfB., The total number of subsets of a set A containing n elements is 2 n ., The collection of all subsets of a set A is called the power set of A, to be, denoted by P(A) ., Let a and b be real numbers such that a b then, (i) closed interval [a, b] {x : xd R, a # x # b},, (ii) open interval (a, b) {x : xd R, a x b},, (iii) right half open interval [a, b) {x : xd R, a # x b},, (iv) left half open interval (a, b] {x : xd R, a x # b}., The union of two sets A and B, denoted by A , B, is the set of all those, elements which are either in A or in B or in both A and B., , A , B {x : xd A or xdB} ., The intersection of two sets A and B, denoted by A + B, is the set of all those, elements which are common to both A and B., , A + B {x : xd A and xdB} ., The difference between two sets A and B, denoted by (A B), is defined as, (A B) {x : xd A and xzB} ., Similarly, (B A) {x : xdB and xz A} ., The symmetric difference between the sets A and B, denoted by A T B, is, defined as A T B (A B),(B A) ., Let A be a subset of the universal set U. Then the complement of A, denoted, by Al, or A c, or U A, is defined as Al {x : xdU and xz A} .
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Sets, , 49, , 20. Various laws of operations on sets, (i) A , A A and A + A A, [Idempotent laws], (ii) A , A and A +U A, [Identity law], [Commutative laws], (iii) A , B B, A and A + B B+ A, (iv) (A , B), C A ,(B, C) and (A + B)+ C A +(B+ C), [Associative laws], (v) I. A ,(B+ C) (A , B)+(A , C), [Distributive law], II. A +(B, C) (A + B),(A + C), [Distributive law], (vi) I. (A , B)l (Al+ Bl), [De Morgan‘s laws], II. (A + B)l (Al, Bl), [De Morgan’s laws], 21. For any two sets A and B, we have, (i) n(A , B) n(A) n(B) n(A + B), È, (ii) n(A B) n(A + B) n(A), A–B A B B–A, (iii) n(B A) n(A + B) n(B), (iv) n(A T B) n(A B) n(B A), A, B, 22. For any three sets A, B and C, we have, n(A , B, C) n(A) n(B) n(C) {n(A + B) n(B+ C) n(A + C)}, n(A + B+ C) .,
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50, , Senior Secondary School Mathematics for Class 11, , 2, , Relations, ORDERED PAIR Two numbers a and b listed in a specific order and enclosed in, parentheses form an ordered pair (a, b) ., In the ordered pair (a, b), we call a as first member (or first component) and b, as second member (or second component)., By interchanging the positions of the components, the ordered pair is, changed., Thus, (a, b) ! (b, a) ., Y, EXAMPLE, , In coordinate geometry, the, position of a point in a plane is, determined by an ordered pair., In the given figure, the ordered, pairs (2, 3) and (3, 2) represent, two different points A and B, respectively., Thus, (2, 3) ! (3, 2) ., , A(2, 3), , 3, , B(3, 2), , 2, , 1, , O, , 1, , 2, , 3, , X, , EQUALITY OF TWO ORDERED PAIRS, , We have (a, b) (c, d) + a c and b d., EXAMPLE 1, , Find a and b, when (a 1, b 5) (2, 3) ., , SOLUTION, , Using the definition of equality of two ordered pairs, we have, (a 1, b 5) (2, 3) & a 1 2 and b 5 3, & a 3 and b 2., Hence, a 3 and b 2., , EXAMPLE 2, , Find a and b, when (2a b, 11) (1, a 3b) ., , SOLUTION, , Using the definition of equality of two ordered pairs, we have, (2a b, 11) (1, a 3b), ... (i), 2a b 1, ), , , ... (ii), a 3b 11., On solving (i) and (ii), we get a 2 and b 3., Hence, a 2 and b 3., 50
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Relations, EXAMPLE 3, SOLUTION, , EXAMPLE 4, SOLUTION, , 51, , 5 1, x, 2, If a 1, y k a , k, find the values of x and y., 3, 3, 3 3, Since the given two ordered pairs are equal, we have, 5, x 5, x, x 2, 2, & a 1k & & x 2, 1, 3, 3, 3, 3, 3, 3 3, 3, 1 2, 2 1, and y & y a k 1 & y 1., 3 3, 3 3, 3, Hence, x 2 and y 1., Express {(x, y) : x 2 y 2 25, where x, y dW} as a set of ordered pairs., It is easy to verify that each of the following ordered pairs of whole, numbers satisfies the given relation x 2 y 2 25:, (5, 0), (0, 5), (3, 4) and (4, 3) ., Hence, the set of required ordered pairs is, {(5, 0), (0, 5), (3, 4), (4, 3)} ., , CARTESIAN PRODUCT OF TWO SETS, , Let A and B be two nonempty sets. Then, the Cartesian product of A and B is the set, denoted by (A# B), consisting of all ordered pairs (a, b) such that ad A and bdB., A# B {(a, b) : ad A and bdB} ., If A or B , we define A# B ., REMARKS, , (i) If n(A) p and n(B) q then n(A# B) pq and n(B# A) pq., (ii) If at least one of A and B is infinite then (A# B) is infinite and (B# A), is infinite., SOLVED EXAMPLES, , EXAMPLE 1, , SOLUTION, , EXAMPLE 2, , SOLUTION, , If A {1, 3, 5} and B {2, 3} then find:, (i) A# B, (ii) B# A, (iii) (A# B)+(B# A), We have, (i) A# B {1, 3, 5}#{2, 3}, {(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)} ., (ii) B# A {2, 3}#{1, 3, 5}, {(2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5)} ., (iii) (A# B)+(B# A) {(3, 3)} ., If A {1, 2, 3}, B {3, 4} and C {4, 5, 6} then find:, (i) A#(B+ C), (ii) (A# B)+(A#C), (iii) A#(B, C), (iv) (A# B),(A#C), We have, (i) B+ C {3, 4}+{4, 5, 6} {4} ., , A#(B+ C) {1, 2, 3}#{4} {(1, 4), (2, 4), (3, 4)} .
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52, , Senior Secondary School Mathematics for Class 11, , (ii) (A# B) {1, 2, 3}#{3, 4}, {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} ., (A#C) {1, 2, 3}#{4, 5, 6}, {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4),, (3, 5), (3, 6)}, , (A# B)+(A#C) {(1, 4), (2, 4), (3, 4)} ., (iii) B, C {3, 4},{4, 5, 6} {3, 4, 5, 6} ., , A#(B, C) {1, 2, 3}#{3, 4, 5, 6}, {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5),, (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)} ., (iv) Also, from (ii), we get, (A# B),(A#C) {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5),, (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)} ., EXAMPLE 3, , Let A {xd N : x 2 5x 6 0}, B {xdW : 0 # x 2} and, C {xd N : x 3}. Verify that, (i) A#(B, C) (A# B),(A#C), (ii) A#(B+ C) (A# B)+(A#C), , SOLUTION, , We have, A {xd N : x 2 5x 6 0} {xd N : (x 2) (x 3) 0} {2, 3};, B {xdW : 0 # x 2} {0, 1} and C {xd N : x 3} {1, 2} ., , A {2, 3}, B {0, 1} and C {1, 2} ., (i) (B, C) {0, 1},(1, 2} {0, 1, 2} ., , A#(B, C) {2, 3}#{0, 1, 2}, {(2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)} ., Now, (A# B) {2, 3}#{0, 1}, {(2, 0), (2, 1), (3, 0), (3, 1)}, and (A#C) {2, 3}#{1, 2}, {(2, 1), (2, 2), (3, 1), (3, 2)} ., , (A# B),(A#C) {(2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)} ., Hence, A#(B, C) (A# B),(A#C) ., (ii) (B+ C) {0, 1}+{1, 2} {1} ., , A#(B+ C) {2, 3}#{1} {(2, 1), (3, 1)} ., And,, (A# B)+(A#C) {(2, 0), (2, 1), (3, 0), (3, 1)}+{(2, 1), (2, 2),, (3, 1), (3, 2)}, {(2, 1), (3, 1)} ., Hence, A#(B+ C) (A# B)+(A#C) .
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Relations, , 53, , EXAMPLE 4, , If (A# B) {(3, 2), (3, 4), (5, 2), (5, 4)}, find A and B., , SOLUTION, , Clearly, we have, A set of all first coordinates of the elements of (A# B), {3, 5} ., B set of all second coordinates of the elements of (A# B), {2, 4} ., Thus, A {3, 5} and B {2, 4} ., , EXAMPLE 5, , A and B are two sets given in such a way that (A# B) contains 6 elements., If three elments of (A# B) be (1, 3), (2, 5) and (3, 3), find its remaining, elements., , SOLUTION, , Since (1, 3), (2, 5) and (3, 3) are in (A# B), it follows that 1, 2, 3 are, elements of A and 3, 5 are elements of B., , (A# B) {1, 2, 3}#{3, 5}, {(1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5)} ., Hence, the remaining elements of (A# B) are (1, 5), (2, 3) and (3, 5)., , EXAMPLE 6, , If A {a, b}, find (A# A) ., , SOLUTION, , We have, (A# A) {a, b}#{a, b}, {(a, a), (a, b), (b, a), (b, b)} ., , EXAMPLE 7, , Let R be the set of all real numbers. What does (R # R) represent?, , SOLUTION, , We have (R # R) {(x, y) : x, y d R} ., Thus, (R # R) represents the set of all coordinates of points in twodimensional space., , EXAMPLE 8, , If (A# A) has 9 elements two of which are (1, 0) and (0, 1), find the set A, and the remaining elements of (A# A)., , SOLUTION, , Clearly, –1, 0 and 1 are elements of A., , A {1, 0, 1}, A # A { 1, 0, 1} # { 1 , 0 , 1 }, {(1, 1), (1, 0), (1, 1), (0, 1), (0, 0), (0, 1), (1, 1),, (1, 0), (1, 1)} ., Hence, the remaining elements of (A# A) are, (–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), (1, 1)., , GRAPHICAL REPRESENTATION OF A # B AND B # A, , Let XlOX and YOYl be the x-axis and y-axis respectively, drawn on a graph, paper., Let A {1, 2, 4} and B {1, 3} . Then,, A# B {(1, 1), (1, 3), (2, 1), (2, 3), (4, 1), (4, 3)} .
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54, , Senior Secondary School Mathematics for Class 11, , These points may be plotted on the graph paper, as shown below:, Y, , X¢, , (–1, 3), , 3, , (–1, 1), , 1, , (2, 3), , (4, 3), , (2, 1), , (4, 1), , 2, , 4, , O, , –1, , Y¢, , A, , X, , B, , This is the graphical representation of A# B., For B# A, we take B {1, 3} along the x-axis and A {1, 2, 4} along the, y-axis and plot the points (1, 1), (1, 2), (1, 4), (3, 1), (3, 2), (3, 4) on the graph, paper., Arrow diagram of A # B, , Let A {1, 2, 4} and B {1, 3} . Then,, A# B {(1, 1), (1, 3), (2, 1), (2, 3), (4, 1), (4, 3)} ., Then, A# B may be represented by arrow diagram, as shown below:, , 1, , –1, 2, , 3, , 4, A, , B, , A #B, , Arrow diagram of B # A, , Again, B# A {(1, 1), (1, 2), (1, 4), (3, 1), (3, 2), (3, 4)} .
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Relations, , 55, , We may exhibit B# A by arrow diagram, as shown below:, –1, , 1, , 2, 4, , 3, B, , A, , B# A, ORDERED TRIPLET Three numbers a, b and c listed in a specific order and, enclosed in parentheses form an ordered triplet (a, b, c) ., Thus, (1, 2, 3) ! (2, 1, 3) ! (3, 2, 1), etc., For any nonempty set A, we define:, (A # A # A) {(a, b, c) : a, b, c d A} ., EXAMPLE 9, SOLUTION, , If A {1, 2}, find (A# A# A) ., A# A {1, 2}#{1, 2} {(1, 1), (1, 2), (2, 1), (2, 2)} ., , A # A # A ( A # A) # A, {(1, 1), (1, 2), (2, 1), (2, 2)}#{1, 2}, {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1),, (2, 1, 2), (2, 2, 1), (2, 2, 2)} ., , Let R be the set of all real numbers. What does (R # R # R) represent?, SOLUTION, (R # R # R) {(a, b, c) : a, b, cd R} ., Thus, (R # R # R) represents the set of all points in threedimensional space., REMARK For any nonempty sets A, B and C, we always have, (A# B)#C A#(B#C) A# B#C., EXAMPLE 10, , EXAMPLE 11, , SOLUTION, , Let A {1, 2}, B {3, 4} and C {4, 5} ., Verify that (A# B)#C A#(B#C) and hence find A# B#C., We have, A# B {1, 2}#{3, 4} {(1, 3), (1, 4), (2, 3), (2, 4)}, , (A# B)#C {(1, 3), (1, 4), (2, 3), (2, 4)}#{4, 5}, {(1, 3, 4), (1, 3, 5), (1, 4, 4), (1, 4, 5), (2, 3, 4),, (2, 3, 5), (2, 4, 4), (2, 4, 5)} ., Again, B#C {3, 4}#{4, 5} {(3, 4), (3, 5), (4, 4), (4, 5)}, , A#(B#C) {1, 2}#{(3, 4), (3, 5), (4, 4), (4, 5)}, {(1, 3, 4), (1, 3, 5), (1, 4, 4), (1, 4, 5), (2, 3, 4),, (2, 3, 5), (2, 4, 4), (2, 4, 5)} ., , (A# B)#C A#(B#C) A# B#C., Hence, (A# B#C) {(1, 3, 4), (1, 3, 5), (1, 4, 4), (1, 4, 5), (2, 3, 4),, (2, 3, 5), (2, 4, 4), (2, 4, 5)} .
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56, , Senior Secondary School Mathematics for Class 11, , EXERCISE 2A, 1. Find the values of a and b, when:, (i) (a 3, b 2) (5, 1), , (ii) (a b, 2b 3) (4, 5), , 5 2, a, 1, (iii) b 1, b l b , l, (iv) (a 2, 2b 1) (b 1, a 2), 3, 3, 3 3, 2. If A {0, 1} and B {1, 2, 3}, show that A# B ! B# A., 3. If P {a, b} and Q {x, y, z}, show that P #Q ! Q # P., 4. If A {2, 3, 5} and B {5, 7}, find:, (i) A# B, (ii) B# A, (iii) A# A, (iv) B# B, , , 5. If A {xd N : x # 3} and B {xdW, x 2}, find (A# B) and (B# A) . Is, (A# B) (B# A)?, 6. If A {1, 3, 5}, B {3, 4} and C {2, 3}, verify that:, (i) A#(B, C) (A# B),(A#C), (ii) A#(B+ C) (A# B)+(A#C), 7. Let A {xdW : x 2}, B {xd N : 1 x # 4} and C {3, 5} . Verify that:, (i) A#(B, C) (A# B),(A#C), (ii) A#(B+ C) (A# B)+(A#C), Hint, , A {0, 1}, B {2, 3, 4} and C {3, 5} ., , 8. If A# B {( 2, 3), ( 2, 4), (0, 3), (0, 4), (3, 3), (3, 4)}, find A and B., 9. Let A {2, 3} and B {4, 5} . Find (A# B) . How many subsets will (A# B), have?, 10. Let A# B {(a, b) : b 3a 2} . If (x, 5) and (2, y) belong to A# B, find the, values of x and y., 11. Let A and B be two sets such that n(A) 3 and n(B) 2., If a ! b ! c and (a, 0), (b, 1), (c, 0) are in A# B, find A and B., 12. Let A { 2, 2} and B {0, 3, 5} . Find:, (i) A# B, (ii) B# A, (iii) A# A, (iv) B# B, Represent each of the above (a) graphically and (b) by arrow diagram., 13. If A {5, 7}, find (i) A# A and (ii) A# A# A., 14. Let A { 3, 1}, B {1, 3} and C {3, 5} . Find:, (i) A# B, (ii) (A# B)#C, (iii) B#C, (iv) A#(B#C), ANSWERS (EXERCISE 2A), , 1., 4., , (i), (i), (ii), (iii), (iv), , a 2, b 3 (ii) a 5, b 1 (iii) a 2, b 1 (iv) a 3, b 2, A# B {(2, 5), (2, 7), (3, 5), (3, 7), (5, 5), (5, 7)}, B# A {(5, 2), (5, 3), (5, 5), (7, 2), (7, 3), (7, 5)}, A# A {(2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)}, B# B {(5, 5), (5, 7), (7, 5), (7, 7)}
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Relations, , 57, , 5. A# B {(1, 0), (1, 1), (2, 0), (2, 1), (3, 0), (3, 1)};, B# A {(0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3)};, No, 8. A { 2, 0, 3} and B {3, 4}, 9. A# B {(2, 4), (2, 5), (3, 4), (3, 5)}; Number of subsets of (A# B) 2 4 16, 10. x 1, y 4, 11. A {a, b, c} and B {0, 1}, 12. (i) A# B {( 2, 0), ( 2, 3), ( 2, 5), (2, 0), (2, 3), (2, 5)}, (ii) B# A {(0, 2), (0, 2), (3, 2), (3, 2), (5, 2), (5, 2)}, (iii) A# A {( 2, 2), ( 2, 2), (2, 2), (2, 2)}, (iv) B# B {(0, 0), (0, 3), (0, 5), (3, 0), (3, 3), (3, 5), (5, 0), (5, 3), (5, 5)}, 13. (i) A# A {(5, 5), (5, 7), (7, 5), (7, 7)}, (ii) A# A# A {(5, 5, 5), (5, 7, 5), (7, 5, 5), (7, 7, 5), (5, 5, 7), (5, 7, 7),, (7, 5, 7), (7, 7, 7)}, 14. (i) A# B {( 3, 1), ( 3, 3), (1, 1), (1, 3)}, (ii) (A# B)#C {( 3, 1, 3), ( 3, 3, 3), (1, 1, 3), (1, 3, 3), ( 3, 1, 5),, ( 3, 3, 5), (1, 1, 5), (1, 3, 5)}, (iii) B#C {(1, 3), (1, 5), (3, 3), (3, 5)}, (iv) A#(B#C) {( 3, 1, 3), ( 3, 1, 5), ( 3, 3, 3), ( 3, 3, 5), (1, 1, 3),, (1, 1, 5), (1, 3, 3), (1, 3, 5)}, , SOME USEFUL RESULTS, THEOREM 1, PROOF, , Let A 3 B and let (a, c) be an arbitrary element of A#C. Then,, (a, c)d A#C & ad A and cdC, & adB and cdC, [a A 3 B], & (a, c)dB#C., Thus, every element of A#C is contained in B#C., , A#C 3 B#C., Hence, A 3 B & A#C 3 B#C for any set C., , THEOREM 2, PROOF, , If A 3 B then prove that A#C 3 B#C for any set C., , If A 3 B and C 3 D then prove that A#C 3 B# D., , Let A 3 B and C 3 D. Then, we have to show that A#C 3 B# D., Let (a, c) be an arbitrary element of A#C. Then,, (a, c)d A#C & ad A and cdC, & adB and cdD, [a A 3 B and C 3 D], & (a, c)dB# D., Thus, every element of A#C is contained in B# D., Hence, A#C 3 B# D.
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58, , Senior Secondary School Mathematics for Class 11, , If A and B are any two nonempty sets, prove that A# B B# A + A B., Let us first assume that A B. Then,, A# B B# B and B# A B# B, [a A B]., , A # B B # A., Thus, A B & A# B B# A., Conversely, let A# B B# A and we have to show that A B., Let a be an arbitrary element of A. Then,, ad A & (a, b)d A# B for some bdB, & (a, b)dB# A, [a A# B B# A], & adB and bd A, & adB (surely)., , A 3 B., Again, let b be an arbitrary element of B. Then,, bdB & (b, a)dB# A for some ad A, & (b, a)d A# B, [a B# A A# B], & bd A and adB, & bd A (surely)., , B 3 A., Thus, A 3 B and B 3 A & A B., , A # B B # A & A B., Hence, A# B B# A + A B., , THEOREM 3, PROOF, , THEOREM 4, PROOF, , For any sets A, B and C, prove that, A#(B, C) (A# B),(A#C) ., Let (a, b) be an arbitrary element of A#(B, C) . Then,, (a, b)d A#(B, C) & ad A and bdB, C, & ad A and (bdB or bdC), & (ad A and bdB) or (ad A and bdC), [a p and (q or r) / (p and q) or (p and r)], & (a, b)d A# B or (a, b)d A#C, & (a, b)d(A# B),(A#C)., , THEOREM 5, , PROOF, , If A 3 B, prove that A# A 3 (A# B)+(B# A) ., Let A 3 B and let (a, a) be an arbitrary element of A# A., Then, (a, a)d A# A & ad A and ad A, & ad A and adB, [a A 3 B], & (ad A, adB) and (adB, ad A), & (a, b)d A# B and (a, b)dB# A, & (a, b)d(A# B)+(B# A) ., , A# A 3 (A# B)+(B# A) ., Hence, A 3 B & A# A 3 (A# B)+(B# A) .
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Relations, , 59, , , … (i), A#(B, C) 3 (A# B),(A#C) ., Again, let (x, y) be an arbitrary element of (A# B),(A#C) . Then,, (x, y)d(A# B),(A#C) & (x, y)d A# B or (x, y)d A#C, & (xd A and y dB) or (xd A and y dC), & xd A and (y dB or y dC), [by distributive law], & xd A and y d(B, C), & (x, y)d A#(B, C)., , … (ii), (A# B),(A#C) 3 A#(B, C) ., Hence, from (i) and (ii), we get, A#(B, C) (A# B),(A#C) ., For any sets A, B and C, prove that, A#(B+ C) (A# B)+(A#C) ., Let (a, b) be an arbitrary element of A#(BkC) . Then,, (a, b)d A#(B+ C) & ad A and bd(B+ C), & ad A and (bdB and bdC), & (ad A and bdB) and (ad A and bdC), & (a, b)d A# B and (a, b)d A#C, & (a, b)d(A# B)+(A#C)., , … (i), A#(B+ C) 3 (A# B)+(A#C) ., Again, let (x, y) be an arbitrary element of (A# B)+(A#C) . Then,, (x, y)d(A# B)+(A#C) & (x, y)d A# B and (x, y)d A#C, & (xd A and y dB) and (xd A and y dC), & xd A and (y dB and y dC), & xd A and y d(B+ C), & (x, y)d A#(B+ C)., , … (ii), (A# B)+(A#C) 3 A#(B+ C) ., Hence, from (i) and (ii), we get, A#(B+ C) (A# B)+(A#C) ., , THEOREM 6, , PROOF, , For any sets A, B and C, prove that, A#(B C) (A# B) (A#C) ., Let (a, b) be an arbitrary element of A#(B C) . Then,, (a, b)d A#(B C) & ad A and bd(B C), & ad A and (bdB and bzC), & (ad A and bdB) and (ad A and bzC), & (a, b)d A# B and (a, b)z A#C, & (a, b)d(A# B) (A#C)., , … (i), A#(B C) 3 (A# B) (A#C) ., , THEOREM 7, , PROOF
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60, , Senior Secondary School Mathematics for Class 11, , Again, let (x, y) be an arbitrary element of (A# B) (A#C) . Then,, (x, y)d(A# B) (A#C) & (x, y)d A# B and (x, y)z A#C, & (xd A and y dB) and (xd A and y zC), & xd A and (y dB and y zC), & xd A and y d(B C), & (x, y)d A#(B C)., , … (ii), (A# B) (A#C) 3 A#(B C) ., From (i) and (ii), we get, A#(B C) (A# B) (A#C) ., For any sets A, B, C and D, prove that, (A# B)+(C # D) (A+ C)#(B+ D) ., Let (a, b) be an arbitrary element of (A# B)+(C # D) . Then,, (a, b)d(A# B)+(C # D) & (a, b)d(A# B) and (a, b)d(C # D), & (ad A and bdB) and (adC and bdD), & (ad A and adC) and (bdB and bdD), & ad(A + C) and bd(B+ D), … (i), & (a, b)d(A + C)#(B+ D) ., Again, let (x, y) be an arbitrary element of (A + C)#(B+ D) . Then,, (x, y)d(A + C)#(B+ D) & xd A + C and y dB+ D, & (xd A and xdC) and (y dB and y dD), & (xd A and y dB) and (xdC and y dD), & (x, y)d A# B and (x, y)dC # D, & (x, y)d(A# B)+(C # D) ., , … (ii), (A + C)#(B+ D) 3 (A# B)+(C # D) ., From (i) and (ii), we get, (A# B)+(C # D) (A+ C)#(B+ D) ., , THEOREM 8, , PROOF, , THEOREM 9, , SOLUTION, , For any sets A and B, prove that, (A# B)+(B# A) (A + B)#(B+ A)., Let (a, b) be an arbitrary element of (A# B)+(B# A) . Then,, (a, b)d(A# B)+(B# A) & (a, b)d A# B and (a, b)dB# A, & (ad A and bdB) and (adB and bd A), & (ad A and adB) and (bdB and bd A), & ad(A + B) and bd(B+ A), & (a, b)d(A + B)#(B+ A) ., , … (i), ( A # B ) + ( B # A ) 3 ( A + B ) # (B + A ) ., Again, let (x, y) be an arbitrary element of (A + B)#(B+ A) . Then,, (x, y)d(A + B)#(B+ A) & xd A + B and y dB+ A, & (xd A and xdB) and (y dB and y d A)
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Relations, , 61, , & (xd A and y dB) and (xdB and y d A), & (x, y)d A# B and (x, y)dB# A, & (x, y)d(A# B)+(B# A) ., , … (ii), ( A + B ) # ( B + A ) 3 ( A # B ) + (B # A ) ., Thus, from (i) and (ii), we get, ( A # B) + (B # A ) (A + B) # (B + A ) ., If A and B are two nonempty sets having n elements in, common then (A# B) and (B# A) have n 2 elements in common., , AN IMPORTANT RESULT, , Let A and B be two nonempty sets such that n(A) 5, n(B) 6 and, n ( A + B ) 3., Find (i) n(A# B), (ii) n(B# A) and (iii) n{(A# B)+(B# A)} ., (i) n(A# B) n(A)# n(B) (5 # 6) 30., (ii) n(B# A) n(B)# n(A) (6 # 5) 30., (iii) Given: n(A + B) 3., , A and B have 3 elements in common., So, (A# B) and (B# A) have 3 2 9 elements in common., Hence, n{(A# B)+(B# A)} 9., , EXAMPLE, , SOLUTION, , If A, B and C be three nonempty sets given in such a way that, A# B A#C then prove that B C., Let A# B A#C and we have to prove that B C., Let bdB. Then,, bdB & (a, b)d A# B for every ad A, & (a, b)d A#C for every ad A, [a A# B A#C], & bdC., , … (i), B 3 C., Again, let cdC. Then,, cdC & (a, c)d A#C for every ad A, & (a, c)d A# B for every ad A, [a A#C A# B], & c d B., , … (ii), C 3 B., From (i) and (ii), we get B C., Hence, A# B A#C & B C., , THEOREM 10, PROOF, , EXERCISE 2B, 1. For any sets A, B and C, prove that:, (i) A#(B, C) (A# B),(A#C), (ii) A#(B+ C) (A# B)+(A#C), (iii) A#(B C) (A# B) (A#C)
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62, , Senior Secondary School Mathematics for Class 11, , 2. For any sets A and B, prove that, ( A # B) + (B # A ) (A + B) # (B + A ) ., 3. If A and B are nonempty sets, prove that, A # B B # A + A B., 4. (i) If A 3 B, prove that A#C 3 B#C for any set C., (ii) If A 3 B and C 3 D then prove that A#C 3 B# D., 5. If A# B 3 C # D and A# B !, prove that A 3 C and B 3 D., 6. If A and B be two sets such that n(A) 3, n(B) 4 and n(A + B) 2 then find:, (i) n(A# B), , (ii) n(B# A), , (iii) n{(A# B)+(B# A}, , 7. For any two sets A and B, show that A# B and B# A have an element in, common if and only if A and B have an element in common., 8. Let A {1, 2} and B {2, 3} . Then, write down all possible subsets of A# B., 9. Let A {a, b, c, d}, B {c, d, e} and C {d, e, f, g}. Then verify each of the, following identities:, (i) A#(B+ C) (A# B)+(A#C), (ii) A#(B C) (A# B) (A#C), (iii) (A# B)+(B# A) (A + B)# (A + B), ANSWERS (EXERCISE 2B), , 6. (i) n(A# B) 12 (ii) n(B# A) 12 (iii) n{(A# B)+(B# A)} 4, 8. , {(1, 2)}, {(1, 3)}, {(2, 2)}, {(2, 3)}, {(1, 2), (1, 3)}, {(1, 2), (2, 2)}, {(1, 2), (2, 3)},, {(1, 3), (2, 2)}, {(1, 3), (2, 3)}, {(2, 2), (2, 3)}, {(1, 2), (1, 3), (2, 2)},, {(1, 2), (1, 3), (2, 3)}, {(1, 3), (2, 2), (2, 3)}, {(1, 2), (2, 2), (2, 3)},, {(1, 2), (1, 3), (2, 2), (2, 3)}, HINTS TO SOME SELECTED QUESTIONS, 6. (i) We know that n (A# B) n (A)# n (B) (ii) n(B# A) n(B)# n(A) ., (iii) Since A and B have 2 elements in common, so (A# B) and (B# A) will have 2 2 4, elements in common., 7. We know that (A# B)+(B# A) (A + B)#(B+ A) (A + B)#(A + B) ., Now, n(A + B) 1 + n{(A + B)#(A + B)} 1 2 1, + n {(A# B)+(B# A)} 1[a (A + B)#(A + B) (A# B)+(B# A)]., 8. A# B {(1, 2), (1, 3), (2, 2), (2, 3)} . So, it will have 2 4 16 subsets., , RELATIONS, Let A and B be two nonempty sets. Then, a relation R from A to B is a, subset of (A# B) ., Thus, R is a relation from A to B + R 3 (A# B) ., If (a, b)d R then we say that ‘a is related to b‘ and we write, a R b., If (a, b)z R then ‘a is not related to b‘ and we write, a R b., , RELATION
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Relations, , 63, , DOMAIN, RANGE AND CO-DOMAIN OF A RELATION, , Let R be a relation from A to B. Then, R 3 (A# B) ., (i) The set of all first coordinates of elements of R is called the domain of R, written, as dom (R)., (ii) The set of all second coordinates of elements of R is called the range of R, denoted, by range (R)., (iii) The set B is called the co-domain of R., Dom (R) {a : (a, b)d R} and Range (R) {b : (a, b)d R} ., Total Number of Relations from A to B, , Let n(A) p and n(B) q. Then, n(A# B) pq., We know that every subset of A# B is a relation from A to B., Total number of subsets of A# B is 2 pq ., total number of relations from A to B 2 pq ., REPRESENTATION OF A RELATION, , Let A and B be two given sets. Then, a relation R 3 A# B can be represented in, any of the forms, given below., (i) ROSTER FORM, , In this form, R is given as a set of ordered pairs., EXAMPLE 1, , SOLUTION, , Let A { 2, 1, 0, 1, 2} and B {0, 1, 4, 9} ., Let R {( 2, 4), (1, 1), (0, 0), (1, 1), (2, 4)}., (i) Show that R is a relation from A to B., (ii) Find dom (R), range (R) and co-domain of R., (i) Since Rf A# B, so R is a relation from A to B., Note that 2R4, 1R1, 0R0, 1R1 and 2R4., (ii) Dom (R) set of first coordinates of elements of R, { 2, 1, 0, 1, 2} ., Range (R) set of second coordinates of elements of R, {0, 1, 4} ., Co-domain of R {0, 1, 4, 9} B., , (ii) SET-BUILDER FORM, , Under this method, for every (a, b)d R, a general relation is being given between a, and b., Using this relation, all the elements of R can be obtained., EXAMPLE 2, , Let A {1, 2, 3, 5} and B {4, 6, 9} ., Define a relation from A to B, given by, R {(a, b) : ad A, bdB and (a b) is odd}., (i) Write R in roster form., (ii) Find dom (R) and range (R).
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64, SOLUTION, , Senior Secondary School Mathematics for Class 11, , (i) Clearly, we have, R {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}., (ii) Dom (R) set of first coordinates of elements of R, {1, 2, 3, 5}., Range (R) set of second coordinates of elements of R, {4, 6, 9} ., , EXAMPLE 3, , Let R {(x, x 3) : x is a prime number less than 10}., (i) Write R in roster form., (ii) Find dom (R) and range (R)., , SOLUTION, , Prime numbers less than 10 are 2, 3, 5, 7., (i) R {(2, 2 3), (3, 3 3), (5, 5 3), (7, 7 3)}, {(2, 8), (3, 27), (5, 125), (7, 343)} ., (ii) Dom (R) {2, 3, 5, 7} ., Range (R) {8, 27, 125, 343} ., , EXAMPLE 4, , SOLUTION, , Let R {(x, y) : x and y are integers and xy 4} ., (i) Write R in roster form., (ii) Find dom (R) and range (R)., Clearly, we have, (i) R {( 4, 1), ( 2, 2), (1, 4), (1, 4), (2, 2), (4, 1)} ., (ii) Dom (R) { 4, 2, 1, 1, 2, 4} ., Range (R) { 4, 2, 1, 1, 2, 4} ., , (iii) ARROW DIAGRAM, , Let R be a relation from A to B. First, we draw two bounded figures to represent, A and B respectively. We mark the elements of A and B in these figures. For, each (a, b)d R, we draw an arrow from a to b. This gives us the required arrow, diagram., EXAMPLE 5, , Let A {1, 2, 3, 4, 5} . Define a relation R from A to A by, R {(x, y) : y 2x 3} ., (i) Depict R using an arrow diagram., (ii) Find dom (R) and range (R)., , SOLUTION, , x 1 & y (2 3) 1z A, x 2 & y ( 4 3) 1,, x 3 & y ( 6 3) 3,, x 4 & y ( 8 3) 5,, x 5 & y (10 3) 7 z A., , R {(2, 1), (3, 3), (4, 5)} .
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Relations, , 65, , (i) We may depict it by arrow diagram, as shown below:, 1, 2, , 1, 2, , 3, , 3, , 4, , 4, , 5, , 5, A, , A, , (ii) We have, dom (R) = {2, 3, 4} and range (R) = {1, 3, 5}., EXAMPLE 6, , SOLUTION, , Let A {1, 2, 3, 4, 5} and B {1, 4, 5} ., Let R be a relation ‘is less than‘ from A to B., (i) List the elements of R., (ii) Find the domain, co-domain and range of R., (iii) Depict the above relation by an arrow diagram., Here, A {1, 2, 3, 4, 5} and B {1, 4, 5} ., (i) R {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)} ., (ii) Dom (R) {1, 2, 3, 4}, range (R) {4, 5}, and co-domain (R) {1, 4, 5} ., (iii) We may represent the above relation by an arrow diagram,, shown below., 1, , 1, , 2, , 4, , 3, , 5, , 4, 5, A, , REMARK, , Let A and B be two nonempty sets. Then, every subset of A# B is a, relation from A to B., Since 1 A# B, so is also a relation from A to B, called an empty or, void relation., , EXAMPLE 7, SOLUTION, , B, , Let A {x, y, z} and B {1, 2} . Find the number of all possible relations, that can be defined from A to B., Here n(A) 3 and n(B) 2. So, n(A# B) (3 # 2) 6., Since a set containing n elements has 2 n subsets., , (A# B) will have 26 64 subsets., But, every subset of A# B is a relation from A to B., Hence, there are in all 64 relations from A to B.
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66, , Senior Secondary School Mathematics for Class 11, , EXERCISE 2C, 1. Let A and B be two nonempty sets., (i) What do you mean by a relation from A to B?, (ii) What do you mean by the domain and range of a relation?, 2. Find the domain and range of each of the relations given below:, (i) R {(1, 1), (1, 1), ( 2, 4), (2, 4), (3, 9)}, 1, (ii) R (bx, x l : x is an integer, 0 x 52, (iii) R {(x, y) : x 2y 8 and x, y d N}, (iv) R {(x, y) : y |x 1| , xdZ and|x|# 3}, 3. Let A {1, 3, 5, 7} and B {2, 4, 6, 8} ., Let R {(x, y) : xd A, y dB and x y} ., (i) Write R in roster form., (ii) Find dom (R) and range (R)., (iii) Depict R by an arrow diagram., 4. Let A {2, 4, 5} and B {1, 2, 3, 4, 6, 8} ., Let R {(x, y) xd A, y dB and x divides y}., (i) Write R in roster form., (ii) Find dom (R) and range (R)., 5. Let A {2, 3, 4, 5} and B {3, 6, 7, 10} ., Let R {(x, y) : xd A, y dB and x is relatively prime to y}., (i) Write R in roster form., (ii) Find dom (R) and range (R)., 6. Let A {1, 2, 3, 5} and B {4, 6, 9} ., Let R {(x, y) : x,d A, y dB and (x y) is odd}., Write R in roster form., 7. Let R {(x, y) : x 3y 12, xd N and y d N} ., (i) Write R in roster form., (ii) Find dom (R) and range (R)., 8. Let A {1, 2, 3, 4, 5, 6} ., Define a relation R from A to A by R {(x, y) : y x 1} ., (i), (ii), (iii), (iv), , Write R in roster form., Find dom (R) and range (R)., What is its co-domain?, Depict R by using arrow diagram., 9. Let R {(x, x 5) : xd{0, 1, 2, 3, 4, 5}} ., (i) Write R in roster form., (ii) Find dom (R) and range (R).
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Relations, , 67, , 10. Let A {1, 2, 3, 4, 6} and let R {(a, b) : a, bd A and a divides b}., (i) Write R in roster form., (ii) Find dom (R) and range (R)., 11. Define a relation R from Z to Z, given by, R {(a, b) : a, bdZ and (a b) is an integer}., Find dom (R) and range (R)., Hint, , The difference of two integers is always an integer., , 12. Let R {(x, y) : x, y dZ and x 2 y 2 # 4} ., (i) Write R in roster form., (ii) Find dom (R) and range (R)., 13. Let A {2, 3} and B {3, 5} ., (i) Find (A# B) and n(A# B) ., (ii) How many relations can be defined from A to B?, 14. Let A {3, 5} and B {7, 9} . Let R {(a, b) : ad A, bdB and (a b) is odd}., Show that R is an empty relation from A to B., Hint, , The difference of two odd numbers cannot be odd., ANSWERS (EXERCISE 2C), , 2., , (i) dom (R) { 2, 1, 1, 2, 3} and range (R) {1, 4, 9}, , 1 1 1, (ii) dom (R) {1, 2, 3, 4} and range (R) '1, , , 1, 2 3 4, (iii) dom (R) {2, 4, 6} and range (R) {3, 2, 1}, (iv) dom (R) { 3, 2, 1, 0, 1, 2, 3} and range (R) {0, 1, 2, 3, 4}, 3. (i) R {(3, 2), (5, 2), (5, 4), (7, 2), (7, 4), (7, 6)}, (ii) dom (R) {3, 5, 7} and range (R) {2, 4, 6}, (iii), 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , A, , B, , (i) R {(2, 2), (2, 4), (2, 6), (2, 8), (4, 4), (4, 8)}, (ii) dom (R) {2, 4} and range (R) {2, 4, 6, 8}, 5. (i) R {(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)}, (ii) dom (R) {2, 3, 4, 5} and range (R) {3, 6, 7, 10}, 6. R {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}, 7. (i) R {(3, 3), (6, 2), (9, 1)}, (ii) dom (R) {3, 6, 9} and range (R) {3, 2, 1}, 4.
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68, , 8., , Senior Secondary School Mathematics for Class 11, , (i) R {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}, (ii) dom (R) {1, 2, 3, 4, 5}, range (R) {2, 3, 4, 5, 6}, (iii) co-domain of R {1, 2, 3, 4, 5, 6}, (iv), 1, 2, , 1, 2, , 3, , 3, , 4, , 4, , 5, 6, , 5, 6, A, , B, , (i) R {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), 5, 10)}, (ii) dom (R) {0, 1, 2, 3, 4, 5} and range (R) {5, 6, 7, 8, 9, 10}, 10. (i) R {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6),, (4, 4), (6, 6)}, (ii) dom (R) {1, 2, 3, 4, 6} and range (R) {1, 2, 3, 4, 6}, 11. dom (R) Z and range (R) Z, 12. (i) R {( 2, 0), (0, 2), (2, 0), (0, 2), (1, 0), (0, 1), (1, 0), (0, 1), (1, 1),, (1, 1), (1, 1), (1, 1)}, (ii) dom (R) { 2, 1, 0, 1, 2} and range (R) { 2, 1, 0, 1, 2}, 13. (i) A# B {(2, 3), (2, 5), (3, 3), (3, 5)} and n(A# B) 4, (ii) number of relations from A to B 2 4 16, 9., , BINARY RELATION ON A SET, , Let A be a nonempty set. Then, every subset of (A# A) is called a binary relation or, simply a relation on A., REMARKS, (i) Since 1 A# A so is relation on A, called the empty or void, relation on A., (ii) Since A# A 3 A# A so (A# A) is a relation on A, called the, universal relation on A., (iii) Let IA {(a, a) : ad A} . Then, clearly, IA 3 (A# A) and therefore, it, is a relation on A, called the identity relation on A., SOLVED EXAMPLES, EXAMPLE 1, SOLUTION, , Let A {1, 2, 3} and R {(1, 2), (2, 2), (3, 1), (3, 2)} . Show that R is a, binary relation on A. Find its domain and range., R {(1, 2), (2, 2), (3, 1), (3, 2)} ., Clearly, R 1 A# A and so R is a relation on A., Dom (R) set of first coordinates of elements of R {1, 2, 3} .
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Relations, , 69, , Range (R) set of second coordinates of elements of R {1, 2} ., Hence, dom (R) {1, 2, 3} and range (R) {1, 2} ., EXAMPLE 2, , SOLUTION, , Let N be the set of all natural numbers. Let R {(a, b) : a, bd N and, 2a b 10} . Show that R is a binary relation on N. Find its domain,, range and co-domain., Here R {(a, b) : a, bd N and 2a b 10} ., Now, 2a b 10 & b (10 2a) ., , (a 1 & b 8), (a 2 & b 6), (a 3 & b 4), (a 4 & b 2) ., , R {(1, 8), (2, 6), (3, 4), (4, 2)} ., Since R 1 N # N, so R is a binary relation on N., Dom (R) = set of 1st coordinates of elements of R, {1, 2, 3, 4} ., Range (R) = set of 2nd coordinates of elements of R, {8, 6, 4, 2} ., Co-domain of R N., , INVERSE RELATION, , Let R be a binary relation on a set A. Then, the inverse of R, denoted by R 1 is a binary, relation on A, defined by R 1 {(b, a) : (a, b)d R} ., Clearly, (a, b)d R + (b, a)d R 1 ., Also, dom (R) range (R 1) and range (R) dom (R 1)., EXAMPLE 3, , SOLUTION, , Let A be the set of first ten natural numbers. Let R be a binary relation on, A, defined by, R {(a, b) : a, bd A and a 2b 10} ., Express R and R 1 as sets of ordered pairs., Show that (i) dom (R) = range (R 1) (ii) range (R) = dom (R 1) ., (10 a), ·, 2, Now, (a 2 & b 4), (a 4 & b 3), (a 6 & b 2),, (a 8 & b 1) ., , R {(2, 4), (4, 3), (6, 2), (8, 1)}, , R 1 {(4, 2), (3, 4), (2, 6), (1, 8)} ., , (i) dom (R) {2, 4, 6, 8} range (R 1) ., (ii) range (R) {4, 3, 2, 1} dom (R 1) ., a 2b 10 & b , , VARIOUS TYPES OF RELATIONS, , Let A be a nonempty set. Then, a relation R on A is said to be:, (i) reflexive, if (a, a)d R for all ad A, i.e., a R a for all ad A;, (ii) symmetric, if (a, b)d R & (b, a)d R for all a, bd A,, i.e., a R b & b R a for all a, bd A;
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70, , Senior Secondary School Mathematics for Class 11, , (iii) transitive, if (a, b)d R and (b, c)d R & (a, c)d R for all a, b, cd A., i.e., a R b, b R c & a R c for all a, b, cd A., EQUIVALENCE RELATION, , A relation which is reflexive, symmetric and transitive is called an equivalence relation., EXAMPLE 4, SOLUTION, , EXAMPLE 5, , SOLUTION, , Let R be a relation on the set Q of all rationals defined by R {(a, b) : a, bdQ, and a bdZ} . Show that R is an equivalence relation., Given: R {(a, b) : a, bdQ and a bdZ} ., (i) Let adQ. Then, a a 0dZ., , (a, a)d R for all adQ., So, R is reflexive., (ii) (a, b)d R & (a b)dZ, i.e., (a b) is an integer, & (a b) is an integer, & (b a) is an integer, & (b, a)d R., Thus, (a, b)d R & (b, a)d R., , R is symmetric., (iii) (a, b)d R and (b, c)d R, &, (a b) is an integer and (b c) is an integer, &, {(a b) (b c)} is an integer, &, (a c) is an integer, &, (a, c)d R., Thus, (a, b)d R and (b, c)d R & (a, c)d R., , R is transitive., Thus, R is reflexive, symmetric and transitive., So, R is an equivalence relation., Let m be a given fixed positive integer., Let R {(a, b) : a, bdZ and (a b) is divisible by m}., Show that R is an equivalence relation on Z., R {(a, b) : a, bdZ and (a b) is divisible by m}., (i) Let adZ. Then,, a a 0, which is divisible by m., , (a, a)d R for all adZ., So, R is reflexive., (ii) Let (a, b)d R. Then,, (a, b)d R & (a b) is divisible by m, & (a b) is divisible by m, & (b a) is divisible by m, & (b, a)d R.
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Relations, , 71, , Thus, (a, b)d R & (b, a)d R., So, R is symmetric., (iii) Let (a, b)d R and (b, c)d R. Then,, (a, b)d R and (b, c)d R., &, (a b) is divisible by m and (b c) is divisible by m, &, {(a b) (b c)} is divisible by m, &, (a c) is divisible by m, &, (a, c)d R., `, (a, b)d R and (b, c)d R & (a, c)d R., So, R is transitive., Thus, R is reflexive, symmetric and transitive., Hence, R is an equivalence relation on Z., EXAMPLE 6, SOLUTION, , EXAMPLE 7, SOLUTION, , EXAMPLE 8, SOLUTION, , Show that the relation ‘is parallel to’ on the set S of all straight lines in a, plane is an equivalence relation., Let S be the set of all straight lines in a plane., Then, the relation, ‘is parallel to’ on S is, (i) reflexive, since every line is parallel to itself,, i.e., L||L for all L in S;, (ii) symmetric, since L||M & M||L for all L, M dS;, (iii) transitive, since for all L, M, N in S, we have, L||M and M||N & L||N., Thus, the given relation is reflexive, symmetric and transitive., Hence, it is an equivalence relation on S., Show that the relation ‘is congruent to’ on the set of all triangles in a, plane is an equivalence relation., Let S be the set of all triangles in a plane., Then, the congruence relation on S is, (i) reflexive, since 3b3 for every 3dS;, (ii) symmetric, since 31 b32 & 32 b31 for all 31, 32 dS, (iii) transitive, since 31 b32 and 32 b33 & 31 b33, for all 31, 32, 33 dS., Hence, the given relation is an equivalence relation., Let R {(a, b) : a, bd N and a b 2} . Show that R satisfies none of, reflexivity, symmetry and transitivity., (i) R is not reflexive, since 2 ! 2 2 and therefore (2, 2)z R., (ii) Since 4 2 2, so (4, 2)d R., But, 2 ! 4 2 . So, (2, 4)z R., Thus, (4, 2)d R but (2, 4)z R., , R is not symmetric.
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72, , Senior Secondary School Mathematics for Class 11, , (iii) Since 16 4 2, so (16, 4)d R., Also, 4 2 2, so (4, 2)d R., But, 16 ! 2 2 & (16, 2)z R., Thus, (16, 4)d R and (4, 2)d R. But (16, 2)z R., , R is not transitive., Hence, R satisfies none of reflexivity, symmetry and, transitivity., EXAMPLE 9, , SOLUTION, , Let S be the set of all real numbers and let R be a binary relation on S, defined by (a, b)d R + 1 ab 0 for all a, bdS. Show that R is reflexive, as well as symmetric. Give an example to show that R is not transitive., (i) Let a be an arbitrary real number., Then, (1 a 2) 0 & (a, a)d R for every adS., , R is reflexive., (ii) Let (a, b)d R. Then,, (a, b)d R & 1 ab 0, & 1 ba 0, & (b, a)d R., , (a, b)d R & (b, a)d R., So, R is symmetric., (iii) In order to show that R is not transitive, consider the real, 2, numbers b l, 1 and 2., 3, 2, 2, 1, 2, Now, b , 1ld R, since (1 b l#12 b1 l 0., 3, 3, 3, 3, And, (1, 2)d R, since {1 (1# 2)} 3 0., 2, 2, 4, 1, But, b , 2lz R, since (1 b l# 22 b1 l 0., 3, 3, 3, 3, 2, 2, Thus, b , 1ld R, (1, 2)d R and b , 2lz R., 3, 3, This shows that R is not transitive., , EXAMPLE 10, , SOLUTION, , Let R be a relation on N # N, defined by, (a, b) R (c, d) + a d b c for all (a, b), (c, d)d N # N., Show that R is an equivalence relation., Here R is a relation on N # N, defined by, (a, b) R (c, d) + a d b c for all (a, b), (c, d)d N # N., We shall show that R satisfies the following properties., (i) Reflexivity:, We know that a b b a for all a, bd N., , (a, b) R (a, b) for all (a, b)d(N # N) ., So, R is reflexive.
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Relations, , 73, , (ii) Symmetry:, Let (a, b) R (c, d) . Then,, (a, b) R (c, d) & a d b c, & cb da, & (c, d) R (a, b)., , (a, b) R (c, d) & (c, d) R (a, b) for all (a, b), (c, d)d N # N., This shows that R is symmetric., (iii) Transitivity:, Let (a, b) R (c, d) and (c, d) R (e, f ) . Then,, (a, b) R (c, d) and (c, d) R (e, f ), &, a d b c and c f d e, &, adcf bcde, &, af be, &, (a, b) R (e, f ) ., Thus, (a, b) R (c, d) and (c, d) R (e, f ) & (a, b) R (e, f ) ., This shows that R is transitive., , R is reflexive, symmetric and transitive., Hence, R is an equivalence relation on N # N., , EXERCISE 2D, 1. What do you mean by a binary relation on a set A?, Define the domain and range of a relation on A., 2. Let A {2, 3, 5} and R {(2, 3), (2, 5), (3, 3), (3, 5)} ., Show that R is a binary relation on A. Find its domain and range., 3. Let A {0, 1, 2, 3, 4, 5, 6, 7, 8} and let R {(a, b) : a, bd A and 2a 3b 12} ., Express R as a set of ordered pairs. Show that R is a binary relation on A., Find its domain and range., 4. If R is a binary relation on a set A, define R 1 on A., Let R {(a, b) : a, bdW and 3a 2b 15}, where W is the set of whole, numbers., Express R and R 1 as sets of ordered pairs., Show that (i) dom (R) = range (R 1) (ii) range (R) = dom (R 1) ., 5. What is an equivalence relation?, Show that the relation of ‘similarity‘ on the set S of all triangles in a plane is, an equivalence relation., 6. Let R {(a, b) : a, bdZ and (a b) is even}., Then, show that R is an equivalence relation on Z., 7. Let A {1, 2, 3} and R {(a, b) : a, bd A and a 2 b 2 # 5., Write R as set of ordered pairs.
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74, , 8., 9., 10., , 11., 12., , 13., , Senior Secondary School Mathematics for Class 11, , Mention whether R is (i) reflexive (ii) symmetric (iii) transitive., Give reason in each case., Let R {(a, b) : a, bdZ and b 2a 4} . If (a, 2)d R and (4, b 2)d R. Then,, write the values of a and b., Let R be a relation on Z, defined by (x, y)d R + x 2 y 2 9. Then, write R, as set of ordered pairs. What is its domain?, Let A be the set of first five natural numbers and let R be a relation on A,, defined by (x, y)d R + x # y., Express R and R 1 as sets of ordered pairs., Find: dom (R 1) and range (R)., Let R {(x, y) : x, y dZ and x 2 y 2 25} ., Express R and R 1 as sets of ordered pairs. Show that R R 1 ., Find R 1, when, (i) R {(1, 2), (1, 3), (2, 3), (3, 2), (4, 5)},, (ii) R {(x, y) : x, y d N, x 2y 8}., Let A {a, b} . List all relations on A and find their number., Hint, , A # A {(a, a), (a, b), (b, a), (b, b)} and every subset of A # A is a relation on A., So, their number 2 4 16., , 14. Let R {(a, b) : a, bd N and a b} ., Show that R is a binary relation on N, which is neither reflexive nor, symmetric. Show that R is transitive., Hint, , Since R 1 N # N, so it is a binary relation on N., ANSWERS (EXERCISE 2D), , 2., 3., 4., 7., 8., 9., 10., , 11., , dom (R) {2, 3} and range (R) {3, 5}, R {(0, 4), (3, 2), (6, 0)}, dom (R) {0, 3, 6} and range (R) {0, 2, 4}, R {(1, 6), (3, 3), (5, 0)}, R 1 {(6, 1), (3, 3), (0, 5)}, R {(1, 1), (2, 2), (3, 3), (2, 1), (1, 2), (3, 2), (2, 3)}., R is reflexive and symmetric but not transitive., a 1, b 4, R {( 3, 0), (0, 3), (3, 0), (0, 3), dom (R) { 3, 0, 3}, R {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4),, (3, 5), (4, 4), (4, 5), (5, 5)}, R 1 {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (2, 2), (3, 2), (4, 2), (5, 2), (3, 3), (4, 3),, (5, 3), (4, 4), (5, 4), (5, 5)}, dom (R 1) {1, 2, 3, 4, 5} range (R), R {( 5, 0), (0, 5), (5, 0), (0, 5), (3, 4), (4, 3), ( 3, 4), (4, 3), ( 3, 4), ( 4, 3)}, R, , 1, , {(0, 5), ( 5, 0), (0, 5), (5, 0), (4, 3), (3, 4), (4, 3), ( 3, 4), ( 4, 3),, ( 3, 4)}
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Relations, , 12. (i) R 1 {(2, 1), (3, 1), (3, 2), (2, 3), (5, 4)}, 13. 16, , 75, , (ii) R 1 {(3, 2), (2, 4), (1, 6)}, , EXERCISE 2E, Very-Short-Answer Questions, 1. Let A and B be two sets such that n(A) 5, n(B) 3 and n(A + B) 2., (i) n(A , B), (ii) n(A# B), (iii) n{(A# B)+(B# A)}, , , , Hint (i) n(A , B) n(A) n(B) n(A + B) ., (ii) n(A # B) n(A) $ n(B), (iii) If n(A + B) m then n{(A # B)+(B # A)} 2 m ., , Find a and b when (a 2b, 13) (7, 2a 3b) ., If A {1, 2}, find A# A# A., If A {2, 3, 4} and B {4, 5}, draw an arrow diagram to represent (A# B) ., If A {3, 4}, B {4, 5} and C {5, 6}, find A#(B#C) ., If A 3 B, prove that A#C B#C., Prove that A# B B# A & A B., If A {5} and B {5, 6}, write down all possible subsets of A# B., Let R {(x, x 2) : x is a prime number less than 10}., (i) Write R in roster form., (ii) Find dom (R) and range (R)., 10. Let A {1, 2, 3} and B {4} ., How many relations can be defined from A to B?, 11. Let A {3, 4, 5, 6} and R {(a, b) : a, bd A and a b} ., Write R in roster form., Find: dom (R) and range (R)., Write R 1 in roster form., 12. Let R {(a, b) : a, bd N, a b} ., Show that R is a binary relation which is neither reflexive, nor symmetric., Show that R is transitive., 2., 3., 4., 5., 6., 7., 8., 9., , ANSWERS (EXERCISE 2E), , 1. (i) 6 (ii) 15 (iii) 4, 2. a 5, b 1, 3. {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}, 4., 2, , 4, , 3, 4, , 5, B, , A, , A #B
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76, , Senior Secondary School Mathematics for Class 11, , 5. {(3, 4, 5), (3, 4, 6), (3, 5, 5), (3, 5, 6), (4, 4, 5), (4, 4, 6), (4, 5, 5), (4, 5, 6)}, 8. , {(5, 5)}, {(5, 6)}, {(5, 5), (5, 6)}, 9. (i) R {(2, 4), (3, 9), (5, 25), (7, 49)}, (ii) dom (R) {2, 3, 5, 7} and range (R) {4, 9, 25, 49}, 10. 8, 11. R {(4, 3), (5, 3), (5, 4), (6, 3), (6, 4), (6, 5), (i) dom (R) {4, 5, 6} and range (R) {3, 4, 5}, R, , 1, , {(3, 4), (3, 5), (4, 5), (3, 6), (4, 6), (5, 6)}, , SUMMARY OF KEY FACTS, 1. Two numbers a and b listed in a specific order and enclosed in parentheses, form an ordered pair (a, b) ., Here a is the first component and b is the second component., In general, (a, b) ! (b, a) ., 2. (a, b) (c, d) + a c and b d., 3. Let A and B be two nonempty sets. Then, the Cartesian product of A and B, is defined as, A# B {(a, b) : ad A and bdB} ., If A or B then A# B ., 4. If n(A) p and n(B) q then n(A# B) pq., 5. Three numbers a, b, c listed in a specific order and enclosed in parentheses, form an ordered triplet (a, b, c) ., (a, b, c) ! (b, a, c) ! (c, a, b), etc., 6. For any nonempty sets A, B, C, we have, (A# B)#C A#(B#C), each denoted by A# B#C., 7. For any sets A, B and C, we have:, (i) A#(B, C) (A# B),(A#C), (ii) A#(B+ C) (A# B)+(A#C), (iii) A#(B C) (A# B) (A#C), (iv) (A# B)+(B# A) (A + B)#(B+ A) (A + B)#(A + B), (v) A# B A#C & B C, (vi) A 1 B & (A# A) 1 (A# B)+(B# A), (vii) A 1 B & (A#C) 1 (B#C), (viii) A 1 B and C 1 D & (A#C) 1 (B# D), (ix) A# B B# A + A B, 8. Let A and B be two nonempty sets and let R 3 A# B., Then, R is called a relation from A to B., If (a, b)d R, we say that ‘a is related to b‘ and we write, a R b.
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Relations, , 77, , Y b., If (a, b)z R, we say that 'a is not related to b‘ and we write, a R, Dom (R) {a : (a, b)d R}, range (R) {b : (a, b)d R} ., 9. We define, R 1 {(b, a) : (a, b)d R},, dom (R) = range (R 1) and range (R) = dom (R 1)., 10. Let A be a nonempty set. Then, every subset of A# A is called a binary, relation on A., 11. Let A be a nonempty set and R be a binary relation on A., Then, R is said to be:, (i) reflexive, if (a, a)d R for all ad A;, i.e., a R a for all ad A., (ii) symmetric, if (a, b)d R & (b, a)d R for all a, bd A;, i.e., a R b & b R a for all a, bd A., (iii) transitive, if (a, b)d R and (b, c)d R & (a, c)d R for all a, b, cd A;, i.e., a R b and b R c & a R c for all a, b, cd R., 12. A relation R on A, which is reflexive, symmetric and transitive, is called an, equivalence relation on A.,
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78, , Senior Secondary School Mathematics for Class 11, , 3, , Functions, Let X and Y be two nonempty sets. Then, a relation f from X to Y is called, a function, if every element in X has a unique image in Y, and we write, f : X " Y., FUNCTION, , Thus, a relation f from X to Y is a function, if dom ( f ) X and no two, distinct ordered pairs in f have the same first coordinate., If (x, y)d f, we write, f (x) y., Here, y is called the image of x under f and x is called the pre-image of y., If f : X " Y then dom ( f ) X and range ( f ) 3 Y., Also, Y is called the co-domain of f., IMAGE OF A SUBSET, , Let f : X " Y and let AfX., Then, the image of A under f is defined as, f ( A ) { f ( x) : x d A } ., Clearly, f (A) 3 Y., EXAMPLE 1, , Let X {1, 2, 3, 4} and Y {1, 4, 9, 16, 25} ., Let f {(x, y) : xdX, y dY and y x 2} ., (i) Show that f is a function from X to Y. Find its domain and range., (ii) Draw a pictorial representation of the above function., (iii) If A {2, 3, 4}, find f (A) ., , SOLUTION, , (i) We have, f {(x, y) : xdX, y dY and y x 2} ., Giving different values to x from the set X and getting the, corresponding value of y x 2, we get, f {(1, 1), (2, 4), (3, 9), (4, 16)} ., Clearly, every element in X has a unique image in Y., Hence, f is a function from X to Y., Dom ( f ) {1, 2, 3, 4} X., Range ( f ) {1, 4, 9, 16} 1 Y., Clearly, 25dY does not have its pre-image in X., (ii) A pictorial representation of the above mapping f is given, below., 78
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Functions, , 79, , f, 1, 2, , 1, 4, , 3, , 9, , 4, , 16, 25, X, , Y, , (iii) Now, let A {2, 3, 4} . Then,, f (2) 2 2 4, f (3) 3 2 9 and f (4) 4 2 16., , f (A) { f (x) : xd A} {4, 9, 16} ., RELATION AS A PARTICULAR CASE OF A FUNCTION, , A relation f from X to Y is a function if dom ( f ) X and no two distinct ordered, pairs in f have the same first coordinate., If (x, y)d f, we write, f (x) y., EXAMPLE 2, , SOLUTION, , EXAMPLE 3, , SOLUTION, , Let X {2, 3, 4, 5} and Y {7, 9, 11, 13, 15, 17} ., Define a relation f from X to Y by:, f {(x, y) : xdX, y dY and y 2x 3} ., (i) Write f in roster form., (ii) Find dom ( f ) and range ( f ) ., (iii) Show that f is a function from X to Y., Here X {2, 3, 4, 5} and Y 2x 3., Now, x 2 & y (2 # 2 3) 7,, x 3 & y ( 2 # 3 3) 9 ,, x 4 & y (2 # 4 3) 11,, x 5 & y (2 # 5 3) 13., (i) , f {(2, 7), (3, 9), (4, 11), (5, 13)} ., (ii) Clearly, dom ( f ) {2, 3, 4, 5} and range ( f ) {7, 9, 11, 13} 1 Y., (iii) It is clear that no two distinct ordered pairs in f have the same, first coordinate., f is a function from X to Y., `, Which of the following relations are functions? Give reasons. In case of a, function, find its domain and range., (i) f {(1, 3), (1, 5), (2, 3), (2, 5)}, (ii) g {(2, 1), (5, 1), (8, 1), (11, 1)}, (iii) h {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6)}, (i) f {(1, 3), (1, 5), (2, 3), (2, 5)} ., Here, one element, namely 1 has two images 3 and 5 under f., , f is not a function.
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80, , Senior Secondary School Mathematics for Class 11, , (ii) g {(2, 1), (5, 1), (8, 1), (11, 1)} ., Clearly, no two distinct ordered pairs in g have the same first, coordinate. So, g is a function., , dom (g) {2, 5, 8, 11} and range g {1} ., (iii) h {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6)} ., Clearly, no two distinct ordered pairs in h have the same first, coordinate. So, h is a function., , dom (h) {2, 4, 6, 8, 10, 12} and range (h) {1, 2, 3, 4, 5, 6} ., EXAMPLE 4, , SOLUTION, , Let A {2, 1, 0, 1, 2} and f : A " Z : f (x) x 2 2x 3., Find (i) range (f ) (ii) pre-images of 6, –3 and 5., We have, f (2) (2) 2 2 #(2) 3 (4 4 3) 5;, f (1) (1) 2 2 #(1) 3 0;, f (0) 3;, f (1) (1 2 3) 4 and f (2) (2 2 2 # 2 3) 3., , f {(2, 5), (1, 0), (0, 3), (1, 4), (2, 3)} ., (i) Range ( f ) {5, 0, 3, 4} ., (ii) f (x) 6 & x 2 2x 3 6, & x 2 2x 9 0 & x , , 2 ! 4 36, (1 ! 10 )z A., 2, , , , 6 has no pre-image in A., f (x) 3 & x 2 2x 3 3, & x 2 2x 0 & x(x 2) 0 & x 0 or x 2., Clearly, 0, 2d A., So, the pre-images of –3 are 0 and 2., f (x) 5 & x 2 2x 3 5, & x 2 2x 8 0 & x 2 4x 2x 8 0, & x(x 4) 2(x 4) 0 & (x 4)(x 2) 0, & x 2 or x 4., Now, 2d A but 4z A., So, the pre-image of 5 is –2., EQUAL FUNCTIONS, , Two functions f and g are said to be equal if, (i) dom ( f ) dom (g), , (ii) co-domain of f co-domain of g, , (iii) f (x) g (x) for every x in their common domain., EXAMPLE 5, , Let A {1, 2} and B {3, 6} and f and g be functions from A to B, defined, by f (x) 3x and g(x) x 2 2. Show that f g.
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Functions, , 81, , SOLUTION, , Clearly, we have, dom ( f ) dom (g) {1, 2} ., Co-domain of f co-domain of g {3, 6} ., Also, f (1) (3 #1) 3, f (2) (3 # 2) 6;, g (1) (1 2 2) 3, g (2) (2 2 2) 6., , f (1) g (1) and f (2) g (2) ., Thus, f (x) g (x) for all xd A., Hence, f g., , EXAMPLE 6, , Let f : Z " Z : f (x) x 2 and g : Z " Z : g (x) |x|2 for all xdZ. Show, that f g., We have, dom ( f ) dom (g) Z, and co-domain of f co-domain of g Z., Also, for all xdZ, we have, f (x) x 2 and g (x) |x|2 x 2 ., , f (x) g (x) for all xdZ., Hence, f g., x2 4, Let f : R " R : f (x) x 2 and g : R {2} " R : g (x) ·, x 2, Show that f ! g. Re-define f and g such that f g., We have, dom ( f ) R and dom (g) R {2} ., , SOLUTION, , EXAMPLE 7, , SOLUTION, , EXAMPLE 8, , SOLUTION, , Since dom ( f ) ! dom (g), so we have f ! g., For every real number x ! 2, we have, x 2 4 (x 2)(x 2) , g (x) , ( x 2), [a (x 2) ! 0]., x 2, ( x 2), Thus, f (x) g (x) for all xd R {2} ., , f g only when, they are re-defined as under:, x2 4, f : R {2} " R : f (x) x 2 and g : R {2} " R : g (x) ·, x 2, Let f {(1, 3), (0, 1), (1, 1), (2, 3)} be a function, described by the, formula, f (x) x . Then, find the values of and . Also, find the, formula., Here f (x) x ., … (i), Also, f (1) 3, f (0) 1, f (1) 1 and f (2) 3., [given], Putting x 1 and f (1) 3 in (i), we get, 3 & 3., … (ii), Putting x 0 and f (0) 1 in (i), we get, … (iii), # 0 1 & 1., Putting 1 from (iii) in (ii), we get 2., , 2 and 1., Hence, f (x) 2x 1 is the required formula.
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82, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 9, , SOLUTION, , Let f : R " R : f (x) x 2 3., Find the pre-images of each of the following under f:, (i) 19, (ii) 28, (iii) 2, 2, , Given: f (x) x 3., (i) Let x be the pre-image of 19. Then,, f (x) 19 & x 2 3 19 & x 2 16 & x !4., , 4 and –4 are the pre-images of 19., (ii) Let x be the pre-image of 28. Then,, f (x) 28 & x 2 3 28 & x 2 25 & x !5., , 5 and –5 are the pre-images of 28., (iii) Let x be the pre-image of 2. Then,, f (x) 2 & x 2 3 2 & x 2 1., But, no real value of x satisfies the equation, x 2 1., , 2 does not have any pre-image under f., , INVERSE OF A FUNCTION, , Let f : X " Y and let y dY., Then, we define, f 1 (y) {xdX : f (x) y} set of pre-images of y., f 1 is called the inverse of f., REMARK, , In above example 9, we have, (i) f 1 {19} { 4, 4}, (ii) f 1 {28} { 5, 5}, , EXAMPLE 10, , SOLUTION, , Let f : R " R : f (x) x 2 1., Find (i) f 1 { 4}, (ii) f 1 {10}, , (iii) f 1 {2} ., , (iii) f 1 {5, 17}., , It is given that f (x) x 2 1., (i) Let f 1 ( 4) x. Then,, f (x) 4 & x 2 1 4 & x 2 5., But, there is no real value of x whose square is –5., , , f 1 { 4} ., 1, (ii) Let f (10) x. Then,, f (x) 10 & x 2 1 10 & x 2 9 & x !3., , f 1 {10} {3, 3} ., 1, (iii) Let f (5) x. Then,, f (x) 5 & x 2 1 5 & x 2 4 & x !2., , f 1 {5} {2, 2}., Let f 1 (17) x. Then,, f (x) 17 & x 2 1 17 & x 2 16 & x !4., , f 1 {17} { 4, 4} ., Hence, f 1 {5, 17} {2, 2, 4, 4} .
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Functions, EXAMPLE 11, , SOLUTION, , EXAMPLE 12, , SOLUTION, , EXAMPLE 13, , SOLUTION, , EXAMPLE 14, , Let f : R " R, defined by, ]Z, f (x) ][1, if xdQ, ]]1, if x Q., z, \, 1, Find (i) f b l, (ii) f (0.34), 2, (v) range ( f ) (vi) f 1 {1}, , 83, , (iii) f ( 2 ), (vii) f, , 1, , (iv) f (), , {1}, , 1, Since each one of and 0.34 is rational, we have, 2, 1, (i) f b l 1 and (ii) f (0.34) 1., 2, Since each one of 2 and is irrational, we have, (iii) f ( 2 ) 1 and (iv) f () 1., (v) range ( f ) { f (x) : xd R}, { f (x) : xdQ},{ f (x) : xd R Q} {1, 1} ., 1, (vi) f {1} {x : f (x) 1} Q., (vii) f 1 {1} {x : f (x) 1} (R Q) ., Let f : R " R, defined by, ]Z]3x 2, x 0, ]], f (x) []1,, x0, ]], ]4x 1, x 0., \, Find (i) f (2), (ii) f (2), (iii) f (0), (iv) f (3.5) ., Clearly, we have, (i) f (2) (4 # 2 1) 9, [a 2 0]., (ii) f (2) {3 #(2) 2} 8, [a 2 0]., (iii) f (0) 1., (iv) f (3.5) (4 # 3.5 1) 15, [a 3.5 0]., Consider the relation, Z] 2, f (x) ][x , 0 # x # 2, ]], 3x, 2 # x # 10., \, Show that f is not a function., We have, dom (f) [0, 10] 1 R., Now, f (2) 2 2 4 and also, f (2) (3 # 2) 6., As such, a single element, namely 2 has two distinct images,, namely 4 and 6., , f is not a function., Consider the relation, Z] 2, g (x) ][x , 0 # x # 3, ]], 3x, 3 # x # 10., \, Show that g is a function.
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84, , Senior Secondary School Mathematics for Class 11, , SOLUTION, , We have, dom (g) [0, 10] 1 R., It is easy to verify that every element in [0, 10] is associated with a, unique real number under g., In particular g (3) 3 2 9 and also g (3) 3 # 3 9, i.e., 3 has a, unique image, namely 9., , g is a function from [0, 10] into R., , EXERCISE 3A, 1. Define a function as a set of ordered pairs., 2. Define a function as a correspondence between two sets., 3. What is the fundamental difference between a relation and a function? Is, every relation a function?, 4. Let X {1, 2, 3, 4}, Y {1, 5, 9, 11, 15, 16}, and f {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)} ., Are the following true?, (i) f is a relation from X to Y (ii) f is a function from X to Y. Justify your, answer in each case., 5. Let X {1, 0, 3, 7, 9} and f : X " R : f (x) x 3 1., Express the function f as set of ordered pairs., 6. Let A {1, 0, 1, 2} and B {2, 3, 4, 5} . Find which of the following are, functions from A to B. Give reason., (i) f {(1, 2), (1, 3), (0, 4), (1, 5)}, (ii) g {(0, 2), (1, 3), (2, 4)}, (iii) h {(1, 2), (0, 3), (1, 4), (2, 5)}, 7. Let A {1, 2} and B {2, 4, 6} . Let f {(x, y) : xd A, y dB and y 2x 1} ., Write f as a set of ordered pairs., Show that f is a relation but not a function from A to B., 8. Let A {0, 1, 2} and B {3, 5, 7, 9} . Let f {(x, y) : xd A, y dB and y 2x 3} ., Write f as a set of ordered pairs., Show that f is a function from A to B., Find dom ( f ) and range ( f ) ., 9. Let A {2, 3, 5, 7} and B {3, 5, 9, 13, 15} ., Let f {(x, y) : xd A, y dB and y 2x 1} ., Write f in roster form. Show that f is a function from A to B., Find the domain and range of f., 10. Let g {(1, 2), (2, 5), (3, 8), (4, 10), (5, 12), (6, 12)} ., Is g a function? If yes, find its domain and range., If no, give reason., 11. Let f {(0, 5), (1, 2), (2, 1), (3, 4), (4, 7)} be a linear function from Z into Z., Write an expression for f., Hint, , Let f (x) ax b.
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Functions, , 85, , { f (5) f (1)}, ·, ( 5 1), { f (1.1) f (1)}, 13. If f (x) x 2, find the value of, ·, (1.1) 1, 12. If f (x) x 3, find the value of, , 14. Let A {12, 13, 14, 15, 16, 17} and f : A " Z : f (x) highest prime factor of x., Find range ( f ) ., 15. Let R be the set of all positive real numbers., Let f : R " R : f (x) log e x., Find (i) range ( f ), (ii) {x : xd R and f (x) 2} ., (iii) Find out whether f (xy) f (x) f (y) for all x, y d R ., 16. Let f : R " R : f (x) 2 x ., Find (i) range ( f ), (ii) {x : f (x) 1}., (iii) Find out whether f (x y) f (x) · f (y) for all x, y d R., 17. Let f : R " R : f (x) x 2 and g : C " C : g (x) x 2, where C is the set of all, complex numbers., Show that f ! g., 18. f, g and h are three functions defined from R to R as follows:, (i) f (x) x 2, (ii) g (x) x 2 1, (iii) h (x) sin x, Then, find the range of each function., 19. Let f : R " R : f (x) x 2 1. Find (i) f 1 {10} (ii) f 1 { 3} ., 20. The function F (x) , units. Find, (i) F (0),, , 9x , 32 is the formula to convert x C to Fahrenheit, 5, , (ii) F (10),, , (iii) the value of x when f (x) 212., , Interpret the result in each case., Hints, , (i) f (0) ', , 9#0, 321 32 & 0%C 32%F., 5, 9 # ( 10), 322 14 & (10)%C 14%F., (ii) F ( 10) (, 5, 9x, 32 212 & x 100 & 212%F 100%C., (iii) F (x) 212 &, 5, ANSWERS (EXERCISE 3A), , 4. (i) Yes (ii) No, 5. f {(1, 0), (0, 1), (3, 28), (7, 344), (9, 730)}, 6. (i) No, since one element, namely –1, has two different images., (ii) No, since dom (g) ! A., (iii) Yes, since each element in A has a unique image in B., 7. f {(1, 4), (1, 6), (2, 6)}, 8. f {(0, 3), (1, 5), (2, 7)}, dom ( f ) {0, 1, 2}, range ( f ) {3, 5, 7}, 9. f {(2, 3), (3, 5), (5, 9), (7, 13)}, dom ( f ) {2, 3, 5, 7}, range ( f ) {3, 5, 9, 13}
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86, , Senior Secondary School Mathematics for Class 11, , 10., 11., 15., 18., 19., , Yes, dom ( f ) {1, 2, 3, 4, 5, 6} and range ( f ) {2, 5, 8, 10, 12}, f (x) 3x 5 12. 31 13. 2.1 14. range ( f ) {3, 13, 7, 5, 2, 17}, (i) R (ii) {e 2} (iii) Yes, 16. (i) R (ii) {0} (iii) Yes, , (i) R {xd R : x $ 0} (ii) {xd R : x $ 1} (iii) {xd R : 1 # x # 1}, (i) {–3, 3} (ii) , 20. (i) 32 (ii) 14 (iii) x 100, HINTS TO SOME SELECTED QUESTIONS, , 4. (i) Since f 1 X #Y, so f is a relation from X to Y., (ii) Dom ( f ) X and range ( f ) {1, 5, 9, 11} 1 Y., But, two different ordered pairs namely (2, 9) and (2, 11) have the same first, coordinate. So, f is not a function., 6. (i) No, since two different ordered pairs (–1, 2) and (–1, 3) have the same first, coordinate., (ii) No, since dom (g) {0, 1, 2} ! A., (iii) Yes, dom (h) A and no two different ordered pairs have the same first coordinate., 11. f (0) 5 & a # 0 b 5 & b 5., f (1) 2 & a #1 b 2 & a 5 2 & a 3., 12., , {f (5) f (1)} {(5) 3 1 3} (125 1) 124, , , , 31., 4, 4, 4, (5 1), , 13., , { f (1.1) f (1)} {(1.1) 2 1 2}, , (1.1 1) 2.1., (1.1 1), (1.1 1), , 14. Highest prime factors of 12, 13, 14, 15, 16, 17 are 3, 13, 7, 5, 2, 17 respectively., 15. (i) For every xd R , we have log e x R. So, range ( f ) R., (ii) f (x) 2 & log e x 2 & x e 2 . So, {x : xd R and f (x) 2} {e 2} ., (iii) f (xy) log e (xy) log e x log e y f (x) f (y) ., 16. (i) f (x) 2 x 0 for every xd R., If xd R , there exists log 2 x such that f (log 2 x) 2 log2 x x., , range ( f ) R ., (ii) f (x) 1 & 2 x 1 20 & x 0. So, {x : f (x) 1} {0} ., (iii) f (x y) 2 x y 2 x # 2 y f (x) $ f (y) ., 17. Dom ( f ) R and dom (g) C & dom ( f ) ! dom (g), 18. (i) Clearly, f (x) x 2 $ 0 for all xd R., (ii) f (x) x 2 1 $ 1 for all xd R., (iii) sin xd[1, 1] ., 19. (i) x 2 1 10 & x 2 9 & x !3., (ii) x 2 1 3 & x 2 4 and there is no real number whose square is –4., , REAL FUNCTIONS, A function which has either R or one of its subsets as its, range, is called a real-valued function., , REAL-VALUED FUNCTIONS
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Functions, , 87, , A function f : X " Y is called a real function, if x 3 R and y 3 R., In general, real functions are described as general expressions or formulae,, without mentioning their domains and co-domains. Following are some, examples of real functions., REAL FUNCTIONS, , SOLVED EXAMPLES, EXAMPLE 1, SOLUTION, , If f (x) 3x 5x 10, find f (x 1) ., We have, f (x) 3x 3 5x 2 10., Replacing x by (x 1) in (i), we get, f (x 1) 3 (x 1) 3 5 (x 1) 2 10, 3 {x 3 1 3x (x 1)} 5 (x 2 2x 1) 10, 3 (x 3 3x 2 3x 1) 5 (x 2 2x 1) 10, 3x 3 14x 2 19x 2., , f (x 1) 3x 3 14x 2 19x 2., 3, , 2, , EXAMPLE 2, , 1, 1, If f (x) x x , show that { f (x)} 3 f (x 3) 3f b x l ·, , SOLUTION, , 1, We have, f (x) x x ·, On cubing both sides of (i), we get, 1, 1, 1, { f (x)} 3 x 3 3 3 # x # x #bx x l, x, dx 3 13 n 3 b1 xl, x, x, SRS, WV, 1 Z]1 1 _b 1, f (x 3) 3f b1 l, SSa f b x l ]][ x 1 bb` b x xlWWW ·, x, ]], b, SS, WW, ], x bb, \, a, T, X, 1, Hence, {f (x)} 3 f (x 3) 3f b x l ·, , EXAMPLE 3, SOLUTION, , EXAMPLE 4, , 1, x1,, x ! 1 then show that f { f (x)} x , where x ! 0., x1, x1, We have, f (x) , where x ! 1., x 1, x1 , ', 11, 1, x1, x, , f { f (x)} f b l , x 1, x1 , ', 11, x1, {(x 1) (x 1)}, ( x 1), , , , 2 1·, #, x, , , ( x 1), {(x 1) (x 1)} 2x, If f (x) , , 1, Hence, f { f (x)} x , where x ! 0., ax b, If y f (x) , and a 2 ! b 2 then prove that x f (y) ., bx a, , … (i), , … (i)
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88, SOLUTION, , Senior Secondary School Mathematics for Class 11, , We have, y f (x) , , , , ax b ·, bx a, , f (y) f { f (x)} f b, , , , ax b , l, ax a, , )a d, , ax b , n b3, bx a, , )b d, , ax b , n a3, bx a, , {(a 2 x ab) (b 2 x ab)}, (bx a), #, , (bx a), {(abx b 2) (abx a 2)}, , (a 2 x b 2 x) (a 2 b 2) x, 2 2 x., (a 2 b 2), (a b ), Hence, x f (y) ., , , EXAMPLE 5, SOLUTION, , EXAMPLE 6, , SOLUTION, , EXAMPLE 7, SOLUTION, , If f (x) , , 3f (x) 1, x1, ·, then prove that f (2x) , x1, f (x) 3, , x1, ·, x1, x1, 3f (x) 1 3 b x 1 l 1, , , ( x 1), f (x) 3, 3, ( x 1), (3x 3) (x 1), (x 1), , , 2x 1 f (2x) ., #, , , (x 1), (x 1) (3x 3) 2x 1, 3f ( x ) 1, ·, Hence, f (2x) , f (x) 3), We have, f (x) , , Let the functions f and g be defined by, ]Z] x 2 9, ]], , when x ! 3, f (x) (x 3) and g (x) [] x 3, ]], when x 3., ]k,, \, Find the value of k such that f (x) g (x) for all xd R., We have, f (x) g (x) for all xd R., , f ( 3) g ( 3), &, [a f (3) (3 3) and g (3) k], (3 3) k, &, k 6., Hence, k 6., 1, 1, If af (x) bf b x l x , where a ! b and x ! 0, find f (x) ., 1, 1, We have, af (x) bf b x l x ·, 1, Replacing x by x in (i), we get, 1, af b x l bf (x) x., , … (i), , … (ii)
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Functions, , 89, , Adding (i) and (ii), we get, 1, 1, 1, a ( f (x) f b x l2 b ( f (x) f b x l2 bx x l, &, , 1, 1, (a b) ( f (x) f b x l2 bx x l, , 1, bx 1x l ·, (a b), On subtracting (ii) from (i), we get, , &, , 1, f (x) f b x l , , … (iii), , 1, 1, 1, a ( f (x) f b x l2 b ( f (x) f b x l2 b x xl, &, , 1, 1, (a b) ( f (x) f b x l2 b x xl, , &, , ( f (x) f b1 l2 1 b1 xl ·, x, (a b) x, , … (iv), , On adding (iii) and (iv), we get, x x, 1, 1, , 2f (x) (, 2(, 2, (a b) (a b), (a b) x (a b) x, (a b) x (a b) x (a b) (a b), , , (a 2 b 2), (a 2 b 2) x, 2, , , 22bx 2 2d 2 2 2 a a bxk ·, (a b ) x, (a b ) x, Hence, f (x) , , 1, a a bxk ·, (a 2 b 2) x, , EXERCISE 3B, 1. If f (x) x 3x 4 and f (x) f (2x 1), find the values of x., 1, 1, x1, 1, ·, 2. If f (x) then show that (i) f b x l f (x) (ii) f b x l , x 1, f ( x), 1, 1, 3. If f (x) x 3 3 then show that f (x) f b x l 0., x, x1, 4. If f (x) then show that f { f (x)} x., x 1, 1, 2x 1 ,, 1, 5. If f (x) , and x !, then prove that f { f (x)} , when it is, 2, 2x 3, (2x 1), 3, ·, given that x !, 2, 1, 6. If f (x) , then show that f [ f { f (x)}] x., (1 x), 2x, 7. If f (x) , then show that f (tan ) sin 2., (1 x 2), 3x 1 ,, 8. If y f (x) , prove that x f (y) ., 5x 3, 2
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90, , Senior Secondary School Mathematics for Class 11, ANSWERS (EXERCISE 3B), , 2, 1. x 1 or x , 3, HINTS TO SOME SELECTED QUESTIONS, 1. f (x) f (2x 1) & x 2 3x 4 (2x 1) 2 3 (2x 1) 4, & x 2 3x 4 4x 2 2x 2 & 3x 2 x 2 0 & (x 1) (3x 2) 0., , 1, 1, 2x 1 ·, 5. f (x) , & f { f (x)} , 2x 3, (2x 1), 1, , 12, (2 #, (2x 1), , , 1, 1, 1 x x 1 [replace x by 1 ], 6. f (x) , & f { f (x)} , x, x, (1 x), (1 x), 1, (1 , 2, (1 x), & f [ f { f (x)}] , , 1 , 12, (, (1 x), (1 x), x #, x., 1, 1, (1 x), (1 x), , 2 tan 2 tan 2 sin , # cos 2 2 sin · cos sin 2., cos , (1 tan 2 ) sec 2 , 3x 1, 3 y 1 3 c 5x 3 m 1, , , , 9x 3 5x 3 14x x., 8. f (y), 5y 3, 15x 5 15x 9 14, 3x 1 , 5c m 3, 5x 3, 7. f (tan ) , , PROBLEMS BASED ON DOMAINS AND RANGES OF REAL FUNCTIONS, SOLVED EXAMPLES, EXAMPLE 1, , Find the domain and the range of the real function, f (x) , , SOLUTION, , We have, f (x) , , 1 ·, ( x 3), , 1 ·, x3, , Clearly, f (x) is defined for all real values of x except that at which, x 3 0, i.e., x 3., , dom ( f ) R {3} ., Let y f (x) . Then,, 1, 1, y & x3 y, x 3, 1, & x d y 3n ·, , … (i), , It is clear from (i) that x assumes real values for all real values of y, except y 0., , range ( f ) R {0} ., Hence, dom ( f ) R {3} and range ( f ) R {0} .
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Functions, EXAMPLE 2, , Find the domain and the range of the real function, f (x) , , SOLUTION, , We have, f (x) , , 91, , x3, ·, x5, , x3, ·, x5, , Clearly, f (x) is defined for all real values of x except that at which, x 5 0, i.e., x 5., , dom ( f ) R {5} ., Let y f (x) . Then,, x3, y & xy 5y x 3, x 5, 5y 3, … (i), & x (y 1) 5y 3 & x ·, y 1, It is clear from (i) that x is not defined when y 1 0, i.e., when, y 1., , range ( f ) R {1} ., Hence, dom ( f ) R {5} and range ( f ) R {1} ., EXAMPLE 3, SOLUTION, , Find the domain and the range of the real function, f (x) , , x2 1 ·, x2 1, , x2 1 ·, x2 1, Clearly, f (x) is defined for all real values of x except those for which, x 2 1 0, i.e., x !1., , dom ( f ) R {1, 1} ., Let y f (x) . Then,, We have, f (x) , , y, , x2 1, & x 2 y y x 2 1 & x 2 (y 1) (y 1), x2 1, y1, y1, ·, & x2 & x !, y 1, y1, , … (i), , It is clear from (i) that x is not defined when y 1 0 or when, y1, 0., y1, Now, y 1 0 & y 1., … (ii), y1, And 0 & (y 1 0 and y 1 0) or (y 1 0 and y 1 0), y 1, & (y 1 and y 1) or (y 1 and y 1), … (iii), & 1 y 1., [a y 1 and y 1 is not possible], Thus, x is not defined when 1 y # 1. [using (ii) and (iii)], , range ( f ) R (1, 1] ., Hence, dom ( f ) R {1, 1} and range ( f ) R (1, 1] .
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92, EXAMPLE 4, , SOLUTION, , Senior Secondary School Mathematics for Class 11, , Find the domain of the real-valued function:, x2 x 1 ·, f (x) 2, x 5x 4, 2, x1, x, ·, We have, f (x) 2, x 5x 4, Clearly, f (x) is defined for all real values of x except those at which, x 2 5x 4 0., But, x 2 5x 4 0 & (x 1)(x 4) 0 & x 1 or x 4., , dom ( f ) R {1, 4} ., , EXAMPLE 5, , Find the domain of each of the following real functions:, (i) f (x) x 3, , SOLUTION, , (ii) g (x) 4 x 2, , (iii) h(x) , , 1 ·, 1x, , (i) We have, f (x) x 3 ., Clearly, f (x) is defined for all real values of x for which, x 3 $ 0, i.e., x $ 3., , dom ( f ) [3, 3) ., (ii) We have, g (x) 4 x 2 ., Clearly, g (x) is defined for all real values of x for which, (4 x 2) $ 0., But, (4 x 2) $ 0 & (4 x 2) # 0, & x 2 4 # 0 & (x 2) (x 2) # 0, & 2 # x # 2 & xd[2, 2] ., , dom (g) [2, 2] ., 1 ·, 1x, Clearly, h(x) is defined for all real values of x for which, (1 x) 0., But, 1 x 0 & 1 x & x 1 & xd(3, 1) ., , dom (h) (3, 1) ., , (iii) We have, h(x) , , EXAMPLE 6, SOLUTION, , Find the domain of the function, f (x) 3 x , , 1, ·, x2 1, , 1, ·, x2 1, Clearly, f (x) is defined for all real values of x for which, We have, f (x) 3 x , , &, &, &, , 3 x $ 0 and x 2 1 0, x 3 # 0 and (x 1)(x 1) 0, x # 3 and (x 1 or x 1), (x # 3 and x 1) or (x # 3 and x 1)
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Functions, , &, &, , EXAMPLE 7, , SOLUTION, , 93, , (x 1) or (1 x # 3), xd(3, 1),(1, 3]., dom ( f ) (3, 1),(1, 3] ., , Find the range of each of the following functions:, (ii) g (x) x 2 2, xd R., (i) f (x) 2 3x, xd R and x 0, (i) We have, f (x) 2 3x, where xd R and x 0., Now, x 0 & 3x 0 & 3x 0, & 3x 2 0 2 & 2 3x 2, & f (x) 2 & f (x)d(3, 2) ., Hence, range ( f ) (3, 2) ., (ii) We have, g (x) x 2 2, xd R., Now, xd R & x 2 $ 0 & x 2 2 $ 0 2, & x 2 2 $ 2 & g (x) $ 2, & g (x)d[2, 3) ., Hence, range (g) [2, 3) ., , EXAMPLE 8, SOLUTION, , x2, Find the domain and the range of the function, f (x) ·, x 3, 2, x, We have, f (x) ·, x 3, Clearly, f (x) is defined for all real values of x for which x 3 ! 0,, i.e., x ! 3., , dom ( f ) R {3} ., Let y f (x) . Then,, x2, y & xy 3y x 2, x 3, & x ( y 1 ) 3y 2, 3y 2, … (i), & x ·, y 1, It follows from (i) that x assumes real values for all y except that for, which y 1 0, i.e., y 1., , range ( f ) R {1} ., Hence, dom ( f ) R {3} and range ( f ) R {1} ., , EXAMPLE 9, SOLUTION, , Find the domain and the range of the real function, f (x) x 3 ., We have, f (x) x 3 ., Clearly, f (x) is defined for all real values of x for which x 3 $ 0,, i.e., x $ 3., , dom ( f ) [3, 3) ., Also, x $ 3 & f (x) x 3 $ 0., , range ( f ) [0, 3) ., Hence, dom ( f ) [3, 3) and range ( f ) [0, 3) .
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94, EXAMPLE 10, , SOLUTION, , Senior Secondary School Mathematics for Class 11, , Find the domain and the range of each of the functions given below., (i) f (x) |x 1|, (ii) g (x) |x|, (i) We have, f (x) |x 1|., Clearly, f (x) is defined for all xd R. So, dom ( f ) R., For all xd R, we have, |x 1|$ 0 & f (x) $ 0., , range ( f ) [0, 3) ., Hence, dom ( f ) R and range ( f ) [0, 3) ., (ii) We have, g (x) |x|., Clearly, g (x) is defined for all xd R. So, dom (g) R., For all xd R, we have, |x|$ 0 & |x|# 0 & g (x) # 0., , range (g) (3, 0] ., Hence, dom (g) R and range (g) (3, 0] ., , EXAMPLE 11, , Find the domain and the range of the real function, f (x) , , SOLUTION, , We have, f (x) , , 1 ·, x2, , 1 ·, x2, , Clearly, f (x) is defined for all real values of x for which x 2 0,, i.e., x 2., , dom ( f ) (2, 3) ., For any real value of x 2, we have, x 2 & x2 0 & x2 0, 1, &, 0 & f (x) 0., x2, , range ( f ) (0, 3) ., Hence, dom ( f ) (2, 3) and range ( f ) (0, 3) ., EXAMPLE 12, , Find the domain and the range of the real function, f (x) 9 x 2 ., , SOLUTION, , We have, f (x) 9 x 2 ., Clearly, f (x) is defined for all real values of x for which (9 x 2) $ 0., And, 9 x 2 $ 0 & x 2 9 # 0 & (x 3)(x 3) # 0, & 3 # x # 3 & xd[3, 3] ., , dom ( f ) [3, 3] ., Let y f (x) . Then,, y 9 x2 & y2 9 x2 & x2 9 y2 & x 9 y2 ., Clearly, x will take real values only when 9 y 2 $ 0., Now, 9 y 2 $ 0 & y 2 9 # 0 & (y 3)(y 3) # 0, & 3 # y # 3 & y d[3, 3]
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Functions, , 95, , & y d[0, 3] {a y 9 x 2 $ 0 for all xd[3, 3]}., , range ( f ) [0, 3] ., Hence, dom ( f ) [3, 3] and range ( f ) [0, 3] ., EXAMPLE 13, , Find the domain and the range of the real function, f (x) , , SOLUTION, , We have, f (x) , , x, ·, (1 x 2), , x, ·, (1 x 2), , Clearly, f (x) is defined for all xd R. So, dom ( f ) R., Let y f (x) . Then,, x, y, & x2 y x y 0, (1 x 2), & x, , 1 ! 1 4y 2, ·, 2y, , … (i), , It is clear from (i) that x will take real values, when, (1 4y 2) $ 0 and y ! 0, &, (4y 2 1) # 0 and y ! 0, &, (2y 1)(2y 1) # 0 and y ! 0, 1, 1, &, by lby l # 0 and y ! 0, 2, 2, 1, 1, # y # and y ! 0, &, 2, 2, 1 1, &, y d; , E {0} ., 2 2, Also, x 0 & y 0., , , 1 1, range ( f ) ; , E ·, 2 2, , 1 1, Hence, dom ( f ) R and range ( f ) ; , E ·, 2 2, EXAMPLE 14, , Find the domain and the range of the real function, f (x) , , SOLUTION, , We have, f (x) , , 3, ·, (2 x 2), , 3, ·, (2 x 2), , Clearly, f (x) is defined for all real values of x except those for which, 2 x 2 0, i.e., x ! 2 ., , dom ( f ) R { 2 , 2 } ., Let y f (x) . Then,, 3, y, & 2 y x 2 y 3 & x 2 y 2y 3, (2 x 2), 2y 3, 2y 3, ·, … (i), & x2 y, & x !, y
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96, , Senior Secondary School Mathematics for Class 11, , It is clear from (i) that x will take real values only when, , 2y 3, y $ 0., , 2y 3, , , y $ 0 + (2y 3 # 0 and y 0) or (2y 3 $ 0 and y 0), 3, 3, + by # and y 0l or by $ and y 0l, 2, 2, 3, + (y 0) or by $ l, 2, 3, + y d(3, 0) or y d; , 3l, 2, 3, + y d(3, 0),; , 3l ·, 2, 3, , range ( f ) (3, 0),; , 3l ·, 2, 3, Hence, dom ( f ) R { 2 , 2 } and range ( f ) (3, 0),; , 3l ·, 2, Z], _b, ], b, x2, Find the domain and the range of the real function f ][ex, 2, o : xd Rb`, ] x 1, b, from R into R., \, a, Now,, , EXAMPLE 15, , SOLUTION, , x2 ,, xd R., x 1, Clearly, x 2 1 ! 0 for any real value of x., , f (x) is defined for all real values of x., , dom ( f ) R., Let y f (x) . Then,, We have, f (x) , , y, , 2, , x2, & x 2 y y x 2 & x 2 ( 1 y) y, (x 2 1), y, y, ·, & x2 & x !, , 1 y, ( 1 y), , It is clear from (i) that x will take real values only when, Now,, , … (i), y, $ 0., , ( 1 y), , y, $ 0 + (y # 0 and 1 y 0) or (y $ 0 and 1 y 0), , ( 1 y), , + (y # 0 and y 1) or (y $ 0 and y 1), + (y $ 0 and y 1), [a y # 0 and y 1 is not possible], + y d[0, 1) ., , range ( f ) [0, 1) ., Hence, dom ( f ) R and range ( f ) [0, 1) ., EXAMPLE 16, , Find the domain and the range of the function, 1, f )dx,, n : xd R and x !!13 ·, 1 x2
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Functions, SOLUTION, , We have, f (x) , , 97, , 1 ,, xd R., (1 x 2), , Clearly, f (x) is defined for all real values of x for which (1 x 2) ! 0., Now, (1 x 2) 0 & (1 x)(1 x) 0 & x 1 or x 1., Thus, f (x) is defined for all values of xd R except !1., , dom ( f ) R {1, 1} ., Let y f (x) . Then,, 1, y, & y x2 y 1 & x2 y y 1, 1 x2, y1, y1, ·, & x2 y & x !, y, , … (i), , It is clear from (i) that x will take real values only when, , y1, y $ 0., , y1, , , y $ 0 + (y 1 # 0 and y 0) or (y 1 $ 0 and y 0), + (y # 1 and y 0) or (y $ 1 and y 0), + (y 0) or (y $ 1), + y d(3, 0) or [1, 3) ., , range ( f ) (3, 0),[1, 3) ., Hence, dom ( f ) R {1, 1} and range ( f ) (3, 0),[1, 3) ., Now,, , EXAMPLE 17, , Find the domain and the range of the real function, f (x) , , SOLUTION, , We have, f (x) , , x 2 25, ·, x5, , x 2 25, ·, x5, , Clearly, f (x) is defined for all real values of x for which x 5 ! 0,, i.e., x ! 5., , dom ( f ) R {5} ., Let y f (x) . Then,, y, , x 2 25, & y x 5, when x 5 ! 0, x5, & y x 5, when x ! 5, & y ! 5 5 & y ! 10., , Then, y can be assigned any real value except 10., , range ( f ) R {10} ., Hence, dom ( f ) R {5} and range ( f ) R {10} ., EXAMPLE 18, , Find the domain and the range of the real function, f (x) , , SOLUTION, , We have, f (x) , , 3x, ·, x3, , 3x, ·, x3
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98, , Senior Secondary School Mathematics for Class 11, , Clearly, f (x) is defined for all real values of x for which x 3 ! 0,, i.e., x ! 3., , dom ( f ) R {3} ., Let y f (x) . Then,, 3x, y & y 1, when x 3 ! 0, x 3, & y 1, when x ! 3., , range ( f ) {1} ., Hence, dom ( f ) R {3} and range ( f ) {1} ., EXAMPLE 19, , Find the domain and the range of the real function, f (x) , , SOLUTION, , We have, f (x) , , |x 3|, ·, ( x 3), , |x 3|, ·, ( x 3), , Clearly, f (x) is defined for all real values of x for which x 3 ! 0,, i.e., x ! 3., , dom ( f ) R {3} ., Now, when x ! 3, we have, ]Z, , {a |x 3| (x 3)}, f (x) ][1, when x 3 0, ]], 1, when x 3 0, {a |x 3| (x 3)}, \, , range ( f ) {1, 1} ., Hence, dom ( f ) R {3} and range ( f ) {1, 1} ., EXAMPLE 20, , Find the domain of the real function, f (x) , , SOLUTION, , We have, f (x) , Now,, &, &, &, &, , , EXAMPLE 21, , 1, ·, , x |x|, , 1, ·, x |x|, , ]Z, |x| ][x, when x $ 0, ]] x, when x 0, \, ]Z , x |x| ][x x, when x $ 0, ]] , x x, when x 0, \Z, ]]2x, when x $ 0, x |x| [, ]]0, when x 0, \, x |x| 0, when x 0, 1, assumes real values only when x |x| 0 and, f (x) , x |x|, this happens only when x 0., dom ( f ) (0, 3) ., , 1, is not defined for any xd R. How will you, x |x|, define dom ( f ) and range ( f ) ?, Show that f (x)
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Functions, , We have, f (x) , , SOLUTION, , Now,, &, &, &, &, , , 99, , 1, ·, , x |x|, , Z], |x| ][x, when x $ 0, ]] x, when x 0, \Z, ] , x |x| ][x x, when x $ 0, ]] , x x, when x 0, Z]\, 0, ], x |x| [ , when x $ 0, ]]2x, when x 0, \, x |x|# 0 for all xd R, 1, is not defined for any xd R., x |x|, dom ( f ) and range ( f ) ., , GREATEST INTEGER FUNCTION (STEP FUNCTION), , The function f : R " R : f (x) [x], where [x] denotes the greatest integer less than or, equal to x, is called the Greatest Integer Function., EXAMPLES, NOTE, , [3.65] = 3, [3.02] = 3, [5] = 5, [–2.6] = –3, [–4] = –4., , x [x] 0 for xdZ and 0 x [x] 1 for xd R Z., , EXAMPLE 22, , Find the domain and the range of the real function, f (x) , , SOLUTION, , We have, f (x) , , 1, ·, x [x], , 1, ·, x [x], , We know that, ]Z]x [x] 0 for all x 0, ]], []x [x] 0, when x 0, ]], ]x [x] 0 for all x 0., \, 1, , is defined only when x [x] 0 and this, f (x) , x [x], happens only when x 0., , dom ( f ) (0, 3) ., Let y f (x) . Then,, 1, 1, … (i), y, & x [x] y ·, x [x], 1, Now, x 0 & x [x] 0 & x [x] 0 & y 0 & y 0., , range ( f ) (0, 3) ., Hence, dom ( f ) (0, 3) and range ( f ) (0, 3) ., EXAMPLE 23, , Find the domain and the range of the real function, f (x) , , 1, ·, x [x]
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100, , Senior Secondary School Mathematics for Class 11, , We have, f (x) , , SOLUTION, , 1, ·, , x [x], , We know that, 0 # x [x] 1 for all xd R, and x [x] 0 for all xdZ., , 0 x [x] 1 for all xd R Z, 1, exists for all xd R Z, &, f (x) , x [x], dom ( f ) R Z., &, Also, 0 x [x] 1 for all xd R Z, &, 0 x [x] 1 for all xd R Z, 1, &, 1, 3 for all xd R Z, x [x], &, 1 f (x) 3 for all xd R Z, range ( f ) (1, 3) ., &, Hence, dom ( f ) R Z and range ( f ) (1, 3) ., , EXERCISE 3C, 1. Find the domain of each of the following real functions:, 3x 5, 2x 3, (ii) f (x) 2, (i) f (x) 2, x 9, x x2, x3 8, x 2 2x 1, (iii) f (x) 2, (iv) f (x) 2, , , x 1, x 8x 12, Find the domain and the range of each of the following real functions:, , 1, 2. f (x) x, 5. f (x) , , 3x 2, x2, , ax b, cx d, 1, 11. f (x) , 2, x 1, 2, x 9, 14. f (x) , x 3, 8. f (x) , , 3. f (x) , , 1, ( x 5), , 6. f (x) , , x 2 16, x4, , 9. f (x) 3x 5, 12. f (x) 1 |x 2|, 15. f (x) , , 4. f (x) , , x3, 2x, , 1, 2x 3, x5, 10. f (x) , 3x, |x 4|, 13. f (x) , x 4, 7. f (x) , , 1, 2 sin 3x, ANSWERS (EXERCISE 3C), , 1. (i) R {3, 3}, (ii) R {1, 2}, (iii) R {2, 6}, 2. dom ( f ) R {0}, range ( f ) R {0}, 3. dom ( f ) R {5}, range ( f ) R {0}, , (iv) R {1, 1}
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Functions, , 101, , 4. dom ( f ) R {2}, range ( f ) R {1}, 5. dom ( f ) R { 2}, range ( f ) R {3}, 6. dom ( f ) R {4}, range ( f ) R {8}, 3, 7. dom ( f ) R b 3, E, range ( f ) R {0}, 2, a, d, 8. dom ( f ) R ' c 1 , range ( f ) R % c /, 5, 9. dom ( f ) ; , 3l, range ( f ) R [0, 3), 3, 10. dom ( f ) (3, 5], range ( f ) [0, 3), 11. dom ( f ) (3, 1),(1, 3), range ( f ) R {0}, 12. dom ( f ) R, range ( f ) (3, 1], 13. dom ( f ) R {4}, range ( f ) {1, 1}, 14. dom ( f ) R {3}, range ( f ) R {6}, 1, 15. dom ( f ) R, range ( f ) ; , 1E, 3, HINTS TO SOME SELECTED QUESTIONS, 1, 2. f (x) x is not defined when x 0. So, dom ( f ) R {0} ., 1, 1, Let y f (x) . Then, y x & x y ·, , , x is not defined when y 0. So, range ( f ) R {0} ., , 1, is not defined when x 5. So, dom ( f ) R {5} ., (x 5), 1, 1, 1, Let y f (x) . Then, y , & x 5 y & x d y 5n ·, (x 5), x is not defined when y 0. So, range ( f ) R {0} ., , 3. f (x) , , x3, is not defined when x 2. So, dom ( f ) R {2} ., 2x, Let y f (x) . Then,, x3, y & 2y xy x 3 & xy x 2y 3, 2 x, 2y 3, & x (y 1) 2y 3 & x ·, y 1, x is not defined when y 1. So, range ( f ) R {1} ., , 4. f (x) , , 3x 2, is not defined when x 2. So, dom ( f ) R {2} ., x2, Let y f (x) . Then,, 3x 2, y , & xy 2y 3x 2 & 3x xy 2y 2, x 2, 2y 2, & x (3 y) 2y 2 & x ·, 3 y, It follows from (i) that x is not defined when y 3., range ( f ) R {3} ., , 5. f (x) , , … (i)
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102, , Senior Secondary School Mathematics for Class 11, x 2 16, is not defined when x 4. So, dom ( f ) R {4} ., x4, Let y f (x) . Then,, , 6. f (x) , , x 2 16 , (x 4), when x ! 4, x4, y ! (4 4) & y ! 8. So, range ( f ) R {8} ., y, , &, , 3, 1, is not defined when 2x 3 # 0, i.e., when x # ·, 2, 2x 3, 3, dom ( f ) R c3, E ·, 2, Let y f (x) . Then,, , 7. f (x) , , 1, 1, 1, 1 1, & y2 , & (2x 3) 2 & x e 2 3o ·, 2 y, (2x 3), 2x 3, y, x is not defined when y 0. So, range ( f ) R {0} ., y, , , , ax b, d, is not defined when cx d 0 & x c ·, cx d, d, dom ( f ) R ' c 1 ·, Let y f (x) . Then,, ax b, y, & cxy dy ax b & cxy ax dy b, cx d, dy b, ·, & x (cy a) dy b & x , cy a, a, Clearly, x is not defined when cy a 0 & y c ·, a, range ( f ) R % c / ·, , 8. f (x) , , 9. f (x) 3x 5 is defined only when 3x 5 $ 0, i.e., when 3x 5 $ 0, i.e., when x $, 5, dom ( f ) ; , 3m ·, 3, 5, x $ & f (x) $ 0 & range ( f ) R [0, 3) ., 3, , , , x5, x5, is defined only when $ 0., 3x, 3 x, x5, Now, 0 & x 5 0 & x 5., 3 x, x5, And, 0 & (x 5 0 and 3 x 0) or (x 5 0 and 3 x 0), 3 x, & (x 5 and 3 x) or (x 5 and 3 x), & (3 x and x 5) or (x 5 and x 3), & 3 x 5 [a x 5 and x 3 cannot hold]., Thus, f (x) is defined when 3 x # 5., dom ( f ) (3, 5] ., x5, Clearly, f (x) , $ 0., 3x, Let y f (x) . Then,, , 10. f (x) , , 5·, 3
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Functions, , 103, , x5, x5, & y 2 & 3y 2 xy 2 x 5, 3x, 3 x, , y, , & x xy 2 3y 2 5 & x (1 y 2) (3y 2 5) & x , , 3y 2 5, 1 y2, , … (i), , ·, , It follows from (i) that x is defined for each real value of y., range ( f ) [0, 3), [a f (x) $ 0]., 11. f (x) , , 1, x2 1, , is defined only when x 2 1 0., , Now, x 2 1 0 & (x 1) (x 1) 0 & (x 1) or (x 1), & xd(3, 1) or xd(1, 3)., dom ( f ) (3, 1),(1, 3) ., Let y f (x) . Then,, y, , 1, x2 1, , & y2 , , 1, (x, , 2, , 1), , & x2 1 , , 1, & x, y2, , 1 , 1, y2, , Clearly, x is not defined when y 0., range ( f ) R {0} ., 12. f (x) 1 |x 2|is defined for all xd R. So, dom ( f ) R., Now, 0 #|x 2| 3 & |x 2|$ 0 and|x 2| 3, & |x 2|# 0 and |x 2| 3, & 1 |x 2|# 1 and 1 |x 2| 3, & 3 1 |x 2|# 1 & f (x)d(3, 1] ., range ( f ) (3, 1] ., |x 4|, is not defined at x 4. So, dom ( f ) R {4} ., x4, Now, when x ! 4, we have, Z], {a |x 4| x 4}, ]1, when x 4 0, f (x) ][, ]1, when x 4 0, {a |x 4| (x 4)} ., \, range ( f ) [1, 1] ., , 13. f (x) , , x2 9, is defined for all real values of x for which x ! 3., x3, dom ( f ) R {3} ., Let y f (x) . Then,, , 14. f (x) , , x2 9, & y x 3, when x ! 3, x3, & y ! 3 3 & y ! 6., range ( f ) R {6} ., 15. Clearly, 1 # sin 3x # 1 for all xd R, y, , &, &, &, , , 2 (1) # 2 ( sin 3x) # 2 1 for all xd R, 1 # 2 sin 3x # 3 for all xd R, 2 sin 3x ! 0 for any xd R., 1, is defined for all xd R., f (x) , (2 sin 3x), , 1 y2, y2, , ·
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104, , Senior Secondary School Mathematics for Class 11, We have, 2 sin 3x $ 1 and 2 sin 3x # 3, 1, 1, 1, 1, 1, , # 1 and , $ & # , # 1., 2 sin 3x, 2 sin 3x 3, 3 2 sin 3x, 1, range ( f ) ; , 1E ·, 3, , GRAPHS OF SOME STANDARD REAL FUNCTIONS, 1. IDENTITY FUNCTION, , The function f : R " R : f (x) x for all xd R is called an identity function on R., Dom ( f ) R and range ( f ) R., EXAMPLE 1, SOLUTION, , Draw the graph of the identity function f : R " R : f (x) x for all xd R., Here f (x) x for all xd R., We may prepare its table as under., x, , –2, , –1, , 0, , 1, , 2, , 3, , f (x) x, , –2, , –1, , 0, , 1, , 2, , 3, , On a graph paper, we draw XlOX and YOYl as the x-axis and the, y-axis respectively., We take 10 small divisions = 1 unit., Now, on this graph paper, we plot the points, A(2, 2), B(1, 1), O(0, 0), C(1, 1), D(2, 2) and E(3, 3) ., Join these points successively to obtain the straight line ABOCDE, as the required graph line, given below., Y, 3, , E, , 2, , D, , 1, , X¢, , –3, , –2, , –1, , B, , A, , C, , O, , 1, , 2, , –1, , –2, Y¢, , Graph of the identity function, f (x) x, , 3, , X
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Functions, , 105, , 2. CONSTANT FUNCTION, , Let c be a fixed real number., Then, the function f : R, , " R : f (x) c for all xdR is called the constant function., , For the constant function f (x) c for all xd R, we have, dom ( f ) R and range ( f ) {c} ., EXAMPLE 2, , SOLUTION, , Draw the graph of each of the following constant functions:, (i) f (x) 2 for all xd R, (ii) f (x) 0 for all xd R, (iii) f (x) 2 for all xd R, (i) Let f (x) 2 for all xd R., Then, dom ( f ) R and range ( f ) {2} ., We may prepare the table given below., x, , –2, , –1, , 0, , 1, , 2, , 3, , f (x) 2, , 2, , 2, , 2, , 2, , 2, , 2, , On a graph paper, we draw XlOX and YOYl as the x-axis and, the y-axis respectively., We take the scale: 10 small divisions = 1 unit., On this graph paper, we plot the points A(–2, 2), B(–1, 2),, C(0, 2), D(1, 2), E(2, 2) and F(3, 2). Join A, B, C, D, E and F, successively to obtain the required graph line ABCDEF, whose, equation is y 2., , Graph of the function, f (x) 2
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106, , Senior Secondary School Mathematics for Class 11, , (ii) Let f (x) 0 for all xd R. Then,, dom ( f ) R and range ( f ) {0} ., Now, we have:, x, , –2, , –1, , 0, , 1, , 2, , f (x) 0, , 0, , 0, , 0, , 0, , 0, , On a graph paper, we draw the horizontal line XlOX as the, x-axis and the vertical line YOYl as the y-axis., We take the scale: 10 small divisions = 1 unit., Now, on this graph paper, we plot the points A(–2, 0), B(–1, 0),, O(0, 0), C(1, 0) and D(2, 0) and join them successively to get the, graph line ABOCD, whose equation is y 0., , Graph of the function, f (x) 0, , (iii) Let f (x) 2 for all xd R. Then,, dom ( f ) R and range ( f ) 2., Now, we have:, x, , –2, , –1, , 0, , 1, , 2, , 3, , f (x) 2, , –2, , –2, , –2, , –2, , –2, , –2, , On a graph paper, we draw the horizontal line XlOX as the, x-axis and the vertical line YOYl as the y-axis., We take the scale: 10 small divisions = 1 unit., On this graph paper, we plot the points A(–2, –2), B(–1, –2),, C(0, –2), D(1, –2), E(2, –2) and F(3, –2) and join these points, successively to get the required graph line ABCDE, as shown, below. Its equation is y 2.
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Functions, , 107, , Graph of the function, f (x) 2, , 3. THE MODULUS FUNCTION, , ]Z, The function f, defined by f (x) |x| ][x, when x $ 0 is called the modulus function., ]]x, when x 0, \, We know that|x|$ 0 for all xd R., , , dom ( f ) R and range ( f ) [0, 3) ., , EXAMPLE 3, , SOLUTION, , Draw the graph of the modulus function, defined by, Z], f : R " R : f (x) |x| ][x, when x $ 0, ]]x, when x 0., \, We have, x, , –3, , –2, , –1, , 0, , 1, , 2, , 3, , 4, , f (x) |x|, , 3, , 2, , 1, , 0, , 1, , 2, , 3, , 4, , On a graph paper draw the horizontal line XlOX as the x-axis and, the vertical line YOYl as the y-axis., Take the scale: 10 small divisions = 1 unit., On this graph paper, plot the points A(–3, 3), B(–2, 2), C(–1, 1),, O(0, 0), D(1, 1), E(2, 2), F(3, 3) and G(4, 4)., Join these points successively to obtain the graph lines ABCO and, ODEFG, as shown below.
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108, , Senior Secondary School Mathematics for Class 11, , Graph of f (x) |x|, 4. THE GREATEST INTEGER FUNCTION, OR STEP FUNCTION, , For any real number x, the symbol [x] denotes the greatest integer less than or equal, to x., EXAMPLES (i) [6.85] = 6, (ii) [8] = 8, (iii) [0.536] = 0, (iv) [–5.8] = –6, (v) [–5] = –5, The function f : R " R : f (x) [x] for all xd R is called the greatest integer, function, or step function., Clearly, dom ( f ) R and range ( f ) Z., EXAMPLE 4, , SOLUTION, , Draw the graph of the greatest integer function:, f : R " R : f (x) [x] for all xd R., Using the definition of [x], we have, [x] 2 for 2 # x 1, [x] 1 for 1 # x 0, [x] 0 for 0 # x 1, [x] 1 for 1 # x 2, [x] 2 for 2 # x 3, [x] 3 for 3 # x 4., Taking the scale: 10 small divisions = 1 unit, the graph of the, function f (x) [x] may be drawn as shown below.
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Functions, , 109, , Graph of greatest integer function [x], 5. THE SMALLEST INTEGER FUNCTION, , For any real number x, the smallest integer greater than or equal to x is denoted by _xi ., For example: _3.8i 4, _6i 6, _0.76i 1, _ 6.2i 6, etc., The function f : R " R : f (x) _xi for all xd R is called the smallest integer, function., Clearly, dom ( f ) R, range ( f ) Z., EXAMPLE 5, SOLUTION, , Draw the graph of the smallest integer function f : R " R : f (x) _xi for, all xd R., Using the definition of _xi, we have, x, , f (x) _xi, , 3 x # 2, , –2, , 2 x # 1, , –1, , 1 x # 0, , 0, , 0 x #1, , 1, , 1 x#2, , 2, , 2 x#3, , 3, , SCALE: 10 small divisions = 1 unit.
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110, , Senior Secondary School Mathematics for Class 11, , Now, the graph may be drawn as shown below., , Graph of smallest integer function _xi, 6. THE FRACTIONAL PART FUNCTION, , Let f : R " R : f (x) x [x] {x} ., Here f (x) {x} denotes the fractional part of x., For example: {2.35} 0.35, {6} 0, {7} 0, { 0.65} 0.35, {3.75} 0.25, etc., Clearly, dom ( f ) R and range ( f ) [0, 1) ., EXAMPLE 6, , Draw the graph of the fractional part function:, f : R " R : f (x) x [x] {x}, , SOLUTION, , We have, f (x) x [x] {x} ., f (3) 0, f (3.1) 0.1, f (3.2) 0.2, …, f (3.9) 0.9, f (4) 0., f (2) 0, f (2.1) 0.1, f (2.2) 0.2, …, f (2.9) 0.9, f (3) 0., f (1) 0, f (1.1) 0.1, f (1.2) 0.2, …, f (1.9) 0.9, f (2) 0., f (0) 0, f (0.1) 0.1, f (0.2) 0.2, …, f (0.9) 0.9, f (1) 0., f (1) 0, f ( 0.9) 0.1, f ( 0.8) 0.2, …, f ( 0.1) 0.9, f (0) 0., f (2) 0, f (1.9) 0.1, f (1.8) 0.2, …, f (1.1) 0.9, f (1) 0., f (3) 0, f (2.9) 0.1, f (2.8) 0.2, …, f (2.1) 0.9, f (2) 0.
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Functions, , 111, , Plotting these points on a graph paper, we may get the graph as, shown below., , Graph of fractional part function, f (x) {x}, 7. THE SIGNUM FUNCTION, , Z] x, ]] , when x ! 0, is called the signum function., The function f : R " R : f (x) []|x|, ]]0, when x 0, \, The above function may be simplified as under:, Z], ]]1, if x 0, ], f (x) []0, if x 0, ]], ]1, if x 0., \, Clearly, dom ( f ) R and range ( f ) {1, 0, 1} ., EXAMPLE 7, , SOLUTION, , Draw the graph of the signum function, f : R " R, defined by, Z], ]]1, if x 0, ]Z] x, ], x, !, 0, ,, when, ], or f (x) ][0, if x 0, f (x) [|x|, ]], ]], ]0, when x 0, ]1, if x 0, \, \, Clearly, we have, x 0 & f (x) 1, x 0 & f ( x) 0, x 0 & f ( x) 1 .
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112, , Senior Secondary School Mathematics for Class 11, , We may now draw the graph as shown below., , Graph of signum function, , 8. POLYNOMIAL FUNCTION, , A function of the form f (x) a0 a1 x a2 x 2 … an x n, where n is a non-negative, integer and a0, a1, a2, …, an d R, is called a polynomial function., EXAMPLES, , (i) f (x) x 3 2x 2 5 is a polynomial function., (ii) g (x) 4x 2 x 2 is a polynomial function., (iii) h(x) 2x 3 x 5 is not a polynomial function., , EXAMPLE 8, , Let f : R " R : f (x) x 2 for all xd R., Find its domain and range., Also, draw its graph., , SOLUTION, , Here, f : R " R : f (x) x 2 for all xd R., Dom ( f ) R and range ( f ) {xd R : x $ 0} [0, 3) ., Now, we have:, x, f (x) x, , 2, , –2, , –1.5, , –1, , 0, , 1, , 1.5, , 2, , 4, , 2.25, , 1, , 0, , 1, , 2.25, , 4, , On a graph paper, we draw XlOX and YOYl as the x-axis and the, y-axis respectively., Now, we plot the points A(–2, 4), B(–1.5, 2.25), C(–1, 1), O(0, 0),, D(1, 1), E(1.5, 2.25) and F(2, 4)., Join these points freehand successively to obtain the required, curve.
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Functions, , Scale: 10 small divisions = 1 unit., , Graph of f (x) x 2, EXAMPLE 9, , Let f : R " R : f (x) x 3 for all xd R., Find its domain and range., Also, draw its graph., , SOLUTION, , Let f : R " R : f (x) x 3 for all xd R., Then,, dom ( f ) R and range ( f ) R., Now, we have:, x, , –2 –1.5 –1 0 1 1.5 2, , f (x) x 3 –8 –3.375 –1 0 1 3.375 8, On a graph paper, we draw XlOX, and YOYl as the x-axis and the y-axis, respectively., We take the scale as, 5 small divisions = 1 unit., Now, we plot the points A(–2, –8),, B(–1.5, –3.375), C(–1, –1), O(0, 0),, D(1, 1), E(1.5, 3.375) and F(2, 8)., We join these points freehand, successively to obtain the required, curve., , 113
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114, , Senior Secondary School Mathematics for Class 11, , 9. RATIONAL FUNCTION, , f (x), , where f (x) and g (x) are polynomial functions of, g ( x), x, defined in a domain, where g (x) ! 0., The functions of the type, , EXAMPLE 10, , SOLUTION, , Let f : (R {0}), , " R : f (x) 1x for all values of xdR {0} ., , Find its domain and range. Also, draw its graph., 1, Let f : (R {0})" R : f (x) x $ Then,, dom ( f ) R {0} and range ( f ) R {0} ., We have:, x, , –4, , –2, , –1, , –0.5 –0.25 0.25 0.5, , 1, , 2, , 4, , 1, –2, –4, 4, 2, 1 0.5 0.25, f (x) x –0.25 –0.5 –1, On a graph paper, let XlOX and YOYl be the x-axis and the y-axis, respectively., We take the scale: 10 small divisions = 1 unit., Now, we plot the points A(–4, –0.25), B(–2, –0.5), C(–1, –1),, D(–0.5, –2), E(–0.25, –4) and F(0.25, 4), G(0.5, 2), H(1, 1), J(2, 0.5) and, K(4, 0.25)., , 1, , Graph of the function, f (x) x
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Functions, , 115, , We join A, B, C, D and E freehand successively to obtain a curve., Also, we join F, G, H, J and K freehand successively to obtain, another curve, as shown above., , EXERCISE 3D, 1. Consider the real function f : R " R : f (x) x 5 for all xd R., Find its domain and range., Draw the graph of this function., 2. Consider the function f : R " R, defined by, Z] , ]]1 x, when x 0, ], f (x) ][1, when x 0, ]], ]x 1, when x 0., \, Write its domain and range., Also, draw the graph of f (x) ., 3. Find the domain and the range of the square root function,, f : R ,{0} " R : f (x) x for all non-negative real numbers., Also, draw its graph., 4. Find the domain and the range of the cube root function,, f : R " R : f (x) x1/3 for all xd R., Also, draw its graph., HINTS TO SOME SELECTED QUESTIONS, 1. Clearly, dom ( f ) R and range ( f ) R.
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116, , Senior Secondary School Mathematics for Class 11, , Now, y x 5., Putting x 0, we get y 5., Putting y 0, we get x 5., On the graph paper, plot the points A(–5, 0) and B(0, 5)., Join AB to get the required graph., 2. x 0 & y 1 x., x 1 & y 1., x 1 & y x 1., Y, A, , G, , x, , +, , y, , B, =, , F, , 1, , –x, , +, , y, , =, , 1, , E, , C, , D, , O, , X¢, , X, , Y¢, , (i) For x 0, we have y 1 x., x, y, , –3, 4, , –2, 3, , –1, 2, , (ii) For x 0, we have y 1., x, y, , 0, 1, , [a x 0 & y 1], , (iii) For x 0, we have y x 1., x, y, , 1, 2, , 2, 3, , 3, 4, , On a graph paper, draw horizontal line XlOX and the vertical line YOYl. Now, plot, the points A (3, 4), B (2, 3), C (1, 2), D (0, 1), E (1, 2), F (2, 3) and G (3, 4) ., Join ABCD and DEFG to get the required graph.
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Functions, , 117, , 3. Let f : R ,{0}" R : f (x) x . Then,, dom ( f ) [0, 3) and range ( f ) [0, 3) ., We have, 0 0, 1 1, 2.25 1.5, 4 2, 6.25 2.5 and 9 3., On a graph paper, we plot the points O(0, 0), A(1, 1), B(2.25, 1.5), C(4, 2), D(6.25, 2.5),, and E(9, 3). Join these points freehand successively to get the required curve., , Graph of the curve, f (x) , , x, , 4. Let f : R " R : f (x) x1/3 for all xd R., Clearly, dom ( f ) R and range ( f ) R., We have, 3, 8 2, 3 5 1.7, 3 3 1.4, 3 1 1, 3 0 0, 3 1 1, 3 3 1.4, 3 5 1.7,, and 3 8 2., On a graph paper, take the horizontal line XlOX as the x-axis and the vertical line, YOYl as the y-axis., Take the scale: 5 small divisions = 1 unit., Now, plot the points A(8, 2), B(5, 1.7), C(3, 1.4), D(1, 1), O(0, 0), E(–1, –1), F(–3, –1.4),, G(–5, –1.7) and H(–8, –2)., Join ABCDOEFGH freehand to get the required curve., Remember: 3 3 1.4 and 3 5 1.7.
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118, , Senior Secondary School Mathematics for Class 11, , OPERATIONS ON FUNCTIONS, (i) SUM OF TWO REAL FUNCTIONS, , Let f : D1 " R and g : D2 " R, where D1 3 R and D2 3 R. Then,, ( f g) : (D1 + D2)" R : ( f g) (x) f (x) g (x) for all xd(D1 + D2)., (ii) DIFFERENCE OF TWO REAL FUNCTIONS, , Let f : D1 " R and g : D2 " R, where D1 3 R and D2 3 R. Then,, ( f g) : (D1 + D2)" R : ( f g) (x) f (x) g (x) for all xd(D1 + D2)., (iii) MULTIPLICATION OF TWO REAL FUNCTIONS, , Let f : D1 " R and g : D2 " R, where D1 3 R and D2 3 R. Then,, ( fg) : (D1 + D2)" R : ( fg) (x) f (x) $ g (x) for all xd(D1 + D2)., NOTE, , This is known as pointwise multiplication of functions., , (iv) QUOTIENT OF TWO REAL FUNCTIONS, , Let f : D1 " R and g : D2 " R, where D1 3 R and D2 3 R., Let D (D1 + D2) {x : g (x) 0} . Then,, f, f, f ( x), e o : D " R : e o ( x) , for all xdD., g, g, g ( x), (v) RECIPROCAL OF A FUNCTION, , Let f : D1 " R and let D D1 {x : f (x) 0} ., Then, the reciprocal of f is the function, , 1,, defined by, f, , 1, 1, 1, for all xdD., : D " R : e o (x) , f, f, f (x), (vi) SCALAR MULTIPLE OF A FUNCTION, , Let f : D " R and let be a scalar (a real number). Then,, (f ) : D " R : (f ) (x) $ f (x) for all xdD., SOLVED EXAMPLES, EXAMPLE 1, , Let f : R " R : f (x) x 2 and g : R " R : g (x) 2x 1., Find (i) ( f g) (x), , SOLUTION, , (ii) ( f g) (x), , (iii) ( fg) (x), , f, (iv) e g o (x), , Here, dom ( f ) R and dom (g) R., , dom ( f )+ dom (g) (R + R) R., (i) ( f g) : R " R is given by, (f g) (x) f (x) g (x) x 2 (2x 1) (x 1) 2 ., (ii) ( f g) : R " R is given by, ( f g) (x) f (x) g (x) x 2 (2x 1) (x 2 2x 1) .
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Functions, , 119, , (iii) ( fg) : R " R is given by, ( fg) (x) f (x) · g (x) x 2 · (2x 1) (2x 3 x 2) ., (iv) {x : g (x) 0} {x : 2x 1 0} ', , 1, 1·, 2, , f, 1, 1, dom e g o R + R ' 1 R ' 1 ·, 2, 2, f, 1, The function g : R ' 1 " R is given by, 2, 2, f, f (x), , e o (x) , x , x! 1 ·, g, 2, g (x) 2x 1, , , EXAMPLE 2, , Let f (x) x and g (x) x be two functions defined over the set of nonnegative real numbers. Find:, f, (i) ( f g) (x), (ii) ( f g) (x), (iii) ( fg) (x), (iv) g (x), , SOLUTION, , Here f : [0, 3)" R : f (x) x and g : [0, 3)" R : g (x) x., , dom ( f ) [0, 3) and dom (g) [0, 3) ., So, dom ( f )+ dom (g) [0, 3)+[0, 3) [0, 3) ., (i) ( f g) : [0, 3)" R is given by, ( f g) (x) f (x) g (x) ( x x) ., (ii) ( f g) : [0, 3)" R is given by, ( f g) (x) f (x) g (x) ( x x) ., (iii) ( fg) : [0, 3)" R is given by, (fg) (x) f (x) · g (x) ( x # x) x 3/2 ., (iv) {x : g (x) 0} {0} ., f, , dom e g o dom ( f )+ dom (g) {x : g (x) 0}, [0, 3)+[0, 3) {0} (0, 3) ., f, So, g : (0, 3)" R is given by, f, f ( x), x, e o (x) , , 1 , x ! 0., g, x, g (x), x, , EXAMPLE 3, , SOLUTION, , 1, and g (x) (x 4) 3 ., ( x 4), f, 1, Find (i) ( f g) (x) (ii) ( f g) (x) (iii) ( fg) (x) (iv) e g o (x) (v) e o (x), f, 1, Clearly, f (x) , is defined for all real values of x except that at, ( x 4), which x 4 0, i.e., x 4., , dom ( f ) R { 4} ., Let f and g be real functions, defined by f (x)
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120, , Senior Secondary School Mathematics for Class 11, , And, g (x) (x 4) 3 is defined for all xd R. So, dom (g) R., , dom ( f )+ dom (g) R { 4}+ R R { 4} ., (i) ( f g) : R { 4} " R is given by, 4, 1 3 1 ( x 4) ·, ( f g ) ( x) f ( x) g ( x) , ( x 4), ( x 4), ( x 4), (ii) ( f g) : R { 4} " R is given by, 4, 1 3 1 ( x 4) ·, ( f g) (x) f (x) g (x) , ( x 4), ( x 4), ( x 4), (iii) ( fg) : R { 4} " R is given by, 1, # ( x 4) 3 ( x 4 ) 2 ., ( fg) (x) f (x) · g (x) , ( x 4), (iv) {x : g (x) 0} {x : (x 4) 3 0} {x : x 4 0} { 4} ., f, , dom e g o dom ( f )+ dom (g) {x : g (x) 0} R { 4} ., f, e o : R { 4} " R is given by, g, 1, f, f (x), ( x 4), 1 ,, e o (x) , , , x ! 4., g, g (x) (x 4) 3 (x 4) 4, (v) Clearly, f (x) ! 0 for any xd R { 4} ., , , 1, : R { 4} " R is given by, f, 1, 1 , e o (x) , f, f (x), , EXAMPLE 4, , Let f and g be real functions defined by f (x) x 1 and g (x) x 1 ., Find (i) ( f g) (x), , SOLUTION, , 1, ( x 4) ., 1, (x 4), , (ii) ( f g) (x), , (iii) ( fg) (x), , f, (iv) e g o (x)., , Clearly, f (x) x 1 is defined for all real values of x for which, x 1 $ 0, i.e., x $ 1. So, dom ( f ) [1, 3) ., Also, g (x) x 1 is defined for all real values of x for which, x 1 $ 0, i.e., x $ 1. So, dom (g) [1, 3) ., , dom ( f )+ dom (g) {1, 3)+[ 1, 3) [1, 3) ., (i) ( f g) : [1, 3)" R is given by, ( f g) (x) f (x) g (x) ( x 1 x 1 ) ., (ii) ( f g) : [1, 3)" R is given by, ( f g) (x) f (x) g (x) ( x 1 x 1 ) ., (iii) ( fg) : [1, 3)" R is given by, ( fg) (x) f (x) · g (x) x 1 # x 1 x 2 1 .
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Functions, , 121, , (iv) {x : g (x) 0} {x : x 1 0} {x : x 1 0} {1} ., , dom ( f )+ dom (g) {x : g (x) 0}, [1, 3)+[1, 3) {1} [1, 3) ., , , f, g "[1, 3)" R is given by, f, f (x), e o (x) , , g, g (x), , EXAMPLE 5, , Let f and g be real functions, defined by f (x) x 2 and g (x) 4 x 2 ., Find (i) ( f g) (x), (ii) ( f g) (x), (iii) ( fg) (x), (iv) ( ff ) (x), (v) (gg) (x), , SOLUTION, , x1, ·, x1, , f, (vi) e g o (x)., , Clearly, f (x) x 2 is defined for all xd R such that, x 2 $ 0, i.e., x $ 2., , dom ( f ) [2, 3) ., Again, g (x) 4 x 2 is defined for all xd R such that, 4 x 2 $ 0., But, 4 x 2 $ 0 & x 2 4 # 0 & (x 2) (x 2) # 0 & xd[2, 2]., , dom (g) [2, 2] ., , dom ( f )+ dom (g) [2, 3)+[2, 2] [2, 2] ., (i) ( f g) : [ 2, 2]" R is given by, ( f g) (x) f (x) g (x) x 2 4 x 2 ., (ii) ( f g) : [2, 2]" R is given by, ( f g) (x) f (x) g (x) x 2 4 x 2 ., (iii) ( fg) : [ 2, 2]" R is given by, ( fg) (x) f (x) · g (x) ( x 2 ) ( 4 x 2 ), ( x 2) 2 ( 2 x) ( x 2 ) (2 x) ., (iv) ( ff ) : [2, 2]" R is given by, ( ff ) (x) f (x) · f (x) ( x 2 ) ( x 2 ) (x 2) ., (v) (gg) : [2, 2]" R is given by, (gg) (x) g (x) · g (x) ( 4 x 2 ) ( 4 x 2 ) (4 x 2) ., (vi) {x : g (x) 0} {x : 4 x 2 0] {x : (2 x) (2 x) 0} {2, 2} ., , , f, dom e g o dom ( f )+ dom {g} {x : g (x) 0}, , , , [2, 2] {2, 2} (2, 2) ., f, , g : ( 2, 2)" R is given by
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122, , Senior Secondary School Mathematics for Class 11, , f, f (x), e o ( x) , , g, g ( x), EXAMPLE 6, , SOLUTION, , x2, 4 x2, , , , 2x, ( 2 x) ( 2 x), , , , 1, ·, ( 2 x), , Let f be the exponential function and g be the logarithmic function. Then,, find:, (i) ( f g) (1), (ii) ( fg) (1), (iii) (4f ) (1), (iv) (3g) (1), Let f : R " R : f (x) e x and g : R " R : g (x) log e x., Then, dom ( f )+ dom (g) R + R R ., (i) ( f g) : R " R is given by, ( f g) (x) f (x) g (x) (e x log e x) ., , , ( f g) (1) (e1 log e 1) (e 0) e., , (ii) ( fg) : R " R is given by, ( fg) (x) f (x) $ g (x) e x (log e x)., , , ( fg) (1) e1 (log e 1) (e # 0) 0., , (iii) (4f ) : R " R is given by, (4f ) (x) 4 # f (x) 4e x ., , (4f ) (1) (4 # e1) 4e., , (iv) (3g) : R " R is given by, (3g) (x) 3 # g (x) 3 #(log e x), , (3g) (1) 3 # g (1) 3 #(log e 1) (3 # 0) 0., EXAMPLE 7, , If f (x) log e (1 x) and g (x) [x] then find:, (i) ( f g) (x), , SOLUTION, , (ii) ( fg) (x), , f, (iii) e g o (x), , g, (iv) e o (x) ., f, , f, g 1, Also find ( f g) (1), ( fg) (0), e g o (1), e ob l ·, f 2, Clearly, log e (1 x) is defined only when 1 x 0, i.e., x 1., , dom ( f ) (3, 1) ., Also, dom (g) R., , dom ( f )+ dom (g) (3, 1)+ R (3, 1) ., (i) ( f g) : (3, 1)" R is given by, ( f g) (x) f (x) g (x) log e (1 x) [x], (ii) ( fg) : (3, 1)" R is given by, ( fg) (x) f (x)# g (x) {log e (1 x)}#[x] ., , (iii) {x : g (x) 0} {x : [x] 0} [0, 1) ., , , f, dom e g o dom ( f )+ dom (g) {x : g (x) 0}, (3, 1)+ R [0, 1) (3, 0) .
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Functions, , 123, , f, , g : ( 3, 0)" R is given by, f, f (x) log e (1 x), e o ( x) , , ·, g, g (x), [x], (iv) {x : f (x) 0} {x : log e (1 x) 0} {0}., , , g, dom e o dom (g)+ dom ( f ) {x : f (x) 0}, f, R +(3, 1) {0} (3, 0),(0, 1)., g, , : (3, 0),(0, 1)" R is given by, f, g, g (x), [x], e o (x) , , ·, f, f (x) log e (1 x), Now, we have:, ( f g) (1) f (1) g (1) [1] log e (1 1) (log e 2) 1., ( fg) (0) f (0)# g (0) log e (1 0)#[0] (log e 1# 0) (0 # 0) 0., , , f, f (1), [1], 1, e o (1) , , , ·, g, g (1) log e (1 1) log e 2, 1, 1, ; E, gb l, g, [0.5], 2, 2, e ob1 l , , , 0., 1, 1, 1, f 2, f b l log e b1 l log e b l, 2, 2, 2, EXAMPLE 8, SOLUTION, , Find the sum and the difference of the identity function and the modulus, function., Let f : R " R : f (x) x be the identity function., And, let g : R " R : g (x) |x|be the modulus function., Then, dom ( f ) R and dom (g) R., , dom ( f )+ dom (g) R + R R., (i) dom ( f g) dom ( f )+ dom (g) R., Now, ( f g) : R " R is given by, ( f g) (x) f (x) g (x), Z], ]x, when x $ 0, x |x| x [], ] x, when x 0, \, Z, Z] , x x, when x $ 0 ]]2x, when x $ 0, ][], [], ]x x, when x 0 ]0, when x 0., \, \ ]Z, ]2x, when x $ 0, Hence, ( f g) (x) [], ]0, when x 0., \, (ii) dom ( f g) dom ( f )+ dom (g) R., , ( f g) (x) f (x) g (x), ]Z]x, when x $ 0, x |x| x ][, ] x, when x 0, \
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124, , Senior Secondary School Mathematics for Class 11, , , , EXAMPLE 9, , SOLUTION, , EXAMPLE 10, SOLUTION, , EXAMPLE 11, SOLUTION, , ]Z]x x, when x $ 0 ]Z]0, when x $ 0, ][, [, ]x x, when x 0 ]]2x, when x 0, \, \Z, ]]0, when x $ 0, ( f g) (x) ][, ]2x, when x 0., \, , Find the sum and the difference of the identity function and the reciprocal, function., 1, Let f : R " R : f (x) x and g : R {0} " R : g (x) x be the identity, function and the reciprocal function respectively., Then, dom ( f )+ dom (g) R + R {0} R {0} ., 1, , ( f g) : R {0} " R : ( f g) (x) f (x) g (x) bx x l ·, 1, Hence, ( f g) (x) bx x l for all xd R {0} ., 1, And, ( f g) : R {0} " R : ( f g) (x) f (x) g (x) bx x l ·, 1, Hence, ( f g) (x) bx x l for all xd R {0} ., Find the product of the identity function by the modulus function., Let f : R " R : f (x) x and g : R " R : g (x) |x| be the identity, function and the modulus function respectively., Then, dom ( fg) dom ( f )+ dom (g) R + R R., , ( fg) : R " R : ( fg) (x) f (x) · g (x) ., Now, ( fg) (x) f (x) · g (x) x ·|x|, ]Z]x, when x $ 0, ]Z]x 2, when x $ 0, x · ][, ][ 2, ]x, when x 0 ]x , when x 0., \, \, ]Z]x 2, when x $ 0, , Hence, ( fg) (x) [] 2, ]x , when x 0., \, Find the product of the identity function by the reciprocal function., 1, Let f : R " R : f (x) x and g : R {0} " R : g (x) x be the identity, function and the reciprocal function respectively., Then, dom ( fg) dom ( f )+ dom (g) R + R {0} R {0} ., , , 1, ( fg) : R {0} " R : ( fg) (x) f (x) · g (x) bx # x l 1., , Hence, ( fg) (x) 1 for all xd R {0} ., EXAMPLE 12, SOLUTION, , Find the quotient of the identity function by the modulus function., Let f : R " R : f (x) x and g : R " R : g (x) |x| be the identity, function and the modulus function respectively.
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Functions, , 125, , f, Now, dom e g o dom ( f )+ dom (g) {x : g (x) 0}, , EXAMPLE 13, SOLUTION, , and {x : g (x) 0} {x :|x| 0} {0}., f, , dom e g o [R + R {0}] {0} R {0} ., Z], f, f, f (x), ]1, when x 0, So, : R {0} " R : e o (x) , x [], g, g, g (x) |x| ]1, when x 0., \, Z], f, Hence, e g o (x) ][1, when x 0, ]]1, when x 0., \, Find the quotient of the identity function by the reciprocal function., 1, Let f : R " R : f (x) and g : R {0} " R : g (x) x be the identity, function and the reciprocal function respectively., f, Now, dom e g o dom ( f )+ dom (g) {x : g (x) 0}, 1, and {x : g (x) 0} 'x : x 01 ., f, , dom e g o [R + R {0}] R {0} ., f, f, f ( x) x, x2 ., So, g : R {0} " R : e g o (x) , g ( x) 1, x, f, Hence, e g o (x) x 2 for all xd R {0} ., , EXERCISE 3E, 1. Let f : R " R : f (x) x 1 and g : R " R : g (x) 2x 3., Find (i) ( f g) (x), , (ii) ( f g) (x), , (iii) ( fg) (x), , f, (iv) e g o (x)., , 2. Let f : R " R : f (x) 2x 5 and g : R " R : g (x) x 2 x., Find (i) ( f g) (x), , (ii) ( f g) (x), , (iii) ( fg) (x), , f, (iv) e g o (x)., , 3. Let f : R " R : f (x) x 3 1 and g : R " R : g (x) (x 1) ., Find (i) ( f g) (x), , (ii) ( f g) (x), , 1, (iii) e o (x), f, , f, (iv) e g o (x)., , x, 4. Let f : R " R : f (x) c , where c is a constant., 1, (ii) (c 2 f ) (x), (iii) b c f l (x)., Find (i) (cf ) (x), 5. Let f : [2, 3)" R : f (x) x 2 and g : [2, 3)" R : g (x) x 2 ., Find (i) ( f g) (x), (ii) ( f g) (x), (iii) ( fg) (x).
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126, , Senior Secondary School Mathematics for Class 11, ANSWERS (EXERCISE 3E), , x1, 2x 3, 2x 5, 2x 3 7x 2 5x (iv) 2, x x, 1, 2 , (iv), x, x 1, x3 1, x, c2, x 2 x 2 for all xd[2, 3), , 1., , (i) 3x 2, , (ii) x 4, , (iii) 2x 2 x 3, , 2., , (i) x 2 3x 5, , (ii) x 2 x 5, , (iii), , 3., , (i) x 3 x 2, , (ii) x 3 x, , (iii), , 4., , (i) x, , (ii) cx, , (iii), , 5., , (i), (iii), , x 2 x 2 for all xd[2, 3), , (ii), , (iv), , x 4 for all xd[2, 3), 2, , HINTS TO SOME SELECTED QUESTIONS, f, 1. (iv) dom e g o dom ( f )+ dom (g) {x : g (x) 0}, R + R '3 1 R '3 1 ·, 2, 2, f, 2. (iv) dom e g o dom ( f )+ dom (g) {x : g (x) 0}, R + R {0, 1} R {0, 1} ., 1, 3. (iii) dom e o dom ( f ) {x : f (x) 0} R {1} ., f, f, (iv) dom e g o dom ( f )+ dom (g) {x : g (x) 0}, R + R {1} R {1} ., , EXERCISE 3F, Very-Short-Answer Questions, 1. Find the set of values for which the function f (x) 1 3x and g (x) 2x 2 1, are equal., 2. Find the set of values for which the functions f (x) x 3 and g (x) 3x 2 1, are equal., 3. Let X {1, 0, 2, 5} and f : X " R : f (x) x 3 1. Then, write f as a set of, ordered pairs., 4. Let A {2, 1, 0, 1, 2} and f : A " Z : f (x) x 2 2x 3. Find f (A) ., 5. Let f : R " R : f (x) x 2 ., Determine (i) range (f ), (ii) {x : f (x) 4}, 6. Let f : R " R : f (x) x 2 1. Find f 1 {10} ., 7. Let f : R " R : f (x) log e x. Find {x : f (x) 2} .
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Functions, , 127, , 8. Let A {6, 10, 11, 15, 21} and let f : A " N : f (n) is the highest prime factor, of n. Find range (f )., 9. Find the range of the function f (x) sin x., 10. Find the range of the function f (x) |x|., 11. Write the domain and the range of the function, f (x) x [x] ., x5, 12. If f (x) then find dom ( f ) and range ( f )., 5 x, 13. Let f {(1, 6), (2, 5), (4, 3), (5, 2), (8, 1), (10, 3)}, and g {(2, 0), (3, 2), (5, 6), (7, 10), (8, 12), (10, 16)} ., f, Find (i) dom ( f g), (ii) dom e g o ·, x1, 1, 14. If f (x) x , find the value of f ( f b x l2 ·, kx, 15. If f (x) , where x ! 1 and f { f (x)} x for x ! 1 then find the value, x 1, of k., x ·, 16. Find the range of the function, f (x) , |x|, 17. Find the domain of the function, f (x) log|x| ., 1, 1, 18. If f bx x l dx 2 2 n for all xd R {0} then write an expression for f (x) ., x, ax b ·, 19. Write the domain and the range of the function, f (x) , bx a, 20. Write the domain and the range of the function, f (x) x 1 ., 21. Write the domain and the range of the function, f (x) |x| ., ANSWERS (EXERCISE 3F), , 1, 4, 1. '2, 1, 2. ' , 11, 2, 3, 4. f (A) {5, 0, 3, 4}, , 3. f {(1, 0), (0, 1), (2, 9), (5, 126)}, 5. (i) [0, 3) (ii) {–2, 2}, , 6. {–3, 3}, , 8. {3, 5, 7, 11}, 9. [–1, 1], 10. [0, 3), 7. {e 2}, 11. dom ( f ) R, range ( f ) [0, 1) 12. dom ( f ) R {5}, range ( f ) {1}, x, 13. (i) {2, 5, 8, 10} (ii) {5, 8, 10} 14. , 15. k 1, x 1, 16. {1, –1}, 17. R {0}, 18. f (x) x 2 2, xd R, a, a, 19. dom ( f ) R ' 1 , range ( f ) R ' 1 20. dom ( f ) [1, 3), range ( f ) R, b, b, 21. dom ( f ) R, range ( f ) (3, 0], HINTS TO SOME SELECTED QUESTIONS, 1. 2x 2 1 1 3x & 2x 2 3x 2 0 & (x 2) (2x 1) 0.
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128, , Senior Secondary School Mathematics for Class 11, , 2. 3x 2 1 x 3 & 3x 2 x 4 0 & (3x 4) (x 1) 0., 5. (i) f (x) x 2 $ 0. So, range ( f ) [0, 3) ., (ii) f (x) 4 & x 2 4 & x !2., 6. f (x) 10 & x 2 1 10 & x 2 9 & x !3., 7. f (x) 2 & log e x 2 & x e 2 ., 8. f (6) 3, f (10) 5, f (11) 11, f (15) 5, f (21) 7., range ( f ) {3, 5, 7, 11} ., , , , 10. f (x) |x|$ 0. So, range ( f ) [0, 3) ., 11. x [x] 0 for xdZ and x [x] 0 for xd R Z., x [x] $ 0 for xd R. So, dom ( f ) R., Also, 0 # x [x] 1 & 0 # x [x] 1. So, range ( f ) [0, 1) ., 12. f (x) is not defined when x 5. So, dom ( f ) R {5} ., , , f (x) , , (x 5), 1. So, range ( f ) {1} ., (5 x), , 13. dom ( f g) dom ( f )+ dom (g) {2, 5, 8, 10}., f, dom e g o dom ( f )+ dom (g) {x : g (x) 0} {2, 5, 8, 10} {2} {5, 8, 10}., 1, (1 x) 1, , x 1 , 1, x x ·, (1 x) & f ( f c x m2 , 1, 1x, 1x x1, x, kx, k# , 1 , kx, x, k2 x, k2 x, x., · So,, 15. f (f (x)) f d n , x 1, kx , kx x 1, kx x 1, 1, x1, x ! x 2 4 (x 1) x ! (x 2), , k 2 kx x 1 & k 2 kx (x 1) 0 & k , 2, 2, & k x 1 or k 1 & k 1 [a k is a constant]., 14. f c x m , 1, , 1, 1 2, 18. f cx x m cx x m 2 & f (x) x 2 2, xd R., , SUMMARY OF KEY FACTS, 1., , (i) Function or Mapping, , Let X and Y be two nonempty sets and let f 3 X #Y be a relation from, X to Y such that, (i) dom ( f ) X and, (ii) no two different ordered pairs in f have the same first coordinate., Then, f is called a function or a mapping from X to Y and we write,, f : X " Y., (ii) Dom ( f ) {x : (x, y)d f } X,, , co-domain ( f ) Y., , range ( f ) {y : (x, y)d f } 3 Y, , and
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Functions, , 129, , (iii) If (x, y)d f, we write, f (x) y., , Here, y is called the image of x under f, while x is called the pre-image, of y., (iv) Two functions f and g having same domain X are said to be equal only, when f (x) g (x) for all xdX., 2., , (i) Function or Mapping, , Let X and Y be two nonempty sets. Then, a rule f which associates to, each xdX, a unique element f (x) of Y, is called a function from X to Y., We write, f : X " Y. Dom ( f ) X, range ( f ) { f (x) : xdX} 3 Y and, co-domain ( f ) Y., (ii) A mapping f : X " Y is one-one if f (x1) f (x2) & x1 x2 ., (iii) Real Functions, , Let R be the set of all real numbers. Then, a function f : R " R is called, a real function., 3., , (i) Constant Function, , Let k be a fixed real number. Then, a function f : R " R : f (x) k for all, xd R is called a constant function. Dom ( f ) R and range ( f ) {k} ., (ii) Identity Function, , The function I : R " R : I (x) x for all xd R is called the identity, function. Dom (I) = Range (I) R., (iii) Modulus Function, , The function f : R " R : f (x) |x| for all xd R is called the modulus, function. Dom ( f ) R and range ( f ) {x : xd R and x $ 0} [0, 3) ., (iv) Greatest Integer Function, or Step Function, , The function f : R " R : f (x) [x], where [x] denotes the greatest integer, less than or equal to x, is called the greatest integer function., Dom ( f ) R, range ( f ) Z., EXAMPLES, , [3.85] = 3, [0.94] = 0, [5.0] = 5, [–7.35] = –8 and [–6.0] = –6, etc., , (v) Smallest Integer Function, , The function f : R " R : f (x) _xi, where _xi denotes the smallest, integer greater than or equal to x. Dom ( f ) R and range ( f ) Z., EXAMPLES [6.8] = 7, [5] = 5, [0.85] = 1, [– 4.5] = – 4, etc., (vi) Fractional Part Function, , The function f : R " R : f (x) {x}, where {x} denotes the fractional part, or decimal part of x. Note that {x} x [x] ., Dom ( f ) R and range ( f ) [0, 1) ., EXAMPLES {6.83} = 0.83, {–3.65} = 0.35, {8} = 0, {–3} = 0, etc., (vii) Signum Function, , ]Z]|x|, ], The function f : R " R : f (x) ][ x , when x ! 0, ]], ]]0, when x 0, \, is called the signum function.
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130, , Senior Secondary School Mathematics for Class 11, , Dom ( f ) R and range ( f ) {1, 0, 1} ., (viii) Exponential Function, , Let a be a real number such that a 0 and a ! 1., Then, f : R " R : f (x) a x for all xd R is called the exponential function., Dom ( f ) R and range ( f ) (0, 3) ., In particular, f : R " R : f (x) e x for all xd R., (ix) Logarithmic Function, , The function f : R " R : f (x) log e x for all xd R is called the, logarithmic function. Dom ( f ) (0, 3) and range ( f ) R., (x) Reciprocal Function, , 1, The function f : R {0} " R : f (x) x for all xd R {0} is called the, reciprocal function. Dom ( f ) R {0} and range ( f ) R {0} ., (xi) Square Root Function, , The function f : R ,{0} " R : f (x) x for all xd R ,{0} is called the, square root function. Dom ( f ) R ,{0} [0, 3), range ( f ) [0, 3) ., (xii) Cube Root Function, , The function f : R " R : f (x) x1/3 for all xd R is called the cube root, function. Dom ( f ) R and range ( f ) R., (xiii) Rational Function, , p(x), where p(x) and q(x) are polynomials, q(x), and q(x) ! 0, is called a rational function. Dom ( f ) R {x : q(x) 0} ., The function f : R " R : f (x) , , 4., , OPERATIONS ON REAL FUNCTIONS, , Let f and g be real functions having domains D1 and D2 respectively. Then,, (i) ( f g) : D1 + D2 " R : ( f g) (x) f (x) g (x) ., (ii) ( f g) : D1 + D2 " R : ( f g) (x) f (x) g (x) ., (iii) ( fg) : D1 + D2 " R : ( fg) (x) f (x) · g (x), (iv) (f ) : D1 " R : (f ) (x) · f (x) ., f, f, f (x), , g (x) ! 0., (v) g : (D1 + D2) {x : g (x) 0} " R : e g o (x) , g ( x),
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Principle of Mathematical Induction, , 131, , 4, , Principle of Mathematical Induction, PRINCIPLE OF MATHEMATICAL INDUCTION, , Let P(n) be a statement involving the natural number n such that, (i) P(1) is true and (ii) P(k 1) is true, whenever P(k) is true, then P(n) is true for all nd N., SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Using the principle of mathematical induction, prove that, 1 3 5 7 … (2n 1) n 2 for all nd N., Let the given statement be P(n) . Then,, P(n) : 1 3 5 7 … (2n 1) n 2 ., Putting n 1 in the given statement, we get, 2, LHS = 1 and RHS 1 1., , LHS = RHS., Thus, P(1) is true., Let P(k) be true. Then,, P ( k ) : 1 3 5 7 … ( 2k 1) k 2 ., Now, 1 3 5 7 … (2k 1) {2(k 1) 1}, {1 3 5 7 … (2k 1)} (2k 1), k 2 (2k 1), , … (i), , [using (i)], , (k 1) 2 ., , P (k 1) : 1 3 5 7 … (2k 1) {2(k 1) 1} (k 1) 2 ., This shows that P(k 1) is true, whenever P (k) is true., Thus, P(1) is true and P (k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, we have, 1 3 5 7 … (2n 1) n 2 for all nd N., EXAMPLE 2, , SOLUTION, , Using the principle of mathematical induction, prove that, 1, 1 4 7 10 … (3n 2) n (3n 1) for all nd N., 2, Let the given statement be P(n) . Then,, 1, P(n) : 1 4 7 10 … (3n 2) n(3n 1) ., 2, 131
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132, , Senior Secondary School Mathematics for Class 11, , Putting n 1 in the given statement, we get, LHS = 1 and RHS , , 1, #1#(3 #1 1) 1., 2, , , LHS = RHS., Thus, P(1) is true., Let P(k) be true. Then,, 1, … (i), k ( 3 k 1) ., 2, Now, 1 4 7 … (3k 2) {3 (k 1) 2}, {1 4 7 … (3k 2)} (3k 1), 1 k (3k 1) (3k 1), [using (i)], 2, 1, 1, (3k 2 k 6k 2) (3k 2 5k 2), 2, 2, 1, 2, (3k 3k 2k 2) 1 · {3k (k 1) 2 (k 1)}, 2, 2, 1, 1, ( k 1) ( 3k 2) ( k 1 ) { 3 ( k 1 ) 1 } ., 2, 2, 1, , P (k 1) : 1 4 7 … {3 (k 1) 2} (k 1) {3 (k 1) 1} ., 2, This shows that P(k 1) is true, whenever P(k) is true., Thus, P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, we have, 1, 1 4 7 … (3n 2) n (3n 1) for all nd N., 2, P (k) : 1 4 7 10 … (3k 2) , , EXAMPLE 3, , Using the principle of mathematical induction, prove that, 12 22 32 … n2 , , SOLUTION, , 1, n (n 1) (2n 1) for all nd N., 6, , Let the given statement be P(n) . Then,, 1, n (n 1)(2n 1) ., 6, Putting n 1 in the given statement, we get, 1, 2, LHS 1 1 and RHS #1# 2 #(2 #1 1) 1., 6, , LHS = RHS., Thus, P(1) is true., Let P(k) be true. Then,, P (n): 1 2 2 2 3 2 … n 2 , , P(k): 1 2 2 2 3 2 … k 2 , , 1, k (k 1)(2k 1) ., 6, , Now, 1 2 3 … k (k 1), {1 2 2 2 3 2 … k 2} (k 1) 2, 1 k (k 1)(2k 1) (k 1) 2, 6, 2, , 2, , 2, , 2, , … (i), , 2, , [using (i)]
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Principle of Mathematical Induction, , 133, , 1 (k 1) · {k (2k 1) 6 (k 1)}, 6, 1 (k 1) (2k 2 7k 6) 1 (k 1) {(2k 2 4k) (3k 6)}, 6, 6, 1 (k 1)(k 2)(2k 3) 1 (k 1)(k 1 1)[2 (k 1) 1] ., 6, 6, , P (k 1): 1 2 2 2 3 2 … (k 1) 2, 1 (k 1)(k 1 1)[2(k 1) 1] ., 6, This shows that P(k 1) is true, whenever P(k) is true., Thus, P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, we have, 12 22 32 … n2 , , 1, n (n 1)(2n 1) for all nd N., 6, , EXAMPLE 4, , Using the principle of mathematical induction, prove that, n(n 1) 2, 2 for all nd N., 13 23 33 … n3 (, 2, , SOLUTION, , Let the given statement be P(n) . Then,, n(n 1) 2, 2 ·, P(n): 1 3 2 3 3 3 … n 3 (, 2, Putting n 1 in the given statement, we get, 3, LHS 1 1 and RHS b, , 1# 2 2 2 , l 1 1., 2, , , LHS = RHS., Thus, P(1) is true., Let P(k) be true for some k d N. Then,, P(k): 1 3 2 3 3 3 … k 3 ), , 2, , k (k 1) 2, 3 ·, 2, , … (i), , Now, 1 3 2 3 3 3 … k 3 (k 1) 3, {1 3 2 3 3 3 … k 3} (k 1) 3, (, , k (k 1) 2, 2 (k 1) 3, 2, , [using (i)], , 2, 2, , (k 1) 2 ( k (k 1)2 (k 1) 2 ( k 4k 4 2, 4, 4, (k 1) 2 (k 2) 2, (k 1) {(k 1) 1} 2, 2, , (, 2, 4, (k 1) {(k 1) 1} 2, F ·, , P(k 1): 1 3 2 3 3 3 … (k 1) 3 <, 2, This shows that P(k 1) is true, whenever P(k) is true.
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134, , Senior Secondary School Mathematics for Class 11, , , P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, we have, n(n 1) 2, 3, 3, 3, 3, 2 for all nd N., 1 2 3 … n (, 2, EXAMPLE 5, , Using the principle of mathematical induction, prove that, 1, 1 · 2 2 · 3 3 · 4 … n(n 1) n(n 1)(n 2) ., 3, , SOLUTION, , Let the given statement be P(n) . Then,, 1, n(n 1)(n 2) ., 3, 1, When n 1, LHS 1 · 2 2 and RHS #1# 2 #(1 2) 2., 3, , LHS = RHS., Thus, the given statement is true for n 1, i.e., P(1) is true., Let P(k) be true. Then,, 1, P(k): 1 · 2 2 · 3 3 · 4 … k (k 1) k (k 1)(k 2) . … (i), 3, Now, 1 · 2 2 · 3 3 · 4 … k (k 1) (k 1)(k 2), {1 · 2 2 · 3 3 · 4 … k (k 1) (k 1)(k 2), 1 k (k 1)(k 2) (k 1)(k 2), [using (i)], 3, 1 · [k (k 1)(k 2) 3(k 1)(k 2)] 1 (k 1)(k 2)(k 3) ., 3, 3, , P (k 1) : 1 · 2 2 · 3 3 · 4 … (k 1)(k 2), 1 (k 1)(k 2)(k 3) ., 3, This shows that P(k 1) is true, whenever P(k) is true., , P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, we have, P(n): 1 · 2 2 · 3 3 · 4 … n(n 1) , , 1 · 2 2 · 3 3 · 4 … n(n 1) , , 1, n(n 1)(n 2) for all nd N., 3, , EXAMPLE 6, , Using the principle of mathematical induction, prove that, 1, 1 · 3 3 · 5 5 · 7 … (2n 1)(2n 1) n(4n 2 6n 1) ., 3, , SOLUTION, , Let the given statement be P(n) . Then,, P(n): 1 · 3 3 · 5 5 · 7 … (2n 1)(2n 1) , When n 1, we have, LHS 1 · 3 3 and RHS , , , , LHS = RHS., , 1, n(4n 2 6n 1) ., 3, , 1, 1, #1#(4 #1 2 6 #1 1) #1# 9 3., 3, 3
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Principle of Mathematical Induction, , 135, , Thus, P(1) is true., Let P(k) be true. Then,, 1, k (4k 2 6k 1) . … (i), 3, Now, 1 · 3 3 · 5 5 · 7 … (2k 1)(2k 1) {2(k 1) 1}{2(k 1) 1}, {1 · 3 3 · 5 5 · 7 … (2k 1)(2k 1)} (2k 1)(2k 3), 1 k (4k 2 6k 1) (2k 1)(2k 3), [using (i)], 3, 1 [(4k 3 6k 2 k) 3(4k 2 8k 3)] 1 (4k 3 18k 2 23k 9), 3, 3, 1, 1, 2, (k 1)(4k 14k 9) (k 1)[4 (k 1) 2 6 (k 1) 1] ., 3, 3, , P(k 1): 1 · 3 3 · 5 5 · 7 … {2(k 1) 1} {2(k 1) 1}, 1 (k 1) {4 (k 1) 2 6 (k 1) 1} ., 3, Thus, P(k 1) is true, whenever P(k) is true., , P(1) is true and P(k 1) is true, whenever P(k) is true., , P(k): 1 · 3 3 · 5 5 · 7 … (2k 1)(2k 1) , , Hence, by the principle of mathematical induction, we have, 1, 1 · 3 3 · 5 5 · 7 … (2n 1)(2n 1) n(4n 2 6n 1) for all nd N., 3, EXAMPLE 7, , Using the principle of mathematical induction, prove that, 1, 1 · 2 · 3 2 · 3 · 4 … n(n 1)(n 2) n(n 1)(n 2)(n 3), 4, for all nd N., , SOLUTION, , Let the given statement be P(n) . Then,, P(n): 1 · 2 · 3 2 · 3 · 4 … n(n 1)(n 2) , When n 1, we have, LHS 1 · 2 · 3 6 and RHS , , 1, n(n 1)(n 2)(n 3) ., 4, , 1, #1# 2 # 3 # 4 6., 4, , , LHS = RHS., Thus, the given statement is true for n 1, i.e., P(1) is true., Let P(k) be true. Then,, P(k): 1 · 2 · 3 2 · 3 · 4 … k (k 1)(k 2), 1 k (k 1)(k 2)(k 3) ., … (i), 4, Now, 1 · 2 · 3 2 · 3 · 4 … k (k 1)(k 2) (k 1)(k 2)(k 3), {1 · 2 · 3 2 · 3 · 4 … k (k 1) (k 2)} (k 1)(k 2)(k 3), 4 (k 1)(k 2)(k 3), 1 k (k 1)(k 2)(k 3) , [using (i)], 4, 4, 1 (k 1)(k 2)(k 3)(k 4) ., 4
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136, , Senior Secondary School Mathematics for Class 11, , P(k 1): 1 · 2 · 3 2 · 3 · 4 … (k 1)(k 2)(k 3), 1 (k 1)(k 2)(k 3)(k 4) ., 4, Thus, P(k 1) is true, whenever P(k) is true., P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, we have, 1, 1 · 2 · 3 2 · 3 · 4 … n(n 1)(n 2) n(n 1)(n 2)(n 3), 4, for all values of nd N., , , EXAMPLE 8, , Using the principle of mathematical induction, prove that, 1 1 1 , 1, n ·, …, 1· 2 2 · 3 3 · 4, n(n 1) (n 1), , SOLUTION, , Let the given statement be P(n) . Then,, P(n):, , 1 1 1 , 1, n ·, …, 1· 2 2 · 3 3 · 4, n(n 1) (n 1), , Putting n 1 in the given statement, we get, 1 1, 1 1·, LHS , and RHS , 1· 2 2, ( 1 1) 2, , LHS = RHS., Thus, P(1) is true., Let P(k) be true. Then,, 1 1 1 , 1, k ·, …, 1· 2 2 · 3 3 · 4, k (k 1) (k 1), 1 1 1 , 1, 1, , Now,, …, 1· 2 2 · 3 3 · 4, k (k 1) (k 1)(k 2), , … (i), , P(k):, , 1, 1, ( 1 1 1 …, 2, 1· 2 2 · 3 3 · 4, k (k 1) (k 1)(k 2), 1, k , (k 1) (k 1)(k 2), , , , [using (i)], , 2, k (k 2) 1, (k 1), (k 1), , , ·, (k 1)(k 2) (k 1)(k 2) (k 2), , P(k 1):, , (k 1), 1 1 1 , 1, , ·, 1 · 2 2 · 3 3 · 4 (k 1)(k 2) (k 2), , Thus, P(k 1) is true, whenever P(k) is true., , P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, we have, 1 1 1 , 1, n for all nd N., …, 1· 2 2 · 3 3 · 4, n(n 1) n 1, EXAMPLE 9, , Using the principle of mathematical induction, prove that, n, 1 1 1 , 1, , ·, …, 3·5 5·7 7·9, (2n 1)(2n 3) 3(2n 3)
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Principle of Mathematical Induction, SOLUTION, , 137, , Let the given statement be P(n) . Then,, n, 1 1 1 , 1, , ·, P(n):, …, 3·5 5·7 7·9, (2n 1)(2n 3) 3 (2n 3), Putting n 1 in the given statement, we get, 1 1, 1, 1 ·, and RHS , 3 · 5 15, 3(2 #1 3) 15, , LHS = RHS., Thus, P(1) is true., Let P (k) be true. Then,, LHS , , k, 1 1 1 , 1, , · … (i), …, 3·5 5·7 7·9, (2k 1)(2k 3) 3(2k 3), 1 1 , 1, 1, , Now,, …, 3·5 5·7, (2k 1)(2k 3) {2(k 1) 1} {2(k 1) 3}, P(k):, , 1, 1, ( 1 1 …, 2, 3·5 5·7, (2k 1)(2k 3) (2k 3)(2k 5), k, 1, , , [using (i)], 3(2k 3) (2k 3)(2k 5), 2, k (2k 5) 3, (2k 5k 3), (k 1)(2k 3), , , , 3(2k 3)(2k 5) 3(2k 3)(2k 5) 3 (2k 3)(2k 5), ( k 1), ( k 1), , , ·, , 3 ( 2k 5) 3 { 2 ( k 1) 3}, 1 1 1 , 1, , P(k 1):, …, 3·5 5·7 7·9, {2(k 1) 1} {2(k 1) 3}, ( k 1), , ·, 3{2(k 1) 3}, This shows that P(k 1) is true, whenever P(k) is true., , P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, we have, n, 1 1 1 , 1, , …, 3·5 5·7 7·9, (2n 1)(2n 3) 3(2n 3), for all values of nd N., EXAMPLE 10, , Using the principle of mathematical induction, prove that, n(n 3), 1 1 , 1, , for all nd N., …, 1· 2 · 3 2 · 3 · 4, n(n 1)(n 2) 4(n 1)(n 2), , SOLUTION, , Let the given statement be P(n) . Then,, P(n):, , n(n 3), 1 1 , 1, , ·, …, , , 1· 2 · 3 2 · 3 · 4, n(n 1)(n 2) 4(n 1)(n 2), , Putting n 1 in the given statement, we get, 1#(1 3), 1 1, 1# 4 1 ·, LHS , and RHS , 1· 2 · 3 6, 4 # ( 1 1) ( 1 2) 4 # 2 # 3 6
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138, , Senior Secondary School Mathematics for Class 11, , , LHS = RHS., Thus, the given statement is true for n 1, i.e., P(1) is true., Let P(k) be true. Then,, k (k 3), 1 1 , 1, , · … (i), …, P(k):, 1· 2 · 3 2 · 3 · 4, k (k 1)(k 2) 4(k 1)(k 2), 1 1 , 1, 1, , Now,, …, 1· 2 · 3 2 · 3 · 4, k (k 1)(k 2) (k 1)(k 2)(k 3), 1, 1, ( 1 1 …, 2, 1· 2 · 3 2 · 3 · 4, k (k 1)(k 2) (k 1)(k 2)(k 3), ), , k ( k 3), 1, 3, , 4(k 1)(k 2) (k 1)(k 2)(k 3), , [using (i)], , 2, 3, 2, k (k 3) 4, (k 6k 9k 4), , 4(k 1)(k 2)(k 3) 4(k 1)(k 2)(k 3), (k 1)(k 1)(k 4), (k 1)(k 4), , , ·, 4(k 1)(k 2)(k 3) 4(k 2)(k 3), 1 1 , 1, , P(k 1):, …, 1· 2 · 3 2 · 3 · 4, (k 1)(k 2)(k 3), (k 1)(k 4), , ·, 4(k 2)(k 3), This shows that P(k 1) is true, whenever P(k) is true., , P(1) is true and P(k 1) is true, whenever P(k) is true., , , , Hence, by the principle of mathematical induction, we have, n(n 3), 1 1 1 , 1, , …, 1· 2 · 3 2 · 3 · 4 3 · 4 · 5, n(n 1)(n 2) 4(n 1)(n 2), for all values of nd N., EXAMPLE 11, , Using the principle of mathematical induction, prove that, 1 · 3 2 · 32 3 · 33 … n · 3n , , SOLUTION, , (2n 1) 3 n, 4, , 1, , 3, , for all nd N., , Let the given statement be P(n) . Then,, , , (2n 1) 3 n 1 3, ·, 4, Putting n 1 in the given statement, we get, , (2 #1 1) 31 1 3 1# 9 3 12, , , 3., LHS 1 · 3 3 and RHS , 4, 4, 4, , LHS = RHS., Thus, P(1) is true., Let P(k) be true. Then,, , (2k 1) 3 k 1 3, ·, … (i), P(k): 1 · 3 2 · 3 2 3 · 3 3 … k · 3 k , 4, , , 1 · 3 2 · 3 2 3 · 3 3 … k · 3 k (k 1) 3(k 1), P(n): 1 · 3 2 · 3 2 3 · 3 3 … n · 3 n
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Principle of Mathematical Induction, , 139, , {1 · 3 2 · 3 2 3 · 3 3 … k · 3 k} (k 1) 3(k 1), , (2k 1) 3 k 1 3, (k 1) 3 k 1, [using (i)], 4, k1, k1, k1, (2k 1) 3 3 4 (k 1) 3, (2k 1 4k 4) 3 3, , , 4, 4, k1, (k 1) 1, , , , , , (6k 3) 3, 3 { 2 ( k 1) 1 } 3, 3, , , ·, 4, 4, , P(k 1): 1 · 3 2 · 3 2 3 · 3 3 … (k 1) · 3 k 1, , , , , , ( k 1) 1, 3, {2(k 1) 1} 3, ·, 4, This shows that P(k 1) is true, whenever P(k) is true., Thus, P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, we have, , (2n 1) 3 n 1 3, 1 · 3 2 · 32 3 · 33 … n · 3n , 4, for all values of nd N., , , , EXAMPLE 12, , Using the principle of mathematical induction, prove that, 1, 1, 1, 1, 1, b1 lb1 lb1 l … b1 l for all nd N., 2, 3, n 1, 4, (n 1), , SOLUTION, , Let the given statement be P(n) . Then,, 1, 1, 1, 1, 1 ·, P (n): b1 lb1 lb1 l … b1 l , 2, 3, n 1, 4, ( n 1), 1, 1, 1 1·, When n 1, LHS b1 l and RHS , 2, 2, ( 1 1) 2, , LHS = RHS., Thus, P(1) is true., Let P(k) be true. Then,, 1, 1, 1, 1 1 ·, … (i), P(k): b1 lb1 lb1 l … b1 , l, 2, 3, 4, k1, ( k 1), 1, 1, 1, 1, 1, Now, (b1 lb1 lb1 l … b1 , l2 · b1 l, 2, 3, 4, k1, k 2, ( k 2) 1, (k 1), G 1 ·, 1, 1 ·=, [using (i)]., (k 1), ( k 2), (k 1) (k 2) (k 2), 1, 1, 1, 1 1 ·, , P(k 1): b1 lb1 lb1 l … b1 , l, 2, 3, 4, k2, ( k 2), This shows that P(k 1) is true, whenever P(k) is true., Thus, P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, we have, 1, 1, 1, 1, 1, b1 lb1 lb1 l … b1 l for all nd N., 2, 3, n 1, 4, (n 1)
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140, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 13, , Using the principle of mathematical induction, prove that, n, a (r 1), 2, n1, , for r 1 and all nd N., a ar ar … ar, (r 1), , SOLUTION, , Let the given statement be P(n) . Then,, a (r n 1), , ·, P(n): a ar ar 2 … ar n 1 , (r 1), When n 1, we have, a ( r 1 1), a., LHS a and RHS , ( r 1), , LHS = RHS., Thus, P(1) is true., Let P(k) be true. Then,, a (r k 1), , ·, … (i), P(k): a ar ar 2 … ar k 1 , (r 1), a ( r k 1), , ar k, [using (i)], Now, (a ar ar 2 … ar k 1) ar k , ( r 1), k1, a (r 1), , ·, (r 1), , a ( r k 1 1), , ·, , P (k 1): a ar ar 2 … ar k 1 ar k , (r 1), This shows that P(k 1) is true, whenever P(k) is true., , P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, we have, n, a (r 1), 2, n1, , for r 1 and all nd N., a ar ar … ar, (r 1), , EXAMPLE 14, , Let a and b be arbitrary real numbers. Using the principle of mathematical, induction, prove that, (ab) n a n b n for all nd N., Let the given statement be P(n) . Then,, P(n): (ab) n a n b n ., When n 1, we have, 1, 1 1, LHS (ab) ab and RHS a b ab., , LHS = RHS., Thus, the given statement is true for n 1, i.e., P(1) is true., Let P(k) be true. Then,, … (i), P(k): (ab) k a k b k ., , SOLUTION, , Now, (ab) k, , 1, , (ab) k (ab) (a k b k)(ab), , [using (i)], , (a · a) (b · b) [by commutativity and associativity of, multiplication on real numbers], k1, k1, (a · b ) ., k, , k
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Principle of Mathematical Induction, , 141, , , , , , P(k 1): (ab) k 1 (a k 1 · b k 1) ., This shows that P(k 1) is true, whenever P(k) is true., , P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, we have, (ab) n a n b n for all xd N., , EXAMPLE 15, SOLUTION, , Using the principle of mathematical induction, prove that (n 2 n) is even, for all nd N., Let P(n): (n 2 n) is even., For n 1, the given expression becomes (1 2 1) 2, which is even., So, the given statement is true for n 1, i.e., P(1) is true., Let P(k) be true. Then,, P(k): (k 2 k) is even, &, , (k 2 k) 2m for some natural number m., , … (i), , Now, (k 1) (k 1) k 3k 2 (k k) 2(k 1), 2m 2(k 1), [using (i)], 2[m (k 1)], which is clearly even., , P(k 1): (k 1) 2 (k 1) is even., Thus, P(k 1) is true, whenever P(k) is true., 2, , , , 2, , 2, , P(1) is true and P(k 1) is true, whenever P(k) is true., , Hence, by the principle of induction, it follows that (n 2 n) is even, for all nd N., EXAMPLE 16, SOLUTION, , Using the principle of mathematical induction, prove that n(n 1)(n 5), is a multiple of 3 for all nd N., Let P(n): n(n 1)(n 5) is a multiple of 3., For n 1, the given expression becomes (1# 2 # 6) 12, which is a, multiple of 3., So, the given statement is true for n 1, i.e., P(1) is true., Let P(k) be true. Then,, P(k): k (k 1)(k 5) is a multiple of 3, ... (i), &, k (k 1)(k 5) 3m for some natural number m., , , , Now, (k 1)(k 2)(k 6), (k 1)(k 2) k 6(k 1)(k 2), k (k 1)(k 5 3) 6(k 1)(k 2), k (k 1)(k 5) 3k (k 1) 6(k 1)(k 2), 3m 3(k 1)(2k 4 k), 3m 3(k 1)(k 4), 3{m (k 1)(k 4)}, which is a multiple of 3., , P(k 1): (k 1)(k 2)(k 6) is a multiple of 3.
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142, , Senior Secondary School Mathematics for Class 11, , Thus, P(k 1) is true, whenever P(k) is true., , P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, it follows that, n(n 1)(n 5) is a multiple of 3 for all nd N., EXAMPLE 17, SOLUTION, , Using the principle of mathematical induction, prove that (7 n 3 n) is, divisible by 4 for all nd N., Let P(n): (7 n 3 n) is divisible by 4., For n 1, the given expression becomes (71 31) 4, which is, divisible by 4., So, the given statement is true for n 1, i.e., P(1) is true., Let P(k) be true. Then,, P(k): (7 k 3 k) is divisible by 4., &, , (7 k 3 k) 4m for some natural number m., , Now, {7, , (k 1), , 3, , (k 1), , … (i), , }, , (k 1), , 7, 7 · 3 k 7 · 3 k 3 ( k 1), 7 ( 7 k 3 k ) 3 k (7 3), , [subtracting and adding 7 · 3 k], , (7 # 4m) 4 · 3 k, , [using (i)], , 4(7m 3 k), which is clearly divisible by 4., , , , , , P(k 1): {7(k 1) 3(k 1)} is divisible by 4., Thus, P(k 1) is true, whenever P(k) is true., , P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, it follows that, (7 n 3 n) is divisible by 4 for all values of nd N., EXAMPLE 18, , SOLUTION, , Using the principle of mathematical induction, prove that (10 2n, divisible by 11 for all nd N., , 1, , 1) is, , , Let P(n): (10 2n 1 1) is divisible by 11., # , For n 1, the given expression becomes {10(2 1 1) 1} 11, which, is divisible by 11., So, the given statement is true for n 1, i.e., P(1) is true., Let P(k) be true. Then,, , P(k): (10 2k 1 1) is divisible by 11, , &, , (10 2k, , Now, {10, , 1, , 1) 11m, for some natural number m., , 2 ( k 1) 1, , 1} {10 · 10, 2, , (2k 1), , … (i), , 1}, , 100 #{10 2k 1 1} 99, (100 #11m) 99, [using (i)], 11#(100m 9), which is divisible by 11., , , , , , P(k 1): {10 2 (k 1) 1 1} is divisible by 11., Thus, P(k 1) is true, whenever P(k) is true.
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Principle of Mathematical Induction, , 143, , , P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, it follows that, , {10 2n 1 1} is divisble by 11 for all nd N., EXAMPLE 19, SOLUTION, , Using the principle of mathematical induction, prove that (2 · 7 n 3 · 5 n 5), is divisible by 24 for all nd N., Let P(n): (2 · 7 n 3 · 5 n 5) is divisible by 24., For n 1, the given expression becomes (2 · 71 3 · 51 5) 24,, which is clearly divisible by 24., So, the given statement is true for n 1, i.e., P(1) is true., Let P(k) be true. Then,, P(k): (2 · 7 k 3 · 5 k 5) is divisible by 24, &, , (2 · 7 k 3 · 5 k 5) 24m, for some md N., k1, , … (i), , k1, , Now, (2 · 7 3 · 5 5), ( 2 · 7 k · 7 3 · 5 k · 5 5), 7 (2 · 7 k 3 · 5 k 5) 6 · 5 k 30, (7 # 24m) 6(5 k 5), [using (i)], (24 #7m) 6 # 5 #(5 k 1 1), RS, VW, k1, (24 #7m) 5 # 24p SSSa (5 1) is divisible by (5 1), i.e., 4WWW, , SS, WW, & (5 k 1 1) 4p, T, X, 24 #(7m 5p), which is clearly divisible by 24., , , , , , P(k 1): (2 · 7 k 1 3 · 5 k 1 5) is divisible by 24., Thus, P(k 1) is true, whenever P(k) is true., , P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, it follows that, (2 · 7 n 3 · 5 n 5) is divisible by 24 for all nd N., EXAMPLE 20, SOLUTION, , Using the principle of mathematical induction, prove that (x n y n) is, divisible by (x y) for all nd N., Let the given statement be P(n) . Then,, P(n): (x n y n) is divisible by (x y) ., When n 1, the given statement becomes: (x1 y1) is divisible by, (x y), which is clearly true., , P(1) is true., Let P(k) be true. Then,, … (i), P(k): (x k y k) is divisible by (x y) ., Now, (x k, , 1, , {x, , y k 1), , k1, , x k y x k y y k 1}, , [on adding and subtracting x k y], , x (x y) y (x y ), which is divisible by (x y) [using (i)]., k, , k, , k
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144, , Senior Secondary School Mathematics for Class 11, , Thus, P(k 1) is true, whenever P(k) is true., , P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, it follows that, (x n y n) is divisible by (x y) for all nd N., EXAMPLE 21, SOLUTION, , Using the principle of mathematical induction, prove that, (1 x) n $ (1 nx) for all nd N, where x 1., Let the given statement be P(n) . Then,, P(n): (1 x) n $ (1 nx), where x 1., When n 1, we have, 1, LHS (1 x) (1 x) and RHS (1 1# x) (1 x) ., , LHS $ RHS is true., Thus, P(1) is true., Let P(k) be true. Then,, P(k): (1 x) k $ (1 kx), where x 1., Now, (1 x) k, , 1, , (1 x) k (1 x), $ (1 kx)(1 x), 1 (k 1) x kx 2, $ 1 ( k 1) x, , … (i), , [using (i)], [a kx 2 $ 0]., , , , P(k 1): (1 x) k 1 $ 1 (k 1) x., Thus, P(k 1) is true, whenever P(k) is true., , P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, we have, (1 x) n $ (1 nx), where x 1 for all nd N., , EXAMPLE 22, SOLUTION, , Using the principle of mathematical induction, prove that n 2 n for all, nd N., Let the given statement be P(n) . Then,, P(n): n 2 n ., When n 1, we get LHS = 1 and RHS 21 2., Clearly, 1 < 2., , P(n) is true for n 1, i.e., P(1) is true., Let P(k) be true. Then,, P(k): k 2 k ., , , Now, k 2 k & 2k 2 k 1, , & (k k ) 2 k 1, , & (k 1) # (k k) 2 k 1, k1, , & (k 1) 2 ., , , P(k 1): (k 1) 2 k 1 ., Thus, P(k 1) is true, whenever P(k) is true., , [a 1 # k]
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Principle of Mathematical Induction, , 145, , , P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, it follows that, n 2 n for all nd N., EXAMPLE 23, , SOLUTION, , Using the principle of mathematical induction, prove that, (2n 7) (n 3) 2 for all values of nd N., Let the given statement be P(n) . Then,, P(n): (2n 7) (n 3) 2 ., When n 1, we have, 2, 2, LHS (2 #1 7) 9 and RHS (1 3) 4 16., , LHS < RHS, [a 9 16]., Thus, P(n) is true for n 1, i.e., P(1) is true., Let P(k) be true. Then,, … (i), P(k): (2k 7) (k 3) 2 ., Now, 2(k 1) 7 (2k 7) 2, (k 3) 2 2 (k 2 6k 11), (k 2 8k 16) (k 4) 2 (k 1 3) 2 ., , [using (i)], , , P(k 1): 2(k 1) 7 (k 1 3) 2 ., Thus, P(k 1) is true, whenever P(k) is true., , P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, it follows that, (2n 7) (n 3) 2 for all nd N., EXAMPLE 24, , Using the principle of mathematical induction, prove that, n3, for all values of nd N., ( 1 2 2 2 … n 2) , 3, , SOLUTION, , Let P(n): (1 2 2 2 … n 2) , , n3 ·, 3, , When n 1, LHS 1 2 1 and RHS , , 13 1 ·, 3 3, , 1, Since 1 , it follows that P(1) is true., 3, Let P(k) be true. Then,, P(k): (1 2 2 2 … k 2) , , k3 ·, 3, , … (i), , Now, 1 2 2 2 … k 2 (k 1) 2, { 1 2 2 2 … k 2 } ( k 1) 2, k3 2, (k 1), 3, 1 · {k 3 3(k 1) 2} 1 · {k 3 3k 2 6k 3}, 3, 3, , , , [using (i)]
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146, , Senior Secondary School Mathematics for Class 11, , 1 [{(k 3 1 3k (k 1)} (3k 2)] 1 · [(k 1) 3 (3k 2)], 3, 3, 1 3, (k 1) ., 3, 1, , P(k 1): 1 2 2 2 … k 2 (k 1) 2 (k 1) 3 ., 3, Thus, P(k 1) is true, whenever P(k) is true., , P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, it follows that, (1 2 2 2 3 2 … n 2) , , n3, for all nd N., 3, , EXAMPLE 25, , Using the principle of mathematical induction, prove that, 1, (1 2 3 … n) (2n 1) 2 for all values of nd N., 8, , SOLUTION, , Let the given statement be P(n) . Then,, P(n): (1 2 3 … n) , , 1, (2n 1) 2 ., 8, , When n 1, we have, LHS = 1 and RHS , , Clearly, 1 , , 9, 1, (2 #1 1) 2 ·, 8, 8, , 9, ·, 8, , , P(1) is true., Let P(k) be true. Then,, P(k): (1 2 3 … k) , , 1, (2k 1) 2 ., 8, , … (i), , Now, 1 2 3 … k (k 1), { 1 2 3 … k } ( k 1), , (2k 1) 2 8 (k 1), 1, 2, [using (i)], (2k 1) (k 1) , 8, 8, 2, 2, 2, (4k 12k 9) (2k 3), {2(k 1) 1}, , , , ·, 8, 8, 8, {2(k 1) 1} 2, ·, , {1 2 3 … k (k 1)} , 8, This shows that P(k 1) is true., Thus, P(k 1) is true, whenever P(k) is true., , P(1) is true and P(k 1) is true, whenever P(k) is true., Hence, by the principle of mathematical induction, it follows that, 1, (1 2 3 … n) (2n 1) 2 for all values of nd N., 8,
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Principle of Mathematical Induction, , 147, , EXERCISE 4, Using the principle of mathematical induction, prove each of the following for, all nd N:, 1, 1. 1 2 3 4 … n n(n 1) ., 2, 2. 2 4 6 8 … 2n n(n 1) ., 1 (3 n 1) ., 2, , 4. 2 6 18 … 2 · 3 n 1 (3 n 1) ., 3. 1 3 3 2 3 3 … 3 n, , 5., , 1, , 111 1 1 ·, …, d1, nn, 2 4 8, 2n, 2, , n(2n 1)(2n 1), ·, 3, , 7. 1 · 2 2 · 2 2 3 · 2 3 … n · 2 n (n 1) 2 n 1 2., 6. 1 2 3 2 5 2 7 2 … (2n 1) 2 , , 12 (6 n 1) ., 5, 1, 1, 1, , …, 2n ·, 9. 1 , ( 1 2) ( 1 2 3), (1 2 3 … n) (n 1), 8. 3 · 2 2 3 2 · 2 3 3 3 · 2 4 … 3 n · 2 n, , 1, , 10., , n, 1 1 1 , 1, , ·, …, 2 · 5 5 · 8 8 · 11, (3n 1)(3n 2) (6n 4), , 11., , n, 1 1 1 , 1, , ·, …, 1 · 4 4 · 7 7 · 10, (3n 2)(3n 1) (3n 1), , 12., , n, 1 1 1 , 1, , ·, …, 1· 3 3 · 5 5 ·7, (2n 1)(2n 1) (2n 1), , 13., , n, 1 1 1 , 1, , ·, …, 2 · 5 5 · 8 8 · 11, (3n 1)(3n 2) (6n 4), , (2n 1), 3, 5, 7, 3 (n 1) 2 ., 14. b1 lb1 lb1 l … )1 , 2, 9, 1, 4, n, 1, 1, 1, 1, 15. b1 lb1 lb1 l … b1 n l (n 1) ., 2, 3, 1, 16. n(n 1)(n 2) is a multiple of 6., 17. (x 2n y 2n) is divisible by (x y) ., 18. (x 2n 1) is divisible by (x 1), where x ! 1., 19. {(41) n (14) n} is divisible by 27., 20. (4 n 15n 1) is divisible by 9., 21. (3 2n, , 2, , 8n 9) is divisible by 8., , 22. (2 1) is a multiple of 7., 3n, , 23. 3 n $ 2 n
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148, , Senior Secondary School Mathematics for Class 11, HINTS TO SOME SELECTED QUESTIONS, , 6. P(k): 1 2 3 2 5 2 … (2k 1) 2 , , k (2k 1)(2k 1), ·, 3, , Now, {1 2 3 2 5 2 … (2k 1) 2} {2(k 1) 1} 2, k (2k 1) (2k 1), (2k 1) 2 1 · {k (2k 1) (2k 1) 3 (2k 1) 2}, 3, 3, 1 (2k 1) {k (2k 1) 3 (2k 1)} 1 (2k 1) (2k 2 5k 3), 3, 3, 1, (k 1) (2k 1) (2k 3) ., 3, , , 7. P(k): 1 · 2 2 · 2 2 3 · 2 3 … k · 2 k (k 1) · 2 k, , 1, , Now, {1 · 2 2 · 2 3 · 2 … k · 2 } (k 1) · 2, 2, , 3, , k, , 2., , … (i), , k1, , {(k 1) · 2 k 1 2} (k 1) · 2 k 1 2k · 2 k 1 2 k · 2 k 2 2., 8. P (k): 3 · 2 2 3 2 · 2 3 … 3 k · 2 k, , 1, , Now, (3 · 2 2 3 2 · 2 3 … 3 k · 2 k, , 12 (6 k 1) ., 5, , 1, , ) 3k, , 1, , · 2k, , 2, , '12 (6 k 1) 3 k 1 · 2 k 1 · 21 1 · {12 (6 k 1) 10 # 6 k 1}, 5, 5, 1, 12, k1, k1, k1, · {2 (6, 6) 10 # 6 } , 1) ., (6, 5, 5, ]Z] (2k 1) b_b, 5, 7, 2, 14. P (k): (1 3) c1 mc1 m … []1 , `b (k 1) ., 9, 4, ], k2 b, \, a, Z, _, 5, 2k 1 ]] 2 (k 1) 1 bb, , , , c, m, Now, (1 3) 1, … e1, [1, o, ], b`, 2, 2, 4, ], k, (k 1) b, \, a, (k 1) 2 (2k 3), (k 1) 2 #, (k 2 4k 4) (k 2) 2 ., (k 1) 2, 1, 1, 1, 15. P (k): (1 1) c1 mc1 m … d1 n (k 1) ., 2, 3, k, 1, 1, 1, 1, Now, )(1 1) c1 mc1 m … d1 n3d1 , n, 2, 3, k, k1, (k 2), (k 1) d1 1 n (k 1) ·, (k 2) ., (k 1), k1, 16. Let k (k 1) (k 2) 6p, , … (i)., , Then,, , (k 1) (k 2) (k 3) (k 1) (k 2) k 3 (k 1) (k 2) k (k 1) (k 2) 3 (k 1) (k 2), 6p 3 (k 1) (k 2), [using (i)], 6p 6q 6 (p q), where (k 1) (k 2) 2q, {a (k 1) and (k 2) being consecutive integers, one, of them must be even and so their product is even}., 17. Let, , (x 2n y 2n), p, (x y), , {x 2 (n, , 1), , … (i)., , Then,, , y 2 (n 1)} (x 2n $ x 2 y 2n $ y 2) x 2n $ x 2 x 2 y 2n x 2 y 2n y 2n $ y 2
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Principle of Mathematical Induction, x 2 (x 2n y 2n) y 2n (x 2 y 2) x 2 p (x y) y 2n (x 2 y 2), , 149, [using (i)], , (x y) [x 2 p y 2n], which is divisible by (x y) ., 18. Let, {x, , (x 2n 1), p., (x 1), , 2 ( n 1), , … (i)., , Then,, , 1} (x 2n · x 2 x 2 x 2 1) x 2 (x 2n 1) (x 2 1), x 2 · p (x 1) (x 2 1), , [using (i)], , (x 1) (px x 1), which is divisible by (x 1) ., 2, , (41) (14) k, p., 27, k, , 19. Let, , {(41), , k1, , (14), , … (i)., , k1, , } (41), , k1, , Then,, , (41) · 14 (41) k · 14 (14) k 1, k, , (41) k (41 14) 14 {(41) k (14) k} 27 #(41) k 14 # 27p, 27 #[(41) k 14] ., 20. Let (4 k 15k 1) 9p, {4, , (k 1), , … (i)., , 15 (k 1) 1} {4, , (k 1), , Then,, , 15k 14} (4 · 4 k 60k 4) 45k 18, , 4 (4 k 15k 1) 9 (5k 2) (4 # 9p) 9 (5k 2), , [using (i)], , 9 #(4p 5k 2), which is divisible by 9., 21. Let (3 2k, {3, , 2, , 2 (k 1) 2, , 8k 9) 8p, , … (i)., , Then,, , 8 (k 1) 9} {3(2k 2) · 3 2 8k 17} (9 · 3(2k 2) 72k 81) (64k 64), 9 · {3(2k 2) 8k 9} 64 (k 1), (9 # 8p) 64 (k 1), 8 #[9p 8 (k 1)], which is divisible by 8., , 22. Let (2 3k 1) 7p, {2, , 3 (k 1), , … (i)., , Then,, , 1} (2 · 2 1) (8 · 2 8) 7 {8 (2 3k 1) 7} (8 #7p) 7, (56p 7) 7 (8p 1), which is divisible by 7., 3k, , 3, , [using (i)], , 3k, , [using (i)], , 23. Clearly, 31 $ 21 . So, the result is true for n 1., Let it be true for n k. Then,, , , 3 k $ 2 k and 3 2 & 3 k · 3 $ 2 k · 2 & 3 k 1 $ 2 k 1 ., So, whenever the result is true for k, then it is also true for (k 1) ., , SUMMARY OF KEY FACTS, PRINCIPLE OF MATHEMATICAL INDUCTION, , Let P(n) be a statement involving the natural number n such that, (i) P(1) is true and (ii) P(k 1) is true, whenever P(k) is true then P(n) is true for, all nd N.,
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150, , Senior Secondary School Mathematics for Class 11, , 5, Complex Numbers and, Quadratic Equations, IMAGINARY NUMBERS, , If the square of a given number is negative then such a number is called an imaginary, number., For example, 1 , 2 , etc., are imaginary numbers., We denote 1 by the Greek letter iota ‘’, which is transliterated as ‘i ’., Thus, 4 2i, 9 3i and, POWERS OF i, , 5 i 5 , etc., , We have, , i 1, i 1 i, i 2 1, i 3 i 2 # i (1)# i i, 0, , and I 4 i 2 # i 2 (1)#(1) 1., Thus, we have, i 0 1, i 1 i, i 2 1, i 3 i, i 4 1., Let us consider i n, where n is a positive integer and n 4., On dividing n by 4, let the quotient be m and the remainder be r. Then,, n 4m r, where 0 # r 4., , , i n i 4m, , EXAMPLES, , r, , i 4 m # i r ( i 4) m # i r i r ., , (i) i 98 i 4 # 24, (ii) i, , 98, , 2, , [a i 4 1], , (i 4 ) 24 # i 2 i 2 1. [a i 4 1], , 2, 2, , 198 198 # i 2 (98i 2) 1 1. [a i 100 1 and i 2 1], 1, i, i, i, i, , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Evaluate:, (i) i 23, (v) ( 1 ) 91, We have, (i) i 23 i (4, , (ii) i 998, , (vi) (i 37 # i 61), # 5) 3, , (ii) i 998 i 4 # 249, , , , (iii) i 998, , (vii) i 1, , (iv) i, , 71, , (i 4) 5 # i 3 i 3 i. [a i 4 1], , 2, , (i 4) 249 # i 2 (1# i 2) i 2 1. [a i 4 1], 150
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Complex Numbers and Quadratic Equations, , 151, , 2, 2, 1, i, i, 1, 998, 1. [a i 1000 (i 4) 250 1], # 2 1000 , 1, i, i, i, i, i, i, i, 1, , (iv) i 71 71 # 72 4 18 i. [a i 4 1], 1, i i, i, (i ), , (iii) i, , 998, , , (v) ( 1 )91 i 91 i 4 # 22 3 (i 4 ) 22 # i 3 1#( i ) i. [a i 3 i ], , (vi) i 37 i 4 # 9, i, , 61, , , (vii) i, , 1, , 1, , (i 4 )9 # i 1# i i., , 3, 3, , , 161 # i 3 i64 4 i16 i i., 1, i, i, i, (i ), , (i 37 i, , 61, , ) i ( i ) 0., , 3, , , 1 1 # i 3 4i i i., 1, i i i, i, , EXAMPLE 2, , Prove that:, , , , (i) i n i n 1 i n 2 i n 3 0, 107, 112, 117, 122, (ii) i i i i 0, 14, (iii) (1 i ) 4 #b1 l 16., i, , SOLUTION, , We have, , , , (i) i n i n 1 i n 2 i n 3, n, 2, 3, i (1 i i i ), i n ( 1 i 1 i ) ( i n # 0) 0., , [a i 2 1 and i 3 i ], , (ii) i 107 i 112 i 117 i 122, i 107 (1 i 5 i 10 i 15 ) i 107 (1 i 4 # i i 8 # i 2 i 12 # i 3 ), i 107 (1 i i 2 i 3 ), , [a i 4 1, i 8 1 and i 12 1], , i 107 (1 i 1 i ) (i 107 # 0) 0., , [a i 2 1, i 3 i], , 14, (iii) (1 i ) 4 #b1 l, i, 4, (1 i ) 4 #b1 1 # i l (1 i ) 4 (1 i ) 4, [a i 2 1], i i, {(1 i ) (1 i )} 4 (1 i 2) 4 {1 ( 1)} 4 2 4 16., EXAMPLE 3, SOLUTION, , 1 29 2, Show that (i 23 b l 2 4., i, We have, # , i 23 i (4 5 3) (i 4 ) 5 # i 3 1#( i ) i. [a i 4 1 and i 3 i], 3, 3, i, i, i, 1 29 1, 1, i. [a i 3 i and i 32 1], b l 29 29 # 3 32 , 1, i, i, i, i, i, , , , 2, , 1 29, (i 23 b l 2 ( i i ) 2 ( 2i ) 2 4i 2 4 #(1) 4., i
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152, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 4, , Simplify:, 1, (i) ( 2i ) b il, 6, , SOLUTION, , (ii) ( i ) (3i ) b, , 1 3, il, 6, , (iii) 4 4 5 9 3 16, , We have, , 1, 1, 1, 1, # ( 1) ·, (i) ( 2i ) b il b 2 # l# i 2 , 6, 6, 3, 3, 1 3, , 1 3, (ii) ( i ) (3i ) b il ( 3i 2 ) b, il, 6, 216, , ( 3)#(1) ; 1 #( i )E, 216, b3 # 1 # il 1 i., 72, 216, , (iii) 4 4 5 9 3 16, (4 # 2i ) (5 # 3i ) (3 # 4i ) [a, (8i 15i 12i) (11) i., EXAMPLE 5, SOLUTION, , [a i 3 i ], , 4 2i, 9 3i, 16 4i ], , , Show that ( 1 ) 4n 3 i, where n is a positive integer., We have, , , ( 1 ) 4n 3 ( i ) 4n 3, [a 1 i ], 4n, ( i ) # ( i ) 3, , {( i ) 4} n #( i 3 ) (1# i ) i., [a ( i ) 4 1 and i 3 ( i ) i ], 4n 3, i., Hence, ( 1 ), EXAMPLE 6, SOLUTION, , Show that the sum (1 i 2 i 4 … i 2n) is 0 when n is odd and 1 when n, is even., Let S 1 i 2 i 4 … i 2n., This is clearly a GP having (n 1) terms with a 1 and r i 2 1., , , , , S, , , , n 1, 2 n 1, a (1 r ) 1#{1 (i ) }, , 2, (1 r), (1 i ), , n1, n1, {1 (1) } {1 (1) }, , 2, 1 (1), Z]1, ]] (1 1) 0, when n is odd, ]][2, ]]1, ]] (1 1) 1, when n is even., ]2, \, , , , AN IMPORTANT RESULT, , For any two real numbers a and b, the result a # b ab is true only, when at least one of the given numbers is either zero or positive., Thus, 2 # 3 ( 2)#( 3) 6 is wrong., In fact, 2 # 3 (i 2 ) (i 3 ) i 2 # 6 6 .
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Complex Numbers and Quadratic Equations, , 153, , EXAMPLE 7, , Explain the fallacy:, 1 (i # i ) 1 # 1 (1)#(1) 1 1., , SOLUTION, , We know that for any real numbers a and b, a # b ab is true, only when at least one of a and b is either 0 or positive., 1 # 1 ! ( 1) # ( 1 ) ., , , EXAMPLE 8, , Evaluate:, (i) 25 # 49, (ii) 36 # 16, We have, (i) 25 # 49 (5i )#(7i ) (35 # i 2) 35 #(1) 35., , SOLUTION, , 36 # 16 (6i )# 4 24i., , (ii), , 16 3 25 36 625 ., Evaluate, We have, 16 3 25 36 625, (4i 3 # 5i 6i 25i ) (4i 15i 6i 25i ) 0., , EXAMPLE 9, SOLUTION, , EXERCISE 5A, 1. Evaluate:, , 19, , (ii) i, (ii) ( 1 ) 93, , 50, , (ii) i, , 2. Evaluate:, , (i) i, (i) ( 1 )192, , 3. Evaluate:, , (i) i, , 4. Evaluate:, , (i) di 41 , , 1, n, i 71, , 62, , (iii) i, (iii) ( 1 ) 30, , 9, , (iii) i, , (ii) di 53 , , 373, , 131, , 1, 53 n, i, , Prove that:, 2, 4, 6, 5. 1 i i i 0., , 7., , 6. 6i, , 11 1 1 , 0., i i2 i3 i4, 46 2, , 21, 9. (i b l 2 2i., , 1, i, , 50, , 5i 33 2i 15 6i 48 7i., , 10, 20, 30, 8. (1 i i i ) is a real number., 25 3, , 18, 10. (i b l 2 2(1 i )., , 1, i, , n, n, 11. (1 i ) b1 l 2 for all values of nd N., , 1n, i, 16 3 25 36 625 0., , 12., , 2, 4, 6, 8, 20, 13. (1 i i i i … i ) 1., , 14. i, , 53, , i 72 i 93 i 102 2i., , 13, , n, n1, 15. / (i i ) (1 i), nd N ., n1, , ANSWERS (EXERCISE 5A), , 1. (i) i, 3. (i) –1, , (ii) –1, (ii) i, , (iii) i, (iii) i, , 2. (i) 1, 4. (i) 2i, , (ii) i (iii) –1, (ii) 0
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154, , Senior Secondary School Mathematics for Class 11, HINTS TO SOME SELECTED QUESTIONS, , 7., , 3, 1 1 i 3 1 1 1 1 i , 1, #, #, i, i, 2 , 1, 3, i and 4 1., 1, i i i3, i, i, i3 i, i, 1 1 1 1, ) 2 3 4 3 ( i 1 i 1) 0., i i, i, i, , 8. (1 i 10 i 20 i 30) {1 i 4, , #2 2, , i 4 # 5 i 4 # 7 2}, , {1 (i 4 ) 2 # i 2 (i 4 ) 5 (i 4 )7 # i 2}, (1 i 2 1 i 2) (1 1 1 1) 0., 11., , 3, 3, 1 1 i i 3, #, i, i., i i i3 i4, , , , 1n, (1 i ) n c1 m (1 i ) n (1 i ) n, i, , ;a, , 1 , iE, i, , {(1 i ) (1 i )} n (1 i 2 ) n 2 n., , [a i 2 1], , 13. This is a GP in which a 1, r i 2 1 and n 11., , , S, , a (1 r n) 1#{1 (1) 11} 2, , 1., 2, (1 r), {1 (1)}, , 14. i 53 i 72 i 93 i 102 (i 4 ) 13 # i (i 4 ) 18 (i 4 ) 23 # i (i 4 ) 25 # i 2, (1# i ) 1 (1# i ) 1#(1) i 1 i 1 2i., 13, , 15. / (i n i n, n1, , 1, , 13, , 13, , ) / i (1 i ) (1 i ) · / i, n, , n1, , n, , n1, , (1 i ) $ {i i 2 i 3 … i 13}, (1 i ), , a (1 r n ), , where a i, r i and n 13, (1 r), , (1 i ) ·, , i (1 i 13 ), (1 i ), i (1 i ) ·, (1 i ) . [a i 13 (i 4 ) 3 # i 1# i i ], (1 i ), (1 i ), , COMPLEX NUMBERS, The numbers of the form (a ib), where a and b are real numbers, and i 1 , are known as complex numbers. The set of all complex numbers is, denoted by C., C {(a ib) : a, bd R} ., , COMPLEX NUMBERS, , 2 5, Each of the numbers (5 8i ), ( 3 2 i ) and b il is a complex, 3 7, number., For a complex number, z (a ib), we have, a real part of z, written as Re(z), , EXAMPLES, , and b imaginary part of z, written as Im(z).
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Complex Numbers and Quadratic Equations, EXAMPLES, , 155, , (i) If z (5 9i ) then Re(z) 5 and Im(z) 9., 2, 2, (ii) If z b 3il then Re(z) and Im(z) 3., 3, 3, 5, 5, (iii) If z b 7 il then Re(z) 7 and Im(z) ·, 8, 8, , PURELY REAL AND PURELY IMAGINARY NUMBERS, , A complex number z is said to be, (i) purely real, if Im(z) 0,, (ii) purely imaginary, if Re(z) 0., Thus, each of the numbers 2, 7, 3 is purely real., 3, And, each of the numbers 2i, ( 3 i ), b il is purely imaginary., 2, CONJUGATE OF A COMPLEX NUMBER, , Conjugate of a complex number z (a ib) is defined as, z (a ib) ., EXAMPLES, , (i) (3 8i ) (3 8i ) (ii) ( 6 2i ) ( 6 2i ) (iii) 3 3., , MODULUS OF A COMPLEX NUMBER, , Modulus of a complex number z (a ib), denoted by|z|, is defined as|z| a 2 b 2 ., EXAMPLES, , (i) If z (2 3i ) then|z| 2 2 3 2 13 ., (ii) If z ( 5 4i ) then|z| ( 5) 2 ( 4) 2 41 ., If z1 a1 ib1 and z 2 a 2 ib 2 then, z1 z 2 + a1 a 2 and b1 b 2 ., , EQUALITY OF COMPLEX NUMBERS, , EXAMPLE, , If 2y (3x y) i (5 2i ), find the values of x and y., , SOLUTION, , Equating the real and imaginary parts, we get, 2y (3x y) i (5 2i ), +, 2y 5 and 3x y 2, 5, 5, and 3x 2, 2, 2, 5, 1, +, y, and x ·, 2, 6, 5, 1, Hence, x and y ·, 6, 2, +, , y, , SUM AND DIFFERENCE OF COMPLEX NUMBERS, , If z1 (a1 ib1) and z 2 (a 2 ib 2) then we define:, (i) z1 z 2 (a1 a 2) i (b1 b 2), (ii) z1 z 2 (a1 a 2) i (b1 b 2) .
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Complex Numbers and Quadratic Equations, , 157, , (a1 ib1) {(a 2 ib 2) (a 3 ib 3)}, z 1 ( z 2 z 3) ., (z1 z 2) z 3 z1 (z 2 z 3) for all z1, z 2, z 3 dC., , , Hence, addition of complex numbers is associative., (iv) EXISTENCE OF ADDITIVE IDENTITY, , For any complex number z, prove that, z 0 0 z z., PROOF, Let z (a ib) and we may write, 0 (0 i 0) . Then,, z 0 (a ib) (0 i0) (a 0) i (b 0) (a ib), and 0 z (0 i0) (a ib) (0 a) i (0 b) (a ib) ., z 0 0 z z for all values of xdC., Thus, 0 is the additive identity for complex numbers., (v) EXISTENCE OF ADDITIVE INVERSE, , For every complex number z, prove that, z ( z) ( z) z 0., PROOF, Let z (a ib) . Then, ( z) ( a) i ( b) ., z ( z) (a ib) {( a) i ( b)}, {a ( a)} i {b ( b)} (0 i0) ., Similarly, ( z) z (0 i0) 0., Hence, z ( z) ( z) z 0., Thus, every complex number z has ( z) as its additive inverse., MULTIPLICATION OF COMPLEX NUMBERS, , Let z1 (a1 ib1) and z 2 (a 2 ib 2) . Then, we define, z1 z 2 (a1 ib1) (a 2 ib 2) (a1 a 2 b1 b 2) i (a1 b 2 b1 a 2) ., , , z1 z 2 {Re (z1) · Re (z 2) Im (z1) · Im (z 2)}, i {Re (z1) · Im (z 2) Re (z 2) · Im (z1)} ., , EXAMPLES, , (i) Let z1 (3 2i ) and z 2 (5 4i ) . Then,, z1 z 2 {(3 # 5) (2 # 4)} i {(3 # 4) (2 # 5)}, (15 8) i (12 10) (7 22i ) ., (ii) Let z1 ( 2 3i ) and z 2 (7 5i ) . Then,, z1 z 2 {( 2)#7 3 #( 5)} i {( 2)#( 5) 3 #7}, (14 15) i (10 21) (1 31i ) ., , PROPERTIES OF MULTIPLICATION OF COMPLEX NUMBERS, (i) CLOSURE PROPERTY, , number., PROOF, , The product of two complex numbers is always a complex, , Let z1 (a1 ib1) and z 2 (a 2 ib 2) be any two complex numbers., Then, a1, a 2 and b1, b 2 are real numbers.
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158, , Senior Secondary School Mathematics for Class 11, , z1 z 2 (a1 ib1) (a 2 ib 2), (a1 a 2 b1 b 2) i (a1 b 2 b1 a 2), which is a complex number., Thus, the product of two complex numbers is always a complex, number., , , (ii) COMMUTATIVE LAW, , For any two complex numbers z1 and z 2, prove that, z1 z 2 z 2 z1 ., PROOF, , Let z1 (a1 ib1) and z 2 (a 2 ib 2) . Then,, z1 z 2 (a1 ib1) (a 2 ib 2) (a1 a 2 b1 b 2) i (a1 b 2 b1 a 2), ( a 2 a 1 b 2 b 1) i ( b 2 a 1 a 2 b 1 ), [by commutativity of multiplication on R], (a 2 ib 2) (a1 ib1) z 2 z1 ., z1 z 2 z 2 z1 for all z1, z 2 dC., Hence, multiplication of complex numbers is commutative., , (iii) ASSOCIATIVE LAW, , For any three complex numbers z1, z 2 and z 3, prove that, ( z 1 z 2) z 3 z 1 ( z 2 z 3 ) ., PROOF, , Let z1 (a1 ib1), z 2 (a 2 ib 2) and z 3 (a 3 ib 3) be any three complex, numbers. Then a1, a 2, a 3 and b1, b 2, b 3 are all real numbers., (z1 z 2) z 3 {(a1 ib1) (a 2 ib 2)} (a 3 ib 3), {(a1 a 2 b1 b 2) i (a1 b 2 b1 a 2)} (a 3 ib 3), {(a1 a 2 b1 b 2) a 3 (a1 b 2 b1 a 2) b 3}, i {(a1 a 2 b1 b 2) b 3 (a1 b 2 b1 a 2) a 3}, {a1 (a 2 a 3 b 2 b 3) b1 (a 2 b 3 b 2 a 3)}, i {a1 (a 2 b 3 a 3 b 2) b1 (a 2 a 3 b 2 b 3)}, (a1 ib1) {(a 2 a 3 b 2 b 3) i (a 2 b 3 a 3 b 2)}, (a1 ib1) {(a 2 ib 2) (a 3 ib 3)} z1 (z 2 z 3) ., Thus, (z1 z 2) z 3 z1 (z 2 z 3) for all z1, z 2, z 3 dC., Hence, multiplication of complex numbers is associative., , (iv) EXISTENCE OF MULTIPLICATIVE IDENTITY, , The complex number (1 i0) is the multiplicative identity in C., Let z (a ib) . Then,, z #1 (a ib) (1 i0) {(a #1) (b # 0)} i {(a # 0) (b #1)} (a ib) z., Similarly, 1# z z., Thus, (z #1) (1# z) z for all zdC., Hence, the complex number 1 (1 i0) is the multiplicative identity.
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Complex Numbers and Quadratic Equations, , 159, , (v) EXISTENCE OF MULTIPLICATIVE INVERSE, , Let z (a ib) . Then,, z, , 1, , (a ib), (a ib), 1 1, 1, , ·, z (a ib) (a ib) # (a ib) (a 2 b 2), , Clearly, z # z, , 1, , z 1 # z 1., , Thus, every z (a ib) has its multiplicative inverse, given by, RS, V, (a ib), z ·, 1, , SSRemember: z 1 z WWW, , z 1 z 2, 2, 2, SS, |z|2WW, (a b ) |z|, T, X, zz |z|2., THINGS TO REMEMBER, , (i) z (a ib) & z (a ib) and|z| 2 (a 2 b 2) ., (ii) z (a ib) & z, , 1, , (a ib), z2 2, ·, 2, |z| (a b ), , (vi) DISTRIBUTIVE LAWS, , For complex numbers z1, z 2, z 3, prove that, z1 · (z 2 z 3) z1 z 2 z1 z 3 ., (z1 z 2) · z 3 z1 z 3 z 2 z 3 ., PROOF, Let z1 (a1 ib1), z 2 (a 2 ib 2) and z 3 (a 3 ib 3) be any three complex, numbers. Then, a1, a 2, a 3 and b1, b 2, b 3 are real numbers., z1 · (z 2 z 3), (a1 ib1) [(a 2 ib 2) (a 3 ib 3)], (a1 ib1) [(a 2 a 3) i (b 2 b 3)], {a1 (a 2 a 3) b1 (b 2 b 3)} i {a1 (b 2 b 3) b1 (a 2 a 3)}, {(a1 a 2 b1 b 2) i (a1 b 2 b1 a 2)} {(a1 a 3 b1 b 3) i (a1 b 3 b1 a 3)}, z1 z 2 z1 z 3 ., Thus, z1 (z 2 z 3) z1 z 2 z1 z 3 for all z1, z 2, z 3 dC., Similarly, we can prove that, (z1 z 2) z 3 z1 z 3 z 2 z 3 ., DIVISION OF TWO COMPLEX NUMBERS, , Let z1 and z 2 be complex numbers such that z 2 ! 0., z1, 1, Then, z z1 · z z1 z 21 ., 2, 2, z1, EXAMPLE, Find z , when z1 (6 3i ) and z 2 (3 i ) ., 2, z1, SOLUTION, We have z z1 z 21 ., 2
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160, , Senior Secondary School Mathematics for Class 11, , Now, z 21 , , z2, , , , 2, , |z 2|, , (3 i ), |3 i|2, , , , (3 i ), {3 (1) 2}, 2, , , , (3 i ), ·, 10, , z1, ( 3 i ) ( 6 3i ) ( 3 i ), 1, , (6 3i ) ·, , z 2 z1 · z 2, 10, 10, , , , 15 (1 i ) 3 (1 i ), , , 15 15i , ·, 2, 10, 10, SOME IDENTITIES ON COMPLEX NUMBERS, THEOREM 1, , For any complex numbers z1 and z 2, prove that:, (i) (z1 z 2) 2 z 12 z 22 2z1 z 2, (ii) (z1 z 2) 2 z 12 z 22 2z1 z 2, (iii) (z 12 z 22) (z1 z 2) (z1 z 2), (iv) (z1 z 2) 3 z 13 z 23 3z1 z 2 (z1 z 2), (v) (z1 z 2) 3 z 13 z 23 3z1 z 2 (z1 z 2), , PROOF, , We have, (i) (z1 z 2) 2 (z1 z 2) (z1 z 2), (z1 z 2) z1 (z1 z 2) z 2, , , , , , z 12, , z 2 z1 z1 z 2 , , , , z 12, , 2z1 z 2 , , , , z 12, , , , z 22, , (z1 z 2) , 2, , z 22, , z 22 2z1 z 2 ., , (ii) (z1 z 2) 2 (z1 z 2) (z1 z 2), (z1 z 2) z1 (z1 z 2) z 2, , , , , , z 12, , z 2 z1 z1 z 2 , , , , z 12, , 2z1 z 2 , , , , 2, z1, , , , (z1 z 2) , 2, , z 22, , z 22, , , , , , , 2, (z 1, , 2, (z 1, , , , , , 2, z 2), , [by distributive law], [a z 2 z1 z1 z 2], , z 22 2z1 z 2 ., , z 2 z1 z1 z 2 , 2, z 2), , [by distributive law], , 2z1 z 2 ., , 2, z1, , (iii) RHS (z1 z 2) (z1 z 2), (z1 z 2) z1 (z1 z 2) z 2, 2, z1, , [by distributive law], [a z 2 z1 z1 z 2], , z 22, , 2z1 z 2., , z 12, , 2, z2, , [by distributive law], , 2, z2, , LHS., , (z1 z 2) (z1 z 2) ., , [by distributive law], [by distributive law], [a z 2 z1 z1 z 2]
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Complex Numbers and Quadratic Equations, , 161, , (iv) (z1 z 2) 3 (z1 z 2) 2 · (z1 z 2), (z 12 z 22 2z1 z 2) (z1 z 2), , , (z 12, , , , z 22, , 2z1 z 2) z1 , , (z 12, , , , [expanding (z1 z 2) 2], z 22, , 2z1 z 2) z 2, , [by distributive law], z 13 z 22 z1 2z 12 z 2 z 12 z 2 z 23 2z1 z 22, z 13 z 23 3z 12 z 2 3z1 z 22 z 13 z 23 3z1 z 2 (z1 z 2) ., , , , , (z1 z 2) 3 z 13 z 23 3z1 z 2 (z1 z 2) ., , (v) LHS (z1 z 2) 3, (z1 z 2) 2 (z1 z 2) (z 12 2z1 z 2 z 22) (z1 z 2), (z 12 2z1 z 2 z 22) z1 (z 12 2z1 z 2 z 22) z 2, z 13 2z 12 z 2 z 22 z1 z 12 z 2 2z1 z 22 z 23, z 13 3z 12 z 2 3z1 z 22 z 23 RHS., This may be written as (z1 z 2) 3 z 13 z 23 3z1 z 2 (z1 z 2) ., SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Simplify:, (i) 3(6 6i ) i (6 6i ), , (ii) (1 i ) ( 3 6i ), , 3, 1 2, (iii) b il b4 il, 2, 3 3, , 4, 1 7, 1, (iv) (b il b6 il2 b il, 5 5, 5, 5, , We have, (i) 3(6 6i ) i (6 6i ) 18 18i 6i 6i2 18 24i 6 12 24i., (ii) (1 i ) ( 3 6i ) 1 i 3 6i 4 7i., 3, 3, 1 2, 1 2, (iii) b il b4 il i 4 i, 2, 2, 3 3, 3 3, , b1 4l b 2 3 l i 11 13 i., 3, 3 2, 3, 6, 4, 1 7, 1, (iv) (b il b6 il2 b il, 5 5, 5, 5, 4, , 1, 7, 1, b i 6 il b il (b1 6l b 7 1 l i2 b 4 il, 5 5, 5, 5, 5, 5 5, 5, , , 29 6 i 4 i b 29 4 l b 6 i il b 5 1 il ·, 5, 5, 5, 5, 5, 5, 5, , EXAMPLE 2, , Express ( 3 2 ) (2 3 i ) in the form (a ib) ., , SOLUTION, , We have, ( 3 2 ) (2 3 i ) ( 3 i 2 ) ( 2 3 i ), 6 i 3 2 6 i 2 i2, ( 6 2 ) 3 (1 2 2 ) i.
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162, EXAMPLE 3, , Senior Secondary School Mathematics for Class 11, , Express each of the following in the form (a ib):, (i) (3 5 ) (3 5 ), (iii) ( 2 3i ), , SOLUTION, , (ii) ( 2 3 ) ( 3 2 3 ), (iv) ( 5 7i ) 2, , 2, , We have, (i) (3 5 ) (3 5 ) (3 5 i ) (3 5 i ) {3 2 ( 5 i ) 2}, (9 5i 2) (9 5) 14 0i., [a (z1 z 2) (z1 z 2) (z 1 z 2 )], 2, , 2, , (ii) ( 2 3 ) ( 3 2 3 ) ( 2 3 i ) ( 3 2 3 i ), (6 4 3 i 3 3 i 6i 2), 7 3 i 0 7 3 i., (iii) ( 2 3i ) ( 2) (3i ) 2 #( 2)# 3i, 2, , 2, , 2, , (4 9i 2 12i ) (4 9 12i ) ( 5 12i ) ., (iv) ( 5 7i ) 2 (5) 2 (7i ) 2 2 # 5 #7i, (25 49i 2 14 5 i ) (25 49 14 5 i ), ( 24 14 5 i ) ., EXAMPLE 4, SOLUTION, , Prove that (1 i ) 4 4., We have, (1 i ) 4 (1 i ) 2 #(1 i ) 2 (1 i 2 2i )#(1 i 2 2i ), ( 2i ) ( 2i ) 4i 2 4 #(1) 4., , EXAMPLE 5, , Express (2 3i ) 3 in the form (a ib) ., , SOLUTION, , We know that (z1 z 2) 3 z 13 z 23 3z1 z 2 (z1 z 2) ., , , [a i 2 1], , (2 3i ) 3 2 3 (3i ) 3 3 # 2 # 3i #(2 3i ), 8 27i 3 36i 54i 2, (8 27i 36i 54), ( 46 9i ) ., , [a i 3 i and i 2 1], , Hence, (2 3i ) 3 ( 46 9i ) ., EXAMPLE 6, , Express (3 5i ) 3 in the form (a ib) ., , SOLUTION, , We know that (z1 z 2) 3 z 13 z 23 3z1 z 2 (z1 z 2) ., , , ( 3 5i ) 3 3 3 ( 5 i ) 3 3 # 3 # 5 i # ( 3 5 i ), 27 125i 3 135i 225i 2, (27 125i 135i 225), (198 10i ) ., , [a i 3 i and i 2 1]
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Complex Numbers and Quadratic Equations, EXAMPLE 7, SOLUTION, , 163, , If z ( 2 3 ), find Re (z), Im (z), z and|z|., The given number may be written as, z 2 i 3., , , Re (z) 2 , Im (z) 3 , z ( 2 i 3 ) ( 2 i 3 ), , and |z|2 {( 2 ) 2 ( 3 ) 2} (2 3) 5 & |z| 5 ., EXAMPLE 8, , SOLUTION, , Write down the modulus of:, (i) 4 3, (ii) 2 5i, , (iii) i, , (iv) (1 3i ) 2, , We know that, if z (a ib) then|z| a 2 b 2 ., (i) Let z 4 3 . Then, z 4 i 3 ., |z|2 {(4) 2 ( 3 ) 2} (16 3) 19 & |z| 19 ., (ii) Let z 2 5i. Then,, |z|2 {2 2 ( 5) 2} (4 25) 29 & |z| 29 ., (iii) Let z 0 i. Then,, |z|2 {0 2 (1) 2} (0 1) 1 & |z| 1 1., (iv) Let z (1 3i ) 2. Then,, z (1) 2 (3i ) 2 2 #(1)# 3i ( 8 6i ) ., |z|2 {( 8) 2 ( 6) 2} (64 36) 100 & |z| 100 10., , EXAMPLE 9, , Write down the conjugate of each of the following:, (i) ( 5 1 ), , SOLUTION, , (ii) ( 6 3 ), , (iii) i 3, , (iv) (4 5i ) 2, , We know that the conjugate of z (a ib) is given as z (a ib) ., (i) Let z 5 1 ( 5 i ) . Then,, z ( 5 i ) ( 5 i ) ., (ii) Let z 6 3 6 i 3 . Then,, z ( 6 i 3 ) ( 6 i 3 ) ., (iii) Let z i 3 i 0 i. Then,, z (0 i ) (0 i ) i., (iv) Let z (4 5i ) 2 (4) 2 (5i ) 2 2 # 4 # 5i, (16 25 40i ) ( 9 40i )., , , EXAMPLE 10, , SOLUTION, , z ( 9 40i ) ( 9 40i ) ., , Find the multiplicative inverse of each of the following:, (ii) 4 3i, (iii) (3i 1) 2, (iv) i, (i) 5 3i, We know that the multiplicative inverse of the complex number z, z, , is given by z 1 2 ·, ||, z
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164, , Senior Secondary School Mathematics for Class 11, , (i) Let z ( 5 3i ) . Then,, 2, 2, 2, z ( 5 3i ) and |z| ( 5 ) 3 (5 9) 14., , , , z, , 1, , ( 5 3i ), 5 3, z2 , e, io ·, 14, 14, 14, |z|, , (ii) Let z 4 3i. Then,, , , , z (4 3i ) (4 3i ) and|z|2 {(4) 2 ( 3) 2} (16 9) 25., (4 3i ), z, , b 4 3 il ·, z 1 2 , 25, 25 25, |z|, , (iii) Let z (3i 1) 2 (9i 2 1 6i ) ( 9 1 6i ) ( 8 6i ) ., , , z ( 8 6i ) ( 8 6i ), , and |z|2 {( 8) 2 ( 6) 2} (64 36) 100., ( 8 6i), , , z, , b 8 6 il b 2 3 il ·, Hence, z 1 2 , 25 50, 100, 100, 100, |z|, (iv) Let z (0 i ) . Then,, 2, 2, 2, z (0 i ) i and |z| 0 (1) 1., , , EXAMPLE 11, , 1, , z 2 i i., |z| 1, , Express each of the following in the form (a ib):, (i), , SOLUTION, , z, , i, (1 i ), , (ii) (1 3 i ), , 1, , (iii), , 5 2i, 1 2i, , We have, (i), , (1 i ) i (1 i ), i i, , #, ( 1 i ) ( 1 i ) ( 1 i ) ( 1 2 i 2), , , 2, (i i ), ( i 1) ( 1 i ), , , b1 1 il ·, , , 2, 2, 2 2, 1 ( 1), , , (ii) (1 3 i ) 1 , , , , (iii), , ( 1 3 i ), 1, #, ( 1 3 i ) ( 1 3 i ), ( 1 3 i ), ( 1 3 i ) ( 1 3 i ), ( 1 3 i ), (1 3i 2), , , , , , ( 1 3 i ), 2, 2, ( 1) ( 3 i ), , ( 1 3 i ), 3, , e 1 , io ·, 4, 4, 4, , 5 2 i (5 2 i ) (1 2 i ) (5 2 i ) (1 2 i ), , , #, 2, 1 2 i (1 2 i ) (1 2 i ), (1 2 i ), (3 6 2 i ), , (1 2 2 i ) ., 3
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Complex Numbers and Quadratic Equations, EXAMPLE 12, , 3 3 2i, 1, Reduce b ld n to the form (a ib) ., 1 2i 1 i 1 3i, , SOLUTION, , We have, , 165, , 3 3 2i, 1, b ld n, 1 2i 1 i 1 3i, (1 i ) (3 6i ) 3 2i, (4 5i ) (3 2i ), 3d, ), n, #, , , , (1 2i ) (1 i ) 1 3i, (3 i ), ( 1 3i ), (4 5i ) (3 2i ) (12 10) (15 8) i (22 7i ) i, , , , #, 10i, i, ( 3 i ) ( 1 3i ), (3 3) (9i i ), , , 2, (7i 22i ), , 10i, , 2, , , 7 22i b 7 22 il b 7 11 il ·, 10, 10 10, 10 5, , EXAMPLE 13, , 1i 1i, Reduce b l to the form (a ib) and hence find its modulus., 1 i 1 i, , SOLUTION, , (1 i ) 2 (1 i ) 2 (4 #1# i ) 4i, 1i 1i, , , 2i., Let z b l , 2, 2, 1 i 1 i, ( 1 i ) (1 i ), (1 i ), Thus, z 0 2i & |z| 0 2 2 2 4 2., Hence, z 0 2i and|z| 2., 1i , (a ib) then show that (a 2 b 2) 1., 1i, , EXAMPLE 14, , If, , SOLUTION, , We have, (1 i), 1i, 1i 1i, , #, 2, 1i, 1i, 1i, 1i, (1 i ), , e 1 1 io, 2, 2, 2, , (a ib) , , 1 2 1 2 1 1 , o e o b 2 2 l 1., 2, 2, , &, , |a ib|2 e, , &, , (a 2 b 2) 1., , Hence, (a 2 b 2) 1., EXAMPLE 15, , 1i m, Find the least positive integer m for which b l 1., 1 i, , SOLUTION, , We have, ( 1 i ) ( 1 i ) ( 1 i ) 2 ( 1 2 i i 2) 2 i, 1i, , , , i., b l, #, 2, 2, 1 i, (1 i ) (1 i ) (1 i 2), 1i m, b l 1 & i m 1., , 1 i, And, we know that 4 is the least positive integer such that i 4 1 and, therefore, m 4.
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Complex Numbers and Quadratic Equations, , 167, , y, x 2 2, 2, , 2, a (a 3b ) and b (3a b ), y, y, d x n 4 (a 2 b 2) and d x n 2(a 2 b 2) ., &, a b, a b, x y, x y, Hence, (i) a 4(a 2 b 2) and (ii) a 2(a 2 b 2) ., b, b, &, , EXAMPLE 19, , Let z be a complex number such that z ! 1 and |z| 1. Then, prove that, z1, b l is purely imaginary. What will be your conclusion when z 1?, z 1, , SOLUTION, , Let z (x iy) be the given complex number such that z ! 1 and, |z| 1. Then,, … (i), |z| 1 & |z|2 & x 2 y 2 1 & x 2 y 2 1 0., , , (z 1) x iy 1 (x 1) iy (x 1) iy, z1, , , b l, #, z 1, (z 1) x iy 1 (x 1) iy (x 1) iy, {(x 1) iy} {(x 1) iy}, , {(x 1) 2 y 2}, , , , (x 2 1 y 2) i {(x 1) y (x 1) y}, {(x 1) 2 y 2}, (x 2 y 2 1) (2y) i, , {(x 1) 2 y 2}, ]Z], b_b, 2yi, b, ]][, ` [using (i)],, ]](x 1) 2 y 2 bbb, \, a, which is purely imaginary., Particular Case When z 1., In this case, z 1 & x iy 1, & (x 1) iy 0, & x 1 0, y 0 & x 1, y 0., z 1 (x iy) 1 (x 1) iy (1 1) i # 0 0 , 0., z 1 (x iy) 1 (x 1) iy (1 1) i # 0 2, z1, Thus, z 1 & is purely real., z 1, , , EXAMPLE 20, SOLUTION, , 3 2i sin , n is purely real., Find the real value of for which d , 1 2i sin , We have, 3 2i sin (3 2i sin ) (1 2i sin ), d, n, #, 1 2i sin , (1 2i sin ) (1 2i sin ), (3 2i sin ) (1 2i sin ), , (1 4i 2 sin 2 )
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168, , Senior Secondary School Mathematics for Class 11, , , , , (3 4 sin 2 ) i (6 sin 2 sin ), (1 4 sin 2 ), (3 4 sin ) i (8 sin ), 2, , (1 4 sin 2 ), , ·, , 3 2i sin , 8 sin , 0., n will be purely real only when, Now, d , 1 2i sin , (1 4 sin 2 ), This happens only when 8 sin 0 + sin 0 + n, nd N., Hence, the required value of is n, where nd N., EXAMPLE 21, SOLUTION, , Show that|1 i|x 2 x has no nonzero integral solution., |1 i|x 2 x & ( 2 ) x 2 x, , [a |1 i| 12 (1) 2 2 ], , 2x , 1, 2 x/2, x, & 2 x/2 1 2 0 & 0 & x 0., 2, Thus, x 0 is the only solution of the given equation., Hence, the given equation has no nonzero integral solution., & 2 x/2 2 x &, , EXAMPLE 22, , SOLUTION, , Solve for x and y:, (i) (3x 7) 2iy 5y (5 x) i, (ii) (x iy) (2 3i ) 4 i, (i) The given equation is, (3x 7) 2iy 5y (5 x) i., Equating the real parts and the imaginary parts of the given, equation separately, we get, 3x 7 5y and 2y 5 x, & 3x 5y 7 and x 2y 5, [on solving 3x 5y 7, x 2y 5]., & x 1 and y 2, Hence, x 1 and y 2., (ii) The given equation is, (x iy) (2 3i ) 4 i, & (2x 3y) i (2y 3x) 4 i, & 2x 3y 4 and 2y 3x 1, [equating real parts and imaginary parts separately], & 2x 3y 4 and 3x 2y 1, 5, 14, [on solving 2x 3y 4, 3x 2y 1]., and y , 13, 13, 5, 14, Hence, x , is the required solution., and y , 13, 13, , &, , x
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Complex Numbers and Quadratic Equations, , EXAMPLE 23, SOLUTION, , 169, , x1 y1, Find the real values of x and y for which e o i., 3 i 3 i, We have, x1 y1 , i, 3i 3i, (x 1) (3 i ) (y 1) (3 i ), , i, #, #, &, (3 i ) (3 i ) (3 i ) (3 i ), ( x 1) ( 3 i ) ( y 1) ( 3 i ), , i, &, (9 i 2), ( 9 i 2), &, &, &, , 3(x 1) i (x 1) 3(y 1) i (y 1), , i, 10, 10, 3(x 1) i (x 1) 3(y 1) i (y 1) 10i, (3x 3 3y 3) i (y 1 x 1) 10i, , &, , (3x 3y 6) i (y x) 10i, 3(x y 2) 0 and y x 10, , &, , [equating real parts and the imaginary parts separately], x y 2 0 and y x 10, , &, , x y 2 and x y 10, , &, , [on solving x y 2, x y 10]., x 4 and y 6, Hence, x 4 and y 6 are the required values., , &, , EXAMPLE 24, , Find the complex number z for which|z 1| z 2 (1 i ) ., , SOLUTION, , Let the required complex number be z (x iy) . Then,, |z 1| z 2 (1 i ), &, |(x iy) 1| (x iy) 2(1 i ), &, , ( x 1) 2 y 2 ( x 2) i ( y 2 ), , &, , (x 1) 2 y 2 (x 2) and y 2 0, [equating real parts and imaginary parts separately], , &, , y 2 and (x 1) 2 ( 2) 2 (x 2), , &, , y 2 and x 2 2x 5 (x 2), , &, , y 2 and (x 2 2x 5) (x 2) 2, , &, , x 2 2x 5 x 2 4x 4 and y 2, 1, 2x 1 and y 2 & x and y 2., 2, , &, , 1, Hence, the required complex number is z b 2il ·, 2
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170, EXAMPLE 25, SOLUTION, , Senior Secondary School Mathematics for Class 11, , Solve the equation|z| z (2 i ) for complex value of z., Let z (x iy) . Then,, |z| z 2 i, &, |x iy| (x iy) 2 i, &, &, , { x 2 y 2 x} iy (2 i ), x 2 y 2 x 2 and y 1, [equating real parts and imaginary parts separately], , &, , y 1 and x 2 1 2 x 2 & y 1 and x 2 1 (2 x), , &, , y 1 and x 2 1 (2 x) 2 & y 1 and x 2 1 x 2 4x 4, , &, , 4x 3 and y 1 & x , , 3, and y 1., 4, , 3, Hence, z b il is the desired solution., 4, EXAMPLE 26, , Solve the equation z 2|z 1| i 0 for complex value of z., , SOLUTION, , Let the required complex number be z x iy. Then,, z 2|z 1| i 0, &, , (x iy) 2|(x iy) 1| i 0, , &, , x 2 { ( x 1) 2 y 2 } ( y 1) i 0, , &, , &, , x 2 { (x 1) 2 y 2 } 0 and y 1 0, [equating real parts and imaginary parts separately, on both sides], y 1 and 2 { (x 1) 2 (1) 2 } ( x), , &, , y 1 and 2 $ x 2 2x 2 ( x), , &, , y 1 and 2 (x 2 2x 2) x 2, , &, , y 1 and x 2 4x 4 0 & (x 2) 2 0 and y 1, x 2 0 and y 1 & x 2 and y 1., , &, , Hence, the required complex number is z ( 2 i ) ., EXAMPLE 27, , SOLUTION, , If z 3 2i, prove that z 2 6z 13 0 and hence deduce that, 3z 3 13z 2 9z 65 0., z 3 2i & z 3 2i & (z 3) 2 4i 2, … (i), & z 2 6z 13 0., Thus, z 2 6z 13 0., Now, 3z 3 13z 2 9z 65 3z (z 2 6z 13) 5 (z 2 6z 13), (3z # 0) (5 # 0) 0, [using (i)]., Hence, 3z 3 13z 2 9z 65 0.
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Complex Numbers and Quadratic Equations, EXAMPLE 28, SOLUTION, , 171, , If z 5 3i, find the value of (z 4 9z 3 26z 2 14z 8) ., We have, z 5 3i & (z 5) 3i, & (z 5) 2 9i 2 & z 2 10z 34 0., , … (i), , Now, z 9z 26z 14z 8, z 2 (z 2 10z 34) z (z 2 10z 34) 2(z 2 10z 34) 60, 4, , 3, , 2, , (z 2 # 0) (z # 0) (2 # 0) 60 60, , [using (i)]., , Hence, the value of z 9z 26z 14z 8 is –60., 4, , EXAMPLE 29, SOLUTION, , 3, , 2, , If z 2 i, prove that z 3 3z 2 9z 8 (1 14i ) ., We have, z 2i & z2 i, & (z 2) 2 i 2 & z 2 4z 5 0., , … (i), , z 3z 2 9z 8, z (z 2 4z 5) 7 (z 2 4z 5) 14z 27, (z # 0) (7 # 0) 14z 27 (14z 27), 14 (2 i ) 27 (1 14i ) ., , , , 3, , [using (i)], , Hence, z 3 3z 2 9z 8 (1 14i ) ., , EXERCISE 5B, 1. Simplify each of the following and express it in the form a ib :, (i) 2(3 4i ) i (5 6i ), (iii) ( 5 6i ) ( 2 i ), , (ii) (3 16 ) (4 9 ), (iv) (8 4i ) ( 3 5i ), , (v) (1 i ) 2 (1 i ) (3 4i ) 2, , (vi) (5 3 ) (5 3 ), , (vii) (3 4i ) (2 3i ), , (viii) ( 2 3 ) ( 3 2 3 ), , 2. Simplify and express each of the following in the form (a ib) :, (i) (2 3 ) 2, 1, (iv) b 2 il, 3, , (vii) (2 i ) 2, , 3, , (ii) (5 2i ) 2, , (iii) ( 3 5i ) 3, , (v) (4 3i ), , 1, , , (vi) ( 2 3 ) 1, , (viii) (1 2i ), , 3, , (ix) (1 i ) 3 (1 i ) 3, , 3. Express each of the following in the form (a ib) :, (5 2 i ), (3 4i ), 1, (i), (ii), (iii), , , ( 4 3i ), (4 5i ), (1 2 i ), , , , ( 2 5i ), (3 4i ), (3 2i ) (2 3i ), (iv), (v), (vi), , , , ( 3 6i ), (4 2i ) (1 i ), (1 2i ) (2 i ), (vii), , ( 2 3i ) 2, (2 i ), , (viii), , (1 i ) 3, (1 i ), 3, , (ix), , (1 2i ) 3, (1 i ) (2 i )
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172, , Senior Secondary School Mathematics for Class 11, , 4. Simplify and express each of the following in the form (a ib) :, 5, 4 5i, 1i, 1, 2, 2, (i) b ld n, (ii) b ld n, 3 2i 1 i 3 2i, 1 4i 1 i 5 3i, 5. Show that, (3 2i ) (3 2i ), 3 is purely real,, , (i) ), (2 3i ) (2 3i ), Z], _, ]]( 7 i 3 ) ( 7 i 3 ) bbb, , ` is purely real., (ii) [], ]]( 7 i 3 ) ( 7 i 3 ) bbb, \, a, 1 i cos , n is purely real., 6. Find the real values of for which d , 1 2i cos , 7. If|z i| |z i| , prove that z is real., 8. Give an example of two complex numbers z1 and z 2 such that z1 ! z 2 and, |z1| |z 2|., 9. Find the conjugate of each of the following:, (i) ( 5 2i ), , (ii), , 3, , (vi), , (v), , 1, (4 3i ), 2, , (iii), , (1 i ) 2, (3 i ), , (vii) 1, , 10. Find the modulus of each of the following:, (ii) ( 3 4i ), (iii) (7 24i ), (i) (3 5 ), ( 3 2i ), ( 4 3i ), , 2, , (v), , (vi), , (2 i ) (1 i ), (1 i ), , (vii) 5, , (iv), , (1 i ) (2 i ), (3 i ), , (viii) (2 5i ) 2, (iv) 3i, (viii) (1 2i ) (i 1), , 11. Find the multiplicative inverse of each of the following:, (2 3i ), ( 1 i ) ( 1 2i ), (i) (1 3 i ), (ii) (2 5i ), (iii), (iv), (1 i ), (1 3i ), 1 i 100 , 12. If b l, (a ib), find the values of a and b., 1 i, 1i 3 1i 3, 13. If b l b l x iy, find x and y., 1 i, 1 i, ib, a, , prove that x 2 y 2 1., 14. If (x iy) , a ib, b, 2c ·, ci, 15. If (a ib) , where c is real, prove that a 2 b 2 1 and a 2, c i, c 1, 1n, 16. Show that (1 i )n b1 l 2 n for all nd N., i, 17. Find the smallest positive integer n for which (1 i ) 2n (1 i ) 2n., 18. Prove that (x 1 i) (x 1 i ) (x 1 i ) (x 1 i ) (x 4 4) ., , 1a, 19. If a (cos i sin ), prove that bcot l i., 2, 1 a
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Complex Numbers and Quadratic Equations, , 20. If z1 (2 i ) and z 2 (1 i ), find, , 173, , z1 z 2 1, ·, z1 z 2 i, , 21. Find the real values of x and y for which:, (ii) (x iy) (3 2i ) (12 5i ), (i) (1 i ) x (1 i ) y 1 3i, (iii) x 4yi ix y 3, (x 3i ), (1 i ), (v), (2 iy), , (iv) (1 i ) y 2 (6 i ) (2 i ) x, (1 i ) x 2i (2 3i ) y i, , i, (vi), (3 i ), (3 i ), , 22. Find the real values of x and y for which (x iy) (3 5i ) is the conjugate of, ( 6 24i ) ., 23. Find the real values of x and y for which the complex numbers ( 3 iyx 2), and (x 2 y 4i ) are conjugates of each other., 24. If z (2 3i ), prove that z 2 4z 13 0 and hence deduce that, 4z 3 3z 2 169 0., 25. If (1 i ) z (1 i ) z then prove that z i z., z1, 26. If b l is purely an imaginary number and z ! 1 then find the value, z 1, of|z|., 27. Solve the system of equations, Re (z 2) 0, |z| 2., 28. Find the complex number z for which|z| z 1 2i., ANSWERS (EXERCISE 5B), , 1. (i) (12 13i ), (v) (9 22i ), , (vi) 28, , 2. (i) (1 4 3 i ), , (ii) (21 20i ), , (v) b, , 4 3, il, 25 25, , (ix) 4i, 3, 4, 3. (i) b il, 25 25, , (ii) (1 7i ), , 3, 2, (vi) e , io, 7, 7, , (iii) ( 3 5i ), (vii) (18 i ), , 22 107, , il, 3, 27, 11, 3, 4, 2 il, (vii) b il (viii) b, 25 25, 125 125, (iii) (198 10i ), , (iv) b, , (iv) b, , (ii) b, , 8 31, il, 41 41, , (iii) (1 2 2 i ), , 1 3, (v) b il, 4 4, 7 1, (ix) b il, 2 2, 5 4, 4. (i) b, il, 13 13, , (vi) b, , 63 16, il, 25 25, , (vii) b, , (ii) b, , 4 5, il, 17 17, , 9. (i) ( 5 2i ), , (ii) b, , 4 3, il, 25 25, , (iii) b, , (vi), , 2, , (vii) i, , (v) 3 i, , (iv) (11 9i ), (viii) 7 3 i, , 4 1, il, 5 15, , 22 19, il (viii) ( 2 0i ), 5, 5, , , 6. (2n 1) , nd I, 2, 1 3, il, 5 5, , 3 4, (iv) b il, 5 5, (viii) ( 21 20i )
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174, , Senior Secondary School Mathematics for Class 11, , 10. (i), , (ii) 5, , 14, 13, (v), 5, , (vi), , (iii) 25, (vii) 5, , 5, , (iv) 3, (viii), , 10, , 5, 5, 1 3 o, 4 3, 1, 2, 11. (i) e ,, (ii) b il, (iii) b il, (iv) b il, i, 29 29, 5 5, 13 13, 4 4, 12. a 1, b 0, 13. x 0, y 2, 17. n 2, 20. 2 2, 21. (i) x 2, y 1, (ii) x 2, y 3, (iii) x 4, y 1, (iv) (x 5, y 2) or (x 5, y 2), (v) x 7, y 5 (vi) x 3, y 1, 22. x 3, y 3, 23. (x 1, y 4) or (x 1, y 4), 3, 26. |z| 1, 27. z 2 (1 ! i ) or z 2 (1 ! i ), 28. b 2il, 2, HINTS TO SOME SELECTED QUESTIONS, 2. (iii) ( 3 5i ) 3 (5i 3) 3 (5i ) 3 3 3 (3 # 5i # 3) (5i 3), ( 125i 27 225 135i ) (198 10i ) ., 1 3, 1 3, 1, 1, (iv) c 2 im ( 1) 3 c2 im (8 i 3 2i c2 im2 ·, 3, 3, 27, 3, , , (, 4, 3, i, ), (, 4, 3, i, ), 1, , , ·, (v) (4 3i ) 1 , #, 25, (4 3i ) (4 3i ), , (vi) ( 2 3 i ) 1 , , (vii) (2 i ), , 2, , (viii) (1 2i ), , , , 3, , ( 2 3 i ), 1, ·, #, ( 2 3 i ) ( 2 3 i ), , (3 4i ), 1, 1, 1, , , ·, #, (2 i ) 2 (4 i 2 4i ) (3 4i ) (3 4i ), , , , (11 2i ), 1, 1, 1, , , ·, #, 3, 3, , , (, ), (11 2i ), i, 11, 2, (1 2i ), {1 8i 6i (1 2i )}, , 5 (1 i ) 2 ( 3 2i ) (4 5i ), 5, 4 5i, 2, 4. (i) d nd n , #, 3 2i 1 i 3 2i, ( 3 2i ) (1 i ), (3 2i ), (1 i ) (4 5i ), (1 i ) (4 5i ), , , (1 5i ) (3 2i ) (1 5i ) (3 2i ), , , (9 i ), (1 i ) (10 8i ), , , c 5 4 im ·, #, 26, 13 13, 13 (1 i ) (1 i ), 5. (i) Given expression , (ii) Given expression , 6. Given expression , , (3 2i ) (2 3i ) (3 2i ) (2 3i ) 0, , 0., 13, (2 3i ) (2 3i ), 2, 2, ( 7 i 3) ( 7 i 3), , ( 7 i 3) ( 7 i 3), , , , 2, 2, 2 {( 7 ) (i 3 ) } 8, , 4·, 10 5, (7 3), , (1 i cos ), (1 2i cos ) (1 2 cos 2 ) i (3 cos ), , ·, #, 2, (1 2i cos ) (1 2i cos ), (1 4 cos ), , This expression will be purely real when 3 cos 0., , Now, 3 cos 0 + cos 0 + (2n 1) , where nd I., 2
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Complex Numbers and Quadratic Equations, , 175, , 7. Let z (x iy) . Then,, |z i|2 |z i|2, &, , |(x iy) i|2 |(x iy) i|2, , &, , |x (y 1) i|2 |x (y 1) i|2 & x 2 (y 1) 2 x 2 (y 1) 2, , 2, 2, & (y 1) (y 1) 0 & 4y 0 & y 0., Hence, z is purely real., 8. Let z1 (3 2i ) and z2 (3 2i ) . Then, z1 ! z2 ., , But,|z| |z2| 13 ., (3 2i ) 2 (9 4i 2 12i ) (5 12i ) (4 3i ) 16 63i, , , , #, 25, (4 3i ), (4 3i ), (4 3i ), (4 3i ), 2, 2, , 16, 63, 256, 3969, 4225, 2, , m c m 3 c, m, & |z| )c, 25, 25, 625 625, 625, , 10. (v) z , , 4225 65 13 ·, 625, 25, 5, (2 i ) (1 i ) (3 i ) (1 i ) (2 4i ), , , (1 2i ) ., (vi) z , #, 2, (1 i ), (1 i ) (1 i ), |z| , , &, , 11. Use the formula, z, (iii) z , , 1, , z2 ·, |z|, , (2 3i ) (1 i ) (5 i ), , c 5 1 im ·, #, 2, 2 2, (1 i ), (1 i ), , (5 i ), 25 1 26 13, 5 1, ·, m, and z c im , 2, 2 2, 2, 4 4, 4, (5 i ), z, , 2 (5 i ) 5 1 ·, c, # , z 1 2 , im, 2, 13, 13, 13 13, |z|, , |z|2 c, , , (1 i ) (1 2i ) (1 3i ) (1 3i ) (8 6i ), , , c 4 3 im ·, #, 5 5, 10, (1 3i ), (1 3i ), (1 3i ), , (, 4, 3, i, ), 16 9, 25 , 4 3, 2, , z c im , and|z| c m , 1., 5 5, 5, 25 25, 25, z, 4 3, 1, , z 2 c im ·, 5 5, |z|, , (iv) z , , 12., , (1 i ) (1 i ) (1 i ) (1 i ) 2 (1 i 2 2i ), , , , i., #, 2, 2, (1 i ) (1 i ) (1 i ), 1i, n ( i ) 100 {( i ) 4} 25 (1) 25 1, 1i, Hence, a 1 and b 0., , , d, , 100, , (1 i ) (1 i ) (1 i ) 2 (1 i 2 2i ) 2i, 1i, , , i., 13. d n , #, 2, 2, 2, 1 i, (1 i ) (1 i ), , 1 i, Similarly, d n i., 1 i, Given expression i 3 ( i ) 3 2i 3 2i 0 ( 2) i., 14. (x iy) , , (a ib), (a ib), a ib, ·, & (x iy) , & (x iy) , a ib, (a ib), (a ib), , [a ( i ) 4 i 4 1].
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176, , Senior Secondary School Mathematics for Class 11, , , (x iy) (x iy) , , (a ib) (a ib), 1 & x 2 y 2 1., #, (a ib) (a ib), , (c i ) (c i ) (c i ) 2 c 2 i 2 2ci (c 2 1), 2, 2c i., , , #, (c i ) (c i ) (c 2 1), (c 2 1), (c 1) (c 2 1), Z], _b, 2, 2, 2, 2, 2, ] J 2 N2, 4c 2 bb (c 1) 4c (c 1) , 2 ] Kc 1O, `b, 1., & |a ib| []KKK 2 OOO 2, 2, 2, 2, 2, ]] c 1, (c 1) bb, (c 1), (c 1) 2, L, P, \, a, 2, 2, & a b 1., (c 2 1), 2c, b, 2c ·, Now, a 2, & a 2, and b 2, (c 1), (c 1), (c 1), , 15. (a ib) , , 2, (i 1) i (i i ) (1 i ), 1, , (1 i ) ., 16. c1 m , # , 2, 1, i, i, i, i, given expression (1 i ) n #(1 i ) n {(1 i ) (1 i )} 2 (1 i 2) n 2 n., , 17., , (1 i ) (1 i ) (1 i ) (1 i ) 2 (1 i 2 2i ) 2i, , , , i., #, 2, 2, 2, (1 i ) (1 i ) (1 i ), , , 2, 2, (1 i ) n (1 i ) n &, , 2, (1 i ) n, , (1 i ), , 2n, , 2n, 1 & d1 i n 1 & i 2n 1., 1i, , Least value of 2n is 4 and so the least value of n is 2., 18. Given expression {(x 1) i } {(x 1) i } {(x 1) i } {(x 1) i }, {(x 1) 2 i 2} {(x 1) 2 i 2} {(x 1) 2 1} {(x 1) 2 1}, {(x 2 2) 2x} {(x 2 2) 2x}, (x 2 2) 2 4x 2 (x 4 4) ., 19., , 1 a (1 cos ) i sin (1 cos ) i sin , #, 1 a (1 cos ) i sin (1 cos ) i sin , , , , 20., , (1 cos 2 sin 2 ) 2i sin , (1 cos ) sin , 2, , 2, , , , 1 (cos 2 sin 2 ) 2i sin , 1 (cos 2 sin 2 ) 2 cos , , 2i sin i sin 2 sin (/2) cos (/2) · , , i ccot m i., 2, 2 (1 cos ) (1 cos ), 2 sin 2 (/2), , z1 z2 1 (2 i ) (1 i ) 1, (1 i ), , 4 #, 2 (1 i ), z1 z2 i (2 i ) (1 i ) i (1 i ) (1 i ), z1 z2 1, 2 # 1 (1) 2 2 2 ., &, z1 z2 i, , 21. (iv) (y 2 6 2x) (y 2 x 1) i 0 & y 2 6 2x 0, y 2 x 1 0, y 2 2x 6 and y 2 x 1., &, On subtracting, we get x 5., , , y 2 5 1 & y 2 4 & y !2., , (v) (x 3i ) (1 i ) (2 iy), &, &, , (x 3i ) (2 y) (y 2) i & (x y 2) i (5 y) 0, x y 2 0 and 5 y 0.
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Complex Numbers and Quadratic Equations, , 177, , (vi) {x (x 2) i } (3 i ) {2y (1 3y) i } (3 i ) i (3 i ) (3 i ), &, {3x (x 2)} {3 (x 2) x} i {6y (1 3y)} {2y (3 9y} i 10i, &, (4x 2 9y 1) (2x 6 3 7y) i 10i, &, (4x 9y 3) (2x 7y 3) i 10i, &, 4x 9y 3 and 2x 7y 13., 22. (x iy) (3 5i ) ( 6 24i ) ( 6 24i ), &, &, , (3x 5y) (5x 3y) i 6 24i, 3x 5y 6 and 5x 3y 24., , 23. ( 3 iyx 2) (x 2 y 4i ), &, , ( 3 iyx 2) (x 2 y 4i ), , &, , (x 2 y 3) (4 yx 2) i 0, , &, , x 2 y 3 0 and 4 yx 2 0, , &, , x 2 y 3, , … (i), , and yx 2 4, , … (ii)., , Putting x ( 3 y) from (i) in (ii), we get, 2, , y ( 3 y) 4 & y 2 3y 4 0 & (y 4) (y 1) 0 & y 4 or y 1., Now, y 1 & x 2 4 & x is imaginary. So, y ! 1., When y 4, we get x 2 ( 3 4) 1 & x !1., (x 1, y 4) or (x 1, y 4) ., 25. (1 i ) z (1 i ) z &, , 2, 2, z (1 i ) (1 i ) (1 i ) (1 i 2i ) , #, i., 2, 2, z (1 i ) (1 i ), , Hence, z i z., 26. Let z x iy. Then,, 2, 2, z 1 (x 1) iy (x 1) iy (x y 1) (x y) i ·, #, 2, 2, z 1 (x 1) iy (x 1) iy, (x 1) y, , Now,, , z1, is purely imaginary + x 2 y 2 1 0 + x 2 y 2 1 + |z| 1., z1, , 27. Let z (x iy) . Then, z 2 (x 2 y 2) 2ixy., Now, Re (z 2) 0 & x 2 y 2 0., |z| 2 & |z|2 4 & x 2 y 2 4., On solving (i) and (ii), we get, x 2 2 and y 2 2., , , x ! 2 and y ! 2 ., , , , z 2 (1 ! i ) or z 2 (1 ! i ) ., , 28. Let the required complex number be z (x iy) . Then,, |z| z 1 2i, &, &, , |x iy| (x iy) 1 2i, x 2 y 2 (x 1) (y 2) i, , … (i), … (ii)
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178, , Senior Secondary School Mathematics for Class 11, , &, , x 2 y 2 (x 1) and y 2 0, , &, , y 2 and x 2 ( 2) 2 (x 1), , &, , y 2 and x 2 4 (x 1) 2 & x , , 3, and y 2., 2, , SOME RESULTS ON MODULUS AND CONJUGATE OF COMPLEX NUMBERS, THEOREM 1, , For any complex numbers z, z1 and z 2 prove that:, (i), (iii), (v), (vii), , (ii), (z) z, (iv), z z 2 Re (z), z z + z is purely real (vi), (viii), z1 z 2 z1 z 2, , (ix) z1 z 2 ( z 2 ) ( z 2 ), PROOF, , zz |z|2 {Re z} 2 {Im (z)} 2, z z 2i · Im (z), z z 0 + z is purely imaginary, z1 z 2 z1 z 2, , z1, z1, (x) d z n , z 2 ! 0, z2, 2, , Let z (a ib), z1 (a1 ib1) and z 2 (a 2 ib 2) . Then,, (i) z (a ib) & z (a ib) (a ib), & ( z ) (a ib) (a ib) z., Hence, ( z ) z., (ii) zz (a ib) (a ib) (a ib) (a ib), (a 2 b 2) {Re (z)} 2 {Im (z)} 2 |z|2., , , zz {Re (z)} 2 {Im (z)} 2 |z|2., , (iii) z z (a ib) (a ib), (a ib) (a ib) 2a 2 Re (z) ., , , z z 2 Re (z) ., , (iv) z z (a ib) (a ib), (a ib) (a ib) 2ib 2i · Im (z) ., , , z z 2i · Im (z) ., , (v) z z + (a ib) (a ib), + (a ib) (a ib) + 2ib 0, + b 0 + Im (z) 0., , , z z + Im (z) 0 + z is purely real., , (vi) z z 0 + (a ib) (a ib) 0, + (a ib) (a ib) 0 + 2a 0 + a 0, + z is purely imaginary., , z z 0 + z is purely imaginary.
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180, PROOF, , Senior Secondary School Mathematics for Class 11, , Let z (a ib), z1 (a1 ib1) and z 2 (a 2 ib 2) . Then,, (i) zz (a ib) (a ib) (a ib) (a ib) (a 2 b 2) |z|2., z z |z|2., , , , (ii) |z| |a ib| a 2 b 2 , |z| |a ib| a 2 ( b) 2 a 2 b 2, and | z| |(a ib)| |( a) i ( b)| ( a) 2 ( b) 2 a 2 b 2 ., |z| |z| | z|., , , , (iii) |z| 0 + |z|2 0 + |a ib|2 0, + a 2 b 2 0 + a 0 and b 0 + z 0., (iv) Let z (a ib) . Then,|z| a 2 b 2 ., Clearly, a 2 b 2 # a # a 2 b 2 ., |z|# Re (z) #|z|., (v) Let z (a ib) . Then,|z| a 2 b 2 ., Clearly, a 2 b 2 # b # a 2 b 2 ., |z|# Im (z) #|z|., (vi) z1 z 2 (a1 ib1) (a 2 ib 2) (a1 a 2 b1 b 2) i (a1 b 2 b1 a 2) ., |z1 z 2| (a1 a 2 b1 b 2) 2 (a1 b 2 b1 a 2) 2, , , , , , a 12 a 22 b 12 b 22 a 12 b 22 b 12 a 22, , , , a 12 (a 22 b 22) b 12 (a 22 b 22) (a 12 b 12) (a 22 b 22), , $ a 12 b 12 .$ a 22 b 22 . |z1|·|z 2|, |z1 z 2| |z1|·|z 2|., Z], _, z1, (a 2 ib 2) bb, ], 1, 1, `, (vii) z z1 · z (a1 ib1) · [], #, ](a 2 ib 2) (a 2 ib 2) bb, 2, 2, \, a, (a1 ib1) (a 2 ib 2) (a1 a 2 b1 b 2) i (b1 a 2 a1 b 2), , , (a 2 ib 2) (a 2 ib 2), (a 22 b 22), , , &, , ( a 1 a 2 b 1 b 2), (a 22, , , , z1, , z2, , , ( a 2 b 1 a 1 b 2), , (a 22 b 22) 2, a1 a 2 b1 b 2 a 2 b1 a1 b 2, 2, , , , i, , (a 22 b 22), Z] a a b b _b2 Z] a b a b _b2, ] 1 2 1 2b ] 2 1 1 2b, [] 2, ` [, `, ] a 2 b 22 bb ]] a 22 b 22 bb, \, a \, a, 2, 2, ( a 1 a 2 b 1 b 2 ) ( a 2 b 1 a 1 b 2), , b 22), , 2, , 2, , 2, , 2, , (a 22 b 22) 2, , 2, , 2, , 2
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182, , Senior Secondary School Mathematics for Class 11, , &, , |z 5i| |z 5i| & |z 5i|2 |z 5i|2, , &, , |(x iy) 5i|2 |(x iy) 5i|2, , &, , |x i (y 5)|2 |x i (y 5)|2, , &, , x 2 (y 5) 2 x 2 (y 5) 2, , [a z (x iy)], [a |x iy|2 (x 2 y 2)], , &, ( y 5) 2 ( y 5 ) 2 0 & 4 # y # 5 0 & y 0 ., , z x i0 & z x, where x is real., Hence, z is a real number., EXAMPLE 2, , If (a ib) , , SOLUTION, , We have, , (x i ) 2, 2, (2x 1), , (a ib) , &, , (a ib) , , &, , |a ib|2 , , &, , then prove that (a 2 b 2) , , (x i ) 2, (2x 1), 2, , (x 2 1), (2x 2 1), , , , (x 2 i 2 2ix), (2x 1), 2, , i·, , (x 2 1), , , , (x 2 1) 2, (2x 2 1) 2, , ·, , (x 2 1) i (2x), 2, (2x 1), , 2x, (2x 2 1), , i·, , 2x, (2x 2 1), , 2, , (2x 2 1), ]Z] (x 2 1) 2, b_b, 2, ], b, 4x, , `, (a 2 b 2) ][ 2, ]](2x 1) 2 (2x 2 1) 2 bbb, \, a, (x 2 1) 2 4x 2 (x 2 1) 2, , , ·, (2x 2 1) 2, (2x 2 1) 2, , Hence, (a 2 b 2) , , EXAMPLE 3, , If (p iq) , , SOLUTION, , We have, , (x 2 1) 2, (2x 2 1) 2, , ·, , ( a 2 1) 2, (a i ) 2, ·, then prove that (p 2 q 2) , 2, (2a i ), (4a 1), , (p iq) , , (a i ) 2 (a 2 i 2 2ai ) (a 2 1) 2ai, , , (2a i ), (2a i ), (2a i ), 2, 2, |(a 1) 2ai|, , &, , |p iq|2 , , &, , ( p 2 q 2) , , |2a i|, , 2, , ( a 2 1) 2, (4a 2 1), , Hence, (p 2 q 2) , , ·, , ( a 2 1) 2, (4a 2 1), , ·, , , , (a 2 1) 2 4a 2, { 4a ( 1 ) }, 2, , 2, , , , ( a 2 1) 2, (4a 2 1)
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Complex Numbers and Quadratic Equations, EXAMPLE 9, , 185, , If (1 i ) (1 2i ) (1 3i ) … (1 ni ) (x iy) then prove that, 2 # 5 #10 #…#(1 n 2) (x 2 y 2) ., , SOLUTION, , (1 i ) (1 2i ) (1 3i ) … (1 ni ) (x iy), &, , |(1 i ) (1 2i ) (1 3i ) … (1 ni )|2 |x iy|2, , &, , |1 i|2 #|1 2i|2 #|1 3i|2 #…#|1 ni|2 (x 2 y 2), , &, , ( 2 ) 2 #( 5 ) 2 #( 10 ) 2 #…#( 1 n 2 ) 2 (x 2 y 2), , &, , 2 # 5 #10 #…#(1 n 2) (x 2 y 2) ., , Hence, 2 # 5 #10 #…#(1 n 2) (x 2 y 2) ., EXAMPLE 10, , If (a ib) (c id) (e if ) (g ih) (A iB) then show that, (a 2 b 2) (c 2 d 2) (e 2 f 2) (g 2 h 2) (A 2 B 2) ., , SOLUTION, , (a ib) (c id) (e if ) (g ih) (A iB), |(a ib) (c id) (e if ) (g ih)| |A iB|, , &, &, , |a ib|·|c id|·|e if|·|g ih| |A iB|, , &, , |a ib|2 ·|c id|2 ·|e if|2 ·|g ih|2 |A iB|2, , &, , (a 2 b 2) (c 2 d 2) (e 2 f 2) (g 2 h 2) (A 2 B 2) ., , Hence, (a 2 b 2) (c 2 d 2) (e 2 f 2) (g 2 h 2) (A 2 B 2) ., EXAMPLE 11, , If and are different complex numbers such that || 1, show that, , 1., 1 , , SOLUTION, , We have, , 1 , , 2, , Z] _b Z] _b, ] b] b, [], `[, `, ]1 bb ]]1 bb, \, a\, a, ( ) ( ), ( ) ( ), , , ·, , , (1 ) (1 ) (1 ) (1 ), , , , , Hence,, , [a |z|2 zz], , , 1 () (), 2, 2, || || ( ), , 1 ( ) ||2 ||2, 2, 1 || ( ), , 1 ||2 ( ), , , 1., 1 , , 1, , [a ||2, ||2], [a || 1].
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186, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 12, , If|z 2 1| |z|2 1 then show that z is imaginary., , SOLUTION, , Let z (x iy) . Then, z 2 (x 2 y 2) 2ixy and|z|2 (x 2 y 2) ., Now,|z 2 1| |z|2 1, &, , |(x 2 y 2 1) i (2xy)| (x 2 y 2 1), , &, , |(x 2 y 2 1) i (2xy)|2 (x 2 y 2 1) 2, , &, , (x 2 y 2 1) 2 4x 2y 2 (x 2 y 2 1) 2, , &, , [x 2 (y 2 1)] 2 [x 2 (y 2 1)] 2 4x 2y 2, , &, , 4x 2 (y 2 1) 4x 2y 2 & 4x 2 {(y 2 1) y 2} 0, , &, 4x 2 0 & x 0., Hence, z 0 iy, which shows that z is imaginary., EXAMPLE 13, , If iz 3 z 2 z i 0 then show that|z| 1., , SOLUTION, , We have, iz 3 z 2 z i 0, &, z 3 iz 2 iz 1 0, &, z 2 (z i ) i (z i ) 0, &, , [on dividing both sides by i], , (z i ) ( z 2 i ) 0, , &, z i or z 2 i., Now, z i & |z| ||, i & |z| 1., And, z 2 i & |z 2| | i| 1 & |z|2 1 & |z| 1., Hence, in either case, we have|z| 1., EXAMPLE 14, , Find all nonzero complex numbers z satisfying z iz 2., , SOLUTION, , Let z x iy. Then, z x iy and z 2 (x 2 y 2) 2ixy., , , z iz 2 & x iy i (x 2 y 2) 2xy, & (x 2xy) i (x 2 y 2 y) 0., , … (i), , On equating real parts and imaginary parts on both sides of (i), separately, we get, … (ii), and x 2 y 2 y 0, … (iii)., x 2xy 0, From (ii), we get, x 2xy 0 & x (1 2y) 0, & x 0 or 1 2y 0 & x 0 or y , Case I, , 1, ·, 2, , When x 0., Putting x 0 in (iii), we get, y 2 y 0 & y (y 1) 0 & y 0 or y 1.
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Complex Numbers and Quadratic Equations, , 187, , Thus, (x 0, y 0) or (x 0, y 1), z (0 i0) or z (0 1 · i ) ., 1, ·, Case II When y , 2, 1, Putting y , in (iii), we get, 2, 1 2 1, 1 1, x2 b l b l 0 & x2 0, 2, 2, 4 2, 3, 3, 1, 1, ·, &, x2 b l & x !, 2, 4 2, 4, , , , , , , , , ex , , 3, 3, , , , y 1 o or ex , , y 1o, 2, 2, 2, 2, , , , , , ze, , 3 1, 3 1, io or z e, io ·, 2, 2, 2, 2, , Hence, the required nonzero complex numbers are, i, e, , 3 1, 3 1, io and e, io ·, 2, 2, 2, 2, , EXAMPLE 15, , Solve the equation, z 2 z, where z is a complex number., , SOLUTION, , Let z x iy. Then,, z 2 z & (x iy) 2 x iy, … (i), & (x 2 y 2) 2ixy x iy., Equating real parts and imaginary parts on both sides of (i), separately, we get, x2 y2 x, From (iii), we get, , … (ii), , and 2xy y, , 2xy y 0 & y (2x 1) 0 & y 0 or x , Case I, , … (iii)., 1, ·, 2, , When y 0., Putting y 0 in (ii), we get, , x 2 x 0 & x (x 1) 0 & x 0 or x 1., , , (x 0, y 0) or (x 1, y 0), Thus, z (0 i0) or z (1 i0) ., 1, ·, Case II When x , 2, 1, Putting x , in (ii), we get, 2, , b, , 3, 1 2, 1, 3, 1 1, ·, l y2 b l & y2 b l & y !, 2, 2, 2, 4 2, 4
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188, , Senior Secondary School Mathematics for Class 11, , , , , , 3, 3, , , ex 1 , y , o or ex 1 , y , o·, 2, 2, 2, 2, , Thus, z e, , 3, 3, 1, 1, , , io or z e, io ·, 2, 2, 2, 2, , 3, 3, 1, 1, , io and e , io are the required roots of, 2, 2, 2, 2, the given equation., , Hence, z 0, 1, e, , EXAMPLE 16, , For all complex numbers z1 and z 2, prove that, , SOLUTION, , |z1 z 2|2 |z1 z 2|2 2 (|z1|2 |z 2|2) ., We have, |z1 z 2|2 (z1 z 2) (z1 z 2), [a (z1 z 2) z1 z 2], , (z1 z 2) (z1 z 2), z1 z1 z 2 z 2 z1 z 2 z 2 z1, |z1|2 |z 2|2 z1 z 2 z 2 z1., , … (i), , |z1 z 2|2 (z1 z 2) (z1 z 2) (z1 z 2) (z1 z 2), z1 z1 z 2 z 2 z1 z 2 z 2 z1, |z1|2 |z 2|2 z1 z 2 z 2 z1 ., , … (ii), , On adding the corresponding sides of (i) and (ii), we get, |z1 z 2|2 |z1 z 2|2 2 (|z1|2 |z 2|2) ., , EXERCISE 5C, 1. Express each of the following in the form (a ib) and find its conjugate:, (2 i ), 1, (i), (ii) (2 3i ) 2, (iii), ( 4 3i ), (1 2i ) 2, (iv), , (1 i ) (1 2i ), ( 1 3i ), , 1 2i 2, (v) b l, 2 i, , (vi), , (2 i ), (3 i ) (1 2i ), , 2. Express each of the following in the form (a ib) and find its multiplicative, inverse:, ( 1 7i ), 4, 1 2i, (i) , (ii), (iii), 1 3i, (1 i 3 ), (2 i ) 2, u v, 3. If (x iy) 3 (u iv) then prove that b x y l 4 (x 2 y 2) ., x y, 4. If (x iy) 1/3 (a ib) then prove that d a n 4 (a 2 b 2) ., b, 3, 5. Express (1 2i ) in the form (a ib) .
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Complex Numbers and Quadratic Equations, , 189, , 6. Find real values of x and y for which, (x 4 2xi ) (3x 2 iy) (3 5i ) (1 2iy)., 7. If z 2 |z| 2 0, show that z is purely imaginary., z1, 8. If is purely imaginary and z ! 1, show that|z| 1., z 1, z1 1, 9. If z1 is a complex number other than –1 such that|z1| 1 and z 2 then, z1 1, show that z 2 is purely imaginary., 10. For all zdC, prove that, 1 , (z z ) Re (z), 2, (iv) (z z ) is real, (i), , 1 , 2, (z z ) Im (z) (iii) zz |z|, 2i, (v) (z z ) is 0 or imaginary., , (ii), , 11. If z1 (1 i ) and z 2 ( 2 4i ), prove that Im d, , z1 z 2, n 2., z1, , 12. If a and b are real numbers such that a 2 b 2 1 then show that a real value, 1 ix, of x will satisfy the equation, (a ib) ., 1 ix, ANSWERS (EXERCISE 5C), , 1. (i) z b, , 4 3 , 4 3, il z b, il, 25 25, 25 25, 2 11, 2 11, (iii) z b il, z b il, 25 25, 25 25, 7, 24, 7 24, (v) z b il, z b il, 25 25, 25 25, , 2. (i) z b, , 1 1, il, z 1 (1 i ), 2 2, , (iii) z (1 i 3 ), z, 5. b, , 1, , (ii) z ( 5 12i ), z ( 5 12i ), 4 3, 4 3, (iv) z b il, z b il, 5 5, 5 5, 3, 3, 1, 1, (vi) z b il, z b il, 10 10, 10 10, (ii) z (1 i ), z, , 1, , , b 1 1 il, 2 2, , 3, , e 1 , io, 4, 4, , 11, 2 il, 125 125, , 1, 6. (x 2, y 3) or bx 2, y l, 3, HINTS TO SOME SELECTED QUESTIONS, , 3. (u iv) (x iy) 3 x 3 iy 3 3ixy (x iy), , &, , (u iv) (x 3 3xy 2) i (3x 2y y 3), , &, , u x 3 3x 2y and v 3x 2y y 3, , &, , c m (x 2 3y 2) (3x 2 y 2) 4 (x 2 y 2) ., , u, x, , v, y, , 4. (x iy) (a ib) 3 a 3 ib 3 3iab (a ib), , &, , (x iy) (a 3 3ab 2) i (3a 2b b 3)
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Complex Numbers and Quadratic Equations, , 191, , POLAR REPRESENTATION OF COMPLEX NUMBERS, COMPLEX PLANE OR ARGAND PLANE, , Let XlOX and YOYl be the mutually perpendicular lines, known as the x-axis, and the y-axis respectively. The complex number (x iy) corresponds to the, ordered pair (x, y) and it can be represented by the point P(x, y) in the x-y plane., The x-y plane is known as the complex plane, or the Argand plane. x-axis is called, the real axis and y-axis is called the imaginary axis., A(3, 5), , Y, B(–4, 3), G(0, 2), , X¢, , F(–3, 0), , O, , E(6, 0) X, , H(0, –1), D(2, –4), , C(–5, –3), Y¢, , Note that every number on the x-axis is a real number, while each on the, y-axis is an imaginary number., The complex numbers represented geometrically in the above diagram are, (3 5i ), ( 4 3i ), ( 5 3i ), (2 4i ),, (6 0i ), ( 3 0i ), (0 2i ) and (0 i ),, represented by the points, A (3, 5), B ( 4, 3), C ( 5, 3), D (2, 4), E (6, 0),, F ( 3, 0), G (0, 2) and H (0, 1) respectively., POLAR FORM OF A COMPLEX NUMBER, , Let the complex number z x iy be, represented by the point P(x, y) in the, complex plane., Let +XOP and|OP| r 0. Then,, P(r, ) are called the polar coordinates, of P., , Y, P(r, q), r, q, X¢, , O, , We call the origin O as pole., Clearly, x r cos and y r sin , , , z r (cos i sin ) ., , y, , Y¢, , x, , X
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192, , Senior Secondary School Mathematics for Class 11, , This is called the polar form, or trigonometric form, or modulus-amplitude, form, of z., Here, r x 2 y 2 |z|is called the modulus of z., And is called the argument, or amplitude of z, written as arg(z), or amp(z)., The value of such that # is called the principal argument of z., METHOD FOR FINDING THE PRINCIPAL ARGUMENT OF A COMPLEX NUMBER, Case I, , When z (x iy) lies on one of the axes:, I. When z is purely real., In this case, z lies on the x-axis., (i) If z lies on positive side of the x-axis then 0., (ii) If z lies on negative side of the x-axis then ., II. When z is purely imaginary., In this case, z lies on the y-axis., ·, 2, , ·, (ii) If z lies on the y-axis and below the x-axis then , 2, (i) If z lies on the y-axis and above the x-axis then , , Case II When z (x iy) does not lie on any axes:, Step 1., , Find the acute angle given by tan , , Step 2., , Find the quadrant in which P(x, y) lies., , Im (z), ·, Re (z), , Then, arg (z) or amp (z) may be obtained as under., (i) When z lies in quad. I;, then, & arg (z) ., (ii) When z lies in quad. II;, then, ( ) & arg (z) ( ) ., (iii) When z lies in quad. III;, then, ( ) or ( ), & arg (z) ( ) or ( ) ., (iv) When z lies in quad. IV;, then, or (2 ), & arg (z) or (2 ) .
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Complex Numbers and Quadratic Equations, , Y, , 193, , Y, , P, , P, =( – ), , X¢, , M, , O, , X, , X¢, , O, , M, , Y¢, , Y¢, , Y, , Y, , M, X¢, , O, , ( – ), , X, , M, , X¢, , O, , X, , P, , P, Y¢, , Y¢, , SUMMARY, , Polar form of z x iy is r (cos i sin ) ., (i) r |z| x 2 y 2 ., y, Im (z), ·, (ii) tan x , Re (z), (iii) When # , then is the principal argument of z., (iv), , X, , Quad. in which z lies, , arg (z), , I, , , , II, , , , III, , ( ), , IV, , or (2 ), , (v) When z is purely real;, then, z lies on the x-axis., (x 0 & 0) and (x 0 & ) ., (vi) When z is purely imaginary;, then, z lies on the y-axis., , , k·, ay 0 & k and ay 0 & , 2, 2
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194, , Senior Secondary School Mathematics for Class 11, , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Convert each of the following complex numbers into polar form:, (i) 3, (ii) –5, (iii) i, (iv) 2i, (i) The given complex number is z 3 0i., Let its polar form be z r (cos i sin ) ., Now, r |z| 3 2 0 2 9 3., Clearly, z 3 0i is represented by the point P(3, 0), which lies, on the positive side of the x-axis., arg (z) 0., Thus, r 3 and 0., Hence, the required polar form of z 3 0i is 3 (cos 0 i sin 0) ., (ii) The given complex number is z 5 0i., Let its polar form be z r (cos i sin ) ., Now, r |z| ( 5) 2 0 2 25 5., Clearly, z 5 0i is represented by the point P( 5, 0), which, lies on the negative side of the x-axis., arg (z) & ., Thus, r 5 and ., Hence, the required polar form of z 5 0i is, 5 (cos i sin ) ., (iii) The given complex number is z 0 i., Let its polar form be z r (cos i sin ) ., Now, r |z| 0 2 1 2 1., Clearly, z 0 i is represented by the point P (0, 1), which lies, on the y-axis and above the x-axis., , , arg (z) & ·, 2, 2, , Thus, r 1 and ·, 2, Hence, the required polar form of z 0 i is, , , , , 1 · acos i sin k, i.e., acos i sin k ·, 2, 2, 2, 2, (iv) The given complex number is z 0 2i., Let its polar form be z r (cos i sin ) ., Now, r |z| 0 2 ( 2) 2 4 2., Clearly, z (0 2i ) is represented by the point P (0, 2), which, lies on the y-axis and below the x-axis., , , ·, arg (z) , & , 2, 2, , ·, Thus, r 2 and , 2
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Complex Numbers and Quadratic Equations, , 195, , Hence, the required polar form of z 0 2i is, , , , , z 2 'cos a k i sin a k1 , i.e., 2 acos i sin k ·, 2, 2, 2, 2, EXAMPLE 2, , SOLUTION, , Find the modulus and argument of each of the complex numbers given below:, (ii) 3 i, (iii) 1 i 3, (i) 1 i, (i) Let z 1 i. Then,|z| 1 2 1 2 2 ., Let be the acute angle given by, Im (z), 1 1 & ·, tan , 1, 4, Re (z), Clearly, the point representing z 1 i is P(1, 1), which lies in, the first quadrant., , arg (z) ·, 4, , Hence,|z| 2 and arg (z) ·, 4, (ii) Let z 3 i. Then,|z| ( 3 ) 2 1 2 4 2., Let be the acute angle given by, Im (z), 1 1 & ·, tan , 6, Re (z), 3, 3, Now, z ( 3 i ) is represented by the point P ( 3 , 1), which, lies in the second quadrant., 5, , ·, arg (z) ( ) a k , 6, 6, 5, ·, Hence,|z| 2 and arg (z) , 6, (iii) Let z 1 i 3 . Then,|z| (1) 2 ( 3 ) 2 4 2., Let be the acute angle given by, tan , , 3, Im (z), , 3 & ·, 1, 3, Re (z), , Now, z (1 i 3 ) is represented by the point P (1, 3 ),, which lies in the third quadrant., 2, , ·, arg (z) ( ) a k , 3, 3, 2, ·, Hence,|z| 2 and arg (z) , 3, EXAMPLE 3, , Convert the complex number (1 i 3 ) into polar form., , SOLUTION, , The given complex number is z (1 i 3 ) ., Let its polar form be z r (cos i sin ) ., Now, r |z| 1 2 ( 3 ) 2 4 2.
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196, , Senior Secondary School Mathematics for Class 11, , Let be the acute angle, given by, Im (z), 3, , 3 & ·, 3, 1, Re (z), Clearly, the point representing z (1 i 3 ) is P (1, 3 ), which lies, in the first quadrant., , , arg (z) ·, 3, , Thus, r |z| 2 and arg (z) ·, 3, , , Hence, the required polar form of z (1 i 3 ) is 2 acos i sin k ·, 3, 3, tan , , EXAMPLE 4, , Convert the complex number (2 3 2i ) into polar form., , SOLUTION, , The given complex number is z (2 3 2i ) ., Let its polar form be z r (cos i sin ) ., Now, r |z| (2 3 ) 2 ( 2) 2 12 4 16 4., Let be the acute angle, given by, Im (z), 2, , 1 & ·, tan , 6, Re (z), 2 3, 3, Clearly, the point representing z (2 3 2i ) is P (2 3 , 2), which, lies in the fourth quadrant., , ·, , arg (z) , 6, , ·, Thus, r |z| 4 and arg (z) , 6, Hence, the required polar form of z (2 3 2i ) is, , , , , 4 ;cos a k i sin a kE 4 acos i sin k ·, 6, 6, 6, 6, , EXAMPLE 5, , Convert the complex number ( 2 2i 3 ) into polar form., , SOLUTION, , The given complex number is z ( 2 2i 3 ) ., Let its polar form be z r (cos i sin ) ., Now, r |z| ( 2) 2 (2 3 ) 2 4 12 16 4., Let be the acute angle, given by, tan , , Im (z), 2 3, , 3 & ·, 2, 3, Re (z), , Clearly, the point representing z ( 2 2i 3 ) is P ( 2, 2 3 ), which, lies in the second quadrant., , 2 ·, , arg (z) ( ) a k , 3, 3, 2, , ·, Thus, r |z| 4 and , 3, 2 , 2, Hence, the required polar form is z 4 bcos, i sin l ·, 3, 3
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Complex Numbers and Quadratic Equations, EXAMPLE 6, , Convert, , SOLUTION, , Let z , , ( 1 7i ), (2 i ) 2, , ( 1 7i ), (2 i ) 2, , 197, , into polar form., , , (1 7i ), (4 i 2 4i ), , , , (1 7i ) (3 4i ), #, (3 4i ) (3 4i ), , (1 7i ) (3 4i ) 25 25i, , ( 1 i ) ., 25, (9 16), Let its polar form be z r (cos i sin ) ., , &, , z, , Now, r |z| (1) 2 1 2 2 ., Let be the acute angle, given by, tan , , Im (z), 1 1 & ·, 1, 4, Re (z), , Clearly, the point representing z (1 i ) is P (1, 1), which lies in, the second quadrant., 3, , ·, , arg (z) ( ) a k , 4, 4, 3, ·, Thus, r |z| 2 and , 4, 3 , 3, Hence, the required polar form is z 2 bcos, i sin l ·, 4, 4, EXAMPLE 7, , SOLUTION, , Convert the complex number, , 1, into polar form., (1 i ), , ( 1 i ) ( 1 i ) (1 i ), 1, , , b1 1 il ·, #, 2, 2 2, (1 i ) (1 i ) (1 i 2), Let its polar form be z r (cos i sin ) ., Let z , , 1 2 1 2, 11 1 1 ·, b l b l , 2, 2, 2, 4 4, 2, Let be the acute angle, given by, Im (z), (1/2), , 1 & ·, tan , 4, Re (z), (1/2), 1, 1, 1 1, Clearly, the point representing z b il is P b , l, which lies, 2 2, 2 2, in the fourth quadrant., , , arg (z) ·, 4, , 1, ·, Thus, r |z| , and , 4, 2, Hence, the required polar form is, , , , , 1, 1, 'cos a k i sin a k1 , i.e.,, z, acos i sin k ·, 4, 4, 4, 4, 2, 2, Now, r |z|
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198, , Senior Secondary School Mathematics for Class 11, , (i 1), , into polar form., i sin k, acos, 3, 3, ( i 1), (i 1), 2 (i 1), , , z, , , i sin k e 1 3 o (1 i 3 ), acos, i, 3, 3, 2, 2, , EXAMPLE 8, , Convert the complex number, , SOLUTION, , Let, , &, , z, , 2 (i 1), (1 i 3 ) (i 1) (1 i 3 ), , #, 2, (1 i 3 ) (1 i 3 ), , ( 3 1) ( 3 1 ), ), , i3 ·, 2, 2, Let its polar form be z r (cos i sin ) ., ( 3 1) 2 ( 3 1) 2, 3 8 2 & r 2., , Now, r 2 |z|2 ), 4, 4, 4, Let be the acute angle, given by, , &, , &, , 1, e1 , o, , 3, Im (z), ( 3 1), (, 3, 1, ), 2, , , , #, tan , 2, Re (z), 1, ( 3 1) ( 3 1), e1 , o, 3, , , atan tan k, 6 , 5, 4, , tan , tan a k tan, , , 12, 4 6, a1 tan · tan k, 6, 4, 5, ·, , 12, , Clearly, the point representing the given complex number is, 31 31, ,, o, which lies in the first quadrant., Pe, 2, 2, 5, 5, ·, & , 12, 12, 5, ·, Thus, r |z| 2 and , 12, 5 , 5, Hence, the required polar form is z 2 bcos, i sin l ·, 12, 12, , , , EXAMPLE 9, SOLUTION, , arg (z) , , Express the complex number (1 i ) in polar form., The given complex number is z 1 i., Let its polar form be z r (cos i sin ) ., Now, r |z| (1) 2 (1) 2 2 ., Let be the acute angle, given by, tan , , Im (z), , 1 1 & ·, 1, 4, Re (z)
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Complex Numbers and Quadratic Equations, , 199, , Clearly, the point representing the complex number z 1 i is, P (1, 1), which lies in the third quadrant., 3, , ·, , arg (z) ( ) a k , 4, 4, 3, ·, Thus, r |z| 2 and , 4, Hence, the required polar form of z (1 i ) is given by, z 2 (cos b, , 3, 3, 3 , 3, l i sin b, l2 , i.e., 2 bcos, i sin l ·, 4, 4, 4, 4, , EXAMPLE 10, , Express the complex number ( 3 i ) in polar form., , SOLUTION, , The given complex number is z ( 3 i ) ., Let its polar form be z r (cos i sin ) ., Now, r |z| ( 3 ) 2 (1) 2 4 2., Let be the acute angle, given by, Im (z), 1, , 1 & ·, tan , 6, Re (z), 3, 3, Clearly, the point representing the complex number z ( 3 i ) is, P ( 3 , 1), which lies in the third quadrant., 5, , ·, , arg (z) ( ) a k , 6, 6, 5, ·, Thus, r |z| 2 and , 6, Hence, the polar form of z ( 3 i ) is given by, z 2 (cos b, , EXAMPLE 11, , SOLUTION, , 5, 5, 5 , 5, l i sin b, l2 , i.e., 2 bcos, i sin l ·, 6, 6, 6, 6, , 1 2i, into polar form., 1 3i, (1 2i ) (1 3i ) (1 2i ) (1 3i ) 5 5i, , , , b 1 1 il ·, Let z , #, 2 2, 10, (1 3i ) (1 3i ), ( 1 9), , Convert the complex number, , Let its polar form be z r (cos i sin ) ., Now, r |z| , , b, , 1 2 1 2, l b l , 2, 2, , 1 1 ·, 2, 2, , Let be the acute angle, given by, Im (z), (1/2), , 1 & ·, (1/2), 4, Re (z), 1 1, 1 1, Clearly, the point representing z b il is P b , l, which lies, 2 2, 2 2, in the second quadrant., tan
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200, , Senior Secondary School Mathematics for Class 11, , 3, , ·, arg (z) ( ) a k , 4, 4, 3, 1, ·, Thus, r |z| , and , 4, 2, , , Hence, the required polar form is z , EXAMPLE 12, SOLUTION, , Express the complex number sin, , 3 , 3, 1, i sin l ·, bcos, 4, 4, 2, , , , i a1 cos k in polar form., 5, 5, , , , i a1 cos k ·, 5, 5, Let its polar form be z r (cos i sin ) ., Let z sin, , , 2, , , , a1 cos k asin 2 cos 2 k 1 2 cos, 5, 5, 5, 5, 5, , , 2, 2 , ·, &, r 2 a1 cos k 4 sin, & r 2 sin, 5, 10, 10, Let be the acute angle, given by, , 2 , 2 sin, 1 cos, Im (z), 5, 10, , , tan & ·, tan , , ·, , 10, 10, Re (z), 2 sin, sin, cos, 5, 10, 10, Clearly, the point representing z lies in the first quadrant as x 0, and y 0., ·, , arg (z) , 10, , , ·, Thus, r 2 sin, and , 10, 10, , , , Hence, the required polar form is 2 sin acos, i sin k ·, 10, 10, 10, Now, r 2 |z|2 sin 2, , EXAMPLE 13, SOLUTION, , Convert 4 (cos 300c i sin 300c) into Cartesian form., 4 (cos 300c i sin 300c) 4 [cos (360c 60c) i sin (360c 60c)], 4 [cos ( 60c) i sin ( 60c)], 3, o, 4 [cos 60c i sin 60c] 4 e 1 i ·, 2, 2, (2 2i 3 ) ., , EXAMPLE 14, , For any complex numbers z, z1 and z 2 prove that:, (i) arg (z) arg (z), (ii) arg (z1 z 2) arg (z1) arg (z 2), (iii) arg (z1 z 2) arg (z1) arg (z 2), z1, (iv) arg d z n arg (z1) arg (z 2), 2, , SOLUTION, , (i) Let z r (cos i sin ) . Then,|z| r and arg (z) ., Now, z r (cos i sin )
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202, , Senior Secondary School Mathematics for Class 11, , EXERCISE 5D, Find the modulus and argument of each of the following complex numbers and, hence express each of them in polar form:, , 1. 4, 5. 1 i, 9. 1 3 i, 1i, 13. , 1 i, 17., , 5i, 2 3i, , 21. 3 i, , 2. 2, 6. 1 i, 10. 2 2i, 14., , 1i, 1i, , 18., , 16, 1 3i, , 22. (i 25) 3, , 24. (sin 120c i cos 120c), , 3. i, 7. 3 i, 11. 4 4 3 i, 1 3i, 15. , 1 2i, 26 3i, 19., 5 3i, (1 i ), 23., , i sin, acos, 3, , 4. 2i, 8. 1 3 i, 12. 3 2 3 2 i, 1 3i, 16. , 1 2i, 20., , 1i, 1i, , , k, 3, , ANSWERS (EXERCISE 5D), , 1. 4, 0, 4 (cos 0 i sin 0), , , , , 'cos a k i sin a k1, 3. 1,, 2, 2, 2, , , , , 2 'cos a k i sin a k1, 5. 2 ,, 4, 4, 4, ,, , , 7. 2,, 2 acos, i sin k, 6, 6, 6, , , , , 2 'cos a k i sin a k1, 9. 2,, 3, 3, 3, 2 ,, 2 , 2, 11. 8,, 8 'cos, i sin 1, 3, 3, 3, ,, , , 13. 1,, i sin k, acos, 2, 2, 2, 3 ,, 3 , 3, 15. 2 ,, 2 bcos, i sin l, 4, 4, 4, ,, , , 17. 2 ,, 2 acos, i sin k, 4, 4, 4, ,, , , 19. 2,, 2 acos, i sin k, 3, 3, 3, 5, 5, 5, , 2 (cos b, l i sin b, l2, 21. 2,, 6, 6, 6, 7 , 7, 7 ,, 23. 2 ,, 2 bcos, i sin l, 12, 12, 12, , 2. 2, , 2 (cos i sin ), , , , 4. 2, , 2 acos i sin k, 2, 2, 2, 3 ,, 3 , 3, 6. 2 ,, 2 bcos, i sin l, 4, 4, 4, 2 ,, 2 , 2, 8. 2,, 2 bcos, i sin l, 3, 3, 3, , , , , 2 2 'cos a k i sin a k1, 10. 2 2 ,, 4, 4, 4, 3 ,, 3 , 3, 12. 6,, 6 bcos, i sin l, 4, 4, 4, , , , , 'cos a k i sin a k1, 14. 1,, 2, 2, 2, 3, 3, 3, , (cos b, l i sin b, l2, 16. 2 ,, 4, 4, 4, 2 ,, 2 , 2, 18. 8,, 8 bcos, i sin l, 3, 3, 3, ,, , , 20. 1,, i sin k, acos, 4, 4, 4, , , , , 'cos a k i sin a k1, 22. 1,, 2, 2, 2, 24. (cos 30c i sin 30c), HINTS TO SOME SELECTED QUESTIONS, , 21. Let z 3 i & r 2 |z| 2 ( 3 ) 2 (1) 2 4 & r |z| 2., 1, 1 & ·, tan , 6, 3, 3
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Complex Numbers and Quadratic Equations, , 203, , The given number represents the point P ( 3 , 1) which lies in the third quadrant., 5, , ·, arg (z) ( ) a k , 6, 6, 5, 5, m i sin c, mF ·, z 2 <cos c, 6, 6, 23. Given number , , 2 (1 i ), (1 3 i ), , #, , (1 3 i ), (1 3 i ), , (1 3 ) (1 3 ), ), , i3, 2, 2, , (1 3 ) 2 (1 3 ) 2 2 (1 3), , , 2 & |z| 2 ., 4, 4, 4, 1, (1 3 ), , , e1 , o, tan tan, 3, ( 3 1), 6, 4, 2, , , , tan , ·, , 1, (1 3 ), ( 3 1), , 1, tan, tan, e1 , o, 6, 4, 3, 2, tan a k tan 5 ·, 12, 4 6, ( 3 1) ( 3 1), ,, o·, The given number z is represented by the point P e, 2, 2, So, it lies in the third quadrant., 7, 5, ·, arg (z) ( ) c m , 12, 12, , , |z|2 , , Hence, z 2 (cos c, , 7, 7, m i sin c, m2 ·, 12, 12, , QUADRATIC EQUATIONS (With Complex Roots), FUNDAMENTAL THEOREM OF ALGEBRA, , A polynomial equation of degree n has at the most n roots., SOLVING A QUADRATIC EQUATION, , Let the given equation be ax 2 bx c 0, where a, b, cd R and a ! 0., Then, ax 2 bx c 0, a ! 0, &, &, &, , ax 2 bx c, c, b, x2 a · x a, b, b 2 c, b 2, x2 a · x b l a b l, 2a, 2a, , &, , bx , , &, , x, , 2, b 2 (b 4ac), l, 2, 2a, 4a, , b 2 4ac, b , !, 2a, 2a, , [adding b, , b 2, l on both sides], 2a
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204, , Senior Secondary School Mathematics for Class 11, , &, , x, , &, , x, , b 2 4ac, b, !, 2a, 2a, , b ! b 2 4ac, ·, 2a, , Hence, the roots of the equation ax 2 bx c 0, a ! 0 are, b b 2 4ac, b b 2 4ac, ·, and, 2a, 2a, The expression (b 2 4ac) is called the discriminant., If (b 2 4ac) 0, the given quadratic equation will have complex roots., Here, we shall consider only the quadratic equations having complex roots., SOLVED EXAMPLES, EXAMPLE 1, SOLUTION, , Solve: x 3 0., We have, x 2 3 0 & x 2 3 & x ! 3 !i 3 ., 2, , , EXAMPLE 2, SOLUTION, , solution set {i 3 , i 3 } ., , Solve: x 2 3x 9 0., The given equation is x 2 3x 9 0., This is of the form ax 2 bx c 0, where a 1, b 3 and c 9., , (b 2 4ac) (3 2 4 #1# 9) (9 36) 27 0., So, the given equation has complex roots., These roots are given by, , , , b ! b 2 4ac 3 ! 27, , [a b 2 4ac 27], 2a, 2 #1, 3 ! i 27 3 ! i 3 3, , , ·, 2, 2, 3 i3 3 3 i3 3, ,, 3, solution set ), 2, 2, 3 3 3 3 3, , , ) 3, i,, i3 ·, 2, 2, 2, 2, , EXAMPLE 3, , Solve: 9x 2 10x 3 0., , SOLUTION, , The given equation is 9x 2 10x 3 0., This is of the form ax 2 bx c 0, where a 9, b 10 and c 3., , (b 2 4ac) {(10) 2 4 # 9 # 3} (100 108) 8 0., So, the given equation has complex roots.
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Complex Numbers and Quadratic Equations, , 205, , These roots are given by, , , , b ! b 2 4ac 10 ! 8, , [a (b 2 4ac) 8], 2a, 2#9, 10 ! i 2 2 5 ! i 2, , , ·, 9, 18, 5 i 2 5 i 2, 2 5, 2, , ,, 3) 5 , , solution set ), i,, i3 ·, 9, 9, 9, 9, 9, 9, 2 x 2 x 2 0., , EXAMPLE 4, , Solve:, , SOLUTION, , The given equation is 2 x 2 x 2 0., This is of the form ax 2 bx c 0, where a 2 , b 1 and c 2 ., , (b 2 4ac) (1 2 4 # 2 # 2 ) (1 8) 7 0., So, the given equation has complex roots., These roots are given by, , , , EXAMPLE 5, SOLUTION, , b ! b 2 4ac 1 ! 7, , [a b 2 4ac 7], 2a, 2 2, 1 ! i 7, , ·, 2 2, Z] , _, ] 1 i 7 1 i 7 bb, ,, solution set [], `b, ] 2 2, 2 2 b, \Z, a, _, ]] 1, 7, 7 bb, 1, [], , , i,, i`b ·, ]2 2 2 2 2 2 2 2 b, \, a, , Solve:, 3 x 2 2 x 3 3 0., The given equation is, 3 x 2 2 x 3 3 0., This is of the form ax 2 bx c 0, where a 3 , b 2 and c 3 3 ., , (b 2 4ac) {( 2 ) 2 4 # 3 # 3 3 } (2 36) 34 0., So, the given equation has complex roots., These roots are given by, b ! b 2 4ac, , 2a, , , , 2 ! 34, , [a b 2 4ac 34], , 2# 3, 2 ! i 34, 2 3, , ·, , ]Z] 2 i 34 2 i 34 b_b, ,, solution set [], `b, ] 2 3, 2 3 b, \Z, a_, ]] 1, 34, 34 bb, 1, [], , , i,, i`b ·, ] 6 2 3, 6 2 3 b, \, a
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206, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 6, SOLUTION, , Solve: 3x 2 8ix 3 0., The given equation is, 3x 2 8ix 3 0., This is of the form ax 2 bx c 0, where a 3, b 8i and c 3., , (b 2 4ac) {(8i ) 2 4 # 3 # 3} ( 64 36) 100 0., So, the given equation has complex roots., These roots are given by, b ! b 2 4ac 8i ! 100, , 2a, 2# 3, 8i ! 10i 4i ! 5i, , , ·, 6, 3, , [a b 2 4ac 100], , Thus, the roots of the given equation are, 4i 5i i, , , and 4i 5i 9i 3i., 3, 3, 3, 3, i, , solution set ' , 3i1 ·, 3, , EXERCISE 5E, Solve:, 1. x 2 2 0, 4. x 2 x 1 0, , 2. x 2 5 0, 5. x 2 x 2 0, , 3. 2x 2 1 0, 6. x 2 2x 2 0, , 7. 2x 2 4x 3 0, 10. 25x 2 30x 11 0, , 8. x 2 3x 5 0, 11. 8x 2 2x 1 0, , 13. 2x 2 3 x 1 0, 20 , 16. 3x 2 4x , 0, 3, 19. x 2 13 4x, , 14. 17x 2 8x 1 0, , 5 x2 x 5 0, 12. 27x 2 10x 1 0, 15. 3x 2 5 7x, , 17. 3x 2 7ix 6 0, , 18. 21x 2 28x 10 0, , 20. x 2 3ix 10 0, , 21. 2x 2 3ix 2 0, , 9., , ANSWERS (EXERCISE 5E), , 1. {i 2 , i 2 }, , 3. ), , 2. {i 5 , i 5 }, , 3 1, 3, 7 1, 7, 1, 1, , 4. ) , i,, i3 5. ) , i, , i3, 2, 2, 2, 2, 2, 2 2, 2, 7. )1 , , 1, 1, i, 1 , i3, 2, 2, , i , i, 3, 2, 2, , 6. {1 i, 1 i }, , 11 3, 11, 3, , 8. ) , i,, i3, 2, 2, 2, 2, , Z], _, 19, 19 bb, 1, ] 1, , , 9. [], i,, i`b, ]2 5 2 5 2 5 2 5 b, \, a, , 2 3, 2, 3, 10. ) , i, , i3, 5, 5 5, 5
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Complex Numbers and Quadratic Equations, , 7 1, 7, 1, , 11. ) , i,, i3, 8, 8, 8, 8, , 2 5, 2, 5, , 12. ) , i,, i3, 27 27 27 27, , 13. ), , 4, 1, 4 1, 14. ' i,, i1, 17 17 17 17, , 3, 5, 3, 5, , , i,, i3, 4, 4, 4, 4, , 11 7, 11, 7, 15. ) , i, , i3, 6, 6, 6, 6, , 2 4 2 4, 16. ' i, i1, 3 3 3 3, , 2, 17. ' i, 3i1, 3, , 14 2, 14, 2, 18. ) , i, , i3, 3 21 3 21, , 19. {2 3i, 2 3i }, , 20. {2i, 5i }, , 207, , 1, 21. ' i, 2i1, 2, HINTS TO SOME SELECTED QUESTIONS, , 17. a 3, b 7i and c 6., (b 2 4ac) {(7i ) 2 4 # 3 # 6} ( 49 72) (121) 0., So, the given equation has complex roots, given by, b ! b 2 4ac 7i ! 121 7i ! 11i, , , ·, 2a, 2# 3, 6, 7i 11i , 7i 11i, 2, 1 ' i, 3i1 ·, Solution set ', 6, 6, 3, , SQUARE ROOTS OF A COMPLEX NUMBER, To Evaluate a ib :, SOLUTION, , Let a ib (x iy) ., On squaring both sides of (i), we get, , … (i), , (a ib) (x iy) 2, … (ii), &, (a ib) (x 2 y 2) i (2xy) ., On comparing real parts and imaginary parts on both sides of (ii),, we get, (x 2 y 2) a and 2xy b, &, , ( x 2 y 2) ( x 2 y 2 ) 2 4 x 2 y 2 a 2 b 2 ., , … (iii) and (x 2 y 2) a 2 b 2, , ( x 2 y 2) a, On solving (iii) and (iv), we get, , … (iv), , 1$ 2 2 ., 1, a b a and y 2 $ a 2 b 2 a. ·, 2, 2, 1, 2, 2, $ a b a. i 1 $ a 2 b 2 a. F ·, Hence, a ib !<, 2, 2, x2 , , REMARK, , Similarly, by assuming that a ib (x iy), we may find a ib .
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208, , Senior Secondary School Mathematics for Class 11, , SOLVED EXAMPLES, 6 8i ., , EXAMPLE 1, , Evaluate, , SOLUTION, , Let 6 8i (x iy) ., On squaring both sides of (i), we get, , … (i), , … (ii), 6 8i (x iy) 2 & 6 8i (x 2 y 2) i (2xy) ., On comparing real parts and imaginary parts on both sides of (ii),, we get, x 2 y 2 6 and 2xy 8, &, , x 2 y 2 6 and xy 4, , &, , (x 2 y 2) (x 2 y 2) 2 4x 2y 2 6 2 4 #16 100 10, , &, , x 2 y 2 6 and x 2 y 2 10, , &, , 2x 2 16 and 2y 2 4, , &, , x 2 8 and y 2 2, , &, x !2 2 and y ! 2 ., Since xy 0, so x and y are of the same sign., , , (x 2 2 and y 2 ) or (x 2 2 and y 2 ) ., , Hence, 6 8i (2 2 2 i ) or ( 2 2 2 i ) ., 5 12i ., , EXAMPLE 2, , Evaluate, , SOLUTION, , Let 5 12i (x iy) ., On squaring both sides of (i), we get, , … (i), , 5 12i (x iy) 2 & 5 12i (x 2 y 2) i (2xy) ., … (ii), On comparing real parts and imaginary parts on both sides of (ii),, we get, x 2 y 2 5 and 2xy 12, &, , x 2 y 2 5 and xy 6, , &, , (x 2 y 2) (x 2 y 2) 2 4x 2y 2 ( 5) 2 4 # 36 169 13, , &, , x 2 y 2 5 and x 2 y 2 13, , &, , 2x 2 8 and 2y 2 18, , &, x 2 4 and y 2 9, &, x !2 and y !3., Since xy 0, so x and y are of the same sign., , (x 2 and y 3) or (x 2 and y 3)., Hence, 5 12i (2 3i ) or ( 2 3i ) .
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Complex Numbers and Quadratic Equations, , 209, , 8 15i ., , EXAMPLE 3, , Evaluate, , SOLUTION, , Let 8 15i (x iy) ., On squaring both sides of (i), we get, (8 15i ) (x iy) 2 & (8 15i ) (x 2 y 2) i (2xy) ., , … (i), , … (ii), On comparing real parts and imaginary parts on both sides of (ii),, we get, x 2 y 2 8 and 2xy 15, , &, , 15, 2, (x 2 y 2) (x 2 y 2) 2 4x 2y 2 64 225 289 17, , &, , x 2 y 2 8 and x 2 y 2 17, , &, , 2x 2 25 and 2y 2 9, 25, 9, x2 , and y 2 , 2, 2, 5, 3, ·, x !, and y !, 2, 2, , &, , &, &, , x 2 y 2 8 and xy , , Since xy 0, so x and y are of the same sign., 5, 5 3, 3 io ·, Hence, 8 15i e, io or e, 2 2, 2, 2, 24 10i ., , EXAMPLE 4, , Evaluate, , SOLUTION, , Let 24 10i (x iy) ., … (i), On squaring both sides of (i), we get, … (ii), ( 24 10i ) (x 2 y 2) i (2xy) ., On comparing real parts and imaginary parts on both sides of (ii),, we get, x 2 y 2 24 and 2xy 10, &, , x 2 y 2 24 and xy 5, , &, , (x 2 y 2) (x 2 y 2) 2 4x 2y 2 ( 24) 2 4 # 25 676 26, , &, , x 2 y 2 24 and x 2 y 2 26, , &, , 2x 2 2 and 2y 2 50, , &, x 2 1 and y 2 25 & x !1 and y !5., Since xy 0, so x and y are of the same sign., , , 24 10i (1 5i ) or ( 1 5i ) ., i ., , EXAMPLE 5, , Evaluate, , SOLUTION, , Let i (x iy) ., On squaring both sides of (i), we get, ( i ) (x iy) 2 & i (x 2 y 2) i (2xy)., , … (i), … (ii)
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210, , Senior Secondary School Mathematics for Class 11, , On comparing real parts and imaginary parts on both sides of (ii),, we get, (x 2 y 2) 0 and 2xy 1., , , (x 2 y 2) (x 2 y 2) 2 4x 2y 2 0 2 1 2 0 1 1 1., , Thus, (x 2 y 2) 0, , … (iii), , and (x 2 y 2) 1, , On solving (iii) and (iv), we get x 2 , , , x !, , … (iv)., , 1, 1, and y 2 ·, 2, 2, , 1, 1 ·, and y !, 2, 2, , Since xy 0, so x and y are of the same sign., 1, , 1 , 1, , y 1o·, , y, ex , o or ex , 2, 2, 2, 2, 1, 1, 1, 1, i·, i·, Hence, i e, o·, o or e, 2, 2, 2, 2, EXAMPLE 6, SOLUTION, , Evaluate, , 4 3 20 4 3 20 ., , 4 3 20 4 6 5 i and 4 3 20 4 6 5 i ., Let 4 6 5 i (x iy)., … (i), On squaring both sides of (i), we get, (4 6 5 i ) (x iy) 2 & (4 6 5 i ) (x 2 y 2) i (2xy). … (ii), On comparing real parts and imaginary parts on both sides of (ii),, we get, (x 2 y 2) 4 and 2xy 6 5 ., , , (x 2 y 2) (x 2 y 2) 2 4x 2y 2 (4) 2 (6 5 ) 2, 16 180 196 14., , Thus, (x 2 y 2) 4, , … (iii), , and (x 2 y 2) 14, , … (iv)., , On solving (iii) and (iv), we get x 9 and y 5., 2, , 2, , , x !3 and y ! 5 ., Since xy 0, so x and y are of the same sign., , , , (x 3, y 5 ) or (x 3, y 5 ) ., 4 3 20 4 6 5 i (3 5 i) or ( 3 5 i ) ., , Similarly, 4 3 20 4 6 5 i (3 5 i) or ( 3 5 i )., Z] , 5 i ) (3 5 i )} 6, ]]{(3, ], or, , 4 3 20 4 3 20 ][, ]], ]{ 3 5 i } { 3 5 i } 6., \, Hence, 4 3 20 4 3 20 6 or 6.
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Complex Numbers and Quadratic Equations, , 211, , EXERCISE 5F, Evaluate:, 1., , 5 12i, , 4., , 1 4 3, , 7 24i, , 5., , 7., , 3 4 7, , 8., , 10., , 15 8i, , 11., , 11 60i, , 13., , 8i, , 14., , 1i, , 2., , 3., , 2 2 3 i, , i, , 6., , 4i, , 16 30i, , 9., , 4 3i, , 12., , 7 30 2, , ANSWERS (EXERCISE 5F), , 1. (3 2i ) or ( 3 2i ), , 2. (3 4i ) or ( 3 4i ), , 3. (1 3 i ) or (1 3 i ), 1, 1 , 5., (1 i ) or, ( 1 i), 2, 2, , 4. (2 3 i ) or ( 2 3 i ), , 7. ( 7 2i ) or ( 7 2i ), , 8. (5 3i ) or ( 5 3i ), , 1, 1 , (1 3i ) or, ( 1 3i ), 2, 2, 11. (5 6i ) or ( 5 6i ), 13. (2 2i ), ( 2 2i ), 21, 3), 2, , 21, 3 i or ), 2, , 2 ( 1 i ), , 10. (1 4i ) or ( 1 4i ), , 9., , 14. ), , 2 (1 i ) or, , 6., , 12. (5 3 2 i ) or ( 5 3 2 i ), 21, 3), 2, , 21, 3i, 2, , HINTS TO SOME SELECTED QUESTIONS, 14. Let 1 i x iy., , … (i), , On squaring both sides, we get, (1 i ) (x iy) 2 (x 2 y 2) i (2xy), , , (x 2 y 2) 1 and 2xy 1, , &, , (x 2 y 2) (x 2 y 2) 2 4x 2y 2 1 2 1 2 2, , &, , x 2 y 2 1 and x 2 y 2 2, , &, , x2 , , 21, and y 2 , 2, , 21, 2, , 21, 21, ·, and y !, 2, 2, Since xy 0, so x and y have the same sign., , &, , x !, , , , )x , , , , 21, , y, 2, , 1i ), , 21, , 2, , 21, 3 or )x , 2, 21, i3 or ), 2, , 21, ,y, 2, 21, i, 2, , 21, 3·, 2, , 21, 3·, 2
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212, , Senior Secondary School Mathematics for Class 11, , EXERCISE 5G, Very-Short-Answer Questions, 1. Evaluate, , 1 ·, i 78, , 2. Evaluate (i 57 i 70 i 91 i 101 i 104 ) ., JKi 180 i 178 i 176 i 174 i 172NO, 3. Evaluate KK 170 168 166 164 162OO ·, Ki i i i i O, P, L, , , 4. Evaluate (i 4n 1 i 4n 1) ., 5. Evaluate ( 36 # 25 ) ., 6. Find the sum (i n i n, , 1, , i n 2 i n 3), where nd N., , 7. Find the sum (i i 2 i 3 i 4 … up to 400 terms) ., 8. Evaluate (1 i 10 i 20 i 30) ., 1, 9. Evaluate di 41 71 n ·, i, , 1i n, 10. Find the least positive integer n for which b l 1., 1 i, 3, , , 11. Express (2 3i ) in the form m (a ib) ., (3 i 5 ) ( 3 i 5 ), in the form (a ib) ., ( 3 2 i) ( 3 2 i), JK3 16 NO, OO in the form (a ib) ., 13. Express KK, K 1 9 O, P, L, 14. Solve for x: (1 i ) x (1 i ) y 1 3i., 12. Express, , 15. Solve for x: x 2 5ix 6 0., 1, ·, 16. Find the conjugate of, (3 4i ), , 17. If z (1 i ), find z 1., , , 18. If z ( 5 3i ), find z 1., 19. Prove that arg (z) arg ( z ) 0., 3, , find z., 4, 21. Find the principal argument of ( 2i ) ., 20. If|z| 6 and arg (z) , , 22., 23., 24., 25., 26., , Write the principal argument of (1 i 3 ) 2., Write 9 in polar form., Write 2i in polar form., Write 3i in polar form., Write z (1 i ) in polar form.
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Complex Numbers and Quadratic Equations, , 213, , 27. Write z (1 i 3 ) in polar form., , 28. If|z| 2 and arg (z) , find z., 4, ANSWERS (EXERCISE 5G), , 1. 1, , 2. i, , 10. n 4, , 3. 1, 11., , 14. x 2, y 1 15., 18. e, , 5 3, io, 14 14, , 20., , 23. 9 (cos i sin ), , 5. 30 6. 0, 7. 0, 7, 12. a 0, b , 46 9i, 2, 3, 4, 16. b il, x 3i or x 2i, 25 25, , 21., z 3 2 ( 1 i ), 2, , , 24. 2 acos, i sin k, 2, 2, 4. 2i, , , , k i sin a k1, 2, 2, 2 , 2, 27. 2 bcos, i sin l, 3, 3, 25. 3 'cos a, , 2 'cos a, , 26., , 8. 0, 9. 2i, 31, 13. b, il, 2 2, 1 1, 17. b il, 2 2, 2, 22., 3, , , , k i sin a k1, 4, 4, , 28. z 2 (1 i ), HINTS TO SOME SELECTED QUESTIONS, , 1., , i2, 1 1, # 1., i 78 i 78 i 2, , [a i 80 (i 4) 20 1], , 2. Given expression (i 48 # i 3 i 68 # i 2 i 88 # i 3 i 100 # i i 104 ), (i 3 i 2 i 3 i 1) ( i 1 i i 1) i., 3. Given expression , , i 172 (i 8 i 6 i 4 i 2 1), i 162 (i 8 i 6 i 4 i 2 1), , i (172 162) i 10 i 8 # i 2 1., , (i 2 1) i 2i, , 1, 4. Given expression (i 4n # i i 4n # i 1) ci m , # 2i., 1, i, i, i, 5. ( 36 # 25 ) (6i # 5i ) 30i 2 30., 6. Given sum i n #(1 i i 2 i 3) i n (1 i 1 i ) 0., 7. Given sum (i i 2 i 3 i 4) (i 5 i 6 i 7 i 8) ……, (i i 2 i 3 1) i 5 (1 i i 2 i 3) … 0., 8. Given sum (1 i 2 1 i 2) 0., 9. ei 41 , , 1 1 1 i , o ei 3 o ei 3 # i o (i i ) 2i., i, i, i 71, , (1 i ) (1 i ) (1 i ) 2 (1 i 2 2i ) 2i, 1i, , , i., 10. d n , #, 2, 2, 1 i, (1 i ) (1 i ) (1 i 2), n, 1i , n, n, 4, d n 1 & i 1 & i i & n 4., 1 i, , [a (1 i i 2 i 3) 0]
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214, , Senior Secondary School Mathematics for Class 11, , 11. (2 3i ) 3 2 3 (3i ) 3 (3 # 2 # 3i ) (2 3i ), (8 27i 3 36i 54) (8 27i 36i 54) (46 9i ) ., (9 5i 2), , , , 14 # i 7i & a 0, b 7 ·, 2 2i i, 2, 2, , , (, 3, 4, i, ), (, 1, 3, i, ), , 3 4i 1 3i, 15 5i 1 (3 i ) ., 13. Given expression # , 2, 10, 1 3i 1 3i, (1 9), 12. Given expression , , 2 2i, , 14. (1 i ) x (1 i ) y 1 3i + (x y) (y x) i 1 3i, x y 1 and y x 3 & x 2, y 1., , , , 15. x 5ix 6 0 & x 2 3ix 2ix 6 0 & x (x 3i ) 2i (x 3i ) 0, & (x 3i ) (x 2i) 0 & x 3i or x 2i., (3 4i ) 3 4i, 1, , c 3 4 im & z c 3 4 im ·, #, 16. z , 25, 25 25, 25 25, (3 4i ) (3 4i ), 2, , 17. z, , 1, , 18. z, , 1, , (1 i ), z2 2, c 1 1 im ·, 2 2, |z| (1 1 2), ( 5 3i ), 5 3, z2 , e, io ·, , 14, 14, (, 5, 9, ), |z|, 3 , 3, , , i sin m 6 'cos a k i sin a k1, 4, 4, 4, 4, 1, 6, , , 1, i·, z 6 a cos i sin k 6 e, (1 i ), o, 4, 4, 2, 2, 2, , 20. z 6 ccos, &, &, , z 3 2 (1 i ) ., , 22. z (1 i 3 ) 2 ( 2 2i 3 ) ., , REVIEW OF FACTS AND FORMULAE, 1., , (i) Imaginary Numbers A given number is said to be imaginary, if its, square is negative, e.g., 1 , 2 , 3 , 4 , etc., (ii) We denote 1 by the Greek letter iota ‘’ transliterated as ‘i ’., (iii) i 0 1, i 1 i, i 2 1, i 3 i and i 4 1., (iv) Let n be a positive integer. On dividing n by 4, let m be the quotient, and r be the remainder., Then, n 4m r, where 0 # r 4., i n i 4m, (v), , r, , ( i 4) m · i r i r ., , [a i 4 1], , 16 # 9 (4i )#(3i ) 12i 2 12., , 2. Complex Numbers, (i) The numbers of the form z (a ib) are called complex numbers,, where a, bd R. Here, Re (z) a and Im (z) b.
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Complex Numbers and Quadratic Equations, , 215, , If b 0 then z is purely real., If a 0 then z is purely imaginary., (ii) Let z1 (a1 ib1) and z 2 (a 2 ib 2) then, I. (z1 z 2) (a1 a 2) i (b1 b 2), II. (z1 z 2) (a1 a 2) i (b1 b 2), III. z1 z 2 (a1 a 2 b1 b 2) i (a1 b 2 a 2 b1) ., (iii) The conjugate of z (a ib) is z (a ib) (a ib) ., (iv) The modulus of z (a ib) is|z| a 2 b 2 ., (v) If z1 (a1 ib1) and z 2 (a 2 ib 2) then z1 z 2 + a1 a 2 and b1 b 2 ., (a ib), z, , ·, (vi) If z (a ib) then z 1 2 2, |z| (a b 2), 3., , (i) z1 z 2 z1 z 2, z1, z, (iv) d z n 1, z2, 2, , (ii) z1 z 2 z1 z 2, , (iii) z1 z 2 z1 z 2, , (v) z z 2 Re (z) and z z 2 Im (z), , (vi) z z |z|2, 4. Polar form of z (x iy) is z r (cos i sin ) ., Here, r x 2 y 2 |z|and the value of such that # is called the, principal argument or amplitude of z., 5. When z is purely real; then, it lies on the x-axis., (i) If z lies on positive side of the x-axis then 0., (ii) If z lies on negative side of the x-axis then ., 6. When z is purely imaginary; then, it lies on the y-axis., , (i) If z lies on the y-axis and above the x-axis then ·, 2, , ·, (ii) If z lies on the y-axis and below the x-axis then , 2, 7. When z (x iy) does not lie on any axes:, Im (z), ·, Re (z), , Step 1., , Find for which tan , , Step 2., , Find the quadrant in which P(x, y) lies., , (i) When z lies in quadrant I;, then, & arg (z) ., (ii) When z lies in quadrant II;, then, & arg (z) ., (iii) When z lies in quadrant III;, then, ( ) & arg (z) ., (iv) When z lies in quadrant IV;, then, & arg (z) .
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216, , 8., , Senior Secondary School Mathematics for Class 11, , (i) Fundamental theorem of algebra, A polynomial equation of degree n has at most n roots., (ii) Let ax 2 bx c 0, where a, b, cd R and a ! 0., Its roots are given by, b b 2 4ac, b b 2 4ac, ·, and , 2a, 2a, Here, D (b 2 4ac) is called the discriminant., , , If D 0, the given quadratic equation has complex roots., 9., , 1, 1, 2, 2, 2, 2, 2, 2, a ib !(x iy), where x $ a b a. and y $ a b a. ·, 2, 2, Similarly, by assuming that a ib !(x iy), we may find a ib .,
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Linear Inequations (In one variable), , 217, , 6, , Linear Inequations, (In one variable), Linear Inequations in One Variable, , Inequalities of the form:, (i) ax b c, (ii) ax b # c, , (iii) ax b c, , (iv) ax b $ c,, , where a, b, c are real numbers, a ! 0 and x is a variable, are called inequations in x., Thus, each of the following is a linear inequation:, (i) 2x 3 7, (ii) 3x 5 # 10, (iii) 3x 2 8, (iv) 4x 3 $ 9, Replacement Set or Domain of the Variable, , A set given to us from which the values of x are replaced in an inequation in x,, is called the replacement set., SOLUTION SET The set of all those values of x taken from the replacement set, which satisfy the given inequation is called the solution set of the inequation., EXAMPLE, SOLUTION, , Write down the solution set of the inequation x 6, when the replacement, set is (i) N, (ii) W, (iii) Z., (i) Solution set {xd N : x 6} {1, 2, 3, 4, 5} ., (ii) Solution set {xdW : x 6} {0, 1, 2, 3, 4, 5} ., (iii) Solution set {xdZ : x 6} {5, 4, 3, 2, 1, 0, 1, 2, 3, …} ., , Rules for Solving an Inequation, RULE 1, RULE 2, RULE 3, RULE 4, , Adding the same number or expression to each side of an inequation, does not change the inequality., Subtracting the same number or expression from each side of an, inequation does not change the inequality., Multiplying (or dividing) each side of an inequation by the same, positive number does not change the inequality., Multiplying (or dividing) each side of an inequation by the same, negative number reverses the inequality., Thus, x 3 & x 3 [on multiplying both sides by –1]., And, 3x # 12 & x $ 4 [on dividing both sides by –3]., , Using Rule 1 and Rule 2, we can drop any term from one side, of an inequation and put it on the other side with the opposite sign., This process is called transposition., , TRANSPOSITION, , 217
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218, , Senior Secondary School Mathematics for Class 11, , Thus, x 3 2 & x 2 3 & x 1., 9, ·, 2, , And, 2x 3 $ 6 & 2x $ 6 3 & 2x $ 9 & x $, SOLVED EXAMPLES, EXAMPLE 1, , Solve 5x 24 when (i) xd N, (ii) xdZ., In each case, represent the solution set on the number line., , SOLUTION, , 5x 24 & x , , 24, 5, & x 4.8., (i) Solution set {xd N : x 4.8}, {1, 2, 3, 4} ., On the number line, we may represent it as shown below., –3, , –2, , 0, , –1, , 1, , 3, , 2, , 5, , 4, , 6, , The darkened circles indicate the natural numbers contained, in the set., (ii) Solution set {xdZ : x 4.8}, {4, 3, 2, 1, 0, 1, 2, 3, …}, {…, 3, 2, 1, 0, 1, 2, 3, 4} ., On the number line, we may represent it as shown below., –4, , –3, , –2, , –1, , 0, , 1, , 2, , 3, , 4, , 5, , 6, , The darkened circles show the integers contained in the set, and three dark dots above the left part of the line show that the, negative integers are continued idefinitely., EXAMPLE 2, , SOLUTION, , 5, Solve 12 1 x # 5 3x when (i) xd N, (ii) xd R., 6, Draw the graph of the solution set in each case., 5, 12 1 x # 5 3x, 6, 11, &, 12 x # 5 3x, 6, [multiplying both sides by 6], &, 72 11x # 30 18x, [adding 72 to both sides], &, 11x # 18x 42, 7x # 42, [adding 18x to both sides], &, [dividing both sides by 7] ., &, x$6, (i) Solution set {xd N : x $ 6}, {6, 7, 8, 9, …} ., The graph of this set is the number line, shown below., –2, , –1, , 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8
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Linear Inequations (In one variable), , 219, , The darkened circles indicate the natural numbers contained, in the set. Three dots above the right part of the line show that, the natural numbers are continued indefinitely., (ii) Solution set {xd R : x $ 6} [6, 3 [., The graph of this set is shown below., –2, , –1, , 0, , 1, , 2, , 3, , 5, , 4, , 6, , 7, , 8, , This graph consists of 6 and all real numbers greater than 6., EXAMPLE 3, SOLUTION, , x4, 0 and draw the graph of the solution set., x2, a, Note that 0 when (a 0 and b 0) or (a 0 and b 0) ., b, , either (x 4 0 and x 2 0) or (x 4 0 and x 2 0) ., Case I When x 4 0 and x 2 0., In this case, x 4 0 and x 2 0, & x 4 and x 2, & x2, & xd(2, 3) ., Case II When x 4 0 and x 2 0., In this case, x 4 0 and x 2 0, & x 4 and x 2, & x 4, & xd(3, 4) ., , solution set (3, 4),(2, 3) ., The graph of the solution set is given below., Solve, , –4, , EXAMPLE 4, SOLUTION, , –3, , –2, , –1, , 0, , 1, , 2, , 3, , 4, , x3, Solve 0 and draw the graph of the solution set., x 4, a, Note that 0 when (a 0 and b 0) or (a 0 and b 0) ., b, , either (x 3 0 and x 4 0) or (x 3 0 and x 4 0) ., Case I, When x 3 0 and x 4 0., In this case, x 3 0 and x 4 0, & x 3 and x 4., , Case II, , This is not possible, as we can never find a real number, which is greater than 3 and less than 4., When x 3 0 and x 4 0., In this case, x 3 0 and x 4 0, & x 3 and x 4, & 4 x 3 & xd( 4, 3) .
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220, , Senior Secondary School Mathematics for Class 11, , , solution set ( 4, 3) ., The graph of the solution set is given below., –4, , –3, , –2, , –1, , 0, , 1, , 2, , 3, , 4, , 5, , x2, 2., x5, , EXAMPLE 5, , Solve, , SOLUTION, , We have, x2, 2, x5, x2 , &, 20, x5, x 2 2x 10, &, 0, x5, (x 12), &, 0, ( x 5), x 12, [on multiplying both sides by 1]., &, 0, x5, , either (x 12 0 and x 5 0) or (x 12 0 and x 5 0) ., Case I, When x 12 0 and x 5 0., In this case, x 12 0 and x 5 0, & x 12 and x 5, & 5 x 12, which is impossible, [a 12 5]., Case II When x 12 0 and x 5 0., In this case, x 12 0 and x 5 0, & x 12 and x 5, & 12 x 5 & xd(12, 5) ., , solution set (12, 5) ., , EXAMPLE 6, , Solve, , SOLUTION, , 5, 3 and represent the solution set on the number line., x2, , 5, 5, 3 & 3 0 [adding 3 to both sides], x2, x 2, 5 3x 6, &, 0, x2, 11 3x, &, 0., x2, , either (11 3x 0 and x 2 0) or (11 3x 0 and x 2 0) ., Case I, When 11 3x 0 and x 2 0., Now, 11 3x 0 and x 2 0, & 3x 11 and x 2, 11, & x, and x 2, 3, 11 ·, … (i), & 2x, 3
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Linear Inequations (In one variable), , 221, , When 11 3x 0 and x 2 0., Now, 11 3x 0 and x 2 0, & 3x 11 and x 2, 11, & x, and x 2., 3, This is not possible, as we cannot find a real number which is, 11, and less than 2., greater than, 3, 11, 11, , solution set 'xd R : 2 x 1 b2, l ·, 3, 3, Case II, , We can represent this set on the number line, as given below., 0, , 1, , 2, , 3, , 11 4, 3, , SOME USEFUL RESULTS, , Let a be a positive real number. Then,, (i) |x| a + a x a + xd( a, a) ., (ii) |x|# a + a # x # a + xd[ a, a] ., (iii) |x| a + x a or x a., (iv) |x|$ a + x # a or x $ a., EXAMPLE 7, , Solve |4 x| 2., , SOLUTION, , We have,|x| a + a x a., , |4 x| 2, 2 4 x 2, +, 2 4 x and 4 x 2, +, 2 4 x and x 2 4, +, 6 x and x 2, +, +, 6 x and x 2, +, 2 x 6., , solution set {xd R : 2 x 6} (2, 6) ., , EXAMPLE 8, SOLUTION, , 1, Solve |3x 2|# , xd R., 2, We have,|x|# a + a # x # a., , , |3x 2|#, , 1, 1, 1, + # 3x 2 #, 2, 2, 2, 1, 1, +, # 3x 2 and 3x 2 #, 2, 2, 1, 1, 2 # 3x and 3x # 2, +, 2, 2
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222, , Senior Secondary School Mathematics for Class 11, , 3, 5, # 3x and 3x #, 2, 2, 5, 1, + # x and x #, 2, 6, 5, 1, #x# ·, 2, 6, 5, 1, 1 5, solution set 'xd R : # x # 1 ; , E ·, 2, 6, 2 6, +, , , EXAMPLE 9, , Solve |x 1| 4, xd R., , SOLUTION, , We have,|x| a + x a or x a., , |x 1| 4 + x 1 4 or x 1 4, + x 4 1 or x 4 1, + x 5 or x 3, + xd(3, 5) or xd(3, 3) ., , solution set (3, 5),(3, 3) ., , EXAMPLE 10, , Solve |3 4x|$ 9, xd R., , SOLUTION, , We have,|x|$ a + x # a or x $ a., , |3 4x|$ 9 + 3 4x # 9 or 3 4x $ 9, + 4x # 9 3 or 4x $ 9 3, + 4x # 12 or 4x $ 6, 3, + x $ 3 or x #, 2, 3, + x#, or x $ 3, 2, 3, + xdb3, E or xd[3, 3) ., 2, 3, , solution set b3, E,[3, 3) ., 2, , EXAMPLE 11, , Solve, , SOLUTION, , Clearly, x 3 ! 0 and therefore, x ! 3., , 2, 5, xd R., |x 3|, , We have,, , 2, 5., |x 3|, , … (i), , Since|x 3|is positive, we may multiply both sides of (i) by|x 3|., This gives, 2 5 |x 3|, 2, +, |x 3|, 5, 2, + |x 3| , 5, 2, 2, +, [a |x| a + a x a], x3 , 5, 5
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Linear Inequations (In one variable), , 223, , 2, 2, x 3 and x 3 , 5, 5, 2, 2, 3 x and x 3, +, 5, 5, 13, 17 ·, x and x , +, 5, 5, 13, 17 ·, x, +, 5, 5, Also, as shown above, x ! 3., +, , , , EXAMPLE 12, SOLUTION, , solution set 'xd R :, , 13, 17, x 1 { 3}, 5, 5, (2.6, 3.4) {3} (2.6, 3),(3, 3.4) ., , |x 2| 1, # 0, xd R., |x 2| 2, We have either x 2 $ 0 or x 2 0., Case I, When x 2 $ 0., In this case, x $ 2 and|x 2| x 2., So, the given inequation becomes, x3, x21, # 0 & # 0., x22, x 4, (x 3 $ 0 and x 4 0) or (x 3 # 0 and x 4 0), Solve, , &, &, &, , (x $ 3 and x 4) or (x 3 and x 4), 3# x 4, [a x 3 and x 4 is not possible], [this includes x $ 2]., xd[3, 4), , When x 2 0., In this case, x 2 and|x 2| (x 2) x 2., So, the given inequation becomes, x 2 1, x 1, x 2 2 #0 & x #0, x1, & x #0, [multiplying num. and denom. by –1], (x 1 # 0 and x 0) or (x 1 $ 0 and x 0), & (x # 1 and x 0) or (x $ 1 and x 0), & 0 x #1, [a x $ 1 and x 0 is not possible], [this includes x 2]., & xd(0, 1], Hence, from the above two cases, we get, solution set (0, 1],[3, 4) ., Case II, , EXAMPLE 13, , Solve, , |x 3| x, 1, xd R., ( x 2)
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224, SOLUTION, , Senior Secondary School Mathematics for Class 11, , We have, |x 3| x, 1, ( x 2), |x 3| x, 1 0, &, ( x 2), |x 3| x x 2, &, 0, ( x 2), |x 3| 2, &, 0., ( x 2), , … (i), , Now, we have either x 3 $ 0 or x 3 0., Case I, When x 3 $ 0., In this case, x $ 3 and|x 3| x 3., (i) becomes, x32, x1, 0 & 0., x2, x 2, (x 1 0 and x 2 0) or (x 1 0 and x 2 0), & (x 1 and x 2) or (x 1 and x 2), & (x 1) or (x 2), & (x 2) or (x 1), & xd(3, 2) or xd(1, 3), & xd[ 3, 2) or xd(1, 3), [a x $ 3], & xd[ 3, 2),(1, 3) ., Case II, , EXAMPLE 14, SOLUTION, , When x 3 0., In this case, x 3 and|x 3| (x 3) x 3., (i) becomes, ( x 5), x 3 2, x5, 0 &, 0 & 0., x2, x 2, ( x 2), (x 5 0 and x 2 0) or (x 5 0 and x 2 0), & (x 5 and x 2) or (x 5 and x 2), & 5 x 2, [a x 5 and x 2 is not possible], & 5 x 3, [a x 3], & xd( 5, 3) ., solution set ( 5, 3),[ 3, 2),(1, 3), ( 5, 2),(1, 3) ., , Solve |x 1||x 2|$ 4, xd R., Putting x 1 0 and x 2 0, we get x 1 and x 2 as the critical, points. These points divide the whole real line into three parts,, namely (3, 1), [1, 2) and [2, 3) . So, we consider the following, three cases.
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Linear Inequations (In one variable), Case I, , , , 225, , When 3 x 1., In this case, x 1 0 and x 2 0., |x 1| (x 1) x 1 and|x 2| (x 2) x 2., Now,|x 1||x 2|$ 4, & x 1 x 2$ 4, 1, & 2x 3 $ 4 & 2x $ 4 3 & 2x $ 1 & x #, 2, 1, & xdb3, E ·, 2, But, 3 x 1., 1, 1, solution set in this case b3, E+(3, 1) b3, E ·, 2, 2, , When 1 # x 2., In this case, x 1 $ 0 and x 2 0., |x 1| x 1 and|x 2| (x 2) x 2., Now,|x 1||x 2|$ 4, & x 1 x 2 $ 4 & 1 $ 4, which is absurd., So, the given inequation has no solution in [1, 2)., Case III When 2 # x 3., In this case, x 2 $ 0 and x 1 0., |x 2| x 2 and|x 1| x 1., Now,|x 1||x 2|$ 4, Case II, , &, , x 1 x 2 $ 4 & 2x 3 $ 4 & 2x $ 7 & x $, , 7·, 2, , Also, in this case, we have x $ 2., 7, 7, solution set in this case ; , 3l+[2, 3) ; , 3l ·, 2, 2, Hence, from all the above cases, we have, 1, 7, solution set b3, E,; , 3l ·, 2, 2, SOLVING SIMULTANEOUS INEQUATIONS IN ONE VARIABLE, METHOD, , Suppose we have to solve two inequations simultaneously. Find the, solution set of each of them. Then, the intersection of these solution, sets is the required solution set., , EXAMPLE 15, SOLUTION, , Solve the inequations 3 # 3 2x 9, xd R. Represent the solution set, on the real line., We have, 3 # 3 2x 9 & 3 # 3 2x and 3 2x 9., Now, 3 # 3 2x & 3 2x # 3, [adding 2x on both sides], [adding 3 on both sides], & 2x # 6, [dividing both sides by 2], & x#3, & xd(3, 3] .
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226, , Senior Secondary School Mathematics for Class 11, , And, 3 2x 9 & 2x 6, [subtracting 3 from both sides], [dividing both sides by 2], & x 3, & xd( 3, 3) ., , solution set (3, 3]+( 3, 3) ( 3, 3] ., We may represent it on the number line as shown below., –4, , EXAMPLE 16, SOLUTION, , –3, , –2, , –1, , 0, , 1, , 2, , 3, , Solve the inequations 2x 3 x 2 # 3x 5, xd R. Draw the graph of, the solution set., We have, 2x 3 x 2 # 3x 5 & 2x 3 x 2 and x 2 # 3x 5., Now, 2x 3 x 2 & x 3 2, [adding x on both sides], [adding 3 on both sides], & x5, & xd(3, 5) ., And, x 2 # 3x 5 & 3x 5 $ x 2 [a a # b & b $ a], [adding x on both sides], & 2x 5 $ 2, [adding 5 on both sides], & 2x $ 3, 3, [dividing both sides by 2], & x$, 2, 3, & xd; , 3l ·, 2, 3, 3, , solution set (3, 5)+; , 3l ; , 5l ·, 2, 2, We may represent it on the number line, as shown below., –2 –3 –1, 2, , EXAMPLE 17, , SOLUTION, , 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7x, 2, xd R., 3, Represent the solution set on the number line., We have, 7x, 7x, 7x, 2x 1 x , 2 & 2x 1 x , and x , 2., 3, 3, 3, 7x, Now, 2x 1 x , & 6x 3 3x 7 x, 3, [multiplying both sides by 3], & 6x 3 2x 7, & 4x 3 7 [adding 2x on both sides], [adding 3 on both sides], & 4x 10, 5, [dividing both sides by 4], & x, 2, 5, & xdb , 3l ·, 2, Solve the system of inequations 2x 1 x
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Linear Inequations (In one variable), , 227, , 7x, [multiplying both sides by 3], 2 & 3x 7 x 6, 3, & 2x 7 6, [adding –7 on both sides], & 2x 1, 1, [dividing both sides by 2], & x, 2, 1, & xdb , 3l ·, 2, 1, 5, 5, , solution set b , 3l+b , 3l b , 3l ·, 2, 2, 2, The solution set on the number line may be represented as shown, below., And, x , , –1, , 0, , 1, , EXAMPLE 18, , Solve the system of inequations 5 #, , SOLUTION, , We have, , 2, , 5 3, 2, , 4, , 2 3x, # 9., 4, , , , , 5 # 2 3x # 9 & 5 # 2 3x and 2 3x # 9., 4, 4, 4, , 2 3x, Now, 5 #, & 20 # 2 3x [multiplying both sides by 4], 4, [adding 2 on both sides], & 22 # 3x, [multiplying both sides by 1], & 22 $ 3x, & 3x # 22, 22, [dividing both sides by 3], & x#, 3, 22, & xdb 3, E ·, 3, , 2 3x, And,, [multiplying both sides by 4], # 9 & 2 3x # 36, 4, & 3x # 34, 34, [dividing both sides by 3], & x$, 3, 34, , 3l ·, & xd;, 3, 34, , 22, , 3l ; 34 , 22E ·, , solution set b3, E+;, 3, 3, 3, 3, EXAMPLE 19, , Solve the system of inequations:, 6x, x, 1, 1, ,, $ ·, 4x 1 2 2x 1 4, , SOLUTION, , The first inequation of the given system is, 6x, 1, , 4x 1 2
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228, , Senior Secondary School Mathematics for Class 11, , , &, , 6x 1, 0, 4x 1 2, 12x 4x 1, 0, 2(4x 1), 8x 1, 8x 1, … (i), 0 &, 0., 4x 1, 2(4x 1), (8x 1 0 and 4x 1 0) or (8x 1 0 and 4x 1 0), (8x 1 and 4x 1) or (8x 1 and 4x 1), , &, , bx , , &, &, &, , &, &, , 1, 1, 1, 1, and x l or bx , and x l, 8, 8, 4, 4, , 1, 1, 1, 1, ;a x , and x is not possibleE, x, 8, 8, 4, 4, 1 1, xdb , l ·, 8 4, , The second inequation of the given system is, x, 1, $, 2x 1 4, x 1, &, $0, 2x 1 4, 2x 1, 4x 2x 1, … (ii), &, $0 &, $ 0., 2x 1, 4 (2x 1), , (2x 1 # 0 and 2x 1 0) or (2x 1 $ 0 and 2x 1 0), &, (2x # 1 and 2x 1) or (2x $ 1 and 2x 1), 1, 1, 1, 1, bx # and x l or bx $ and x l, 2, 2, 2, 2, 1, 1, bx l or bx $ l, &, 2, 2, 1, 1, &, xdb3, l or xd; , 3l, 2, 2, 1, 1, &, xdb3, l,; , 3l, 2, 2, 1 1, 1, 1, , solution set b , l+ (b3, l,; , 3l2 ., 8 4, 2, 2, Hence, the given system of inequations has no solution., &, , EXAMPLE 20, , Solve the system of inequations: |x 1|# 5, |x|$ 2., , SOLUTION, , The first inequation is|x 1|# 5., Using|x|# a + a # x # a, we get, |x 1|# 5 & 5 # x 1 # 5, & 4 # x # 6 & xd[ 4, 6] ., The second inequation is|x|$ 2., Using|x|$ a + x # a or x $ a, we get
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Linear Inequations (In one variable), , , , 229, , |x|$ 2 & x # 2 or x $ 2, & xd(3, 2],[2, 3) ., solution set for the given system is, {(3, 2],[2, 3)}+[ 4, 6] [ 4, 2],[2, 6] ., 6, , –4, , –2, , 2, , WORD PROBLEMS, EXAMPLE 1, SOLUTION, , EXAMPLE 2, SOLUTION, , EXAMPLE 3, , SOLUTION, , Find all pairs of consecutive odd positive integers both of which are, smaller than 10 such that their sum is more than 11., Let the required consecutive odd positive integers be x and x 2., Then,, x 2 10 and x (x 2) 11, &, x 8 and 2x 2 11, &, x 8 and 2x 9, 9, &, x 8 and x , 2, 9, &, x 8 & 4.5 x 8., 2, , x can take the odd integral values 5 and 7., Hence, the required pairs of odd integers are (5, 7) and (7, 9)., Find all pairs of consecutive even positive integers both of which are larger, than 5 such that their sum is less than 23., Let the required consecutive even positive integers be x and x 2., Then,, x 5 and x (x 2) 23, &, x 5 and 2x 21, &, 5 x and x 10.5, &, 5 x 10.5., , x can take the even integral values 6, 8 and 10., Hence, the required pairs of even integers are (6, 8), (8, 10) and, (10, 12)., The longest side of a triangle is 3 times the shortest side and the third side, is 2 cm shorter than the longest side. If the perimeter of the triangle is at, least 61 cm, find the minimum length of the shortest side., Let the minimum length of the shortest side be x cm., Then, the longest side 3x cm and the third side (3x 2) cm., , x 3x (3x 2) $ 61 & 7x $ 63 & x $ 9., Hence, the minimum length of the shortest side is 9 cm.
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230, EXAMPLE 4, , SOLUTION, , EXAMPLE 5, , SOLUTION, , EXAMPLE 6, , SOLUTION, , Senior Secondary School Mathematics for Class 11, , The cost and revenue functions of a product are given by C (x) 20x 4000, and R (x) 60x 2000 respectively, where x is the number of items, produced and sold. How many items must be sold to realise some profit?, Let the required number of items sold be x and let P(x) be the profit, function. Then,, profit function = (revenue function) – (cost function), &, P (x) (60x 2000) (20x 4000), &, P(x) 40x 2000., For some profit, we have P(x) 0., , 40x 2000 0, &, 40x 2000, &, x 50., Hence, the manufacturer must sell more than 50 items to realise, some profit., A man wants to cut three lengths from a single piece of cloth of length, 91 cm. The second length is to be 3 cm longer than the shortest and the, third length is to be twice as long as the shortest. What are the possible, lengths of the shortest piece of cloth if the third piece is to be at least 5 cm, longer than the second., Let the length of the shortest piece be x cm. Then, second length, (x 3) cm and third length 2x cm., , x (x 3) 2x # 91 and 2x $ (x 3) 5, &, 4x 3 # 91 and 2x $ x 8, &, 4x # 91 3 and 2x x $ 8, &, 4x # 88 and x $ 8, &, x # 22 and x $ 8, &, 8 # x # 22., Hence, the length of the shortest piece is to be greater than or equal, to 8 but less than or equal to 22., In drilling world’s deepest hole it was found that the temperature, T (x) in degree Celsius, x km below the earth’s surface was given by, T (x) 30 25 (x 3), 3 # x # 15. At what depth will the temperature be, between 155C and 205C?, We have,, T (x) 30 25 (x 3), &, T (x) 30 25x 75, … (i), &, T (x) 25x 45., Let x km below the earth’s surface, the temperature lies between, 155C and 205C. Then,, 155 T (x) 205, &, 155 25x 45 205, &, 155 25x 45 and 25x 45 205
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Linear Inequations (In one variable), , 231, , &, 155 45 25x and 25x 205 45, &, 200 25x and 25x 250, &, 8 x and x 10, &, 8 x 10., Clearly, 8 x 10 lies in the range 3 # x # 15., Hence, the required depth is more than 8 km and less than 10 km., EXAMPLE 7, , SOLUTION, , A manufacturer has 460 litres of a 9% acid solution. How many litres, of a 3% acid solution must be added to it so that the acid content in the, resulting mixture be more than 5% but less than 7%?, Let x litres of a 3% acid solution be added to 460 litres of 9% acid, solution. Then,, total quantity of mixture (460 x) litres., Total acid content in (460 x) litres of mixture, (b460 # 9 l bx # 3 l2 litres b 207 3x l litres., 5, 100, 100, 100, Now, the acid content in the resulting mixture must be more than, 5% and less than 7%., 207 3x, , 5% of (460 x) b, l 7% of (460 x), 5, 100, 5, 4140 3x, 7, #(460 x) , #(460 x), &, , 100, 100, 100, &, 5(460 x) 4140 3x 7 (460 x), &, 2300 5x 4140 3x 3220 7x, &, 2300 5x 4140 3x and 4140 3x 3220 7x, &, 5x 3x 4140 2300 and 4140 3220 7x 3x, &, 2x 1840 and 920 4x, &, x 920 and 230 x, &, 230 x 920., Hence, the required quantity of 3% acid solution to be added must, be more than 230 litres and less than 920 litres., , EXAMPLE 8, , SOLUTION, , A solution is to be kept between 40C and 45C. What is the range, of temperature in degree Fahrenheit, if the conversion formula is, 9, F C 32?, 5, 9, 5, … (i), F C 32 & C (F 32)., 5, 9, , &, &, , 40 C 45, 5, 40 (F 32) 45, 9, 5, 5, 40 (F 32) and (F 32) 45, 9, 9
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232, , Senior Secondary School Mathematics for Class 11, , 9, 9, F 32 and F 32 45 #, 5, 5, &, 72 F 32 and F 32 81, &, 72 32 F and F 81 32, &, 104 F and F 113, &, 104 F 113., Hence, the solution is to be kept between 104F and 113F., &, , EXAMPLE 9, , 40 #, , m, The IQ of a person is given by the formula, IQ c #100, where m is the, mental age and c is the chronological age. If 80 # IQ # 140 for a group of, 12-year children, find the range of their mental age., , SOLUTION, , 25m, m, When c 12, we have IQ a #100k , $, 3, 12, 25m, , 80 # IQ # 140 & 80 #, # 140, 3, 25m, 25m, & 80 #, and, # 140, 3, 3, 3, 3, # 80 # m and m # 140 #, &, 25, 25, 48, 84, &, # m and m #, 5, 5, & 9.6 # m # 16.8., Hence, the required mental age for a group of 12-year children is, 9.6 years or more and 16.8 years or less., , EXERCISE 6A, 1. Fill in the blanks with correct inequality sign (, , $, #) ., (i) 5x 20 & x …… 4, (ii) 3x 9 & x …… 3, (iii) 4x 16 & x …… 4, (iv) 6x # 18 & x …… 3, (v) x 3 & 2x …… 6, a, b, (vi) a b and c 0 & c …… c, (vii) p q 3 & p …… q, (viii) u v 2 & u …… v, Solve each of the following inequations and represent the solution set on the, number line., , 2., 3., 4., 5., , where, 6x # 25,, 2x 5,, where, 3x 8 2, where, 5x 2 17, where, , (i) xd N,, (i) xdZ,, (i) xdZ,, (i) xdZ,, , (ii) xdZ., (ii) xd R., (ii) xd R., (ii) xd R.
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Linear Inequations (In one variable), , 6. 3x 4 x 6, where xd R., 7. 3 2x $ 4x 9, where xd R., 8., 9., 10., 11., 12., 13., , 5x 8 4x 7 ,, where xd R., $, 3, 2, , 5x 4x 1, 1, where xd R., 3, 4, 1 2 , 1, b x 1l $ (x 2), where xd R., 3, 4 3, , , 2x 1 x 1 3x 1 ,, where xd R., , 3, 12, 4, x ( 5x 2 ) ( 7x 3 ) ,, where xd R., , 3, 5, 4, (2x 1) (3x 2) (2 x), , where xd R., , $, 3, 5, 4, , Solve:, , x3, 0, xd R, x1, 2x 3, 16., 0, xd R, 3x 7, 3, 18. 2, xd R, x 2, 5x 8, 20. 2, xd R, 4 x, 14., , 22. |5 2x|# 3, xd R, , x3, 0, xd R, x4, x7, 17. $ 0, xd R, x 2, 1, 19. # 2, xd R, x 1, 15., , 21. |3x 7| 4, xd R, 1, 23. |4x 5|# , xd R, 3, , , |x 2| x, 25., 2, xd R, x, , 1, 1, # , xdR, |x| 3 2, 2x 1, 26., 2, xd R, x1, |x| 1, 28., $ 0, xd R { 2, 2}, |x| 2, , 27., , |x 3|, 0, xd R, x3, , 29., , 1, $ 1, xd R { 2, 2}, 2 |x|, , 30. |x 1||x| 3, xd R, , 31., , 2, 1, x ! 4, x4, , 24., , Solve the following systems of linear inequations:, 6, 4, #3# , x 0, x1, x 1, 3x, 34. 5x 7 3 (x 3), 1 , $x 4, 2, 4x, 36. 3x 2 x , 3, 3, 3x, 38. 12 4 # 2, 5, 32., , 33. 11 # 4x 3 # 13, 6 5x, 35. 2 , 7, 4, 7x 1 3x 8 , 37., 3,, 11 0, 2, 5, 39. 1 #|x 2|# 3, , 233
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234, , Senior Secondary School Mathematics for Class 11, , 40. Find all pairs of consecutive odd positive integers, both of which are, smaller than 18 such that their sum is more than 20., 41. Find all pairs of consecutive even positive integers both of which are larger, than 8 such that their sum is less than 25., 42. A company manufactures cassettes. Its cost and revenue functions are, C(x) 26000 30x and R(x) 43x respectively, where x is the number of, cassettes produced and sold in a week. How many cassettes must be sold, by the company to realise some profit?, 43. The water acidity in a pool is considered normal when the average pH, reading of three daily measurements is between 8.2 and 8.5. If the first two, pH readings are 8.48 and 8.35, find the range of the pH values for the third, reading that will result in the acidity level being normal., 44. A manufacturer has 640 litres of a 8% solution of boric acid. How many, litres of a 2% boric acid solution be added to it so that the boric acid content, in the resulting mixture will be more than 4% but less than 6%?, 45. How many litres of water will have to be added to 600 litres of the 45%, solution of acid so that the resulting mixture will contain more than 25%,, but less than 30% acid content?, 46. To receive grade A in a course one must obtain an average of 90 marks or, more in five papers, each of 100 marks. If Tanvy scored 89, 93, 95 and 91, marks in first four papers, find the minimum marks that she must score in, the last paper to get grade A in the course., ANSWERS (EXERCISE 6A), , 1. (i) < (ii) < (iii) >, 2. (i) {1, 2, 3, 4}, , (iv) $ (v) <, , –2, , (vi) >, , (vii) <, , (viii) >, , 0, , 1, , 2, , 3, , 4, , –4, , –3, , –2, , –1, , 0, , –2, , –1, , 0, , –1, , 0, , 1, , 2, , 3, , 4, , –2, , –1, , 0, , –1, , (ii) {…, –3, –2, –1, 0, 1, 2, 3, 4}, , 3. (i) {–3, –4, –5, –6, …}, –6, , –5, , (ii) (3, 2.5), –3, 2.5, , 4. (i) {–1, 0, 1, 2, 3, 4, …}, (ii) ( 2, 3)
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Linear Inequations (In one variable), , 5. (i) {2, 1, 0, –1, –2, …}, –2, , –1, , 0, , 1, , 0, , 1, , 2, , 3, , 2, , (ii) (3, 3), 6. (5, 3), 0, , 3, , 2, , 1, , 4, , 5, , 7. (3, 2], 0, , 1, , 2, , 5, 8. b3, E, 2, 5, 2, , 9. (3, 8), –8 –7 –6 –5 –4 –3 –2 –1 0, , 10. (3, 5.5], 11. (0, 3), 0, , 12. (4, 3), 0 1 2 3 4, , 13. (3, 2], 0 1 2, , 3, 7, 16. b3, l,b , 3l, 2, 3, 3, 19. (3, 1),; , 3l, 2, , 14. (1, 3), , 15. (3, 4),(3, 3), , 17. (3, 2),[7, 3), , 7, 18. b2, l, 2, , 20. (3, 0),(4, 3), , 21. (3, 1),b, , 7 4, 23. ; , E, 6 3, , 24. (3, 5],( 3, 3),[5, 3), , 25. (3, 2),(1, 3), , 3, 26. b , 1l,(1, 3), 4, , 11 ,, 3l, 3, , 22. [1, 4], , 27. (3, 3), , 28. [1, 1],(3, 2),(2, 3), , 29. ( 2, 1],[1, 2), , 30. (3, 2),(1, 3), , 31. (2, 4),(4, 6), , 33. 2 # x # 4, , 34. 3 x # 2, , 1, # x #1, 3, 35. 4.4 x 2.8, 32., , 235
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236, , Senior Secondary School Mathematics for Class 11, , 80, 10, 5, 37. 3 x 21, 38., x#, x3, 3, 3, 2, 39. [1, 1],[3, 5], 40. (11, 13), (13, 15), (15, 17), 41. (10, 12), 42. More than 2000, 43. Between 7.77 and 8.67, 44. More than 320 litres and less than 1280 litres, 45. More than 300 litres but less than 480 litres, 46. 82 marks, 36., , HINTS TO SOME SELECTED QUESTIONS, 1. (ii) 3x 9 & x 3., (iii) 4x 16 & x 4., (iv) 6x # 18 & x $ 3., (v) x 3 & 2x 6., a b·, (vii) p q 3 & p q., (vi) a b and c 0 & c c, 5, 3. 2x 5 & x , & x 2.5., 2, (i) All integers less than 2.5 are 3, 4, 5, 6, … ., (ii) Set of all real numbers less than 2.5 are (3, 2.5) ., 4. 3x 6 & x 2., (i) All integers greater than 2 are 1, 0, 1, 2, 3, 4, … ., (ii) Set of all real numbers greater than 2 is ( 2, 3) ., 7. 3 2x $ 4x 9 & 6x $ 12 & x # 2., , , solution set (3, 2] ., , 5, 8. 10x 16 $ 12x 21 & 2x $ 5 & x # ·, 2, 5, solution set c3, E ·, 2, 9. 15x 4 (4x 1) 12 & 15x 16x 4 12 & x 8 & x 8., , 10., , solution set (3, 8) ., , 1 · (2x 3) x 2, 11, $, & 2x 3 $ 4x 8 & 2x $ 11 & x # ·, 3, 3, 2, 4, solution set (3, 5.5] ., , 11. (2x 1) 4 (x 1) 3 (3x 1) & 2x 3 9x 3 & 11x 0 & x 0., , , solution set (0, 3) ., , 12. 15x 20 (5x 2) 12 (7x 3) & 15x 100x 40 84x 36, , , solution set (4, 3) ., , & 15x 16x 4 & x 4 & x 4., , 13. 20 (2x 1) $ 15 (3x 2) 12 (2 x) & 40x 20 $ 45x 30 24 12x, , , solution set (3, 2] ., , & 17x $ 34 & x # 2., , 14. (x 3 0 and x 1 0) or (x 3 0 and x 1 0), &, &, , (x 3 and x 1) or (x 3 and x 1), 1 x 3 & xd(1, 3) ., , [x $ 3 and x 1 is not possible]
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Linear Inequations (In one variable), , 237, , 15. (x 3 0 and x 4 0) or (x 3 0 and x 4 0), &, &, , (x 3 and x 4) or (x 3 and x 4), (x 4) or (x 3) & xd(3, 4),(3, 3) ., , 16. (2x 3 0 and 3x 7 0) or (2x 3 0 and 3x 7 0), &, &, , cx , , 3, 3, 7, 7, and x m or cx and x m, 2, 3, 2, 3, 3, 3, 7, 7, cx m or cx m & xdc3, m,c , 3m ·, 2, 3, 2, 3, , 17. (x 7 # 0 and x 2 0) or (x 7 $ 0 and x 2 0), &, &, 18., , 19., , 20., , (x # 7 and x 2) or (x $ 7 and x 2), (x 2) or (x $ 7) & xd(3, 2),[7, 3) ., , 3 , 3 2x 4, 7 2x, 20 &, 0 & 0, x2, x2, x 2, & (7 2x 0 and x 2 0) or (7 2x 0 and x 2 0), & (7 2x and x 2) or (7 2x and x 2), 7, 7, & cx and x 2m or cx and x 2m, 2, 2, 7, 7 ·, 7, & 2 x & xdc2, m, [x and x 2 cannot hold], 2, 2, 2, 3 2x, 1 2x 2, 1 , 2#0 &, #0 & #0, x1, x1, x 1, & (3 2x # 0 and x 1 0) or (3 2x $ 0 and x 1 0), & (3 # 2x and x 1) or ( 2x $ 3 and x 1), 3, 3, & cx $ and x 1m or cx # and x 1m, 2, 2, 3 , 3, & (x 1) or cx $ m xd(3, 1),; , 3m ·, 2, 2, 5x 8 , 5x 8 8 2x, 7x, 20 &, 0 & 0, 4x, 4x, 4 x, & (7x 0 and 4 x 0) or (7x 0 and 4 x 0), & (x 0 and 4 x) or (x 0 and x 4), & (x 0 and x 4) or (x 0 and x 4), & (x 0) or (x 4) & xd(3, 0),(4, 3) ., , 21. Using|x| a & x a or x a, we get, |3x 7| 4 & 3x 7 4 or 3x 7 4, & 3x 3 or 3x 11 & x 1 or x , , 11, 3, , 11, & xd(3, 1),c , 3m ·, 3, 22. Using|x|# a + a # x # a, we get, |5 2x|# 3 &, &, &, &, , 3 # 5 2x # 3, 3 # 5 2x and 5 2x # 3, 2x # 8 and 2x $ 2 & x # 4 and x $ 1, 1 # x # 4 & xd[1, 4] .
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238, , Senior Secondary School Mathematics for Class 11, , 23. Using|x|# a + a # x # a, we get, |4x 5|#, , 24. Case I, , When x $ 0., Then,|x| x., 2x3, 5x, 1, 1, 1, 1, , # & #0 &, # 0 & # 0., x 3 2, x 3, 2 (x 3), |x| 3 2, &, &, , (5 x # 0 and x 3 0) or (5 x $ 0 and x 3 0), (x $ 5 and x 3) or (x # 5 and x 3), (x $ 5) or (x 3), , &, &, , (0 # x 3) or (x $ 5), xd[0, 3),[5, 3) ., , , , Case II, , 25., , 16, 1, 1, 1, 14, & # 4x 5 # &, # 4x and 4x #, 3, 3, 3, 3, 3, 7, 4, 7, 4, 7 4, & # x and x # & # x # & xd; , E ·, 6, 3, 6, 3, 6 3, , When x 0., Then,|x| x., 1, 1, 1, 1, 1, 1, 1, 1, , # & # & # & $0, x 3 2, x 3 2, 2 x 3, |x| 3 2, x32, x5, &, $ 0 & $ 0., x3, x 3, (x 5 $ 0 and x 3 0) or (x 5 # 0 and x 3 0), & (x $ 5 and x 3) or (x # 5 and x 3), & (x 3) or (x # 5) ., But, x 0., ( 3 x 0) or (x # 5) & xd( 3, 0),(3, 5] ., Hence, xd(3, 5],( 3, 0),[0, 3),[5, 3), &, xd(3, 5],( 3, 3),[5, 3)., , |x 2| x, |x 2| x 2x, |x 2| 3x, 2 0 &, 0 &, 0., x, x, x, Case I When x 2 $ 0., Then, x $ 2 and|x 2| x 2., |x 2| 3x, x 2 3x, 2 2x, , 0 &, 0 &, x, x, x 0, 2, 2, & x 2 0 & x 2 & 2x 2 & x 1., Case II, , (x $ 2 and x 1) & x 1 & xd(1, 3) ., When x 2 0., Then, x 2 and|x 2| (x 2) ., |x 2| 3x, x 2 3x, 4x 2, , 0 &, 0 &, 0, x, x, x, 4x 2, 2, 2 , , &, x 0 & 4 x 0 & x 4, 1, ·, & 4x 2 & x , 2, 1, , x 2 and x , & x 2 & xd(3, 2) ., 2, Hence, solution set (3, 2),(1, 3) ., , [a x $ 0]
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Linear Inequations (In one variable), , 239, , 26. Using|x| a + x a or x a, we have, 2x 1, 2x 1 , 2x 1, 2 +, 2 or, 2., x1, x1, x1, 2x 1 , Case I When, 2., x1, 4x 3, 2x 1, 2x 1 2x 2, Then, 2 0 &, 0 &, 0., x 1, x1, x1, (4x 3 0 and x 1 0) or (4x 3 0 and x 1 0), 3, 3, & cx and x 1m or cx and x 1m, 4, 4, 3, 3, x 1 & xdc , 1m ·, &, 4, 4, Case II, , 27., , 2x 1, 2., x1, 2x 1, 2x 1 2x 2, 1, Then, 2 0 &, 0 & 0, x 1, x1, x 1, & x 1 0 & x 1 & xd(1, 3) ., 3, solution set c , 1m,(1, 3) ., 4, When, , |x 3|, 0 + x 3 0 + x 3 + xd(3, 3) ., x3, , 28. Case I, , When x $ 0., Then,|x| x and so, , |x| 1, x1, $ 0 & $ 0., x 2, |x| 2, , (x 1 $ 0 and x 2 0) or (x 1 # 0 and x 2 0), & (x $ 1 and x 2) or (x # 1 and x 2), & (x # 1) or (x 2) & (0 # x # 1) or (x 2), & xd[0, 1],(2, 3) ., Case II When x 0., |x| 1, x 1, x1, Then,|x| x and so, $ 0 & $ 0 & $ 0., x 2, x 2, |x| 2, (x 1 $ 0 and x 2 0) or (x 1 # 0 and x 2 0), & (x $ 1 and x 2) or (x # 1 and x 2), & (x $ 1) or (x 2), & (1 # x 0) or (x 2) & xd[1, 0),(3, 2)., , 29. Case I, , [a x $ 0], , [a x 0], , solution set [1, 0),[0, 1],(3, 2),(2, 3), [1, 1],(3, 2),(2, 3) ., , When x $ 0., , x1, 1, 1, $ 1 & 1 $ 0 & $ 0., 2 x, 2 x, 2 |x|, (x 1 $ 0 and 2 x 0) or (x 1 # 0 and 2 x 0), & (x $ 1 and x 2) or (x # 1 and x 2), & (1 # x 2) & xd[1, 2) ., Case II When x 0., 1 x, 1x, 1, 1, Then,|x| x. So,, $ 1 & 1$ 0 &, $0 & #0, 2 x, 2x, 2 x, 2 |x|, Then,|x| x and so
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240, , Senior Secondary School Mathematics for Class 11, , & (1 x # 0 and 2 x 0) or (1 x $ 0 and 2 x 0), & (x # 1 and x 2) or (x $ 1 and x 2), & ( 2 x # 1) & xd( 2, 1] ., solution set ( 2, 1],[1, 2) ., 30. Putting x 1 0 and x 0, we get x 1 and x 0 as the critical points. These points, divide the whole real line into three parts, namely (3, 1), [1, 0) and [0, 3) ., Case I When 3 x 1., In this case, x 1 0 and x 0., |x 1| (x 1) x 1 and|x| x., So,|x 1||x| 3 & x 1 x 3 & 2x 4 & x 2., Case II, , solution set in this case (3, 2) ., [a 3 x 1], When 1 # x 0., In this case, x 1 $ 0 and x 0., |x 1| x 1 and|x| x., So,|x 1||x| 3 & x 1 x 3 & 1 3, which is absurd., , Case III When 0 # x 3., , In this case, x 1 0 and x $ 0., |x 1| x 1 and|x| x., So,|x 1||x| 3 & x 1 x 3 & 2x 2 & x 1., solution set in this case (1, 3) ., Hence, solution set (3, 2),(1, 3) ., 31. Using|x| a + x a or x a, we get, 2, 2, 1 or 1, x4, x 4, 2 , 2, &, 1 0 or 1 0, x4, x 4, 6x, x2, 0 or 0., &, x4, x 4, x2, Case I When, 0., x4, Then, (x 2 0 and x 4 0) or (x 2 0 and x 4 0), & (x 2 and x 4) or (x 2 and x 4), & 2 x 4 & xd(2, 4) ., Case II, , 32., , 6x, 0., x4, Then, (6 x 0 and x 4 0) or (6 x 0 and x 4 0), & (x 6 and x 4) or (x 6 and x 4), & 4 x 6 & xd(4, 6) ., solution set (2, 4),(4, 6) ., When, , 1 3x, 4, 4, # 3 & 3 # 0 & # 0., x1, x 1, x 1, , (1 3x # 0 and x 1 0) or (1 3x $ 0 and x 1 0), 1, 1, cx $ and x 1m or cx # and x 1m, &, 3, 3
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Linear Inequations (In one variable), , 241, , 1, 1, or x 1 & xd; , 3m,(3, 1), 3, 3, 6, 6, 3 3x, 1x, Again, $ 3 & 3 $ 0 &, $ 0 & $ 0., x 1, x 1, x1, 1 x, (1 x $ 0 and 1 x 0) or (1 x # 0 and 1 x 0), & (x # 1 and x 1) or (x $ 1 and x 1), &, , x$, , &, , 1 x # 1 & xd(1, 1] ., , , , 1, 1, solution set ((3, 1),; , 3m2+(1, 1] ; , 1E ·, 3, 3, , 39. (i) |x 2|$ 1 & (x 2) # 1 or (x 2) $ 1, [a |x|$ a & x # a or x $ a], & x # 1 or x $ 3 & xd(3, 1],[3, 3) ., (ii) |x 2|# 3 & 3 # x 2 # 3 & 1 # x # 5 & xd[1, 5] ., , required solution set {(3, 1],[3, 3)}+[1, 5] [1, 1],[3, 5] ., 40. Let the required odd integers be x and x 2., Then, x 2 18 and x (x 2) 20, & x 16 and 2x 2 20, & x 16 and x 1 10, & x 9 and x 16 & 9 x 16., So, the required values of x are 11, 13 and 15., Hence, the required pairs are (11, 13), (13, 15) and (15, 17)., 42. P (x) R (x) C (x) & P (x) 43x (26000 30x), & P (x) (13x 26000) ., Now, P (x) 0 & 13x 26000 0 & x 2000., Hence, more than 2000 cassettes must be sold., 43. Let the third reading be x. Then,, 8.48 8.35 x, 8.2 , 8.5, 3, & 24.6 16.83 x 25.5, & 24.6 16.83 x 25.5 16.83 & 7.77 x 8.67., 44. Let x litres of 2% boric acid be added to 640 litres of 8% solution of boric acid. Then,, total mixture (640 x) litres., 640 # 8 x # 2, m litres., Boric acid content in this mixture c, 100, 100, 640 # 8 x # 2, 6, 4, 1, #(640 x), #(640 x) ', , 100, 100, 100, 100, & 2560 4x 5120 2x 3840 6x, & 2x 2560 and 4x 1280 & 320 x 1280., 45. Let x litres of water may be added. Then,, 600 # 45, 30% of (600 x), 100, 600 x, 1800 3x, 270 , 10, 4, 600 x, 1800 3x, 270 and, 270, 10, 4, 600 x 1080 and 1800 3x 2700 & 300 x 480., 25% of (600 x) , , &, &, &
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242, , Senior Secondary School Mathematics for Class 11, , 46. Let the required marks be x. Then,, (89 93 95 91 x) $ (90 # 5) & x $ 82., , EXERCISE 6B, Very-Short-Answer Questions, 1, 0., x2, Find the solution set of the inequation|x 1| 2., Find the solution set of the inequation|2x 3| 1., |x 2|, Find the solution set of the inequation, 0, x ! 2., (x 2), x1, Find the solution set of the inequation 1., x 2, Solve the system of inequations x 2 $ 0, 2x 5 # 3., Solve 4x 16, when xdZ., Solve x 5 4x 10, when xd R., 3, Solve 1, when xd R., x 2, x, 1, Solve , when xd R., x 5 2, Solve|x| 4, where xd R., Solve|x| 4, where xd R., , 1. Find the solution set of the inequation, 2., 3., 4., 5., 6., 7., 8., 9., 10., 11., 12., , ANSWERS (EXERCISE 6B), , 1., 5., 9., 12., , 2. (1, 3), (3, 2), 6. xd[2, 4], (3, 2), xd(3, 2),(5, 3), xd(3, 4),(4, 3), , 3. (3, 1),(2, 3), 4. (2, 3), 7. xd{ 5, 6, 7, …}, 8. xd(3, 5), 10. xd(3, 5),(5, 3) 11. xd( 4, 4), , HINTS TO SOME SELECTED QUESTIONS, 1., , 1, 0 & x 2 0 & x 2 & xd(3, 2) ., (x 2), , 2. Using|x| a + a x a, we get, |x 1| 2 & 2 x 1 2 & 1 x 3 & xd(1, 3) ., 3. Using|x| a + x a or x a, we get, |2x 3| 1 & 2x 3 1 or 2x 3 1, & x 1 or x 2 & xd(3, 1),(2, 3) ., 4. Clearly, (x 2) 0 & x 2 & xd(2, 3) .
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Linear Inequations (In one variable), , 243, , 1, x1 , x1x2, 10 &, 0 & 0 & x2 0, x2, x2, x 2, & x 2 & xd(3, 2) ., 6. (x 2 $ 0 and 2x 5 # 3), 5., , &, , (x $ 2 and x # 4) & 2 # x # 4 & xd[2, 4] ., , 7. 4x 16 & x 4 & xd[ 5, 6, 7, 8, …} ., 8. x 5 4x 10 & 3x 15 & x 5 & xd(3, 5) ., 9., , 3 , 3x2, 5x, 10 &, 0 & 0., x2, x2, x 2, (5 x 0 and x 2 0) or (5 x 0 and x 2 0), & (x 5 and x 2) or (x 5 and x 2), & (x 5 or x 2) & xd(3, 2),(5, 3) ., , 10., , 2x x 5, x5, x 1, 0 &, 0 & 0., x5 2, x 5, 2 (x 5), , (x 5 0 and x 5 0) or (x 5 0 and x 5 0), & (x 5 and x 5) or (x 5 and x 5), & (x 5) or (x 5) & xd(3, 5),(5, 3) ., 11. Using|x| a + a x a, we get, |x| 4 & 4 x 4 & xd( 4, 4) ., 12. Using|x| a + x a or x a, we get, |x| 4 + x 4 or x 4, + xd(3, 4),(4, 3) ., , KEY FACTS AND FORMULAE, 1., , (i) Linear Inequations in One Variable:, Inequalities of the form:, (i) ax b c, (ii) ax b # c, (iii) ax b c, (iv) ax b $ c, where a, b, cd R, a ! 0 and x is a variable, are called inequations in x., (ii) Replacement Set A set from which the values are replaced in an, inequation in x, is called the replacement set., (iii) Solution Set The set of all those values of x taken from the, replacement set which satisfy the given inequation is called the, solution set of the inequation., , 2. Rules for Solving an Inequation, Adding the same number or expression to each side of an inequation, does not change the inequality., RULE 2, Subtracting the same number or expression from each side of an, inequation does not change the inequality., RULE 3, Multiplying (or dividing) each side of an inequation by the same, positive number does not change the inequality., RULE 1
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244, RULE 4, , Senior Secondary School Mathematics for Class 11, , Multiplying (or dividing) each side of an inequation by the same, negative number reverses the inequality., Thus, x 3 & x 3 [on multiplying both sides by –1]., And, 3x # 12 & x $ 4 [on dividing both sides by –3]., , Using Rule 1 and Rule 2, we can drop any term from one side, of an inequation and put it on the other side with the opposite sign., 3. Some Useful Results, Let a be a positive real number. Then,, (i) |x| a + a x a + xd( a, a) ., (ii) |x|# a + a # x # a + xd[ a, a] ., (iii) |x| a + x a or x a., (iv) |x|$ a + x # a or x $ a., 4. Solving Two Simultaneous Linear Inequations in x, Suppose we have to solve two inequations in x simultaneously. Then, we, find the solution set for each one of them. The intersection of these solution, sets is the required solution set., , TRANSPOSITION, ,
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Linear Inequations (In two variables), , 245, , 7, , Linear Inequations, (In two variables), The statements of inequalities of the form ax by c 0,, ax by c $ 0, ax by c 0 and ax by c # 0 are called linear inequations in two, variables x and y., The set of all ordered pairs (x, y) which satisfy the given inequation is, called the solution set of the inequation., LINEAR INEQUATIONS, , GRAPH OF A LINEAR INEQUATION Let an inequation of any of the types given, above be given., Then, in order to solve it, we proceed according to steps given below., Step 1 Consider the equation, ax by c 0., , Step 2, , Draw the graph of this equation, which is a line., In case of strict inequation > or < draw the line dotted, otherwise make, it thick., This line divides the plane into two equal parts., Choose a point [if possible (0, 0)], not lying on this line. If this point, satisfies the given inequation then shade the part of the plane, containing this point, otherwise shade the other part., The shaded portion represents the solution set of the given inequation., The dotted line is not a part of the solution, while the thick line is a, part of it., SOLVED EXAMPLES, , EXAMPLE 1, , Solve 3x 4y # 12 graphically., , SOLUTION, , Consider the equation, 3x 4y 12., The values of (x, y) satisfying 3x 4y 12 are:, x, , 4, , 0, , y, , 0, , 3, , Plot the points A(4, 0) and B(0, 3) on a graph paper., Then, line AB represents 3x 4y 12., This line divides the plane of the paper in two equal parts., 245
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246, , Senior Secondary School Mathematics for Class 11, , Clearly, the point (0, 0) satisfies 3x 4y # 12., So, we shade that part of the plane divided by AB which contains, (0, 0)., The shaded part of the plane together with all the points on line AB, represents 3x 4y # 12., , Y, , B(0, 3), , A(4, 0), X¢, , O, , X, , Y¢, , EXAMPLE 2, , Draw the graph of the solution set of the inequation 2x y $ 1., , SOLUTION, , Consider the equation, 2x y 1., The values of (x, y) satisfying 2x y 1 are:, x, , 2, , 0, , y, , 3, , –1, , Take a graph paper and plot the points A(2, 3) and B(0, 1) ., Then, line AB represents 2x y 1., This line divides the plane of the paper in two equal parts., Clearly, the point (0, 0) does not lie on 2x y 1., Also, (0, 0) does not satisfy 2x y $ 1., So, we shade that part of the plane divided by line AB which does, not contain (0, 0), as shown below., The shaded part of the plane together with all points on the line AB, constitutes the solution set of the inequation, 2x y $ 1.
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Linear Inequations (In two variables), , 247, , Y, , A, (2, 3), , X¢, , O, B, (0, –1), , Y¢, , EXAMPLE 3, SOLUTION, , Solve 2x 3y 6 graphically., Consider the equation, 2x 3y 6., x y , 1., 3 2, Thus, it cuts the x-axis at A(3, 0) and the y-axis at B(0, 2) ., , AB represents the line, 2x 3y 6., This equation may be written as,, , X
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248, , Senior Secondary School Mathematics for Class 11, , Clearly, the point (0, 0) satisfies 2x 3y 6., , (0, 0) does not satisfy 2x 3y 6., So, we shade that part of the plane divided by AB which does not, contain O(0, 0) ., The shaded part of the plane excluding line AB forms the solution, set of 2x 3y 6, as shown above., EXAMPLE 4, SOLUTION, , Solve x y 5 graphically., Consider the equation, x y 5., xy, 1., 5 5, Thus, it cuts the x-axis at A(5, 0) and the y-axis at B(0, 5) ., , AB represents the line x y 5., Clearly, (0, 0) satisfies the inequation x y 5., So, we shade that part of the plane divided by AB which contains, O(0, 0) ., The shaded part of the plane excluding line AB forms the solution, set of x y 5, as shown below., This equation may be written as, , Y, , B(0, 5), , x, +, y, =, 5, A(5, 0), X¢, , O, , X, , Y¢, , EXAMPLE 5, SOLUTION, , Solve y 2 graphically., The graph of y 2 is a straight line AB drawn at a distance of, 2 units above the x-axis and parallel to the x-axis., Clearly, the point (0, 0) satisfies the inequation, y 2.
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Linear Inequations (In two variables), , 249, , So, we shade that part of the plane divided by AB which contains, the origin O(0, 0), excluding AB., The shaded part of the plane forms the solution set of the inequation,, y 2, as shown below., , EXAMPLE 6, SOLUTION, , Solve the inequation x 3 graphically., We know that the graph of x 3 is a straight line AB drawn, parallel to the y-axis at a distance of 3 units to the left of the y-axis., Y, , x = –3, , A, , X¢, , B, , O, , X, , Y¢
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250, , Senior Secondary School Mathematics for Class 11, , Clearly, the point (0, 0) does not satisfy the inequation, x 3., So, we shade that part of the plane divided by AB which does not, contain O(0, 0), excluding AB., The shaded part of the plane is the solution set of x 3, as shown, above., EXAMPLE 7, , SOLUTION, , Exhibit graphically the solution set of the linear inequations:, x 2y # 10, x y # 6, x # 4, x $ 0 and y $ 0., (i) First, we draw the graph of x 2y # 10., Consider the line, x 2y 10., Clearly, the points A(0, 5) and B(10, 0) satisfy, x 2y 10., Then, line AB represents x 2y 10., Clearly, (0, 0) satisfies the inequation x 2y # 10., Thus, the line AB and part of the plane containing O(0, 0), represent the solution set of x 2y # 10., (ii) Next, we draw the graph of x y # 6., Consider the line, x y 6., Clearly, the points C(0, 6) and D(6, 0) satisfy x y 6., So, line CD represents x y 6., Clearly, (0, 0) satisfies the inequation, x y # 6., So, the line CD and part of the plane containing O(0, 0), represent the solution set of x y # 6., (iii) Next, we draw the graph of x # 4., We know that x 4 is the line EF parallel to the y-axis and, passing through the point E(4, 0).
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Linear Inequations (In two variables), , 251, , Clearly, (0, 0) satisfies the inequation, x # 4., Thus, the line EF and part of the plane containing O(0, 0), represent the solution set of x # 4., (iv) x $ 0 is represented by the y-axis and the plane on its right., (v) y $ 0 is represented by the x-axis and the plane above the x-axis., The intersection of all these planes is the shaded part, which, together with its boundary represents the solution of the given, system of inequations., EXAMPLE 8, , SOLUTION, , Draw the graph of the solution set of the inequations:, 2x y $ 2, x y # 1, x 2y # 8, x $ 0 and y $ 0., (i) First, we draw the graph of 2x y $ 2., Consider the line, 2x y 2., x y, Now, 2x y 2 & 1., 1 2, This line meets the axes at A(1, 0) and B(0, 2)., Clearly, the line AB represents 2x y 2., Now, (0, 0) does not satisfy 2x y $ 2., Thus, the line AB and part of the plane separated by AB and, not containing O(0, 0), represent the solution set of 2x y $ 2., (ii) Next, we draw the graph of x y # 1., Consider the line, x y 1., y, x, 1., xy 1 & , 1 (1), This line meets at the axes at A(1, 0) and C(0, –1)., Clearly, the line AC represents x y 1., Also, (0, 0) satisfies, x y # 1., Thus, the line AC and part of the plane separated by AC and, containing O(0, 0) represent the solution set of x y # 1., (iii) Now, we draw the graph of x 2y # 8., Consider the line, x 2y 8., x y, Now, x 2y 8 & 1., 8 4, This line meets the axes at D(8, 0) and E(0, 4)., So, the line DE represents x 2y 8., Also, (0, 0) satisfies the inequation, x 2y # 8., Thus, the line DE and part of the plane containing O(0, 0), represent the solution set of x 2y # 8., (iv) x $ 0 is represented by the y-axis and the plane on its right., (v) y $ 0 is represented by the x-axis and the plane above the x-axis.
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252, , Senior Secondary School Mathematics for Class 11, , The intersection of all these planes is the required shaded part,, representing the solution of the given system., Y, , E(0, 4), , B, (0, 2), , X¢, , O, , A(1, 0), , D, (8, 0), , X, , C(0, –1), , Y¢, , EXAMPLE 9, , SOLUTION, , Solve the following system of inequations graphically:, 5x 4y # 40, x $ 2, y $ 3., (i) First, we draw the graph line 5x 4y 40., x y , Now, 5x 4y 40 & , 1., 8 10, This line meets the axes at A(8, 0) and B(0, 10) ., Clearly, the line AB represents 5x 4y 40., Now, (0, 0) satisfies the inequation 5x 4y # 40., Thus, the line AB and part of the plane containing O(0, 0) and, separated by AB represent the solution set of the inequation,, 5x 4y # 40., (ii) Clearly, x 2 is the line CD parallel to the y-axis at a distance, of 2 units to its right., Clearly, (0, 0) does not satisfy the inequation, x $ 2., So, x $ 2 consists of the line CD and part of the plane on right, of CD., (iii) Also, y 3 is the line EF parallel to the x-axis lying above it at, a distance of 3 units from it.
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Linear Inequations (In two variables), , 253, , Clearly, (0, 0) does not satisfy the inequation, y $ 3., So, y $ 3 consists of the line EF and part of the plane above it., The intersection of all these planes is the required shaded part,, representing the solution of the given system, as shown below., Y, B(0, 10), , C, , y=3, E, , 5x, , x=2, , F, , +, 4y, =, 40, A(8, 0), , X¢, , O, , X, Y¢, , EXAMPLE 10, , SOLUTION, , D, , Show that the following system of linear inequations has no solution:, x 2y # 3, 3x 4y $ 12, x $ 0, y $ 1., (i) First, we draw the graph of x 2y # 3., Consider the line, x 2y 3., Clearly, the points A(1, 1) and B(–1, 2) lie on this line., Then, line AB represents x 2y 3., Now, (0, 0) satisfies x 2y # 3., So, the line AB and part of the plane containing O(0, 0), represent the solution set of x 2y # 3., (ii) Next, we draw the graph of 3x 4y $ 12., Consider the line, 3x 4y 12., x y, Now, 3x 4y 12 & 1., 4 3
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254, , Senior Secondary School Mathematics for Class 11, , So, this line meets the axes at C(4, 0) and D(0, 3)., Thus, CD represents the line 3x 4y $ 12., Clearly, (0, 0) does not satisfy 3x 4y $ 12., Therefore, the line CD and part of the plane not containing, O(0, 0) represent the solution set of 3x 4y $ 12., (iii) Clearly, x $ 0 is represented by the y-axis and the plane on its, right., (iv) y $ 1 represents a line parallel to the x-axis at a distance of 1, unit above the x-axis and the plane above this line., , x=0, , Y, , D(0, 3), , x+, , 2y, , B, , 2), , (–1, , 3x, =3, , +4, , y=, , 12, , A(1, 1), y=1, C(4, 0), , X¢, , O, , X, Y¢, , Clearly, all these planes have no common intersection., Hence, the given system of inequations has no solution., EXAMPLE 11, , SOLUTION, , Show that the solution set of the following linear inequations is an, unbounded region 2x y $ 8, x 2y $ 10, x $ 0, y $ 0., (i) First, we draw the graph of 2x y $ 8., Consider the line, 2x y 8., x y, Now, 2x y 8 & 1., 4 8, This line meets the axes at A(4, 0) and B(0, 8)., Clearly, the line AB represents 2x y 8., It is also clear that (0, 0) does not satisfy 2x y $ 8., Thus, the line AB and the part of the plane not containing, O(0, 0) represent the solution of the inequation 2x y $ 8.
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Linear Inequations (In two variables), , 255, , (ii) Next, we draw the graph of x 2y $ 10., Consider the line, x 2y 10., Clearly, the points C(0, 5) and D(2, 4) lie on this line., Also, (0, 0) does not satisfy x 2y $ 10., So, the line CD and the part of the plane not containing O(0, 0), represent the solution set of x 2y $ 10., (iii) x $ 0 is represented by the y-axis and the plane to its right., (iv) y $ 0 is represented by the x-axis and the plane above the, x-axis., , It is clear from the above graph that the solution set of the, given inequations is an unbounded region, shown by the, shaded region.
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256, EXAMPLE 12, , SOLUTION, , Senior Secondary School Mathematics for Class 11, , Find the linear inequations for which the shaded region in the figure given, below, is the solution set., , (i) Consider the line, 3x 4y 18., We observe here from the graph that the shaded region and, the origin O(0, 0) lie on the same side of this line., But, (0, 0) satisfies the inequation 3x 4y # 18., So, the inequation corresponding to this line is 3x 4y # 18., (ii) Consider the line, 2x 3y 3., We observe here from the graph that the shaded region and, the origin O(0, 0) lie on the opposite sides of this line., Here, (0, 0) satisfies the inequation 2x 3y # 3., So, the inequation corresponding to this line is 2x 3y $ 3., (iii) Consider the line, 7x 4y 14., We observe here from the graph that the shaded region and, the origin O(0, 0) lie on the same side of this line., Here, (0, 0) satisfies the inequation 7x 4y # 14., So, the inequation corresponding to this line is 7x 4y # 14., (iv) Consider the line, x 6y 3., We observe here from the graph that the shaded region and, the origin O(0, 0) lie on the same side of this line.
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Linear Inequations (In two variables), , 257, , Also, the point (0, 0) satisfies the inequation x 6y # 3., So, the inequation corresponding to this line is x 6y # 3., (v) It is clear from the given figure that the whole shaded region, lies in the first quadrant only. So, x $ 0 and y $ 0., Hence, the linear inequations corresponding to the given solution, set are, 3x 4y # 18, 2x 3y $ 3, 7x 4y # 14, x 6y # 3, x $ 0 and y $ 0., EXAMPLE 13, , Find the linear inequations for which the shaded region of the figure given, below, is the solution set., , x=5, , Y, (0, 8), , y=5, , (0, 4), , x, +, y, =, 4, , x, , X¢, , O, , (4, 0), , +, , y, , =, , 8, , X, , Y¢, , SOLUTION, , (i) Consider the graph line, x 5., It is clear that the shaded region and the origin O(0, 0) lie on, the same side of this line., And, the point (0, 0) satisfies the inequation x # 5., So, the inequation corresponding to this line is x # 5., (ii) Consider the graph line, y 5., Clearly, the shaded region and the origin O(0, 0) lie on the, same side of this line.
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258, , Senior Secondary School Mathematics for Class 11, , And, the point (0, 0) satisfies the inequation y # 5., So, the inequation corresponding to this line is y # 5., (iii) Consider the graph line, x y 8., Clearly, the shaded region and the origin O(0, 0) lie on the, same side of this line., And, the point (0, 0) satisfies the inequation x y # 8., So, the inequation corresponding to this line is x y # 8., (iv) Consider the graph line, x y 4., It is clear from the graph that the shaded region and the origin, O(0, 0) lie on the opposite sides of this line. Also, the point, (0, 0) satisfies the inequation x y # 4., So, the inequation corresponding to this line is x y $ 4., (v) It is clear from the given figure that the whole shaded region, lies in the first quadrant only., So, x $ 0 and y $ 0., Hence, the linear inequations corresponding to the given, solution set are, x # 5, y # 5, x y # 8, x y $ 4, x $ 0 and y $ 0., EXAMPLE 14, , Find the linear inequations for which the shaded region of the figure given, below, is the solution set., Y, , x, , –, , y, , =, , 1, , x+, , X¢, , (1, 0), , O, , 2y, , =8, , X, , 2x, +y, =2, , Y¢, , SOLUTION, , (i) Consider the graph line, x 2y 8., It is clear that the shaded region and the origin O(0, 0) lie, on the same side of the line x 2y 8 and (0, 0) satisfies the, inequation x 2y # 8., So, the inequation corresponding to this line is x 2y # 8.
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Linear Inequations (In two variables), , 259, , (ii) Consider the graph line, 2x y 2., It is clear that the shaded region and the origin O(0, 0) lie on, the opposite sides of the line 2x y 2., And, (0, 0) does not satisfy the inequation 2x y $ 2., So, the inequation corresponding to this line is 2x y $ 2., (iii) Consider the graph line, x y 1., It is clear that the shaded region and the origin O(0, 0) lie on, the same side of the line x y 1., And, (0, 0) satisfies the inequation x y # 1., So, the inequation corresponding to this line is x y # 1., (iv) It may be noted here that the shaded region is above the x-axis, and is on the right side of the y-axis., So, we must have x $ 0 and y $ 0., Hence, the linear inequations corresponding to the given solution, set are, x 2y # 8, 2x y $ 2, x y # 1, x $ 0, y $ 0., EXAMPLE 15, , A small manufacturing firm produces two types of gadgets A and B,, which are first processed in the foundry and then sent to another machine, for finishing. The number of man-hours of labour required in each shop for, the production of each unit of A and B and the number of man-hours for, the firm available per week are as follows:, Foundry, , SOLUTION, , Machine shop, , Man-hours for 1 unit, of gadget A, , 10, , 5, , Man-hours for 1 unit, of gadget B, , 6, , 4, , Firm’s capacity per, week in man-hours, , 1000, , 600, , Formulate it in the form of linear inequations. Draw the graph showing, the solution of all these inequations., Suppose that the firm manufactures x units of A and y units of B., , number of man-hours for A and B in the foundry, 10x 6y,, and number of man-hours for A and B in the machine shop, 5x 4y., , 10x 6y # 1000 and 5x 4y # 600., Clearly, x $ 0 and y $ 0 [ number of units cannot be negative]., SCALE 1 small division = 2 units.
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260, , Senior Secondary School Mathematics for Class 11, , Y, , 10, x+, , D, , 6y, =1, 00, 0, , B, , 5x, +, 4y, , A, , =, 60, 0, , X¢ O, , C, , X, , Y¢, , (i) First we draw the graph of 10x 6y # 1000., Consider the equation, 10x 6y 1000., 10x 6y 1000 & 5x 3y 500., The values of (x, y) satisfying 5x 3y 500 are:, x, , 70, , 40, , y, , 50, , 100, , Plot the points A(70, 50) and B(40, 100) on a graph paper and, join them by the thick line AB., Consider (0, 0). It does not lie on 5x 3y 500., Also, (0, 0) satisfies 10x 6y # 1000., Therefore, the line AB and the part of the plane separated by, AB, containing (0, 0), represent the graph of 10x 6y # 1000., (ii) Now, we draw the graph of 5x 4y # 600., Consider the line, 5x 4y 600.
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Linear Inequations (In two variables), , 261, , x y , 1., 120 150, This line meets the axes at C(120, 0) and D(0, 150)., Plot these points on the same graph paper as above and join, them by the thick line CD., Now, consider (0, 0). It does not lie on 5x 4y 600., Also, (0, 0) satisfies the inequation 5x 4y # 600., So, the line CD and that part of the plane separated by CD,, containing (0, 0), represent the graph of 5x 4y # 600., x $ 0 is represented by the y-axis and the part of the plane on, its right side., y $ 0 is represented by the x-axis and the plane above the x-axis., Clearly, the shaded region together with its boundary, represents the solution set of the given inequations., 5x 4y 600 &, , EXERCISE 7, Draw the graph of the solution set of each of the following inequations:, 1. x y $ 4, 4. x $ y 2, , 2. x y # 3, 5. 3x 2y 6, , 3. y 2 # 3x, 6. 3x 5y 15, , Solve the following systems of inequations graphically:, 7. x $ 2, y $ 3, 9. x y # 6, x y $ 4, , 8. 3x 2y # 12, x $ 1, y $ 2, 10. 2x y $ 6, 3x 4y # 12, , 11. x y # 9, y x, x $ 0, , 12. 2x y 1, x 2y 1, , 13. 5x 4y # 20, x $ 1, y $ 2, , 14. 3x 4y # 60, x 3y # 30, x $ 0, y $ 0, , 15. 2x y $ 4, x y # 3, 2x 3y # 6, 16. x 2y # 10, x y $ 1, x y # 0, x $ 0, y $ 0, 17. 4x 3y # 60, y $ 2x, x $ 3, x $ 0, y $ 0, 18. x 2y # 2, x y $ 3, 2x y # 4, x $ 0, y $ 0, 19. x 2y # 100, 2x y # 120, x y # 70, x $ 0, y $ 0, 20. x 2y # 2000, x y # 1500, y # 600, x $ 0, y $ 0, 21. Show that each of the following systems of linear inequations has no, solution:, (i) 3x 2y $ 24, 3x y # 15, x $ 4, (ii) 2x y # 2, x 2y $ 0, x $ 0, y $ 0, 22. Show that the solution set of the following system of inequations is an, unbounded set:, 3x y $ 12, x y $ 9, x $ 0, y $ 0.
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262, , Senior Secondary School Mathematics for Class 11, , 23. Find the linear inequations for which the shaded area is the solution set in, the figure given below., Y, C (0, 4), , 4x, +, 3y, , A, (0, 3), , =1, 2, 3x, , X¢, , O, Y¢, , +, , 4y, , D, (3, 0), , =1, , 2, B, (4, 0), , X, , 24. Find the linear inequations for which the shaded area in the figure given, below, is the solution set.
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Linear Inequations (In two variables), , 263, , 25. A furniture dealer deals in only two items: tables and chairs. He has, ` 30000 to invest and a space to store at most 60 pieces. A table costs him, ` 1500 and a chair ` 300. Formulate the data in the form of inequations and, draw a graph representing the solution of these inequations., Hint, , Suppose he makes x tables and y chairs. Then,, 1500x 300y # 30000, x y # 60, x $ 0 and y $ 0., , 26. If a young man rides his motorcycle at 40 km per hour, he has to spend ` 6, per km on petrol and if he rides it at 50 km per hour, the petrol cost rises to, ` 10 per km. He has ` 500 to spend on petrol and wishes to find the, maximum distance he can travel within one hour. Formulate the data in the, form of inequations and draw a graph representing the solution of these, inequations., Hint, , Suppose he rides x km at 40 km/hr and y km at 50 km/hr., y, x, , Then, 6x 10y # 500,, # 1, x $ 0, y $ 0., 40 50, ANSWERS (EXERCISE 7), , 23. 3x 4y # 12, 4x 3y # 12, x $ 0, y $ 0, 24. x y # 4, x 5y $ 4, 6x 2y $ 8, x # 3, y # 3, x $ 0, y $ 0,
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264, , Senior Secondary School Mathematics for Class 11, , 8, , Permutations, THE FACTORIAL, Let n be a positive integer. Then, the continued product of first, n natural numbers is called factorial n, to be denoted by n !, or n . Also, we define 0 ! 1., When n is negative or a fraction, n ! is not defined., Thus, n ! n (n 1) (n 2) … 3 · 2 · 1., , FACTORIAL NOTATION, , EXAMPLE, , 6 ! (6 # 5 # 4 # 3 # 2 #1) 720, 3 ! 3 # 2 #1 6 and 1 ! 1., , DEDUCTION, , n ! n (n 1) (n 2) (n 3) … 3 · 2 · 1, n [(n 1) (n 2) (n 3) … 3 · 2 · 1], n [(n 1) !] ., Thus, 9 ! 9#(8 !), 6 ! 6 #(5 !), 3 ! 3 #(2 !) ., Also, 1 ! 1#(0 !) & 0 ! 1., Thus, we have, (i) n ! n (n 1) (n 2) … 3 # 2 #1., (ii) n ! n #(n 1) !., (iii) 0 ! 1., SOLVED EXAMPLES, , EXAMPLE 1, , SOLUTION, , Compute:, 10 !, (i), (7 !)#(3 !), , (ii), , 30 !, 28 !, , (iii), , 11 ! 10 !, 9!, , We have, (i), , 10 # 9# 8 #(7 !) 10 # 9 # 8, 10 !, , , 120., 3 # 2 #1, (7 !)#(3 !) (7 !)#(3 # 2 #1), , 30 ! 30 # 29 #(28 !) , 30 # 29 870., 28 !, (28 !), 11 ! 10 ! 11#10 #(9 !) 10 #(9 !) (110 10) (9 !) , (iii), 100., 9!, 9!, (9 !), (ii), , EXAMPLE 2, SOLUTION, , x 1 1 ,, find the value of x., 10 ! 8 ! 9 !, On multiplying both sides by 10 !, we get, If, , 264
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Permutations, , 265, , 10 # 9 #(8 !) 10 #(9 !), 10 ! 10 !, , & x, 8!, 9!, (8 !), (9 !), & x (10 # 9) 10 (90 10) 100., Hence, x 100., x, , EXAMPLE 3, , SOLUTION, , Express each of the following products in factorials:, (i) 5 # 6 #7 # 8 # 9, (ii) 2 # 4 # 6 # 8 #10, We have, (1# 2 # 3 # 4)#(5 # 6 #7 # 8 # 9), (i) 5 # 6 #7 # 8 # 9 , (1# 2 # 3 # 4), [multiplying and dividing the given, product by (1# 2 # 3 # 4)], 1# 2 # 3 # 4 # 5 # 6 #7 # 8 # 9 9 ! ·, 1# 2 # 3 # 4, 4!, (ii) 2 # 4 # 6 # 8 #10 (2 #1)#(2 # 2)#(2 # 3)#(2 # 4)#(2 # 5), 2 5 #(1# 2 # 3 # 4 # 5) 2 5 #(5 !) ., , EXAMPLE 4, SOLUTION, , EXAMPLE 5, SOLUTION, , EXAMPLE 6, SOLUTION, , EXAMPLE 7, , Find the LCM of (5 !, 6 !, 7 !) ., We have, LCM (5 !, 6 !, 7 !) LCM (5 !, 6 # 5 !, 7 # 6 # 5 !), (5 !)# LCM (1, 6, 42) (5 !)# 42, 7 # 6 # 5 ! 7 !., If (n 1) ! 90 #(n 1) !, find n., We have, (n 1) ! 90 #(n 1) !, &, (n 1)# n #(n 1) ! 90 #(n 1) !, &, (n 1) n 90, &, (n 1) n 10 # 9, [writing 90 as product of two consecutive integers], &, n 9., Hence, n 9., If (n 2) ! 60 #(n 1) !, find n., We have, (n 2) ! 60 #(n 1) !, &, (n 2)#(n 1)# n #(n 1) ! 60 #(n 1) !, &, (n 2) (n 1) n 60, &, (n 2) (n 1) n 5 # 4 # 3, [writing 60 as product of 3 consecutive integers], &, n 3., Hence, n 3., Prove that (n ! 1) is not divisible by any natural number between 2, and n.
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266, , Senior Secondary School Mathematics for Class 11, , SOLUTION, , Since n ! n (n 1) (n 2) … 3 · 2 · 1, it follows that n ! is divisible by, every natural number between 2 and n. So, (n ! 1), when divided, by any natural number between 2 and n, leaves 1 as remainder., Hence, (n ! 1) is not divisible by any natural number between 2, and n., , EXAMPLE 8, , Prove that n(n 1) (n 2) … (n r 1) , , SOLUTION, , We have, n (n 1) (n 2) … (n r 1), n (n 1) (n 2) … (n r 1) · (n r) !, , (n r) !, [multiplying num. and denom. by (n r) !], n! ·, (n r) !, n! ·, Hence, n (n 1) (n 2) … (n r 1) , (n r) !, , EXAMPLE 9, SOLUTION, , EXAMPLE 10, SOLUTION, , n! ·, (n r) !, , (2n) !, 2 n #{1# 3 # 5 #…#(2n 1)} ., Prove that, n!, We have, (2n) ! {1# 2 # 3 # 4 # 5 #…#(2n 2)#(2n 1)# 2n}, , n!, n!, {1# 3 # 5 #…#(2n 1)}#{2 # 4 # 6 #…#(2n 2)# 2n}, , n!, {1# 3 # 5 #…# (2n 1)}# 2 n #{1# 2 # 3 #…# n}, , n!, {1# 3 # 5 #…#(2n 1)}# 2 n #(n !), , n!, 2 n #{1# 3 # 5 #…#(2n 1)} ., (2n) !, 2 n #{1# 3 # 5 #…#(2n 1)} ., Hence,, n!, Prove that (n !) 2 # (n !)# n n (2n) ! for all nd N., We have, (n !) 2 (n !)#(n !) (n !)#(1# 2 # 3 #…# n), [a r # n for each r 1, 2, 3, …, n]., # (n !)# n n, … (i), , (n !) 2 # (n !)# n n., Again, we have, (2n) ! (1# 2 # 3 #…# n)#(n 1)#(n 2)#…# 2n, (n !)# n n., [a (n r) n for each r 1, 2, …, n]., Thus, (n !)# n n (2n) !., … (ii), Hence, from (i) and (ii), we have, (n !) 2 # (n !)# n n (2n) !.
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Permutations, , 267, , EXERCISE 8A, 1. Compute:, 9!, (i), (5 !)#(3 !), , (ii), , 32 !, 29 !, , (iii), , (12 !) (10 !), 9!, , 2. Prove that LCM {6 !, 7 !, 8 !} 8 !., 1 1 1 145 ·, 10 ! 11 ! 12 ! 12 !, 1 1 x, 4. If, , find the value of x., 6! 7! 8!, 3. Prove that, , 5. Write the following products in factorial notation:, (i) 6 #7 # 8 # 9 #10 #11#12, (ii) 3 # 6 # 9 #12 #15, 6. Which of the following are true or false?, (i) (2 3) ! 2 ! 3 !, (ii) (2 # 3) ! (2 !)#(3 !), , , , 7. If (n 1) ! 12 #(n 1) !, find the value of n., 8. If (n 2) ! 2550# n !, find the value of n., 9. If (n 3) ! 56 #(n 1) !, find the value of n., n!, n!, 2 : 1, find the value of n., :, (2 !)#(n 2) ! (4 !)#(n 4) !, (2n) !, n!, 44 : 3, find the value of n., 11. If, :, (3 !)#(2n 3) ! (2 !)#(n 2) !, n!, 12. Evaluate, , when n 15 and r 12., (r !)#(n r) !, , 10. If, , 13. Prove that (n 2)#(n !) (n !) (n 1) !., 14. Prove that:, n! , (i), n (n 1) (n 2) … (r 1), r!, n!, n!, (ii) (n r 1) ·, (n r 1) ! (n r) !, (iii), , (n 1) !, n!, n!, , , r ! (n r) ! (r 1) ! (n r 1) ! r ! (n r 1) !, ANSWERS (EXERCISE 8A), , 1. (i) 504, 12 !, 5. (i), 5!, , 8. n 49, , (ii) 29760, (ii) 3 5 (5 !), 9. n 5, , (iii) 1310, 6. (i) False, 10. n 5, , 4. x 64, (ii) False, 11. n 6, , 7. n 3, 12. 455, , HINTS TO SOME SELECTED QUESTIONS, 1 , 1, 1 (11 1) , 1, 3. LHS (, 2, 10 ! 11#(10 !) 12 ! 11#(10 !) 12 #(11 !)
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268, , Senior Secondary School Mathematics for Class 11, (144 1) 145, 1, 12 , , , RHS., 11 ! 12 #(11 !) 12 #(11 !) 12 !, , 7. (n 1) n#(n 1) ! 12 #(n 1) ! & (n 1)# n 12 4 # 3 & n 3., 8. (n 2) (n 1)#(n !) 51# 50 #(n !) & (n 2) (n 1) 51# 50, & n 1 50 & n 49., 9. (n 3) (n 2)#(n 1) ! 8 #7 #(n 1) ! & (n 3) (n 2) 8 #7 & n 2 7 & n 5., 10., , 24 #(n 4) ! 2, n!, , #, 1, n!, 2 #(n 2) (n 3)#(n 4) !, , (n 2) (n 3) 6 3 # 2 & n 3 2 & n 5., (2n) (2n 1) (2n 2)#(2n 3) !, 2 #(n 2) !, 44, 11., #, 3, 6 #(2n 3) !, n (n 1)#(n 2) !, 8 # n#(n 1)#(2n 1) 44, 6, 44, , # & 2n 1 11 & n 6., &, & 2n 1 , 3, 3, 8, 6 # n#(n 1), &, , 12., , 15 #14 #13 #(12 !), 15 !, , 455., (12 !)#(3 !), 6 #(12 !), , 13. RHS (n !) (n 1)#(n !) (n 2)#(n !) LHS., n (n 1) (n 2) … (r 1)#(r !) n !, , ·, r!, r!, (ii) We know that (n r 1) ! (n r 1) · (n r) !., , 14. (i) RHS , , n!, 1, 1, ' , 1, (n r) ! · (r 1) ! r n r 1, (n 1), (n 1) !, n!, , , RHS., ·, (n r) ! · (r 1) ! r (n r 1) (r !) · (n r 1) !, , (iii) LHS , , FUNDAMENTAL PRINCIPLES OF COUNTING, If there are two operations such that, one of them can be performed in m ways, and when it has been performed in anyone of, these m ways, the second operation can be performed in n ways then the two operations, in succession can be performed in (m # n) ways., , 1. FUNDAMENTAL PRINCIPLE OF MULTIPLICATION, , EXAMPLE 1, , SOLUTION, , In a class there are 15 boys and 12 girls. The teacher wants to select 1 boy, and 1 girl to represent the class for a function. In how many ways can the, teacher make the selection?, Here the teacher is to perform two operations:, (i) selecting 1 boy out of 15 boys,, and (ii) selecting 1 girl out of 12 girls., The first of these can be done in 15 ways and the second in 12 ways., So, by the fundamental principle of multiplication, the required, number of ways (15 #12) 180., Hence, the teacher can make the selection of 1 boy and 1 girl in, 180 ways.
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Permutations, , 269, , 2. FUNDAMENTAL PRINCIPLE OF ADDITION If there are two operations such that they, can be performed independently in m and n ways respectively then either of the two, operations can be performed in (m n) ways., EXAMPLE 2, , SOLUTION, , In a class there are 16 boys and 9 girls. The teacher wants to select either, a boy or a girl as a class representative. In how many ways can the teacher, make the selection?, Here the teacher is to perform either of the following two, operations:, (i) selecting 1 boy out of 16 boys,, and (ii) selecting 1 girl out of 9 girls., The first of these can be performed in 16 ways and the second in, 9 ways., By the fundamental principle of addition, either of the two, operations can be performed in (16 9) ways = 25 ways., Hence, the teacher can make the selection of 1 boy or 1 girl in, 25 ways., ILLUSTRATIVE EXAMPLES, , EXAMPLE 1, SOLUTION, , EXAMPLE 2, SOLUTION, , EXAMPLE 3, , SOLUTION, , In a cinema hall, there are three entrance doors and two exit doors. In how, many ways can a person enter the hall and then come out?, Clearly, a person can enter the hall through any of the three, entrance doors. So, there are 3 ways of entering the hall., After entering the hall, the person can come out through any of the, two exit doors. So, there are 2 ways of coming out., Hence, by the fundamental principle of multiplication, the number, of ways in which a person can enter the hall and then come out, ( 3 # 2) 6., Three persons enter a railway carriage, where there are 5 vacant seats. In, how many ways can they seat themselves?, Clearly, the first person can occupy any of the 5 seats., So, there are 5 ways in which the first person can seat himself., Now, the second person can occupy any of the remaining 4 seats., So, the second person can be seated in 4 ways., Similarly, the third person can occupy a seat in 3 ways., Hence, by the fundamental principle of multiplication, the required, number of ways (5 # 4 # 3) 60., The flag of a newly formed forum is in the form = = = of three blocks, each, to be coloured differently. If there are six different colours on the whole to, choose from, how many such designs are possible?, The first block of the flag can be coloured by anyone of the 6 given, colours. So, there are 6 ways to colour the first block.
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270, , Senior Secondary School Mathematics for Class 11, , Now, the second block can be coloured by anyone of the remaining, five colours. So, there are 5 ways to colour the second block., The third block can now be coloured by anyone of the remaining, four colours. So, there are 4 ways to colour the third block., Hence, by the fundamental principle of multiplication, the required, number of designs (6 # 5 # 4) 120., EXAMPLE 4, , SOLUTION, , EXAMPLE 5, , SOLUTION, , There are 4 routes between Delhi and Patna. In how many different ways, can a man go from Delhi to Patna and return, if for returning, (i) any of the routes is taken;, (ii) the same route is taken;, (iii) the same route is not taken?, We have the following three cases., Case (i) When any of the routes is taken for returning:, The man may take any route for going from Delhi to, Patna., So, there are 4 ways of going from Delhi to Patna., When done so, he may return by any of the 4 routes., So, there are 4 ways of returning from Patna to Delhi., Hence, by the fundamental principle of multiplication,, the total number of ways for going to Patna and returning, back to Delhi (4 # 4) 16., Case (ii) When the same route is taken for returning:, In this case, there are 4 ways of going to Patna and only, 1 way of returning, namely by the same route., Hence, the required number of ways (4 #1) 4., Case (iii) When the same route is not taken for returning:, In this case, there are 4 ways of going to Patna., But, the man does not return by the same route., So, there are 3 ways of returning back to Delhi., Hence, the required number of ways (4 # 3) 12., There are six multiple-choice questions in an examination. Find the total, number of ways of answering these questions, if the first three questions, have 5 choices each and the next three have 4 choices each., Each of the first three questions has 5 choices., So, each one of them can be answered in 5 ways., Each of the next three questions has 4 choices., So, each one of them can be answered in 4 ways., Hence, by the fundamental principle of multiplication,, the total number of ways of answering the six questions, (5 # 5 # 5 # 4 # 4 # 4) 8000.
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Permutations, EXAMPLE 6, SOLUTION, , 271, , Four flags of different colours are given. How many different signals can, be generated, if a signal requires the use of two flags, one below the other?, The total number of signals is equal to the number of ways of filling, two places, , in succession by four flags of different colours., , The upper place can be filled in 4 different ways by any of the four, flags. Following it, the lower space can be filled in 3 different ways, by anyone of the remaining three flags., Hence, by the fundamental principle of multiplication, the required, number of signals (4 # 3) 12., EXAMPLE 7, , SOLUTION, , EXAMPLE 8, SOLUTION, , Find the number of 4-letter words, with or without meaning, which can be, formed out of the letters of the word, 'NOSE‘, when:, (i) the repetition of the letters is not allowed,, (ii) the repetition of the letters is allowed., Case (i) When the repetition of the letters is not allowed:, In this case, the total number of words is the same as the, number of ways of filling 4 places = = = = by 4 different, letters chosen from N, O, S, E., The first place can be filled by any of the 4 letters. Thus,, there are 4 ways of filling the first place. Since the, repetition of letters is not allowed, so the second place can, be filled by any of the remaining 3 letters in 3 different, ways., Following it, the third place can be filled by any of the two, remaining letters in 2 different ways. And, the fourth, place can be filled by the remaining one letter in 1 way., So, by the fundamental principle of multiplication, the, required number of 4-letter words (4 # 3 # 2 #1) 24., Hence, required number of words = 24., Case (ii) When the repetition of the letters is allowed:, In this case, each of the four different places can be filled, in succession in 4 different ways., So, the required number of 4-letter words, (4 # 4 # 4 # 4) 256., How many words (with or without meaning) of three distinct letters of the, English alphabet are there?, Clearly, we have to fill up three places by distinct letters of the, English alphabet., The first place can be filled by any of the 26 letters. Thus, there are, 26 ways to fill the first place., The second place can be filled by any of the remaining 25 letters., So, there are 25 ways to fill the second place., The third place can be filled by any of the remaining 24 letters.
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272, , Senior Secondary School Mathematics for Class 11, , Thus, there are 24 ways to fill the third place., Hence, by the fundamental principle of multiplication, the required, number of words (26 # 25 # 24) 15600., EXAMPLE 9, SOLUTION, , EXAMPLE 10, SOLUTION, , EXAMPLE 11, SOLUTION, , EXAMPLE 12, , How many numbers are there between 100 and 1000 in which all the, digits are distinct?, Clearly, every number between 100 and 1000 is a 3-digit number., So, we have to form all possible 3-digit numbers with distinct, digits. We cannot have 0 at the hundred‘s place., So, the hundred‘s place can be filled with any of the nine digits,, namely 1, 2, 3, 4, 5, 6, 7, 8, 9., So, there are 9 ways of filling the hundred‘s place. After filling, the hundred‘s place, nine digits are left including 0. So, there are, 9 ways of filling the ten‘s place., Now, the unit’s place can be filled by any of the remaining 8 digits., So, there are 8 ways of filling the unit‘s place., Hence, the required number of numbers (9 # 9 # 8) 648., How many 9-digit numbers of different digits can be formed?, The ten-crore‘s place of a 9-digit number cannot be 0., So, this place can be filled up by any of the digits from 1 to 9. So,, there are 9 ways of filling the ten-crore‘s place., The crore‘s place can now be filled with 0 or any of the remaining, 8 digits. Thus, there are 9 ways of filling it., The ten-lakh’s place can now be filled with any of the remaining, 8 digits. Thus, there are 8 ways of filling it., Similarly, we can fill lakh‘s place in 7 ways; ten thousand‘s place, in 6 ways; thousand‘s place in 5 ways; hundred‘s place in 4 ways;, ten‘s place in 3 ways and unit’s place in 2 ways., Hence, by the fundamental principle of multiplication, the required, number of numbers (9 # 9 # 8 #7 # 6 # 5 # 4 # 3 # 2) 3265920., How many numbers between 2000 and 3000 can be formed from the digits, 2, 3, 4, 5, 6, 7 when repetition of digits is not allowed?, Clearly, a number between 2000 and 3000 will have 2 at the, thousand‘s place. So, this place can be filled in 1 way only., Now, the hundred‘s place can be filled by any of the remaining five, digits. So, there are 5 ways of filling the hundred‘s place., Similarly, the ten‘s place can be filled in 4 ways and the unit’s place, can be filled in 3 ways., Hence, by the fundamental principle of multiplication, the total, number of required numbers (1# 5 # 4 # 3) 60., How many 3-digit odd numbers can be formed by using the digits 1, 2, 3,, 4, 5, 6 when, (i) the repetition of digits is not allowed?, (ii) the repetition of digits is allowed?
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Permutations, , 273, , SOLUTION, , For a number to be odd, we must have 1, 3 or 5 at the unit’s place., So, there are 3 ways of filling the unit’s place., Case (i) When the repetition of digits is not allowed:, In this case, after filling the unit’s place, we may fill the, ten‘s place by any of the remaining five digits. So, there, are 5 ways of filling the ten‘s place., Now, the hundred‘s place can be filled by any of the, remaining 4 digits. So, there are 4 ways of filling the, hundred‘s place., So, by the fundamental principle of multiplication, the, required number of odd numbers (3 # 5 # 4) 60., Case (ii) When the repetition of digits is allowed:, Since the repetition of digits is allowed, so after filling the, unit’s place, we may fill the ten‘s place by any of the given, six digits. So, there are 6 ways of filling the ten‘s place., Similarly, the hundred‘s place can be filled by any of the, given six digits. So, it can be filled in 6 ways., Hence, by the fundamental principle of multiplication,, the required number of odd numbers (3 # 6 # 6) 108., , EXAMPLE 13, , How many odd numbers less than 1000 can be formed by using the digits, 0, 2, 5, 7 when the repetition of digits is allowed?, , SOLUTION, , Since each number is less than 1000, required numbers are the, 1-digit, 2-digit and 3-digit numbers., One-digit numbers: Clearly, there are two one-digit odd numbers,, namely 5 and 7, formed of the given digits., Two-digit numbers: Since we are to form 2-digit odd numbers, we, may put 5 or 7 at the unit’s place. So, there are 2 ways of filling the, unit’s place., Now, we cannot use 0 at the ten’s place and the repetition of digits, is allowed. So, we may fill up the ten’s place by any of the digits, 2, 5, 7. Thus, there are 3 ways of filling the ten’s place., Hence, the required type of 2-digit numbers (2 # 3) 6., Three-digit numbers: To have an odd 3-digit number, we may put, 5 or 7 at the unit’s place. So, there are 2 ways of filling the unit’s, place., We may fill up the ten’s place by any of the digits 0, 2, 5, 7. So, there, are 4 ways of filling the ten’s place., We cannot put 0 at the hundred’s place. So, the hundred’s place, can be filled by any of the digits 2, 5, 7 and so it can be done in, 3 ways., , the required number of 3-digit numbers (2 # 4 # 3) 24.
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274, , Senior Secondary School Mathematics for Class 11, , Hence, the total number of required type of numbers, (2 6 24) 32., EXAMPLE 14, SOLUTION, , EXAMPLE 15, SOLUTION, , How many numbers are there between 100 and 1000 such that 7 is in the, unit‘s place?, Clearly, the numbers between 100 and 1000 are 3-digit numbers., So, we have to form 3-digit numbers with 7 at the unit‘s place., Clearly, 0 cannot be there at the hundred‘s place., So, the hundred‘s place can be filled with any of the digits from, 1 to 9., Thus, the hundred‘s place can be filled in 9 ways., Now, the ten‘s place can be filled with any of the digits from 0 to 9., So, there are 10 ways of filling the ten‘s place., The unit’s place can be filled with 7, i.e., in 1 way only., Hence, the number of required type of numbers (9 #10 #1) 90., How many numbers are there between 100 and 1000 such that at least one, of their digits is 7?, Clearly, the numbers between 100 and 1000 are 3-digit numbers., So, we have to form 3-digit numbers such that at least one of their, digits is 7., Case I, Three-digit numbers with 7 at unit’s place:, The number of ways to fill the hundred‘s place = 9, [by any digit from 1 to 9]., The number of ways to fill the ten‘s place = 10, [by any digit from 0 to 9]., The number of ways to fill the unit‘s place = 1 [by 7 only]., the number of such numbers (9 #10 #1) 90., Case II Three-digit numbers with 7 at ten‘s place:, The number of ways to fill the hundred‘s place = 9, [by any digit from 1 to 9]., The number of ways to fill the ten‘s place = 1 [by 7 only]., The number of ways to fill the unit place = 10, [by any digit from 0 to 9]., the number of such numbers (9 #1#10) 90., Case III Three-digit numbers with 7 at hundred‘s place:, The number of ways to fill the hundred‘s place = 1, [by 7 only]., The number of ways to fill the ten‘s place = 10, [by any digit from 0 to 9]., The number of ways to fill the unit’s place = 10, [by any digit from 0 to 9].
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Permutations, , 275, , , the number of such numbers (1#10 #10) 100., Hence, the total number of required numbers (90 90 100) 280., EXAMPLE 16, SOLUTION, , EXAMPLE 17, SOLUTION, , EXAMPLE 18, , How many numbers are there between 100 and 1000, which have exactly, one of their digits as 7?, Clearly, the numbers between 100 and 1000 are 3-digit numbers., So, we have to form 3-digit numbers having exactly one of their, digits as 7., Case I, 3-digit numbers having 7 at unit’s place only:, Number of ways to fill the unit’s place =1 [with 7 only]., Number of ways to fill the ten‘s place = 9, [any digit from 0 to 9 except 7]., Number of ways to fill the hundred‘s place = 8, [any digit from 1 to 9 except 7]., So, the number of such numbers (1# 9 # 8) 72., Case II 3-digit numbers having 7 at ten‘s place only:, Number of ways to fill the ten‘s place = 1 [with 7 only]., Number of ways to fill the unit’s place = 9, [any digit from 0 to 9 except 7]., Number of ways to fill the hundred‘s place = 8, [any digit from 1 to 9 except 7]., So, the number of such numbers (1# 9 # 8) 72., Case III 3-digit numbers having 7 at hundred‘s place only:, Number of ways to fill the hundred‘s place = 1, [with 7 only]., Number of ways to fill the ten‘s place = 9, [any digit from 0 to 9 except 7]., Number of ways to fill the unit’s place = 9, [any digit from 0 to 9 except 7]., So, the number of such numbers (1# 9 # 9) 81., Hence, the total number of required numbers (72 72 81) 225., How many numbers are there between 100 and 1000 such that every digit, is either 2 or 9?, Clearly, the numbers between 100 and 1000 are 3-digit numbers., So, we have to form 3-digit numbers by using 2 and 9., Clearly, the repetition of digits is allowed., Each one of the unit‘s, ten‘s and hundred‘s places can be filled in, 2 ways., Hence, the required number of numbers (2 # 2 # 2) 8., How many 3-digit numbers can be formed without using the digits 0, 2,, 3, 4, 5 and 6?
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276, , Senior Secondary School Mathematics for Class 11, , SOLUTION, , Clearly, we have to form 3-digit numbers by using the digits 1, 7, 8, and 9 while the repetition of digits is allowed., Clearly, each one of the unit‘s, ten‘s and hundred‘s places can be, filled by any of the digits 1, 7, 8, 9, i.e., in four ways., Hence, the number of required numbers (4 # 4 # 4) 64., , EXERCISE 8B, 1. There are 10 buses running between Delhi and Agra. In how many ways, can a man go from Delhi to Agra and return by a different bus?, 2. A, B and C are three cities. There are 5 routes from A to B and 3 routes from, B to C. How many different routes are there from A to C via B?, 3. There are 12 steamers plying between A and B. In how many ways could, the round trip from A be made if the return was made on (i) the same, steamer? (ii) a different steamer?, 4. In how many ways can 4 people be seated in a row containing 6 seats?, 5. In how many ways can 5 ladies draw water from 5 taps, assuming that no, tap remains unused?, 6. In a textbook on mathematics there are three exercises A, B and C consisting, of 12, 18 and 10 questions respectively. In how many ways can three, questions be selected choosing one from each exercise?, 7. In a school, there are four sections of 40 students each in XI standard. In, how many ways can a set of 4 student representatives be chosen, one from, each section?, 8. In how many ways can a vowel, a consonant and a digit be chosen out of, the 26 letters of the English alphabet and the 10 digits?, 9. How many 8-digit telephone numbers can be constructed using the digits 0, to 9 if each number starts with 270 and no digit appears more than once?, 10., , (a) A coin is tossed three times and the outcomes are recorded. How, many possible outcomes are there?, (b) How many possible outcomes if the coin is tossed:, (i) four times?, (ii) five times?, (iii) n times?, , 11. Find the number of different signals that can be generated by arranging at, least 2 flags in order (one below the other) on a vertical staff, if five different, flags are available., 12. How many 4-letter codes can be formed using the first 10 letters of the, English alphabet, if no letter can be repeated?, 13. Given, A {2, 3, 5} and B {0, 1} . Find the number of different ordered pairs, in which the first entry is an element of A and the second is an element of B., 14. How many arithmetic progressions with 10 terms are there whose first, term is in the set {1, 2, 3} and whose common difference is in the set {2, 3, 4}?
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Permutations, , 277, , 15. There are 6 items in column A and 6 items in column B. A student is asked, to match each item in column A with an item in column B. How many, possible (correct or incorrect) answers are there to this question?, 16. A mint prepares metallic calendars specifying months, dates and days in, the form of monthly sheets (one plate for each month). How many types of, February calendars should it prepare to serve for all the possibilities in the, future years?, 17. From among the 36 teachers in a school, one principal and one viceprincipal are to be appointed. In how many ways can this be done?, 18. A sample of 3 bulbs is tested. A bulb is labelled as G if it is good and D if it, is defective. Find the number of all possible outcomes., 19. For a set of five true or false questions, no student has written the all, correct answer and no two students have given the same sequence of, answers. What is the maximum number of students in the class for this, to be possible?, 20. In how many ways can the following prizes be given away to a class of, 20 students:, first and second in mathematics; first and second in chemistry; first in, physics and first in English?, 21. Find the total number of ways of answering 5 objective-type questions,, each question having 4 choices., 22. A gentleman has 6 friends to invite. In how many ways can he send, invitation cards to them, if he has 3 servants to carry the cards?, 23. In how many ways 6 rings of different types can be worn in 4 fingers?, 24. In how many ways can 5 letters be posted in 4 letter boxes?, 25. How many 3-letter words can be formed using a, b, c, d, e if, (i) repetition of letters is not allowed?, (ii) repetition of letters is allowed?, 26. How many 4-digit numbers are there, when a digit may be repeated any, number of times?, 27. How many numbers can be formed from the digits 1, 3, 5, 9 if repetition of, digits is not allowed?, 28. How many 3-digit numbers are there with no digit repeated?, 29. How many 3-digit numbers can be formed by using the digits 0, 1, 3, 5, 7, while each digit may be repeated any number of times?, 30. How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7, 9, when no digit is repeated? How many of them are divisible by 10?, 31. How many natural numbers less than 1000 can be formed from the digits, 0, 1, 2, 3, 4, 5 when a digit may be repeated any number of times?, 32. How many 6-digit telephone numbers can be constructed using the digits, 0 to 9, if each number starts with 67 and no digit appears more than once?
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278, , Senior Secondary School Mathematics for Class 11, , 33. In how many ways can three jobs I, II and III be assigned to three persons, A, B and C if one person is assigned only one job and all are capable of, doing each job?, 34. A number lock on a suitcase has three wheels each labelled with ten digits, 0 to 9. If opening of the lock is a particular sequence of three digits with no, repeats, how many such sequences will be possible? Also, find the number, of unsuccessful attempts to open the lock., 35. A customer forgets a four-digit code for an automated teller machine (ATM), in a bank. However, he remembers that this code consists of digits 3, 5, 6,, 9. Find the largest possible number of trials necessary to obtain the correct, code., 36. In how many ways can 3 prizes be distributed among 4 girls, when, (i) no girl gets more than one prize?, (ii) a girl may get any number of prizes?, (iii) no girl gets all the prizes?, ANSWERS (EXERCISE 8B), , 1., 7., 11., 17., 22., 27., 32., , 90, 2560000, 320, 1260, 36, 64, 1680, , 2., 8., 12., 18., 23., 28., 33., , 15, 1050, 5040, 8, 46, 648, 6, , 3., 9., 13., 19., 24., 29., 34., , (i) 12 (ii) 132, 2520, 10., 6, 14., 31, 20., 25., 45, 100, 30., 720, 719, 35., , 4. 360, 5. 120, 6. 2160, (a) 8 (b) (i) 16 (ii) 32 (iii) 2 n, 9, 15. 720 16. 14, 57760000, 21. 4 5, (i) 60 (ii) 125, 26. 9000, 600, 120, 31. 215, 24, 36. (i) 24 (ii) 64 (iii) 60, , HINTS TO SOME SELECTED QUESTIONS, 3. (i) In this case the round trip from A is to be made by the same steamer. So, there are, 12 ways of making the round trips., (ii) In this case, number of ways to go from A to B = 12., Number of ways to go from B to A = 11., Hence, the required number of ways (12 #11) 132., 4. 1st person can be seated in 6 ways, 2nd person in 5 ways, 3rd person in 4 ways and, 4th person in 3 ways., Required number of ways (6 # 5 # 4 # 3) 360., 5. 1st lady can draw water from any of the 5 taps., So, first lady can draw water in 5 ways, 2nd in 4 ways, 3rd in 3 ways, 4th in 2 ways, and 5th in 1 way., Required number of ways (5 # 4 # 3 # 2 #1) 120., 6. 1 question out of 12 can be chosen in 12 ways., 1 question out of 18 can be chosen in 18 ways., 1 question out of 10 can be chose in 10 ways., Required number of ways (12 #18 #10) 2160.
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Permutations, , 279, , 7. From each of the 4 sections 1 representative out of 40 can be chosen in 40 ways., So, by the fundamental principle of multiplication, the required number of ways, (40 # 40 # 40 # 40) 2560000., 8. There are 5 vowels, 21 consonants and 10 digits., Number of ways of choosing 1 vowel out of 5 vowels = 5., Number of ways of choosing 1 consonant out of 21 consonants = 21., Number of ways of choosing 1 digit out of 10 digits = 10., Required number of ways (5 # 21#10) 1050., 9. The remaining digits to be used are 1, 3, 4, 5, 6, 8, 9., And, we have to select 5 digits out of these eight., First digit can be anyone of these 7 digits., So, there are 7 ways of choosing the 1st digit., Similarly, we may choose 2nd digit in 6 ways, 3rd digit in 5 ways, 4th digit in 4 ways, and 5th digit in 3 ways., By the fundamental principle of multiplication, the required number of ways, (7 # 6 # 5 # 4 # 3) 2520., 10. (a) In a single throw of a coin there are 2 possible outcomes namely Head (H) and, Tail (T)., So, by the fundamental principle of multiplication, the total number of possible, outcomes in 3 tosses (2 # 2 # 2) 2 3 8., (b) Similarly, we have, (i) total number of possible outcomes in 4 tosses (2 # 2 # 2 # 2) 2 4 16., (ii) total number of possible outcomes in 5 tosses (2 # 2 # 2 # 2 # 2) 2 5 32., (iii) total number of possible outcomes in n tosses 2 n., 11. A signal can consist of either 2 flags, 3 flags, 4 flags or 5 flags., 2-flag signals:, 1st place can be filled by any of the 5 flags in 5 ways., 2nd place can be filled by any of the remaining 4 flags in 4 ways., Total number of 2-flag signals (5 # 4) 20., 3-flag signals:, 1st place can be filled in 5 ways, 2nd in 4 ways and 3rd in 3 ways., Total number of 3-flag signals (5 # 4 # 3) 60., 4-flag signals:, 1st place can be filled in 5 ways, 2nd in 4 ways, 3rd in 3 ways and 4th in 2 ways., Total number of 4-flag signals (5 # 4 # 3 # 2) 120., 5-flag signals:, 1st place can be filled in 5 ways, 2nd in 4 ways, 3rd in 3 ways, 4th in 2 ways and 5th, in 1 way only., Total number of 5-flag signals (5 # 4 # 3 # 2 #1) 120., Hence, total number of signals (20 60 120 120) 320.
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280, , Senior Secondary School Mathematics for Class 11, , 12. We have to fill up 4 places = = = = by first 10 letters of the English alphabet. The first, place can be filled by any of the 10 letters in 10 ways., Similarly, 2nd place can be filled in 9 ways; 3rd place can be filled in 8 ways and 4th, place can be filled in 7 ways., Required number of 4-letter codes (10 # 9# 8 #7) 5040., 13. If n (A) p and n (B) q, then n (A# B) pq., Here n (A) 3 and n (B) 2, then n (A# B) (3 # 2) 6., 14. Required number of AP‘s (3 # 3) 9., 15. First item in column A can be matched with any of the 6 items in column B., Thus, the first item of column A can be matched in 6 ways., Similarly, 2nd item of column A can be matched in 5 ways., Similarly, 3rd, 4th, 5th and 6th items of column A can be matched in 4, 3, 2 and 1 ways, respectively., By the fundamental principle of multiplication, the required number of ways, (6 # 5 # 4 # 3 # 2 #1) 720., 16. There are two types of Februaries, one having 28 days and the other having 29 days., So, there are 2 ways of preparing calendars for February., Also, the month may start from any day from Monday to Sunday. So, there are 7 ways, of preparing it for various days., Hence, the required number of plates (2 #7) 14., 17. Number of ways of choosing 1 principal out of 36 teachers = 36., Number of ways of choosing 1 vice-principal out of 35 teachers = 35., Hence, the required number of ways (36 # 35) 1260., 18. All possible outcomes are, GGG, GGD, GDG, DGG, GDD, DGD, DDG, DDD., Clearly, there are 8 possible outcomes., 19. Clearly there are 2 ways of answering each of the 5 questions, i.e., true or false., So, the total number of different sequences of answers (2 # 2 # 2 # 2 # 2) 32., There is only one all correct answer., Maximum number of sequences of answers leaving aside the all correct answer, (32 1) 31., Hence, the maximum possible number of students = 31., 20. 1st and 2nd prizes in mathematics can be given in (20 #19) ways;, 1st and 2nd prizes in chemistry can be given in (20 #19) ways;, 1st prize in physics can be given in 20 ways;, 1st prize in English can be given in 20 ways., Required number of ways in all (20 #19)#(20 #19)# 20 # 20 57760000., 21. Since each question can be answered in 4 ways, total number of ways of answering, 5 questions (4 # 4 # 4 # 4 # 4) 4 5., 22. Each friend can be invited by any of the 3 servants, i.e., in 3 ways. So, the required, number of ways (3 # 3 # 3 # 3 # 3 # 3) 3 6., 23. The first ring can be worn in any of the 4 fingers. So, there are 4 ways of wearing it., Similarly, each of the other rings can be worn in 4 ways., So, the required number of ways (4 # 4 # 4 # 4 # 4 # 4) 4 6.
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Permutations, , 281, , 24. Since each letter can be posted in anyone of the 4 letter boxes, so each letter can be, posted in 4 ways., So, the required number of ways (4 # 4 # 4 # 4 # 4) 4 5., 25. (i) Total number of 3-letter words is equal to the number of ways of filling 3 places., First place can be filled in 5 ways by any of the given five letters. Second place, can be filled in 4 ways by any of the remaining 4 letters and the third place can be, filled in 3 ways by any of the remaining 3 letters., So, the required number of 3-letter words (5 # 4 # 3) 60., (ii) When repetition of letters is allowed, each place can be filled by any of the 5 letters, in 5 ways., , the required number of ways (5 # 5 # 5) 125., 26. In a 4-digit number, 0 cannot be placed at the thousand‘s place. So, thousand‘s place, can be filled by any digit from 1 to 9 in nine ways. Now, each of hundred‘s, ten‘s and, one‘s places can be filled by any of the digits from 0 to 9 in ten ways., So, the required number of numbers (9 #10 #10 #10) 9000., 27. One-digit numbers:, Clearly, there are four 1-digit numbers., , Two-digit numbers:, , We may fill the unit’s place by any of the four given digits., Thus, there are 4 ways to fill the unit’s place., The ten‘s place may now be filled by any of the remaining three digits. So, there are, 3 ways to fill the ten‘s place., Number of 2-digit numbers (4 # 3) 12., Three-digit numbers:, Number of ways to fill the unit‘s, ten‘s and hundred‘s places are 4, 3 and 2 respectively., Number of 3-digit numbers (4 # 3 # 2) 24., Four-digit numbers:, Number of ways to fill the unit‘s, ten‘s, hundred‘s and thousand‘s places are 4, 3, 2, and 1 respectively., Number of 4-digit numbers (4 # 3 # 2 #1) 24., Hence, the number of required numbers (4 12 24 24) 64., 28. The hundred‘s place cannot contain 0. So, it can be filled by any digit from 1 to 9., Thus, there are 9 ways to fill this place. Now, the ten‘s digit can be filled by any of the, remaining 9 digits out of 0 to 9. So, there are 9 ways to fill the ten‘s place. And, there, are 8 ways to fill the unit’s digit., Required number of numbers (9 # 9 # 8) 648., 29. The hundred‘s place cannot contain 0. So, it can be filled by any of the digits 1, 3, 5, 7., So, there are 4 ways to fill it., Now, each of the ten‘s and unit‘s digits can be filled by any of the given five digits., So, each can be filled in 5 ways., the required number of numbers (4 # 5 # 5) 100., 30. The lakh‘s place cannot be 0. So, it can be filled by any of the digits 1, 3, 5, 7, 9. So,, there are 5 ways to fill it. The ten-thousand‘s place can be filled by 0 or the remaining, four digits, i,e., in 5 ways. The thousand‘s, hundred‘s, ten‘s and unit’s digits can be, filled in 4, 3, 2 and 1 ways respectively., Required number of numbers (5 # 5 # 4 # 3 # 2 #1) 600.
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282, , Senior Secondary School Mathematics for Class 11, Particular case Let us put 0 at unit’s place, i.e., it is filled in 1 way., Now ten‘s, hundred‘s, thousand‘s, ten-thousand‘s and lakh‘s place can be filled in, 5, 4, 3, 2 and 1 ways respectively., So, the number of numbers having 0 as unit’s digit (5 # 4 # 3 # 2 #1) 120., So, 120 numbers are divisible by 10., , 31. There are 5 one-digit natural numbers out of the given digits., For getting 2-digit natural numbers we cannot put 0 at the ten‘s place. So, this place, can be filled by any of the given five nonzero digits in 5 ways., The unit’s digit can be filled by any of the given six digits in 6 ways., So, the number of 2-digit natural numbers (5 # 6) 30., Similarly, to get a 3-digit number, we cannot put 0 at the hundred’s place. So, this, place can be filled in 5 ways, each of the ten‘s and unit‘s places can be filled in 6 ways., the number of 3-digit numbers (5 # 6 # 6) 180., Required number of numbers (5 30 180) 215., 32. Out of 10 digits from 0 to 9, we have used 2 digits, namely 6 and 7., So, the 3rd, 4th, 5th and 6th digit may be filled in (10 2) 8, 7, 6 and 5 ways, respectively., Hence, the required number of numbers (8 #7 # 6 # 5) 1680., 33. A can be assigned any of the three jobs. So, there are 3 ways of assigning a job to A., B can now be assigned any of the two remaining jobs in 2 ways and C can now be, assigned the job in 1 way., So, the required number of ways (3 # 2 #1) 6., 34. The first term of the sequence may be anyone of the digits from 0 to 9. So, there are, 10 ways of getting the first term., So, the 2nd and 3rd terms may be obtained in 9 ways and 8 ways respectively., Required number of sequences (10 # 9 # 8) 720., Number of unsuccessful attempts (720 1) 719., 35. 1st, 2nd, 3rd, 4th digits may be chosen in 4 ways, 3 ways, 2 ways and 1 way, respectively., Number of largest possible trials (4 # 3 # 2 #1) 24., 36. (i) The 1st prize can be given to anyone of the 4 girls in 4 ways., Similarly, the second prize can be given in 3 ways and the 3rd prize can be given, in 2 ways., , the required number of ways (4 # 3 # 2) 24., (ii) 1st prize can be given in 4 ways, 2nd prize can be given in 4 ways and the 3rd, prize can be given in 4 ways., , the required number of ways (4 # 4 # 4) 64., (iii) Number of ways in which a girl does not get all the prizes, = (number of ways in which a girl may get any number of prizes), – (number of ways in which a girl gets all the prizes), (64 4) 60.
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Permutations, , 283, , PERMUTATIONS, The different arrangements which can be made out of a given number, of things by taking some or all at a time, are called permutations., REMARK In permutations, the order of arrangement of the objects is taken into, consideration. When the order is changed, a different permutation is, obtained., PERMUTATIONS, , ILLUSTRATIVE EXAMPLES, EXAMPLE 1, , All permutations of the letters A, B, C taking two at a time are:, AB, BA, AC, CA, BC, CB., Thus, we obtain 6 different permutations., , EXAMPLE 2, , All permutations of the letters A, B, C taking all at a time are:, ABC, ACB, BAC, BCA, CAB, CBA., Thus, the 3 letters A, B, C taking all at a time give 6 permutations., , EXAMPLE 3, , All permutations of the letters A, B, C, D taking 3 at a time are:, ABC, ACB, BAC, BCA, CAB, CBA;, ABD, ADB, BAD, BDA, CAD, CDA;, BCD, BDC, CBD, CDB, DBC, DCB;, ACD, ADC, CAD, CDA, DAC, DCA., Thus, the 4 letters A, B, C, D taking 3 at a time give 24 permutations., , PERMUTATIONS WHEN ALL THE OBJECTS ARE DISTINCT, , The number of permutations of n different objects taken r at a time, where, 0 r # n and the objects do not repeat, is given by, P (n, r) or nPr n (n 1) (n 2) … (n r 1) ., The number of permutations of n different objects taken r at a time is, the same as the number of ways of filling in r vacant places by the n, objects., The first place can be filled up by anyone of these n objects., Thus, there are n ways of filling up the first place., Now, we are left with (n 1) objects., The second place can be filled up by anyone of these (n 1) objects., Thus, there are (n 1) ways of filling up the second place., Now, we are left with (n 2) objects., The third place can be filled up by any of these (n 2) objects., Thus, there are (n 2) ways of filling up the third place., Similarly, the 4th place can be filled in (n 3) ways, the 5th place can, be filled in (n 4) ways and so on, and the rth place can be filled in, {n (r 1)} (n r 1) ways., , THEOREM 1, , PROOF
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284, , Senior Secondary School Mathematics for Class 11, , Thus, the total number of ways of filling in r vacant places by n, different objects in succession is {n (n 1) (n 2) … (n r 1)} ., Hence, nPr n (n 1) (n 2) … (n r 1) ., We have, n, Pr n (n 1) (n 2) … (n r 1), n (n 1) (n 2) … (n r 1) (n r) (n r 1) … 3 · 2 · 1, , (n r) (n r 1) … 3 · 2 · 1, n, !, , where 0 r # n., , (n r) !, , REMARK, , The number of all permutations of n different objects taken all at a time is, given by nPn n !., The number of all permutations of n different objects taken all at a, time is the same as the number of ways of filling n vacant places by n, different objects., Proceeding as in Theorem 1, we have, n, Pn n (n 1) (n 2) (n 3) …# 3 # 2 #1 n !., , THEOREM 2, PROOF, , Prove that 0 ! 1., We have, n!, n, Pr , (n r) !, n!, & nPn , 0!, n, !, & n! , 0!, n! , & 0! , 1., n!, Hence, 0 ! 1., , THEOREM 3, PROOF, , REMARK, , Thus, we have nPr , , … (i), [putting r n in (i)], [a nPn n !], , n! ,, where 0# r # n., (n r) !, , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Evaluate:, (i) 12P4, , (ii) 75P2, , (iii) 8P8, , We have, 12 ! 12 ! 12 #11#10 # 9 #(8 !) , 11880., 8!, (12 4) ! 8 !, 75 ! 75 ! 75 #74 #(73 !) , (ii) 75P2 , (75 #74) 5550., 73 !, (75 2) ! 73 !, (i), , P4 , , 12, , (iii) 8P8 8 ! (8 #7 # 6 # 5 # 4 # 3 # 2 #1) 40320.
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Permutations, EXAMPLE 2, , SOLUTION, , Find the value of n such that, (i) nP5 42 # nP3, n 4, , 285, n, , (ii), , P4, , 5 , n 4., P4 3, , n1, , We know that nPr n (n 1) (n 2) … (n r 1) ., (i) We have, n, P5, n (n 1) (n 2) (n 3) (n 4), 42 &, 42, n, n (n 1) (n 2), P3, [as n (n 1) (n 2) ! 0], & (n 3) (n 4) 42, & n 2 7n 30 0 & (n 10) (n 3) 0, & n 10, [a n ! 3, as n cannot be negative]., Hence, n 10., n (n 1) (n 2) (n 3), 5 &, 5, , , , , 3, 3, (, n, ), (, n, ), (, n, ), (, n, ), 1, 2, 3, 4, P4, & 3n 5 (n 4) [as (n 1) (n 2) (n 3) (n 4) ! 0], & 2n 20 & n 10., Hence, n 10., n, , (ii), , EXAMPLE 3, SOLUTION, , P4, , n1, , If 5 # 4Pr 6 # 5Pr 1, find r., We have, 5 # 4Pr 6 # 5Pr 1, 5!, 4! , &, 5#, 6#, (4 r) !, {5 (r 1)} !, 5 ! 6 #(5 !), &, [a 5 #(4 !) 5 !], (4 r) ! (6 r) !, 6, 6, 1, 1, , , &, &, (4 r) ! (6 r) !, (4 r) ! (6 r) (5 r) {(4 r) !}, 6, 1 & (6 r) (5 r) 6 0, &, (6 r) (5 r), r 2 11r 24 0 & (r 3) (r 8) 0, &, [a r ! 8, as 4P8 is not defined]., r3, Hence, r 3., &, , EXAMPLE 4, SOLUTION, , If nP4 2 # 5P3, find n., We have, n, P4 2 # 5P3, &, n (n 1) (n 2) (n 3) 2 # 5 # 4 # 3, &, n (n 3) (n 1) (n 2) 120, &, &, &, &, , (n 2 3n) (n 2 3n 2) 120, 2, m (m 2) 120 0, where m n 3n, m 2 2m 120 0 & (m 12) (m 10) 0, m 12 or m 10.
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286, , Senior Secondary School Mathematics for Class 11, , This gives, n 2 3n 12, or n 2 3n 10, 2, &, n 3n 12 0 or n 2 3n 10 0, 3 ! 9 48, &, n, or (n 5) (n 2) 0, 2, 3 ! i 39, or n 5 or n 2, 2, [neglecting the negative and imaginary values of n]., &, n5, Hence, n 5., n, , &, , EXAMPLE 5, , SOLUTION, , (i) If 15Pr 2730, find the value of r., (ii) If 10Pr 5040, find the value of r., We have, (i) 15Pr 2730, Pr 15 #182, &, Pr 15 #14 #13, & r 3., Hence, r 3., 15, , &, , 15, , (ii), , (up to 3 factors), , Pr 5040, , 10, , Pr 10 # 504, , &, , 10, , &, , 10, , Pr 10 # 9 # 56, , &, Pr 10 # 9# 8 #7 (up to 4 factors), & r 4., Hence, r 4., 10, , EXAMPLE 6, SOLUTION, , If 56Pr 6 : 54Pr 3 (30800 : 1), find r., We have, 56 !, 56, 56 ! ·, Pr 6 , [56 (r 6)] ! (50 r) !, 54 !, 54 ! ·, And, 54Pr 3 , [54 (r 3)] ! (51 r) !, 56, , , , Pr 6, , 54, , Pr 3, , , , (51 r) !, 56 !, #, , 54 !, (50 r) !, , 56 # 55 #(54 !) (51 r) · [(50 r) !], #, 54 !, (50 r) !, 56 # 55 #(51 r) ., , , 56, , Thus,, , Pr 6, , 54, , Pr 3, , 30800 & 56 # 55 #(51 r) 30800, 1, , Hence, r 41., , & (51 r) 10 & r 41.
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Permutations, EXAMPLE 7, SOLUTION, , 287, , , , If 2n 1Pn 1 : 2n 1Pn 3 : 5, find n., We have, , 2n 1, Pn 1 : 2n 1Pn 3 : 5, (2n 1) !, (2n 1) !, 3, &, :, {(2n 1) (n 1)} ! {(2n 1) n} ! 5, (2n 1) ! (2n 1) ! 3, , &, :, 5, (n 2) ! (n 1) !, , , (2n 1) !, (n 1) !, 3, #, &, (n 2) !, (2n 1) ! 5, (2n 1)# 2n #[(2n 1) !], (n 1) !, 3, #, &, (n 2)#(n 1)# n #[(n 1) !] (2n 1) ! 5, &, 10 (2n 1) 3 (n 2) (n 1), , &, , 20n 10 3 (n 2 3n 2), , 3n 2 11n 4 0, 1, (n 4) (3n 1) 0 & n 4 [a n ! , as n cannot be negative]., 3, Hence, n 4., , &, &, , EXAMPLE 8, SOLUTION, , If 22Pr 1 : 20Pr 2 11 : 52, find r., We have, 22, 20, Pr 1 : Pr 2 11 : 52, &, &, &, , 20 !, 22 !, 11 : 52, :, {22 (r 1)} ! {20 (r 2)} !, 20 ! 11, 22 !, :, (21 r) ! (18 r) ! 52, (18 r) ! 11, 22 !, , #, 52, 20 !, (21 r) !, , 22 # 21#(20 !), (18 r) ! 11, , #, 52, 20 !, (21 r) (20 r) (19 r)#[(18 r) !], 22 # 21, 11, , &, (21 r) (20 r) (19 r) 52, &, (19 r) (20 r) (21 r) 2 # 21# 52 2 # 3 #7 # 4 #13, &, (19 r) (20 r) (21 r) 12 #13 #14, &, 19 r 12 & r (19 12) 7., Hence, r 7., &, , EXAMPLE 9, , SOLUTION, , Prove that:, (i) nPn nPn 1, , , , (ii) nPr n · n 1Pr 1, , By using the formula for Pr , we have, (i) nPn 1 , , , , , , (iii) n 1Pr r · n 1Pr 1 nPr, , n, , n!, n ! n ! nPn ., {n (n 1)] ! 1 !
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288, , Senior Secondary School Mathematics for Class 11, , (ii) nPr , (iii), , n · ( n 1) !, n! , n · n 1Pr 1 ., , , (n r) ! [(n 1) (r 1)] !, , n1, , , , Pr r · n 1Pr 1, (n 1) !, (n 1) !, 3, ), r·, (n 1 r) !, [(n 1) (r 1)] !, (n 1) !, (n 1) !, 3, ), r·, (n 1 r) !, (n r) !, (n 1) !, (n 1) !, 3, ), r·, (n r 1) !, (n r) · [(n r 1) !], (n 1) !, n · [(n 1) !], r, , '1 , 1, (n r 1) !, (n r), (n r) · [(n r 1) !], n ! nPr ., (n r) !, , , , , Hence, n 1Pr r · n 1Pr 1 nPr ., EXAMPLE 10, SOLUTION, , EXAMPLE 11, SOLUTION, , If r s # n then prove that nPs is divisible by nPr ., We have, n, Ps n (n 1) (n 2) … (n r 1) (n r) … (n s 1), {n (n 1) (n 2) … (n r 1)}#{(n r) (n r 1) … (n s 1)}, nPr #{(n r) (n r 1) … (n s 1)},, which is clearly divisible by nPr ., n, n, Hence, Ps is divisible by Pr ., Find the number of permutations of 7 objects, taken 3 at a time., Number of permutations of 7 objects, taken 3 at a time, 7P3 7 # 6 # 5 [up to three terms], 210., , EXERCISE 8C, 1. Evaluate:, (i) 10P4, 2. Prove that, , (ii), , 62, , P3, , P3 3 # 9P2 10P3 ., , 9, , 3., , (i) If nP5 20 # nP3, find n., , (ii) If 16 # nP3 13 # n 1P3, find n., 2n, n, (iii) If P3 100 # P2, find n., , 4., , (i) If 5Pr 2 # 6Pr 1, find r., (ii) If 20Pr 13 # 20Pr 1, find r., (iii) If 11Pr 12Pr 1, find r., , 5., , (i) If nP4 : nP5 1 : 2, find n., , , (ii) If n 1P3 : n 1P3 5 : 12, find n., , (iii) 6P6, , (iv) 9P0
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Permutations, , 289, , 6. If 15Pr 1 : 16Pr 2 3 : 4, find r., , , , , 7. If 2n 1Pn : 2n 1Pn 1 22 : 7, find n., 11 , , , 8. Find n, if n 5Pn 1 , (n 1) · n 3Pn ., 2, , 9. Prove that 1 1 · 1P1 2 · 2P2 3 · 3P3 … n · nPn n 1Pn 1 ., 10. Find the number of permutations of 10 objects, taken 4 at a time., ANSWERS (EXERCISE 8C), , 1. (i) 5040 (ii) 226920 (iii) 720 (iv) 1 3. (i) n 8 (ii) n 15 (iii) n 13, 4. (i) r 3 (ii) r 8 (iii) r 9, 5. (i) n 6 (ii) n 8, 6. r 14, 7. n 10, 8. n 6 or n 7, 10. 5040, HINTS TO SOME SELECTED QUESTIONS, 5. (ii), , 6., , (n 1) (n 2) (n 3) 5, , & 12 (n 2) (n 3) 5 (n 1) n, 12, (n 1) n (n 1), & 7n 2 65n 72 0 & 7n 2 56n 9n 72 0, & 7n (n 8) 9 (n 8) 0 & (n 8) (7n 9) 0 & n 8., , 15 !, 16 !, 3 : 4 & 15 ! : 16 ! 3 : 4., :, {15 (r 1)} ! {(16 (r 2)} !, (16 r) ! (18 r) !, (18 r) ! 3, (18 r) (17 r) · [(16 r) !] 3, 15 !, 15, !, &, ·, , #, #, 4, 16 !, 4, (16 r) !, (16 r) !, 16 #(15 !), , (18 r) (17 r) 12 & r 2 35r 294 0 & (r 14) (r 21) 0 & r 14, [a r ! 21]., (2n 1) !, (2n 1) !, 22, , 7., :, 7, (2n 1 n) ! {(2n 1) (n 1)} !, , , &, , (2n 1) !, (n 2) !, (2n 1) ! (n 2) (n 1) n · [(n 1) !] 22, 22 &, , #, #, 7, 7, (n 1) !, (2n 1) !, (n 1) !, (2n 1) (2n) · [(2n 1) !], , 2, & 7 (n 2) (n 1) 44 (2n 1) & 7n 67n 30 0 & (n 10) (7n 3) 0., Hence, n 10., 9. 1 1 · (1 !) 2 · (2 !) 3 · (3 !) … n · (n !), , 1 (2 1) (1 !) (3 1) · (2 !) (4 1) · (3 !) … {(n 1) 1} · (n !), {1 2 · (1 !) 3 · (2 !) 4 · (3 !) … (n 1) · (n !)} {1 ! 2 ! 3 ! … n !}, {1 2 ! 3 ! 4 ! … n ! (n 1) !} {1 ! 2 ! 3 ! … n !} (n 1) !., , WORD PROBLEMS ON PERMUTATIONS, EXAMPLE 1, SOLUTION, , In how many ways can 6 persons stand in a queue?, Required number of ways, = number of arrangements of 6 different things taking all at, a time, 6P6 6 ! (6 # 5 # 4 # 3 # 2 #1) 720.
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290, EXAMPLE 2, SOLUTION, , EXAMPLE 3, SOLUTION, , EXAMPLE 4, SOLUTION, , EXAMPLE 5, SOLUTION, , Senior Secondary School Mathematics for Class 11, , How many different signals can be made by 4 flags from 6 flags of different, colours?, Required number of signals, = number of arrangements of 6 flags taking 4 at a time, 6P4 (6 # 5 # 4 # 3) 360., Eight athletes are participating in a race. In how many ways can the first, 3 prizes be won by them?, Total number of ways of winning first 3 prizes by 8 athletes, = number of arrangements of 8 athletes taking 3 at a time, 8P3 (8 #7 # 6) 336., In how many ways can 3 different rings be worn in 4 fingers with at most, one in each finger?, Required number of ways, = number of ways of arranging 4 fingers taking 3 at a time, 4P3 (4 # 3 # 2) 24., 3 men have 4 coats, 6 waistcoats and 5 caps. In how many ways can they, wear them?, Total number of ways in which 3 men can wear 4 coats, = number of arrangements of 4 coats taking 3 at a time, 4P3 (4 # 3 # 2) 24., Total number of ways in which 3 men can wear 6 waistcoats, = number of arrangements of 6 waistcoats taking 3 at a time, 6P3 (6 # 5 # 4) 120., Total number of ways in which 3 men can wear 5 caps, 5P3 (5 # 4 # 3) 60., Hence, by the fundamental principle of multiplication, the required, number of ways (24 #120 # 60) 172800., , EXAMPLE 6, SOLUTION, , How many different signals can be given by using any number of flags, from 4 flags of different colours?, These signals can be made by using 1 or 2 or 3 or 4 flags at a time., Number of signals made by 4 flags:, (i) taking 1 at a time 4P1 4;, (ii) taking 2 at a time 4P2 (4 # 3) 12;, (iii) taking 3 at a time 4P3 (4 # 3 # 2) 24;, (iv) taking 4 at a time 4P4 (4 # 3 # 2 #1) 24., Hence, by the fundamental principle of addition, the total number, of signals (4 12 24 24) 64.
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Permutations, EXAMPLE 7, , SOLUTION, , EXAMPLE 8, SOLUTION, , EXAMPLE 9, , SOLUTION, , 291, , A code is to consist of two distinct letters followed by two distinct numbers, between 1 and 9. For example, CA 35, DQ 72, etc., are codes., (i) How many such codes are there?, (ii) How many of them end with an even integer?, (i) We know that there are 26 letters in the English alphabet., Number of ways of choosing 2 letters out of 26, 26P2 (26 # 25) 650., Number of ways of choosing 2 numbers out of 9, 9P2 (9 # 8) 72., By the fundamental principle of multiplication, the total, number of codes (650 #72) 46800., (ii) Particular case When each code ends with an even integer, In this case, it can have any of the numbers 2, 4, 6, 8 at the, extreme right position. This position can thus be filled in, 4 ways., The next left position can be filled with any of the remaining, 8 numbers. So, there are 8 ways of filling this position., Number of ways of choosing pairs of even numbers, (4 # 8) 32., Number of ways of choosing 2 letters out of 26, 26P2 (26 # 25) 650., Hence, the number of such codes which end with an even, integer (32 # 650) 20800., Find the number of ways in which 5 boys and 3 girls can be arranged in a, row so that no two girls are together., In order that no two girls are together, we must arrange the 5 boys,, each denoted by B, as under:, XBXBXBXBXBX, Here, B denotes the position of a boy and X that of a girl., Number of ways in which 5 boys can be arranged at 5 places, 5P5 5 ! (5 # 4 # 3 # 2 #1) 120., Number of ways in which 3 girls can be arranged at 6 places (each, marked X) 6P3 (6 # 5 # 4) 120., Hence, by the fundamental principle of multiplication, the required, number of ways (120 #120) 14400., There are 2 English, 3 Sanskrit and 4 Hindi books. In how many ways can, they be arranged on a shelf so as to keep all the books of the same language, together?, Let us make 1 packet for each of the books on the same language., Number of ways of arranging the 3 packets 3P3 3 ! 6.
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292, , Senior Secondary School Mathematics for Class 11, , Number of ways of arranging 2 English books 2 ! 2., Number of ways of arranging 3 Sanskrit books 3 ! 6., Number of ways of arranging 4 Hindi books 4 ! 24., Hence, the required number of ways (2 # 6 # 24)# 6 1728., EXAMPLE 10, SOLUTION, , EXAMPLE 11, , SOLUTION, , EXAMPLE 12, , SOLUTION, , In how many ways can 7 examination papers be arranged so that the best, and the worst papers are never together?, Let us tie the best paper (b) and the worst paper (w) together and, consider (bw) as one paper., Now, this (bw) and 5 other papers may be arranged in, 6, P6 6 ! 720 ways., Also, these two papers may be arranged among themselves in, 2 ! 2 ways., Total number of arrangements with best and worst papers together, (720 # 2) 1440., Total number of ways of arranging 7 papers, 7P7 7 ! 5040., Number of arrangements with best and worst paper never together, (5040 1440) 3600., In how many ways can 6 balls of different colours, namely black, white,, blue, red, green and yellow be arranged in a row in such a way that the, black and white balls are never together?, Let us tie the black ball (b) and white ball (w) together and consider, (bw) as one ball., Now, this (bw) and 4 other balls may be arranged in, 5, P5 5 ! 120 ways., Also, these two balls may be arranged among themselves in, 2 ! 2 ways., Total number of arrangements with black and white balls together, (120 # 2) 240., Number of ways of arranging 6 balls among themselves, 6P6 6 ! 720., Number of ways of arranging 6 balls such that black and white, balls are never together (720 240) 480., The principal wants to arrange 5 students on a platform such that the boy, Sajal occupies the second position and the girl Tanvy is always adjacent, to the girl Aditi. How many such arrangements are possible?, Let the five seats be arranged as shown below., I II III IV V, = S = = =, Keep Sajal fixed at the second position.
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Permutations, , 293, , Since Tanvy and Aditi are to sit together, none of them can occupy, the first seat., The first seat can be occupied by any of the two remaining students, in 2 ! 2 ways., Number of ways in which 2 seats namely III, IV or IV, V may be, occupied by Tanvy and Aditi (2 !) (2 !) (2 2) 4., The remaining seat may now be occupied by the 5th student in, 1 way only., Hence, the required number of arrangements (2 # 4 #1) 8., EXAMPLE 13, SOLUTION, , EXAMPLE 14, , SOLUTION, , EXAMPLE 15, , SOLUTION, , How many words (with or without meaning) can be formed by using all, the letters of the word, ‘DELHI‘ using each letter exactly once?, The word, ‘DELHI‘ contains 5 different letters., Total number of words formed by using all the letters of the given, word 5P5 5 ! (5 # 4 # 3 # 2 #1) 120., Hence, the required number of words = 120., How many 4-letter words (with or without meaning) can be formed out, of the letters of the word, ‘LOGARITHMS‘, if repetition of letters is not, allowed?, The word, ‘LOGARITHMS‘ contains 10 different letters., Number of 4-letter words formed out of 10 given letters, 10P4 (10 # 9# 8 #7) 5040., Hence, the required number of 4-letter words = 5040., How many words (with or without meaning) can be formed from the, letters of the word, ‘DAUGHTER‘, so that, (i) all vowels occur together?, (ii) all vowels do not occur together?, The given word, ‘DAUGHTER‘ contains 3 vowels A, U, E and 5, consonants D, G, H, T, R., Case (i) When all vowels occur together:, Let us assume (AUE) as a single letter., Then, this letter (AUE) along with 5 other letters can be, arranged in 6P6 (6 !) ways (6 # 5 # 4 # 3 # 2 #1) ways, = 720 ways., These 3 vowels may be arranged among themselves in, 3 ! 6 ways., Hence, the required number of words with vowels together, (6 !)#(3 !) (720 # 6) 4320., Case (ii) When all vowels do not occur together:, Number of words formed by using all the 8 letters of the, given word
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294, , Senior Secondary School Mathematics for Class 11, , 8P8 8 ! (8 #7 # 6 # 5 # 4 # 3 # 2 #1) 40320., Number of words in which all vowels are never together, = (total number of words) – (number of words with, all vowels together), (40320 4320) 36000., EXAMPLE 16, , SOLUTION, , EXAMPLE 17, , SOLUTION, , How many words (with or without meaning) can be made from the letters, of the word ‘MONDAY‘, assuming that no letter is repeated, when:, (i) 4 letters are used at a time?, (ii) all letters are used at a time?, (iii) all letters are used but the first letter is a vowel?, The given word ‘MONDAY‘ contains 6 letters out of which two, letters namely O and A are vowels., Case (i) When 4 letters are used at a time:, Then, the required number of words, = number of arrangements of 6 letters taking 4 at a time, 6P4 (6 # 5 # 4 # 3) 360., Case (ii) When all the letters are used at a time:, Then, the required number of words, = number of arrangements of 6 letters taking all at a time, 6P6 6 ! (6 # 5 # 4 # 3 # 2 #1) 720., Case (iii) When all the letters are used and the first letter is a vowel:, The given word contains 2 vowels, namely O and A., So, the first letter can be chosen 1 out of 2 vowels., Thus, there are 2 ways of choosing the first letter., Number of ways of arranging the remaining 5 letters, among themselves 5 ! (5 # 4 # 3 # 2 #1) 120., Hence, the required number of words (2 #120) 240., (i) How many words can be formed from the letters of the word,, ‘TRIANGLE‘?, (ii) How many of them will begin with T and end with E?, (i) The given word ‘TRIANGLE‘ contains 8 letters., Total number of words formed with these 8 letters, 8P8 8 ! 40320., Hence, the required number of words = 40320., (ii) By fixing T in the beginning and E at the end, the remaining, 6 letters can be arranged among themselves in 6P6 6 ! 720, ways., Hence, the total number of words, each beginning with T and, ending with E 720.
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Permutations, EXAMPLE 18, SOLUTION, , EXAMPLE 19, SOLUTION, , EXAMPLE 20, SOLUTION, , EXAMPLE 21, , SOLUTION, , 295, , How many words can be formed with the letters of the word ‘ORDINATE‘, so that the vowels occupy odd places?, The given word consists of 8 letters, out of which there are 4 vowels, and 4 consonants., Let us mark out the positions to be filled up, as shown below:, (1 ) ( 2 ) ( 3 ) ( 4 ) ( 5 ) ( 6 ) ( 7 ) ( 8 ) ., Since vowels occupy odd places, they may be placed at 1, 3, 5, 7., Number of ways of arranging 4 vowels at 4 odd places, 4P4 4 ! 24., The remaining 4 letters of the given word are consonants, which, can be arranged at 4 even places marked 2, 4, 6, 8., Number of ways of arranging 4 consonants at 4 even places, 4P4 4 ! 24., Hence, the total number of words in which the vowels occupy odd, places (24 # 24) 576., How many 3-digit numbers can be formed by using the digits 1 to 9, if no, digit is repeated?, Required number of numbers, = number of permutations of 9 digits taking 3 at a time, 9P3 (9 # 8 #7) 504., Hence, the required number of numbers = 504., Find the number of 4-digit numbers that can be formed using the digits, 1, 2, 3, 4, 5 if no digit is repeated. How many of them will be even?, Required number of 4-digit numbers, = number of arrangements of 5 digits taking 4 at a time, 5P4 (5 # 4 # 3 # 2) 120., Particular case To get an even number, the unit’s digit is 2 or 4., Thus, there are 2 ways of filling the unit’s digit., Number of ways of filling remaining 3 places with the remaining 4, digits 4P3 (4 # 3 # 2) 24., Hence, the required number of even numbers (2 # 24) 48., How many 6-digit telephone numbers can be constructed with the digits, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 if each number starts with 35 and no digit appears, more than once?, Out of the given 10 digits, the two digits namely 3 and 5 are, reserved for the two extreme left places of the desired numbers., The remaining 8 digits can be arranged in the remaining four, positions in 8P4 (8 #7 # 6 # 5) 1680 ways., Hence, 1680 telephone numbers can be constructed.
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296, , Senior Secondary School Mathematics for Class 11, , EXERCISE 8D, 1. In how many ways can 5 persons occupy 3 vacant seats?, 2. In how many ways can 7 people line up at a ticket window of a cinema, hall?, 3. In how many ways can 5 children stand in a queue?, 4. In how many ways can 6 women draw water from 6 wells, if no well, remains unused?, 5. In how many ways can 4 different books, one each in chemistry, physics,, biology and mathematics, be arranged on a shelf?, 6. Six students are contesting the election for the presidentship of the students’, union. In how many ways can their names be listed on the ballot papers?, 7. It is required to seat 5 men and 3 women in a row so that the women occupy, the even places. How many such arrangements are possible?, 8. There are 6 items in column A and 6 items in column B. A student is asked, to match each item in column A with an item in column B. How many, possible, correct or incorrect answers are there to this question?, 9. Five letters F, K, R, T and V one in each were purchased from a plastic, warehouse. How many ordered pairs of letters, to be used as initials, can, be formed from them?, 10. Ten students are participating in a race. In how many ways can the first, three prizes be won?, 11. If there are 6 periods on each working day of a school, in how many ways, can one arrange 5 subjects such that each subject is allowed at least one, period?, 12. In how many ways can 6 pictures be hung from 4 picture nails on a wall?, 13. Find the number of words formed (may be meaningless) by using all the, letters of the word ‘EQUATION‘, using each letter exactly once., 14. Find the number of different 4-letter words (may be meaningless) that can, be formed from the letters of the word ‘NUMBERS‘., 15. How many words can be formed from the letters of the word ‘SUNDAY‘?, How many of these begin with D?, 16. How many words beginning with C and ending with Y can be formed by, using the letters of the word ‘COURTESY‘?, 17. Find the number of permutations of the letters of the word ‘ENGLISH‘. How, many of these begin with E and end with I ?, 18. In how many ways can the letters of the word ‘HEXAGON‘ be permuted?, In how many words will the vowels be together?, 19. How many words can be formed out of the letters of the word ‘ORIENTAL‘, so that the vowels always occupy the odd places?, 20. In how many ways can the letters of the word ‘FAILURE‘ be arranged so, that the consonants may occupy only odd positions?, 21. In how many arrangements of the word ‘GOLDEN‘ will the vowels never, occur together?
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Permutations, , 297, , 22. Find the number of ways in which the letters of the word ‘MACHINE‘ can, be arranged such that the vowels may occupy only odd positions., 23. How many permutations can be formed by the letters of the word ‘VOWELS‘,, when, (i) there is no restriction on letters;, (ii) each word begins with E;, (iii) each word begins with O and ends with L;, (iv) all vowels come together;, (v) all consonants come together?, 24. How many numbers divisible by 5 and lying between 3000 and 4000 can, be formed by using the digits 3, 4, 5, 6, 7, 8 when no digit is repeated in any, such number?, 25. In an examination, there are 8 candidates out of which 3 candidates have to, appear in mathematics and the rest in different subjects. In how many ways, can they be seated in a row, if candidates appearing in mathematics are not, to sit together?, 26. In how many ways can 5 children be arranged in a line such that, (i) two of them, Rajan and Tanvy, are always together?, (ii) two of them, Rajan and Tanvy, are never together?, 27. When a group photograph is taken, all the seven teachers should be in the, first row and all the twenty students should be in the second row. If the, two corners of the second row are reserved for the two tallest students,, interchangeable only between them, and if the middle seat of the front row, is reserved for the principal, how many arrangements are possible?, 28. Find the number of ways in which m boys and n girls may be arranged in a, row so that no two of the girls are together, it being given that m n., ANSWERS (EXERCISE 8D), , 1., 6., 11., 16., 21., 24., , 60, 2., 720, 7., 600, 12., 720, 17., 480, 22., 12, 25., (m !)#(m 1) !, 28., (m 1 n) !, , 5040, 7200, 360, 5040, 120, 576, 14400, , 3., 8., 13., 18., 23., 26., , 120, 4. 720, 5. 24, 720, 9. 20, 10. 720, 40320, 14. 840, 15. 720, 120, 5040, 720, 19. 576, 20. 576, (i) 720 (ii) 120 (iii) 24 (iv) 240 (v) 144, (i) 48 (ii) 72 27. (18 !)#(6 !)# 2, , HINTS TO SOME SELECTED QUESTIONS, 7. 5 men can be arranged at 5 places in 5P5 5 ! 120 ways., Starting from man, we leave alternate places for women, as shown below., M=M=M=M=M=
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298, , Senior Secondary School Mathematics for Class 11, Now, 3 women can be arranged at these 5 places, marked =, in, 5, P3 (5 # 4 # 3) 60 ways., Required number of ways (120 # 60) 7200., , 9. Number of ordered pairs 5P2 (5 # 4) 20., 10. Required number of ways 10P3 ., 11. Out of 6 periods, 5 may be arranged for 5 subjects in 5P5 ways., Remaining 1 period may be arranged for anyone of the five subjects in 5P1 ways., Required number of ways (5P5 # 5P1) ., 18. The given word contains 7 letters, which may be arranged among themselves in, 7 ! 5040 ways., The given word contains 3 vowels and 4 consonants., Taking the 3 vowels EAO as one letter, this letter and 4 more letters can be arranged, in 5 ! 120 ways., The 3 vowels can be arranged among themselves in 3 ! 6 ways., Required number of arrangements with vowels together (120 # 6) 720., 19. The given word contains 4 vowels and 4 consonants., Let us fix the 8 places as shown below., (1) (2) (3) (4) (5) (6) (7) (8), The 4 vowels can be arranged at 4 places (namely 1, 3, 5, 7) in 4P4 4 ! 24 ways., And, the 4 consonants can be arranged at 4 places (namely 2, 4, 6, 8) in 4P4 4 !, = 24 ways., Required number of ways (24 # 24) 576., 20. The given word ‘FAILURE‘ has 3 consonants and 4 vowels, which can be arranged at, seven places, shown below., (1) (2) (3) (4) (5) (6) (7), Now, 3 consonants may be placed at any of the 3 places out of the 4 marked 1, 3, 5, 7., Number of ways of arranging the consonants 4P3 24., And, the 4 vowels can be arranged at the remaining 4 places in 4P4 24 ways., Required number of ways (24 # 24) 576., 21. The given word ‘GOLDEN‘ contains 6 letters which may be arranged among, themselves in 6 ! 720 ways., It has 2 vowels and 4 consonants., Taking the 2 vowels OE as a single letter, this letter and 4 consonants together may, be arranged in 5 ! 120 ways., Also, O and E may be arranged among themselves in 2 ! ways., Number of arrangements with vowels together (120 # 2) 240., Number of arrangements with vowels never together (720 240) 480., 22. There are 7 letters in the given word, out of which there are 3 vowels and 4 consonants., Let us mark out the positions to be filled up as follows:, (1) (2) (3) (4) (5) (6) (7), Now, the 3 vowels can be placed at any of the three places out of the four, marked, 1, 3, 5, 7., So, the number of ways of arranging the vowels 4P3 4 # 3 # 2 24.
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Permutations, , 299, , Also, the 4 consonants at the remaining 4 positions may be arranged in 4P4 4 !, = 24 ways., the requisite number of ways (24 # 24) 576., 24. Fix up 3 at the thousand‘s place and 5 at the unit’s place. Now, the remaining 2 places, can be filled by the remaining 4 digits in 4P2 (4 # 3) 12 ways., 25. Clearly, there are 5 candidates not appearing in mathematics., Let us arrange these 5 in a row, each shown by X., They can be arranged in 5 ! 120 ways., On both sides of each X, we put an M, as shown below:, MXMXMXMXMXM, Now, 3 candidates in mathematics can be arranged at 6 places in 6P3 ways (6 # 5 # 4), ways = 120 ways., Total number of arrangements (120 #120) 14400., 26. Total number of ways of arranging 5 children in a row 5 ! 120., (i) Considering Rajan + Tanvy as one child, this one and 3 more can be arranged in, 4 ! 24 ways., Rajan and Tanvy can be arranged among themselves in 2 ! 2 ways., Required number of ways (24 # 2) 48., (ii) When Rajan and Tanvy are never together, the number of arrangements, (120 48) 72., 27. Keeping the middle seat of the front row fixed for the principal, the remaining 6 seats, can be occupied by the remaining 6 teachers in 6 ! ways., The two corner seats may be occupied by the two tallest students in 2 ways. The, remaining 18 seats may be occupied by the remaining 18 students in 18 ! ways., 28. We may arrange m boys at m places in m ! ways, leaving a seat between each pair of, boys as shown below:, # B# B#…# B#., Now, n girls may be arranged in (m 1) places, shown by cross marks., , , The number of such arrangements is m 1Pn ., , PERMUTATIONS OF OBJECTS NOT ALL DIFFERENT, THEOREM 1, , PROOF, , Let there be n objects, of which m objects are alike of one kind, and the, remaining (n m) objects are alike of another kind. Then, the total number, of mutually distinguishable permutations that can be formed from these, objects is, n!, ·, (m !)#(n m) !, , Let the required number of permutations be x., Consider any of these x permutations., Now, replace m like things in this permutation by m different things., These m different things may be arranged amongst themselves in, m ! ways.
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300, , Senior Secondary School Mathematics for Class 11, , Similarly, if we replace (n m) like things by (n m) different things,, then they can be arranged amongst themselves in (n m) ! ways., Thus, if both the replacements are done simultaneously then each one, of the x permutations gives rise to (m !)#(n m) ! permutations., x permutations give rise to x #(m !)#(n m) ! permutations., Now, each of these x #(m !)#(n m) ! permutations is a permutation, of n different things, taken all at a time., x #(m !)#(n m) ! n !., Hence, x , REMARK 1, , n!, ·, (m !)#(n m) !, , Let there be n things of which p1 are alike of one kind, p 2 are alike of, another kind, p 3 are alike of 3rd kind, …, pr are alike of rth kind such that, p1 p 2 … pr n. Then, the number of permutations of these n things is, n!, ·, (p1 !)#(p 2 !)#… (pr !), , REMARK 2, , The number of permutations of n things of which p are alike of one kind, q, n!, ·, are alike of another kind and remaining all are distinct is, (p !)#(q !), SOLVED EXAMPLES, , EXAMPLE 1, SOLUTION, , EXAMPLE 2, SOLUTION, , EXAMPLE 3, , SOLUTION, , Find the number of different permutations of the letters of the word,, ‘BANANA'., The given word ‘BANANA‘ contains 6 letters, out of which three are, alike of one kind (3 A‘s), two are alike of second kind (2 N‘s) and, one of its own kind (1 B)., 6!, 60., Total number of their permutations , (3 !)#(2 !)#(1 !), How many words can be formed using the letter A thrice, the letter B, twice and the letter C thrice?, We are given 8 letters, namely AAABBCCC. Out of these letters, three are alike of one kind (3 A‘s), two are alike of second kind, (2 B‘s) and three are alike of third kind (3 C‘s)., Hence, the required number of permutations, 8!, , 560., (3 !)#(2 !)#(3 !), (i) Find how many arrangements can be made with the letters of the, word ‘MATHEMATICS‘., (ii) In how many of them are the vowels together?, (i) There are 11 letters in the word ‘MATHEMATICS‘. Out of these, letters M occurs twice, A occurs twice, T occurs twice and the, rest are all different.
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Permutations, , 301, , Hence, the total number of arrangements of the given letters, 11 !, , 4989600., (2 !)#(2 !)#(2 !), (ii) The given word contains 4 vowels, namely A, E, A, I. Treating, these 4 vowels AEAI as one letter, we have to arrange 8 letters, MTHMTCS + AEAI , out of which M occurs twice, T occurs, twice and the rest are all different., 8!, 10080., So, the number of all such arrangements , (2 !)#(2 !), Now, out of 4 vowels, A occurs twice and the rest are all, distinct., 4! , So, the number of arrangements of these vowels , 12., 2!, Hence, the number of arrangements in which 4 vowels are, together (10080 #12) 120960., EXAMPLE 4, , SOLUTION, , I. Find the number of arrangements of the letters of the word,, ‘INDEPENDENCE‘., II. In how many of these arrangements:, (i) do the words start with P?, (ii) do all the vowels occur together?, (iii) do the vowels never occur together?, (iv) do the words begin with I and end in P?, I. The given word contains 12 letters out of which N occurs, 3 times, E occurs 4 times, D occurs 2 times and the rest are all, different., Hence, the required number of arrangements, , , 12 !, 1663200., (3 !)#(4 !)#(2 !), , II. (i) After fixing P at the extreme left position, there are 11, letters consisting of 3 N‘s, 4 E‘s, 2 D‘s, 1 I and 1 C., So, the number of words beginning with P, = number of arrangements of 11 letters out of, which there are 3 N‘s, 4 E‘s, 2 D‘s, 1 I and 1 C, 11 !, , 138600., (3 !)#(4 !)#(2 !), (ii) There are 5 vowels in the given word, namely 4 E‘s and 1 I., Let us treat them as a single letter EEEEI ., This letter with 7 remaining letters will give us 8 letters in, which there are 3 N‘s, 2 D‘s, 1 P, 1 C and 1 letter EEEEI ., Number of all such arrangements , , 8!, 3360., (3 !)#(2 !)
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302, , Senior Secondary School Mathematics for Class 11, , Number of arrangements of 5 vowels, namely 4 E‘s and 1 I, 5 ! 5., 4!, Hence, the number of words in which vowels are together, (3360 # 5) 16800., (iii) Number of arrangements in which vowels do not occur, together, = (total number of arrangements) – (number of, arrangements in which vowels occur together), (1663200 16800) 1646400., (iv) Let us fix I at the left end and P at the right end of each, arrangement., Then, we are left with 10 letters, out of which there are, 3 N‘s, 4 E‘s, 2 D‘s and 1 C., Hence, the number of words which begin with I and end, 10 !, 12600., in P , (3 !)#(4 !)#(2 !), EXAMPLE 5, , SOLUTION, , (i) How many different words can be formed by using all the letters of, the word, ‘ALLAHABAD‘?, In how many of them:, (ii) both L‘s do not come together?, (iii) the vowels occupy the even positions?, (i) The given word, ‘ALLAHABAD‘ contains 9 letters consisting of, 4 A‘s, 2 L‘s, 1 H, 1 B and 1 D., Hence, the number of different words formed by using all the, 9!, 7560., letters of the given word , (4 !)#(2 !), (ii) Let us take both L together and we treat LL as 1 letter., Then, we will have to arrange 8 letters, namely LL , 4 A‘s,, 1 H, 1 B and 1 D., 8! , So, the number of words having both L together , 1680., 4!, Hence, the number of words with both L not occurring together, (7560 1680) 5880., (iii) There are 4 vowels and all are alike, i.e., 4 A‘s., Also, there are 4 even places, namely 2nd, 4th, 6th and 8th., So, we arrange A‘s at these 4 places, as shown below., = A = A = A = A =, 4! , Number of arrangements of 4 A‘s at 4 places , 1., 4!, Now, we are left with 5 letters consisting of 2 L‘s, 1 H, 1 B and, 1 D.
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Permutations, , Number of arrangements of these 5 letters at 5 places , , 303, , 5! , 60., 2!, , Hence, the number of words in which vowels occupy even, places (1# 60) 60., EXAMPLE 6, SOLUTION, , In how many of the distinct permutations of the letters in ‘MISSISSIPPI‘, do the 4 I‘s not come together?, The given word ‘MISSISSIPPI‘ contains 11 letters, out of which there, are 4 I‘s, 4 S‘s, 2 P‘s and 1 M., Total number of words formed by the letters of the given word, , , 11 !, 34650., (4 !)#(4 !)#(2 !), , There are 4 I‘s in the given word and we treat I I I I as 1 letter., This single letter with remaining 7 letters will give us 8 letters out, of which there are 4 S‘s, 2 P‘s, 1 M and 1 I I I I ., So, the number of all such arrangements in which 4 I‘s are together, , , 8!, 840., (4 !)#(2 !), , Thus, the number of words formed when 4 I‘s are together = 840., Hence, the number of words formed when 4 I‘s are not together, (34650 840) 33810., EXAMPLE 7, , SOLUTION, , (i) How many words can be formed with the letters of the word,, ‘HARYANA‘?, How many of these, (ii) have H and N together?, (iii) begin with H and end with N?, (iv) have 3 vowels together?, (i) The given word ‘HARYANA‘ consists of 7 letters, out of which, there are 1 H, 3 A‘s, 1 R, 1 Y and 1 N., Total number of words formed by all the letters of the given, 7! , word , 840., 3!, (ii) Let us consider HN as a single letter., Now, HN + ARYAA will give us 6 letters out of which there, are 3 A‘s, 1 R, 1 Y and 1 HN ., 6! , Total number of all such arrangements , 120., 3!, But, H and N can be arranged amongst themselves in 2 ! ways., Hence, the number of words having H and N together, (120 # 2) 240.
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304, , Senior Secondary School Mathematics for Class 11, , (iii) After fixing H in first place and N in last place, we have, 5 letters, out of which there are 3 A‘s, 1 R and 1 Y., Hence, the number of words beginning with H and ending, 5! , with N , 20., 3!, (iv) The given word contains 3 vowels AAA and let us treat AAA1, as 1 letter., Now, we have to arrange 5 letters HRYN + AAA at 5 places., Hence, total number of words formed having all vowels, together 5 ! (5 # 4 # 3 # 2 #1) 120., EXAMPLE 8, , SOLUTION, , In how many ways can the letters of the word ‘PERMUTATIONS‘ be, arranged, if, (i) the words start with P and end with S?, (ii) the vowels are all together?, (i) Let us fix P at the left end and S at the right end of each, arrangement., Then, we are left with 10 letters, out of which T occurs 2 times, and the rest are all different., 10 ! , Hence, the required number of arrangements , 1814400., 2!, (ii) The given word contains 5 vowels, namely E, U, A, I, O., Treating these five vowels EUAIO as one letter, we have to, arrange 8 letters, namely EUAIO + PRMTTNS, out of which T, occurs 2 times and the rest are all different., Number of all such arrangements , , 8! , 20160., 2!, , Now, the 5 vowels are all different. They can be arranged, among themselves in 5 ! 120 ways., Hence, the number of arrangements in which 5 vowels are, together (20160 #120) 2419200., EXAMPLE 9, , SOLUTION, , If the different permutations of the word ‘EXAMINATION‘ are listed as, in a dictionary, how many items are there in the list before the first word, starting with E?, In a dictionary, the words at each stage are arranged in an, alphabetical order., When the letters of the word ‘EXAMINATION‘ are arranged in the, dictionary, the first word would be AAEIIMNNOTX., Starting with A, we arrange the remaining 10 letters AEIIMNNOTX, in which there are 2 I‘s, 2 N‘s and the rest are all different, in all, possible ways.
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Permutations, , 305, , Thus, the number of words formed beginning with A, , , 10 !, 907200., (2 !)#(2 !), , In the dictionary, the next word would be EAAIIMNNOTX., Hence, 907200 words are there in the dictionary before the first, word starting with E., EXAMPLE 10, , SOLUTION, , (i) Find the number of words which can be made using all the letters of, the word, ‘AGAIN‘., (ii) If these words are written as in a dictionary, what will be the 50th, word?, (i) The given word ‘AGAIN‘ consists of 5 letters, out of which, there are 2 A‘s and the rest are all distinct., Hence, the number of words formed by using all the letters of, 5! , the given word , 60., 2!, (ii) When the letters of the word ‘AGAIN‘ are listed in a dictionary,, the first word would be AAGIN., Starting with A and arranging the remaining 4 letters A, G, I,, N, we obtain 4 ! words, i.e., 24 words., The 25th word would be GAAIN, which starts with G., Now, starting with G arrange the remaining 4 letters AAIN,, (in which A occurs 2 times and the rest are different) in all, possible ways., 4! , Number of such arrangements , 12., 2!, Thus, we obtain 12 new words., The 37th word would be IAAGN, which starts with I., Now, starting with I and arranging the remaining 4 letters,, 4! , AAGN, we get, 12 words., 2!, The 49th word would be NAAGI and hence the 50th word is, NAAIG., , EXAMPLE 11, SOLUTION, , How many numbers greater than a million can be formed with the digits, 2, 3, 0, 3, 4, 2, 3?, Any number greater than one million will contain all the seven, digits., Now, we have to arrange seven digits, out of which 2 occurs twice,, 3 occurs thrice and the rest are distinct., Number of such arrangements , , 7!, 420., (2 !)#(3 !)
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306, , Senior Secondary School Mathematics for Class 11, , These arrangements will also include those which contain 0 at the, million‘s place., Keeping 0 fixed at the million‘s place, the remaining 6 digits out of, which 2 occurs twice, 3 occurs thrice and the rest are distinct can be, 6!, 60 ways., arranged in , (2 !)#(3 !), Hence, the number of required numbers (420 60) 360., EXAMPLE 12, , How many numbers greater than 400000 can be formed by using the, digits 0, 2, 2, 4, 4, 5?, , SOLUTION, , We are being given six digits and each number greater than 400000, is a six-digit number, having 4 or 5 in the extreme left position., When 4 occupies the extreme left position, the remaining 5 places, can be filled with the digits 0, 2, 2, 4, 5 (in which 2 occurs 2 times, and rest are all different)., 5! , Number of such numbers , 60., 2!, When 5 occupies the extreme left position, the remaining 5 places, can be filled with the digits 0, 2, 2, 4, 4 (in which 2 occurs twice,, 4 occurs twice and the rest are all distinct)., Number of such numbers , , 5!, 30., (2 !)#(2 !), , Hence, the required number of numbers (60 30) 90., EXAMPLE 13, , SOLUTION, , How many numbers can be formed using the digits 1, 2, 3, 4, 3, 2, 1 so, that the odd digits always occupy the odd places?, We have been given seven digits, namely 1, 2, 3, 4, 3, 2, 1., So, we have to form 7-digit numbers, so that odd digits occupy, odd places., Every 7-digit number has 4 odd places., Given four odd digits are 1, 3, 3, 1 out of which 1 occurs 2 times and, 3 occurs 2 times., 4!, 6., So, the number of ways to fill up 4 odd places , (2 !)#(2 !), Given three even digits are 2, 4, 2 in which 2 occurs 2 times and, 4 occurs 1 time., 3! , So, the number of ways to fill up 3 even places , 3., 2!, Hence, the required number of numbers (6 # 3) 18.
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Permutations, , 307, , EXERCISE 8E, 1. Find the total number of permutations of the letters of each of the words, given below:, (i) APPLE, (ii) ARRANGE, (iii) COMMERCE, (iv) INSTITUTE, (v) ENGINEERING, (vi) INTERMEDIATE, 2. In how many ways can the letters of the expression x 2y 3z 4 be arranged, when written without using exponents?, 3. There are 3 blue balls, 4 red balls and 5 green balls. In how many ways can, they be arranged in a row?, 4. A child has three plastic toys bearing the digits 3, 3, 5 respectively. How, many 3-digit numbers can he make using them?, 5. How many different signals can be transmitted by arranging 2 red, 3 yellow, and 2 green flags on a pole, if all the seven flags are used to transmit a signal?, 6. How many words can be formed by arranging the letters of the word, ‘ARRANGEMENT’, so that the vowels remain together?, 7. How many words can be formed by arranging the letters of the word, ‘INDIA’, so that the vowels are never together?, 8. Find the number of arrangements of the letters of the word ‘ALGEBRA’, without altering the relative positions of the vowels and the consonants., 9. How many words can be formed from the letters of the word ‘SERIES‘,, which start with S and end with S?, 10. In how many ways can the letters of the word ‘PARALLEL’ be arranged so, that all L’s do not come together?, 11. How many different words can be formed with the letters of the word, ‘CAPTAIN’? In how many of these C and T are never together?, 12. In how many ways can the letters of the word ‘ASSASSINATION’ be, arranged so that all S’s are together?, 13. (i) How many arrangements can be made by using all the letters of the, word ‘MATHEMATICS’?, (ii) How many of them begin with C?, (iii) How many of them begin with T?, 14. In how many ways can the letters of the word ‘INTERMEDIATE’ be arranged, so that:, (i) the vowels always occupy even places?, (ii) the relative orders of vowels and consonants do not change?, 15. (i) Find the number of different words formed by using all the letters of, the word, ‘INSTITUTION’., In how many of them, (ii) are the three T’s together?, (iii) are the first two letters the two N’s?, 16. How many five-digit numbers can be formed with the digits 5, 4, 3, 5, 3?
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308, , Senior Secondary School Mathematics for Class 11, , 17. How many numbers can be formed with the digits 2, 3, 4, 5, 4, 3, 2 so that, the odd digits occupy the odd places?, 18. How many 7-digit numbers can be formed by using the digits 1, 2, 0, 2, 4,, 2, 4?, 19. How many 6-digit numbers can be formed by using the digits 4, 5, 0, 3, 4, 5?, 20. The letters of the word ‘INDIA’ are arranged as in a dictionary. What are the, 1st, 13th, 25th, 49th and 60th words?, ANSWERS (EXERCISE 8E), , 1., 2., 7., 12., 14., 16., 20., , (i) 60, (ii) 1260, (iii) 5040, (iv) 30240, (v) 277200, (vi) 19958400, 1260, 3. 27720, 4. 3, 5. 210, 6. 60480, 42, 8. 72, 9. 12, 10. 3000, 11. 2520, 1800, 151200, 13. (i) 4989600 (ii) 453600 (iii) 907200, (i) 21600 (ii) 21600, 15. (i) 554400 (ii) 30240 (iii) 10080, 30, 17. 18, 18. 360, 19. 150, (i) ADIIN (ii) DAIIN (iii) IADIN (iv) NADII (v) NIIDA, HINTS TO SOME SELECTED QUESTIONS, , 2. There are 2 x’s, 3 y’s and 4 z’s., , , total number of arrangements , , 9!, ·, (2 !)#(3 !)#(4 !), , 3. We have to arrange 12 balls, out of which 3 are of one kind, 4 are of second kind and, 5 are of third kind., 12 !, ·, Required number of arrangements , (3 !)#(4 !)#(5 !), 4. We have to arrange 3 objects in a row, out of which 2 are alike of one kind and the, third one of its own kind., 3! , 3., 2!, 7!, ·, 5. Required number of signals , (2 !)#(3 !)#(2 !), Total number of arrangements , , 6. Let us assume that the vowels AAEE form a single letter., Now, AAEE + RRNGMNT gives us 8 letters, out of which there are 2 R’s, 2 N’s, 1 G,, 1 M and 1 T., 8!, 10080., Number of these arrangements , (2 !)#(2 !), Now, AAEE has 4 letters, out of which there are 2 A’s and 2 E’s., 4!, 6., (2 !)#(2 !), Required number of words formed (10080 # 6) 60480., Number of their arrangements , , 7. The word ‘INDIA’ contains 2 I’s, 1 A, 1 N and 1 D., Number of permutations of the letters of the given word , Let us assume that the vowels IIA form 1 letter., , 5! , 60., 2!
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Permutations, , 309, , Now, IIA + ND gives 3 letters, all distinct., Number of their arrangements 3 ! 6., Now, IIA has 3 letters, out of which there are 2 I’s and 1 A., 3! , Number of their arrangements , 3., 2!, Number of words having all vowels together (6 # 3) 18., Number of words in which all vowels are not together (60 18) 42., 8. In the given word, there are 3 vowels, namely A, E, A out of which there are 2 A’s and, 1 E., 3! , Number of arrangements of the vowels , 3., 2!, The given word contains 4 consonants, namely L, G, B, R., Number of arrangements of the consonants 4 ! 24., Required number of arrangements (3 # 24) 72., 9. After fixing S at the beginning and S at the end of each word, we have to arrange, 4 letters out of which there are 2 E’s, 1 R and 1 I., Number of their arrangements , , 4! , 12., 2!, , 10. The given word ‘PARALLEL’ has 8 letters, out of which there are 2 A’s, 3 L’s, 1 P, 1 R, and 1 E., 8!, 3360., Number of their arrangements , (2 !)#(3 !), Let us assume LLL as 1 letter., Then, LLL + PARAE has 6 letters, out of which there are 2 A’s and the rest are all, distinct., 6! , Number of their arrangements , 360., 2!, Number of arrangements in which 3 L’s are not together (3360 360) 3000., 11. The given word ‘CAPTAIN’ has 7 letters, out of which there are 2 A’s and the rest are, all different., 7! , Number of their arrangements , 2520., 2!, Let us assume CT as 1 letter., Then, CT + APAIN has 6 letters, out of which there are 2 A’s and the rest are all, distinct., 6! , Number of their arrangements , 360., 2!, Also, CT can be arranged among themselves in 2 ! 2 ways., Number of arrangements in which C and T are together (360 # 2) 720., Number of arrangements in which C and T are never together (2520 720) 1800., 12. The given word ‘ASSASSINATION’ has 3 A’s, 2 I’s, 2 N’s, 4 S’s, 1 T and 1 O., We consider SSSS as one letter., Then, SSSS + AAAIINNTO consists of 10 letters, out of which there are 3 A’s, 2 I’s,, 2 N’s, 1 T, 1 O and 1 SSSS ., 10 !, 151200., Number of their arrangements , (3 !)#(2 !)#(2 !)
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310, , Senior Secondary School Mathematics for Class 11, , 13. (i) The given word ‘MATHEMATICS’ has 11 letters, out of which there are 2 M’s,, 2 A’s, 2 T’s and the rest are all distinct., 11 !, 4989600., Number of their arrangements , (2 !)#(2 !)#(2 !), (ii) Let each arrangement begin with C. Then, out of the remaining 10 letters, there, are 2 M’s, 2 A’s, 2 T’s and the rest are all distinct., 10 !, 453600., Number of all such arrangements , (2 !)#(2 !)#(2 !), (iii) Let each arrangement begin with T. Then, out of the remaining 10 letters, there are, 2 M’s, 2 A’s and the rest are all distinct., 10 !, 907200., Number of all such arrangements , (2 !)#(2 !), 14. (i) The given word ‘INTERMEDIATE’ has 12 letters, namely 6 vowels and 6 consonants., Since vowels occupy even places, we have, N = T = R = M = D = T =, type of arrangements, where vowels occupy even places, as shown by boxes., It has 6 consonants, namely 2 T’s, 1 N, 1 R, 1 M and 1 D., 6! , Number of arrangements of 6 consonants , 360., 2!, The given word has 6 vowels, namely 2 I’s, 3 E’s and 1 A., 6!, 60., Number of arrangements of 6 vowels , (2 !)#(3 !), , total number of words formed (360 # 60) 21600., (ii) It is given that we may replace a vowel by another vowel and we can replace a, consonant by a consonant, given herewith., 6! , Number of arrangements of consonants , 360., 2!, 6!, 60., Number of arrangements of vowels , (3 !)#(2 !), Total number of words formed (360 # 60) 21600., 15. (i) The given word ‘INSTITUTION’ has 11 letters, out of which there are 3 I’s, 3 T’s,, 2 N’s, 1 S, 1 U and 1 O., 11 !, 554400., Total number of words , (3 !)#(3 !)#(2 !), (ii) Treating TTT as one letter, the given word is expressed as TTT + INSIUION,, which consists of 9 letters out of which there are 3 I’s, 2 N’s, 1 S, 1 U and 1 O., Number of words in this case , , 9!, 30240., (3 !)#(2 !), , (iii) Keeping NN as first letter, we arrange the remaining 9 letters, out of which, there are 3 I’s, 3 T’s, 1 S, 1 U and 1 O., 9!, 10080., Number of words formed , (3 !)#(3 !), 16. We have to form 5-digit numbers in which 5 occurs 2 times; 3 occurs 2 times and, 4 occurs 1 time., Required number of numbers , , 5!, 30., (2 !)#(2 !)#(1 !)
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Permutations, , 311, , 17. There are three odd digits, namely 3, 5, 3 out of which 3 occurs twice and 5 occurs, once., 3! , Number of ways to fill the odd places , 3., 2!, There are four even digits namely 2, 4, 4, 2 out of which 2 occurs twice and 4 occurs, twice., 4!, 6., Number of ways to fill the even places , (2 !)#(2 !), Required number of numbers (3 # 6) 18., 6!, 7!, , 18. Required number of numbers (, 2 (420 60) 360., (3 !)#(2 !) (3 !)#(2 !), 6!, 5!, , 19. Required number of numbers (, 2 (180 30) 150., (2 !)#(2 !) (2 !)#(2 !), 20. The word ‘INDIA’ contains 5 letters out of which there are 2 I’s, 1 N, 1 D and 1 A., Number of arrangements of its letters , , 5! , 60., 2!, , In the dictionary, the first word is ADIIN., Starting with A, we arrange the letters D, I, I, N in, , 4! , 12 ways., 2!, , The 13th word is DAIIN., Starting with D, we arrange the letters A, I, I, N in, , 4! , 12 ways., 2!, , The 25th word is IADIN., Starting with I, we arrange the letters A, D, I, N in 4 ! 24 ways., The 49th word is NADII., 4! , Starting with N, we arrange the letters A, D, I, I in, 12 ways., 2!, The 60th word is NIIDA., , PERMUTATIONS WITH REPETITIONS, The number of permutations of n different objects, taken r at a time when, each may be repeated any number of times in each arrangement, is n r., PROOF, We know that the number of all permutations of n objects, taken r at, a time is the same as the number of ways of filling up r places with n, different objects., Clearly, the first place can be filled in n ways. Since each object may be, repeated, the second place can be filled in n ways., Similarly, each of the 3rd, 4th, …, rth places can be filled in n ways., the requisite number of permutations n # n # n #… r times n r., Corollary When r n,, i.e., the number of permutations of n different objects, taken all, at a time when each may be repeated any number of times in each, arrangement, is n n., THEOREM 1
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312, , Senior Secondary School Mathematics for Class 11, SUMMARY, , (i) Number of permutations of n different objects, taken r at a, time when each may be repeated any number of times in, each arrangement, is n r., (ii) Number of permutations of n different objects, taken all at, a time when each may be repeated any number of times in, each arrangement, is n n., SOLVED EXAMPLES, EXAMPLE 1, SOLUTION, , EXAMPLE 2, SOLUTION, , EXAMPLE 3, SOLUTION, , EXAMPLE 4, , SOLUTION, , In how many ways can 4 letters be posted in 3 letter boxes?, Each letter can be posted in any of the 3 letter boxes., So, each letter can be posted in 3 ways., , 4 letters can be posted in (3 # 3 # 3 # 3) 3 4 81 ways., In how many ways can 6 different rings be worn in 4 fingers of the hand?, Each ring may be worn in any of the 4 fingers., So, each ring may be worn in 4 ways., , 6 rings may be worn in (4 # 4 # 4 # 4 # 4 # 4) 4 6 4096 ways., In how many ways can 5 apples be distributed among 6 boys, there being, no restriction to the number of apples each boy may get?, Each apple may be given to any of the 6 boys., So, each apple may be given in 6 ways., , 5 apples may be given in (6 # 6 # 6 # 6 # 6) 6 5 7776 ways., In how many ways can 3 prizes be distributed among 4 boys, when, (i) no boy gets more than 1 prize;, (ii) a boy may get any number of prizes;, (iii) no boy gets all the prizes?, Case (i) When no boy gets more than 1 prize., The first prize may be given to any of the 4 boys in 4 ways., The 2nd prize may be given in 3 ways, since the boy who, got the 1st prize cannot receive it., The 3rd prize may be given to any of the remaining 2 boys, in 2 ways., Required number of ways (4 # 3 # 2) 24., Case (ii) When a boy may get any number of prizes., The 1st prize may be given to any of the 4 boys in 4 ways., Since a boy may get any number of prizes, so 2nd prize, may be given to any of the 4 boys in 4 ways., Similarly, the 3rd prize may be given in 4 ways., Required number of ways (4 # 4 # 4) 64.
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Permutations, , 313, , Case (iii) When no boy gets all the prizes., , The number of ways in which a boy gets all the prizes is 4,, as anyone of the 4 boys may get all the prizes., Hence, the number of ways in which a boy does not get all, the prizes (64 4) 60., EXAMPLE 5, SOLUTION, , EXAMPLE 6, SOLUTION, , How many four-digit numbers can be formed with the digits 1, 2, 3, 4, 5, 6, when a digit may be repeated any number of times in any arrangement?, Clearly, the thousand’s place can be filled in 6 ways as anyone of, the 6 digits may be placed at it., Since there is no restriction on repetition of digits, each one of the, hundred’s, ten’s and unit’s digits can be filled in 6 ways., Hence, the required number of numbers (6 # 6 # 6) 216., How many 4-digit numbers are there when a digit may be repeated any, number of times?, Clearly, 0 cannot be placed at the thousand’s place., So, this place can be filled with any digit from 1 to 9., Thus, there are 9 ways of filling the thousand’s place., Since repetition of digits is allowed, each one of the remaining, 3 places can be filled in 10 ways, i.e., with any digit from 0 to 9., So, the required number of numbers (9 #10 3) 9000., , EXERCISE 8F, 1. A child has 6 pockets. In how many ways can he put 5 marbles in his, pocket?, 2. In how many ways can 5 bananas be distributed among 3 boys, there being, no restriction to the number of bananas each boy may get?, 3. In how many ways can 3 letters be posted in 2 letter boxes?, 4. How many 3-digit numbers are there when a digit may be repeated any, number of times?, 5. How many 4-digit numbers can be formed with the digits 0, 2, 3, 4, 5 when, a digit may be repeated any number of times in any arrangement?, 6. In how many ways can 4 prizes be given to 3 boys when a boy is eligible for, all prizes?, 7. There are 4 candidates for the post of a chairman and one is to be elected by, votes of 5 men. In how many ways can the votes be given?, , ANSWERS (EXERCISE 8F), , 1. 6, , 5, , 2. 3, , 5, , 3. 8, , 4. 900, , 5. 500, , 6. 81, , 7. 1024
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314, , Senior Secondary School Mathematics for Class 11, HINTS TO SOME SELECTED QUESTIONS, , 2. Each banana may be given to any of the 3 boys., So, each banana can be given in 3 ways., 3. Each letter may be posted in 2 ways., 4. We may fill hundred’s digit by any digit from 1 to 9, i.e., in 9 ways. Each of the ten’s, and unit’s digits may be filled in 10 ways., Required number of numbers (9 #10 #10) 900., 5. Required number of numbers (4 # 5 # 5 # 5) 500., 6. Each prize can be distributed in 3 ways., 7. Each man can vote in 4 ways., , CIRCULAR PERMUTATIONS, So far we have been considering the arrangements of objects in a line. Such, permutations are known as linear permutations., Instead of arranging the objects in a line, if we arrange them in the form of, a circle, we call them circular permutations., In circular permutations, what really matters is the position of an object, relative to the others., If we consider the linear permutations, ABCD, BCDA, CDAB and DABC, then, clearly, they are distinct., Now, we arrange A, B, C, D along the circumference of a circle as, shown below:, A, , D, B, , D, C, , C, A, , C, B, , B, D, , B, A, , C, , A, D, , If we consider the position of an object relative to others then we find that, the above four arrangements are the same., For example, if four persons A, B, C, D sit round a table as shown in the first, arrangement and then each one shifts to the next chair to the left, we obtain the, second arrangement., In this arrangement each person has the same neighbour to the left, and the, same neighbour to the right, as he had in the first arrangement., Thus, the first and the second arrangements give rise to the same circular, permutation.
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Permutations, , 315, , Same is the case with the third and fourth arrangements given above., Thus, in circular permutations, we fix the position of one of the objects and, then arrange the other objects in all possible ways., There are two types of circular permutations:, (i) The circular permutations in which the clockwise and the, anticlockwise arrangements give rise to different permutations, e.g.,, seating arrangements of persons round a table., (ii) The circular permutations in which the clockwise and the anticlockwise, arrangements give rise to same permutations, e.g., arranging some, beads to form a necklace., Look at the circular permutations, given below:, A, , A, B, , D, C, , D, , B, C, , Suppose A, B, C, D are the four beads forming a necklace., They have been arranged in clockwise and anticlockwise directions in, the first and second arrangements respectively., Now, if the necklace in the first arrangement be given a turn, from, clockwise to anticlockwise, we obtain the second arrangement. Thus,, there is no difference between the above two arrangements., , RESULTS ON CIRCULAR PERMUTATIONS, The number of circular permutations of n different objects is (n 1) !., Fixing the position of an object can be done in n ways, as the position, of anyone of them may be fixed., Thus, each circular permutation corresponds to n linear permutations,, depending upon where from we start., n!, Since there are n ! linear permutations, it follows that there are n , i.e.,, (n 1) ! circular permutations., , THEOREM 1, PROOF, , The number of ways in which n persons can be seated round a table is, (n 1) !., Let us fix the position of one person and then arrange the remaining, (n 1) persons in all possible ways. Clearly, this can be done in (n 1) !, ways., Hence, the required number of ways (n 1) !., , THEOREM 2, PROOF, , THEOREM 3, , Show that the number of ways in which n different beads can be arranged, 1, to form a necklace is (n 1) !., 2
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316, PROOF, , Senior Secondary School Mathematics for Class 11, , Fixing the position of one bead, the remaining (n 1) beads can be, arranged in (n 1) ! ways., In case of arranging the beads, there is no distinction between the, clockwise and the anticlockwise arrangements., 1, So, the required number of ways (n 1) !., 2, ILLUSTRATIVE EXAMPLES, , EXAMPLE 1, , SOLUTION, , EXAMPLE 2, , SOLUTION, , EXAMPLE 3, SOLUTION, , In how many ways can 8 students be arranged in, (i) a line,, (ii) a circle?, (i) The number of ways in which 8 students can be arranged in a, line 8 ! (8 #7 # 6 # 5 # 4 # 3 # 2 #1) 40320., (ii) The number of ways in which 8 students can be arranged in a, circle (8 1) ! 7 ! (7 # 6 # 5 # 4 # 3 # 2 #1) 5040., If 20 persons were invited for a party, in how many ways can they and the, host be seated at a circular table?, In how many of these ways will two particular persons be seated on either, side of the host?, Clearly, there are 21 persons, to be seated round a circular table., (i) Let us fix the seat of one person, say the host. The remaining 20, persons can now be arranged in 20 ! ways., Hence, the number of ways in which these 21 persons can be, seated round a circular table = 20 !., (ii) The two particular persons can be seated on either side of, the host in 2 ways and for each way of their taking seats, the, remaining 18 persons can be seated at the circular table in 18 !, ways., the number of ways of seating 21 persons at a circular, table with two particular persons on either side of the host, (2 #18 !) ., In how many ways can a party of 4 men and 4 women be seated at a, circular table so that no two women are adjacent?, The 4 men can be seated at the circular table, M, W, W, such that there is a vacant seat between every, pair of men., M, M, The number of ways in which these 4 men can, W, W, M, be seated at the circular table = 3 ! = 6., Now, the 4 vacant seats may be occupied by 4 women in, 4, P4 4 ! 24 ways., , , the required number of ways (6 # 24) 144.
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Permutations, EXAMPLE 4, , SOLUTION, , EXAMPLE 5, SOLUTION, , EXAMPLE 6, , SOLUTION, , EXAMPLE 7, SOLUTION, , 317, , A round table conference is to be held between delegates of 20 countries. In, how many ways can they be seated if two particular delegates may wish to, sit together?, Regarding these two delegates as one, the 19 delegates can be, arranged at a circular table in 18 ! ways., These two delegates can be arranged among themselves in 2 !, i.e.,, 2 ways., , the required number of ways 2 #(18 !) ., In how many ways can 7 persons sit around a table so that all shall not, have the same neighbours in any two arrangements?, 7 persons sit at a round table in 6 ! ways., But, in clockwise and anticlockwise arrangements, each person, will have the same neighbours., So, the required number of ways (1/2)#(6 !) 360., 3 boys and 3 girls are to be seated around a table in a circle. Among them,, the boy X does not want any girl neighbour and the girl Y does not want, any boy neighbour. How many such arrangements are possible?, Fix the positions of X and Y as shown in the figure., X, B2, Now, the boys B 1, B 2 and the girls G 1, G 2 may have B1, their neighbours X and Y respectively., G1, G2, Y, The boys B 1 and B 2 may be arranged among, themselves in 2 ! ways. And, the girls G 1 and G 2 may be arranged, among themselves in 2 ! ways., Hence, the required number of arrangements (2 # 2) 4., Find the number of ways in which 8 different beads can be arranged to, form a necklace., Fixing the position of one bead, the remaining beads can be, arranged in 7 ! ways. But, there is no distinction between the, clockwise and anticlockwise arrangements., So, the required number of arrangements (1/2)#(7 !) 2520., , EXERCISE 8G, 1. In how many ways can 6 persons be arranged in, (i) a line,, (ii) a circle?, 2. There are 5 men and 5 ladies to dine at a round table. In how many ways, can they seat themselves so that no two ladies are together?, 3. In how many ways can 11 members of a committee sit at a round table so, that the secretary and the joint secretary are always the neighbours of the, president?, 4. In how many ways can 8 persons be seated at a round table so that all shall, not have the same neighbours in any two arrangements?
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318, , Senior Secondary School Mathematics for Class 11, , 5. In how many different ways can 20 different pearls be arranged to form a, necklace?, 6. In how many different ways can a garland of 16 different flowers be made?, ANSWERS (EXERCISE 8G), , 1. (i) 6 ! 720 (ii) 120, 4. 2520, , 2. (4 !)#(5 !) 2880, 1, 5. #(19 !), 2, , 3. 80640, 1, 6. #(15 !), 2, , EXERCISE 8H, Very-Short-Answer Questions, 1. If (n 1) ! 12 #[(n 1) !], find the value of n., 1 1 x, , find the value of x., 4! 5! 6!, How many 3-digit numbers are there with no digit repeated?, How many 3-digit numbers above 600 can be formed by using the digits, 2, 3, 4, 5, 6, if repetition of digits is allowed?, How many numbers divisible by 5 and lying between 4000 and 5000 can be, formed from the digits 4, 5, 6, 7, 8 , if repetition of digits is allowed?, In how many ways can the letters of the word ‘CHEESE’ be arranged?, In how many ways can the letters of the word ‘PERMUTATIONS’ be arranged, if each word starts with P and ends with S?, How many different words can be formed by using all the letters of the, word ‘ALLAHABAD’?, How many permutations of the letters of the word ‘APPLE’ are there?, How many words can be formed by the letters of the word ‘SUNDAY’?, In how many ways can 4 letters be posted in 5 letter boxes?, In how many ways can 4 women draw water from 4 taps, if no tap remains, unused?, How many 5-digit numbers can be formed by using the digits 0, 1 and 2?, In how many ways can 5 boys and 3 girls be seated in a row so that each girl, is between 2 boys?, A child has plastic toys bearing the digits 4, 4 and 5. How many 3-digit, numbers can he make using them?, In how many ways can the letters of the word ‘PENCIL’ be arranged so that, N is always next to E?, , 2. If, 3., 4., 5., 6., 7., 8., 9., 10., 11., 12., 13., 14., 15., 16., , ANSWERS (EXERCISE 8H), , 1. n 3, 6. 120, , 2. x 36, 7. 1814400, , 3. 648, 8. 7560, , 4. 25, 9. 60, , 5. 25, 10. 720
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Permutations, , 11. 625, 16. 120, , 12. 24, , 13. 162, , 319, , 14. 24, , 15. 3, , HINTS TO SOME SELECTED QUESTIONS, 1. (n 1)# n#[(n 1) !] 12 #[(n 1) !] & (n 1)# n 12 4 # 3 & n 3., 2., , 6, x, x 1 6, 1 , 1, , c1, m, &, & x c # 30m 36., 30, 5, 5, 5, 4 ! 5 #(4 !) (6 # 5)#(4 !), , 3. The hundred’s, ten’s and unit’s digits can be filled in 9, 9 and 8 ways respectively., Hence, the required number of numbers (9 # 9 # 8) 648., 4. Given digits are 2, 3, 4, 5, 6 and repetition of digits is allowed., The hundred’s place can be filled by 6 in 1 way only., Each of the ten’s and unit’s digits can be filled by any of the given five digits in 5, ways., Required number of numbers (1# 5 # 5) 25., 5. Given digits are 4, 5, 6, 7, 8 and repetition of digits is allowed., Clearly, the required number must have 4 at the thousand’s place., So, this place can be filled in 1 way., Since each number is divisible by 5, so we must have 5 at the unit’s place. So, unit’s, place can be filled in 1 way., Each of the hundred’s and ten’s places can be filled in 5 ways., the required number of numbers (1#1# 5 # 5) 25., 6. The given word contains 6 letters, out of which E is repeated 3 times and all others are, different., 6! , So, the required number of ways , 120., 3!, 7. The given word contains 12 letters. Keeping P and S at first and last positions, 10, letters remain there out of which, there are 2 T’s and the rest are different., Required number of ways , , 10 ! , 1814400., 2!, , 8. The given word contains 9 letters out of which there are 4 A’s, 2 L’s and others are all, distinct., 9!, 7560., Required number of words , (4 !)#(2 !), 9. The given word contains 5 letters, out of which there are 2 P’s and the rest are all, different., 5! , Required number of permutations , 60., 2!, 10. The given word contains 6 letters, all distinct., Required number of words 6 ! 720., 11. Every letter can be posted in 5 ways., Required number of ways 5 4., 12. 1st tap can be used by any of the 4 women in 4 ways., 2nd tap can be used by any of the remaining 3 women.
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320, , Senior Secondary School Mathematics for Class 11, 3rd tap can be used by any of the remaining 2 women., 4th tap can be used by the remaining 1 woman., Required number of ways 4 # 3 # 2 #1 24., , 13. The ten-thousand’s place can be filled by 1 or 2, i.e., in 2 ways., Each of the other 4 digits can be filled in 3 ways., Required number of numbers (2 # 3 # 3 # 3 # 3) 162., 14. Let us fix 5 boys, keeping a space between every 2 boys, as shown below:, B = B = B = B = B., Required number of ways 4P3 (4 # 3 # 2) 24., 15. There are 3 digits out of which 4 is repeated 2 times and 5 occurs only once., Required number of numbers , , 3 ! 3 # 2 #1 , 3., 2, 2!, , 16. Keeping EN as 1 letter, this letter and 4 other letters of the given word can be, arranged among themselves in 5 ! 120 ways., , KEY FACTS AND FORMULAE, 1. (i) n ! n #(n 1)#(n 2)#…# 3 # 2 #1, (ii) n ! n #[(n 1) !], (iii) 0 ! 1, 2. Fundamental Principle of Multiplication, If there are two operations such that one of them can be performed in m, ways and when it has been performed in anyone of these m ways, the, second operation can be performed in n ways, then the two operations in, succession can be performed in (m # n) ways., 3. Fundamental Principle of Addition, If there are two operations such that they can be performed independently, in m and n ways respectively, then either of the two operations can be, performed in (m n) ways., 4. Permutations The different arrangements which can be made out of a given, number of things by taking some or all at a time, are called permutations., 5. (i) Number of permutations of n different objects taking r at a time,, where 0 r # n and the objects do not repeat, is given by, n! ·, n, Pr n (n 1) (n 2) … (n r 1) , (n r) !, (ii) nPn n !., (iii) Let there be n things of which m are alike of one kind and the remaining, (n m) are alike of second kind, then, n!, ·, total number of permutations , (m !)#[(n m) !]
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Permutations, , 321, , (iv) The number of permutations of n objects, where p1 objects are of one, kind, p 2 are of second kind, …, p k are of kth kind and the rest, if any,, are distinct, is, n!, ·, (p1 !)#(p 2 !)#…#(p k !), (v) Number of permutations of n different objects, taken r at a time when, each may be repeated any number of times in each arrangement, is n r., 6. The number of circular permutations of n different objects is (n 1) !., 7. The number of ways in which n persons can be seated round a table is, (n 1) !., 8. The number of ways in which n different beads can be arranged to form a, 1, necklace is (n 1) !., 2,
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322, , Senior Secondary School Mathematics for Class 11, , 9, , Combinations, , Each of the different groups or selections that can be made out of a, given number of things by taking some or all of them at a time, irrespective of their, arrangements, is called a combination., , COMBINATIONS, , ILLUSTRATIVE EXAMPLES, EXAMPLE 1, , All combinations of three persons A, B, C taken two at a time are, AB, AC, BC., Clearly, AB and BA represent the same combination but they, represent different permutations., , EXAMPLE 2, , All combinations of four persons A, B, C, D taken two at a time are, AB, AC, AD, BC, BD, CD., , EXAMPLE 3, , All combinations of four persons A, B, C, D taken three at a time are, ABC, ABD, ACD, BCD., , DIFFERENCE BETWEEN A PERMUTATION AND A COMBINATION, , In a combination, only a group is made and the order in which the objects are, arranged is immaterial., On the other hand, in a permutation, not only the group is formed, but also, an arrangement in a definite order is considered., In general, we use the word arrangements for permutations and the, word selections for combinations., EXAMPLE 4, (i) AB and BA are two different permutations but each represents, the same combination., (ii) ABC, ACB, BAC, BCA, CAB, CBA are six different permutations, but each one represents the same combination, namely ABC., REMARK, , NOTATION, , The number of all combinations of n things, taken r at a time, is denoted, by nCr, or C(n, r), where n and r are integers such that n 0, r $ 0 and, n $ r., , Combinations of n Different Objects, THEOREM 1, , The number of all combinations of n distinct objects, taken r at a time, is, given by, n!, n, $, Cr , (r !)#(n r) !, 322
Page 334 : Combinations, PROOF, , 323, , Let the number of all combinations of n objects, taken r at a time, be x., Then, nCr x., Now, each combination contains r objects, which may be arranged, among themselves in r ! ways., Thus, each combination gives rise to r ! permutations., x combinations will give rise to x #(r !) permutations., So, the number of permutations of n things, taken r at a time, is x #(r !) ., Consequently, nPr x #(r !) nCr #(r !) ., n, , , , Cr , , n, , Pr, n!, n!, n, , <a Pr F ·, r ! (r !)#(n r) !, (n r) !, , n, Corollary 1 Prove that C r , , PROOF, , n(n 1) (n 2) … to r factors, ·, r!, , We have, n!, (r !)#(n r) !, n(n 1) (n 2) … (n r 1) (n r) (n r 1) … 3 · 2 · 1, , (r !)#[(n r) (n r 1) … 3 · 2 · 1], n(n 1) (n 2) … (n r 1), , r!, n(n 1) (n 2) … to r factors, , ·, r!, n (n 1) (n 2) … to r factors, ·, Hence, nCr , r!, Cr , , n, , 11, , EXAMPLE 1, , Evaluate, , SOLUTION, , 11#10 # 9 # 8 11#10 # 9 # 8 , 11, C4 , 330., 4!, 4 # 3 # 2 #1, , C4 ., , n, n, Corollary 2 Prove that C n C0 1., , PROOF, , We have, Cr , , n, , n!, ·, (r !)#(n r) !, , … (i), , Putting r n in (i), we get, Cr , , n, , n!, 1 1, (n !)#(n n) ! 0 !, , [a 0 ! 1]., , Putting r 0 in (i), we get, C0 , , n, , n!, n! 1, (0 !)#(n 0) ! 1#(n !), , Hence, nCn nC0 1., , [a 0 ! 1].
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324, , Senior Secondary School Mathematics for Class 11, n, , n, Corollary 3 Prove that C r , , We have, , PROOF, , Pr, ·, r!, n, , P, n!, 1 ·( n! 2 r ·, r!, (r !)#(n r) ! (r !) (n r) !, , Cr , , n, , EXAMPLE 2, , If nPr 720 and nCr 120, find r., , SOLUTION, , We have, n, , Cr , , n, , Pr, 720 , 720 , &, 120 & r ! , 6 (3 # 2 #1) 3 !, 120, r!, r!, & r 3., , Hence, r 3., SUMMARY OF THE ABOVE RESULTS, , Cr , , n(n 1) (n 2) … (n r 1), n!, , , r!, (r !)#(n r) !, , Cr , , n(n 1) (n 2) … to r factors, r!, , (i), , n, , (ii), , n, , (iii), , n, , (iv), , n, , Cn nC0 1, n, , Cr , , Pr, r!, , n, , PROPERTIES OF C r, , For 0# r # n, prove that nCr nCn r ., We have, n!, n!, n, , nCr ., Cn r , (n r) !#{n (n r)} ! (n r) !#(r !), Hence, nCr nCn r ., , THEOREM 1, PROOF, , 100, , EXAMPLE 1, , Evaluate, , SOLUTION, , Using the result nCr nCn r, we get, , C 98 ., , C 98 100C(100 98) 100C 2 , , 100, , THEOREM 2, PROOF, , 100 # 99 , 4950., 2, , If nC x nC y and x ! y then prove that x y n., , We have, n, C x nC y nCn y, & x y or x n y, & x ny, & x y n., Hence, nC x nC y and x ! y then x y n., , [a nCr nCn r], [a x ! y (given)]
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Combinations, EXAMPLE 2, , 325, , (i) If nC18 nC12 then find the value of 32Cn ., (ii) If, , 10, Cr Cr 4 then find the value of Cr ., 5, , 10, , (i) C18 C12 & n (18 12) 30., n, , SOLUTION, , n, , 32 # 31 , 496., 2!, 10, 10, (ii) Cr Cr 4 & r (r 4) 10 & r 3., 5# 4 , 5Cr 5C 3 5C 5 3 5C 2 , 10., 2!, Cn 32C 30 32C 32 30 32C 2 , , , , 32, , Let 1# r # n. Then, prove that, , n, Cr nCr 1 n 1Cr . [Pascal‘s rule], We have, n, Cr nCr 1, n!, n!, , , r ! · (n r) ! (r 1) ! · {n (r 1)} !, n!, n!, , , r ! · (n r) ! (r 1) ! · (n r 1) !, (n !) · (n r 1), n!· r, , , r ! · (n r 1) ! r ! · (n r 1) !, n!, (, 2 · {n r 1 r}, r ! · (n r 1) !, (n 1) · (n !), , (r !) · (n r 1) !, (n 1) !, , (r !) (n 1 r) !, , THEOREM 3, , PROOF, , n 1Cr ., , , Hence, nCr nCr 1 n 1Cr ., THEOREM 4, PROOF, , , , If 1# r # n, prove that n # n 1Cr 1 (n r 1)# nCr 1 ., , We have, n#, , n1, , (n 1) !, (r 1) !#[(n 1) (r 1)] !, n #(n 1) !, , (r 1) !#(n r) !, (n !)#(n r 1), , , (r 1) !#(n r) !#(n r 1), [multiplying num. and denom. by (n r 1)], n!, ( n r 1) #, (r 1) !#(n r 1) !, (n r 1)# nCr 1 ., , Cr 1 n#, , , , Hence, n # n 1Cr 1 (n r 1)# nCr 1 .
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326, REMARK, , Senior Secondary School Mathematics for Class 11, , The above result may be written as, n, Cr 1, n, , $, n1, Cr 1 (n r 1), , THEOREM 5, , If n and r are positive integers such that 1# r # n then prove that:, n, n, Cr, Cr, , n, n r 1, (i) n, (ii), n1, r, r, C, C, r1, , r 1, , (i) We know that, , PROOF, , n!, n!, ·, and nCr 1 , (r !)#(n r) !, (r 1) !#(n r 1) !, , Cr , , n, , n, , , , Cr, , , , , , n, , Cr 1, , (r 1) !#(n r 1) !, n!, #, , n!, (r !)#(n r) !, , (r 1) !#(n r 1)#(n r) !, r #(r 1) !#(n r) !, , , n r 1·, r, , , n, , Cr, , Hence,, , n, , Cr 1, , , n r 1$, r, , (ii) We have, Cr , , n, , n!, (r !)#(n r) !, , Cr 1 , , and, , n1, , ( n 1) !, (n 1) !, , ·, (r 1) !#{(n 1) (r 1)} ! (r 1) !#(n r) !, , n, , Cr, , , , n1, , Cr 1, , , , (r 1) !#(n r) ! n, n!, ·, #, r, (r !)#(n r) !, (n 1) !, , n, , Cr, , Hence,, , n1, , Cr 1, , n·, r, , SUMMARY OF THE ABOVE RESULTS, , Cr Cn r, , (i), , n, , (ii), , n, , (iii), , n, , n, , C x nC y and x ! y & x y n, , , Cr nCr 1 n 1Cr, n, , Cr 1, , (iv), , n1, , (vi), , n, , Cr 1, , n, , Cr, , Cr 1, , , , n, (n r 1), , , r 1, nr, , [Pascal‘s rule], n, , (v), , Cr, , n1, , Cr 1, , n, r
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Combinations, , 327, , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Evaluate:, (i) 10C 3, We have, , (ii) 11C8, , (iii) 50C 48, , (iv) 63C63, , 10 # 9 # 8 10 # 9 # 8 , 120., 3 # 2 #1, 3!, 11#10 # 9 11#10 # 9 , (ii) 11C8 11C(11 8) 11C 3 , 165, 3 # 2 #1, 3!, [a nCr nCn r]., 50 # 49 , (iii) 50C 48 50C(50 48) 50C 2 , 1225., 2, 63, n, (iv) C63 1, [a Cn 1]., (i), , EXAMPLE 2, SOLUTION, , C3 , , 10, , If nC8 nC6, find nC 3 ., We know that nC p nCq and p ! q & n p q., , , , C8 nC6 & n (8 6) 14., 14 #13 #12 14 #13 #12 , n, C 3 14C 3 , 364., 3 # 2 #1, 3!, n, , EXAMPLE 3, , If nCr 1 : nCr : nCr 1 3 : 4 : 5, find the values of n and r., , SOLUTION, , We have,, , n, , Cr, , n, , Cr 1, , , r 1$, nr, , … (i), , n, , Cr, , 4 in (i), we get, Cr 1 5, r1 4, & 5r 5 4n 4r & 4n 9r 5., nr 5, Replacing r by (r 1) in (i), we get, , Putting, , n, , … (ii), , n, , Cr 1, , 3, r, r, & , nr1, n r 1 4, & 3n 3r 3 4r & 3n 7r 3., On solving (ii) and (iii), we get n 62 and r 27., n, , Cr, , , , EXAMPLE 4, , If 2nC 3 : nC 3 11 : 1, find n., , SOLUTION, , We have, , 2n(2n 1) (2n 2) 2n(n 1) (2n 1), , 3, 3!, , , , , n, (, n, 1, ), (, n, 2, ), n, (, n, 1, ), (, n, 2), , ·, and nC 3 , 6, 3!, 2n, C 3 2n ( n 1) ( 2n 1), 4(2n 1), 6, , , ·, , #, n, 3, n(n 1) (n 2), (n 2), C3, C3 , , 2n, , 2n, , Now,, , C3, , n, , C3, , 4(2n 1) 11, 11 &, , 1, 1, (n 2), , … (iii)
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328, , Senior Secondary School Mathematics for Class 11, , & 8n 4 11n 22 & 3n 18 & n 6., , Hence, n 6., EXAMPLE 5, SOLUTION, , EXAMPLE 6, SOLUTION, , EXAMPLE 7, SOLUTION, , Prove that the product of r consecutive positive integers is divisible by r !., Let the r consecutive positive integers be (n 1), (n 2), …, (n r) ., Then, their product (n 1) (n 2) (n 3) … (n r), (n !) (n 1) (n 2) (n 3) … (n r), , n!, (n r) !, (n r) !, , (r !) ·, n!, (r !)#[(n r) r] !, nr, (r !)# Cr, which is divisible by r !,, , since n rCr is a positive integer., If nPr nPr 1 and nCr nCr 1, find n and r., n! , n!, (n r) ! (n r 1) !, & (n r) ! (n r 1) !, & (n r)#(n r 1) ! (n r 1) !, & n r 1., n, Cr, n, n, 1, Cr Cr 1 & n, Cr 1, RS, VW, n, Cr, SS, nr1 , n r 1WW, , &, 1, a, S n, r, r WW, S, Cr 1, T, X, & n 2r 1., Solving (i) and (ii), we get n 3 and r 2., n, Pr Pr 1 &, , n, , 2 n #[1 $ 3 $ 5 … (2n 1)], ·, n!, (2n) !, (2n) !, 2n, , Cn , , (n !) (2n n) ! (n !) 2, Prove that, , , , , , , , Cn , , 2n, , (2n) (2n 1) (2n 2) … 4 · 3 · 2 · 1, (n !) 2, [(2n) (2n 2) (2n 4) … 4 · 2]#[(2n 1) (2n 3) … 5 · 3 · 1], (n !) 2, 2 n [n (n 1) (n 2) … 2 · 1]#[(2n 1) (2n 3) … 5 · 3 · 1], (n !) 2, 2 #(n !)#[1 · 3 · 5 … (2n 3) (2n 1)], n, , (n !) 2, n, 2 #[1 · 3 · 5 … (2n 3) (2n 1)], ·, (n !), , … (i), , … (ii)
Page 340 : Combinations, , 329, , 2 n #{1# 3 # 5 #…#(2n 1)}, $, n!, , Hence, 2nCn , EXAMPLE 8, , Prove that, 4n, C 2n : 2nCn {1# 3 # 5 #…# (4n 1)} : {1# 3 # 5 #…#(2n 1)} 2., , SOLUTION, , We have, 4nC, (4n) !, (n !)#(n !), 2n, , #, 2nC, (2n) !#(2n) !, (2n) !, n, , , , (4n) !#(n !) 2, (2n) !#[(2n) !] 2, [1 · 2 · 3 · 4 … (4n 1) (4n)]#(n !) 2, (2n) !#[1 · 2 · 3 · 4 … (2n 2) (2n 1) (2n)] 2, , , , , [1 · 3 · 5 … (4n 1)]#[2 · 4 · 6 … (4n)]#(n !) 2, (2n) !#[1 · 3 · 5 … (2n 1)] 2 #[2 · 4 · 6 … (2n 2) (2n)] 2, [1 · 3 · 5 … (4n 1)]# 2 2n #[1 · 2 · 3 … (2n)]#(n !) 2, , (2n) !#[1 · 3 · 5 … (2n 1)] 2 # 2 2n #(n !) 2, [1 · 3 · 5 … (4n 1)], , ·, [1 · 3 · 5 … (2n 1)] 2, C 2n : 2nCn [1 · 3 · 5 … (4n 1)} : {1 · 3 · 5 … (2n 1)} 2., , , , 4n, , , , , EXAMPLE 9, , If n 2C8 : n 2P4 57 : 16, find n., , SOLUTION, , We have, , (n 2) !, (n 2) !, , , , (8 !) (n 2 8) ! (8 !) (n 6) !, (n 2) !, (n 2) !, , , and n 2P4 , (n 2 4 ) ! (n 6) !, n2, , C8 , , n2, , C8, , <a Cr , n, , n!, F, (r !) · (n r) !, , <a Pr , n, , n! ·, F, (n r) !, , 57, P4 16, (n 2) !, (n 6) ! 57, , #, &, (8 !) (n 6) ! (n 2) ! 16, (n 2) (n 1) n (n 1) · (n 2) !, &, (n 2) !, b 57 # 8 #7 # 6 # 5 # 4 # 3 # 2 #1l, 16, &, (n 2) (n 1) n(n 1) 21# 20 #19 #18, &, n 1 18 & n 19., Hence, n 19., Now,, , n2
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330, , Senior Secondary School Mathematics for Class 11, , EXERCISE 9A, 1. Evaluate:, (i) 20C 4, (iv), , (ii), , 71, , (v), , C71, , 16, , (iii), , C13, , 90, , C88, , 6, , n1, , (vi), , Cn, , / 6Cr, , r1, , 2. Verify that:, (i) 15C8 15C 9 15C6 15C7 0, 3. (i) If nC7 nC 5, find n., (ii) If nC14 nC16, find nC 28 ., (iii) If nC16 nC14, find nC 27 ., , (ii), , C 4 10C 3 11C 4, , 10, , (i) If 20Cr 20Cr 6, find r., (ii) If 18Cr 18Cr 2, find rC 5 ., 5. If nCr 1 nC 3r, find r., 4., , 6. If 2nC 3 : nC 3 12 : 1, find n., 7. If 15Cr : 15Cr 1 11 : 5, find r., 8. If nPr 840 and nCr 35, find the value of r., 9. If nCr 1 36, nCr 84 and nCr 1 126, find n and r., , , , 10. If n 1Cr 1 : nCr 11 : 6 and nCr : n 1Cr 1 6 : 3, find n and r., 11. How many different teams of 11 players can be chosen from 15 players?, 12. If there are 12 persons in a party and if each two of them shake hands with, each other, how many handshakes are possible?, 13. How many chords can be drawn through 21 points on a circle?, 14. From a class of 25 students, 4 are to be chosen for a competition. In how, many ways can this be done?, , ANSWERS (EXERCISE 9A), , 1. (i) 4845, 3. (i) n 12, , (ii) 560, (ii) 435, , (iv) 1, (v) n 1, , 4. (i) r 7 (ii) 56, , (iii) 4005, (iii) 4060, , 1 , (n 1) 6. n 5, 4, 10. n 10, r 5 11. 1365, 5. r , , 7. r 5, 12. 66, , 8. r 4, 13. 210, , (vi) 63, 9. n 9, r 3, 14. 12650, , HINTS TO SOME SELECTED QUESTIONS, 2n(2n 1) (2n 2), 3 # 2 #1, 12, 6., #, 3 # 2 #1, n(n 1) (n 2) 1, & 4 (2n 1) 12 (n 2) & 2n 1 3n 6 & n 5., n, , 7., , Cr, , n, , , n r 1 &, r, , Cr 1, Find r., , 15, , Cr, , 15, , Cr 1, , , , , 15 r 1 16 r & 16 r 11 ·, r, r, r, 5
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Combinations, , 331, , n, , Pr, ·, r!, , 8. Use nCr , n, , 9., , n, , Cr, , r, r 1·, and n, nr1, Cr, Cr 1 n r, r, 36 3 & 3n 3r 3 7r & 3n 10r 3, , n r 1 84 7, 84 2, r1, and , & 2n 2r 3r 3 & 2n 5r 3., n r 126 3, Cr 1, n, , , , n, , … (i), … (ii), , n1, , C , , n and n r 1 n 1 ·, r1, Cr, Cr 1 r, n62, n 1 11, , & 6n 6 11r 11 & 6n 11r 5., r 3 1 and r 1, 6, Solve n 2r 0 and 6n 11r 5., , 10. Use, , Cr, , n1, , 11. Required number of teams 15C11 15C(15 11) 15C 4 ., 12. Number of handshakes 12C2 ., 13. Number of chords 21C2 ., 14. Required number of ways 25C 4 ., , PRACTICAL PROBLEMS ON COMBINATIONS, SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , In how many ways can a cricket eleven be chosen out of a batch of 15, players, if, (i) there is no restriction on the selection?, (ii) a particular player is always chosen?, (iii) a particular player is never chosen?, (i) Total number of ways of selecting 11 players out of 15, 15C11 15C15 11 15C 4 15 #14 #13 #12 1365., 4 # 3 # 2 #1, (ii) When a particular player is always chosen, then we will have, to choose 10 players out of remaining 14., required number of ways, 14C10 14C14 10 14C 4 14 #13 #12 #11 1001., 4 # 3 # 2 #1, (iii) When a particular player is never chosen, then we will have to, choose 11 players out of remaining 14., required number of ways, 14C11 14C14 11 14C 3 14 #13 #12 364., 3 # 2 #1, , EXAMPLE 2, , In how many ways can a committee of 5 members be selected from 6 men, and 5 ladies, consisting of 3 men and 2 ladies?
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332, , Senior Secondary School Mathematics for Class 11, , SOLUTION, , We have to select (3 men out of 6) and (2 ladies out of 5)., , required number of ways, ( 6C 3 # 5C 2) b 6 # 5 # 4 # 5 # 4 l (20 #10) 200., 3 # 2 #1 2 #1, , EXAMPLE 3, , Out of 5 men and 2 women, a committee of 3 is to be formed. In how many, ways can it be formed if at least one woman is to be included?, The committee can be formed by choosing, (i) 1 woman and 2 men;, or (ii) 2 women and 1 man., Now, the number of ways of choosing (1 woman out of 2) and, (2 men out of 5), (2C1 # 5C 2) b2 # 5 # 4 l (2 #10) 20., 2 #1, And, the number of ways of choosing (2 women out of 2) and, (1 man out of 5), (2C 2 # 5C1) (1# 5) 5., Hence, the total number of ways of forming the committee, (20 5) 25., , SOLUTION, , EXAMPLE 4, , SOLUTION, , A committee of 5 is to be formed out of 6 men and 4 ladies. In how many, ways can this be done, when, (a) at least 2 ladies are included;, (b) at most 2 ladies are included?, (a) We have to make a selection of, (i) (2 ladies out of 4) and (3 men out of 6), or (ii) (3 ladies out of 4) and (2 men out of 6), or (iii) (4 ladies out of 4) and (1 man out of 6)., The number of ways of these selections are:, C 2 # 6C 3 6 # 20 120., , Case I, , 4, , Case II, , 4, , C 3 # 6C 2 4 #15 60., , C 4 # 6C1 1# 6 6., Hence, the required number of ways (120 60 6) 186., (b) We have to make a selection of, (i) (1 lady out of 4) and (4 men out of 6), or (ii) (2 ladies out of 4) and (3 men out of 6)., The number of ways of these selections are:, 4, Case I, C1 # 6C 4 4 #15 60., Case III, , 4, , C 2 # 6C 3 6 # 20 120., Hence, the required number of ways (60 120) 180., Case II, , EXAMPLE 5, , 4, , An examination paper containing 12 questions consists of two parts,, A and B. Part A contains 7 questions and part B contains 5 questions.
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Combinations, , SOLUTION, , 333, , A candidate is required to attempt 8 questions, selecting at least 3 from, each part. In how many ways can the candidate select the questions?, Clearly, the candidate may select the questions as under:, (i) (3 out of 7 from A) and (5 out of 5 from B), or (ii) (4 out of 7 from A) and (4 out of 5 from B), or (iii) (5 out of 7 from A) and (3 out of 5 from B)., The number of ways of these selections are:, C 3 # 5C 5 35 #1 35., , Case I, , 7, , Case II, , 7, , C 4 # 5C 4 35 # 5 175., , C 5 # 5C 3 21#10 210., Hence, the required number of ways (35 175 210) 420., Case III, , EXAMPLE 6, , SOLUTION, , EXAMPLE 7, SOLUTION, , 7, , For the post of 5 teachers, there are 23 applicants. 2 posts are reserved for, SC candidates and there are 7 SC candidates among the applicants. In, how many ways can the selection be made?, Clearly, there are 7 SC candidates and 16 non-SC candidates., So, we have to make a selection of (2 out of 7) and (3 out of 16)., Hence, the number of ways of making the selection, 7C 2 # 16C 3 b7 # 6 # 16 #15 #14 l 11760., 2, 3 # 2 #1, How many diagonals are there in a polygon of n sides?, A polygon of n sides has n vertices., By joining any two of these vertices, we obtain either a side or a, diagonal of the polygon., Number of all straight lines obtained by joining 2 vertices at a time, nC 2 1 n (n 1) ., 2, These straight lines include n sides of the polygon., Hence, the number of diagonals of the polygon, ;1 n(n 1) nE 1 n(n 3) ., 2, 2, , EXAMPLE 8, , SOLUTION, , There are 10 points in a plane, no three of which are in the same straight, line, except 4 points, which are collinear., Find (i) the number of lines obtained from the pairs of these points;, (ii) the number of triangles that can be formed with vertices as these, points., (i) Number of lines formed by joining the 10 points, taking 2 at, a time, 10C 2 b10 # 9 l 45., 2 #1, Number of lines formed by joining the 4 points, taking 2 at, a time
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334, , Senior Secondary School Mathematics for Class 11, , 4C 2 b 4 # 3 l 6., 2 #1, But, 4 collinear points, when joined pairwise, give 1 line., Hence, the required number of straight lines, (45 6 1) 40., (ii) Number of triangles formed by 10 points, taking 3 at a time, 10C 3 10 # 9 # 8 120., 3 # 2 #1, Number of triangles formed by 4 points, taking 3 at a time, 4C 3 4C 4 3 4C1 4., But, these 4 points being collinear, so no triangle is formed., Hence, the required number of triangles (120 4) 116., EXAMPLE 9, SOLUTION, , EXAMPLE 10, SOLUTION, , A box contains 6 red and 5 white balls. In how many ways can 6 balls be, selected so that there are at least two balls of each colour?, We may select, (i) (2 red balls out of 6) and (4 white balls out of 5), or (ii) (3 red balls out of 6) and (3 white balls out of 5), or (iii) (4 red balls out of 6) and (2 white balls out of 5)., The number of ways of these selections are:, 6# 5, 6, 5, 6, 5, Case I ( C 2 # C 4) ( C 2 # C1) b, # 5l 75., 2, 6# 5# 4 5# 4 , 6, 5, 6, 5, Case II ( C 3 # C 3) ( C 3 # C 2) b, #, l 200., 3 # 2 #1 2 #1, 6# 5 5# 4 , 6, 5, 6, 5, Case III ( C 4 # C 2) ( C 2 # C 2) b, #, l 150., 2 #1 2 #1, Hence, the required number of ways (75 200 150) 425., How many committees of 5 persons with a chairperson can be selected, from 12 persons?, Number of ways of selecting a chairperson out of 12 persons, 12C1 12., Number of ways of selecting 4 persons out of remaining 11, 11C 4 11#10 # 9 # 8 330., 4 # 3 # 2 #1, Hence, the required number of ways (12 # 330) 3960., , EXAMPLE 11, , SOLUTION, , A box contains 3 black, 2 white and 4 red balls. In how many ways can, 3 balls be drawn from the box, if at least one black ball is to be included in, the draw?, In all, we have 3 black and 6 non-black balls., We may select:, (i) (1 black ball out of 3) and (2 non-black balls out of 6), or (ii) (2 black balls out of 3) and (1 non-black ball out of 6), or (iii) (3 black balls out of 3).
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Combinations, , 335, , The number of ways of these selections are:, 6# 5 , 3, 6, Case I ( C1 # C 2) b3 #, l 45., 2 #1, 3, 6, 3, 6, Case II ( C 2 # C1) ( C1 # C1) (3 # 6) 18., C 3 3 ! (3 # 2 #1) 6., Hence, the required number of ways (45 18 6) 69., Case III, , EXAMPLE 12, , SOLUTION, , 3, , A group consists of 7 boys and 4 girls. In how many ways can a team of, 5 members be selected, if the team has, (i) no girls?, (ii) at least 1 boy and 1 girl?, (iii) at least 3 girls?, (i) When no girl is taken:, Then, we have to select 5 boys out of 7., required number of ways, 7C 5 7C7 5 7C 2 7 # 6 21., 2 #1, (ii) When at least 1 boy and 1 girl are taken:, In this case, we may choose:, (1 boy and 4 girls) or (2 boys and 3 girls), or (3 boys and 2 girls) or (4 boys and 1 girl)., required number of ways, (7C1 # 4C 4) (7C 2 # 4C 3) (7C 3 # 4C 2) (7C 4 # 4C1), (7 #1) (7C 2 # 4C1) (7C 3 # 4C 2) (7C 3 # 4C1), 7 b7 # 6 # 4l b7 # 6 # 5 # 4 # 3 l b7 # 6 # 5 # 4l, 3 # 2 #1 2 #1, 3 # 2 #1, 2 #1, (7 84 210 140) 441., (iii) When at least 3 girls are taken:, Then, we may choose:, (3 girls and 2 boys) or (4 girls and 1 boy)., required number of ways, (4C 3 # 7C 2) (4C 4 # 7C1), (4C1 # 7C 2) (1# 7C1) b4 # 7 # 6 l (1#7), 2 #1, (84 7) 91., , EXAMPLE 13, , A bag contains 6 white marbles and 5 red marbles. Find the number of, ways in which 4 marbles can be drawn from the bag, if, (a) they can be of any colour;, (b) 2 must be white and 2 red;, (c) they must all be of the same colour.
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336, SOLUTION, , Senior Secondary School Mathematics for Class 11, , (a) Drawing 4 marbles of any colour:, Total number of marbles (6 5) 11., In this case, we have to draw 4 marbles out of 11., required number of ways, 11C 4 11#10 # 9 # 8 330., 4 # 3 # 2 #1, (b) Drawing 2 white and 2 red marbles:, We may draw:, (2 white marbles out of 6) and (2 red marbles out of 5)., required number of ways, (6C 2 # 5C 2) b 6 # 5 # 5 # 4 l (15 #10) 150., 2 #1 2 #1, (c) Drawing 4 marbles of the same colour:, We may draw:, (4 white marbles out of 6) or (4 red marbles out of 5) [Note it], required number of ways, (6C 4 5C 4) (6C 2 5C1) b 6 # 5 5l 20., 2 #1, , EXAMPLE 14, SOLUTION, , We wish to select 6 persons from 8, but if the person A is chosen, then B, must be chosen. In how many ways can the selection be made?, We have two cases: (either A is chosen) or (A is not chosen)., Case I, When A is chosen:, In this case A is chosen and B is chosen., So, we must choose 4 persons out of remaining 6., , , required number of ways 6C 4 6C 2 , , 6# 5 , 15., 2 #1, , When A is not chosen:, In this case, we have to choose 6 persons out of 7., required number of ways 7C6 7C1 7., Combining both the cases, we have:, total number of ways (15 7) 22., Case II, , EXAMPLE 15, , Find the number of ways of choosing 4 cards from a pack of 52 playing, cards. In how many of these, (i) 4 cards are of the same suit?, (ii) 4 cards belong to four different suits?, (iii) 4 cards are face cards?, (iv) 2 are red cards and 2 are black cards?, (v) cards are of the same colour?
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Combinations, SOLUTION, , 337, , Number of ways of choosing 4 cards out of 52, 52C 4 52 # 51# 50 # 49 270725., 4 # 3 # 2 #1, (i) Choosing 4 cards of the same suit:, We have to draw:, (4 cards from 13 cards of diamond), or (4 cards from 13 cards of spade), or (4 cards from 13 cards of club), or (4 cards from 13 cards of heart)., required number of ways, (13C 4 13C 4 13C 4 13C 4) 4 # 13C 4, 4 #13 #12 #11#10 2860., 4# 3# 2# 1, (ii) Choosing 4 cards from 4 different suits:, We have to draw:, (1 card from 13 cards of diamond), and (1 card from 13 cards of spade), and (1 card from 13 cards of club), and (1 card from 13 cards of heart)., required number of ways, (13C1 # 13C1 # 13C1 # 13C1) (13 #13 #13 #13) (13) 4., (iii) Choosing 4 cards out of 12 face cards:, We have to choose 4 cards out of 12 face cards., required number of ways, 12C 4 12 #11#10 # 9 495., 4 # 3 # 2 #1, (iv) Choosing 2 red cards and 2 black cards:, We have to choose 2 red cards out of 26 and 2 black cards out, of 26., required number of ways, (26C 2 # 26C 2) b 26 # 25 l (325 # 325) 105625., 2, 2, , (v) Choosing all the 4 cards of the same colour:, We have to choose 4 cards out of 26 black cards or 4 cards out, of 26 red cards., required number of ways, (26C 4 26C 4) (2 # 26C 4) b2 # 26 # 25 # 24 # 23 l, 4 # 3 # 2 #1, 29900., EXAMPLE 16, , From a class of 25 students, 10 are to be chosen for an excursion party., There are 3 students who decide that either all of them will join or none of, them will join. In how many ways can they be chosen?
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338, SOLUTION, , Senior Secondary School Mathematics for Class 11, , Either 3 particular students join or none of them joins., When 3 particular students join the party:, In this case, we have to choose 7 students out of 22., This can be done in 22C7 ways., Case II When 3 particular students do not join the party:, In this case, we have to choose 10 students out of 22., This can be done in 22C10 ways., Hence, the required number of ways (22C7 22C10) 817190., Case I, , EXAMPLE 17, , SOLUTION, , EXAMPLE 18, SOLUTION, , A boy has 3 library tickets and 8 books of his interest in the library. Of these, 8, he does not want to borrow Mathematics Part II, unless Mathematics, Part I is also borrowed. In how many ways can he choose the 3 books to be, borrowed?, Clearly, the boy may borrow Mathematics Part II or he may not, borrow it., Case I, When the boy borrows Mathematics Part II:, In this case, he borrows Mathematics Part I also., Since he has 3 library tickets in all,, he may now borrow 1 book out of the remaining 6 books., This can be done in 6C1 6 ways., Case II When the boy does not borrow Mathematics Part II:, Then, he may borrow 3 books out of the remaining 7 books., 7 #6# 5 , This can be done in 7C 3 , 35 ways., 3 # 2 #1, Hence, the required number of ways (6 35) 41., A convex polygon has 44 diagonals. Find the number of its sides., Let the given convex polygon have n sides., Then, the number of its diagonals (nC 2 n) ., n(n 1), n 44, Now, nC 2 n 44 &, 2, & n 2 3n 88 0 & (n 11) (n 8) 0, & n 11 [a n ! 8]., Hence, the given polygon has 11 sides., , EXAMPLE 19, SOLUTION, , If m parallel lines in a plane are intersected by a family of n parallel lines,, find the number of parallelograms formed., A parallelogram is formed by choosing two straight lines from, the set of m parallel lines and two straight lines from the set of n, parallel lines., Number of ways of choosing 2 lines out of m parallel lines mC 2 ., Number of ways of choosing 2 lines out of n parallel lines nC 2 .
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Combinations, , 339, , Hence, the required number of parallelograms formed, m(m 1) n(n 1) mn(m 1) (n 1), (mC 2 # nC 2) , , #, $, 2, 2, 4, EXAMPLE 20, SOLUTION, , In an examination, a candidate has to pass in each of the 5 subjects. In how, many ways can he fail?, The candidate can fail by failing in 1 or 2 or 3 or 4 or 5 subjects out, of 5 in each case., , the total number of ways in which he can fail, 5C1 5C 2 5C 3 5C 4 5C 5, 5C1 5C 2 5C 2 5C1 5C 5, (5 10 10 5 1) 31., , EXAMPLE 21, SOLUTION, , [a nCr nCn r], , Determine the number of 5-card combinations out of a deck of 52 cards if, there is exactly one ace in each combination., Number of ways of selecting 1 ace out of 4 aces 4C1 4., Number of ways of selecting 4 cards out of remaining 48, 48C 4 48 # 47 # 46 # 45 194580., 4 # 3 # 2 #1, Hence, the required number of combinations, (194580 # 4) 778320., , EXAMPLE 22, , SOLUTION, , EXAMPLE 23, , SOLUTION, , In how many ways can one select a cricket team of eleven from 17 players, in which only 5 players can bowl, if each cricket team of 11 must include, exactly 4 bowlers?, Number of ways of selecting 4 bowlers out of 5, 5C 4 5C(5 4) 5C1 5., Number of ways of selecting 7 batsmen out of 12, 12C7 12C(12 7) 12C 5 12 #11#10 # 9 # 8 792., 5 # 4 # 3 # 2 #1, Hence, the number of ways of selecting the team (5 #792) 3960., A committee of 12 persons is to be formed from 9 women and 8 men. In, how many ways can this be done if at least 5 women have to be included, in a committee? In how many of these committees are (a) the women in, majority? (b) the men in majority?, There are 9 women and 8 men in all. In order to form a committee, consisting of at least 5 women, we may choose, (i) 5 women and 7 men or (ii) 6 women and 6 men, or (iii) 7 women and 5 men or (iv) 8 women and 4 men, or (v) 9 women and 3 men., Hence, the total number of ways of forming the committee, (9C 5 # 8C7) (9C6 # 8C6) (9C7 # 8C 5) (9C8 # 8C 4) (9C 9 # 8C 3)
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340, , Senior Secondary School Mathematics for Class 11, , (9C 4 # 8C1) (9C 3 # 8C 2) (9C 2 # 8C 3) (9C1 # 8C 4) (1# 8C 3), ' 9 # 8 #7 # 6 # 81 ' 9 # 8 #7 # 8 #7 1 '9 # 8 # 8 #7 # 6 1, 3 # 2 #1 2 #1, 2 #1 3 # 2 #1, 4 # 3 # 2 #1, #, #, #, 8 #7 # 6, 8, 7, 6, 5 , '9 #, 1 b1#, l, 3 # 2 #1, 4 # 3 # 2 #1, (1008 2352 2016 630 56) 6062., (a) Clearly, the women are in majority in (iii), (iv) and (v) cases., So, the number of committees in which the women are in, majority, (9C7 # 8C 5) (9C8 # 8C 4) (9C 9 # 8C 3), (9C 2 # 8C 3) (9C1 # 8C 4) (1 8C 3), (2016 630 56) 2702., (b) Clearly, the men are in majority in (i) only., So, the number of committees in which the men are in majority, (9C 5 # 8C7) (9C 4 # 8C1) b 9 # 8 #7 # 6 # 8l 1008., 4 # 3 # 2 #1, EXAMPLE 24, , SOLUTION, , EXAMPLE 25, , SOLUTION, , In a village, there are 87 families of which 52 families have at most, 2 children. In a rural development programme, 20 families are to be chosen, for assistance, of which at least 18 families must have at most 2 children., In how many ways can the choice be made?, It is given that 52 families have at most 2 children and (87 52) 35, families have more than 2 children., We have to select 20 families of which at least 18 must have at most, 2 children., The selection can be made as under:, (i) 18 families out of 52 and 2 families out of 35, or (ii) 19 families out of 52 and 1 family out of 35, or (iii) 20 families out of 52., So, the number of ways in which these choices can be made, (52C18 # 35C 2) (52C19 # 35C1) (52C 20) ., In how many ways can 19 identical books on English and 17 identical, books on Hindi be placed in a row on a shelf so that two books on Hindi, may not be together?, In order that two books on Hindi are never together, we must, place these books as under:, XEXEXEX … XEX,, where E denotes the position of an English book and X that of a, Hindi book., Since there are 19 books on English, the number of cross marks are, therefore, 20.
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Combinations, , 341, , Since all books on English are identical, they can be placed in only, 1 way at the places marked E., Number of ways of placing 17 identical books on Hindi at 20 places,, 20 #19 #18 , each marked X 20C17 20C 3 , 1140., 3 # 2 #1, Hence, the required number of ways (1#1140) 1140., , EXERCISE 9B, 1. In how many ways can 5 sportsmen be selected from a group of 10?, 2. A bag contains 5 black and 6 red balls. Find the number of ways in which, 2 black and 3 red balls can be selected., 3. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls, and 4 blue balls if each selection consists of 3 balls of each colour., 4. How many different boat parties of 8 consisting of 5 boys and 3 girls can be, made from 20 boys and 10 girls?, 5. In how many ways can a student choose 5 courses out of 9 courses if, 2 specific courses are compulsory for every student?, 6. A sports team of 11 students is to be constituted, choosing at least 5 from, class XI and at least 5 from class XII. If there are 20 students in each of these, classes, in how many ways can the team be constituted?, 7. From 4 officers and 8 clerks, in how many ways can 6 be chosen (i) to, include exactly one officer, (ii) to include at least one officer?, 8. A cricket team of 11 players is to be selected from 16 players including, 5 bowlers and 2 wicketkeepers. In how many ways can a team be selected, so as to consist of exactly 3 bowlers and 1 wicketkeeper?, 9. In how many ways can a cricket team be selected from a group of 25 players, containing 10 batsmen, 8 bowlers, 5 all-rounders and 2 wicketkeepers,, assuming that the team of 11 players requires 5 batsmen, 3 all-rounders,, 2 bowlers and 1 wicketkeeper?, 10. A question paper has two parts, part A and part B, each containing 10, questions. If the student has to choose 8 from part A and 5 from part B, in, how many ways can he choose the questions?, 11. In an examination, a student has to answer 4 questions out of 5. Questions, 1 and 2 are compulsory. Find the number of ways in which the student can, make the choice., 12. In an examination, a student has to answer 10 questions, choosing at least, 4 from each of part A and part B. If there are 6 questions in part A and 7 in, part B, in how many ways can these questions be chosen?, 13. In an examination, a candidate is required to answer 7 questions out of, 12, which are divided into two groups, each containing 6 questions. One, cannot attempt more than 5 questions from either group. In how many, ways can he choose these questions?
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342, , Senior Secondary School Mathematics for Class 11, , 14. Out of 6 teachers and 8 students, a committee of 11 is to be formed. In how, many ways can this be done, if the committee contains, (i) exactly 4 teachers?, (ii) at least 4 teachers?, 15. A committee of 7 has to be formed from 9 boys and 4 girls. In how many, ways can this be done when the committee consists of, (i) exactly 3 girls?, (ii) at least 3 girls?, (iii) at most 3 girls?, 16. A committee of three persons is to be constituted from a group of 2 men, and 3 women. In how many ways can this be done? How many of these, committees would consist of 1 man and 2 women?, 17. A committee of 5 is to be formed out of 6 gents and 4 ladies. In how many, ways can this be done, when, (i) at least 2 ladies are included?, (ii) at most 2 ladies are included?, 18. From a class of 14 boys and 10 girls, 10 students are to be chosen for a, competition, at least including 4 boys and 4 girls. The 2 girls who won the, prizes last year should be included. In how many ways can the selection, be made?, 19. Find the number of 5-card combinations out of a deck of 52 cards if at least, one of the five cards has to be a king., 20. Find the number of diagonals of, (i) a hexagon,, (ii) a decagon,, (iii) a polygon of 18 sides., 21. How many triangles can be obtained by joining 12 points, four of which are, collinear?, 22. How many triangles can be formed in a decagon?, 23. How many different selections of 4 books can be made from 10 different, books, if, (i) there is no restriction?, (ii) two particular books are always selected?, (iii) two particular books are never selected?, 24. How many different products can be obtained by multiplying two or more, of the numbers 3, 5, 7, 11 without repetition?, 25. Find the number of ways in which a committee of 2 teachers and 3 students, can be formed out of 10 teachers and 20 students. In how many of these, committees, (i) a particular teacher is included?, (ii) a particular student is included?, (iii) a particular student is excluded?, 26. There are 18 points in a plane of which 5 are collinear. How many straight, lines can be formed by joining them?, ANSWERS (EXERCISE 9B), , 1. 252, , 2. 200, , 3. 800, , 4. 1860480, , 5. 35, , 6. 2(20C 5 # 20C6)
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344, , Senior Secondary School Mathematics for Class 11, , 16. Total number of committees 5C3 5C2 10., Number of committees consisting of 1 man and 2 women, (2C1 # 3C2) (2 # 3C1) (2 # 3) 6., 17. (i) (4C2 # 6C3) (4C3 # 6C2) (4C 4 # 6C1)., (ii) (6C 5) (6C 4 # 4C1) (6C3 # 4C2)., 18. 2 girls who won the prize last year are surely to be taken. So, we have to make a, selection of 8 students out of 14 boys and 8 girls, choosing at least 4 boys and at least, 2 girls., Thus, we may choose:, (4 boys, 4 girls) or (5 boys, 3 girls) or (6 boys, 2 girls)., Required number of ways, (14C 4 # 8C 4) (14C 5 # 8C3) (14C6 # 8C2) ., 19. Required number of combinations, = (total number of 5-card combinations), – (number of 5-card combinations having no king), (52C 5 48C 5) ., 20. (i) Number of diagonals of a hexagon (6C2 6) (15 6) 9., (ii) Number of diagonals of a decagon (10C2 10) (45 10) 35., (iii) Number of diagonals of a polygon of 18 sides (18C2 18) 135., 21. Number of triangles (12C3 4C3) (220 4) 216., 22. Number of triangles in a decagon 10C3 120., 23. (i) Number of ways of selecting 4 books out of 10 10C 4 ., (ii) After choosing 2 particular books, we have to choose 2 books out of remaining 8., , required number of ways 8C2 ., (iii) In this case, we have to choose 4 books out of remaining 8., , required number of ways 8C 4 ., 24. Required number of products, = number of ways of selecting 2 or 3 or 4 numbers out of 3, 5, 7, 11, (4C2 4C3 4C 4) (6 4 1) 11., 25. Number of ways of forming a committee of 2 teachers and 3 students (10C2 # 20C3) ., (i) Number of committees in which a particular teacher is included (9C1 # 20C3) ., (ii) Number of committees in which a particular student is included (10C2 #19C2) ., (iii) Number of committees in which a particular student is excluded (10C2 # 19C3) ., 26. Required number of straight lines (18C2 5C2 1) ., , MIXED PROBLEMS ON PERMUTATIONS AND COMBINATIONS, SOLVED EXAMPLES, EXAMPLE 1, , Out of 7 consonants and 4 vowels, how many words of 3 consonants and, 2 vowels can be formed?
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Combinations, , 345, , SOLUTION, , Total number of ways of choosing (3 consonants out of 7) and, (2 vowels out of 4), ( 7C 3 # 4C 2) b7 # 6 # 5 # 4 # 3 l 210., 3 # 2 #1 2 #1, Number of groups, each containing 3 consonants and 2 vowels, = 210., Each group contains 5 letters., Number of ways of arranging 5 letters amongst themselves, 5 ! (5 # 4 # 3 # 2 #1) 120., Hence, the required number of words (210 #120) 25200., , EXAMPLE 2, , The English alphabet has 21 consonants and 5 vowels. How many words, with two different consonants and two different vowels can be formed, from the alphabet?, Total number of ways of choosing, (2 consonants out of 21) and (2 vowels out of 5), 21C 2 # 5C 2 b 21# 20 # 5 # 4 l 2100., 2 #1, 2 #1, Number of groups, each containing 2 consonants and 2 vowels, = 2100., Each group contains 4 letters., Number of ways of arranging 4 letters amongst themselves, 4 ! (4 # 3 # 2 #1) 24., Hence, the required number of words (2100 # 24) 50400., , SOLUTION, , EXAMPLE 3, SOLUTION, , EXAMPLE 4, , SOLUTION, , How many words with or without meaning, each of 2 vowels and, 3 consonants can be formed from the letters of the word ‘DAUGHTER’?, The given word ‘DAUGHTER’ contains 3 vowels and 5 consonants., Total number of ways of choosing, (2 vowels out of 3) and (3 consonants out of 5), (3C 2 # 5C 3) (3C1 # 5C 2) b3 # 5 # 4 l 30., 2 #1, Number of groups, each containing 2 vowels and 3 consonants = 30., Each group contains 5 letters., Number of ways of arranging 5 letters amongst themselves, 5 ! (5 # 4 # 3 # 2 #1) 120., Hence, the required number of words (30 #120) 3600., How many words with or without meaning, can be formed using all the, letters of the word ‘EQUATION’ at a time so that vowels and consonants, occur together?, The given word ‘EQUATION’ contains 5 vowels and 3 consonants., All vowels can be put together in 5 ! ways., All consonants can be put together in 3 ! ways.
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346, , Senior Secondary School Mathematics for Class 11, , Considering all vowels as 1 letter and all consonants as 1 letter,, vowels and consonants can be arranged in 2 ! ways., , required number of words, (5 !)#(3 !)#(2 !), (5 # 4 # 3 # 2 #1)#(3 # 2 #1)#(2 #1) 1440., EXAMPLE 5, SOLUTION, , In how many ways can 5 girls and 3 boys be seated in a row so that no two, boys are together?, Since no two boys are to be together, let us arrange 5 girls as under:, XGXGXGXGXGX,, where G denotes the position of a girl and X that of a boy., These girls may be arranged at 5 places in 5 ! ways., Now, 3 boys can be seated at 6 places, each marked X., Number of ways of seating 3 boys at 6 places 6P3 ., , , required number of ways, (5 !)# 6P3 (5 !)# 6 !, 3!, ((5 # 4 # 3 # 2 #1)#b 6 # 5 # 4 # 3 # 2 #1 l2 14400., 3 # 2 #1, , Hence, the required number of ways is 14400., , EXERCISE 9C, 1. Out of 12 consonants and 5 vowels, how many words, each containing, 3 consonants and 2 vowels, can be formed?, 2. How many words, each of 3 vowels and 2 consonants, can be formed from, the letters of the word ‘INVOLUTE’?, 3. The English alphabet has 21 consonants and 5 vowels. How many words, with two different consonants and three different vowels can be formed, from the alphabet?, 4. In how many ways can 4 girls and 3 boys be seated in a row so that no two, boys are together?, 5. How many words, with or without meaning, can be formed from the letters, of the word, ‘MONDAY’, assuming that no letter is repeated, if, (i) 4 ketters are used at a time?, (ii) all letters are used at a time?, (iii) all letters are used but first letter is a vowel?, ANSWERS (EXERCISE 9C), , 1. 264000, 2. 2880, 5. (i) 360 (ii) 720 (iii) 240, , 3. 252000, , 4. 480
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Combinations, , 347, , HINTS TO SOME SELECTED QUESTIONS, 5. (i) Number of 4-letter words from the letters of ‘MONDAY’ 6P4 360., (ii) Number of words formed from all the letters of ‘MONDAY’ 6P6 6 ! 720., (iii) The given word contains two vowels, namely A and O., So, first letter of the word can be chosen in 2 ways., Remaining 5 places can be filled in 5 ! ways., , required number of words 2 #(5 !) 240., , EXERCISE 9D, Very-Short-Answer Questions, 1. If 20Cr 20Cr 10, , then find the value of 17Cr ., , 2. If 20Cr 1 20Cr 1 then find the value of 10Cr ., 3. If nC12 nC8, , then find the value of 22Cn ., , 4. If 35Cn 7 35C 4n 2 then find the value of n., 5. Find the values of (i) 200C198,, 6. If mC1 nC 2 prove that m , , (ii) 76C0,, , (iii) 15C15 ., , 1, n (n 1) ., 2, , 7. Write the value of (5C1 5C 2 5C 3 5C 4 5C 5) ., , , 8. If n 1C 3 2 (nC 2), find the value of n., 9. If nPr 720 and nCr 120 then find the value of r., 2, , 2, , , , 10. If (n n)C 2 (n n)C 4 then find the value of n., 11. How many words are formed by 2 vowels and 3 consonants, taken from, 4 vowels and 5 consonants?, 12. Find the number of diagonals in an n-sided polygon., 13. Three persons enter a railway compartment having 5 vacant seats. In how, many ways can they seat themselves?, 14. There are 12 points in a plane, out of which 3 points are collinear. How, many straight lines can be drawn by joining any two of them?, 15. In how many ways can a committee of 5 be made out of 6 men and 4 women,, containing at least 2 women?, 16. There are 13 cricket players, out of which 4 are bowlers. In how many ways, can a team of 11 be selected from them so as to include at least 3 bowlers?, 17. How many different committees of 5 can be formed from 6 men and, 4 women, if each committee consists of 3 men and 2 women?, 18. How many parallelograms can be formed from a set of 4 parallel lines, intersecting another set of 3 parallel lines?
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348, , Senior Secondary School Mathematics for Class 11, ANSWERS (EXERCISE 9D), , 1. 136, , 2. 1, , 3. 231, , 4. n 6, , 7. 31, , 8. n 5, , 9. r 3, , 10. n 3, , 13. 60, , 14. 64, , 15. 186, , 16. 72, , 5. (i) 19900 (ii) 1 (iii) 1, n(n 3), 11. 60, 12., 2, 17. 120, 18. 12, , HINTS TO SOME SELECTED QUESTIONS, 1. r r 10 20 & r 15. So, 17Cr 17C15 17C2 , , 17 #16 , 136., 2 #1, , 2. r 1 r 1 20 & r 10. So, 10Cr 10C10 1., 3. n 12 8 20 &, , Cn 22C20 22C2 , , 22, , 4. n 7 4n 2 35 & 5n 30 & n 6., 200 #199 , 5. (i) 200C198 200C2 , 19900., 2, (ii) 76C0 1., C15 1., , 22 # 21 , 231., 2, , 15, , (iii), , 6. mC1 nC2 & m , , n(n 1), ·, 2, , 7. 5C1 5C2 5C3 5C 4 5C 5 (5C1 5C2 5C2 5C1 5C 5) (5 10 10 5 1) 31., (n 1) n (n 1) n (n 1), , # 2 & n 1 6 & n 5., 6, 2, 720 , 9. nPr (nCr)#(r !) & r ! , 6 3 # 2 #1 3 ! & r 3., 120, 8., , n1, , C3 2 (nC2) &, , 10. n 2 n (2 4) & n 2 n 6 0 & (n 3) (n 2) 0 & n 3., 11. Number of words (4C2 # 5C3) (4C2 # 5C2) (6 #10) 60., 2, n (n 1), , n n n 2n, 2, 2, 2, 3n n (n 3), n, , , ·, 2, 2, 13. First can occupy any of the 5 seats, 2nd may occupy any of 4 seats and third may, occupy any of 3 seats., Required number of ways (5 # 4 # 3) 60., , 12. Required number of diagonals (nC2 n) , , 14. Required number of straight lines (12C2 3C2 1) 64., 15. (2 women and 3 men) or (3 women and 2 men) or (4 women and 1 man), (4C2 # 6C3) (4C3 # 6C2) (4C 4 # 6C1) (120 60 6) 186., 16. There are 4 bowlers and 9 batsmen., We may choose (3 bowlers and 8 batsmen) or (4 bowlers and 7 batsmen)., Required number of ways (4C3 # 9C8) (4C 4 # 9C7) 36 36 72., 17. Number of committees (6C3 # 4C2) (20 # 6) 120., 18. Number of parallelograms (4 # 3) 12.
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Combinations, , 349, , KEY FACTS AND FORMULAE, 1., , Each of the different groups or selections that can be made, out of a given number of things by taking some or all of them at a time,, irrespective of their arrangements, is called a combination., 2. We use the word arrangements for permutations and selections for, combinations., 3. The number of combinations of n things taken r at a time is denoted by nCr ,, or C(n, r), where n 0, r $ 0 and n $ r., n(n 1) (n 2) … to r factors, n!, , ·, 4. (i) nCr , r!, (r !)#(n r) !, (ii) nCn nC0 1., COMBINATION, , n, , (iii) nCr , 5., , Pr, ·, r!, , (i) nCr nCn r ., (ii) nC x nC y and x ! y & x y n., , , (iii) nCr nCr 1 n 1Cr, n, , (iv), , Cr, , n1, , Cr 1, , n, , (v), , Cr, , n, , Cr 1, , n and, r, , [Pascal’s rule]., n, , Cr 1, , n1, , Cr 1, , , , n, ·, (n r 1), , , r 1·, nr, ,
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350, , Senior Secondary School Mathematics for Class 11, , 10, , Binomial Theorem, The expression of the form (a b) n is called a binomial. By, direct multiplication it is easy to expand (a b) 2, (a b) 3, etc. The ancient, mathematicians knew about the expansion of (a b) n for 0 # n # 7. Around, 1660, B Pascal introduced Pascal’s triangle for the coefficients in the expansion, of (a b) n and in the same year he gave the present form of binomial theorem., HISTORY, , BINOMIAL THEOREM (For Non-negative Integers), , , , , (a b) n nC0 a n nC1 a n 1b nC 2 a n 2b 2 … nCn 1 ab n, , 1, , nCn b n., , By direct multiplication, we have, [a x 0 1 for every real x], , (i) (a b) 0 1., , (ii) (a b) 1 a b 1C0 a 1C1 b., (iii) (a b) 2 a 2 2ab b 2 2C0 a 2 2C1 ab 2C 2 b 2., (iv) (a b) 3 a 3 3a 2b 3ab 2 b 3 3C0 a 3 3C1 a 2b 3C 2 ab 2 3C 3 b 3., (v) (a b) 4 a 4 4a 3b 6a 2b 2 4ab 3 b 4, 4C0 a 4 4C1 a 3b 4C 2 a 2b 2 4C 3 ab 3 4C 4 b 4., Given below is the Pascal’s Triangle showing the coefficients of various, terms in a binomial expansion., Index, , Coefficients, , 0, , 1, , 1, , 1, , 2, , 1, , 3, , 1, , 4, , 1, , 5, , 1, , 6, , 1, , 7, 8, , 1, 1, , 8, , 3, , 5, , 7, , 6, , 15, , 1, 4, , 10, 20, , 35, 56, , 1, 3, , 10, , 21, 28, , 2, , 4, , 6, , 1, , 5, 15, , 35, 70, , 1, 6, 21, , 56, , PASCAL’S TRIANGLE, , 350, , 1, 1, 7, 28, , 1, 8, , 1
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352, , Senior Secondary School Mathematics for Class 11, , Thus, the coefficients of the terms equidistant from the beginning and, the end in a binomial expansion are equal., (iii) General term in the expansion of (a b) n is the (r 1)th term given by, , Tr 1 nCr a n rb r., (iv) In the expansion of (a b) n, there are (n 1) terms., n, I. When n is even, the middle term a 1kth term., 2, II. When n is odd, two middle terms are, 1, 1 , (n 1)th term and ' (n 1) 11th term., 2, 2, (v) pth term from the end in (a b) n, (n 1 p 1)th term from the beginning, (n p 2)th term from the beginning., (vi) Results on binomial coefficients, We have the binomial expansion:, , , , (a b) n nC0 a n nC1 a n 1b nC 2 a n 2b 2 … nCn 1 ab n 1 nCn b n … (A), Putting a b 1 in (A), we get, n, … (B), C 0 nC 1 nC 2 nC 3 … nC n 2 n ., Putting a 1 and b 1 in (A) and using (B), we get, 2n n1, n, C0 nC 2 nC 4 … nC1 nC 3 nC 5 … , 2 ., 2, , Hence, nC0 nC 2 nC 4 … nC1 nC 3 nC 5 … 2 n 1., DEDUCTIONS, , (i) Putting a 1 and b x in the binomial expansion of (a b) n, we get, (1 x) n nC0 nC1 x nC 2 x 2 … nCn x n., n, , Thus, (1 x) n / nCr x r., r0, r, , And, Tr 1 nCr x ., , (ii) Replacing b by b in the binomial expansion of (a b) n, we get, , , , , , , (a b) n nC0 a n nC1 a n 1b nC 2 a n 2b 2 nC 3 a n 3b 3 …, n, , &, , , , , (a b) n / (1) r · nCr a n rb r., r0, , Tr 1 in the expansion of (a b) n is given by, , Tr 1 (1) r · nCr a n rb r., , (iii) Putting a 1 and b x in the above expansion, we get, (1 x) n nC0 nC1 x nC 2 x 2 nC 3 x 3 … (1) n · nCn x n, n, , &, , (1 x) n / (1) r · nCr x r., , , , Tr 1 in the expansion of (1 x) n is given by, Tr 1 (1) r · nCr x r., , r0
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Binomial Theorem, , 353, , SUMMARY, , 1. We have, n, , , , , , n, n, n r r, n, n r r, (i) (a b) / Cr a b and Tr 1 Cr a b ., r0, n, , , , , , (ii) (a b) n / (1) r · nCr a n rb r and Tr 1 ( 1) r · nCr a n rb r., r0, , (iii) When n is even, then, n, the middle term a 1kth term., 2, (iv) When n is odd, then, 1, 1, the middle terms are (n 1)th and ' (n 1) 11th., 2, 2, (v) pth term from the end (n p 2)th term from the beginning., 2., , n, , (i) (1 x) n / nCr x r and Tr 1 nCr x r., r0, n, , (ii) (1 x) / (1) r · nCr x r and Tr 1 (1) r · nCr x r., n, , r0, , 3., , (i) nC0 nC1 nC 2 … nCn 2 n., (ii) nC0 nC 2 nC 4 … nC1 nC 3 nC 5 … 2 n 1., SOLVED EXAMPLES, , EXAMPLE 1, SOLUTION, , Expand (x 2y) using binomial expansion., Using the binomial expansion, we have, (x 2 2y) 5 5C0 (x 2) 5 5C1 (x 2) 4 (2y) 5C 2 (x 2) 3 (2y) 2, 2, , 5, , 5C 3 (x 2) 2 (2y) 3 5C 4 x 2 (2y) 4 5C 5 (2y) 5, x 10 10x 8y 40x 6y 2 80x 4y 3 80x 2y 4 32y 5., EXAMPLE 2, SOLUTION, , 2 6, 2 n using binomial expansion., x, Using the binomial expansion, we get, Expand dx 3 , , 3, 2, 2 6 6, 6, 3 3 2, 3 6 6, 3 5 2, 6, 3 4 2, 3, dx 2 n C0 (x ) C1 (x ) d 2 n C 2 (x ) d 2 n C 3 (x ) d 2 n, x, x, x, x, 6, 5, 4, 6C 4 (x 3) 2 d 22 n 6C 5 x 3 d 22 n 6C6 d 22 n, x, x, x, x 18 12x 13 60x 8 160x 3 240, 192, 64 ·, x2, x 7 x 12, , EXAMPLE 3, SOLUTION, , Expand (1 x x 2) 3 using binomial expansion., Put 1 x y. Then,, ( 1 x x 2) 3 ( y x 2 ) 3
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Binomial Theorem, , 355, , 2 [x 8 (6x 4 6x 6) (1 2x 2 x 4)], 2 [x 8 6x 6 7x 4 2x 2 1], 2x 8 12x 6 14x 4 4x 2 2., EXAMPLE 6, SOLUTION, , 1 12, Find the 10th term in the expansion of b2x 2 x l ·, The general term in the given expansion is given by, 1 r, , Tr 1 12Cr #(2x 2) (12 r) #b x l ·, , T10 T(9 1), 12C 9 #(2x 2) (12 9) #b1 l, x, 1, 12, 2 3, C 3 # ( 2x ) # d 9 n, [a, x, d12 #11#10 # 8x 6 # 19 n 1760, ·, 3 # 2 #1, x, x3, 9, , [putting r 9 in (i)], , SOLUTION, , C 9 C(12 9) C 3], , 12, , Hence, the 10th term in the given expansion is, EXAMPLE 7, , … (i), , 12, , 12, , 1760, ·, x3, , Find the 6th term in the expansion of b, , 4x 5 9 ·, l, 5 2x, The general term in the given expansion is given by, Tr 1 (1) r # 9Cr #b, , , T6 T(5 1), , 5 r, 4x (9 r), l #b l ·, 5, 2x, , … (i), , (9 5), 5 5, (1) 5 # 9C 5 #b 4x l, #b l, [putting r 5 in (i)], 5, 2x, 4, 5, 9, 9, 9, 9C 4 #b 4x l #b 5 l, [a C 5 C(9 5) C 4], 5, 2x, RS, 5 V, 8 4, W , SSS9 # 8 #7 # 6 # 2 x4 # 55 5WWW 5040 ·, x, #, #, #, 4, 3, 2, 1, S, 2 x W, 5, T, X, 5040, Hence, the 6th term in the given expansion is x ·, , EXAMPLE 8, SOLUTION, , N9, JKx 3, 2O, Find the 4th term from the end in the expansion of KK 2OOO ·, K2 x, P, L, The general term in the given expansion is given by, Tr 1 (1) r # Cr #d, 9, , 3 (9 r ), , x n, 2, , #d, , 2 r·, 2n, x, , … (i), , Now, pth term from the end, (n p 2)th term from the beginning., , 4th term from the end, (9 4 2)th term from the beginning [a n 9, p 4], 7th term from the beginning
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356, , Senior Secondary School Mathematics for Class 11, , T7 T(6 1), 3 (9 6), , (1) 6 # 9C6 #d x n, 2, , #d, , 2 6, 2n, x, , [putting r 6 in (i)], , 3 3, , 6, 9C 3 #d x n #d 22 n, 2, x, JK9 # 8 #7 x 9 64 ON 672, KK, # # 12OOO 3 ·, K3 # 2 #1, 8, x, x, P, L, 672, Hence, the 4th term from the end is 3 ·, x, , [a 9C6 9C(9 6) 9C 3], , EXAMPLE 9, , If the ratio of the 5th term from the beginning to the 5th term from the end, 1 n, in the expansion of e4 2 4 o is 6 : 1 then find the value of n., 3, , SOLUTION, , The given expression may be written as (2 4 3, The general term in this expansion is given by, , 1, , 1, , Tr 1 nCr #(2 4 ) (n, , r), , # (3, , (n r ), 4, , b, , 1 n, 4, , ) ., , 7, 1 r7, 4, , r 77, l, , ), , n, #3 4 ., &, Tr 1 Cr # 2, Now, pth term from the end, (n p 2)th term from the beginning., , 5th term from the end, (n 5 2)th term from the beginning, (n 3)th term from the beginning., Now, T5 T(4 1), , nC 4 # 2, nC 4 # 2, , (n 4), 4, (n 4), 4, , #3, , b, , 4 7 7, l, 4, , , , 77, , # 3 1., , … (i), , [putting r 4 in (i)], … (ii), , And, Tn 3 T(n 4) 1, nC n 4 # 2 1 # 3, nC 4 # 2 # 3, , , (n 4), 4, , ., , T5, 6, , Tn 3, 1, n, , , , (n 4), 4, , C4 #2, , n, , (n 4), 4, , C4 # 2# 3, , #3, , 1, , (n 4), 4, , , , 6, 1, , [putting r (n 4) in (i)], … (iii) [a nCn r nCr]
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Binomial Theorem, , , , , 2, 2, , (n 4), (, 12, 4, (n 8), 4, , #3, , #3, , (n 4), (, 12, 4, , (n 8), 4, , 357, , 6, , 6 1/2, , (n 8), 4, , 6 1/2, ( 2 # 3), n8 1, , & n 8 2 & n 10., 2, 4, Hence, n 10., , , EXAMPLE 10, , Find the term independent of x in the expansion of d, , SOLUTION, , Let Tr 1 be independent of x., Here, Tr 1 ( 1) r # 15Cr #d, , 2 (15 r), , 3x n, 2, , #b, , , , 15, , 3x 2 1 n, ·, 2, 3x, , 1 r, l, 3x, , , , , , , … (i), Tr 1 ( 1) r # 15Cr # 3 (15 2r) # 2 (r 15) # x (30 3r)., Now, Tr 1 will be independent of x only when the power of x in it, is 0., , 30 3r 0 & 3r 30 & r 10 & r 1 11., Thus, T11 is independent of x., Now, T11 T(10 1), (1) 10 # 15C10 # 3 (15 20) # 2 (10 15) # x 0 [putting r 10 in (i)], 15C, , 8a 15C10 15C(15 10) 15C 5B, 5, 6, 1001, 1, b15 #14 #13 #12 #11 #, ·, l, 36 # 36 # 6, 2592, 5 # 4 # 3 # 2 #1, 1001, ·, Hence, the term independent of x in the given expansion is, 2592, 15C 5 # 3 5 # 2 5 , , EXAMPLE 11, SOLUTION, , 5, , 10, JK x, 3 NO, 2OO ·, Find the term independent of x in the expansion of KKK, 3 2x O, P, L, Let Tr 1 be independent of x., r, x (10 r) JKK 3 NOO, Here, Tr 1 10Cr #b, #KK 2OO, l, 3, 2x, L P, , &, , (10 r), 2, , x, Tr 1 10Cr #b l, 3, 10Cr # x, , r, , #3 2 #, , r, b5 2rl, 2, , 1, , #, 3, , 10Cr # x, , b5 , , 5r, l, 2, , 1, r 2r, 2x, , # 3 (r, , r #3, b5 l, 2, , 5), , , , # 2 r., , r, 2, , #2, , r, , … (i), , Now, Tr 1 will be independent of x only when the power of x in it, is 0.
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358, , Senior Secondary School Mathematics for Class 11, , 5r , 5r , 0 &, 5 & r 2 & r 1 3., 2, 2, Thus, T3 is independent of x., Putting r 2 in (i), we get, T3 T(2 1), , , 5, , 10C 2 # 3 (2 5) # 2 2 # x 0 b10 # 9 # 3 3 # 2 2l, 2 #1, 45, 45, 5, e 3, o b 27 # 4 l 12 ·, 3 #2 2, 5, ·, Hence, the term independent of x in the given expansion is, 12, EXAMPLE 12, , SOLUTION, , If the term free from x in the expansion of d x , , m 10, 2 n is 405, find the value, x, , of m., The general term in the given expansion is given by, m r, , Tr 1 (1) r # 10Cr #( x ) (10 r) #d 2 n, x, (1) r # 10Cr # x, (1) r # 10Cr # x, (1) r # 10Cr # x, , r, b5 l, 2, , r, , #, , r, b5 2rl, 2, , b5 , , 5r, l, 2, , m, 2r, x, #m r, , # m r., , … (i), , Let Tr 1 be free from x., Then, the power of x in Tr 1 must be 0., 5r , 5r , , 5, 0 &, 5 & r 2 & r 1 3., 2, 2, So, T3 will be free from x., Now, T3 T(2 1), (1) 2 # 10C 2 # x 0 # m 2, [putting r 2 in (i)], #, 10, 9, 2, 2, b, # m l 45m ., 2, But, it is given that the term free from x is 405., , 45m 2 405 & m 2 9 & m !3., Hence, m !3., EXAMPLE 13, SOLUTION, , 1 15, 1, in the expansion of dx 4 3 n ·, 17, x, x, The general term in the given expansion is given by, 1 r, , Tr 1 (1) r # 15Cr #(x 4) (15 r) #d 3 n, x, , , … (i), Tr 1 (1) r # 15Cr # x (60 7r)., Find the coefficients of x 32 and
Page 370 : Binomial Theorem, , 359, , Putting 60 7r 32, we get 7r 28 & r 4 & r 1 5., , Now, T5 T(4 1) (1) 4 # 15C 4 # x (60 28) 15C 4 # x 32., , , coefficient of x 32 15C 4 b, , 15 #14 #13 #12 , l 1365., 4 # 3 # 2 #1, , 1 17 ·, x F, 17, x, , Then, Ts 1 (1) s # 15C s # x (60 7s)., … (ii) [putting r s in (i)], Putting 60 7s 17, we get s 11 and therefore, s 1 12., , Thus, the 12th term contains x 17., Now, T12 T(11 1), Let Ts 1 be the term containing x, , <a, , 17, , (1) 11 # 15C11 # x (60 77), C4 # x, 15, , [putting s 11 in (ii)], , 17, , ., , 15C 4 b15 #14 #13 #12 l 1365., 4 # 3 # 2 #1, , So, the coefficient of x 17 in the given expansion is –1365., , Hence, the coefficient of x 32 is 1365 and that of x 17 is –1365., , , EXAMPLE 14, SOLUTION, , coefficient of x, , 17, , 2 18, Prove that there is no term containing x 10 in the expansion of bx 2 x l ·, The general term in the given expansion is given by, Tr 1 (1) r # 18Cr #(x 2) (18, , , r), , Tr 1 (1) r # 18Cr # 2 r # x (36, , 2r, #b x l, , 3r), , ., , 10, , Let Tr 1 contain x . Then,, 26, ·, 3, Since the value of r cannot be a fraction, so there is no term, containing x 10., b 12, Find the middle term in the expansion of eax 2 o ·, x, The general term in the given expansion is given by, 36 3r 10 & 3r 26 & r , , EXAMPLE 15, SOLUTION, , Tr 1 (1) r # 12Cr #(ax) (12, , , r), , #e, , b r, o, x2, , , , … (i), Tr 1 (1) r # 12Cr # a (12 r) # b r # x (12 3r)., Since the given binomial contains an even power, so it has only one, middle term., 12, Middle term b 1lth term 7th term., 2, Now, T7 T(6 1), ( 1) 6 # 12C6 # a (12 6) # b 6 # x (12 3# 6) [putting r 6 in (i)], 12C6 # a 6b 6 # x 6
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360, , Senior Secondary School Mathematics for Class 11, 6 6, 6 6N, JK, O, KK12 #11#10 # 9 # 8 #7 # a b6 OO 924a6 b ·, K 6 # 5 # 4 # 3 # 2 #1, x O, x, P, L, 924a 6b 6, ·, Hence, the middle term in the given expansion is, x6, , EXAMPLE 16, SOLUTION, , 7, , x3n ·, 6, The general term in the given expansion is given by, , Find the middle terms in the expansion of d3x , r, 7, 7 r, Tr 1 (1) # Cr #(3x) #d, , , , , r, , x3n, 6, , , , , , , Tr 1 (1) r # 7Cr # 3 (7 2r) # 2 r # x (7 2r)., Clearly, the given expansion has 8 terms., 8, So, it has two middle terms, namely b lth, i.e., 4th and 5th., 2, Now, T4 T(3 1), (1) 3 # 7C 3 # 3 (7 6) # 2 3 # x 13, d 7 # 6 # 5 # 3 1 # 13 # x 13n, 3 # 2 #1, 2, , … (i), , [putting r 3 in (i)], , 13, 13, , e 35 # 3 # x o 105x ·, 8, 8, And, T5 T(4 1), , (1) 4 # 7C 4 # 3 (7 8) # 2 4 # x 15, 1, , 4, , C3 # 3 # 2 # x, 7, , [putting r 4 in (i)], [a 7C 4 7C(7 4) 7C 3], , 15, , 15, d 7 # 6 # 5 # 1 # 14 # x 15n 35x ·, 3 # 2 #1 3 2, 48, , Hence, the required middle terms are, EXAMPLE 17, , SOLUTION, , 105x 13, 35x 15, ·, and, 8, 48, , Show that the middle term in the expansion of (1 x) 2n is, 1 · 3 · 5 … (2n 1) n n, · 2 · x , where nd N., n!, Clearly, the number of terms in the expansion of (1 x) 2n is (2n 1) ., 2n, , middle term b 1lth term (n 1)th term Tn 1 ., 2, (2n) !, · xn, Now, Tn 1 2nCn · x n , (n !)#(n !), 1 · 2 · 3 · 4 … (2n 2) (2n 1) (2n) n, , ·x, (n !)#(n !), [1 · 3 · 5 … (2n 3) (2n 1)]#[2 · 4 · 6 … (2n 2) (2n)] n, , ·x, (n !) (n !)
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Binomial Theorem, , 361, , [1 · 3 · 5 … (2n 1)]# 2 #[1 · 2 · 3 … (n 1) · n] n, ·x, (n !)#(n !), n, n, [1 · 3 · 5 … (2n 1)]# 2 # x, , ·, (n !), n, , , , Hence, the middle term in the given expansion is, n, n, 1 · 3 · 5 … (2n 1)# 2 # x, ·, (n !), EXAMPLE 18, , 1 2n, If x p occurs in the expansion of bx 2 x l , prove that its coefficient is, (2n) !, 4n p, 2n p, n !3# )d, n !3, )d, 3, 3, , SOLUTION, , ·, , 1 2n, The general term in the expansion of bx 2 x l is given by, Tr 1 2nCr #(x 2) (2n, , r), , 1 r, #b x l, , , , , Tr 1 2nCr # x (4n 3r)., This term contains x p only when 4n 3r p., And, 4n 3r p & r , , … (i), , (4n p), ·, 3, , Putting 4n 3r p in (i), we get, coefficient of x p 2nCr, where r , , (4n p), 3, , (2n) !, Z], _, , ], 4n p, (4n p) bb, n !3# []=2n , )d, G !`b, ], b, 3, 3, \, a, (2n) !, , ·, 4n p, 2n p, n !3# )d, n !3, )d, 3, 3, , , , (2n) !, , (r !)#(2n r) !, , 1 2n, Hence, the coefficient of x p in the expansion of bx 2 x l is, (2n) !, ·, , 4n p, 2n p, n !3# )d, n !3, )d, 3, 3, EXAMPLE 19, , SOLUTION, , Show that the coefficient of the middle term in the expansion of (1 x) 2n, is the sum of the coefficients of two middle terms in the expansion of, , (1 x) 2n 1., The expansion of (1 x) 2n contains (2n 1) terms and therefore, the, (n 1)th term is the middle term.
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362, , Senior Secondary School Mathematics for Class 11, , , , the middle term in the expansion of (1 x) 2n is given by, tn 1 2nCn x n , , (2n) !, (2n) ! n, xn , 2x ., (n !) · (n !), (n !), , Thus, the coefficient of the middle term of (1 x) 2n is, , , (2n) !, (n !) 2, , ·, , Again, the expansion of (1 x) 2n 1 contains 2n terms., So, the middle terms are the nth and the (n 1)th terms., , , Now, tn t(n 1) 1 2n 1Cn 1 x n, , , 1, , and tn 1 , , 2n 1, , n, , Cn x ., , , the sum of the coefficients of the middle terms of (1 x) 2n 1, 2n 1Cn 1 2n 1Cn, (2n 1) !, (2n 1) !, , , , (n 1) !#(n !) (n !)#(n 1) !, 2 #(2n 1) !# n, , [multiplying num. and denom. by n], (n !)# n #(n 1) !, (2n) · (2n 1) ! (2n) !, , , (n !) · (n !), (n !) 2, = the coefficient of the middle term of (1 x) 2n., , EXAMPLE 20, , SOLUTION, , If a1, a 2, a 3, a 4 be the coefficients of four consecutive terms in the expansion, of (1 x) n then prove that, a3, a1, 2a 2, , , ·, (a1 a 2) (a 3 a 4) (a 2 a 3), Let a1, a 2, a 3 and a 4 be the coefficients of the rth, (r 1)th, (r 2)th, and (r 3)th terms in the expansion of (1 x) n. Then,, a1 nCr 1; a 2 nCr, a 3 nCr 1 and a 4 nCr 2 ., , , Now, a1 a 2 ( nCr 1 nCr) n 1Cr;, , , a 3 a 4 ( nCr 1 nCr 2) n 1Cr 2;, and a 2 a 3 ( nCr nCr 1) , , n1, , Cr 1 ., , n, n, a3, C, C, a1, , n 1r 1 n 1 r 1, , (a1 a 2) (a 3 a 4), Cr, Cr 2, n, n, Cr 1, Cr 1, n, , , ;a nCr a k · n 1Cr 1E, r, 1 n, n1 ·n, n, b, l, b l Cr 1, r · Cr 1, r 2, ( r 2 ) 2 ( r 1), r , , ·, , ( n 1) ( n 1) ( n 1), n, n, 2# C, 2 # Cr, 2a 2, 2 ( r 1), n1 r , , ·, And,, , ( a 2 a 3), ( n 1), n, 1, n, Cr 1 b, l · Cr, , r 1, a3, a1, 2a 2, , , ·, Hence,, (a1 a 2) (a 3 a 4) (a 2 a 3)
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Binomial Theorem, EXAMPLE 21, SOLUTION, , 363, , , In the binomial expansion of (1 x) m n, prove that the coefficients of x m, n, and x are equal., , In the binomial expansion of (1 x) m n, the general term is given by, mn, r, … (i), Tr 1 , Cr # x ., m, mn, , , coefficient of x in the expansion of (1 x), (m n) !, (m n) !, m nC m , , ·, {(m n m) !}#(m !) (n !)#(m !), , , coefficient of x n in the expansion of (1 x) m n, (m n) !, (m n) !, m nC n , , ·, {(m n n) !}#(n !) (m !)#(n !), Hence, the coefficients of x m and x n are equal., EXAMPLE 22, , Find the value of r when it is being given that the coefficients of (2r 4)th, and (r 2)th terms in the expansion of (1 x) 18 are equal., , SOLUTION, , In the expansion of (1 x) 18, we have Tp 1 18C p x p., , , T(2r 4) T(2r 3) 1 18C 2r 3 · x 2r, , and T(r 2) T(r 3) 1 C(r 3) · x, 18, , r3, , 3, , ., , … (i), … (ii), , It is being given that the coefficients of T(2r 1) and T(r 2) are equal., So, from (i) and (ii), we get, 18C, 18Cr 3 & (2r 3) (r 3) or (2r 3) (r 3) 18, 2r 3, [a nC p nCq & p q or p q n]., Now, (2r 3) (r 3) or (2r 3) (r 3) 18, , r 6 or r 6, , [a the negative value of r is not permissible]., r6, Hence, r 6., EXAMPLE 23, , Find the coefficient of x 6y 3 in the expansion of (x 2y) 9., , SOLUTION, , The general term in the expansion of (x 2y) 9 is given by, , Tr 1 9Cr # x (9 r) #(2y) r, , , , Tr 1 9Cr # 2 r # x (9 r) # y r., We have to find the coefficient of x 6y 3., Putting r 3 in (i), we get, T(3 1) 9C 3 # 2 3 #(x 6y 3) ., , , , … (i), , coefficient of x 6y 3 in the given expansion, ( 9C 3 # 2 3) b 9 # 8 #7 # 8l 672., 3 # 2 #1, , Hence, the coefficient of x 6y 3 in the given expansion is 672., EXAMPLE 24, , , Write down the binomial expansion of (1 x) n 1 when x 8. Deduce that, , (9 n 1 8n 9) is divisible by 64, where n is a positive integer.
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364, SOLUTION, , Senior Secondary School Mathematics for Class 11, , Using the binomial expansion, we get, , , , , , , , (1 x) n 1 n 1C0 n 1C1 x n 1C 2 x 2 n 1C 3 x 3 … n 1Cn 1 x n 1., Putting x 8 in the above expansion, we get, , , , 9 n 1 1 (n 1)# 8 n 1C 2 #(8) 2 n 1C 3 #(8) 3 , , , , , (9 n, , 1, , , , (9 n, , 1, , Hence, (9, EXAMPLE 25, SOLUTION, , … n 1Cn 1 #(8) n, , 1, , 8n 9) (8) 2 #{n 1C 2 n 1C 3 # 8 … (8) n 1}, 8n 9) 64 #(an int eger) ., , n1, , 8n 9) is exactly divisible by 64., , Using the binomial theorem, prove that (6 n 5n) always leaves the, remainder 1 when divided by 25., Using binomial expansion, we get, (6 n 5n) (1 5) n 5n, {nC0 nC1 # 5 nC 2 #(5) 2 nC 3 #(5) 3 nC 4 #(5) 4 , … nCn # 5 n} 5n, {1 5n nC 2 #(5) 2 nC 3 #(5) 3 nC 4 #(5) 4 , … nCn # 5 n} 5n, {1 nC 2 #(5) 2 nC 3 #(5) 3 nC 4 #(5) 4 , (5) 2 #{nC 2 nC 3 # 5 nC 4 #(5) 2 , 25 #(an integer) 1., , … nC n # 5 n}, , … n C n # 5 (n, , 2), , }1, , Hence, (6 n 5n) when divided by 25 leaves the remainder 1., EXAMPLE 26, SOLUTION, , , Show that 2 4n 4 15n 16, where nd N is divisible by 225., We have, , , , 2 4n 4 15n 16 2 4 (n 1) 15n 16 16 (n 1) 15n 16., (n 1), n1, 15n 16 (1 15), 15n 16, Now, 16, , n 1C0 n 1C1 #15 n 1C 2 #(15) 2 n 1C 3 #(15) 3 , , , … n 1Cn 1 #(15) n 1 15n 16, 1 (n 1)#15 n 1C 2 #(15) 2 n 1C 3 #(15) 3 , , … (15) n 1 15n 16, n 1C 2 #(15) 2 n 1C 3 #(15) 3 … (15) n 1, (15) 2 #[n 1C 2 n 1C 3 #15 … (15) n 1], 225#(an integer) ., Hence, (2 4n, EXAMPLE 27, , 4, , 15n 16) is divisible by 225., , If a and b are distinct integers then prove that (a b) is a factor of, (a n b n), whenever n is a positive integer.
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Binomial Theorem, SOLUTION, , 365, , We may write, a n (a b b) n {(a b) b} n, nC0 #(a b) n nC1 #(a b) (n 1) # b nC 2 #(a b) (n 2) # b 2 , , &, , , … nCn 1 #(a b)# b n 1 nCn # b n, (a b) n nb (a b) n 1 1 n (n 1) b 2 (a b) n 2 , 2, , … nb n 1 (a b) b n, 1, , , (a n b n) (a b) n nb (a b) n 1 n (n 1) b 2 (a b) n 2 , 2, … nb n (a b), , (a b)#[(a b) n 1 nb (a b) n 2, 1 n (n 1) b 2 (a b) n 3 nb n], 2, [taking (a b) common throughout], (a b)#(an int eger) ., Hence, (a b) is a factor of (a n b n) ., EXAMPLE 28, SOLUTION, , The second, third and fourth terms in the binomial expansion of (x a) n, are 240, 720 and 1080 respectively. Find x, a and n., By the binomial expansion, we have, , , , (x a) n nC0 x n nC1 x n 1a nC 2 x n 2a 2 nC 3 x n 3a 3 …, 1, , , &, (x a) n x n nx n 1a n (n 1) x n 2a 2, 2, 1 n (n 1) (n 2) x n 3a 3 … ., 6, n1, , … (i), T2 240 & nx a 240,, , T3 720 & n (n 1) x n 2a 2 1440,, , T4 1080 & n (n 1) (n 2) x n 3a 3 6480., , … (ii), , … (iii), On dividing (ii) by (i), we get, (n 1) a 1440, 6, a, , ·, … (iv), & x, x, 240, ( n 1), On dividing (iii) by (ii), we get, (n 2) a 6480, 9, a, , ·, … (v), & x, x, 1440, 2 ( n 2), a, Equating the values of x from (iv) and (v), we get, 6 , 9, & 12 (n 2) 9 (n 1) & 3n 15 & n 5., (n 1) 2 (n 2), Putting n 5 in (i), we get x 4a 48., 3, a 6, a 3, Putting n 5 in (iv), we get x & x & a x., 2, 2, 4, , … (vi)
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366, , Senior Secondary School Mathematics for Class 11, , 3, x in (vi), we get, 2, 3, 2, x 4 # x 48 & x 5 48 # 32 2 5 & x 2., 2, 3, 3, , a # 2 3., 2, Hence, x 2, a 3, and n 5., Putting a , , EXAMPLE 29, SOLUTION, , If the coefficients of (r 1)th, rth and (r 1)th terms in the expansion of, (x 1) n are in the ratio 1 : 3 : 5, find the values of n and r., We have, ( x 1) n ( 1 x) n, nC 0 nC 1 x nC 2 x 2 … nC r 2 x r 2 nC r 1 x r 1, nC r x r … ., , , … (i), T(r 1) T(r 2) 1 nCr 2 x r 2,, , , … (ii), , Tr 1 nCr x r., It is given that, coeff. of T(r 1) : coeff. of Tr : coeff. of T(r 1) 1 : 3 : 5, , … (iii), , Tr T(r 1) 1 nCr 1 x r 1,, , C(r 2) : nC(r 1) : nCr 1 : 3 : 5, , n, , &, , n, , Cr 2, , &, , n, , n, C, 1 and n r 1 3 ·, 3, 5, Cr, , Cr 1, Cr 2 1, ( r 1) ! · ( n r 1) ! 1, n!, &, , But, n, #, , , , ·, 3, 3, n!, ( r 2) ! ( n r 2) !, Cr 1, n, , ( r 1), 1, (n r 2) 3, & n 4r 5., , &, , … (i), , n, , Cr 1, , r, 3 &, 3, 5, ( n r 1) 5, & 3n 8r 3., Solving (i) and (ii), we get n 7 and r 3., Hence, n 7 and r 3., And,, , n, , Cr, , , , … (ii), , , , EXAMPLE 30, , If the coefficients of x r 1, x r and x r 1 in the binomial expansion of (1 x) n, are in AP, prove that, n 2 n (4r 1) 4r 2 2 0., , SOLUTION, , We know that, , (1 x) n nC0 nC1 x nC 2 x 2 … nCr 1 x r 1 nCr x r, , , So, the coefficients of x r 1, x r and x r, Cr 1, nCr , and nCr 1 respectively., , n, , 1, , nC r 1 x r 1 … ., , in the given expansion are
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Binomial Theorem, , 367, , It is given that nCr 1, nCr , and nCr 1 are in AP., , 2 # nC r nC r 1 nC r 1, n, , &, &, , Cr 1, n, , Cr, , n, , , , Cr 1, n, , Cr, , 2, , (r !) · (n r) !, n!, #, n!, ( r 1) ! · ( n r 1) !, , , &, , &, , (r !) · (n r) !, n!, 2, #, n!, ( r 1) ! · ( n r 1 ) !, , (n r), r, , 2, , , (n r 1) (r 1), [a (n r 1) ! (n r 1) · {(n r) !} and r ! r · {(r 1) !}], r (r 1) (n r) (n r 1) 2 (r 1) (n r 1), , n 2 n (4r 1) 4r 2 2 0., Hence, n 2 n (4r 1) 4r 2 2 0., , &, , EXAMPLE 31, , SOLUTION, , Simplify the expression, (1 x) 1000 x (1 x) 999 x 2 (1 x) 998 … x 1000, and find the coefficient of x 50., Clearly, the given series is a geometric series in which, a (1 x) 1000, r , , , given sum , , x (1 x) 999, (1 x) 1000, , , , x, and n 1001., (1 x), , a (1 r n ), ( 1 r), , x 1001, (1 x) 1000 # (1 b l 2, 1 x, (1 x) 1001 x 1001, , x, b1 l, 1 x, 1 1001C1 x 1001C 2 x 2 … 1001C1000 x 1000, x 1001 x 1001, 1001, 1001, 2, 1001, 1000, 1 C1 x C 2 x … C1 x ., , , coefficient of x 50 in the above expansion, (1001) !, (1001) !, 1001C 50 , , ·, , ·, (50) ! (1001 50) ! (50) ! · (951) !, , EXAMPLE 32, , If n is a positive integer, find the coefficient of x, 1 n, (1 x) n b1 x l ·, , SOLUTION, , We have, , 1, , in the expansion of, , (1 x) n (1 x) 2n, 1 n, , ·, (1 x) n b1 x l (1 x) n #, xn, xn
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368, , Senior Secondary School Mathematics for Class 11, , Coefficient of x, , 1, , 1 n, in the expansion of (1 x) n b1 x l, , (1 x) 2n, xn, 1, n, = coefficient of (x # x ) in the expansion of (1 x) 2n, , = coefficient of x n 1 in the expansion of (1 x) 2n, = coefficient of x, , 1, , in the expansion of, , 2n, , 2n, 2n, r, =a (1 x) / Cr x G ·, , 2nCn 1, EXAMPLE 33, SOLUTION, , r0, , Find the coefficient of x 4 in the expansion of (1 x x 2 x 3) 11., We have, (1 x x 2 x 3) 11 {(1 x) x 2 (1 x)} 11 {(1 x) (1 x 2)} 11, (1 x) 11 #(1 x 2) 11, (11C0 x 0 11C1 x 11C 2 x 2 11C 3 x 3 11C 4 x 4 …), #[11C0 (x 2) 0 11C1 (x 2) 11C 2 (x 2) 2 …], (1 11x 55x 2 165x 3 330x 4 …), #(1 11x 2 55x 4 …) ., , , coefficient of x 4 in the given expansion, (1# 55) (55 #11) (330 #1), (55 605 330) 990., Hence, the coefficient of x 4 in the given expansion is 990., EXAMPLE 34, SOLUTION, , Find the coefficient of x 4 in the product (1 2x) 4 #(2 x) 5., Using the binomial expansion, we get, (1 2x) 4 4C0 4C1 (2x) 4C 2 (2x) 2 4C 3 (2x) 3 4C 4 (2x) 4, 1 8x 24x 2 32x 3 16x 4, and (2 x) 5 2 5 5C1 (2 4) x 5C 2 (2 3) x 2 5C 3 (2 2) x 3, 5C 4 (2) x 4 5C 5 x 5, 32 80x 80x 2 40x 3 10x 4 x 5, 4, , (1 2x) #(2 x) 5, (1 8x 24x 2 32x 3 16x 4), #(32 80x 80x 2 40x 3 10x 4 x 5)., Sum of the terms containing x 4 in the given product, (1#10x 4) 8x ( 40x 3) (24x 2) (80x 2), (32x 3) ( 80x) (16x 4 # 32), 10x 4 320x 4 1920x 4 2560x 4 512x 2, (10 320 1920 2560 512) x 4 438x 4., Hence, the coefficient of x 4 in the given product is 438.
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Binomial Theorem, EXAMPLE 35, , SOLUTION, , 369, , If P be the sum of all odd terms and Q that of all even terms in the, expansion of (x a) n, prove that, (i) (P 2 Q 2) (x 2 a 2) n,, (ii) 4PQ {(x a) 2n (x a) 2n},, (iii) 2 (P 2 Q 2) [(x a) 2n (x a) 2n] ., By the binomial expansion, we have, , , , , (x a ) n n C 0 x n n C 1 x n 1 n C 2 x n 2 n C 3 x n 3 n C 4 x n 4 , … nC n a n., (T1 T2 T3 T4 T5 … Tn Tn 1), (T1 T3 T5 …) (T2 T4 T6 …), P Q., … (I), , , , And, (x a) n nC0 x n nC1 x n 1a nC 2 x n 2a 2 nC 3 x n 3a 3 , … ( 1) n nCn a n, (T1 T2 T3 T4 …), (T1 T3 T5 …) (T2 T4 T6 …), P Q., … (II), (i) On multiplying (I) and (II), we get, , (x 2 a 2) n (P 2 Q 2) ., (ii) On squaring (I) and (II) and subtracting, we get, {(x a) 2n (x a) 2n} {(P Q) 2 (P Q) 2} 4PQ., (iii) On squaring (I) and (II) and adding, we get, {(x a) 2n (x a) 2n} {(P Q) 2 (P Q) 2} 2 (P 2 Q 2) ., EXAMPLE 36, SOLUTION, , Using binomial theorem, find the value of (103) 4., By the binomial expansion, we have, (103) 4 (100 3) 4, 4C0 · (100) 4 4C1 #(100) 3 # 3 4C 2 #(100) 2 # 3 2, 4C 3 #100 # 3 3 4C 4 # 3 4, (100) 4 12 #(100) 3 54 #(100) 2 108 #100 81, 100000000 12 #1000000 540000 10800 81, 100000000 12000000 540000 10800 81, 112550881., Hence, (103) 4 112550881., , EXAMPLE 37, SOLUTION, , Using binomial theorem, find the value of (99) 4., By the binomial expansion, we have, (99) 4 (100 1) 4, 4C0 #(100) 4 4C1 #(100) 3 #1 4C 2 #(100) 2 #1 2, 4C 3 #100 #1 3 4C 4 #1 4
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370, , Senior Secondary School Mathematics for Class 11, , (100) 4 4 #(100) 3 6 #(100) 2 4 #100 1, 100000000 4000000 60000 400 1, 96059601., Hence, (99) 4 96059601., EXAMPLE 38, SOLUTION, , Using binomial theorem, find the value of (0.99) 15 up to four places of, decimal., We have, (0.99) 15 (1 0.01) 15, 1 15C1 #(0.01) 15C 2 #(0.01) 2 15C 3 #(0.01) 3 …, [neglecting higher powers of 0.01], 1 15 #(0.01) 105 #(0.0001) 455 #(0.000001) …, 1 0.15 0.0105 0.000455 0.860045., Hence, (0.99) 15 0.860045., , EXAMPLE 39, , Using binomial theorem, prove that (101) 50 (100 50 99 50) ., , SOLUTION, , Putting (101) 50 a and (100 50 99 50) b, we get, (a b) (101) 50 (100) 50 (99) 50, (101) 50 (99) 50 (100) 50, (100 1) 50 (100 1) 50 (100) 50, 2 #[50C1 #100 49 50C 3 #100 47 … 50C 49 #100] (100) 50, [2 # 50C 3 #100 47 2 # 50C 5 #100 45 … 2 # 50C 49 #100], = (a positive integer)., Thus, a b 0 & a b & (101) 50 (100 50 99 50) ., , EXAMPLE 40, SOLUTION, , Which is larger, (1.01) 1000000 or 10000?, We have, (1.01) 1000000 10000, (1.01) p 10000, where 1000000 p, (1 0.01) p 10000, pC0 pC1 # 0.01 pC 2 #(0.01) 2 … pC p #(0.01) p 10000, 1 (p # 0.01) [pC 2 #(0.01) 2 … (0.01) p] 10000, 1 (1000000 # 0.01) [1000000C 2 #(0.01) 2 … (1.01) 1000000], 10000, 1 10000 [a positive real number] 10000, 1 (a positive real number) 0, Hence, (1.01) 1000000 10000.
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Binomial Theorem, , 371, , EXERCISE 10A, Using binomial theorem, expand each of the following:, 1. (1 2x) 5, , 2. (2x 3) 6, , 3. (3x 2y) 5, , 4. (2x 3y) 4, , 5. b, , 27, 6. bx 2 x l, , 5, , 1, 7. dx y n, , 2x 3 6, l, 3 2x, , 8. ( x y ) 8, , 10. (1 2x 3x 2) 4, , 11. b1 , , 9. ( 3 x 3 y ) 6, , x2 4, l , x!0, 2 x, , 12. (3x 2 2ax 3a 2) 3, , Evaluate:, 13. ( 2 1) 6 ( 2 1) 6, , 14. ( 3 1) 5 ( 3 1) 5, , 15. (2 3 ) 7 (2 3 ) 7, , 16. ( 3 2 ) 6 ( 3 2 ) 6, , n, , 17. Prove that / nCr · 3 r 4 n., r0, , 18. Using binomial theorem, evaluate each of the following:, (i) (101) 4, , (ii) (98) 4, , (iii) (1.2) 4, 19. Using binomial theorem, prove that (2 7n 1) is divisible by 49, where, nd N., 20. Prove that (2 x ) 4 (2 x ) 4 2 (16 24x x 2) ., 3n, , 21. Find the 7th term in the expansion of b, , 4x 5 8 ·, l, 5 2x, a, b 12, 22. Find the 9th term in the expansion of e 2 o ., b 2a, 23. Find the 16th term in the expansion of ( x y ) 17., 24. Find the 13th term in the expansion of e9x , , 1, 3 x, , 18, , o , x ! 0., , xn, 25. If the coefficients of x 7 and x 8 in the expansion of b2 l are equal then find, 3, the value of n., 26. Find the ratio of the coefficient of x 15 to the term independent of x in the, 2 15, expansion of bx 2 x l ·, 27. Show that the ratio of the coefficient of x 10 in the expansion of (1 x 2) 10 and, 2 10, the term independent of x in the expansion of bx x l is 1 : 32., 3, 1 9, 28. Find the term independent of x in the expansion of (1 x 2x 3) b x 2 l ·, 2, 3x, 29. Find the coefficient of x in the expansion of (1 3x 7x 2) (1 x) 16.
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372, , Senior Secondary School Mathematics for Class 11, , 30. Find the coefficient of, (i) x 5 in the expansion of (x 3) 8., 1 9, (ii) x 6 in the expansion of b3x 2 l ·, 3x, a 10, , (iii) x 15 in the expansion of d3x 2 3 n ·, 3x, (iv) a 7b 5 in the expansion of (a 2b) 12., 31. Show that the term containing x 3 does not exist in the expansion of, 1 8, b3x l ·, 2x, 1 20, 32. Show that the expansion of b2x 2 x l does not contain any term, involving x 9., 1 12, 33. Show that the expansion of bx 2 x l does not contain any term, , involving x 1., 34. Write the general term in the expansion of (x 2 y) 6., 1 12, 35. Find the 5th term from the end in the expansion of bx x l ·, 36. Find the 4th term from the end in the expansion of b, , 4x 5 9·, l, 5 2x, , 37. If the 7th terms from the beginning and end in the expansion of e 3 2 , are equal, find the value of n., 38. Find the middle term in the expansion of:, 8, x, (i) (3 x) 6, (ii) b 3yl, 3, x a 10, 2 10, (iii) a a x k, (iv) bx 2 x l, , 3, , 1 n, o, 3, , 39. Find the two middle terms in the expansion of:, (i) (x 2 a 2) 5, p x 9, (iii) d x p n, , 1 11, n, x3, 3 9, x, (iv) d3x n, 6, (ii) dx 4 , , 40. Find the term independent of x in the expansion of:, 1 9, 2n, 3x, 1 3n, (iii) dx 2 n, x, (i) d2x , , 6, , 2, , 3x 1 n, (ii) d, 2, 3x, (iv) d3x , , 2 15, 2n, x, , 41. Find the coefficient of x 5 in the expansion of (1 x) 3 (1 x) 6., 42. Find numerically the greatest term in the expansion of (2 3x) 9, where
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376, , Senior Secondary School Mathematics for Class 11, Putting a 2 and b x , we get, , (2 x ) 4 (2 x ) 4 2 [2 4 6 # 2 2 #( x ) 2 ( x ) 4] 2 (16 24x x 2) ., xr, , , , 25. We have, Tr 1 nCr 2 n r #c m nCr # 2 n r # 3 r # x r., 3, , , , , Coeff. of x 7 coeff. of x 8 & nC7 # 2 n 7 # 3 7 nC8 # 2 n 8 # 3 8., 8, n8, (8 !)#(n 8) ! 1, n!, 2 n 7 # 3 7 1 1 &, ·, #, 2# 3 6, 6, (7 !) (n 7) !, (n !), C8 2, #3, 8 1, , & n 8 48 & n 56., n8 6, , , 2r, 26. Tr 1 15Cr (x 2) 15 r #c x m 15Cr # 2 r # x 30 3r, [30 3r 15 & 3r 15 & r 5] and [30 3r 0 & 3r 30 & r 10] ., Find (T6 : T11) (T5 1 : T10 1) ., n, , , , C7, , n, , 3, 1 9, 28. In the expansion of E c x 2 m , we have, 2, 3x, 3 (9 r ) 1 r, c m, Tr 1 (1) r · 9Cr · c x 2m, 2, 3x, (9 2r), , &, , , , , 3, r 9, (18 3r), ., Tr 1 (1) · Cr · (9 r) · x, 2, RS, VW, 1, 1, (1 x 2x 3) SSSea0 # 3 a1 # x a2o from EWWW, S, W, x, T, _bWV, _bX Z], SRSZ]], 5, 3, bW, b ], (1 x 2x 3) SS][(1) 7 · 9C7 · 3 2 # 13 b` ][(1) 6 · 9C6 · 3 3 # x 0b`WW, SS], bWW, b, ], 2, x, 2, aX, a \, T\, x 1 & 18 3r 1 & r is fraction, =, G, 18 3r 0 & r 6 and 18 3r 3 & r 7, , (1 x 2x 3) = 1 3 7 G, 18, 27x, required term c, , 2 7, m 17 ·, 27 18, 54, , 29. In the expansion of E (1 x) 16, we have, r 16, r, Tr 1 (1) · Cr x ., Product of given expressions (1 3x 7x 2) (1 16x …) ., Terms containing x [1#(16x) ( 3x)#1] (19x) ., coefficient of x 19., a r, , 30. (iii) Tr 1 (1) r #10Cr (3x 2) (10 r) #d 3 n, 3x, , , Tr 1 (1) r #10Cr # 3 (10 2r) # x (20 5r) # a r., &, Now, 20 5r 15 & 5r 35 & r 7., , (iv) Tr 1 (1) r #12Cr # a (12, , r), , #(2b) r, , , Tr 1 (1) r #12Cr # 2 r # a (12 r)b r., &, Put r 5 and get T6 ., , &, , r, , ·c 1 m, 2x, r 8, (8 r ) 1, , , ·, ·, · r · x (8 2r)., Tr 1 ( 1) Cr 3, 2, , 31. Tr 1 (1) r · 8Cr · (3x) (8, , r)
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Binomial Theorem, , 377, , 5, Now, 8 2r 3 & r , which is a fraction., 2, r, 6, (12 2r), # y r., 34. Tr 1 (1) # Cr # x, 35. pth term from the end (n p 2)th term from the beginning., 5th term from the end (12 5 2)th = 9th term., 37. 7th term from the end (n 7 2)th term (n 5)th term., , , T7 Tn 5 & n 5 7 & n 12., , 40. (iii) Tr 1 (1) r · 3nCr x (3n, , r), , #e, , Tr 1 (1) r · 3nCr x (3n, &, Put 3n 3r 0 & r n., So, Tn 1 is free from x., And, Tn 1 (1) n · 3nCn ., , 1 r, o, x2, , 3r ), , ., , 41. (1 x) 3 (1 x) 6 (3C0 3C1 x 3C2 x 2 3C3 x 3), #( 6C0 6C1 x 6C2 x 2 6C3 x 3 6C 4 x 4 6C 5 x 5 6C6 x 6), (1 3x 3x x )#(1 6x 15x 2 20x 3 15x 4 6x 5 x 6) ., Coeff. of x in the given product ( 6 45 60 15) 6., 2, , 3, , 5, , 42. We have, (2 3x) 9 2 9 c1 , Tr 1 2 9 # 9Cr c, , , 3x 9 ·, m, 2, , 3x r, 3x r 1, m and Tr 2 9 # 9Cr 1 c m ·, 2, 2, , r, r1, Tr 1, 2 9 # 9Cr # 3 r # x r # 2 r 1 # r1 1 # 9 19, Tr, 2, 2 # C, 3, x, , r1, , C, (9 !), (r 1) !, 3 # x # 9 r c 3 xm#, #, #(10 r) !, 2, 2, (r !) · (9 r) !, (9 !), C, 9, , r1, , (10 r) 9 (10 r), c 3 # 3 m#, , r, 2 2, 4r, , 3, ;a x E ·, 2, , Let the greatest term be Tr 1 ., Then, Tr 1 $ Tr &, , Tr 1, 9 (10 r), $1 &, $1, Tr, 4r, , & 4r # 90 9r & 13r # 90 & r # 6, , 12 ·, 13, , Thus, the maximum value of r is 6 for which Tr 1 is greatest., , , , 36, 3, # x 6, where x ·, 2, 26, 3, 36 36, 9, 9, numerical value of greatest term at x 2 # C3 # 6 # 6, 2, 2, 2, greatest term T6 1 2 9 # 9C6 #, , 12, 13, 9# 8 #7 # 3 3 e 7 # 3 o ·, 2, 3 # 2 #1 2, , 43. It is given that 2nC1, 2nC2 and 2nC3 are in AP., , , 2 # 2nC2 2nC1 2nC3
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Binomial Theorem, , 379, , EXERCISE 10B, Very-Short-Answer Questions, 1 10, 1. Show that the term independent of x in the expansion of bx x l is 252., 2. If the coefficients of x 2 and x 3 in the expansion of (3 px) 9 are the same then, 9, prove that p ·, 7, 1 11, , 3. Show that the coefficient of x 3 in the expansion of bx x l is –330., NO10, JK2x 2, 3 2OO is 252., 4. Show that the middle term in the expansion of KK, K 3, 2x O, P, L, x 3 10 405, ·, 5. Show that the coefficient of x 4 in the expansion of d 2 n is, 2 x, 256, 3 11, 6. Prove that there is no term involving x 6 in the expansion of b2x 2 x l ·, 7. Show that the coefficient of x 4 in the expansion of (1 2x x 2) 5 is 212., 8. Write the number of terms in the expansion of ( 2 1) 5 ( 2 1) 5., 1 9, 2n ?, 3x, 10. Write the coefficient of the middle term in the expansion of (1 x) 2n., 9. Which term is independent of x in the expansion of dx , , 11. Write the coefficient of x 7y 2 in the expansion of (x 2y) 9., 12. If the coefficients of (r 5)th and (2r 1)th terms in the expansion of (1 x) 34, are equal, find the value of r., 7, KJ 3 x 3ON, 13. Write the 4th term from the end in the expansion of KK 2 OOO ·, Kx, 6, P, L, n, n, 14. Find the coefficient of x in the expansion of (1 x) (1 x) ., 15. In the binomial expansion of (a b) n, the coefficients of the 4th and 13th, terms are equal to each other. Find the value of n., 16. Find the positive value of m for which the coefficient of x 2 in the expansion, of (1 x) m is 6., ANSWERS (EXERCISE 10B), , 8. Three, 13., , 35x, 48, , 6, , 9. 4th, 14. (1) n #(n 1), , 10., , 2n, , Cn, , 15. n 15, , 11. 144, , 12. 14, , 16. m 4, , HINTS TO SOME SELECTED QUESTIONS
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380, , Senior Secondary School Mathematics for Class 11, r, , · c 1 m (1) r · 10Cr x (10 2r)., x, Putting 10 2r 0, we get r 5., 10 # 9# 8 #7 # 6 , T6 (1) 5 · 10C 5 , 252., 5 # 4 # 3 # 2 #1, , 1. Tr 1 (1) r · 10Cr x (10, , 2. Tr 1 9Cr 3 (9, , r), , r), , #(px) r 9Cr 3 (9, , r), , · p rx r., , Coeff. of x 2 coeff. of x 3 & 9C2 # 3 7 # p 2 9C3 # 3 6 # p 3., r, , · c 1 m (1) r · 11Cr x (11 2r)., x, Now, 11 2r 3 & 2r 14 & r 7., 11#10 # 9 # 8 3 , 3, 3, 3, 7 11, 11, T8 (1) · C7 x C 4 x , x, 330x ., 4 # 3 # 2 #1, , 3. Tr 1 (1) r · 11Cr x (11, , 2 10 r, , r), , 3 r 10, , , 10 2r, # 3 2r 10 # x (20 4r)., Cr # 2, 2o, 2x, The expansion has 11 terms. So, middle term = 6th term., 10 # 9# 8 #7 # 6 , m 252., T6 T5 1 10C 5 # 2 0 # 3 0 # x 0 c, 5 # 4 # 3 # 2 #1, 3 r, x (10 r), , , 5. Tr 1 (1) r · 10Cr · c m, #e 2 o (1) r · 10Cr · 3 r · 2 r 10 · x (10 3r)., 2, x, Putting 10 3r 4, we get 3r 6 & r 2., 405 4, 8, 2 10, 2, 4, T3 T(2 1) (1) · C2 # 3 # 2 # x , x ., 256, r, 3, , , , 6. Tr 1 (1) r · 11Cr (2x 2) (11 r) · c x m (1) r · 11Cr 2 (11 r) # 3 r · x 22 3r, 16, 22 3r 6 & 3r 16 & r , which is a fraction., 3, 4. Tr 1 10Cr e, , 2x o, 3, , #e, , 7. (1 2x x 2) 5 {(1 x) 2} 5 (1 x) 10., 10 # 9# 8 #7 , 4 10, 210., Tr 1 10Cr x r. Coefficient of x C 4 , 4 # 3 # 2 #1, 8. Each expansion gives 6 terms. Their sum will contain 3 terms, as 3 terms are cancelled., r, · e 1 2 o (1) r · 9Cr · x (9 3r) · 1r ·, 3, 3x, Putting 9 3r 0, we get r 3., Hence, T4, i.e., 4th term is independent of x., 2n, 10. Middle term c 1mth term (n 1)th term Tn 1 ., 2, Now, Tr 1 2nCr x r & Tn 1 2nCn x n. Its coefficient is 2nCn ., , 9. Tr 1 (1) r · 9Cr · x 9, , 11. Tr 1 9Cr x (9, , r), , r, , · (2y) r 9Cr # 2 r # x (9 r) # y r., , Putting r 2, we get the coefficient of x 7y 2 as 9C2 # 2 2 144., 12. Tm T(m 1) 1 34Cm 1 # x m, , , Tr 5 34Cr 6 x r, , Now, Cr 6 , , 6, , 1, , ., , and T2r 1 34C2r 2 x 2r, , 2, , ., , C2r 2 & (r 6 2r 2) or (r 6 2r 2 34), & r 4 or r 14 & r 14., , 13. rth term from the end (n r 2)th term., 34, , , , 34, , 4th term from the end (7 4 2)th term = 5th term.
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Binomial Theorem, , 381, , r, , 3 (7 r ) e x 3 o r 7, , , r 7, (7 2r), ·, # 2 r # x 5r 614., Tr 1 (1) · Cr e 2 o, ( 1) # C r # 3, 35, x, 1, 61, 4, 4x 7, 6 7, 6, ·, T5 T(4 1) (1) # C 4 # 3 # 2 # x C3 #, x , 48, (3 # 2 4), 14. Tr 1 in (1 x) n (1) r, , 1 n, , · C r · x r., , , , (1 x) (1 x) (1 x) [1 nC1 x … (1) n # nCn 1 x n 1 ( 1) n 1 · nCn x n] ., n, , Coeff. of x n in the given product (1 x) (1 x) n [(1) n, 15. Tr 1 Cr a, n, , nr, , (1), , 1 n, , n1, , · Cn (1) n · nCn 1], , (1) n # n (1) n #(n 1) ., , r, , ·b ., , Coeff. of T4 coeff. of T13 & nC3 nC12 & (3 12) n & n 15., 16. (1 x) m mC0 mC1 x mC2 x 2 … mCm x m., m (m 1), 6 & m 2 m 12 0 & (m 4) (m 3) 0 & m 4., mC2 6 &, 2!, , REVIEW OF FACTS AND FORMULAE, 1. Binomial Theorem, If x and a are real numbers then for all nd N, we have, , , , (x a) n nC0 x na 0 nC1 x n 1a nC 2 x n 2a 2 … nCr x n ra r … nCn x 0a n, n, , , , & (x a) n / nCr x n ra r., r0, , In this expansion, we have, (i) total number of terms (n 1) ., (ii) sum of the powers of x and a in each term is n., (iii) coefficients of terms equidistant from the beginning and the end, are equal., , (iv) general term is given by Tr 1 nCr x n ra r., 2. (1 x) n nC0 nC1 # x nC 2 # x 2 … nCr # x r … nCn x n, n, , & (1 x) n / nCr x r., r0, , In this expansion, we have, (i) total number of terms (n 1) ., (ii) general term, Tr 1 nCr x r., , , , , , , 3. (x a) n nC0 x n nC1 x n 1a nC 2 x n 2a 2 … (1) r · nCr x n ra r …, (1) n · nCn x n, n, n, r n, nr r, & (x a) / (1) · Cr x a ., r0, , In this expansion, we have, (i) total number of terms (n 1) ., (ii) sum of the powers of x and a in each term is n., , (iii) general term is given by Tr 1 (1) r · nCr x n ra r.
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382, , Senior Secondary School Mathematics for Class 11, , 4. (1 x) n nC0 nC1 x nC 2 x 2 nC 3 x 3 … (1) r · nCr x r … (1) n· nCn x n, n, , & ( 1 x) n / ( 1 ) r · n C r x r ., r0, , In this expansion, we have, (i) total number of terms (n 1) ., (ii) general term, Tr 1 (1) r · nCr x r., 5., , , , , , (i) (x a) n (x a) n 2 [nC0 x na 0 nC 2 x n 2a 2 nC 4 x n ra 4 …]., , , , , , , (ii) (x a) n (x a) n 2 [nC1 x n 1a nC 3 x n 3a 3 nC 5 x n 5a 5 …]., 6. Middle Terms, n, (i) If n is even then a 1kth term is the middle term., 2, 1 , 1, (ii) If n is odd then (n 1)th and (n 3)th terms are the middle terms., 2, 2,
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Arithmetic Progression, , 383, , 11, , Arithmetic Progression, ARITHMETIC PROGRESSION, A succession of numbers arranged in a definite order according to a, certain given rule is called a sequence., , SEQUENCE, , The number occurring at the nth place of a sequence is called its nth term or, the general term, to be denoted by an ., A sequence is said to be finite or infinite according as the number of terms in it is, finite or infinite respectively., By adding the terms of a sequence, we get a series., A series is said to be finite or infinite according as the number of terms in it is, finite or infinite respectively., EXAMPLE 1, SOLUTION, , Write first five terms of the sequence given by the rule an (2n 1) and, obtain the corresponding series., We have, an (2n 1) ., ... (i), Putting n 1, 2, 3, 4, 5, … successively in (i), we get, a1 (2 #1 1) 3; a 2 (2 # 2 1) 5; a 3 (2 # 3 1) 7;, a 4 (2 # 4 1) 9 and a 5 (2 # 5 1) 11., Hence, the required sequence is 3, 5, 7, 9, 11, ... ., The corresponding series is 3 5 7 9 11 … ., , EXAMPLE 2, , Write first four terms of the sequence given by an , the corresponding series., , SOLUTION, , 1, (2n 3) and obtain, 6, , 1, (2n 3) ., 6, Putting n 1, 2, 3, 4, … successively in (i), we get, We have, an , , 1, 1, 1, 1, (2 #1 3) , ; a (2 # 2 3) ;, 6, 6 2 6, 6, 3, 5, 1, 1, 1, a 3 (2 # 3 3) ; a 4 (2 # 4 3) ·, 6, 6 2, 6, 6, 1 1 1 5, , , , ,… ., Hence, the required sequence is, 6 6 2 6, 1 1 1 5, The corresponding series is … ., 6 6 2 6, a1 , , 383, , ... (i)
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384, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 3, , The Fibonacci sequence is defined by, 1 a1 a 2 and an an 1 an 2, n 2., an 1, Find a for n 1, 2, 3, 4, 5., n, , SOLUTION, , We have, a1 1, a 2 1 and an an 1 an 2 for n 2., , a 3 (a 2 a1) (1 1) 2, a 4 (a 3 a 2) (2 1) 3,, a 5 (a 4 a 3) (3 2) 5., a3 2, a5 5, a2, a4 3, , , , , , a1 1, a 2 1 2, a 3 2 and a 4 3 ·, , PROGRESSIONS, , Sequences following certain patterns are called progressions., ARITHMETIC PROGRESSION (AP), , It is a sequence in which each term except the first one differs from its preceding term, by a constant., This constant difference is called the common difference of the AP., In an AP we usually denote the first term by a, the common difference by d, and the nth term by Tn ., SOME EXAMPLES OF AP, EXAMPLE 1, SOLUTION, , EXAMPLE 2, , Show that the sequence defined by Tn 3n 5 is an AP. Find its common, difference., We have, Tn 3n 5., … (i), Replacing n by (n 1) in (i), we get, … (ii), Tn 1 3 (n 1) 5 & Tn 1 3n 2., Subtracting (ii) from (i), we get, (Tn Tn 1) (3n 5) (3n 2) 3, which is constant., Hence, the given sequence is an AP with common difference 3., Show that the sequence log a, log e, AP. Find its common difference., , SOLUTION, , KJa 4ON, a2o, a3, , log e 2 o, log KKK 3OOO, … forms an, b, b, b, L P, , By symmetry, we find that, JK n 1NO, an, a, and Tn 1 log KKK n 2OOO ·, n1o, b, b, P, L, JK n 1NO, n, a, a, , (Tn Tn 1) log e n 1 o log KK n 2OO, Kb O, b, P, L, JK n, n 2N, O, log KK an 1 # b n 1OO log b a l constant., Kb, b, a O, L, P, a, Hence, the given sequence in an AP with common difference log b l ·, b, Tn log e
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Arithmetic Progression, , 385, , Show that the sequence defined by Tn 3n 2 2 is not an AP., We have, 2, 2, 2, Tn 3n 2 and Tn 1 3(n 1) 2 & Tn 1 3n 6n 5., , EXAMPLE 3, SOLUTION, , 2, 2, , (Tn Tn 1) (3n 2) (3n 6n 5) & (Tn Tn 1) (6n 3) ., This shows that (Tn Tn 1) is not independent of n and therefore, it, is not constant., Hence, the given sequence is not an AP., , GENERAL TERM OF AN AP, , Show that the nth term of an AP with first term a and common difference, d is given by Tn a (n 1) d., Let us consider an AP with first term a and common difference d., Then, the given AP is, a, (a d), (a 2d), (a 3d), (a 4d), … ., In this AP, we have, first term, T1 a (a 0 # d) a (1 1) d;, second term, T2 (a d) a (2 1) d;, third term, T3 (a 2d) a (3 1) d;, ……, ……, ……, ……, ……, ……, ……, ……, ……, ……, nth term, Tn a (n 1) d., Hence, Tn a (n 1) d., This is called the general term of the AP., , THEOREM 1, PROOF, , In an AP with first term = a and common difference = d, we have, Tn a (n 1) d., SOME SIMPLE PROPERTIES OF AN AP, , (i) If a constant is added to each term of an AP then the resulting, progression is an AP., (ii) If a constant is subtracted from each term of an AP then the resulting, progression is an AP., (iii) If each term of an AP is multiplied by the same nonzero number then, the resulting progression is an AP., (iv) If each term of an AP is divided by the same nonzero number then the, resulting progression is an AP., SOLVED EXAMPLES, EXAMPLE 1, , Show that the progression 7, 12, 17, 22, 27, … is an AP. Find its general, term and the 14th term.
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386, , Senior Secondary School Mathematics for Class 11, , SOLUTION, , We have, (12 7) (17 12) (22 17) (27 22) 5, which is constant., So, the given progression is an AP in which a 7 and d 5., Its general term, Tn {a (n 1) d} 7 (n 1)# 5 & Tn (5n 2) ., , 14th term, T14 (5 #14 2) 72., , EXAMPLE 2, , Show that the progression log a, log (ab), log (ab 2), log (ab 3), … is an, AP. Find its general term and the 10th term., We have, log (ab) log a log a log b log a log b,, log (ab 2) log (ab) (log a 2 log b) (log a log b) log b,, , SOLUTION, , log (ab 3) log (ab 2) (log a 3 log b) (log a 2 log b) log b., (T2 T1) (T3 T2) (T4 T3) log b constant., , , So, the given progression is an AP., Let A be the first term and d be the common difference of this AP., Then, A log a and d log b., , general term, Tn A (n 1) d, &, Tn log a (n 1) log b., And, 10th term, T10 log a (10 1) log b log a 9 log b, &, T10 log a log (b 9) & T10 log (ab 9) ., EXAMPLE 3, SOLUTION, , EXAMPLE 4, SOLUTION, , Show that the progression 21, 16, 11, 6, 1, … is an AP. Write its first term, and the common difference. Which term of this AP is –54?, The given progression is 21, 16, 11, 6, 1, … ., We have, (16 21) (11 16) (6 11) (1 6) 5, which is constant., So, the given progression is an AP., Its first term, a 21 and common difference, d 5., Let its nth term be –54. Then,, Tn 54 & a (n 1) d 54, & 21 (n 1)#( 5) 54, & 5n 26 54 & 5n 80 & n 16., , T16 54., Hence, the 16th term of the given AP is –54., Show that the progression 11, 7, 3, 1, 5, …, 161 is an AP. How many, terms does it have?, We have, ( 7) (11) ( 7 11) 4, ( 3) ( 7) ( 3 7) 4,, 1 ( 3) (1 3) 4 and (5 1) 4., , {( 7) (11)} {( 3) ( 7)} {1 ( 3)} (5 1) 4, which is, constant.
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Arithmetic Progression, , 387, , So, the given progression is an AP in which a 11 and d 4., Let it have n terms. Then,, Tn 161 & a (n 1) d 161, & (11) (n 1)# 4 161, & 4n 176 & n 44., , T44 161 and hence there are 44 terms in the given AP., EXAMPLE 5, SOLUTION, , EXAMPLE 6, SOLUTION, , EXAMPLE 7, SOLUTION, , Is 319 a term of the AP 11, 17, 23, 29, 35, …?, In the given AP, we have a 11 and d (17 11) 6., If possible, let the nth term of this AP be 319. Then,, Tn 319 & a (n 1) d 319, & 11 (n 1)# 6 319, & 6n 5 319 & 6n 314, 157 1, & 3n 157 & n , 52 ·, 3, 3, Since the number of terms cannot be a fraction, so it follows that, 319 is not a term of the given AP., 3, 1, 1, Which term of the AP 30, 29 , 28 , 27 , … is the first negative term?, 2, 4, 4, We have, 57 117 111 57 3, 117 , ·, 30l b , l b, l, b, 2, 2, 4, 4, 4, 4, 3, ·, , a 30 and d , 4, Let its nth term be the first negative term. Then,, Tn 0 & a (n 1) d 0, 3, & 30 (n 1)#b l 0, 4, & 123 3n 0 & 123 3n, & 3n 123 & n 41., Hence, the 42nd term of the given AP is the first negative term., 3, 3, ·, And, T42 a (42 1) d 30 41#b l , 4, 4, 3, ·, Hence, the first negative term of the given AP is, 4, Find the 10th common term between the arithmetic series, 3 7 11 15 … and 1 6 11 16 … ., Clearly, the first common term in the two series is 11., LCM of the common differences of the two series = LCM (4, 5) = 20., Let us now consider an AP in which a 11 and d 20., Every term of this AP is a common term of the two given series.
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388, , Senior Secondary School Mathematics for Class 11, , So, the required 10th common term, {11 (10 1)# 20} (11 180) 191., Hence, 191 is the 10th common term of the two given series., EXAMPLE 8, , SOLUTION, , If a1, a 2, a 3, …, an are in AP with common difference, d (where d ! 0) then prove that the sum of the series, sin d (cosec a1 cosec a 2 cosec a 2 cosec a 3 … cosec an 1 cosec an) is, equal to (cot a1 cot an) ., Since a1, a 2, a 3, …, an are in AP with common difference d, we have, (a 2 a1) (a 3 a 2) (a 4 a 3) … (an an 1) d., , sin d (cosec a1 cosec a 2 cosec a 2 cosec a 3 , … cosec an 1 cosec an), sin d, sin d, sin d, sin d, , , …, sin a1 sin a 2 sin a 2 sin a 3 sin a 3 sin a 4, sin an 1 sin an, sin (an an 1), sin (a 2 a1) sin (a 3 a 2) sin (a 4 a 3), , , , …, sin a1 sin a 2 sin a 2 sin a 3 sin a 3 sin a 4, sin an 1 sin an, [a d (a 2 a1) (a 3 a 2) … (an an 1)], sin a 2 cos a1 cos a 2 sin a1 sin a 3 cos a 2 cos a 3 sin a 2, , , , sin a1 sin a 2, sin a 2 sin a 3, sin an cos an 1 cos an sin an 1, …, sin an 1 sin an, , , cos an 1 cos an, cos a1 cos a 2, cos a 2 cos a 3, , , , n, o…d, ne, sin a1 sin a 2, sin a 2 sin a 3, sin an 1 sin an, (cot a1 cot a 2) (cot a 2 cot a 3) (cot a 3 cot a 4) , … (cot an 1 cot an), (cot a1 cot an) ., Hence, sin d (cosec a1 cosec a 2 cosec a 2 cosec a 3 , … cosec an 1 cosec an) (cot a1 cot an) ., d, , EXAMPLE 9, , If a1, a 2, a 3, …, an are in AP, where ai 0 for each i, show that, (n 1), 1, 1, 1, , …, , ·, a1 a 2, a2 a3, an 1 an, a1 a n, , SOLUTION, , Let d be the common difference of the given AP. Then,, (a 2 a1) (a 3 a 2) … (an an 1) d., Now, LHS , , , 1, 1, 1, , …, , , , ( a2, a1 ) ( a 3, a2 ), ( an, an 1 ), ( a 2 a1 ), ( a 2 a1 ) ( a 2 a1 ), , , , ( a3 a2 ), ( a3 a2 ) ( a3 a2 ), , …, , , , ( an an 1 ), ( an an 1 ) ( an an 1 )
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Arithmetic Progression, , , , , 389, , ( an an 1 ), ( a 2 a1 ) ( a 3 a 2 ), …, , (a 2 a1), (a 3 a 2), (an an 1), ( an an 1 ), ( a 2 a1 ) ( a 3 a 2 ), , …, d, d, d, [a (a 2 a1) (a 3 a 2) (a 4 a 3) … (an an 1) d], , 1 · {( a 2 a1 ) ( a 3 a 2 ) ( a 4 a 3 ) , d, … ( an an 1 )}, _, ]Z], ( an a1 ) bbb, 1 · ( an a1 ) 1 · ][]( an a1 )#, `b, d, d ]], ( an a1 ) bb, \, a, ]Z] (an a1) b_b 1 ]Z] a1 (n 1) d a1 b_b, 1 · [], `b · [], `b, d ]( an a1 ) b d ], a1 a n b, \, a, \, a, [a an a1 (n 1) d], Z] (n 1) d _b, ( n 1), ], b, 1 · [], RHS., `, d ] a1 an bb ( a1 an ), \, a, Thus, LHS = RHS and hence the result follows., EXAMPLE 10, , If a1, a 2, a 3, …, an be an AP of nonzero terms then prove that, 1 1 1 1 (n 1) ·, a1 a 2 a 2 a 3 a 3 a 4 … a n 1 a n, a1 a n, , SOLUTION, , Let d be the common difference of the given AP. Then,, (a 2 a1) (a 3 a 2) (a 4 a 3) … (an an 1) d., 1 1 1 1, , a1 a 2 a 2 a 3 a 3 a 4 … a n 1 a n, 1 ·( d d d … d 2, an 1 an, d a1 a 2 a 2 a 3 a 3 a 4, , , , (a a ), ( a a ) (a a ) (a a ), 1 ·) 2 1 3 2 4 3 … n n 1 3, a, a, a, a, a, a, an 1 an, d, 1 2, 2 3, 3 4, [a (a 2 a1) (a 3 a 2) … (an an 1) d], 1 · )d 1 1 n d 1 1 n d 1 1 n … d 1 1 n3, a1 a 2, a2 a3, a3 a4, an 1 an, d, , (a a ), 1 ·d1 1 n 1 · n 1, a1 a n, d a1 a n, d, {a (n 1) d} a1, G, 1 ·= 1, a1 a n, d, (n 1) d (n 1), 1·, , a1 a n, a1 a n ·, d, Hence, the result follows., , [a an a1 (n 1) d]
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390, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 11, SOLUTION, , In an AP, the pth term is q and (p q)th term is 0. Then, prove that its qth, term is p., Let a be the first term and d be the common difference of the given, AP. Then,, … (i), Tp q & a (p 1) d q, and Tp q 0 & a (p q 1) d 0., … (ii), On subtracting (i) from (ii), we get, qd q & d 1., Putting d 1 in (i), we get a (q p 1) ., , Tq a (q 1) d (q p 1) (q 1)#(1) & Tq p, [a a (q p 1) and d 1]., Hence, the qth term is p., , TO FIND THE nTH TERM FROM THE END OF AN AP, , Show that the nth term from the end of an AP with the first term a,, common difference d and the last term l is given by {l (n 1) d} ., Let a be the first term, d be the common difference and l be the last, term of a given AP., Then, the AP is given by, a, a d, a 2d, …, (l 2d), (l d), l., last term l l (1 1) d, 2nd term from the end (l d) l (2 1) d, 3rd term from the end (l 2d) l (3 1) d, … … … … … … … … …, … … … … … … … … …, nth term from the end l (n 1) d., Hence, nth term from the end l (n 1) d., , THEOREM 2, PROOF, , EXAMPLE 12, SOLUTION, , Find the 17th term from the end of the AP 36, 31, 26, 21, …, 79., Here a 36, d { 31 ( 36)} 5 and l 79., , 17th term from the end l (n 1) d, 79 (17 1)# 5 (79 80) 1., Hence, the 17th term from the end is –1., , SOME RESULTS ON AP, , If the nth term of a progression is a linear expression in n then show that, it is an AP., Let the nth term of a progression be a linear expression in n., Then, Tn an b,, … (i), where a and b are constants., Replacing n by (n 1) in (i), we get, … (ii), Tn 1 a (n 1) b., , THEOREM 1, PROOF
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Arithmetic Progression, , 391, , On subtracting (ii) from (i), we get, (Tn Tn 1) a, which is constant., This shows that the difference between any two consecutive terms of, the given progression is constant., Hence, the given progression is an AP., If the mth term of an AP be (1/n) and its nth term be (1/m) then show that, its (mn)th term is 1., In the given AP, let the first term a and the common difference d., 1, 1, Then, Tm n and Tn m, 1, 1, … (i), and a (n 1) d m ·, … (ii), & a (m 1) d n, On subtracting (ii) from (i), we get, (m n), 1 1, (m n) d b n m l & (m n) d mn, 1, & d mn ·, 1, Putting d mn in (i), we get, (m 1) 1, 1 (m 1), a mn n & a (n mn 2, m (m 1), 1, & a, & a mn ·, mn, 1, 1, Thus, we get a mn and d mn ·, (mn)th term a (mn 1) d, (mn 1), 2 mn 1., ( 1 , mn, mn, mn, Hence, the (mn)th term of the given AP is 1., , THEOREM 2, PROOF, , If the mth term of a given AP is n and its nth term is m then show that its, pth term is (n m p) ., In the given AP, let the first term = a and the common difference = d., Then, Tm n and Tn m, … (i), and a (n 1) d m., … (ii), & a (m 1) d n, On subtracting (ii) from (i), we get, (m n) d (n m) & d 1., Putting d 1 in (i), we get a (n m 1) ., pth term a (p 1) d, (n m 1) (p 1)#(1) [a a (n m 1) and d 1], (n m p) ., Hence, the pth term of the given AP is (n m p) ., , THEOREM 3, PROOF
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392, , Senior Secondary School Mathematics for Class 11, , If m times the mth term of an AP is equal to n times its nth term, show that, the (m n)th term of the AP is zero., In the given AP, let the first term = a and the common difference = d., Now, m · Tm n · Tn, & m · [a (m 1) d] n · [a (n 1) d], & [(m 2 n 2) (m n)]# d (n m)# a, & ( m n 1) d a, & a (m n 1) d 0, & Tm n 0., Hence, the (m n)th term of the given AP is zero., , THEOREM 4, PROOF, , If the pth, qth and rth terms of an AP be a, b, c respectively, show that, (q r) a (r p) b (p q) c 0., Let x be the first term and d be the common difference of the given AP., Then, Tp a, Tq b and Tr c, Z], ]] x (p 1) d a, … (i), ], … (ii), & ][] x (q 1) d b, ]], … (iii), x (r 1) d c., \, On subtracting (ii) from (i), we get, (a b), ·, (p q) d (a b) & d , (p q), , THEOREM 5, PROOF, , Again, subtracting (iii) from (ii), we get, (b c), ·, (q r) d (b c) & d , (q r), (a b) (b c), , , (p q) (q r), & (q r) a (q r) b (p q) b (p q) c, & (q r) a (r p) b (p q) c 0., Hence, (q r) a (r p) b (p q) c 0., REMARK, , The above result may also be written as, (a b) r (b c) p (c a) q 0., Show that the sum of (m n)th and (m n)th terms of an AP is equal to, twice the mth term., Let a be the first term and d be the common difference of the given AP., Then,, Tm n Tm n {a (m n 1) d} {a (m n 1) d}, 2a 2(m 1) d, 2 {a (m 1) d} 2 #Tm ., Hence, Tm n Tm n 2 #Tm ., , THEOREM 6, PROOF, , [each equal to d]
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Arithmetic Progression, , 393, , Show that in an AP, the sum of the terms equidistant from the beginning, and end is always equal to the sum of the first and last terms., Let a1, a 2, a 3, …, an be an AP with common difference d. Then,, (mth term from the beginning) + (mth term from the end), {a1 (m 1) d} {an (m 1) d} (a1 an), = sum of the first and last terms of the given AP., , THEOREM 7, PROOF, , PROBLEMS BASED ON AP, For solving problems based on AP it is always convenient to make following, choices:, (i) 3 numbers in AP as (a d), a, (a d);, (ii) 4 numbers in AP as (a 3d), (a d), (a d), (a 3d);, (iii) 5 numbers in AP as (a 2d), (a d), a, (a d), (a 2d);, (iv) 6 numbers in AP as (a 5d), (a 3d), (a d), (a d), (a 3d), (a 5d) ., SOLVED EXAMPLES, EXAMPLE 13, SOLUTION, , The sum of three numbers in AP is 24 and their product is 440. Find the, numbers., Let the required numbers be (a d), a and (a d) . Then,, (a d) a (a d) 24 & 3a 24 & a 8., Thus, the numbers are (8 d), 8 and (8 d) ., But their product is 440., , (8 d)# 8 #(8 d) 440, &, (8 d) (8 d) 55, &, 64 d 2 55 & d 2 64 55 9 & d !3., Hence, the required numbers are 5, 8, 11 or 11, 8, 5., , EXAMPLE 14, SOLUTION, , The product of three numbers in AP is 224 and the largest number is 7, times the smallest. Find the numbers., Let the required numbers be (a d), a and (a d) ., Then, largest number (a d) and smallest number (a d) ., 3a, ·, , … (i), a d 7 (a d) & 6a 8d & d , 4, And, (a d)# a #(a d) 224, &, , a(a 2 d 2) 224, 2, , 9a o , 16, 224 & a 3 b224 # l 512 8 3 & a 8., 7, 16, 3, Putting a 8 in (i), we get d b # 8l 6., 4, , the required numbers are (8 6), 8 and (8 6), i.e., 2, 8 and 14., , &, , a ea 2
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394, EXAMPLE 15, SOLUTION, , Senior Secondary School Mathematics for Class 11, , Find four numbers in AP whose sum is 20 and the sum of whose squares, is 120., Let the required numbers be, (a 3d), (a d), (a d) and (a 3d) ., Then, (a 3d) (a d) (a d) (a 3d) 20 & 4a 20 & a 5., And (a 3d) 2 (a d) 2 (a d) 2 (a 3d) 2 120, &, , 2(a 2 9d 2) 2(a 2 d 2) 120, , &, 4a 2 20d 2 120 & a 2 5d 2 30, &, 25 5d 2 30 & 5d 2 5 & d 2 1 & d !1., Hence, the required numbers are 2, 4, 6, 8 or 8, 6, 4, 2., EXAMPLE 16, SOLUTION, , Divide 32 into four parts which are in AP such that the product of extremes, is to the product of means as 7 : 15., Let the required parts be (a 3d), (a d), (a d) and (a 3d) ., Then, (a 3d) (a d) (a d) (a 3d) 32 & 4a 32 & a 8., So, the required parts are (8 3d), (8 d), (8 d) and (8 3d) ., But product of extremes is to the product of means as 7 : 15., (8 3d) (8 3d) 7, , , 15, (8 d) (8 d), &, , (64 9d 2), (64 d 2), , 7, 15, , 15(64 9d 2) 7 (64 d 2), 512 , &, 128d 2 512 & d 2 , 4 & d !2., 128, Thus, (a 8 and d 2) or (a 8 and d 2)., &, , Hence, the required parts are 2, 6, 10, 14 or 14, 10, 6, 2., , EXERCISE 11A, 1. Write first 4 terms in each of the sequences:, (2n 3), , , (i) an (5n 2), (ii) an , (iii) an (1) n 1 # 2 n 1, 4, 2. Find first five terms of the sequence, defined by, a1 1, an an 1 3 for n $ 2., 3. Find first 5 terms of the sequence, defined by, an 1, a1 1, an n for n $ 2., 4. Find the 23rd term of the AP 7, 5, 3, 1, –1, –3, … ., 5. Find the 20th term of the AP 2 , 3 2 , 5 2 , 7 2 , … ., 6. Find the nth term of the AP 8, 3, –2, –7, –12, … ., 5 2 1, 7. Find the nth term of the AP 1, , , , … ., 6 3 2
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Arithmetic Progression, , 395, , 8. Which term of the AP 9, 14, 19, 24, 29, … is 379?, 9. Which term of the AP 64, 60, 56, 52, 48, … is 0?, 10. How many terms are there in the AP 11, 18, 25, 32, 39, …, 207?, 5 1 1 1 , 5 , …, 1, 11. How many terms are there in the AP 1 , 1 , ,, 16 ?, 6 6 2 6 6, 6, 12. Is –47 a term of the AP 5, 2, 1, 4, 7, …?, 13. The 5th and 13th terms of an AP are 5 and –3 respectively. Find the AP and, its 30th term., 3 1, 1, 14. The 2nd, 31st and the last terms of an AP are 7 , and 6 respectively., 2, 4 2, Find the first term and the number of terms., 15. If the 9th term of an AP is 0, prove that its 29th term is double the 19th term., 16. The 4th term of an AP is three times the first and the 7th term exceeds twice, the third term by 1. Find the first term and the common difference., 17. If 7 times the 7th term of an AP is equal to 11 times its 11th term, show that, the 18th term of the AP is zero., 18. Find the 28th term from the end of the AP 6, 9, 12, 15, 18, …, 102., 19. Find the 16th term from the end of the AP 7, 2, 3, 8, 13, …, 113., 20. How many 3-digit numbers are divisible by 7?, 21. How many 2-digit numbers are divisible by 3?, 22. If 1, 2, 3, …, n are in AP whose common difference is d, show that, (tan n tan 1), ·, sec 1 sec 2 sec 2 sec 3 … sec n 1 sec n , sin d, T4 2, T, · Find 7 ·, 23. In an AP, it is being given that, T7 3, T10, 24. Three numbers are in AP. If their sum is 27 and their product is 648, find, the numbers., 25. The sum of three consecutive terms of an AP is 21 and the sum of the, squares of these terms is 165. Find these terms., 26. The angles of a quadrilateral are in AP whose common difference is 10c., Find the angles., 27. The digits of a 3-digit number are in AP and their sum is 15. The number, obtained by reversing the digits is 594 less than the original number. Find, the number., 28. Find the number of terms common to the two arithmetic progressions, 5, 9, 13, 17, …, 217 and 3, 9, 15, 21, …, 321., 29. We know that the sum of the interior angles of a triangle is 180. Show that, the sum of the interior angles of polygons with 3, 4, 5, 6, … sides form an, arithmetic progression. Find the sum of the interior angles for a 21-sided, polygon., 30. A side of an equilateral triangle is 24 cm long. A second equilateral triangle, is inscribed in it by joining the midpoints of the sides of the first triangle. The
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396, , Senior Secondary School Mathematics for Class 11, , process is continued. Find the perimeter of the sixth inscribed equilateral, triangle., 31. A man starts repaying a loan as the first instalment of ` 10000. If he, increases the instalments by ` 500 every month, what amount will he, pay in 30th instalment?, ANSWERS (EXERCISE 11A), , 1. (i) 7, 12, 17, 22, , (ii), , 1 1 3 5, , , ,, 4 4 4 4, , 1 1 1 1, ,, ,, ,, 3. 1,, 4. –37, 2 6 24 120, 1, 7. Tn (7 n) 8. 75th, 6, 12. No, , (iii) 4, –8, 16, –32, 5. 39 2, 9. 17th, , 13. (9, 8, 7, 6, …), T30 20, , 16. a 3, d 2, , 18. 21, 3, 21. 30, 23., 4, 26. 75c, 85c, 95c, 105c, 30. 2.25 cm, 31. ` 24500, , 2. 1, 4, 7, 10, 13, , 6. Tn (13 5n), 10. 29, , 11. 28, , 14. a 8, n 59, , 19. –38, , 20. 128, , 24. 6, 9, 12, , 25. 4, 7, 10, , 27. 852, , 28. 18, , 29. 3420, , HINTS TO SOME SELECTED QUESTIONS, 14. Let a be the first term and d be the common difference of the given AP., 31, 31, & ad , & 4a 4d 31, 4, 4, 1, 1, and T31 & a 30d & 2a 60d 1., 2, 2, 1, ·, On solving (i) and (ii), we get a 8 and d , 4, 13, 13, 13, 1, · Then, a (n 1) d , Let Tn , & 8 (n 1)#c m , & n 59., 2, 2, 2, 4, Then, T2 , , … (i), … (ii), , 15. T9 0 & a 8d 0 & a 8d., T29 a 28d 8d 28d 20d., T19 a 18d 8d 18d 10d., Hence, T29 2 #T19 ., 16. T4 3a and T7 2T3 1, & a 3d 3a and (a 6d) 2 (a 2d) 1, … (i), and a 2d 1., & 2a 3d 0, On solving (i) and (ii), we get a 3 and d 2., 20. The given numbers are 105, 112, 119, …, 994., This is an AP for which a 105, d 7 and l 994., Let Tn 994. Then, 105 (n 1)#7 994 & n 128., 22. Let 2 1 3 2 4 3 … n n 1 d., sec 1 sec 2 sec 2 sec 3 … sec n 1 sec n, , … (ii)
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Arithmetic Progression, , 397, , sin d, sin d, sin d, 1, , …, ), cos n 1 cos n 3, sin d cos 1 cos 2 cos 2 cos 3, Z], _, sin ( n n 1) bb, ] sin ( 2 1) sin ( 3 2), 1 · [], , …, `, cos n 1 cos n bb, sin d ]cos 1 cos 2 cos 2 cos 3, \, a, sin 2 cos 1 cos 2 sin 1 sin 3 cos 2 cos 3 sin 2, 1 ·), , , cos 1 cos 2, cos 2 cos 3, sin d, sin n cos n 1 cos n sin n 1, 3, …, cos n 1 cos n, 1 · {(tan 2 tan 1) (tan 3 tan 2) … (tan n tan n 1)}, sin d, (tan n tan 1), 1 · (tan n tan 1) , ·, sin d, sin d, a 3d 2, 23., & 3a 9d 2a 12d & a 3d., a 6d 3, T7 a 6d 3d 6d 9d, , , , 3·, , T10 a 9d 3d 9d 12d 4, , , 26. Let the required angles be ac, (a 10)c, (a 20)c, (a 30)c . Then,, a a 10 a 20 a 30 360 & 4a 300 & a 75., 27. Let the hundred’s, ten’s and unit digit’s be a d, a, a d., Then, a d a a d 15 & 3a 15 & a 5., [100 (a d) 10a (a d)] [100 (a d) 10a (a d)] 594, & 99d 99d 594 & 198d 594 & d 3., Hence, the required number is 852., 28. The first common term in two APs is 9., LCM of their common differences = LCM (4, 6) = 12., New AP of common elements is 9, 21, 33, 45, … ., Let it contain n terms. Then,, Tn # 216 & 9 (n 1)#12 # 216 & (n 1)#12 # 207, & (n 1)# 12 204 & n 18., So, there are 18 common terms in two APs., 29. The sum of the interior angles of a triangle is 180. A polygon of 4 sides can be divided, into two triangles and therefore, the sum of its angles is 360. A polygon of 5 sides can, be divided into 3 triangles and therefore, the sum of its angles is (3 #180c) 540c and, so on. Thus, a polygon of 21 sides is divided into (21 2) 19 triangles and therefore,, the sum of its angles is (19 #180c) 3420c., 30. We know that the line segment joining midpoints of the opposite sides of a triangle is, parallel to the third side and equal to half of it., Perimeters of 1st, 2nd, 3rd, 4th, 5th and 6th triangle are respectively 72 cm, 36 cm,, 9, 9, 18 cm, 9 cm, cm, cm., 2, 4, 31. The instalments form the AP, 10000 + 10500 + 11000 + 11500 + … up to 30 terms., Here a 10000, d 500 and we have to find T30 .
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398, , Senior Secondary School Mathematics for Class 11, , SUM OF n TERMS OF AN AP, THEOREM 1, , PROOF, , Prove that the sum of n terms of an AP with first term a and common, difference d is given by:, n, I. Sn (a l), where l is the last term., 2, n, II. Sn · [2a (n 1) d] ., 2, , Let us consider an AP containing n terms with the first term a, the, common difference d and the last term l., Let the sum of these n terms be Sn . Then,, ... (i), Sn a (a d) (a 2d) … (l 2d) (l d) l., Writing the above series in a reverse order, we get, ... (ii), Sn l (l d) (l 2d) … (a 2d) (a d) a., Adding the corresponding terms of (i) and (ii), we get, 2Sn {(a l) (a l) … to n times} n(a l) ., n, Sn (a l) ., 2, But, l nth term a (n 1) d., n, n, Sn · {a a (n 1) d} · {2a (n 1) d} ., 2, 2, n, Hence, Sn · {2a (n 1) d} ., 2, , SUMMARY, , (i) Sum of n terms of an AP with the first term a and the last term l is, given by, n, Sn (a l) ., 2, (ii) Sum of n terms of an AP with the first term a and the common difference, d is given by, n, Sn · {2a (n 1) d} ., 2, SOLVED EXAMPLES, EXAMPLE 1, , Find the sum of 23 terms of the AP 5, 9, 13, 17, … ., , SOLUTION, , Here, a 5, d (9 5) 4 and n 23., n, Now, Sn #{2a (n 1)# d} ., 2, 23, , #{2 # 5 (23 1)# 4}, S 23 , 2, b 23 # 98l 1127., 2, Hence, the sum of 23 terms of the given AP is 1127.
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Arithmetic Progression, , 399, , EXAMPLE 2, , The first, second and the last terms of an AP are a, b, c respectively. Prove, (a c) (b c 2a), ·, that the sum of the AP is, 2 (b a ), , SOLUTION, , Let d be the common difference of the given AP and let it contain n, terms. Then, d (b a) ., Now, Tn c & a (n 1) d c, & a (n 1) (b a) c, [a d (b a)], ( c a), & (n 1) , (b a), (c a ), (b c 2a), 13 , ·, & n), , (b a ), (b a ), , , EXAMPLE 3, , SOLUTION, , n , (a l), where l is the last term, 2, (b c 2a), , #(a c), 2 (b a), (b c 2a), =a n , and l cG, 2 (b a), (a c) (b c 2a), , ·, 2 (b a ), , sum of the given AP , , 9, 11, How many terms of the AP 6,, , 5,, , … are needed to give the, 2, 2, sum 25? Explain the double answer., Let the required number of terms be n. Then,, 11, , ( 6)1 b 11 6l 1 and Sn 25., a 6, d ', 2, 2, 2, n, , Sn 25 & · [2a (n 1) d] 25, 2, n, 1, & · ;2 #( 6) (n 1) · E 25, 2, 2, n · n 25 , n (n 25) , &, 25 & ·, 25, b, l, 2 2 2, 2, 2, 2, & n 25n 100 0 & (n 5) (n 20) 0, & n 5 or n 20., This shows that the sum of first five terms is –25 and that the sum, of first 20 terms is also –25., It means that the sum of all the terms from 6th to 20th is zero., , EXAMPLE 4, , Find the value of x when 1 6 11 16 … x 148., , SOLUTION, , The given series is an arithmetic series in which a 1 and, d ( 6 1) 5 ., Let it contain n terms. Then, we have to find the nth term.
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400, , Senior Secondary School Mathematics for Class 11, , Now, Sn 148, n, &, · [2a (n 1) d] 148, 2, n·, &, [2 #1 (n 1)# 5] 148, 2, &, n(5n 3) 296 & 5n 2 3n 296 0, 5n 2 40n 37n 296 0, 5n(n 8) 37 (n 8) 0 & (n 8) (5n 37) 0, 37, E·, &, n 8 ;a n !, 5, , T8 1 (8 1)# 5 36 [a Tn a (n 1) d]., Hence, x 36., &, &, , EXAMPLE 5, SOLUTION, , EXAMPLE 6, SOLUTION, , Find the sum of all odd integers from 1 to 1001., The odd integers from 1 to 1001 are 1, 3, 5, 7, …, 999, 1001., This is an AP in which a 1, d (3 1) 2 and l 1001., Let the number of terms be n. Then,, Tn 1001 & a (n 1) d 1001, & 1 (n 1)# 2 1001 & n 501., Now, a 1, l 1001 and n 501., 501, n, · (1 1001) (501# 501) 251001., , Sn (a l) , 2, 2, Hence, the required sum is 251001., Find the sum of all natural numbers lying between 100 and 1000, which, are multiples of 5., Required numbers are 105, 110, 115, …, 995., This is an AP in which a 105, d (110 105) 5 and l 995., Let the number of terms in this AP be n. Then,, Tn 995 & a (n 1) d 995, & 105 (n 1)# 5 995, 890 , & (n 1) , 178 & n 179., 5, 179, 179, n, #(105 995) b, #1100l, , Sn (a l) , 2, 2, 2, (179 # 550) 98450., Hence, the required sum is 98450., , EXAMPLE 7, , Find the sum of integers from 1 to 100 that are divisible by 2 or 5., , SOLUTION, , Required sum, = (sum of integers from 1 to 100, divisible by 2), + (sum of integers from 1 to 100, divisible by 5), – (sum of integers from 1 to 100, divisible by 10)
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Arithmetic Progression, , 401, , (2 4 6 … 100) (5 10 15 … 100), (10 20 30 … 100), 50, 20, 10, , (2 100) (5 105) (10 100), 2, 2, 2, (2550 1050 550) 3050., EXAMPLE 8, SOLUTION, , EXAMPLE 9, SOLUTION, , EXAMPLE 10, , Find the sum of all two-digit numbers which when divided by 4 yield 1 as, remainder., Required sum, 13 17 21 25 … 97., This is an AP in which a 13, d (17 13) 4 and l 97., Let the number of terms be n. Then,, Tn 97 & a (n 1) d 97, & 13 (n 1)# 4 97 & n 22., n, 22, #(13 97) (11#110) 1210., , required sum (a l) , 2, 2, Hence, the required sum is 1210., The sum of n terms of two arithmetic progressions are in the ratio, (3n 8) : (7n 15) . Find the ratio of their 12th terms., Let a1, a 2 and d1, d 2 be the first terms and common differences of, the first and second AP respectively. Then, by the given condition,, we have, sum to n terms of first AP 3n 8, sum to n terms of sec ond AP 7n 15, n·, [2a1 (n 1) d1], , 2, 3n 8, &, n·, , [2a 2 (n 1) d 2] 7n 15, 2, 2a1 (n 1) d1, , 3n 8 ·, … (i), &, 2a 2 (n 1) d 2 7n 15, 12th term of first AP a1 11d1, Now,, 12th term of sec ond AP a 2 11d 2, 2a 22d1 2a1 (23 1) d1, 1, , 2a 2 22d 2 2a 2 (23 1) d 2, , 3 # 23 8, [putting n 23 in (i)], 7 # 23 15, 77 7 ·, 176 16, Hence, the ratio of the 12th terms of given APs is 7 : 16., In an AP, the first term is 2 and the sum of first five terms is one-fourth of, the sum of next five terms. Show that its 20th term is –112 and the sum of, its first 20 terms is –1100.
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402, SOLUTION, , Senior Secondary School Mathematics for Class 11, , We have, 1, (T T T T T ), 4 6 7 8 9 10, 1 · [(T1 T2 … T5 T6 … T10), 4, (T1 T2 T3 T4 T5)], , (T1 T2 T3 T4 T5) , , &, &, &, &, &, &, , , , 1·, (S10 S 5), where Sn denotes the sum of first n terms, 4, 4S 5 S10 S 5, 5S 5 S10, , S5 , , 5, 10, 5 # #[2 # 2 (5 1) d] #[2 # 2 (10 1) d], 2, 2, 50 (1 d) (20 45d), 5d 30 & d 6., a 2 and d 6., T20 [2 (20 1)#( 6)] 112, , and S 20 , EXAMPLE 11, SOLUTION, , EXAMPLE 12, , SOLUTION, , 20, #[2 # 2 (20 1)#( 6)] 1100., 2, , The difference between any two consecutive interior angles of a polygon is, 5. If the smallest angle is 120, find the number of sides of the polygon., Let the number of sides of the polygon be n., Then, the sum of its interior angles (2n 4) right angles, {(n 2)#180}c ., … (i), Since the difference between any two consecutive interior angles of, the polygon is constant, its angles are in AP., In this AP, we have a 120 and d 5., Now, Sn (n 2)#180, [from (i)], n·, &, {2 #120 (n 1)# 5} (n 2)#180, 2, 235n 5n 2 , &, 180n 360 & 5n 2 125n 720 0, 2, 2, &, n 2 25n 144 0 & (n 9) (n 16) 0, &, n 9 or n 16., But, when n 16, we have, last angle {120 (16 1)# 5}c 195c, which is not possible., , n 9., Hence, the number of sides of the given polygon = 9., The income of a man is ` 400000 in the first year and he receives an, increase of ` 10000 to his income per year for 19 years. Find the total, amount he received in 20 years., Here, we have an AP with a 400000, d 10000 and n 20.
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Arithmetic Progression, , Using the formula, Sn , , 403, , n, #[2a (n 1) d], we have, 2, , 20, #[2 # 400000 19 #10000] 10 # 990000 9900000., 2, Hence, the man received ` 9900000 as the total amount in 20 years., S 20 , , EXAMPLE 13, , SOLUTION, , EXAMPLE 14, , SOLUTION, , EXAMPLE 15, , SOLUTION, , A man saved ` 660000 in 20 years. In each succeeding year after the first, year he saves ` 2000 more than what he saved in the previous year. How, much did he save in the first year?, Here, we have an AP in which S 20 660000, d 2000 and we have, to find the value of a., n, Using the formula, Sn #[2a (n 1) d], we get, 2, 20, #[2a (20 1)# 2000], S 20 , 2, &, 10 #[2a 19 # 2000] 660000, [a S 20 660000], &, 20a 380000 660000 & 20a 280000 & a 14000., Hence, the man saved ` 14000 in the first year., In a potato race, 20 potatoes are placed in a line at intervals of 4 metres, with the first potato 24 metres from the starting point. A contestant is, required to bring the potatoes back to the starting place, one at a time., How far would he run in bringing back all the potatoes?, Distances moved in bringing back the 1st, 2nd, 3rd, …, 20th potato, to the starting point are (2 # 24) m, (2 # 28) m, (2 # 32) m, …, and, so on., Total distance moved {48 56 64 72 … to 20 terms}, 20 #[2 # 48 (20 1)# 8] m, 2, [a a 48, d 8 and n 20], 10 #(96 152) m 2480 m., Hence, total distance covered = 2480 m., In a cricket tournament 16 school teams participated. A sum of ` 16000, is to be awarded among themselves as prize money. If the team in the last, place is awarded ` 550 in prize money and the award increases by the, same amount for successive finishing places, how much amount will the, team in the first place receive?, Let the first place team receive ` a and let the award increase by, ` d. Then, these prizes form an AP in which first term a, common, difference d, T16 550 and S16 16000., Now, T16 550 & a 15d 550., … (i), 16, #[2a 15d] 16000, And, S16 16000 &, 2, … (ii), & 2a 15d 2000., On subtracting (i) from (ii), we get a 1450., Hence, the team in the first place receives ` 1450.
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404, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 16, , The first term of an AP is a and the sum of the first p terms is zero. Show, a (p q) q, ·, that the sum of the next q terms is, ( p 1), , SOLUTION, , Let the common difference of the given AP be d. Then,, p, S p 0 & #[2a (p 1) d] 0, 2, 2a, ·, & 2a ( p 1 ) d 0 & d , ( p 1), (p q), · [2a (p q 1) d], , Sp q , 2, R, V, (p q) SS, ( 2a) WWW, , · SS2a (p q 1)#, 2, (p 1)WW, S, TR, X, SS (p q 1)VWW a (p q) q, W, a(p q) · SS1 , ·, S, (p 1) WW, (p 1), X, T, a (p q) q, ·, Hence, the required sum is, ( p 1), , … (i), , [using (i)], , EXAMPLE 17, , If the mth term of an AP is a and its nth term is b, show that the sum of its, (m n), (a b), 3·, · )a b , (m n) terms is, 2, (m n), , SOLUTION, , Let the first term and the common difference of the given AP be A, and d respectively. Then,, … (i), Tm a & A (m 1) d a, and Tn b & A (n 1) d b., … (ii), On subtracting (ii) from (i), we get, (a b), ·, … (iii), (m n) d (a b) & d , (m n), Adding (i) and (ii), we get, 2A (m n 2) d a b, &, 2A (m n 1) d d a b, ab, … (iv), &, 2A (m n 1) d (a b d) ba b l, m n, [using (iii)]., (m n), · [2A (m n 1) d]., , Tm n , 2, (m n), , · ;a b a b E, [using (iv)]., &, Tm n , m n, 2, , EXAMPLE 18, , If the ratio of the sums of m and n terms of an AP be (m 2 : n 2), show that, the ratio of their mth and nth terms is (2m 1) : (2n 1) ., Let the first term and the common difference of the given AP be a, and d respectively. Then,, , SOLUTION
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Arithmetic Progression, , 405, , m·, [2a (m 1) d], 2, Sm m 2, 2 & 2, m2, n·, Sn n, n, [2a (n 1) d], 2, 2a (m 1) d m, , &, 2a (n 1) d n, & n [2a (m 1) d] m [2a (n 1) d], & 2a (m n) d (m n) & d 2a., Tm a (m 1) d a 2 (m 1) a 1 2m 2 2m 1, , , , , ·, , Tn a (n 1) d, a 2 (n 1) a 1 2n 2 2n 1, Hence, the ratio of their mth and nth terms is (2m 1) : (2n 1) ., SOME RESULTS ON SUMMATION OF AP, , If the sum of first n terms of a progression is a quadratic expression in n,, show that it is an AP., Let Sn an 2 bn c, … (i), where a, b, c are constants and a ! 0., On replacing n by (n 1) in (i), we get, , THEOREM 1, PROOF, , , , Sn 1 a (n 1) 2 b (n 1) c., Tn (Sn Sn 1), (an 2 bn c) {a (n 1) 2 b (n 1) c}, , a {n 2 (n 1) 2} b {n (n 1)}, a (2n 1) b 2an (b a) ., Thus, Tn 2an (b a) ., Replacing n by (n 1) in (ii), we get, Tn 1 2a (n 1) (b a) ., , … (ii), … (iii), , On subtracting (iii) from (ii), we get, (Tn Tn 1) 2a, which is constant., Hence, the given progression is an AP., If there are (2n 1) terms in an AP, prove that the sum of odd terms and, the sum of even terms bear the ratio (n 1) : n., Since the given AP contains (2n 1) terms, so it has (n 1) odd terms, and n even terms., Let a be the first term and d be the common difference of the given AP., Let S1 and S 2 be the sum of all odd terms and the sum of all even terms, respectively. Then,, S1 T1 T3 T5 … T2n 1, (n 1), , · [T1 T2n 1], [sum of (n 1) odd terms], 2, (n 1), , · [a {a (2n 1 1) d}], [a T2n 1 a (2n 1 1) d], 2, (n 1) (a nd) ., , THEOREM 2, PROOF
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406, , Senior Secondary School Mathematics for Class 11, , S 2 T2 T4 T6 … T2n, n · (T2 T2n), 2, n · [(a d) {a (2n 1) d}], 2, n (a nd) ., S1 (n 1) (a nd), S1 n 1, , , , &, n ·, S2, S2, n (a nd), Hence, the required ratio is (n 1) : n., , [sum of n even terms], [a T2n a (2n 1) d], , 1, 1, If the mth term of an AP is b n l and its nth term is b m l then show that the, 1, sum of its mn terms is (mn 1) ., 2, Let a be the first term and d be the common difference of the given AP., 1, 1, Then, Tm n and Tn m ·, 1, 1, a (m 1) d n, … (i), and a (n 1) d m ·, … (ii), On subtracting (ii) from (i), we get, (m n), 1 1, 1, (m n) d b n m l mn & d mn ·, 1, Putting d mn in (i), we get, (m 1) 1, 1 (m 1), 1, a mn n & a (n mn 2 mn ·, mn ·, Smn , [2a (mn 1) d], 2, (mn 1), 2, mn · ( 2 , ;a a 1 and d 1 E, mn, mn, mn, mn, 2, 1, (mn 1) ., 2, 1, Hence, the sum of mn terms of the given AP is (mn 1) ., 2, , THEOREM 3, , PROOF, , If the sum of first p terms of an AP is the same as the sum of its first q, terms then show that the sum of its first (p q) terms is zero., Let a be the first term and d be the common difference of the given AP., Then, S p Sq, p, q, · [2a (p 1) d] · [2a (q 1) d], &, 2, 2, & 2(p q) a {p (p 1) q (q 1)} d 0, , THEOREM 4, PROOF, , &, &, &, &, , 2(p q) a {(p 2 q 2) (p q)} d 0, 2(p q) a (p q) (p q 1) d 0, (p q) {2a (p q 1) d} 0, 2a (p q 1) d 0., , … (i)
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Arithmetic Progression, , (p q), · [2a (p q 1) d] 0, 2, Hence, the sum of the first (p q) term is zero., , , Sp q , , 407, , [using (i)]., , If the sum of m terms of an AP be n and the sum of its n terms be m, show, that the sum of its (m n) terms is (m n) ., Let a be the first term and d be the common difference of the given AP., m·, Then, Sm n &, {2a (m 1) d} n, 2, … (i), & 2am m (m 1) d 2n., n, And, Sn m & · {2a (n 1) d} m, 2, … (ii), & 2an n (n 1) d 2m., On subtracting (ii) from (i), we get, 2a (m n) {(m 2 n 2) (m n)} · d 2(n m), & 2a (m n) (m n) (m n 1) d 2(n m), & (m n) {2a (m n 1) d} 2(m n), … (iii), & 2a (m n 1) d 2., (m n), · {2a (m n 1) d}, Sm n , 2, (m n), , · ( 2) (m n), [using (iii)]., 2, Hence, the sum of (m n) terms is (m n) ., , THEOREM 5, PROOF, , THEOREM 6, , PROOF, , The sum of the first p, q and r terms of an AP are a, b, c respectively., Show that, a, b, c, , p · (q r) q · (r p) r · (p q) 0., , Let x be the first term and d be the common difference of the given AP., Then, S p a, Sq b and Sr c., p, Now, S p a & · [2x (p 1) d] a, 2, a, d, … (i), & p x (p 1) · ·, 2, q, Sq b & · [2x (q 1) d] b, 2, b, d, … (ii), & q x (q 1) · ·, 2, r, And, Sr c & · [2x (r 1) d] c, 2, c, d, … (iii), & r x ( r 1) · ·, 2, On multiplying (i), (ii) and (iii) by (q r), (r p) and (p q) respectively, and adding, we get
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408, , Senior Secondary School Mathematics for Class 11, , a, b, c, , p · (q r) q · (r p) r · (p q), x · [(q r) (r p) (p q)], d · [(p 1) (q r) (q 1) (r p) (r 1) (p q)], 2, d, (x # 0) b # 0l 0., 2, a b c , Hence, p (q r) q (r p) r (p q) 0., If the sums of the first n, 2n, 3n terms of an AP are S1, S 2 and S 3, respectively, prove that S 3 3(S 2 S1) ., Let a be the first term and d be the common difference of the given AP., n, 1, … (i), Then, S1 · {2a (n 1) d} & S1 na n (n 1) d,, 2, 2, 2n ·, … (ii), S2 , {2a (2n 1) d} & S 2 2na n (2n 1) d, 2, 3n, 3, · {2a (3n 1) d} & S 3 3na n (3n 1) d., and S 3 , … (iii), 2, 2, 1, (S 2 S1) na nd '(2n 1) (n 1)1, 2, na n · (3n 1) d n · {2a (3n 1) d}., 2, 2, 3, n, · {2a (3n 1) d} S 3 ., 3 ( S 2 S 1) , 2, Hence, S 3 3 (S 2 S1) ., , THEOREM 7, PROOF, , If S1, S 2, S 3, …, Sm be the sums of n terms of m arithmetic progressions, whose first terms are 1, 2, 3, …, m respectively and the common, differences are 1, 3, 5, …, (2m 1) respectively then show that, 1, S1 S 2 S 3 … Sm mn (mn 1) ., 2, Clearly, we have, n, S1 #{2 #1 (n 1)#1}, [a a 1, d 1], 2, n, S 2 #{2 # 2 (n 1)# 3}, [a a 2, d 3], 2, n, S 3 #{2 # 3 (n 1)# 5}, [a a 3, d 5], 2, … … … … … … …, … … … … … … …, n, Sm #{2 # m (n 1) (2m 1)}, [a a m, d (2m 1)], 2, Adding columnwise, we get, n, S1 S 2 S 3 … Sm #[2 #(1 2 3 … m)], 2, n(n 1), , #[1 3 5 … (2m 1)], 2, , THEOREM 8, , PROOF
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Arithmetic Progression, , 409, , n(n 1) m, %n # m (1 m)/ , # #[1 (2m 1)], 2, 2, 2, %n # m(m 1)/ %n # m 2 (n 1)/, 2, 2, mn #[(m 1) m (n 1)] 1 mn (mn 1) ., 2, 2, 1, Hence, S1 S 2 S 3 … Sm mn (mn 1) ., 2, THEOREM 9, , PROOF, , If the sum of m terms of an AP is equal to the sum of either the next n, terms or the next p terms then prove that, 1 1, 1 1, (m n) d m p n (m p) b m n l ·, , Let us consider an AP with first term a and common difference d., Then, (T1 T2 T3 … Tm) (Tm 1 Tm 2 Tm 3 … Tm n), & 2(T1 T2 T3 … Tm) (T1 T2 … Tm Tm 1 … Tm n), [adding (T1 T2 T3 … Tm) on both sides], (m n), m, #[2a (m n 1) d], & 2 # #[2a (m 1) d] , 2, 2, (m n), #(x nd), where [2a (m 1) d] x, & mx , 2, & 2mx (m n) x (m n) nd, … (i), & (m n) x (m n) nd., Similarly, (m p) x (m p) pd., … (ii), On dividing (i) by (ii), we get, (m n) (m n) n, , (m p) (m p) p, [by cross multiplication], & (m n) (m p) n (m p) (m n) p, &, , 1 1, 1 1, (m n) d p m n (m p) b n m l [on dividing both sides by mnp], , &, , 1 1, 1 1, (m n) d m p n (m p) b m n l ·, , 1 1, 1 1, Hence, (m n) d m p n (m p) b m n l ·, THEOREM 10, , PROOF, , The first term of an AP is a and the sum of its first p terms is zero. Then,, a (p q) q, ·, show that the sum of its next q terms is, ( p 1), , Let d be the common difference of the given AP. Then,, p, S p 0 & [2a (p 1) d] 0, 2, 2a ·, … (i), & 2a (p 1) d 0 & d , (1 p), , [a p ! 0]
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410, , Senior Secondary School Mathematics for Class 11, , Let S be the sum of next q terms. Then,, S Tp 1 Tp 2 … Tp q, (T1 T2 … Tp Tp 1 Tp 2 … Tp q) (T1 T2 … Tp), [a S p 0], , Sp q Sp Sp q 0 Sp q, (p q), · [2a (p q 1) d], 2, (p q), , · =2a (p q 1)# 2a G, 2, ( 1 p), SRS (p q 1)WVW a (p q) q, W, (p q) a · SS1 , ·, S, (1 p) WW, ( p 1), X, T, a (p q) q, ·, Hence, the sum of next q terms is, ( p 1), , , THEOREM 11, , PROOF, , [using (i)], , If the sum of n terms of an AP is (cn dn 2), where c and d are constants, then show that the common difference of the AP is 2d., , Let Sn cn dn 2, … (i),, where c and d are constants., Replacing n by (n 1) in (i), we get, Sn 1 c (n 1) d (n 1) 2, & Sn 1 dn 2 n (c 2d) (d c) ., On subtracting (ii) from (i), we get, (Sn Sn 1) 2dn (c d), & Tn 2dn (c d) ., Replacing n by (n 1) in (iii), we get, Tn 1 2d (n 1) (c d) ., On subtracting (iv) from (iii), we get, (Tn Tn 1) 2d, which is constant., Hence, the common difference of the given AP is 2d., , … (ii), , … (iii), … (iv), , EXERCISE 11B, 1. Find the sum of 23 terms of the AP 17, 12, 7, 2, –3, … ., 1 2, 2. Find the sum of 16 terms of the AP 6, 5 , 4 , 4, … ., 3 3, 3. Find the sum of 25 terms of the AP 2 , 2 2 , 3 2 , 4 2 , … ., 4. Find the sum of 100 terms of the AP 0.6, 0.61, 0.62, 0.63, … ., 5. Find the sum of 20 terms of the AP (x y), (x y), (x 3y), … ., x y 3x 2y 5x 3y, ,, ,… ., 6. Find the sum of n terms of the AP ,, x y xy, xy, 7. Find the sum of the series 2 5 8 11 … 191., 8. Find the sum of the series 101 99 97 95 … 43.
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Arithmetic Progression, , 9., 10., 11., 12., 13., 14., 15., 16., 17., 18., 19., 20., 21., 22., 23., 24., 25., 26., , 411, , Find the value of x such that 1 4 7 10 … x 715., Find the value of x such that 25 22 19 16 … x 112., Find the rth term of an AP, the sum of whose first n terms is (3n 2 2n) ., Find the sum of n terms of an AP whose rth term is (5r 1) ., If the sum of a certain number of terms of the AP 27, 24, 21, 18, … is –30, find, the last term., How many terms of the AP 26, 21, 16, 11, … are needed to give the sum 11?, How many terms of the AP 18, 16, 14, 12, … are needed to give the sum 78?, Explain the double answer., 1, 2, How many terms of the AP 20, 19 , 18 , … must be taken to make the sum, 3, 3, 300? Explain the double answer., The sums of n terms of two arithmetic progressions are in the ratio, (7n 5) : (5n 17) . Show that their 6th terms are equal., If the ratio between the sums of n terms of two arithmetic progressions is, (7n 1) : (4n 27), find the ratio of their 11th terms., Find the sum of all odd integers from 1 to 201., Find the sum of all even integers between 101 and 199., Find the sum of all integers between 101 and 500, which are divisible by 9., Find the sum of all integers between 100 and 600, each of which when, divided by 5 leaves 2 as remainder., The sum of first 7 terms of an AP is 10 and that of next 7 terms is 17. Find, the AP., If the sum of n terms of an AP is (3n 2 5n) and its mth term is 164, find the, value of m., Find the sum of all natural numbers from 1 and 100 which are divisible by, 4 or 5., 1, If the sum of n terms of an AP is 'nP n (n 1) Q1 , where P and Q are, 2, constants then find the common difference., , 27. If Sm m 2p and Sn n 2p, where m ! n in an AP then prove that S p p 3., 28. A carpenter was hired to build 192 window frames. The first day he made, 5 frames and each day, thereafter he made 2 more frames than he made the, day before. How many days did he take to finish the job?, ANSWERS (EXERCISE 11B), , 1. –874, 2. 16, n, · {n (2x y) y}, 6., 2 ( x y), 10. x 7, 14. 11, , 3. 325 2, , 4. 109.5, , 5. 20 (x 18y), , 7. 6176, , 8. 2160, , 9. x 64, , 1, 13. –30, n (5n 7), 2, 15. 6 and 13, sum of all terms from 7th to 13th is zero, 11. Tr (6r 1) 12. Sn
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412, , Senior Secondary School Mathematics for Class 11, , 16. 25 and 36, sum of all terms from 26th to 36th is zero, 19. 10201, 20. 7350, 21. 13266, 22. 34950, 3, 1, 2, 23. 1 1 1 1 …, 24. m 27, 25. 2050, 7, 7, 7, 28. 12 days, , 18. 4 : 3, 26. Q, , HINTS TO SOME SELECTED QUESTIONS, 5. a (x y) and d (x y) (x y) 2y., 20, #[2a (20 1) d] 10 #[2 (x y) 19 #( 2y)] 20 (x 18y) ., 2, 1 ·, 6. Required sum , {(x y) (3x 2y) (5x 3y) … to n terms], (x y), S20 , , 1 ·n·, [2 (x y) (n 1) (2x y)], (x y) 2, n, , · {n (2x y) y} ., 2 (x y), , , [a a (x y) and d (2x y)], , 9. Here, a 1 and d 3. Let it contain n terms. Then,, n·, n, [2a (n 1) d] 715 & · [2 #1 (n 1)# 3] 715, 2, 2, & n (3n 1) 1430 & 3n 2 n 1430 0 & 3n 2 66n 65n 1430 0, &, , , 3n (n 22) 65 (n 22) 0 & (n 22) (3n 65) 0 & n 22., x T22 (a 21d) (1 21# 3) 64., , 10. 3n 2 21n 32n 224 0., 11. Sn 3n 2 2n & Sn 1 3 (n 1) 2 2 (n 1) & Sn 1 3n 2 4n 1., , , Tn (Sn Sn 1) (3n 2 2n) (3n 2 4n 1) (6n 1) & Tr (6r 1) ., , 13. Here, a 27, d (24 27) 3 and Sn 30. Then,, n·, n, [2a (n 1) d] 30 & · [54 (n 1)#( 3)] 30, 2, 2, 3n 2 57n 60 0 & n 2 19n 20 0 & n 20., 20, n, Now, Sn 30 & (a l) 30 &, (27 l) 30 & l 30., 2, 2, 2, 58, · Let Sn 300. Then,, 16. Here, a 20 and d c 20m , 3, 3, 2, n, n, #[2a (n 1) d] 300 & #<40 (n 1)#c mF 300, 2, 2, 3, n · (122 2n) , 2, , 300 & n (61 n) 900 & n 61n 900 0, 2, 3, 2, & n 36n 25n 900 0 & n (n 36) 25 (n 36) 0 & (n 25) (n 36) 0., S25 300 and S36 300., Sum of all terms from 26th to 36th is zero., n·, [2a1 (n 1) d1], 2a1 (n 1) d1, , , 2, 7n 5 &, 7n 5, 17. n, , 5, n, 17, 2, a2 (n 1) d2 5n 17, · [2a2 (n 1) d2], 2
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Arithmetic Progression, , 413, , n1, m d1, a 5d1, , , 2, 7n 5 & 1, 7 #11 5, 5n 17, (n 1), a2 5d2 5 #11 17, a2 , d2, 2, , n1 , ;, 5 & n 11E, 2, , a1 c, , &, , 72 1., 72, , Hence, their 6th terms are equal., 18. Proceeding as above, we get, 1, a1 (n 1) d1, a 10d1, , , 2, 7n 1 & 1, 7 # 21 1, 27, 10d2 4 # 21 27, 1, 4, n, a, 2, , , a2, (n 1) d2, 2, 148 4 ·, 111 3, 19. Required sum 1 3 5 7 … 201., Tn 201 & 1 (n 1)# 2 201 & n 101., 101, Sn , #(1 201) (101#101) 10201., 2, 20. Required sum 102 104 106 … 198., 21. Required sum 108 117 126 … 495., 22. Required sum 102 107 112 … 597., 23. S7 10 and S14 (17 10) 27., 7, , #[2a 6d] 10 & 7a 21d 10, 2, 14, #[2a 13d] 27 & 14a 91d 27., and, 2, Solve for a and d., , n1 , ;, 10 & n 21E, 2, , … (i), … (ii), , 24. Tn (Sn Sn 1) (3n 2 5n) (3n 2 n 2) (6n 2) ., Tm 164 & 6m 2 164 & 6m 162 & m 27., 25. Required numbers = (numbers divisible by 4) + (numbers divisible by 5), – (numbers divisible by 20), (4 8 12 … 100) (5 10 15 … 100), (20 40 60 … 100), 25, 20, 5, , #(4 100) #(5 100) #(20 100), 2, 2, 2, (1300 1050 300) 2050., 1, 1, 26. Sn nP n (n 1) Q & Sn 1 (n 1) P (n 1) (n 2) Q., 2, 2, Tn (Sn Sn 1) P (n 1) Q & Tn 1 P (n 2) Q., d (Tn Tn 1) Q., 27. Sm m 2p and Sm 1 (m 1) 2p & Sm Sm 1 {m 2 (m 1) 2} p (2m 1) p., Tm (2m 1) p & T1 p, T2 3p, T3 5p, … & a p and d 2p., p, p, S p #[2p (p 1)# 2p] d # 2p 2n p 3., 2, 2, 28. We have 5 7 9 … 192, n, #[10 (n 1)# 2] 192 & n (n 4) 192 & n 2 4n 192 0, 2, & (n 16) (n 12) 0 & n 12.
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414, , Senior Secondary School Mathematics for Class 11, , WORD PROBLEMS ON AP, , EXERCISE 11C, 1. The interior angles of a polygon are in AP. The smallest angle is 52 and the, common difference is 8. Find the number of sides of the polygon., 2. A circle is completely divided into n sectors in such a way that the angles, of the sectors are in AP. If the smallest of these angles is 8 and the largest is, 72, calculate n and the angle in the fifth sector., 3. There are 30 trees at equal distances of 5 metres in a line with a well, the, distance of the well from the nearest tree being 10 metres. A gardner waters, all the trees separately starting from the well and he returns to the well, after watering each tree to get water for the next. Find the total distance the, gardner will cover in order to water all the trees., 4. Two cars start together from the same place in same direction. The first, goes with uniform speed of 60 km/hr. The second goes at a speed of, 48 km/hr in the first hour and increases the speed by 1 km each succeeding, hour. After how many hours will the second car overtake the first car if, both cars go non-stop?, 5. Arun buys a scooter for ` 44000. He pays ` 8000 in cash and agrees to pay, the balance in annual instalments of ` 4000 each plus 10% interest on the, unpaid amount. How much did he pay for it?, 6. A man accepts a position with an initial salary of ` 26000 per month. It is, understood that he will receive an automatic increase of ` 250 in the very, next month and each month thereafter., Find his (i) salary for the 10th month, (ii) total earnings during the first, year., 7. A man saved ` 660000 in 20 years. In each succeeding year after the first, year he saved ` 2000 more than what he saved in the previous year. How, much did he save in the first year?, 8. 150 workers were engaged to finish a piece of work in a certain number of, days. Four workers dropped the second day, four more workers dropped, the third day, and so on. It takes 8 more days to finish the work now. Find, the number of days in which the work was completed., 9. A man saves ` 4000 during first year, ` 5000 during second year and in this, way he increases his savings by ` 1000 every year. Find in what time his, savings will be ` 85000., 10. A man arranges to pay off a debt of ` 36000 by 40 annual instalments which, form an AP. When 30 of the instalments are paid, he dies, leaving one-third, of the debt unpaid. Find the value of first instalment., 11. A manufacturer of TV sets produced 6000 units in the third year and, 7000 units in the seventh year. Assuming that the production increases, uniformly by a fixed number every year, find the production (i) in the first, year, (ii) in the 10th year, (iii) in 7 years.
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Arithmetic Progression, , 415, , 12. A farmer buys a used tractor for ` 180000. He pays ` 90000 in cash and, agrees to pay the balance in annual instalments of ` 9000 plus 12% interest, on the unpaid amount. How much did the tractor cost him?, ANSWERS (EXERCISE 11C), , 1. 3, 2. n 9, 5th angle = 40 3. 4795 m, 4. 25 hours, 6. (i) ` 28250 (ii) ` 328500, 7. ` 14000, 8. 25 days, 10. ` 510, 11. (i) 5500 (ii) 7750 (iii) 43750 12. ` 239400, , 5. ` 62000, 9. 10 years, , HINTS TO SOME SELECTED QUESTIONS, 1. Let the number of sides of the polygon be n. Then,, sum of its interior angles (2n 4)# 90c (180n 360)c ., 52 60 68 … to n terms (180n 360), n, #[2 # 52 (n 1)# 8] 180n 360, &, 2, 2, & n 33n 90 0 & (n 3) (n 30) 0 & n 3 or n 30., When n 30, then last angle [52 29 # 8]c 284c, which is a reflex angle. So, n ! 30., Hence, n 3., 2. Let the number of sectors be n. Then,, 720 , n , (a l) 360 & n (8 72) 720 & n , 9., 2, 80, Now, a 8, Tn 72 and n 9., 8 8d 72 & 8d 64 & d 8., So, the AP is 8, 16, 24, 32, … . So, 5th angle (a 4d) 40c., 3. Total distance covered in metres, 2 #{10 15 20 … up to 29 terms} + 30th term, 2 # 29 #{20 28 # 5} (10 29 # 5), 2, (4640 155) 4795 m., 4. Suppose the second car overtakes the first car in n hours., Distance covered by the first car in n hours (60n) km., Distance covered by the second car in n hours, [48 49 50 … to n terms], n #[2 # 48 (n 1)#1] n (95 n) ., 2, 2, n, , (95 n) 60n & 95 n 120 & n 25 hours., 2, 5. Total cost = ` 44000. Paid in cash = ` 8000. Balance = ` 36000., Amount of each instalment = ` 4000., Number of instalments 36000 ' 4000 9., 10, #1m ` 7600., 1st instalment ` c4000 36000 #, 100, 10, 2nd instalment ` c4000 32000 #, #1m ` 7200., 100, 10, 3rd instalment ` c4000 28000 #, #1m ` 6800, and so on., 100
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416, , Senior Secondary School Mathematics for Class 11, Total payment ` {8000 (7600 7200 6800 … to 9 terms)}, ` '8000 9 (15200 8 #( 400)}1 ` 62000., 2, , 6. The amounts form an AP with a = ` 26000 and d ` 250., (i) T10 (a 9d) ` (26000 9 # 250) ` 28250., 12, (ii) Total earning during 1 year ` ' #(2 # 26000 11# 250)1 ` 328500., 2, 7. Suppose that the man saved ` x in first year. Then,, x (x 2000) (x 4000) … to 20 terms = 660000., This is an AP in which a x, d 2000 and n 20., 20, S20 660000 &, #(2x 19 # 2000) 660000 & x 14000., 2, 8. Suppose that the work was completed in n days., Total number of workers engaged for 1 day each, 150 146 142 … to n terms, n #[2 #150 (n 1)#( 4)] n (152 2n) ., 2, If 150 workers are engaged each day, time taken (n 8) days., Number of workers engaged for 1 day 150 (n 8) ., , , n (152 2n) 150 (n 8) & n 2 n 600 0, , & (n 25) (n 24) 0, , & n 25, [a n ! 24]., 9. Let the total savings be ` 85000 in n years. Then,, n, #[2 # 4000 (n 1)#1000] 85000, 2, 2, & n 7n 170 0 & (n 17) (n 10) 0 & n 10 years., 10. Let the first instalment be ` a and let it increase by ` d each year., 2, Then, S30 c # 36000m 24000 and S 40 36000, 3, 30, 40, #(2a 29d) 24000 and, #(2a 39d) 36000, &, 2, 2, , , … (i), and 2a 39d 1800., … (ii), & 2a 29d 1600, On solving (i) and (ii), we get a 510 and d 20., first instalment = ` 510., 11. Let the production in first year be a, increasing uniformly at the rate of d per year., Then,, a 2d 6000 and a 6d 7000., On solving, we get a 5500 and d 250., (i) Production in the 1st year a 5500., (ii) Production in the 10th year (a 9d) (5500 9 # 250) ., 7, 7, (iii) Production in 7 years #[2a (7 1) d] #[2 # 5500 6 # 250] ., 2, 2, 12. Total cost = ` 180000. Paid in cash = ` 90000. Balance = ` 90000., Amount of each instalment = ` 9000., 90000 , Number of instalments , 10., 9000
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Arithmetic Progression, , 417, , 1st instalment ` c9000 90000 #, , 12, #1m ` 19800., 100, 12, 2nd instalment ` c9000 81000 #, #1m ` 18720., 100, 12, 3rd instalment ` c9000 72000 #, #1m ` 17640, and so on., 100, Total payment ` [90000 (19800 18720 17640 … up to 10 terms], ` ;90000 10 · {39600 9 #(1080)}E, 2, ` (90000 149400) ` 239400., , ARITHMETIC MEAN, ARITHMETIC MEAN, , If a, A, b are in AP then we say that A is the arithmetic mean (AM) between a and b., INSERTION OF A SINGLE ARITHMETIC MEAN BETWEEN a AND b, , Let a and b be two given numbers and let A be the arithmetic mean between a, and b. Then,, a, A, b are in AP, & Aa bA, ab ·, & 2A a b & A , 2, ab ·, Hence, the arithmetic mean between a and b is, 2, EXAMPLE 1 Find the arithmetic mean between, (i) 14 and –6,, (ii) (a b) and (a b)., SOLUTION, (i) Arithmetic mean between 14 and –6, 14 ( 6) 8, , 4., 2, 2, (ii) Arithmetic mean between (a b) and (a b), (a b) (a b) 2a, , , a., 2, 2, EXAMPLE 2, , SOLUTION, , an bn, o is the AM between a and b then find the value of n., bn1, a, (a b), ·, We know that the AM between a and b is, 2, n, n, a b, , e n1, o is the AM between a and b, bn1, a, an bn a b, &, n1, 2, bn1, a, , , n, n, n, &, 2a 2b a a n 1b b n 1a b n, n, n, n1, n1, &, a b a bb a, If e, , n1
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418, , Senior Secondary School Mathematics for Class 11, , , a n a n 1 b b n 1a b n, n1, n1, a (a b) b (a b), , , [on dividing both sides by (a b)], &, an 1 bn 1, a n1 a 0, &, 1 b l & n 1 0 & n 1., b l, b, b, Hence, the required value of n is 1., , &, &, , n ARITHMETIC MEANS BETWEEN a AND b, , If a, A1, A 2, …, An, b are in AP then we say that A1, A 2, …, An are the n arithmetic, means between a and b., INSERTION OF n ARITHMETIC MEANS BETWEEN a AND b, , Let a and b be two given numbers and let A1, A 2, …, An be the n arithmetic, means between a and b. Then,, a, A1, A 2, …, An, b are in AP., In this AP, we have first term a, last term b, number of terms (n 2) ., Let the common difference be d. Then,, b Tn 2 & b a (n 2 1) d, ba, & d ·, n 1, A1 (a d), A 2 (a 2d), A 3 (a 3d), …, An (a nd), , … (i), , (b a), 2 (b a), n (b a ), 3, A 2 )a , 3, …, An )a , 3·, ( n 1), ( n 1), ( n 1), These are the required n arithmetic means between a and b., , &, , A1 )a , , EXAMPLE 3, SOLUTION, , Insert six arithmetic means between 15 and –13., Let A1, A 2, A 3, A 4, A 5, A6 be the six arithmetic means between 15, and –13. Then,, 15, A1, A 2, A 3, A 4, A 5, A6, 13 are in AP., (13 15) 28, (b a) ( 13 15), G·, =a d , , 4, , Now, d , 7, (6 1), (n 1), ( 6 1), , A1 (15 d) (15 4) 11, A 2 (15 2d) (15 8) 7,, A 3 (15 3d) (15 12) 3, A 4 (15 4d) (15 16) 1,, A 5 (15 5d) (15 20) 5, A6 (15 6d) (15 24) 9., Hence, the required six AMs between 15 and –13 are, 11, 7, 3, –1, –5 and –9., , EXAMPLE 4, SOLUTION, , If m arithmetic means are inserted between 1 and 31 so that the ratio of the, 7th and (m 1)th means is 5 : 9 then find the value of m., Let A1, A 2, …, Am be the m arithmetic means between 1 and 31., Then, 1, A1, A 2, …, Am, 31 are in AP., (31 1), (b a), G·, =a d , 30, , d, , , (m 1) m 1, (m 1)
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Arithmetic Progression, , 419, , 7 # 30 (m 211), ·, l, m1, (m 1), (m 1)# 30, 3, (m 1)th mean Am 1 Tm a (m 1) d )1 , (m 1), (31m 29), , ·, (m 1), A7, (m 211), ( m 1), 5 &, 5, Now,, #, , 9, Am 1, ( m 1), (31m 29) 9, (m 211), 5 & 9m 1899 155m 145, &, (31m 29) 9, 7th mean A7 T8 (a 7d) b1 , , & 146m 2044 & m , Hence, m 14., EXAMPLE 5, SOLUTION, , EXAMPLE 6, SOLUTION, , 2044 , 14., 146, , If x, y, z are in AP and A1 is the AM between x and y while A 2 is the AM, between y and z then prove that the AM between A1 and A 2 is y., Since x, y, z are in AP, we have x z 2y., … (i), xy, ·, … (ii), Since A1 is the AM between x and y, we have A1 , 2, yz, ·, … (iii), Since A 2 is the AM between y and z, we have A 2 , 2, 1, 1 x y y z2, AM between A1 and A 2 (A1 A 2) (, , 2, 2 2, 2, x 2y z 4y, , , y, [using (i)]., 4, 4, Hence, the AM between A1 and A 2 is y., If the AM between pth and qth terms of an AP be equal to the AM between, its rth and sth terms then prove that p q r s., Let a be the first term and d be the common difference of the given, AP. Then,, 1, 1, (T T ) (Tr Ts), 2 p q, 2, &, Tp Tq Tr Ts, &, { a ( p 1 ) d} { a ( q 1 ) d } { a ( r 1 ) d } { a ( s 1 ) d }, &, 2a (p q 2) d 2a (r s 2) d, &, ( p q 2) d ( r s 2) d, &, p q 2 r s 2 & p q r s., Hence, p q r s., , EXAMPLE 7, , The n arithmetic means between 20 and 80 are such that the, first mean : last mean 1 : 3. Find the value of n.
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420, SOLUTION, , Senior Secondary School Mathematics for Class 11, , Let A1, A 2, …, An be n AMs between 20 and 80., Then, 20, A1, A 2, …, An, 80 are in AP., Let their common difference be d. Then,, Tn 2 80 & 20 (n 2 1) d 80 & d , First mean, A1 T2 (20 d) b20 , , 60, l, n1, (20n 80), , ·, (n 1), Last mean, An Tn 1 [20 (n 1 1) d], , 60, ·, (n 1), , … (i), [using (i)], , (20 n # 60 2, [using (i)], (n 1), (80n 20), , ·, (n 1), A1 1, (20n 80), (n 1), &, 1, , #, An 3, (n 1), (80n 20) 3, & 3 (20n 80) (80n 20) & 20n 220 & n 11., Hence, n 11., EXAMPLE 8, SOLUTION, , EXAMPLE 9, SOLUTION, , If n arithmetic means are inserted between two numbers, prove that the, sum of the means equidistant from the beginning and the end is constant., Let A1, A 2, …, An be n AMs between a and b., Then, a, A1, A 2, …, An, b are in AP., , mth mean from the beginning + mth mean from the end, (m 1)th term from the beginning, + (m 1)th term from the end, {a (m 1 1) d} {b (m 1 1) d} (a b) constant., Hence, the sum of the means equidistant from the beginning and, the end is constant., Show that the sum of n arithmetic means between two numbers is n times, the single arithmetic mean between them., Let a and b be two given numbers and let A be the single AM, ab ·, between them. Then, A , 2, Let A1, A 2, …, An be the n arithmetic means between a and b., Then a, A1, A 2, …, An, b are in AP., n, n, Now, (A1 A 2 … An) (A1 An), 9Sn (a l)C, 2, 2, n (a b), [a a, A1, A 2, …, An, b is an AP, 2, & a b A1 A n ], n# (AM between a and b)., Hence, the sum of n arithmetic means between a and b is n times, the AM between a and b.
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Arithmetic Progression, , 421, , EXERCISE 11D, 1. Find the arithmetic mean between, (i) 9 and 19,, (ii) 15 and –7,, (iii) –16 and –8., 2. Insert four arithmetic means between 4 and 29., 3. Insert three arithmetic means between 23 and 7., 4. Insert six arithmetic means between 11 and –10., 5. There are n arithmetic means between 9 and 27. If the ratio of the last mean, to the first mean is 2 : 1, find the value of n., 6. Insert arithmetic means between 16 and 65 such that the 5th AM is 51. Find, the number of arithmetic means., 7. Insert five numbers between 11 and 29 such that the resulting sequence is, an AP., 8. Prove that the sum of m arithmetic means between the two numbers is to, the sum of n arithmetic means between them is m : n., ANSWERS (EXERCISE 11D), , 1. (i) 14 (ii) 4 (iii) –12, 5. n 5, , 2. 9, 14, 19, 24, 6. six, , 3. 19, 15, 11, 4. 8, 5, 2, –1, –4, –7, 7. 14, 17, 20, 23, 26, , HINTS TO SOME SELECTED QUESTIONS, 5. Let A1, A2, …, An be the n arithmetic means between 9 and 27., Then, 9, A1, A2, …, An, 27 are in AP., 18, 9 (n 2 1) d 27 & d ·, n 1, An 2, T , 9 (n 1 1) d 2, & n 12 &, , T2, 1, 1, A1 1, 9 (2 1) d, , , 9, & 9 nd 18 2d & d ·, n 2, 18 9, & 18 (n 2) 9 (n 1) & n 5., n1 n2, , 6. 16, A1, A2, A3, A 4, A 5, …, An, 65 are in AP., Then, T6 51 & a 5d 51 & 16 5d 51 & d 7., Tn 2 65 & 16 (n 2 1) d 65 & 16 (n 1)#7 65 & n 6., 8. Sum of m AMs between a and b, m #(AMs between a and b) m #, Sum of n AMs between a and b, n #(AMs between a and b) n#, , , required ratio m : n., , (a b), ·, 2, , (a b), ·, 2
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422, , Senior Secondary School Mathematics for Class 11, , PROPERTIES OF AP, , If to each term of an AP a fixed nonzero number is added then the resulting, progression is an AP., Let a, a d, a 2d, a 3d, … be a given AP and let k be a fixed nonzero, number. Then,, (a k), {(a d) k}, {(a 2d) k}, … is an AP, in which first term (a k) and common difference = d., , THEOREM 1, PROOF, , THEOREM 2, PROOF, , If each term of a given AP is multiplied or divided by a given nonzero fixed, number k then the resulting progression is an AP., (i) Let a, a d, a 2d, … be the given AP and let k be a given nonzero, number. Then,, ak, (a d) k, (a 2d) k, … is clearly an AP, in which the first term ak and the common difference dk., (ii) Let a, a d, a 2d, … be the given AP and let k be a given nonzero, number. Then,, a , (a d) , (a 2d) , …, is clearly an AP, k, k, k, d, a, in which the first term and the common difference ·, k, k, , SUMMARY, , If a1, a 2, a 3, a 4, … are in AP and k is a nonzero number then, (i) (a1 k), (a 2 k), (a 3 k), (a 4 k), … are in AP., (ii) (a1 k), (a 2 k), (a 3 k), (a 4 k), … are in AP., (iii) a1 k, a 2 k, a 3 k, a 4 k, … are in AP., a1 a 2 a 3 a 4, , , , , … are in AP., (iv), k k k k, SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , If a, b, c, are in AP, show that, (i) (b c), (c a) and (a b) are in AP., (ii) a 2 (b c), b 2 (c a) and c 2 (a b) are in AP., Since a, b, c are in AP, we have, 2b (a c) ., (i) (b c), (c a), (a b) will be in AP, if (c a) (b c) (a b) (c a), i.e., if a b b c, i.e., if 2b a c, which is true by (i)., a, b, c are in AP & (b c), (c a), (a b) are in AP., , … (i)
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Arithmetic Progression, , 423, , (ii) a 2 (b c), b 2 (c a), c 2 (a b) will be in AP, if b 2 (c a) a 2 (b c) c 2 (a b) b 2 (c a), i.e.,, i.e.,, i.e.,, i.e.,, , , if c (b 2 a 2) ab (b a) a (c 2 b 2) bc (c b), if (b a) (ab bc ca) (c b) (ab bc ca), if (b a) (c b), if 2b (a c), which is true by (i)., a, b, c are in AP & a 2 (b c), b 2 (c a), c 2 (a b) are in AP., , EXAMPLE 2, , If a, b, c are in AP, show that, 1, 1, 1, ,, ,, are in AP., ( b c ) ( c a) ( a b), , SOLUTION, , Since a, b, c are in AP, we have, … (i), 2b (a c) ., 1, 1, 1, ,, ,, will be in AP, Now,, ( b c ) ( c a) ( a b), 1, 1, 1, 1, , , , if, , , , , ( c, a) ( b, c) ( a, b) ( c, a), ( b c ) ( c a) ( c a) ( a b), , i.e., if, ( c a) ( b c ), ( a b) ( c a), ( b a), ( c b), , i.e., if, , , , ( c, a) ( b, c) ( a, b) ( c a), ( b a) ( c b), , i.e., if, ( b c ) ( a b), i.e., if b a c b, i.e., if 2b a c, which is true by (i)., 1, 1, 1, ,, ,, are in, , a, b, c are in AP &, , , , ( b, c) ( c, a) ( a, b), AP., , EXAMPLE 3, , If a, b, c are in AP, show that, [(b c) 2 a 2], [(c a) 2 b 2], [(a b) 2 c 2] are in AP., Since a, b, c are in AP, we have, 2b (a c) ., , SOLUTION, , … (i), , Now, [(b c) 2 a 2], [(c a) 2 b 2], [(a b) 2 c 2] will be in AP, i.e.,, i.e.,, i.e.,, i.e.,, , if [(c a) 2 b 2] [(b c) 2 a 2] [(a b) 2 c 2] [(c a) 2 b 2], if (a b c) (c a b) (a b c) (b c a), (a b c) (a b c) (a b c) (c a b), , , , , if (a b c) [(c a b) (b c a)], (a b c) [(a b c) (c a b)], if 2(a b) 2(b c), if (a b) (b c)
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424, , Senior Secondary School Mathematics for Class 11, , i.e.,, , &, EXAMPLE 4, , SOLUTION, , if 2b (a c), which is true by (i)., a, b, c are in AP, [(b c) 2 a 2], [(c a) 2 b 2], [(a b) 2 c 2] are in AP., , If a, b, c are in AP, show that, 1, 1, 1, are in AP., (i), bc ca ab, 1 1, 1 1, 1 1, (ii) a b c l, b b c a l, c b a l are in AP., b, b, (i) Since a, b, c are in AP, the terms obtained by dividing each term, of this AP by abc are also in AP., 1 1 1, are in AP., Consequently, , ca ,, bc, ab, (ii) a, b, c are in AP, 1, 1, 1, are in AP, [dividing each term by abc], &, bc ca ab, ab bc ca , d(ab bc ca n, ab bc ca, n, d, n are in AP, & d, ca, bc, ab, [multiplying each term by (ab bc ca)], &, , &, &, &, EXAMPLE 5, , SOLUTION, , ab bc ca , b ab bc ca l, ab bc ca , 1n, 1 d, 1n are in AP, ca, bc, ab, [adding –1 to each term], (ac ab) (ab bc) (bc ca), ,, ,, are in AP, ca, bc, ab, c b , a a c k, b a, n b ca c d, n are in AP, ad, bc, ab, 1 1, 1 1, 1 1, a b c l, b b c a l, c b a l are in AP., b, b, , d, , If a 2, b 2, c 2 are in AP, prove that, 1 , 1 , 1, are in AP., (b c) (c a) (a b), 1 , 1 , 1, will be in AP, (b c) (c a) (a b), 1 1 1 1, if, (c a) (b c) (a b) (c a), i.e.,, i.e.,, i.e.,, , (b c) (c a) (c a) (a b), , (c a) (b c), (a b) (c a), (b a), (c b), , if, (c a) (b c) (a b) (c a), (b a) (c b), , if, (b c) (a b), if
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Arithmetic Progression, , i.e.,, , if (b a) (b a) (c b) (c b), , i.e.,, i.e.,, , if b 2 a 2 c 2 b 2, if a 2, b 2, c 2 are in AP., , Hence, a 2, b 2, c 2 are in AP &, EXAMPLE 6, , SOLUTION, , if b 2 a 2 c 2 b 2, if a 2, b 2, c 2 are in AP., , Hence, a 2, b 2, c 2 are in AP &, , SOLUTION, , 1 , 1 , 1, are in AP., (b c) (c a) (a b), , If a 2, b 2, c 2 are in AP, prove that, a , b , c, are in AP., bc ca ab, a , b , c, will be in AP, bc ca ab, a , b, c , if, 1, 1,, 1 are in AP, c a, bc, ab, [on adding 1 to each term], abc, abc, abc, i.e., if, are in AP, ca, bc, ab, 1 , 1 , 1, i.e., if, are in AP, bc ca ab, [on dividing each term by (a b c)], 1 1 1 1, i.e., if , c a bc ab ca, (b c) (c a) (c a) (a b), , i.e., if, (c a) (b c), (a b) (c a), (b a) (c b), , i.e., if, (b c) (a b), i.e., if (b a) (b a) (c b) (c b), i.e.,, i.e.,, , EXAMPLE 7, , 425, , a , b , c, are in AP., (b c) (c a) (a b), , Show that (a 2 ab b 2), (c 2 ac a 2) and (b 2 bc c 2) are in AP, if, a, b, c are in AP., The terms (a 2 ab b 2), (c 2 ac a 2), (b 2 bc c 2) will be in AP, i.e.,, i.e.,, , if (c 2 ac a 2) (a 2 ab b 2) (b 2 bc c 2) (c 2 ac a 2), if c 2 ac ab b 2 b 2 bc ac a 2, if (a 2 c 2 2ac) b 2 b 2 bc ab, , i.e., if (a c) 2 b 2 b (a b c), i.e., if (a c b) (a c b) b (a b c), i.e., if a c b b, i.e., if a c 2b, i.e., if a, b, c are in AP., Hence, a, b, c are in AP & (a 2 ab b 2), (c 2 ac a 2) and (b 2 bc c 2)
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426, , Senior Secondary School Mathematics for Class 11, , are in AP., EXAMPLE 8, , If (b c) 2, (c a) 2, (a b) 2 are in AP, prove that, 1 , 1 , 1, are in AP., (b c) (c a) (a b), , SOLUTION, , (b c) 2, (c a) 2, (a b) 2 are in AP, (c a) 2 (b c) 2 (a b) 2 (c a) 2, … (i), (b a) (2c a b) (c b) (2a b c) ., 1 , 1 , 1, will be in AP, Now,, (b c) (c a) (a b), 1 1 1 1, if, (c a) (b c) (a b) (c a), (b c) (c a) (c a) (a b), , i.e., if, (c a) (b c), (a b) (c a), (a b 2c), ( b c 2a ), , i.e., if, (c a) (b c) (a b) (c a), i.e., if (a b) (a b 2c) (b c) (b c 2a), i.e., if (b a) (2c a b) (c b) (2a b c), which is true by (i)., , &, &, , Hence, (b c) 2, (c a) 2, (a b) 2 are in AP, 1 , 1 , 1, are in AP., &, (b c) (c a) (a b), EXAMPLE 9, , If (a 2 2bc), (b 2 2ac), (c 2 2ab) are in AP, show that, 1 , 1 , 1, are in AP., (b c) (c a) (a b), , SOLUTION, , a 2 2bc, b 2 2ac, c 2 2ab are in AP, &, , (a 2 2bc) (ab bc ac), (b 2 2ac) (ab bc ac), (c 2 2ab), (ab bc ac) are in AP, [by adding (ab bc ac) to each term], , (a 2 bc ab ac), (b 2 ac ab bc), (c 2 ab bc ac) are in AP, (a b) (a c), (b c) (b a), (c b) (c a) are in AP, (a b) (c a), (a b) (b c), (b c) (c a) are in AP, 1, 1, 1, ,, ,, are in AP, &, (b c) (c a) (a b), [on dividing each term by (a b) (b c) (c a)], 1 , 1 , 1, are in AP, &, (b c) (c a) (a b), [on multiplying each term by –1]., Hence, the result follows., &, &, &, , EXAMPLE 10, , If, , (b c a) (c a b) (a b c), 1 1 1, ,, ,, are in AP, prove that a , , c are in AP., a, c, b, b
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Arithmetic Progression, SOLUTION, , EXAMPLE 11, SOLUTION, , (b c a) (c a b) (a b c), are in AP, ,, ,, a, c, b, (b c a ), (c a b), (a b c), (, 22 , (, 22 , (, 22 are in AP, &, a, c, b, [adding 2 to each term], (a b c) (a b c) (a b c), ,, ,, are in AP, &, a, c, b, 1, 1, 1, [on dividing each term by (a b c)]., &, a b c are in AP, If a 2 (b c), b 2 (c a), c 2 (a b) are in AP then show that either a, b, c are, in AP or ab bc ca 0., a 2 (b c), b 2 (c a), c 2 (a b) are in AP, &, , b 2 (c a) a 2 (b c) c 2 (a b) b 2 (c a), (b 2a a 2b) (b 2c a 2c) (c 2b b 2c) (c 2a b 2a), , &, &, &, &, &, &, &, , ab (b a) c (b 2 a 2) bc (c b) a (c 2 b 2), (b a) (ab bc ca) (c b) (ab bc ca), (ab bc ca) {(b a) (c b)} 0, (ab bc ca) (2b a c) 0, ab bc ca 0 or 2b a c 0, ab bc ca 0 or 2b a c, ab bc ca 0 or a, b, c are in AP., , &, , EXAMPLE 12, SOLUTION, , 427, , If a, b, c are in AP, prove that (a c) 2 4 (b 2 ac) ., 1, Since a, b, c are in AP, we have b (a c) ., 2, 1, 2, 2, , RHS 4 (b ac) 4 # ' (a c) ac1, 4, (a c) 2 4ac (a c) 2 LHS., Hence, (a c) 2 4 (b 2 ac) ., , EXAMPLE 13, , If a, b, c are in AP, prove that a 3 4b 3 c 3 3b (a 2 c 2) ., , SOLUTION, , Since a, b, c are in AP, we have b , , 1 , (a c) ., 2, , 3, 1 , (a c) (a 2 c 2) (a 3 c 3 ac 2 a 2c), 2, 2, ac 3 3 3 3 3 2 2, l c, and LHS a 3 4 · b, (a c ac a c) ., 2, 2, , LHS = RHS., , , RHS 3 ·, , EXERCISE 11E, 1. If a, b, c are in AP, prove that, (i) (a c) 2 4 (a b) (b c), (iii) a 3 c 3 6abc 8b 3, , (ii) a 2 c 2 4ac 2 (ab bc ca)
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428, , Senior Secondary School Mathematics for Class 11, , 2. If a, b, c are in AP, show that, (a 2b c) (2b c a) (c a b) 4abc., 3. If a, b, c are in AP, show that, (i) (b c a), (c a b), (a b c) are in AP., (ii) (bc a 2), (ca b 2), (ab c 2) are in AP., 1 1 1, 4. If a , , c are in AP, prove that, b, (b c) (c a) (a b), ,, ,, (i), a, c are in AP., b, (b c a) (c a b) (a b c), ,, ,, (ii), are in AP., a, c, b, 1 1, 1 1, 1 1, 5. If a b c l, b b c a l, c b a l are in AP, prove that a 2 (b c), b 2 (c a), c 2 (a b), b, b, are in AP., a (b c) b (c a) c (a b), ,, ,, 6. If a, b, c are in AP, show that, are also in AP., ca, bc, ab, HINTS TO SOME SELECTED QUESTIONS, 1 1 1, 4. a , , c are in AP, b, &, &, &, , &, , abc, abc, abc, are in AP, a, c, b, bc, ca, ab, 1 a ,1, 1, c are in AP, b, , , , b c, c a, a b, a, c are in AP. This proves (i)., b, bc ca ab , 1,, 1, c, 1 are in AP, a, b, bca, cab, abc, are in AP. This proves (ii)., a, c, b, , 1 1, 1 1, 1 1, 5. a d c n, b c c a m, c d a n are in AP, b, b, a (b c) b (c a) c (a b), ,, ,, are in AP, &, ca, bc, ab, 2, 2, 2, , , , a (b c) b (c a) c (a b), ,, ,, are in AP, &, abc, abc, abc, 2, 2, 2, & a (b c), b (c a), c (a b) are in AP., 6., , a (b c) b (c a) c (a b), will be in AP, , ca ,, bc, ab, a (b c), b (c a), c (a b), 1,, 1,, 1 are in AP, if, ca, bc, ab, , , , , , , ab bc ca , ab bc ca , ab bc ca, i.e., if, are in AP, ca, bc, ab, 1, 1, 1, i.e., if, are in AP, bc ca ab
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Arithmetic Progression, , 429, , abc , abc , abc, are in AP, bc ca ab, i.e., if a, b, c are in AP., i.e., if, , EXERCISE 11F, Very-Short-Answer Questions, 1. If the sum of n terms of an AP is given by Sn (2n 2 3n) then find its, common difference., 2. If 9 times the 9th term of an AP is equal to 13 times the 13th term, show that, its 22nd term is 0., 3. In an AP it is given that Sn qn 2 and Sm qm 2. Prove that Sq q 3., 4. Find three arithmetic means between 6 and –6., 5. The 9th term of an AP is 0. Prove that its 29th term is double the 19th term., 6. How many terms are there in the AP 13, 16, 19, …, 43?, 7. Find the 8th term from the end of the AP 7, 9, 11, …, 201., 8. How many 2-digit numbers are divisible by 7?, 9. If 7th and 13th terms of an AP be 34 and 64 respectively then find its 18th, term., 10. What is the 10th common term between the APs 3, 7, 11, 15, 19, … and, 1, 6, 11, 16, …?, 11. The first and last terms of an AP are 1 and 11 respectively. If the sum of its, terms is 36, find the number of terms., 12. In an AP, the pth term is q and (p q)th term is 0. Show that its qth term is p., 13. If, , 3 5 7 9 … up to 35 terms, 7, find the value of n., 5 8 11 … up to n terms, , 14. Write the sum of first n even natural numbers., 15. Write the sum of first n odd natural numbers., 1, 16. The sum of n terms of an AP is an 2 bn. Find the common difference., 2, 17. If the sums of n terms of two APs are in the ratio (2n 3) : (3n 2), find the, ratio of their 10th terms., ANSWERS (EXERCISE 11F), , 1. 4, 9. 89, 15. n 2, , 4. 3, 0, –3, 10. 191, 16. a, , 6. 11, 11. 6, 17. 41 : 59, , 7. 197, 13. n 10, , 8. 13, 14. n (n 1)
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430, , Senior Secondary School Mathematics for Class 11, HINTS TO SOME SELECTED QUESTIONS, , 1. Tn (Sn Sn 1) (2n 2 3n) {2 (n 1) 2 3 (n 1)}, (2n 2 3n) (2n 2 n 1) 4n 1., d (T2 T1) (4 # 2 1) (4 #1 1) (9 5) 4., 2. 9 (a 8d) 13 (a 12d) & 4a 84d 0 & a 21d 0 & T22 0., 3. Sn qn 2 & S1 q and S2 4q, & T1 q and T1 T2 4q & T1 q and T2 3q., , , Sq , , q, q, #[2q (q 1)# 2q] d # 2q 2n q 3., 2, 2, , 4. Let the required AM be x1, x2, x3 . Then,, 6, x1, x2, x3, 6 are in AP., Now, T5 6 & a 4d 6 & 6 4d 6 & 4d 12 & d 3., x1 6 d (6 3) 3, x2 6 2d (6 6) 0, x3 6 3d (6 9) 3., 5. T9 0 & a 8d 0 & a 8d., T19 a 18d 8d 18d 10d and T29 a 28d 8d 28d 20d., T29 2 #T19 ., 6. Let it contain n terms. Then, Tn 43 & a (n 1) d 43., 13 (n 1)# 3 43 & n 11., 7. Here, a 7, d (9 7) 2 and l 201., Let Tn 201. Then, a (n 1) d 201 & 7 (n 1)# 2 201 & n 98., 8th term from the end l (n 1) d (201 7 # 2) (201 14) 197., 8. 2-digit numbers divisible by 7 are 14, 21, 28, …, 98., Let their number be n. Then, Tn 98., 14 (n 1)#7 98 & n 13., 9. a 6d 34, … (i), and, , a 12d 64., , On solving (i) and (ii), we get a 4 and d 5., T18 a 17d (4 17 # 5) 89., 10. The first common term = 11., Common difference of new AP = LCM (4, 5) = 20., So, the new AP is 11, 31, 51, … ., Its 10th term (11 9 # 20) 191., n , n, (a l) 36 & #(1 11) 36 & n 6., 2, 2, 35, · {6 34 # 2}, 2, 7 & 3n 2 7n 370 0, 13. n, · {10 (n 1)# 3}, 2, & (3n 37) (n 10) 0 & n 10., 11., , 14. Given sum 2 4 6 8 … 2n, n (2 2n) n (n 1) ., 2, 15. Given sum 1 3 5 … to n terms, n #[2 #1 (n 1)# 2] n 2., 2, , … (ii)
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Arithmetic Progression, , 431, , 16. Tn (Sn Sn 1), '1 an 2 bn1 '1 a (n 1) 2 b (n 1)1, 2, 2, 1, 1, 2, ' an bn1 ' an 2 an 1 a bn b1 can b 1 am ·, 2, 2, 2, 2, 1, 1, Tn 1 a (n 1) b a can b am a., 2, 2, d (Tn Tn 1) a., n·, [2a1 (n 1) d1], 2a1 (n 1) d1 2n 3, , 2, 2n 3 &, , ·, 17. n, , 3, n, 2, 2a2 (n 1) d2 3n 2, · [2a2 (n 1) d2], 2, n1, m d1, a1 c, a 9d1 2 #19 3, , 2, n1 , 2n 3 & 1, , ;putting, , 9 & n 19E, , , 3, n, 2, 2, a2 9d2 3 #19 2, n 1, , m d2, a2 c, 2, a1 9d1 41, , ·, &, a2 9d2 59, Hence, the ratio of their 10th terms is 41 : 59., , REVIEW OF FACTS AND FORMULAE, 1., , (i) Sequence A succession of numbers arranged in a definite order, according to a certain given rule is called a sequence., The number occurring at the nth place of a sequence is called its nth, term, denoted by an ., (ii) By adding the terms of a sequence, we get a series., (iii) A series is said to be finite or infinite according as the number of terms, in it is finite or infinite respectively., 2. (i) Progressions Sequences following certain patterns are called, progressions., (ii) Arithmetic Progression (AP), The progression of the form a, (a d), (a 2d), … is called an AP with, first term a and common difference d., We have, Tn a (n 1) d, called the general term., Clearly, Tn Tn 1 d., (iii) If l is the last term of the AP then nth term from the end l (n 1) d., (iv) If an AP contains n terms then, pth term from the end (n p 1)th term from the beginning., 3. Sum to n terms of an AP is given by, n, (i) Sn · {2a (n 1) d}, 2, n, (ii) Sn (a l), 2, (iii) S1 T1, (S 2 S1) T2, Sn Sn 1 Tn
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432, , Senior Secondary School Mathematics for Class 11, , (i) 3 numbers in AP are taken as (a d), a, (a d) ., (ii) 4 numbers in AP are taken as (a 3d), (a d), (a d), (a 3d) ., (iii) 5 numbers in AP are taken as (a 2d), (a d), a, (a d), (a 2d) ., 5. If a1, a 2, a 3, a 4, … are in AP and k is a nonzero number then, (i) (a1 k), (a 2 k), (a 3 k), (a 4 k), … are in AP., (ii) (a1 k), (a 2 k), (a 3 k), (a 4 k), … are in AP., (iii) a1 k, a 2 k, a 3 k, a 4 k, … are in AP., a1 a 2 a 3 a 4, , , , , … are in AP., (iv), k k k k, 4., , 6., , (i) Arithmetic Mean (AM), , If a, A, b are in AP then A is the arithmetic, 1 , (a b) ., 2, (ii) If a, A1, A 2, …, An, b are in AP then A1, A 2, …, An are n arithmetic, means between a and b., mean between a and b and A , ,
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Geometrical Progression, , 433, , 12, , Geometrical Progression, GEOMETRICAL PROGRESSION (GP), , A sequence a1, a 2, a 3, …, an is called a geometrical progression, if each term is nonzero, ak 1, and a r (constant) for all k $ 1., k, The constant ratio is called its common ratio., A geometrical progression is abbreviated as GP., In a GP we usually denote the first term by a, the common ratio by r and the, nth term by Tn ., The nth term of a GP is called its general term., SOME EXAMPLES OF GP, EXAMPLE 1, , SOLUTION, , EXAMPLE 2, , SOLUTION, , EXAMPLE 3, , SOLUTION, , Show that the progression 6, 18, 54, 162, … is a GP. Write down its first, term and the common ratio., 18 54 162 , We have, 3 (constant)., 6 18, 54, So, the given progression is a GP in which the first term = 6 and the, common ratio = 3., 1, Show that the progression 16, 4, 1, , … is a GP. Write down its first, 4, term and the common ratio., 1 (1/4) 1, 4, , , We have , (constant)., 1, 16, 4, 4, 1, ·, So, the given progression is a GP in which a 16 and r , 4, 1 , 1 , 2 , 4 , …, is a GP. Write down its first, 2 3 9 27, term and find the common ratio., We have, 1, 1, , 2 2, , 1, , ' b 1 l (2 # 3 2 2 ,, b l ' b # 2l , 3, 3, 2, 3, 3 9, 3, 9 (1), 4, 2 4 9 2 ·, b l', b # l, 27, 9, 27 2, 3, 2, 1, ·, So, the given progression is a GP in which a and r , 2, 3, , Show that the progression, , 433
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434, EXAMPLE 4, SOLUTION, , Senior Secondary School Mathematics for Class 11, , Show that the sequence given by Tn (2 # 3 n) for all nd N is a GP. Find, its first term and the common ratio., We have, , Tn (2 # 3 n) and Tn 1 (2 # 3 n 1) ., , , Tn 1 2 # 3 n 1, , n 3 (constant) for all nd N., Tn, 2# 3, , Also, T1 (2 # 3 1) (2 # 3) 6., , the first term = 6 and the common ratio = 3., GEOMETRIC SERIES If a1, a 2, a 3, …, a n, … is a GP then the sum a1 a 2 a 3 … an …, is called a geometric series., A geometric series is finite or infinite according as the corresponding GP, is finite or infinite., , GENERAL TERM OF A GP, THEOREM 1, PROOF, , Prove that the nth term of a GP with first term a and common ratio r is, , given by Tn ar n 1., Let us consider a GP in which first term a and common ratio r., Then, the GP is a, ar, ar 2, … ., Its first term, T1 a ar (1, , 1), , ;, , second term, T2 ar ar (2, third term, T3 ar ar, …, …, …, …, …, …, …, …, 2, , , , nth term, Tn ar (n, , Hence, Tn ar, , n1, , 1), , 1), , ;, , (3 1), , ;, …, …, , ., , (known as the general term of the GP)., , REMEMBER, , In a GP with first term a and common ratio r, we have, , nth term, Tn ar n 1., SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Find the 10th term and the general term of the progression, 1 , 1 , , 1, 2, 4, … ., 4 2, In the given progression, we have, 1, 1, 1, 1, b l ' b # 4l 2, 1 ' b l 1#( 2) 2,, 2, 2, 2, 4, ( 2) ' 1 2 and 4 ' ( 2) 2., 1, So, the given progression is a GP in which a and r 2., 4
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Geometrical Progression, , , , the 10th term, T10 ar (10, , The general term, Tn ar (n, EXAMPLE 2, , , ar 9 1 #( 2) 9 512 128., 4, 4, , 1 #( 2) (n 1) (1) (n 1) # 2 (n 3)., 4, , Show that the progression, 1,, , SOLUTION, , 1), , 1), , 435, , ( 2 1) (3 2 2 ) (5 2 7), ,, ,, ,…, 12, 2 3, 24 3, , is a GP. Find its 5th term., In the given progression, we have, Z, _, T2 ]]]( 2 1) 1 bbb ( 2 1) T3 (3 2 2 ), ( 2 1), 2 3, ,, ,, [], , , # `b , #, T1 ]] 2 3, T2, 12, 1 bb, 2 3, 2 3, ( 2 1), \, a, T4 (5 2 7), ( 2 1), 12, , , ·, #, T3, 24 3, (3 2 2 ), 2 3, , , T2 T3 T4, ( 2 1), , , …, (constant)., T1 T2 T3, 2 3, , So, the given progression is a GP in which a 1 and r , , , the 5th term, T5 ar (5, , Hence, T5 , , 1), , (17 12 2 ), ·, 144, , ( 2 1), , JK 2 1NO4, O, ar 4 1#KK, K 2 3 OO, P, L, ( 2 1) 4 (3 2 2 ) 2, , , 144, 144, (17 12 2 ), , ·, 144, , EXAMPLE 3, , Which term of the GP 2, 8, 32, 128, … is 131072?, , SOLUTION, , In the given GP, we have a 2 and r , , 8, 4., 2, , Let its nth term be 131072. Then,, Tn 131072 & ar (n, , 1), , 131072, , (n 1), , 131072, , & 4 (n 1) 65536 4 8, & n 1 8 & n 9., Hence, the 9th term of the given GP is 131072., & 2# 4, , 1, 1 , 1 , 1 ,…, is, ?, 729 3, 3 3 3 9 3, , EXAMPLE 4, , Which term of the GP 3 ,, , SOLUTION, , In the given GP, we have a 3 and r e, , 1, 1, 1, # o ·, 3, 3, 3, , 2 3, , ·
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436, , Senior Secondary School Mathematics for Class 11, , Let its nth term be, , 1 ·, Then,, 729 3, , 1, 1, , & ar (n 1) , 729 3, 729 3, 1 (n 1) 1, & 3 #b l, 3, 729 3, 1 , 1, 1 1, & (n 1), (729 # 3) 2187 3 7, 3, & n 1 7 & n 8., 1 ·, Hence, the 8th term of the given GP is, 729 3, Tn , , EXAMPLE 5, SOLUTION, , If the 4th and 9th terms of a GP are 54 and 13122 respectively, find the GP., Also, find its general term., Let a be the first term and r be the common ratio of the given GP., , Then, T4 54 & ar (4 1) 54 & ar 3 54., … (i), , And, T9 13122 & ar (9 1) 13122 & ar 8 13122., On dividing (ii) by (i), we get, , … (ii), , 8, ar 13122, 5, 5, & r 243 3 & r 3., 3, 54, ar, Putting r 3 in (i), we get, a # 3 3 54 & 27a 54 & a 2., Thus, a 2 and r 3., So, the required GP is 2, 6, 18, 54, … ., The general term of the GP is given by, , , Tn ar (n 1) {2 # 3 (n 1)} ., , EXAMPLE 6, SOLUTION, , The first term of a GP is 1 and the sum of its 3rd and 5th terms is 90. Find, the common ratio of the GP., It is given that the first term of the GP is 1. Let r be the common ratio, of this GP. Then,, , , T3 1# r (3 1) r 2 and T5 1# r (5 1) r 4., , , (T3 T5) 90 & r 2 r 4 90, , & r 4 r 2 90 0, & t 2 t 90 0, where r 2 t, & (t 10) (t 9) 0, & t 10 or t 9, [a t r 2 ! 10], & t9, 2, & r 9, & r 3 or r 3., Hence, the common ratio of the given GP is 3 or –3.
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Geometrical Progression, EXAMPLE 7, SOLUTION, , 437, , The 4th, 7th and 10th terms of a GP are a, b, c respectively. Prove that, b 2 ac., Let A be the first term and r be the common ratio of the given GP., Then,, , , , a Ar (4 1) Ar 3; b Ar (7 1) Ar 6 and c Ar (10 1) Ar 9., , , ac (Ar 3)#(Ar 9) A 2r 12 (Ar 6) 2 b 2., , Hence, b 2 ac., EXAMPLE 8, , If a, b, c are three consecutive terms of an AP and x, y, z are three, consecutive terms of a GP, then prove that, , , , x (b c) · y (c a) · z (a b) 1., , SOLUTION, , It is given that a, b, c are in AP. Let d be the common difference of, this AP. Then,, (b c) (c b) d, (c a) {(c b) (b a)} 2d, and (a b) (b a) d., Also, x, y, z are in GP. So, y xz ., , , x (b, , c), , · y (c, , a), , · z (a, , b), , d, , x #( xz ) 2d # z d, [a (b c) d, (c a) 2d, (a b) d and y xz ], x d #(xz) d # z d, x d # x d # z d # z d x ( d d) # z ( d d) (x 0 # z 0) 1., Hence, x (b, EXAMPLE 9, , SOLUTION, , c), , · y (c, , a), , · z (a, , b), , 1., , If a, b, c, d are in GP, prove that, (a 2 b 2 c 2) (b 2 c 2 d 2) (ab bc cd) 2., Let r be the common ratio of the GP a, b, c, d., Then, b ar, c ar 2 and d ar 3., 2, 2, 2, 2, 2, 2, , LHS (a b c ) (b c d ), (a 2 a 2r 2 a 2r 4) (a 2r 2 a 2r 4 a 2r 6), a 4r 2 (1 r 2 r 4) 2., And, RHS (ab bc cd) 2 (a 2r a 2r 3 a 2r 5) 2, a 4r 2 (1 r 2 r 4) 2., Hence, (a 2 b 2 c 2) (b 2 c 2 d 2) (ab bc cd) 2., , EXAMPLE 10, , SOLUTION, , If a, b, c, d are in GP, prove that, (b c) 2 (c a) 2 (d b) 2 (a d) 2., Let r be the common ratio of the GP a, b, c, d., Then, b ar, c ar 2 and d ar 3., 2, 2, 2, , LHS (b c) (c a) (d b), (ar ar 2) 2 (ar 2 a) 2 (ar 3 ar) 2
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438, , Senior Secondary School Mathematics for Class 11, , {a (r r 2)} 2 {a (r 2 1)} 2 {a (r 3 r)} 2, a 2 (r r 2) 2 a 2 (r 2 1) 2 a 2 (r 3 r) 2, a 2 · {(r r 2) 2 (r 2 1) 2 (r 3 r) 2}, a 2 · {(r 2 r 4 2r 3) (r 4 1 2r 2) (r 6 r 2 2r 4)}, a 2 · (r 6 2r 3 1) a 2 (1 r 3) 2, (a ar 3) 2 (a d) 2 RHS., Hence, (b c) 2 (c a) 2 (d b) 2 (a d) 2., EXAMPLE 11, , If the pth, qth and rth terms of a GP be a, b, c respectively then prove that, , , , a (q r) · b (r p) · c (p q) 1., , SOLUTION, , Let A be the first term and R be the common ratio of the given GP., Then,, , Tp a & a AR (p 1), Tq b & b AR (q, Tr c & c AR (r, , , a, , (q r ), , (r p ), , 1), , 1), , ., , (p q), , ·b, ·c, {AR (p 1)} (q r) · {AR (q 1)} (r p) · {AR (r 1)} (p q), A (q r) · R (p 1)(q r) · A (r p) · R (q 1)(r p) · A (p q) · R (r 1)(p q), A {(q r) (r p) (p q)} · R {(p 1)(q r) (q 1)(r p) (r 1)(p q)}, A 0 · R p (q r) q (r p) r (p q) (r q) (p r) (q p), (1# R 0) (1#1) 1., , Hence, a (q, EXAMPLE 12, SOLUTION, , r), , · b (r, , p), , · c (p, , q), , 1., , If the first and the nth terms of a GP are a and b respectively and P is the, product of its first n terms then prove that P 2 (ab) n., Let r be the comon ratio of the given GP. Then,, 1, , , , b, b (n 1) ·, , , Tn b & ar n 1 b & r n 1 a & r b a l, P product of first n terms of the GP, , … (i), , a · ar · ar 2 … · ar n 1, 1, , a n · r {1 2 3 … (n 1)} a n · r 2 n (n 1), [taking the sum of the AP], 1, Z], 1 _, b, , n, n, 1, (, ), ] (n 1)b2, `b, a n · []]b b l, [using (i)], b, a, \, a, n, 1, 1, # n (n 1)2, (, , 2, 2, a n · b b l (n 1), a n · bb l, a, a, bn, bn, P 2 a 2n · b a l da 2n · n n a nb n (ab) n., a, 2, n, , Hence, P, (ab) ., , &
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Geometrical Progression, EXAMPLE 13, , 439, , The (m n)th and the (m n)th terms of a GP are p and q respectively., q (m/2n), Show that the mth and the nth terms of the GP are pq and p · d p n, , SOLUTION, , respectively., Let a be the first term and r be the common ratio of the given GP., Then,, Tm n p and Tm n q, &, , ar (m, , n 1), , p, , … (i), , and ar (m, , n 1), , q, , … (ii), , ( m n 1), , &, , p, ar, , , ar (m n 1) q, , &, , r {(m, , &, , r, , 2n, , &, , 1 d q n2n ·, r, p, , , , Tm ar (m, , n 1) (m n 1)}, , , , p, q, 1, , p, p 2n, & r d n, q, q, 1, , … (iii), , 1), , ar (m n 1) #b1 l, r, _n, 1b, ]Z], ] q 2nbb, p # ][]d n b`b, ]p b, \ 1a, 1, q b2n # nl, q 2, p #d n, p #d n pq ., p, p, n, , And, Tn ar (n, , [using (i) and (iii)], , 1), , ar (m n 1) #b1 l, r, _m, m, 1b, ]Z], ] q 2nbb, q 2n, [, `, d, n, d, n, ], b, p#], p#, ] p bb, p, \, a, m, q 2n, d, n, Hence, Tm pq and Tn p # p ·, m, , [using (i) and (iii)]., , EXAMPLE 14, , The pth, qth and rth terms of an AP as well as those of a GP are a, b, c,, respectively. Prove that, , , , a b c · b c a · c a b 1., , SOLUTION, , Let x be the first term and d be the common difference of the AP., Then,, a x (p 1) d, b x (q 1) d and c x (r 1) d., , … (i),, … (ii),, (b c) (q r) d, ( c a) ( r p) d, and (a b) (p q) d, … (iii).
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440, , Senior Secondary School Mathematics for Class 11, , Now, let A be the first term and R be the common ratio of the GP., Then,, , , … (iv),, … (v),, a AR (p 1), b AR (q 1), , ( r 1), … (vi)., and c AR, , , , , a (b c) · b (c a) · c (a b), {AR (p 1)} (q r) d #{AR (q 1)} (r p) d #{AR (r 1)} (p q) d, A {(q r) d (r p) d (p q) d} # R {(p 1)(q r) d (q 1)(r p) d (r 1)(p q) d}, A 0 # R 0 (1#1) 1, RS, VW, SSa (q r) d (r p) d (p q) d 0, WW, W, S, Sand p (q r) d q (r p) d r (p q) d (q r) d (r p) d (p q) d 0W ., T, X, , , , Hence, a (b c) · b (c a) · c (a b) 1., EXAMPLE 15, SOLUTION, , If x, 2y, 3z are in AP, where the distinct numbers x, y, z are in GP then, find the common ratio of the GP., It is given that the distinct numbers x, y, z are in GP., Let r be the common ratio of this GP. Then,, y xr and z xr 2., Now, x, 2y, 3z are in AP, &, &, &, &, , x, 2xr, 3xr 2 are in AP, 2xr x 3xr 2 2xr, 3xr 2 4xr x 0, 3r 2 4r 1 0, , &, , 1, (r 1) (3r 1) 0 & r , 3, , [a x ! 0, as it is a term of GP], , Hence, the common ratio of the GP is, EXAMPLE 16, SOLUTION, , [a x ! y ! z & r ! 1] ., 1·, 3, , In a GP of positive terms, if any term is equal to the sum of the next two, terms then show that the common ratio of the GP is 2 sin 18c., Let us consider a GP of positive terms with first term a and, common ratio r. Then, a 0 and r 0., It is given that Tn Tn 1 Tn 2 ., Now, Tn Tn 1 Tn 2, &, &, &, &, &, , , , ar n 1 ar n ar n 1, , 2, [dividing both sides by ar n 1], 1 rr, r2 r 1 0, 1 ! 1 4 1 ! 5, , r, 2, 2, 1 5, 51, o 2 sin 18c, 2e, r, [a r 0] ., 2, 4, , Hence, the common ratio of the GP is 2 sin 18c.
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Geometrical Progression, EXAMPLE 17, SOLUTION, , 441, , If p, q, r are in AP, prove that the pth, qth and rth terms of any GP are, in GP., Since p, q, r are in AP, we have (q p) (r q) ., … (i), Let us consider a GP with first term A and common ratio R., , , , Then, Tp AR (p 1), Tq AR (q 1) and Tr AR (r 1)., , , Tq, Tp, , , , (q 1), AR (p 1) R (q p) R (r q), AR, , [using (i)], , Tr AR (r 1), , R (r q )., Tq AR (q 1), Tr Tr, , Thus,, and therefore, Tp , Tq , Tr are in GP., Tp Tq, and, , Hence, the pth, qth and rth terms of any GP are in GP., EXAMPLE 18, , SOLUTION, , Let S be the sum, P be the product and R be the sum of the reciprocals of, three terms of a GP. Then, prove that P 2R 3 S 3., a, Let the three terms of the given GP be r , a and ar. Then,, 2, a, 1 r r n, a, 3, S a r a ark a d, P a r # a # ark a, r, 2, 2, r 1 1, 1, 1, 1 r r 1 n 1 d1 r r n ·, and R b a a ar l a br 1 r l a d, r, a, r, , , , 2 3, 6, P R a #, , (1 r r ), , 2 3, , 3 3, , a r, , 3, , 2, a 3 d 1 r r n S 3., r, , Hence, P R S ., 2, , 3, , 3, , EXAMPLE 19, , For a, b, c to be in GP, what is the value of, , SOLUTION, , For a, b, c to be in GP, we must have, bc, bc1, a b r or a b r ·, b c, Case I, When a r., b, In this case, b ar and c br., a b a ar a (1 r) , , b c b br b (1 r), b c 1, Case II When a r ·, b, In this case, a br and b cr., a b br b b (r 1) , , b c cr c c (r 1), Hence, the value of d, , a, b, ab, n is or c ·, b, bc, , ab, ?, bc, , a·, b, , b·, c
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442, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 20, , If a, b, c are in GP and a 1/x b 1/y c 1/z, prove that x, y, z are in AP., , SOLUTION, , Since a, b, c are in GP, we have, b 2 ac., 1/x, Let a b 1/y c 1/z k (say)., Then, a k x, b k y and c k z., Putting these values in (i), we get, , … (i), , , (k y) 2 (k x)#(k z) & k 2y k (x z)., , 2y x z., Hence, x, y, z are in AP., , EXAMPLE 21, SOLUTION, , EXAMPLE 22, , SOLUTION, , 1 1 1, If a, b, c are in AP; b, c, d are in GP and c , , e are in AP, then show that, d, a, c, e are in GP., a, b, c are in AP & 2b a c., … (i), b, c, d are in GP & c 2 bd., … (ii), 1, 1, 1, 2 1 1 (c e), c d e are in AP & d b c e l, ce, 2ce ·, , … (iii), & d, (c e), (a c) 2ce, ·, [from (i) and (iii)], Now, c 2 bd & c 2 , 2, (c e), (a c) e, & c, (c e), & c (c e) (a c) e, & c 2 ae, & a, c, e are in GP., Hence, a, c, e are in GP., If p, q, r are in GP and the equations, px 2 2qx r 0 and dx 2 2ex f 0, d e f, have a common root then show that p , q , r are in AP., Since p, q, r are in GP, we have, q 2 pr., , … (i), , On solving px 2qx r 0, we get, 2, , 2q ! 4q 2 4pr 2q q, , , p, 2p, 2p, q, 2, Thus, x p is a repeated root of px 2qr r 0., q, , x p is also a root of dx 2 2ex f 0, q 2, q, &, d · d p n 2e d p n f 0, x, , [using (i)].
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Geometrical Progression, , &, &, , dq 2 2eqp fp 2 0, d 2e fp , p q q2 0, , &, , d 2e f , p q r 0, , &, , d f 2e ·, p r, q, , 443, , … (ii), [on dividing (ii) by pq 2], [a q 2 pr], , d e f, Hence, p , q , r are in AP., EXAMPLE 23, SOLUTION, , Find all sequences which are simultaneously AP and GP., Let a1, a 2, a 3, …, an, … be a sequence which is both an AP as well as, a GP., Consider three consecutive terms an, an 1, an 2 of this AP and GP., Then, 2an 1 an an 2, … (i), [a an, an 1, an 2 are in AP]., And, if r be the common ratio of the GP, then, , , an a1 r n 1, an 1 a1 r n and an 2 a1 r n 1., Putting these values in (i), we get, , , 2a1 r n a1 r n 1 a1 r n 1 & r 2 2r 1 0 & (r 1) 2 0 & r 1., Putting r 1, we get an a1, an 1 a1 and an 2 a1 ., Thus, we get a constant sequence a1, a1, a1, a1, … ., Hence, a constant sequence is the only sequence which is both AP, and GP., , EXAMPLE 24, SOLUTION, , If the pth, qth, rth and sth terms of an AP be in GP then prove that, (p q), (q r), (r s) are in GP., Let a be the first term and d be the common difference of the given, AP. Then,, Tp a (p 1) d, Tq a (q 1) d,, Tr a (r 1) d, Ts a (s 1) d., It is given that Tp , Tq , Tr and Ts are in GP., Let the first term of this GP be A and its common ratio be R. Then,, … (i), Tp A & a (p 1) d A, Tq AR & a (q 1) d AR, Tr AR & a (r 1) d AR, 2, , … (ii), 2, , and Ts AR 3 & a (s 1) d AR 3., On subtracting (ii) from (i), we get, (p q) d A (1 R) ., On subtracting (iii) from (ii), we get, (q r) d AR (1 R) ., , … (iii), … (iv), … (v), … (vi)
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444, , Senior Secondary School Mathematics for Class 11, , On subtracting (iv) from (iii), we get, (r s) d AR 2 (1 R) ., , … (vii), Now, we have, (q r) 2d 2 A 2R 2 (1 R) 2 and (p q) d #(r s) d A 2R 2 (1 R) 2., , , (q r) 2d 2 (p q) d #(r s) d, , &, (q r) 2 (p q)#(r s), &, (p q), (q r) and (r s) are in GP., Hence, (p q), (q r), (r s) are in GP., nth TERM FROM THE END OF A GP, , Show that the nth term from the end of a GP with the first term a, the, l, common ratio r and the last term l is given by (n 1) ·, r, Let a be the first term, r be the common ratio and l be the last term of a, given GP. Then,, l, l, 2nd term from the end r (2 1) ;, r, l, l, 3rd term from the end 2 (3 1) ;, r, r, … … … … … … … …, … … … … … … … …, , THEOREM 2, , PROOF, , nth term from the end , , l, , r, , (n 1), , ·, , Hence, the nth term from the end , EXAMPLE 25, SOLUTION, , l, , r, , (n 1), , ·, , Find the 8th term from the end of the GP 3, 6, 12, 24, …, 12288., Here r 2 and l 12288., l, l 12288 12288 , , 8th term from the end (8 1) 7 , 96., 7, 128, r, 2, r, Hence, the 8th term from the end = 96., , EXERCISE 12A, 1. Find the 6th and nth terms of the GP 2, 6, 18, 54, … ., 2. Find the 17th and nth terms of the GP 2, 2 2 , 4, 8 2 , … ., 3. Find the 7th and nth terms of the GP 0.4, 0.8, 1.6, … ., 3 1 1 2, , ,, , ,… ., 4. Find the 10th and nth terms of the GP, 4 2 3 9, 5. Which term of the GP 3, 6, 12, 24, … is 3072?, 1 1 , 1, …, is –128?, 6. Which term of the GP ,, 4 2, 7. Which term of the GP 3 , 3, 3 3 , … is 729?
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Geometrical Progression, , 445, , 8. Find the geometric series whose 5th and 8th terms are 80 and 640, respectively., 1, 1, 9. Find the GP whose 4th and 7th terms are, and, respectively., 18, 486, 10. The 5th, 8th and 11th terms of a GP are a, b, c respectively. Show that b 2 ac., 11. The first term of a GP is –3 and the square of the second term is equal to its, 4th term. Find its 7th term., 1 ·, 12. Find the 6th term from the end of the GP 8, 4, 2, …,, 1024, 2 , 2, 2,, 13. Find the 4th term from the end of the GP, …, 162., 27 9 3, 14. If a, b, c are the pth, qth and rth terms of a GP, show that, (q r) log a (r p) log b (p q) log c 0., 15. The third term of a GP is 4. Find the product of its first five terms., 16. In a finite GP, prove that the product of the terms equidistant from the, beginning and end is the product of first and last terms., 17. If, , a bx b cx c dx, (x ! 0) then show that a, b, c, d are in GP., a bx b cx c dx, , 18. If a and b are the roots of x 2 3x p 0 and c and d are the roots of, x 2 12x q 0, where a, b, c, d form a GP, prove that (q p) : (q p) 17 : 15., ANSWERS (EXERCISE 12A), , 1. 486, 2 # 3 (n, , 1), , 2. 512, ( 2 ) (n, , 5. 11th, 6. 10th, 3 1 1 1, , ,, , ,…, 9., 2 2 6 18, , 1), , 3. 25.6,, , 2n, 5, , 7. 12th, 11. –2187, , 2 (n 1), 128 , 3, #b l, 3, 6561 4, 8. 5 10 20 40 80 …, 1, 12., 13. 6, 32, 4., , 15. 1024, HINTS TO SOME SELECTED QUESTIONS, 14. Let A be the first term and R be the common ratio of the given GP. Then,, a AR (p, b AR, , & log a log A (p 1) log R, , … (i), , (q 1), , & log b log A (q 1) log R, , … (ii), , 1), , & log c log A (r 1) log R., , … (iii), , c AR (r, , , 1), , (q r) log a (r p) log b (p q) log c, (q r) log [AR (p 1)] (r p) log {AR (q 1)] (p q) log [AR (r 1)], (log A) {(q r) (r p) (p q)} (log R) {(p 1) (q r) (q 1) (r p), (r 1) (p q)}, (log A)# 0 (log R)# 0 0.
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446, , Senior Secondary School Mathematics for Class 11, , 15. Let a be the first term and r be the common ratio of the GP. Then, the first five terms, are a, ar, ar 2, ar 3, ar 4., Also, T3 4 & ar 2 4., Product of first five terms (a # ar # ar 2 # ar 3 # ar 4 ) a 5r 10 (ar 2) 5 (4) 5 1024., 16. Let a be the first term, r be the common ratio and l be the last term of a GP containing, n terms. Then,, (pth term from the beginning) # (pth term from the end), (pth term from the beginning) #{(n p 1)th term from the beginning}, Tp #T(n p 1) ar p 1 # ar (n p 1) 1 ar (p 1) # ar (n p), a #(ar n 1) (T1 #Tn) (first term # last term) ., (a bx) (a bx) (b cx) (b cx), a bx b cx, b c, , &, & a ·, a bx b cx, (a bx) (a bx) (b cx) (b cx), b, b cx c dx, cd·, Similarly,, &, b cx c dx, b c, bcd, a, & a, b, c, d are in GP., b c, 18. Let r be the common ratio of the given GP., , 17., , Then, b ar, c ar 2 and d ar 3., Also, a b 3, ab p, c d 12 and cd q., Now, a b 3, c d 12 & a (1 r) 3 and ar 2 (1 r) 12, ar 2 (1 r), 4 & r2 4 & r 2, a (1 r), & a (1 2) 3 & a 1., &, , p ab a # ar a 2r (1 2 # 2) 2 and q cd ar 2 # ar 3 a 2r 5 (1) 2 # 2 5 32., q p 32 2 34 17, , , ·, Hence, , q p 32 2 30 15, , , PROBLEMS ON GP, For solving problems on GP, it is always convenient to take:, a, (i) 3 numbers in GP as r , a ar;, a a, (ii) 4 numbers in GP 3 , r , ar, ar 3;, r, a a, (iii) 5 numbers in GP as 2 , r , a, ar, ar 2;, r, (iv) the terms as a, ar, ar 2, … when their product is not given., SOLVED EXAMPLES, EXAMPLE 1, , 13, The sum of first three terms of a GP is, and their product is –1. Find, 12, these terms.
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Geometrical Progression, SOLUTION, , 447, , a, Let the first three terms of the given GP be r , a and ar., a, Then, r # a # ar 1 & a 3 1 & a 1., 1, 13, 13, a, And, r a ar , & r 1r , [a a 1], 12, 12, 1, 13 , & r r , 1, 12, 1 r 2 25, , &, r, 12, , , & 25r, 12 12r 2, & 12r 2 25r 12 0, 2, & 12r 16r 9r 12 0, & 4r (3r 4) 3 (3r 4) 0, 3, 4, ·, or r , & (3r 4) (4r 3) 0 & r , 3, 4, So, the required terms are, , i.e.,, EXAMPLE 2, SOLUTION, , 1, , 1,, (, ( 4/3), 3, 4, 1, or, 3, 4, , 4, 2 or, 3, 4 , 1,, 3, , 1, , 1, 3 2, (, 4, ( 3/4), 3, ·, 4, , Find three numbers in GP whose sum is 13 and the sum of whose squares, is 91., a, Let the required numbers be r , a and ar. Then,, a , r a ar 13, 2, a 2 2 2 , and, a a r, 91., 2, r, On squaring both sides of (i), we get, , … (i), … (ii), , a ar a ark 169, 2, , a2 2 2 2 da2 2 2 n , a a r o 2 r a a r 169, r2, a, &, 91 2a a r a ark 169, &, 91 26a 169, &, 26a 78 & a 3., Putting a 3 in (i), we get, &, , e, , [using (ii)], [using (i)], , 3 , 3 , 2, , , r 3 3r 13 & r 3r 10 & 3r 10r 3 0, & (r 3) (3r 1) 0 & r 3 or r , Hence, the required numbers are 1, 3, 9 or 9, 3, 1., , 1·, 3
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448, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 3, , Find three numbers in GP whose sum is 52 and the sum of whose products, in pairs is 624., , SOLUTION, , Let the required numbers be a, ar, ar 2. Then,, (a ar ar 2) 52 & a (1 r r 2) 52, , … (i), , and a · ar ar · ar a · ar 624 & a r (1 r r ) 624., On dividing (ii) by (i), we get, 12, ar 12 and therefore, a r ·, 12, Putting a r in (i), we get, 12 · 2 , 2 , r (1 r r ) 52 & 3 (1 r r ) 13r, & 3r 2 10r 3 0, & (3r 1) (r 3) 0, 2, , 2, , & r, , 2, , 2, , … (ii), , 1, or r 3., 3, , , a 36 or a 4., Hence, the required numbers are 36, 12, 4 or 4, 12, 36., EXAMPLE 4, , SOLUTION, , If the product of three numbers in GP is 216 and the sum of their products, in pairs is 156, find the numbers., a, Let the required numbers be r , a and ar. Then,, a, 3, , 216 6 3 & a 6., r # a # ar 216 & a, a, a, And, r # a a # ar r # ar 156, 1, &, a 2 b r r 1l 156 & (6 2) (1 r r 2) 156r, [a a 6], &, , 36 (r 2 r 1) 156r & 3 (r 2 r 1) 13r, , &, , 3r 2 10r 3 0 & (3r 1) (r 3) 0 & r , , 1, or r 3., 3, , So, the required numbers are 18, 6, 2 or 2, 6, 18., EXAMPLE 5, , Find four numbers in GP such that the third term is greater than the first, by 9 and the second term is greater than the fourth by 18., , SOLUTION, , Let the required numbers be a, ar, ar 2 and ar 3. Then,, T3 T1 9 & ar 2 a 9 & a (r 2 1) 9, , … (i), , and T2 T4 18 & ar ar 3 18 & ar (1 r 2) 18., On dividing (ii) by (i), we get r 2., Putting r 2 in (i), we get a (4 1) 9 & a 3., , … (ii), , Hence, the required numbers are 3, –6, 12 and –24.
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Geometrical Progression, , 449, , EXAMPLE 6, , The sum of three numbers in GP is 56. If we subtract 1, 7, 21 from these, numbers in that order, we get an AP. Find the numbers., , SOLUTION, , Let the required numbers be a, ar and ar 2. Then,, a ar ar 2 56., Also, (a 1), (ar 7) and (ar 2 21) are in AP., , … (i), , , 2 (ar 7) (a 1) (ar 2 21) & a ar 2 2ar 8., Using (ii) in (i), we get, , … (ii), , 16, ar (2ar 8) 56 & 3ar 48 & ar 16 & r a ·, 16, Putting r a in (i), we get, 256, a 16 a 56 & a 2 16a 256 56a, & a 2 40a 256 0, & a 2 8a 32a 256 0, & a(a 8) 32 (a 8) 0, & (a 8) (a 32) 0, & a 8 or a 32., 16, Now, a 8 & r , & r 2., 8, 16, 1, And, a 32 & r , & r ·, 32, 2, Hence, the required numbers are (8, 16, 32) or (32, 16, 8)., , EXERCISE 12B, , 2, , x, 7 in GP?, 7, 2, 2. For what values of x are the numbers (x 9), (x 6) and 4 in GP?, , 1. For what values of x are the numbers, , 39, 3. The sum of three numbers in GP is, and their product is 1. Find the, 10, numbers., 13, and their product is –1. Find, 4. The sum of first three terms of a GP is, 12, the GP., 5. Find three numbers in GP whose sum is 38 and whose product is 1728., 6. Find three numbers in GP whose sum is 65 and whose product is 3375., 7. The sum of three numbers in GP is 21 and the sum of their squares is 189., Find the numbers., 8. The product of three numbers in GP is 216. If 2, 8, 6 be added to them in that, order, we get an AP. Find the numbers., 9. The product of three numbers in GP is 1000. If 6 is added to the second, number and 7 is added to the third number, we get an AP. Find the, numbers.
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450, , Senior Secondary School Mathematics for Class 11, ANSWERS (EXERCISE 12B), , 5, 5, 2, 2, 1, or , 1,, 2, 2, 5, 5, , 1. x 1 or x 1, , 2. x 0 or x 16, , 3., , 3, 4, 3, 4, 1, or , 1,, 3, 3, 4, 4, 7. 12, 6, 3 or 3, 6, 12, , 5. 18, 12, 8 or 8, 12, 18, , 6. 45, 15, 5 or 5, 15, 45, , 8. 18, 6, 2 or 2, 6, 18, , 9. 5, 10, 20 or 20, 10, 5, , 4., , SUM OF n TERMS OF A GP, THEOREM, , PROOF, , Prove that the sum of n terms of a GP with the first term a and the common, ratio r is given by, ]Z]na, when r 1;, ]], ]] a (1 r n ), ], ,, Sn ][] (1 r) when r 1;, ]] n, ]] a (r 1), ]], ,, ] (r 1) when r 1., \, Let us consider a GP with the first term a and the common ratio r., Then,, , … (i), Sn a ar ar 2 … ar n 1., Case 1, , If r 1, we have, Sn a a … to n terms na., , Case 2, , If r ! 1, we have, rSn ar ar 2 … ar n, , 1, , ar n., , … (ii), , On subtracting (ii) from (i), we get, (1 r) Sn (a ar n ) a (1 r n ), n, n, a (1 r ), a ( r 1), ·, or Sn , (1 r), ( r 1), Z], n, ]] a (1 r ), , when r 1, ]], S ][ (1 r), n, ]]] a (r n 1), ]], ,, ] (r 1) when r 1, \, If a GP contains n terms with first term a, common ratio r and last, term l then, , &, , REMARK, , Sn , , , , l ar n 1., Case 1, , … (i), , When r 1, we have, Sn , , a (1 r ) (a ar ) (a lr), , , (1 r), (1 r), (1 r), n, , n, , [using (i)].
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Geometrical Progression, Case 2, , When r 1, we have, n, n, a (r 1) (ar a) (lr a), , , Sn , ( r 1), (r 1), ( r 1), , 451, , [using (i)]., , SUMMARY, , In a GP with first term a and common ratio r, we have, ]Z]na, when r 1;, ]], ]] a (1 r n ) (a lr), ], , when r 1;, , Sn ][] (1 r), ( 1 r), ]] n, ]] a (r 1) (lr a), ]], , when r 1., , ] (r 1), ( r 1), \, SOLVED EXAMPLES, EXAMPLE 1, , Find the sum of 8 terms of the GP 3, 6, 12, 24, … ., , SOLUTION, , Here a 3, r 2 1 and n 8., n, a (r 1), , we get, Using the formula, Sn , ( r 1), S8 , , EXAMPLE 2, SOLUTION, , EXAMPLE 3, SOLUTION, , 8, 3 # ( 2 1), 3 #(256 1) 3 # 255 765., (2 1), , Find the sum of the geometric series 1 , , 111, … to 12 terms., 2 4 8, , 1, 1 and n 12., 2, n, a (1 r ), , we get, Using the formula, Sn , (1 r), 1, 1n, 1# (1 b l 2 d1 12 n, 12, 2, (2 1) 4095, 2, , , , ·, S12 , 11, 1, 1, 2048, 2, b1 l, b l, 2, 2, Here a 1, r , , How many terms of the geometric series 1 4 16 64 … will make the, sum 5461?, Let the required number of terms be n., Here, a 1, r 4 1 and Sn 5461., n, n, a ( r 1), 1#(4 1), 5461, , &, Sn , (r 1), (4 1), n, &, (4 1) 16383, &, 4 n 16384 4 7, &, n 7., Hence, the required number of terms is 7.
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452, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 4, , Find the sum of the series 2 6 18 54 … 4374., , SOLUTION, , Clearly, the given series is a geometric series in which a 2, r 3 1, and l 4374., (lr a) (4374 # 3 2) 13120, , , 6560., , the required sum , 2, ( r 1), (3 1), Hence, the sum of the given series is 6560., , EAMPLE 5, SOLUTION, , In a GP, it is being given that T1 3, Tn 96 and Sn 189. Find the value, of n., Here, a 3, l 96 and Sn 189., Let the common ratio of the given GP be r., (lr a), (96r 3), 189, Then, Sn , &, (r 1), (r 1), & (96r 3) (189r 189), & 93r 186 & r 2., Now, l ar n, , 1, , , & 3 # 2 n 1 96, n1, 32 2 5 & n 1 5 & n 6., & 2, , Hence, n 6., EXAMPLE 6, SOLUTION, , Sum the series 5 55 555 … to n terms., We have, 5 55 555 … to n terms, 5 #{1 11 111 … to n terms}, 5 #{9 99 999 … to n terms}, 9, 5 #{(10 1) (10 2 1) (10 3 1) … to n terms}, 9, 5 #{(10 10 2 10 3 … to n terms) n}, 9, Z], _b, ]10 #(10 n 1), b, 5 # [], n`b 5 #(10 n 1 9n 10) ., ], b 81, , 9, (10 1), \, a, 5, , Hence, the required sum is #(10 n 1 9n 10) ., 81, , EXAMPLE 7, , Sum the series .4 .44 .444 … to n terms., , SOLUTION, , We have, .4 .44 .444 … to n terms, 4 #{.1 .11 .111 … to n terms}, 4 #{.9 .99 .999 … to n terms}, 9, 4 #{(1 .1) (1 .01) (1 .001) … to n terms}, 9
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Geometrical Progression, , 453, , 4 #{(1 1 … to n terms) (.1 .01 .001 … to n terms)}, 9, RS, n _, n V, ]Z], a (1 r ) bb, S .1#{1 (.1) }WWW, [a Sn , 4 #SSn , W, ], b`, ], 9 S, (1 .1) W, (1 r) b, \, a, TR, XV, SS, 1 · 1 WW, 1, W, SS, 10 ) (10) n 3WW, S, WW, 4 #SSn , 9 SS, 1, W, b1 l WW, SS, W, 10, X, T, n, (10 1), 4, 1, 1, 3, 4 # )n , , #, n (1 n 2G, 9, 9 = 9, 9 · 10 n, 10, 4 #=9n (1 1 n 2G 4 # (9n 1 1 n 2 ·, 81, 81, 10, 10, Hence, the required sum is, EXAMPLE 8, SOLUTION, , 4, 1, # 9n 1 n n ·, 81 d, 10, , The sum of some terms of a GP is 315. Its first term is 5 and the common, ratio is 2. Find the number of its terms and the last term., Let the given GP contain n terms. Then,, a 5, r 2 1 and Sn 315., , , Sn 315 &, &, , n, a (r 1), 315, ( r 1), , 5 #(2 n 1), 315, ( 2 1), , & (2 n 1) 63 & 2 n 64 2 6 & n 6., , , Last term ar (n 1) 5 # 2 (6 1) (5 # 2 5) (5 # 32) 160., Hence, the given GP contains 6 terms and its last term is 160., , EXAMPLE 9, SOLUTION, , The sum of first three terms of a GP is 16 and the sum of its next three, terms is 128. Find the sum of n terms of the GP., Let a be the first term and r be the common ratio of the given GP., Then, a ar ar 2 16 and ar 3 ar 4 ar 5 128, &, , a (1 r r 2) 16, , … (i), , and, , ar 3 (1 r r 2) 128, , On dividing (ii) by (i), we get r 8 2 & r 2., Putting r 2 in (i), we get, 3, , a (1 2 4) 16 & a , , 16, ·, 7, , Thus, in the given GP, we have a , , , 3, , 16, and r 2 1., 7, , 16, n, # (2 n 1), a ( r 1), 7, , 16 (2 n 1) ., Sn , 7, (r 1), (2 1), , … (ii).
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454, EXAMPLE 10, SOLUTION, , Senior Secondary School Mathematics for Class 11, , In a GP, the sum of first two terms is –4 and the 5th term is 4 times the, 3rd term. Find the GP., Let a be the first term and r be the common ratio of the given GP., Then,, T1 T2 4 and T5 4 #T3, a ar 4 and ar 4 4 # ar 2, a (1 r) 4 and r 2 4, … (i) and r !2., a (1 r) 4, 4, ·, Putting r 2 in (i), we get a , 3, Putting r 2 in (i), we get a 4., 4, 4 8 16, ,, ,, ,…, , a, and r 2 gives the GP, 3 3, 3, 3, and, a 4 and r 2 gives the GP 4, 8, 16, … ., 4 8 16, ,, , …1 or {4, 8, 16, …} ., Hence, the required GP is ' ,, 3 3, 3, , &, &, &, , EXAMPLE 11, , SOLUTION, , In an increasing GP, the sum of the first and last terms is 66, the product, of the second and the last but one is 128 and the sum of the terms is 126., How many terms are there in this GP?, Let the given GP contain n terms. Let a be the first term and r be the, common ratio of this GP., Since the given GP is increasing, we have r 1., Now, T1 Tn 66 & a ar (n, , 1), , 66., , And, T2 #Tn 1 128 & ar # ar, , ( n 2), , 2 (n 1), , & a r, , … (i), , 128, , 128 & ar (n 1) 128 ·, a, , … (ii), , Using (ii) in (i), we get, 128, a a 66 & a 2 66a 128 0, & a 2 2a 64a 128 0, & a (a 2) 64 (a 2) 0, & (a 2) (a 64) 0, & a 2 or a 64., Putting a 2 in (ii), we get, 128, 128 32., 2, 4, a, Putting a 64 in (ii), we get, r (n, , r, , 1), , (n 1), , … (iii), , 128, 128 1 , which is rejected, since r 1., 2, 64 # 64 32, a, , Thus, a 2 and r (n, , 1), , 32.
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Geometrical Progression, , Now, Sn 126 &, , 455, , n, a ( r 1), 126, ( r 1), , n, n, r 1, r 1, & 2 d n 126 & 63, r 1, r 1, (n 1), , 32r 1 , #r 1 , 63, 63 &, r1, r1, & 32r 1 63r 63 & 31r 62 & r 2., (n 1), 32 2 5 & n 1 5 & n 6., , r, Hence, there are 6 terms in the given GP., &, , REMARK, , r, , Consider two GPs a, ar, ar 2, ar 3, … and A, AR, AR 2, AR 3, … ., Taking the products of their corresponding elements, we get, aA, aA (rR), aA (rR) 2, aA (rR) 3, …, which is clearly a GP with first term aA and common ratio rR., Based upon this result, we take the following example., , EXAMPLE 12, , Find the sum of the products of the corresponding terms of finite, geometrical progressions, 1, 2, 4, 8, 16, 32 and 128, 32, 8, 2, ·, 2, , SOLUTION, , Taking the products of the corresponding terms of two given GPs,, we get a new progression, 1, (2 #128), (4 # 32), (8 # 8), (16 # 2), b32 # l,, 2, i.e., 256, 128, 64, 32, 16, which is clearly a GP in which a 256 and, 16 1, r, 1., 32 2, 1 5, (1 b l 2, #, 256, 5, 2, a (1 r ), , , the required sum , 1, (1 r), b1 l, 2, 256 #e1 , , , 1, b l, 2, , 1, o, 25, , 256 # 2 #b1 1 l, 32, , b256 # 2 # 31 l 496., 32, EXAMPLE 13, , Find the sum of n terms of the sequence given by an (3 n 5n), nd N,, , SOLUTION, , Let the sum of n terms of the given sequence be Sn . Then,, S n a1 a 2 a 3 … a n, (3 1 5 #1) (3 2 5 # 2) (3 3 5 # 3) … (3 n 5 # n), (3 3 2 3 3 … 3 n ) (5 #1 5 # 2 5 # 3 … 5 # n)
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456, , Senior Secondary School Mathematics for Class 11, , (3 3 2 3 3 … 3 n ) 5 #(1 2 3 … n), 3 ( 3 n 1), 5 # n (1 n), 2, ( 3 1), 3, 5, n, (3 1) # n (n 1) ., 2, 2, 3, 5, Hence, the required sum is (3 n 1) n (n 1) ., 2, 2, , , EXAMPLE 14, , If S1, S 2 and S 3 be respectively the sum of n, 2n and 3n terms of a GP then, prove that S1 (S 3 S 2) (S 2 S1) 2., , SOLUTION, , Let a be the first term and r be the common ratio of the given GP., Then,, Z, _, a (1 r n ) ]] a (1 r 3n ) a (1 r 2n ) bb, [, `b, , , , # ]], S1 (S 3 S 2), (1 r), (1 r), (1 r) b, \, a, a (1 r n ) (a ar 3n a ar 2n ), , #, ( 1 r), (1 r), , , , a (1 r n ) ar 2n (1 r n ), #, (1 r), ( 1 r), a r (1 r ), 2 2n, , n 2, , ( 1 r), , ·, , 2, , Z], _2, ] a (1 r 2n ) a (1 r n ) bb, [, `b, , And, (S 2 S1) ]], (1 r), (1 r) b, \, a, (a ar 2n a ar n ) 2, , 2, (1 r), 2, , , , , {ar n (1 r n )} 2, ( 1 r), , 2, , a r (1 r ), 2 2n, , ( 1 r), , n 2, , 2, , ·, , Hence, S1 (S 3 S 2) (S 2 S1) 2., EXAMPLE 15, , If S be the sum, P be the product and R be the sum of the reciprocals of n, S n, terms in a GP, prove that P 2 b l ·, R, , SOLUTION, , Let the given GP be a, ar, ar 2, …, ar (n, S, , a (1 r ), (1 r), , 1), , . Then,, , n, , and P [a # ar # ar 2 #…# ar, , … (i), (n 1), , ]
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Geometrical Progression, 2 3 … (n 1)], , &, , P a n r [1, , &, , P a nr 2 (n 1) n, , P 2 a 2nr (n 1) n., , 1, , &, , 457, , 1, , a nr 2 (n 1) n, , , , … (ii), , 1, ( n 12, r, 1, 1, 1, 1, And, R ( a ar … n 1 2 b a l ·, ar, '1 11, r, (1 r ), r, R aak ·, n, (1 r) · r, n, , &, , R, , (1 r ), n, , ·, (n 1), a (1 r) · r, On dividing (i) by (iii), we get, , &, , … (iii), , n, (n 1), S a (1 r ) a (1 r) · r, a 2r (n 1), ·, n, R, (1 r), (1 r ), , S n, , , b l {a 2r (n 1)} n a 2n · r (n 1) n P 2, R, S n, Hence, P 2 b l ·, R, , , [using (ii)]., , EXERCISE 12C, 1. Find the sum of the GP:, (i) 1 3 9 27 … to 7 terms, (ii) 1 3 3 3 3 … to 10 terms, (iii) 0.15 0.015 0.0015 … to 6 terms, 1 1 1, (iv) 1 … to 9 terms, 2 4 8, 1 1 , (v) 2 , … to 8 terms, 2 2 2, 2 1 1 3, (vi) … to 6 terms, 9 3 2 4, 2. Find the sum of the GP:, (i), , 7 21 3 7 … to n terms, , 1 1 1 , … to n terms, 3 32 33, (iii) 1 a a 2 a 3 … to n terms (a ! 1), (iv) x 3 x 5 x 7 … to n terms, (ii) 1 , , (v) x (x y) x 2 (x 2 y 2) x 3 (x 3 y 3) … to n terms
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458, , Senior Secondary School Mathematics for Class 11, , 3. Find the sum to n terms of the sequence:, 1 2, 1 2, 1 2, (i) bx x l , dx 2 2 n , dx 3 3 n , … to n terms, x, x, (ii) (x y), (x 2 xy y 2), (x 3 x 2y xy 2 y 3), … to n terms, 4. Find the sum:, 3 4 3 4 , … to 2n terms., 5 52 53 54, 5. Evaluate:, 10, , (i) / (2 3 n ), n1, , n, , (ii) / [2 k 3 (k, k1, , 1), , ], , 8, , (iii) / 5 n, n1, , 6. Find the sum of the series:, (i) 8 88 888 … to n terms, (ii) 3 33 333 … to n terms, (iii) 0.7 0.77 0.777 … to n terms, 7. The sum of n terms of a progression is (2 n 1) . Show that it is a GP and find, its common ratio., 8. In a GP, the ratio of the sum of first three terms is to that of first six terms is, 125 : 152. Find the common ratio., 9. Find the sum of the geometric series 3 6 12 … 1536., 10. How many terms of the series 2 6 18 … must be taken to make the sum, equal to 728?, 11. The common ratio of a finite GP is 3 and its last term is 486. If the sum of, these terms is 728, find the first term., 1 ·, 12. The first term of a GP is 27 and its 8th term is, Find the sum of its first, 81, 10 terms., 1, 1, respectively. Find the sum of, 13. The 2nd and 5th terms of a GP are, and, 2, 16, the GP up to 8 terms., 1, 1, respectively. Find the sum of, 14. The 4th and 7th terms of a GP are, and, 27, 729, n terms of the GP., 15. A GP consists of an even number of terms. If the sum of all the terms is, 5 times the sum of the terms occupying the odd places, find the common, ratio of the GP., 16. Show that the ratio of the sum of first n terms of a GP to the sum of the terms, 1, from (n 1)th to (2n)th term is n ·, r, ANSWERS (EXERCISE 12C), , 1. (i) 1093, 255 2, (v), 128, , (ii) 121 ( 3 1), 133, (vi), 144, , 333333, (iii), 2000000, , (iv), , 171, 256
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Geometrical Progression, , 2. (i), , 7, ( 3 1) (3 n/2 1), 2, 3, , (iv), 3., 4., 6., 7., 12., , x # (x, , 2n, , 1), , (ii), (v), , (x 2 1), , {3 n (1) n}, , (iii), , , , 4 # 3 (n 1), x 2 (x 2n 1), , 459, , , , {1 ( a ) n }, (1 a ), , xy (x ny n 1), (xy 1), , ( x 2 1), ]Z, J n Nb_, n, JKx 2n 1NO, 1 · ]][ 2 d x 1 n 2 KKKy 1OOObb`, OOdx 2 1 n 2n (ii), (i) KK 2, y, x, ], K y 1 Obbb, n, Kx 1O, x1, (x y) ]], x2, L, Pa, L, P, \, 2, n, N, J, 19 KK5 1OO, 1, , ·, 5. (i) 29544, (ii) [2 (n 2) 3 n 5], (iii) 488280, 2, 24 KK 5 2n OO, P, L, 8, 1, 7, 1, , , (ii), (i), [10 (n 1) 10 9n], [10 (n 1) 10 9n] (iii), 9n 1 n n, 27, 81, 81 d, 10, 3, 8. r , 9. 3069, 10. 6, 11. a 2, r2, 5, 81 1, 3, 85, 1, 13., 14. d1 n n, 15. r 4, 1, 2 d 3 10 n, 2, 128, 3, HINTS TO SOME SELECTED QUESTIONS, , 1. (i) S7 , , 1#(3 7 1) (2187 1) 2186, , , 1093., 2, 2, (3 1), , (ii) S10 , , 10, 1#{( 3 ) 1}, , ( 3 1), , , , 5, (3 1), , ( 3 1), , #, , ( 3 1), ( 3 1), , , , 242 ( 3 1), 121 ( 3 1) ., (3 1), , 15 3, 0.015 1, and r , ·, 100 20, 0.15 10, 3, 1, # 1, 20 e 10 6 o, (1000000 1) 1, c 3 # 10 m ·, # 999999 333333 ·, , S6 , 20, 9, 6 1000000 2000000, 1000000, 1, c1 m, 10, , (iii) a , , 1, and n 9., 2, 1 9, 1#<1 c m F, 2, 2 #e1 19 o c 2 # 513 m 171 ·, , S9 , 3, 3 512, 256, 1, 2, c1 m, 2, 1, (v) a 2 , r and n 8., 2, 1, 2 e1 8 o, 2, 255 2, c 2 # 255 # 2m , ·, , S8 , 256, 128, 1, c1 m, 2, , (iv) a 1, r , , 3, 1 9, 2, (vi) a , r c # m , and n 6., 9, 3, 2, 2
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460, , Senior Secondary School Mathematics for Class 11, , , , 2 ( c 3 m62 2, 729, #1, 1, # '1 , 9, 2, 9, 64, , , , c 2 # 2 m#c 665 m 133 ·, S6 , 3, 5, 9, 5, 64, 144, c1 m, c m, 2, 2, , 2. (i) a 7 and r , , , Sn , , 7, , 3., , n, 7 #{( 3 ) 1}, , Sn , , ( 3 1), , (ii) a 1 and r , , , , 21, , #, , ( 3 1), ( 3 1), , , , 7, · ( 3 1) (3 n/2 1) ., 2, , 1, ·, 3, 1 n, Z], _, m2, ] (1) n bb {3 n (1) n}, 3, 3 # ][1 n b` , ·, , 1, 4 ], 3 b, 4 # 3 (n 1), c1 m, \, a, 3, , 1# (1 c, , (iii) A 1 and R a., n, n, A#{1 R n} 1#{1 ( a) } {1 ( a) }, , , ·, , Sn , (1 R), (1 a), (1 a), (iv) a x 3 and r , , , Sn , , x5 2, x ., x3, , x 3 #{(x 2) n 1}, (x 1), 2, , , , x 3 #(x 2n 1), (x 2 1), , ·, , (v) Given expression, {x 2 x 4 x 6 … to n terms} + {xy x 2y 2 x 3y 3 … to n terms}, x 2 (1 x 2 x 4 … to n terms} + xy (1 xy x 2y 2 … to n terms}, , , x 2 #1#{(x 2) n 1}, (x 2 1), , , , xy #1#{(xy) n 1} x 2 (x 2n 1) xy (x ny n 1), , , ·, (xy 1), (xy 1), (x 2 1), , 3. (i) Given sum, 2, 2, 2, 2, cx 1 m ex 2 12 o ex 3 13 o … dx n 1n n, x, x, x, x, , ex 2 12 2o ex 4 14 2o ex 6 16 2o … ex 2n 12n 2o, x, x, x, x, (x 2 x 4 x 6 … x 2n) e 12 14 16 … 12n o (2 2 … n times), x, x, x, x, Z], n_, b, 1 ·] 1 b, [1 e 2 o b`, 2 ], 2, 2 n, x b, x 2 (x 2n 1), (x 2n 1), x [(x ) 1] x ], , , 2n 2, 2n, \, a 2n , 2, 2, 1, (x 1), x (x 1), (x 1), e1 2 o, x, JKx 2n 1NO, JKx 2n 1NO, JK 2n 1NO, O 1 K, O, O 2 1 2n., Kx, x 2 KK 2, o, K x 1 OO x 2n KK x 2 1 OO 2n KK x 2 1 OOex, x 2n, L, P, L, P, L, P
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Geometrical Progression, , 461, , (ii) Multiplying and dividing each term by (x y), we get the given sum as, , 1 · 2 2 3 3 4 4 , {(x y ) (x y ) (x y ) … to n terms}, ( x y), , , 1 · 2 3 4, 2, 3, 4, {(x x x … to n terms) (y y y … to n terms)}, ( x y), , Z], _, n, KJKy n 1ONObbb, ]], , OOb` ·, · ][]x 2 d x 1 n y 2 KK, x1, y 1 bb, ], L, Pa, \, 3, 3, 3, 4, 4, 4. Given sum ) 3 5 … to n terms3 ) 2 4 … to n terms3, 5 5, 5, 5, 5, Z, _, Z, _, 3 ]] 1 nbb 4 ]] 1 nbb, # [1, # [1, `, `, 5 ]] e 5 2 o bb 5 2 ]] e 5 2 o bb, , \, a, \, a, 1, 1, e1 2 o, e1 2 o, 5, 5, , , 1, (x y), , 2n, 2n, (5 1), 25 (5 1), 4, c 3 # 25 m, , c, m, #, 2n, 5 24, 25 24, 5 2n, 5, 2n, 2n, J, N, J, N, K, O, K, O, , , c 5 1 mKK5 2n 1OO 19 · KK5 2n 1OO ·, O, 8 6 K 5, O 24 K 5, L, P, L, P, 10, , 5. (i) / (2 3 n) (2 3 1) (2 3 2) (2 3 3) … (2 3 10 ), n1, , (2 2 2 … taken 10 times) (3 3 2 3 3 … 3 10 ), ]Z], b_, 10, 1 b, o`b, (2 #10) []3 #e 3, 31 b, ], \, a, 1, 59049, 20 c, m (20 29524) 29544., 2, n, , (ii) / [2 k 3 (k, k1, , 1), , ] (2 1 3 0) (2 2 3 1) (2 3 3 2) … up to n terms, (2 1 2 2 2 3 … to n terms) (1 3 3 2 … to n terms), n, n, , , 2d 2 1 n 1 · d 3 1 n, , 2 1, 31, 2 (2 n 1) 1 (3 n 1) 2 (n 1) 2 1 · 3 n 1, 2, 2, 2, 2 (n 1) 1 · 3 n 5 2 (n 1) 1 (3 n 5) 1 [2 (n 2) 3 n 5] ., 2, 2, 2, 2, , 8, , (iii) / 5 n [5 5 2 5 3 … to 8 terms], n1, , 8, 5 (390625 1) 5 # 390624, , , 5e 5 1 o , 51, 4, 4, (5 # 97656) 488280., , 6. (i) Given sum 8 #[1 11 111 … to n terms], 8 #[9 99 999 … to n terms], 9, 8 #[(10 1) (10 2 1) (10 3 1) … to n terms], 9
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462, , Senior Secondary School Mathematics for Class 11, 8 #{(10 10 2 10 3 … to n terms) n}, 9, n, , 8 # )10 #d10 1 n n3 8 [10 (n 1) 10 9n] ., 9, 10 1, 81, , 7 77 777 , … to n terms, 10 100 1000, 7 # ' 1 11 111 … to n terms1, 10 100 1000, 7, # ' 9 99 999 … to n terms1, 9, 10 100 1000, Z], _b, 7, ], # [c1 1 m e1 1 2 o e1 1 3 o … to n termsb`, ], bb, 9 ], 10, 10, 10, \Z, a, ], b_, 7 # ][n e 1 1 2 1 3 … to n termsob`, bb, 9 ]], 10 10, 10, \Z, a, _, ]], 1 1 bb, 1, b, ]], d, n, n, 10, 10 bb 7, ], 1, 7 # ][n , b`b 81 #d9n 1 n n ·, 9 ]], 1, 10, b, , c, m, 1, b, ]], 10 b, a, \, n, (n 1), 1] ., 7. Sn (2 1) and Sn 1 [2, (iii) Given series , , , Tn (Sn Sn 1) [2 n 2 (n 1)] & T1 1, T2 2, T3 4, T4 8, … ., Clearly, 1, 2, 4, 8, … is a GP with common ratio = 2., 15. Let us consider a GP with first term a and common ratio r and having 2n terms. Then,, T1 T2 T3 T4 … T2n 5 (T1 T3 T5 … T2n 1), , , a ar ar 2 … ar (2n 1) 5 [a ar 2 ar 4 … ar (2n 2)], R, V, 2, n, 2n, SS1 (r ) WWW, 1r, & a e o 5a SSS, W & 1 r 5 & r 4., 1 r, S 1 r 2 WW, T, X, 16. Required ratio Sn : (S2n Sn) ., , &, , Now, Sn , , , a (1 r n), a (1 r 2n) a (1 r n) (1 r n), , and S2n , (1 r), (1 r), (1 r), , (S2n Sn) , , a (1 r n ) (1 r n ) a (1 r n ) a (1 r n) (1 r n 1) a (1 r n ) r n, , , , ·, (1 r), (1 r), (1 r ), (1 r), , Hence, the required ratio is, , 1 ·, rn, , WORD PROBLEMS ON GP, EXAMPLE 1, , SOLUTION, , A man has 2 parents, 4 grandparents, 8 great-grandparents, and so on., Find the number of his ancestors during the ten generations preceding, his own., Required number of ancestors, 2 4 6 8 … up to 10 terms, (2 10 1), 2#, 2 (2 10 1), [GP with a 2, r 2 and n 10], ( 2 1)
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Geometrical Progression, , 463, , 2 (1024 1) (2 #1023) 2046., Hence, the required number of the man’s ancestors is 2046., EXAMPLE 2, , SOLUTION, , EXAMPLE 3, SOLUTION, , A man writes a letter to four of his friends. He asks each one of them to, copy the letter and mail to four different persons with the instruction that, they move the chain similarly. Assuming that the chain is not broken and, it costs ` 4 to mail one letter, find the amount spent on postage when 6th, set of letters is mailed., Successive number of letters are 4, 16, 64, … ., This is a GP with a 4 and r 4., , Number of letters in the 6th set ar (6 1) (4 # 4 5) 4 6 4096., Cost of postage on these leters ` (4096 # 4) ` 16384., What will ` 10000 amount to in 4 years after its deposit in a bank which, pays annual interest at the rate of 10% per annum compounded annually?, Original sum = ` 10000., 10, Amount after 1 year ` 10000 ` b10000 #, #1l ` 11000,, 100, 10, amount after 2 years ` 11000 ` b11000 #, #1l ` 12100,, 100, 10, amount after 3 years ` 12100 ` b12100 #, #1l ` 13310,, 100, and so on., Thus, these amounts form a GP, 10000, 11000, 12100, 13310, …, 11000 12100 13310 11, ·, 10000 11000 12100 10, 11 ·, In this GP, we have a 10000, r , 10, , Amount after 4 years T5 ar (5 1) ar 4, such that, , ` <10000 #b11 l F ` 14641., 10, 4, , Hence, the amount after 4 years is ` 14641., Alternative method, , Here, P ` 10000, R 10% p.a. and T 4 years., , , amount after 4 years ` (P #b1 , , R T, l2, 100, , 4, ` (10000 #b1 10 l 2, 100, , ` (10000 #b11 l 2 ` 14641., 10, 4, , Hence, the amount after 4 years is ` 14641.
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464, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 4, , A manufacturer reckons that the value of a machine which costs him, ` 125000 will depreciate each year by 20%. Find the estimated value of, the machine at the end of 5 years., , SOLUTION, , Short-cut method, , Value of the machine after 5 years ` (125000 #b1 , , 20 5, l2, 100, , ` (125000 #b 4 l 2, 5, #, 125000, 1024, `b, l ` 40960., 3125, 5, , Hence, the value of the machine after 5 years is ` 40960., Alternative method, , Initial value of the machine = ` 125000., Value of the machine after 1 year = 80% of ` 125000, ` b125000 # 80 l ` 100000., 100, Value of the machine after 2 years = 80% of ` 100000, ` b100000 # 80 l ` 80000., 100, Thus, the yearwise values of the machine are, ` 125000, ` 100000, ` 80000, … ., 100000 4 , 80000 4 ,, Here,, etc., 125000 5 100000 5, 4, Thus, these amounts form a GP with a 125000 and r ·, 5, , Value after 5 years = 6th term T6 ar (6 1) ar 5, ` (125000 #b 4 l 2, 5, 5, , ` b125000 #1024 l ` 40960., 3125, EXAMPLE 5, , SOLUTION, , The number of bacteria in a certain culture doubles every hour. If there, were 30 bacteria present in the culture originally, how many bacteria, would be present at the end of 2nd hour, 4th hour and nth hour?, The bacteria present in the culture originally, at the end of 1st hour,, at the end of 2nd hour, at the end of 3rd hour and so on, are, 30, 60, 120, 240, … ., 60 , This is a GP with a 30 and r , 2., 30, Number of bacteria at the end of 2nd hour, T3 ar 2 (30 # 2 2) 120., Number of bacteria at the end of 4th hour, T5 ar 4 (30 # 2 4) 480.
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Geometrical Progression, , 465, , Number of bacteria at the end of nth hour, Tn 1 ar (n 1 1) 30 # 2 n., EXAMPLE 6, , SOLUTION, , The inventor of the chessboard suggested a reward of one grain of wheat for, the first square; 2 grains for the second; 4 grains for the third; and so on,, doubling the number of grains for subsequent squares. How many grains, would have to be given to the inventor? (Note that there are 64 squares in, the chessboard.), Required number of grains, 1 2 1 2 2 2 3 … to 64 terms, 1 (2 1 2 2 2 3 … 2 63 ), 1, , EXAMPLE 7, , SOLUTION, , 2 (2 63 1), (1 2 64 2) (2 64 1) ., ( 2 1), , The lengths of three unequal edges of a rectangular solid block are in GP., The volume of the block is 216 cm 3 and the total surface area is 252 cm 2., Find the length of its longest edge., a, Let the lengths of its edges be r cm, a cm and ar cm., a, Then, its volume a r # a # ark cm 3 a 3 cm 3., , a 3 216 6 3 & a 6., 6, So, the edges are 6r cm, 6 cm and r cm., , surface area 2 [lb bh lh], 2 ;6r # 6 6 # 6 6r # 6E cm 2, r, r, 36, 2, 2 ;36r , E, 72 #;r 1 1E cm 2., r 36 cm, r, , , 1, 72 #br r 1l 252, , 2 (r 2 1 r) 7r & 2r 2 5r 2 0, 1, &, (r 2) (2r 1) 0 & r 2 or r ·, 2, Each value of r gives the longest edge = 12 cm., &, , EXERCISE 12D, 1. What will ` 15625 amount to in 3 years after its deposit in a bank which, pays annual interest at the rate of 8% per annum, compounded annually?, 2. The value of a machine costing ` 80000 depreciates at the rate of 15% per, annum. What will be the worth of this machine after 3 years?, 3. Three years before the population of a village was 10000. If at the end of, each year, 20% of the people migrated to a nearby town, what is its present, population?
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466, , Senior Secondary School Mathematics for Class 11, , 4. What will ` 5000 amount to in 10 years, compounded annually at 10% per, annum? [Given (1.1) 10 2.594], 5. A manufacturer reckons that the value of a machine which costs him, ` 156250, will depreciate each year by 20%. Find the estimated value at the, end of 5 years., 6. The number of bacteria in a certain culture doubles every hour. If there, were 50 bacteria present in the culture originally, how many bacteria, would be present at the end of (i) 2nd hour, (ii) 5th hour and (iii) nth hour?, Hint, , At the end of 2nd hour means 3rd term ar 2 (50 # 2 2) ., ANSWERS (EXERCISE 12D), , 1. ` 19683, , 2. ` 49130, , 5. ` 51200, , 6. (i) 200, , 3. 5120, (ii) 1600, , (iii) 50 # 2, , 4. ` 12970, n, , PROPERTIES OF GP, 2, , 3, , Let a, ar, ar , ar , … be a GP. Then,, (i) 1 , 1 , 1 2 , 1 3 , … is a GP., a ar ar ar, (ii) ka, k (ar), k (ar 2), k (ar 3), … is a GP, where k ! 0., k k k, k, (iii) a , ar , 2 , 3 , … is a GP, where k ! 0., ar ar, (iv) a k, (ar) k, (ar 2) k, (ar 3) k, … is a GP, where k ! 0., SOLVED EXAMPLES, EXAMPLE 1, SOLUTION, , EXAMPLE 2, SOLUTION, , If a, b, c are in GP, prove that log a, log b, log c are in AP., Let a, b, c be in GP. Then,, b 2 ac & log b 2 log (ac), & 2 log b log a log c, & log a, log b, log c are in AP., If a, b, c, d are in GP then prove that a b, b c, c d are also in GP., Let a, b, c, d be in GP with common ratio r. Then,, b ar, c ar 2 and d ar 3., , (a b) (a ar) a (1 r),, (b c) (ar ar 2) ar (1 r),, (c d) (ar 2 ar 3) ar 2 (1 r) ., , , (b c) 2 a 2r 2 (1 r) 2 and (a b) (c d) a 2r 2 (1 r) 2.
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Geometrical Progression, , 467, , Consequently, (b c) 2 (a b) (c d) ., Hence, (a b), (b c) and (c d) are in GP., EXAMPLE 3, SOLUTION, , If a, b, c, d are in GP then prove that a n b n, b n c n, c n d n are also, in GP., Let a, b, c, d be in GP with common ratio r. Then,, b ar, c ar 2 and d ar 3., , (a n b n) (a n a nr n) a n (1 r n),, (b n c n) (a nr n a nr 2n) a nr n (1 r n), and (c n d n) (a nr 2n a nr 3n) a nr 2n (1 r n) ., , , (b n c n) 2 a 2nr 2n (1 r n) 2 and (a n b n)(c n d n) a 2nr 2n (1 r n) 2., , Consequently, (b n c n) 2 (a n b n) (c n d n) ., Hence, (a n b n), (b n c n) and (c n d n) are in GP., EXAMPLE 4, SOLUTION, , If (a 2 b 2), (ab bc), (b 2 c 2) are in GP then prove that a, b, c are also, in GP., Let (a 2 b 2), (ab bc) and (b 2 c 2) be in GP. Then,, (ab bc) 2 (a 2 b 2) (b 2 c 2), &, a 2b 2 b 2c 2 2ab 2c a 2b 2 a 2c 2 b 4 b 2c 2, &, b 4 a 2c 2 2ab 2c 0, &, (b 2 ac) 2 0 & b 2 ac 0 & b 2 ac., Hence, a, b, c are in GP., , EXAMPLE 5, SOLUTION, , If mth, nth and pth terms of a GP form three consecutive terms of a GP, then prove that m, n, p are three consecutive terms of a AP., Let a be the first term and r be the common ratio of the given GP., Then,, , , , Tm ar (m 1), Tn ar (n 1) and Tp ar (p 1)., Since Tm, Tn, Tp are in GP, we have, T n2 Tm #Tp, &, , [ar (n, , 1) 2, , , , ] [ar (m 1) # ar (p 1)], , , , &, a 2r (2n 2) a 2r (m p 2), (2n 2), (m p 2), r, &, r, &, 2n 2 m p 2, &, 2n m p & m, n, p are in AP., Hence, m, n, p are three consecutive terms of an AP., , EXAMPLE 6, SOLUTION, , If the 4th, 10th and 16th terms of a GP are x, y, z respectively, prove that, x, y, z are in GP., Let a be the first term and r be the common ratio of the given GP., Then, T4 x, T10 y and T16 z.
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468, , Senior Secondary School Mathematics for Class 11, , &, , ar 3 x, ar 9 y and ar 15 z., , , , y 2 (ar 9) 2 a 2r 18 and xz (ar 3) (ar 15) a 2r 18., , Consequently, we have y 2 xz., Hence, x, y, z are in GP., EXAMPLE 7, SOLUTION, , Three numbers are in AP and their sum is 15. If 1, 3, 9 be added to them, respectively, they form a GP. Find the numbers., Let the required numbers be (a d), a and (a d) . Then,, (a d) a (a d) 15 & 3a 15 & a 5., So, the numbers are (5 d), 5 and (5 d) ., Adding 1, 3, 9 respectively to these numbers, we get the numbers, (6 d), 8 and (14 d) ., These numbers are in GP., , 8 2 (6 d) (14 d) & 84 8d d 2 64, & d 2 8d 20 0, & d 2 10d 2d 20 0, & d (d 10) 2 (d 10) 0, & (d 10) (d 2) 0, & d 10 or d 2., So, the required numbers are (3, 5, 7) or (15, 5, –5)., , EXERCISE 12E, 1. If p, q, r are in AP then prove that pth, qth and rth terms of any GP are in GP., 2. If a, b, c are in GP then show that log a n, log b n, log c n are in AP., 3. If a, b, c are in GP then show that, , 1 , 1 , 1, are in AP., log a m log b m log c m, , 4. Find the values of k for which k 12, k 6 and 3 are in GP., 5. Three numbers are in AP and their sum is 15. If 1, 4, 19 be added to them, respectively then they are in GP. Find the numbers., 6. Three numbers are in AP and their sum is 21. If the second number is, reduced by 1 and the third is increased by 1, we obtain three numbers in, GP. Find the numbers., 7. The sum of three numbers in GP is 56. If 1, 7, 21 be subtracted from them, respectively, we obtain the numbers in AP. Find the numbers., 8. If a, b, c are in GP, prove that, , a 2 ab b 2 b a ·, ab bc ca c b, , 9. If (a b), (b c), (c a) are in GP then prove that (a b c) 2 3 (ab bc ca) ., 10. If a, b, c are in GP, prove that, (i) a (b 2 c 2) c (a 2 b 2), 1, 1, 1 , (ii) 2, (a b 2) b 2 (b 2 c 2)
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Geometrical Progression, , 469, , (iii) (a 2b 2c) (a 2b 2c) a 2 4c 2, 1, 1 1, (iv) a 2b 2c 2 d 3 3 3 n a 3 b 3 c 3., a, b, c, 11. If a, b, c, d are in GP, prove that, (i) (b c) (b d) (c a) (c d), ab cd a c, (ii) 2 2 , b, b c, (iii) (a b c d) 2 (a b) 2 2 (b c) 2 (c d) 2., 12. If a, b, c are in GP, prove that, , 1 , 1 , 1, are in AP., (a b) 2b b c, , 13. If a, b, c are in GP, prove that a 2, b 2, c 2 are in GP., 14. If a, b, c are in GP, prove that a 3, b 3, c 3 are in GP., 15. If a, b, c are in GP, prove that (a 2 b 2), (ab bc), (b 2 c 2) are in GP., 16. If a, b, c, d are in GP, prove that (a 2 b 2), (b 2 c 2), (c 2 d 2) are in GP., 17. If a, b, c, d are in GP then prove that, 1, , 2 1 2 , 2 1 2 are in GP., 2, 2, (a b ) (b c ) (c d ), 18. If (p 2 q 2), (pq qr), (q 2 r 2) are in GP then prove that p, q, r are in GP., 19. If a, b, c are in AP and a, b, d are in GP, show that a, (a b) and (d c) are, in GP., 20. If a, b, c are in AP, and a, x, b and b, y, c are in GP then show that x 2, b 2, y 2 are, in AP., ANSWERS (EXERCISE 12E), , 4. (k 0) or (k 15), 7. 8, 16, 32, , 5. (2, 5, 8) or (26, 5, –16) 6. (12, 7, 2) or (3, 7, 11), , HINTS TO SOME SELECTED QUESTIONS, 1. We have, 2q p r., Let Tp , Tq , Tr be the given terms of a GP with first term A and common ratio R., Then,, (Tq) 2 {AR (q, , 1) 2, , , } A 2 R ( 2 q 2), , and (Tp #Tr) {AR (p, , , 1), , # AR (r, , 1), , } A 2R (p, , r 2), , A 2R (2q 2)., , (Tq) 2 (Tp #Tr) and hence Tp , Tq , Tr are in GP., , 2. a, b, c are in GP & b 2 ac & (b 2) n (ac) n & b 2n a nc n., , , 2 log b n log a n log c n & log a n, log b n, log c n are in AP., , 3. a, b, c are in GP & b 2 ac, , , 2 log m b log m a log m c & 2 #, , log b, log a log c, , , log m log m log m
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470, , Senior Secondary School Mathematics for Class 11, , , 2#, , 1 1 1, 1 , 1 , 1, are in AP., &, log b m log a m log c m, log a m log b m log c m, , k6 3, & (k 6) 2 3k 36, k 12 k 6, & k 2 12k 3k & k 2 15k 0 & k (k 15) 0, k 0 or k 15., 6. Let the required numbers be (a d), a, (a d) . Then, 3a 21 & a 7., So, these numbers are (7 d), 7, (7 d) ., (7 d), 6 and (8 d) are in GP., , 4., , 6 (8 d), & (7 d) (8 d) 36 & d 2 d 20 0, 6, (7 d), & (d 5) (d 4) 0 & d 5 or d 4., Hence, the required numbers are (12, 7, 2) or (3, 7, 11)., 8. Let r be the common ratio of the given GP. Then,, So,, , b ar and c ar 2., , , 2, 2, a 2 a 2r a 2r 2 a (1 r r ) 1, a 2r a 2r 3 a 2r 2 a 2r (1 r r 2) r, ar a a (1 r) 1 ·, RHS 2, ar ar ar (1 r) r, , LHS , , Hence, LHS = RHS., (b c) (c a), , 9., & (b c) 2 (a b) (c a), (a b) (b c), & b 2 c 2 2bc ac a 2 bc ab, & a 2 b 2 c 2 ab bc ac, [adding 2 (ab bc ac) on both sides]., & (a b c) 2 3 (ab bc ac), 12. Let b ar and c ar 2. Then,, 1 1 , 1 , 1 1 ,, 1 , 1, ·, and, a b a ar a (1 r) 2b 2ar, (b c) ar (1 r), (1 r), 1 1 , 1, 1, , , 1 2 #d 1 n ·, , 2b, (a b) (b c) a (1 r) ar (1 r) ar (1 r) ar, Hence,, , 1 , 1 , 1, are in AP., (a b) 2b (b c), , 13. Let b ar and c ar 2. Then, a 2c 2 a 2 (ar 2) 2 a 2 # a 2r 4 a 4r 4 (b 2) 2., 19. (a c) 2b and b 2 ad., We have to prove that (a b) 2 a (d c) ., Now, (a b) 2 a 2 b 2 2ab a 2 ad a (a c), 2, & (a b) ad ac a (d c) ., Hence, a, (a b), (d c) are in GP., , 20. x 2 ab, y 2 bc and a c 2b., , , x 2 y 2 (ab bc) b (a c) 2b 2., , Hence, x 2, b 2, y 2 are in AP.
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Geometrical Progression, , 471, , GEOMETRIC MEAN, Let a and b be two given numbers. We say that G is the geometric, mean (GM) between a and b if a, G, b are in GP., G is the GM between a and b + a, G, b are in GP, G b, + a , G, + G 2 ab + G ab ., , GEOMETRIC MEAN, , So, GM between a and b ab ., REMARKS, , (i) The GM between two positive numbers is positive., (ii) The GM between two negative numbers is negative., (iii) The GM between two numbers of opposite signs does not exist., SOLVED EXAMPLES, , EXAMPLE 1, , SOLUTION, , EXAMPLE 2, SOLUTION, , Find the geometric mean between, (i) 6 and 24, (ii) –9 and –25, (iii) –6 and 9, (i) The GM between 6 and 24 6 # 24 144 12., (ii) The GM between –9 and 25 ( 9)#( 25) 225 15., (iii) Since the GM between two numbers of opposite signs does not, exist, so the GM between –6 and 9 does not exist., Find two positive numbers a and b whose AM and GM are 34 and 16, respectively., We have, ab , 34 and ab 16, 2, &, a b 68 and ab 256, (a b) (a b) 2 4ab (68) 2 4 # 256 3600 !60, , &, , &, a b 68 and a b !60, &, (a b 68, a b 60) or (a b 68, a b 60), &, (a 64, b 4) or (a 4, b 64), Hence, the required numbers are (a 64, b 4) or (a 4, b 64) ., EXAMPLE 3, , Find the value of n so that, between a and b, where a ! b., , SOLUTION, , a, , a, , (n 1), , b (n 1), may be the geometric mean, a bn, n, , (n 1), , b (n 1), is GM between a and b, a bn, n, , (n 1), , &, , a, , &, , [ a (n, , 1 1, b (n 1), a2 b2, n, a b, n, , 1), , b (n 1)] (a n b n) `a 2 b 2j, 1, , 1
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472, , Senior Secondary School Mathematics for Class 11, 1), , b (n 1)] a, , 1, 1, bn l, 2 b2, , 1, , a2 b, , 1, bn l, 2, , &, , [ a (n, , &, , $a (n 1) a bn 2 l b 2 . a 2 b bn 2 l b (n 1), , &, , a, , &, , a, , 1, , 1, 1, bn l`, 2 a2, , 1, , b 2j b, 1, , 1, 1 JK, bn l, bn l KK, 2 b, 2 2K, L, , 1, , 1, 1, bn l`, 2 a2, , NO, OO, O, P, , 1, , b 2j, 1, , 9a `a 2 b 2j ! 0, since a ! bC, 1, , 1, , a bn 2 l a 0, 1 b l, b l, b, b, 1, 1, &, n 0 & n ·, 2, 2, 1, Hence, n ·, 2, 1, , &, , EXAMPLE 4, SOLUTION, , If A and G are respectively the arithmetic and geometric means between, two distinct positive numbers a and b then prove that A G., We have, ab, A, and G ab, 2, a b 2 ab 1, ab , ( a b)2 0, &, AG , ab , 2, 2, 2, &, A G., Hence, AM GM., , EXAMPLE 5, , If A and G are respectively the arithmetic and geometric means between, two positive numbers a and b then prove that the quadratic equation, having a, b as its roots is given by x 2 2Ax G 2 0., , SOLUTION, , We have, , ab, and G ab ., 2, The equation having a, b as its roots is given by, A, , x 2 (a b) x ab 0, , ab, and G abE ·, 2, Hence, the quadratic equation having a, b as its roots is, x 2 2Ax G 2 0., , &, , EXAMPLE 6, , SOLUTION, , x 2 2Ax G 2 0, , ;a A , , If A and G be the AM and GM between two positive numbers then prove, that the numbers are A ! A 2 G 2 ., Let the required numbers be a and b. Then,, ab, A, and G ab ., 2, The quadratic equation having a, b as its roots is given by, x 2 (a b) x ab 0, &, , x 2 2Ax G 2 0
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Geometrical Progression, , &, , x, , 473, , 2A ! 4A 2 4G 2, 2, 2, & x A! A G ., 2, , Hence, the required numbers are A ! A 2 G 2 ., EXAMPLE 7, , If x, y, z are distinct, (x y) (y z) (z x) 8xyz., , positive, , numbers, , SOLUTION, , Using AM > GM, we have, xy, yz, zx, xy ,, yz and, zx, 2, 2, 2, &, x y 2 xy , y z 2 yz and z x 2 zx, , then, , prove, , that, , &, (x y) (y z) (z x) 2 xy # 2 yz # 2 zx, &, (x y) (y z) (z x) 8xyz., Hence, (x y) (y z) (z x) 8xyz., EXAMPLE 8, SOLUTION, , EXAMPLE 9, SOLUTION, , If a, b, c, d are four distinct positive numbers in GP then prove that, a d b c., a, b, c, d are in GP, a, b, c are in GP, &, b is the GM between a and c., &, ac, is the AM between a and c., Also, we know that, 2, But, AM > GM., ac, … (i), , b & a c 2b., 2, Again, a, b, c, d are in GP, b, c, d are in GP, &, c is the GM between b and d., &, bd, is the AM between b and d., Also, we know that, 2, But, AM > GM., bd, … (ii), , c & b d 2c., 2, From (i) and (ii), we get, a c b d 2b 2c., Hence, a d b c., The sum of two numbers is 6 times their geometric mean. Show that the, numbers are in the ratio (3 2 2 ) : (3 2 2 ) ., Let the required numbers be a and b. Then,, a b 6 ab, ab 3, &, 2 ab 1, a b 2 ab 3 1 2, , , [by componendo and dividendo], &, a b 2 ab 3 1 1
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474, , Senior Secondary School Mathematics for Class 11, , &, , &, , &, , &, , ( a b)2, ( a b), , 2, , , , ( 2) 2, 1, , 2, , &, , ( a b) ( a b), , ( a b) ( a b), , a b, 2, , 1, , a, b, 21, 21, [by componendo and dividendo], , JK a NO2 ( 2 1) 2, 21, & KK OOO , K b, b, 21, ( 2 1) 2, L P, a 32 2 ·, b 32 2, a, , , , Hence, a : b (3 2 2 ) : (3 2 2 ) ., EXAMPLE 10, SOLUTION, , If the AM and GM of the roots of a quadratic equation be 8 and 5, respectively then obtain the equation., Let the required roots be a and b., ab, ab , … (i), Then, AM , &, 8 & a b 16, 2, 2, and GM ab & ab 5 & ab 25., … (ii), , sum of roots = 16 and product of roots = 25., Hence, the required equation is x 2 16x 25 0., , EXAMPLE 11, , If the AM and GM of two positive numbers a and b are in the ratio m : n,, show that, a : b (m m 2 n 2 ) : ( m m 2 n 2 ) ., , SOLUTION, , Let A and G be respectively the AM and GM of a and b. Then,, ab, A, and G ab, 2, &, a b 2A and G 2 ab., Now, the equation having roots a and b is, x 2 (a b) x ab 0, &, , x 2 2Ax G 2 0, , &, , x, , [a a b 2A and ab G 2], , 2A ! 4A 2 4G 2, A! A2 G2, 2, , a A A 2 G 2 and b A A 2 G 2 ., Now, A : G m : n, (given)., Let A km and G kn for some constant k. Then,, , &, , 2, 2, 2 2, 2 2, 2, 2, a A A G km k m k n m m n ·, b A A 2 G 2 km k 2m 2 k 2n 2 m m 2 n 2, , Hence, a : b (m m 2 n 2 ) : (m m 2 n 2 ) .
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Geometrical Progression, EXAMPLE 12, SOLUTION, , 475, , Find two positive numbers whose difference is 12 and whose AM exceeds, the GM by 2., Let the required positive numbers be a and b, where a b., Then, a b 12., … (i), Also, AM GM 2, ab , &, ab 2, 2, &, a b 2 ab 4, &, , ( a b)2 4, , &, a b 2., Now, a b 12, &, ( a b ) ( a b ) 12, &, , ( a b )# 2 12 &, , … (ii), [from (i)], , a b 6., , … (iii), , [using (ii)], , On solving (ii) and (iii), we get a 4 and b 2., , a 16 and b 4., Hence, the required numbers are 16 and 4., EXAMPLE 13, SOLUTION, , If xd R, find the minimum value of 3 x 3 (1, We know that AM GM., , for all xd R, we have, , &, , x), , ., , x, (1 x), 3 3, , 3 x # 3 (1 x), 2, x, (1 x), 3 3, 3, 2, , {3 x 3 (1 x)} 2 3 ., , &, , Hence, the minimum value of 3 x 3 (1 x) is 2 3 for any xd R., n GEOMETRIC MEANS BETWEEN TWO GIVEN NUMBERS, , Let a and b be two given numbers. We say that G1, G 2, G 3, …, Gn are n geometric means, between a and b, if a, G1, G 2, G 3, …, Gn, b are in GP., EXAMPLE 14, SOLUTION, , Insert n geometric means between a and b., Let G1, G 2, G 3, …, Gn be the n geometric means between a and b., Then a, G1, G 2, G 3, …, Gn, b are in GP., This GP contains (n 2) terms., Let the common ratio of this GP be r. Then,, Tn 2 b & ar (n, , 1), 1, , , , b (n 1), G1 ar a · b a l, ;, 2, , 1, , , b & r b b l(n 1) ·, a, , b (n 1), G 2 ar 2 a · b a l, ;
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476, , Senior Secondary School Mathematics for Class 11, 3, , b (n 1), G 3 ar 3 a · b a l, ;, … … … … …, … … … … …, n, , b (n 1) ·, Gn ar n a · b a l, EXAMPLE 15, SOLUTION, , Insert two numbers between 3 and 81 so that the resulting sequence is a GP., Let the required numbers be G1 and G 2 . Then,, 3, G1, G 2, 81 are in GP., Let the common ratio of this GP be r. Then,, , T4 81 & ar (4 1) 81, & 3 # r 3 81 & r 3 27 (3) 3 & r 3., , G1 (3 # r) (3 # 3) 9 and G 2 (3 # r 2) (3 # 9) 27., Hence, the required numbers are 3 and 27., , EXAMPLE 16, SOLUTION, , Insert three numbers between 1 and 256 so that the resulting sequence, is a GP., Let the required numbers be G1, G 2, G 3 . Then,, 1, G1, G 2, G 3, 256 are in GP., Let r be the common ratio of this GP. Then,, , T5 256 & ar (5 1) 256, & 1# r 4 256 & r 4 256 (4) 4 & r 4., , G1 (1# 4) 4, G 2 (1# 4 2) 16 and G 3 (1# 4 3) 64., Hence, the required numbers are 4, 16, 64., , EXAMPLE 17, , SOLUTION, , If G be the GM between two given numbers and A1, A 2 be the two AMs, between them, prove that, G 2 (2A1 A 2) (2A 2 A1) ., Let a, b be the two given numbers. Then,, … (i), G 2 ab., Also, a, A1, A 2, b are in AP, &, 2A1 a A 2 and 2A 2 A1 b, &, (2A1 A 2) a and (2A 2 A1) b, &, (2A1 A 2) (2A 2 A1) ab, &, , (2A1 A 2) (2A 2 A1) G 2, , [using (i)]., , Hence, G (2A1 A 2) (2A 2 A1) ., 2, , EXAMPLE 18, , If A is the arithmetic mean and G1, G 2 be the two geometric means, between any two numbers then prove that, 2A , , G 12 G 22, , ·, G 2 G1
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Geometrical Progression, SOLUTION, , 477, , Let the given numbers be a and b. Then,, ab, A, & a b 2A., 2, And, a, G1, G 2, b are in GP., , , G1 G 2, , b, a, G1 G 2, , &, , a, , &, , ab , , &, , 2A , , … (i), , G 12, G 22, and b , G2, G1, G 12 G 22, , G 2 G1, , G 12 G 22, , G 2 G1, , Hence, 2A , , [using (i)]., , G 12 G 22, , ·, G 2 G1, , EXERCISE 12F, 1. Find two positive numbers a and b, whose, (i) AM = 25 and GM = 20, (ii) AM = 10 and GM = 8., 2. Find the GM between the numbers, 9, (iii) 0.15 and 0.0015, (i) 5 and 125, (ii) 1 and, 16, (iv) –8 and –2, (v) –6.3 and –2.8, (vi) a 3b and ab 3, 3. Insert two geometric means between 9 and 243., 1, 4. Insert three geometric means between and 432., 3, 5. Insert four geometric means between 6 and 192., 6. The AM between two positive numbers a and b (a b) is twice their GM., Prove that a : b (2 3 ) : (2 3 ) ., 7. If a, b, c are in AP; x is the GM between a and b; y is the GM between b and c;, then show that b 2 is the AM between x 2 and y 2., 8. Show that the product of n geometric means between a and b is equal to the, nth power of the single GM between a and b., 9. If AM and GM of the roots of a quadratic equation are 10 and 8 respectively, then obtain the quadratic equation., ANSWERS (EXERCISE 12F), , 1. (i) (a 40 and b 10) or (a 10 and b 40), (ii) (a 16 and b 4) or (a 4 and b 16), 3, 2. (i) 25 (ii), (iii) 0.015 (iv) –4 (v) –0.42 (vi) a 2b 2, 4, 3. 27, 81, 4. 2, 12, 72, 5. 12, 24, 48, 96 9. x 2 20x 64 0
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478, , Senior Secondary School Mathematics for Class 11, HINTS TO SOME SELECTED QUESTIONS, , ab 2, 1, 6. AM 2 (GM) + (a b) 2 ab +, 2, 2 ab 1, (, a b 2 ab 2 1, , +, a b 2 ab 2 1, (, ( a, a b, 3, , +, +, 1, a b, ( a, +, , +, , a, b, , , , a b)2, , , , ( 3) 2, , a b), (1) 2, b) ( a b), 31, , b) ( a b), 31, 2, , 2, 31, a ( 3 1) 2 3 ·, + , b ( 3 1) 2 2 3, 31, , 7. (a c) 2b, x ab and y bc, &, , x 2 y 2 (ab bc) b (a c) 2b 2., , 8. Let G1, G2, …, Gn be n GMs between a and b, and let r be its common ratio. Then,, a, G1, G2, …, Gn, b are in GP., 2 1), , 1, , , b & r (n 1) c b m & r c b m(n 1) ·, a, a, , , , Tn 2 b & ar (n, , , , G1 #G2 #G3 #…# Gn (ar)#(ar )#…# (ar ), 2, , 1, , a n # r (1 2 … n ) a n # r 2, n/2, , a n #c b m, a, , … (i), , n, , n (n 1), , a n/2 # b n/2, , [using (i)], , ( ab ) G , where G ab ., 9. Let a and b be the roots of the quadratic equation., n, , n, , Then, x 2 (a b) x ab 0., ab , Now,, 10 and ab 8 & a b 20 and ab 64., 2, Hence, the required equation is x 2 20x 64 0., , INFINITE GEOMETRIC SERIES, THEOREM, , Prove that the sum of an infinite GP with first term a and common ratio r,, where|r| 1, is given by, a ·, S, (1 r), , PROOF, , Let us consider an infinite GP with first term a and common ratio r,, where 1 r 1, i.e.,|r| 1., The sum of n terms of a GP is given by, n, n, a (1 r ), a ar 3 ·, … (i), & Sn ), Sn , (1 r), (1 r) (1 r), Since|r| 1, so when n increases then r n decreases., Thus, when n "3 then r n " 0.
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Geometrical Progression, , , , 479, , n, , lim, , n "3, , r, 0., (1 r), , Hence, the sum of an infinite GP is given by, n, , a ar 3, S lim Sn lim ), (1 r), n "3, n " 3 (1 r), n, , a , ar , a a, ', lim, 01, (1 r) n "3 (1 r), ( 1 r), ( 1 r), [using (i)]., a, ,, This gives, S , when|r| 1., (1 r), , , REMARKS, , 1. We shall denote the sum of an infinite GP by S., 2. If r $ 1 then sum of an infinite GP is S 3., SUMMARY, , Sum of an infinite GP with the first term a and the common, a ·, ratio r, where|r| 1, is given by S , (1 r), SOLVED EXAMPLES, EXAMPLE 1, SOLUTION, , Find the sum of the infinite geometric series b1 , , 11 1 , …3l ·, 3 9 27, , In the given infinite geometric series, we have, 1, 1, a 1 and r such that|r| 1., 3, 3, Hence, the sum of the given infinite series is, 1 3·, 2, 2, b l, 3, 3, Hence, the required sum is ·, 2, S, , EXAMPLE 2, SOLUTION, , EXAMPLE 3, , a , (1 r), , 1, , 1, b1 l, 3, , 1 1 1 , … 3n ·, 3 32 33, The given series is an infinite geometric series in which, 1, 1, a 1, r and |r| 1., 3, 3, Hence, the sum of the given infinite geometric series is, a 1 1 3·, S, 1, 4, 4, (1 r), b1 l b l, 3, 3, Find the sum of the infinite geometric series d1 , , Find the sum of the infinite geometric series b, , 5 5, , 5 …3l ·, 4 16 64
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480, , Senior Secondary School Mathematics for Class 11, , SOLUTION, , In the given infinite geometric series, we have, 5, 5, 4, 1, 1, a, , r b # l and |r| 1., 5, 16, 4, 4, 4, Hence, the sum of the given infinite geometric series is, a ( 5/4) 5 4 , b # l, S, 1., 5, 1, 4, (1 r), b1 l, 4, , EXAMPLE 4, , Find the sum of the infinite geometric series, ( 2 1) 1 ( 2 1) …3., We have, ( 2 1) ( 2 1), 1, 1, , , ·, #, 1, ( 2 1) ( 2 1) ( 2 1), , SOLUTION, , So, the given series is an infinite geometric series in which, a ( 2 1) and r ( 2 1) 1., Hence, the sum of the given infinite geometric series is, S, , a ( 2 1) ( 2 1) (2 2 ), #, , (1 r) {1 ( 2 1)} (2 2 ) (2 2 ), , , EXAMPLE 5, , SOLUTION, , 4 3 2 (4 3 2 ), , ·, 2, (4 2), , Find the sum of the infinite series:, 1 1, 1, 1, 1, 1, (2 2 3 4 5 6 … 32 ·, 3, 2, 2, 3, 3, The given series can be expressed as the sum of two infinite, geometric series, shown below., 1 1, 1, 1, 1, 1, (2 2 3 4 5 6 … 32, 3, 2, 2, 3, 3, (1 13 15 … 32 ( 12 14 16 … 32, 2 2, 2, 3, 3, 3, 1, 1, 'an inf inite GP with a and r 1, 2, 4, 1, 1, 'an inf inite GP with a and r 1, 9, 9, (1/2), (1/9), (1/2) (1/9), , , , 1, 1, (, 3/4) (8/9), b1 l b1 l, 9, 4, 1, 4, 1, b # l b # 9 l b2 1 l, 2 3, 9 8, 3 8, , 16 3 19 ·, 24, 24,
Page 492 : Geometrical Progression, EXAMPLE 6, SOLUTION, , 481, , Prove that 6 1/2 · 6 1/4 · 6 1/8 · … 3 6., 1 1 1, We observe here that b …3l is an infinite geometric, 2 4 8, 1, 1 2, 1, , series in which a, and r b # l such that|r| 1., 2, 2, 4 1, So, this sum is given by, 1, 1, b l, b l, 2, 2, , 1., S, 1, 1, b1 l b l, 2, 2, , , Hence,, EXAMPLE 7, , SOLUTION, , 1, , 1, , 1, , 1, ·64, , 1, ·68, , 6 2 · 6 4 · 6 8 ·…3 6, 1, 62, , 1 1 1, b … 3l, 2 4 8, , … (i), , 61 6, , [using (i)]., , · … 3 6., , a a, b b, If |r| 1, x da r 2 … 3n, y eb r 2 … 3o, r, r, xy ab, c, c, and z dc 2 4 … 3n, prove that z c ·, r, r, Clearly, each one of the given series is an infinite geometric series., Taking the sum of each series, we get, a ar, ;a 1st term a and common ratio 1E,, x, r, ( r 1), 1, b1 r l, , b br, ;a 1st term b and common ratio 1E, y, r, (r 1), 1, b1 r l, and z , , c, d1 , , 1, n, r2, , , , 2, , cr, 1, <a 1st term c and common ratio 2F, (r 1), r, 2, , 2, xy, (r 1) ab, ar # br #, , 2, z, c ·, ( r 1) ( r 1 ), cr, xy ab, Hence, z c ·, , , , EXAMPLE 8, , If y x x 2 x 3 … 3, where|x| 1, prove that x , , SOLUTION, , By summing the given infinite GS, we get, x, & y (1 x) x, y, (1 x), & y xy x & x xy y, y, ·, x ( 1 y) y & x , ( 1 y), y, ·, Hence, x , (1 y), , y, ·, (1 y)
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482, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 9, , If x 1 a a 2 … 3, where|a| 1, and y 1 b b 2 … 3, where|b| 1,, xy, ·, prove that (1 ab a 2b 2 … 3) , , (x y 1), , SOLUTION, , By summing the given infinite GS, we get, 1, 1 ·, x, and y , (1 a ), (1 b), , , xy, , ( x y 1), , 1, 1, ( ·, 2, (1 a) (1 b), 1, 1 , 12, ( , (1 a) (1 b), , (1 a) (1 b), 1, 3, ), #, , , , , (1 a) (1 b) (1 b) (1 a) (1 a) (1 b), 1, (1 ab) 1, , (1 ab), (1 ab a 2b 2 … 3) ., xy, ·, Hence, (1 ab a 2b 2 … 3) , (x y 1), EXAMPLE 10, , If x (1 r a r 2a … 3) and y (1 r b r 2b … 3), prove that, 1, , y1 b, x1 a, rb x l e y o ·, 1, , SOLUTION, , By summing the given infinite geometric series, we get, 1, 1, x, and y , (1 r a), (1 r b ), 1, 1, &, (1 r a ) x and (1 r b ) y, &, , 1, 1, r a b1 x l and r b d1 y n, , &, , y1 b, x1 a, r b x l and r e y o ·, , 1, , 1, , 1, , y1 b, x1 a, Hence, r b x l e y o ·, 1, , EXAMPLE 11, SOLUTION, , Use geometric series to express 0.555… 0.5 as a rational number., We have, 0.5 0.5555…, 0.5 0.05 0.005 0.0005 … 3, 5 52 53 54 …3, 10 10, 10, 10
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Geometrical Progression, , b, , 5, l, 10, 5, , 9, 1, b1 l, 10, 5, Hence, 0.5 ·, 9, EXAMPLE 12, SOLUTION, , EXAMPLE 13, SOLUTION, , ;a S , , Find the rational number whose decimal form is 1.345., We have, 1.345 1.3454545 … 3, 1.3 0.045 0.00045 …3, 1.3 ) 453 455 … 33, 10, 10, 45, o, 10 3, 1.3 , 1, d1 2 n, 10, 1.3 45 # 100 13 45, 99 10 990, 1000, , 143 5 148 74 ·, 110, 110 55, 74, ·, Hence, 1.345 , 55, , SOLUTION, , a, in an infinite GSE ·, (1 r), , Find the rational number whose decimal form is 0.142., We have, 0.142 0.142424242 … ., 0.1 0.042 0.00042 0.0000042 …3, 1 ( 423 425 427 … 32, 10 10, 10, 10, 42, R, d 3 n SSthis being a GS in whichVWW, 10, WW, SS, 1, 3N, JK, , , WW, SS, 42, 42 10 OO 1, 1, 10, K, , , #, a, and, r, , 1, OO, KK 5, WW, d1 2 n SS, 3, 2, 42, 10, 10, 10, W, S, 10, P, L, X, T, 1 b 42 # 100 l b 1 42 l 141 ·, 99, 990, 10 1000, 10 990, 141, ·, Hence, 0.142 , 990, , e, , EXAMPLE 14, , 483, , The sum of an infinite GP is, , RSthis being a GS in whichVW, WW, SS, W, SS, 45, 1, SSa 3 and r 2 1WWW, W, S 10, 10, X, T, b13 1 l, 10 22, , 4, 80, · Find its, and its common ratio is, 5, 9, , first term., Let a be the first term of the given infinite GP., 4, 4, 4 1., Here, r , and|r| , 5, 5, 5
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484, , Senior Secondary School Mathematics for Class 11, , The sum of this infinite GP is S , , 80, ·, 9, , 80, 9, 4, b1 l, 5, 5a 80, &, & 5a 80 & a 16., 9, 9, Hence, the first term is 16., , , EXAMPLE 15, SOLUTION, , a, &, (1 r), , S, , a, , The sum of first two terms of an infinite geometric series is 15 and each term, of the series is equal to the sum of all the terms following it. Find the series., , Let the series be a ar ar 2 … ar n ar (n 1) … 3., Then, first term a and common ratio r., Clearly, a ar 15 & a (1 r) 15., … (i), Also, ar (n, , 1), , ar n ar (n 1) … 3, , [a an an 1 an 2 … 3], , n, ar, (1 r), , a RHS is a GS in which first term ar n, G, =, and common ratio r, 1, &, 1 r r & 2r 1 & r ·, 2, 3, 1, 2, Putting r in (i), we get a 15 & a b15 # l 10., 2, 2, 3, 1, Thus, a 10 and r ·, 2, 5 5, Hence, the required series is b10 5 …3l ·, 2 4, &, , EXAMPLE 16, SOLUTION, , ar, , (n 1), , The sum of an infinite geometric series is 8. If its second term is 2, find its, common ratio., Let the infinite geometric series be a, ar, ar 2, …, 3. Then,, … (i), ar 2, a , … (ii), and, 8., (1 r), 2, Putting r a from (i) in (ii), we get, a, , b1 2a l, , 8 &, , 2, , a, 8 & a 2 8a 16 0, (a 2), , & (a 4) 2 0 & a 4 0 & a 4., 1, 2, Putting a 4 in (i), we get r & r ·, 2, 4, 1, Hence, the common ratio of the given geometric series is ·, 2, EXAMPLE 17, SOLUTION, , The sum of an infinite geometric series is 15 and the sum of the squares of, these terms is 45. Find the series., Let a be the first term and r be the common ratio.
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Geometrical Progression, , 485, , Then, the series is a ar ar 2 … 3., Now, (a ar ar 2 … 3) 15 &, , a , 15, (1 r), , and (a 2 a 2r 2 a 2r 4 … 3) &, , a, 45., 2, , (1 r ), , … (i), , 2, , On squaring both sides of (i), we get, , … (ii), , a2 , 225., ( 1 r) 2, , … (iii), , On dividing (iii) by (ii), we get, (1 r 2) 225, a2, 1r, , #, & 5, 2, 2, 1 r, 45, , (1 r), a, & 1 r 5 5r, & 6r 4 & r , , 2·, 3, , 2, in (i), we get, 3, 1, 2, a 15 #b1 l b15 # l 5., 3, 3, 2, Thus, a 5 and r ·, 3, 10 20 40 , Hence, the required series is b5 , …3l ·, 3, 9 27, Putting r , , EXAMPLE 18, , If S1, S 2, S 3, …, S p denote the sums of infinite geometric series whose, , SOLUTION, , first terms are 1, 2, 3, …, p respectively and whose common ratios are, 1 , 1 , 1 , …, 1, respectively, then show that, 2 3 4, ( p 1), 1, S1 S2 S3 … Sp p (p 3) ., 2, By the summation of an infinite geometric series, we get, S1 , Sp , , 1, , 1, b1 l, 2, p, d1 , , 2; S 2 , , 1, n, p1, , 2, , 1, b1 l, 3, , 3; S 3 , , 3, , 1, b1 l, 4, , 4, …,, , (p 1) ., , S1 S 2 S 3 … S p [2 3 4 … (p 1)], p, · {2 (p 1)} 1 p (p 3) ., 2, 2, 1, , , , , , , Hence, S1 S 2 S 3 … S p, p (p 3) ., 2, , , EXAMPLE 19, , The side of a given square is 10 cm. The midpoints of its sides are joined to, form a new square. Again, the midpoints of the sides of this new square are, joined to form another square. This process is continued indefinitely. Find, (i) the sum of the areas and (ii) the sum of the perimeters of the squares.
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486, SOLUTION, , Senior Secondary School Mathematics for Class 11, , Let ABCD be the given square with each side equal to 10 cm., Let E, F, G, H be the midpoints of the sides AB, BC, CD and DA, respectively. Let P, Q, R, S be the midpoints of the sides EF, FG, GH, and HE respectively., G, D, C, BE BF 5 cm, &, , , &, , EF BE 2 BF 2 25 25 cm, 50 cm 5 2 cm., , R, , Q, , H, , 5 2, S, 5, 1, EF , cm , cm, 2, 2, A, 2, 25, 25, 2, 2, , PQ FP FQ , cm 25 cm 5 cm., 2, 2, , F, P, , FQ FP , , E, , B, , Thus, the sides of the squares are 10 cm, 5 2 cm, 5 cm, … ., (i) Sum of the areas of the squares formed, {(10) 2 (5 2 ) 2 5 2 … 3} cm 2, (100 50 25 … 3) cm 2, RStaking the sum of infinite GPVW, WW, S, 100 cm 2 200 cm 2 SS, WW ·, 1, 1, S, , , with, a, 100, and, r, WW, SS, b1 l, 2, 2, X, T, (ii) Sum of perimeters of the squares formed, (40 20 2 20 …) cm, , , ( 2 1), 40 2, 40, #, cm , cm (80 40 2 ) cm., 1, ( 2 1) ( 2 1), e1 , o, 2, , EXAMPLE 20, , Each side of an equilateral triangle is 18 cm. The midpoints of its sides, are joined to form another triangle whose midpoints, in turn, are joined, to form still another triangle. The process is continued indefinitely. Find, the sum of the (i) perimeters of all the triangles and (ii) areas of all the, triangles., , SOLUTION, , Let3ABC be the given triangle having each, side 18 cm. Let D, E, F be the midpoints of, BC, CA, AB respectively to form 3DEF. Let, G, H, I be the midpoints of DE, EF and FD, respectively to form 3GHI., We continue this process indefinitely., 9, cm, …, and so on., 2, (i) Sum of the perimeters of all triangles so formed, 3 '18 9 9 9 …31 cm, 2 4, 9 1, a, '3 #, 1 cm, where a 18 and r , 18 2, (1 r), , The sides of these triangle are 18 cm, 9 cm,
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Geometrical Progression, , 487, , ]Z], 18 b_b, b cm (3 # 36) cm 108 cm., ][]3 #, 1 `, b1 lbbb, ]], 2, \, a, (ii) Sum of the areas of all the triangles so formed, , , 3, 92 92, · ((18) 2 (9) 2 b l b l … 32 cm 2, 2, 4, 4, , , , 3, 81 81 , · '324 81 , … 31 cm 2, 4, 4 16, , 3, 81 1, · a cm 2, where a 324 and r , 324 4, 4 (1 r), ]Z] 3, b_b, ]] # 324 bb cm 2 108 3 cm 2., 1 `, ][] 4, b1 lbbb, 4 b, ]], \, a, 4, After striking a floor a certain ball rebounds of the height from which it, 5, has fallen. Find the total distance that it travels before coming to rest, if it, , , EXAMPLE 21, , is gently dropped from a height of 120 metres., SOLUTION, , Initially, the ball falls from a height of 120 m. After striking the, 4, floor, it rebounds and goes to a height of b #120l m., 5, Now, it falls from this height and after striking the floor, it rebounds, 4, 4, 42, and goes to a height of of b #120l m (b l #1202 m., 5, 5, 5, This process continues indefinitely till the ball comes to rest., Total distance covered by the ball in metres, 2, 3, 120 2 #=' 4 #1201 (b 4 l #1202 (b 4 l #1202 … 3G, 5, 5, 5, , a, 4, 4, 1 , where a b #120l 96 and r , 5, 5, (1 r), ]Z, b_, ]]120 2 # 96 bb {120 (192 # 5)} (120 960) 1080., 4, [], `, b1 lbbb, ]], 5, \, a, Hence, the total distance covered by the ball is 1080 m., 120 '2 #, , EXAMPLE 22, , If, , 3, , 2n, , n0, , 0, SOLUTION, , 3, , x / cos , y / sin , 2n, , n0, , and, , 3, , z / cos sin ,, n0, , , then prove that xz yz z xy., 2, , We have, 3, , x / cos 1 cos cos … 3, n0, , 2n, , 2, , 4, , 2n, , 2n, , where
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488, , Senior Secondary School Mathematics for Class 11, , , , a ,, where a 1 and r cos 2 [sum of an infinite GP], (1 r), , 1, 1 ·, (1 cos 2 ) sin 2 , 1, sin 2 x ·, , , , , 3, , … (i), , y / sin 1 sin sin … 3, 2n, , 2, , 4, , n0, , a ,, where a 1 and r sin 2 , (1 r), 1, 12 ·, , 2, , (1 sin ) cos , 1, 2, cos y ·, , , , , 3, , [sum of an infinite GP], , … (ii), , z / cos sin 1 cos sin cos sin … 3, 2n, , 2n, , 2, , 2, , 4, , 4, , n0, , a ,, where a 1 and r cos 2 sin 2 , (1 r), 1, 1, , , (1 cos 2 sin 2 ) 1 (1 sin 2 ) (1 cos 2 ), 1, 1, , , [using (i) and (ii)], 1, 1, 1, 1 1, 1 b1 x ld1 y n 1 d1 x y xy n, xy, 1, , , ·, ( x y 1), 1, 1, 1, d n, x y xy, xy, , z & xz yz z xy., x y 1, Hence, xz yz z xy., , , EXERCISE 12G, Find the sum of each of the following infinite series:, 1. 8 4 2 4 2 2 …3, 3., 5., , 1 1 1 , 2, …3, 2 2 2 4 2, , 2. 6 1.2 0.24 …3, 4. 10 9 8.1 … 3, , 2 3 2 3 , …3, 5 52 53 54, , 6. Prove that 9 1/3 # 9 1/9 # 9 1/27 #… 3 3., 7. Find the rational number whose decimal expansion is given below:, (i) 0.3, , (ii) 0.231, , (iii) 3.52
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Geometrical Progression, , 489, , 8. Express the recurring decimal 0.125125125… 0.125 as a rational number., 9. Write the value of 0.423 in the form of a simple fraction., 10. Write the value of 2.134 in the form of a simple fraction., 11. The sum of an infinite geometric series is 6. If its first term is 2, find its, common ratio., 12. The sum of an infinite geometric series is 20 and the sum of the squares of, these terms is 100. Find the series., 13. The sum of an infinite GP is 57 and the sum of their cubes is 9747. Find, the GP., ANSWERS (EXERCISE 12G), , 1. 8 (2 2 ), 7. (i), 11., , 1, 3, , 2 2, 3, 125, 8., 999, , 2. 7.5, 231, 999, , 3., , 317, 90, 24, 72 …3l, 12. b8 , 5 25, , (ii), , 2, 3, , (iii), , 100, 19, 419, 9., 990, , 13, 24, 2113, 10., 990, 38 , 76 , 152 ,, … 3l, 13. b19,, 3 9 27, 4., , 5., , HINTS TO SOME SELECTED QUESTIONS, 1. a 8 and r , , , S, , 4 2, 1 ·, 8, 2, , a , (1 r), , 8, 1, e1 , o, 2, , , , ( 2 1), 8 2, 8 (2 2 ) ., #, , ( 2 1) ( 2 1, , 1.2 , 0.2., 6, a 6 60 15 , S, 7. 5., 8, 2, (1 r) 0.8, 1, 1 1, ·, 3. a 2 and r , #, 2, 2, 2, 2. a 6 and r , , , , S, , a , (1 r), , 2, 2 2, , ·, 3, 1, 2, , c1 m, , 9, 0.9., 10, a 10, 10 100 ·, S, (1 r) (1 0.9) 1.9 19, , 4. a 10 and r , , , 3, 3, 3, 2 2, 2, 5. S e 3 5 … 3o e 2 4 6 … 3o, 5 5, 5, 5, 5, 5, a1, a2, , where ca1 2 , r1 1 m and ca2 3 , r2 1 m ·, , , 5, 25, 25, 25, (1 r1) (1 r2)
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490, , Senior Secondary School Mathematics for Class 11, , 6. 9 1/3 # 9 1/9 # 9 1/27 #… 3, 9, , , 1 1 1, b … 3l, 3 9 27, , , (1/3), 1, b1 l, 9 3, , a, , 9 (1 r) , where a , , 1, 1, and r , 3, 3, , 1, , 9 2 9 3., , 7. (ii) x 0.231231231… and 1000x 231.231231231… ., 231 ·, On subraction, we get 999x 231 & x , 999, 231 ·, Hence, 0.231 , 999, (iii) x 3.5222… & 10x 35.222… and 100x 352.222… ., 317 ·, , 90x 317 & x , 90, 8. x 0.125125… & 1000x 125.125125… ., 125 ·, 999x 125 & x , 999, 125 ·, Hence, 0.125125… , 999, 9. x 0.42323… & 10x 4.232323… and 1000x 423.2323… ., 419, ·, 990x (423 4) 419 & x , 990, 419 ·, Hence, 0.423 , 990, 10. x 2.1343434… & 10x 21.343434… and 1000x 2134.3434 … ., 2113 ·, 990x (2134 21) 2113 & x , 990, 2113 ·, Hence, 2.134 , 990, a , 1, 1, 2 , 2, 11. , 6 & , 6 & 1 r & r c1 m ·, 3, 3, 3, 1 r, 1 r, 12. Let a be the first term and r be the common ratio of the given series., 2, , Then,, , a , a, 400., 20 &, (1 r), (1 r) 2, , … (i), , And,, , a2 , 100., 2, (1 r ), , … (ii), , (1 r 2) 400, 3, 1r, a2, , #, & 4 & r ·, 2, 5, 100, 1 r, (1 r), a2, a , , 20 & a 8., 3, c1 m, 5, 24 72 , Hence, the required series is c8 , … 3m ·, 5 25, , , 13. Let a be the first term and r be the common ratio of the given GP., Then,, , 3, , a , a, (57) 3., 57 &, 3, 1r, (1 r), , … (i)
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Geometrical Progression, , 491, , And, a 3 a 3r 3 a 3r 6 … 3 9747, &, , a3 , 9747., (1 r 3), , , , (1 r 3) (57) 3, (1 r 3), 2, a3, , 19 & 1 r r2 19, #, &, 3, 3, 3, 9747, (1 r), a, (1 r), (1 r), , … (ii), , & 1 r r 2 19r 2 38r 19, & 18r 2 39r 18 0, & 18r 2 27r 12r 18 0, & (3r 2) (6r 9) 0 & r , , 2, 3, , 3, ;a r ! , as 1 r 1 for infinite GPE ·, 2, , , a, 2, c1 m, 3, , 57 & a c57 # 1 m 19., 3, , Hence, the required GP is 19,, , 38 , 76 , 152 ,, … 3., 3 9 27, , EXERCISE 12H, Very-Short-Answer Questions, 1. If the 5th term of a GP is 2, find the product of its first nine terms., 2. If the (p q)th and (p q)th terms of a GP are m and n respectively, find its, pth term., 3. If 2nd, 3rd and 6th terms of an AP are the three consecutive terms of a GP, then find the common ratio of the GP., 4. Write the quadratic equation, the arithmetic and geometric means of whose, roots are A and G respectively., 5. If a, b, c are in GP and a 1/x b 1/y c 1/z then prove that x, y, z are in AP., 6. If a, b, c are in AP and x, y, z are in GP then prove that the value of, , , , x b c · y c a · z a b is 1., 7. Prove that d1 , , 3, 1 1 1 1 , … 3n ·, 3 32 33 34, 4, , 8. Express 0.123 as a rational number., 9. Express 0.6 as a rational number., 10. Express 0.68 as a rational number., 11. The second term of a GP is 24 and its fifth term is 81. Find the sum of its first, five terms., 12. The ratio of the sum of first three terms is to that of first six terms of a GP is, 125 : 152. Find the common ratio., 39, and their product is 1. Find the, 13. The sum of first three terms of a GP is, 10, common ratio and these three terms.
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492, , Senior Secondary School Mathematics for Class 11, ANSWERS (EXERCISE 12H), , 1. 512, , 2., , 4. x 2 2Ax G 2 0, , 3. 3, , mn, , 123, 31, 2, 9., 10., 999, 3, 45, 5, 5, 5, 2, 2, 2, 13. r or ; , 1, or , 1,, 2, 2, 5, 2, 5 5, 8., , 11. 211, , 12., , 3, 5, , HINTS TO SOME SELECTED QUESTIONS, 1. Let first term a and common ratio r. Then,, T5 2 & ar 4 2., Required product, (a # ar # ar 2 #…# ar 8), a 9 # r (1 2 3 … 8) a 9r 36 (ar 4 ) 9 2 9 512., 2. Let first term a and common ratio r. Then,, Tp q m & ar (p, Tp q n & ar, , , 2 2 (p 1), , mn a r, , q 1), , (p q 1), , {ar, , m, n, , ( p 1) 2, , }, , & ar (p, , 1), , mn & Tp mn ., , 3. (a d), (a 2d), (a 5d) are in GP., a 2d a 5d, , & (a 2d) 2 (a d) (a 5d), ad, a 2d, & d 2 2ad 0 & d (d 2a) 0 & d 2a, 3a, So, these terms are a, 3a, 9a. Its common ratio 3., a, , [a d ! 0]., , 4. Let the roots of the quadratic equation be and ., Then, the equation is x 2 ( ) x 0., , , A, , , & 2A., 2, , G & G 2., Hence, the required equation is x 2 2Ax G 2 0., 5. Let a 1/x b 1/y c 1/z m. Then,, a m x, b m y and c m z., Since a, b, c are in GP, we have, b 2 ac & (m y ) 2 (m x # m z ), , , , & m 2y m x z & 2y x z., x, y, z are in AP., , 6. a, b, c are in AP & 2b a c., x, y, z are in GP & y 2 xz., , , x (b, , c), , · y (c, , a), , · z (a, , b), , x (b c) · ( xz ) (c a) · z (a b), , [a y xz ]
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Geometrical Progression, 1, , x (b c) · x 2, , , (c a), , 1, , · z2, , (c a), , · z (a, , b), , 1, 1 , (b c) (c a), (c a) (a b), 2, x, · z2, 1, , x2, , {2b (a c)}, , 1, , {c a 2b}, , x 0 # z 0 1., 1, ·, 7. This is an infinite GP in which a 1 and r , 3, a 1 3·, S3 , 1, 4, (1 r), c1 m, 3, · z2, , 8. x 0.123123… and 1000x 123.123123… ., 123 ·, 999x 123 & x , 999, 10. x 0.688… & 10x 6.888… and 100x 68.888 … ., 62 31 ·, 90x 62 & x , 90 45, 11. ar 24 and ar 4 81., , , , 3, ar 4 81 27, 27 3 3, 3, c m & r ·, & r , ar, 8, 8, 2, 2, 24, 3, 2, a#, 24 & a c24 # m 16., 2, 3, 5, a (r 1), , S5 , (r 1), , 12., , 35, 16 # )c m 13, 5, 5, 2, 32 #(3 2 ), , (243 32) 211., 5, 3, 2, c 1m, 2, , a (r 3 1), (r 1), 125 & 3 1, 125 & (r 3 1) 152 ·, #, 125, (r 1), a (r 6 1) 152, (r 1) 152, , 152 27 3 3, 3, 3, c m & r ·, r c, 1m, 5, 5, 125, 125, a,, 13. Let the three terms of the GP be r a and ar. Then,, a, 3, , 1 & a 1., r # a # ar 1 & a, 1,, So, the given terms are r 1 and r., 1 39, 1 39 29, , r 1 r 10 & r r c 10 1m 10, , , & 10r 2 10 29r, & 10r 2 29r 10 0, , & (2r 5) (5r 2) 0, 5, 2, or r ·, 2, 5, 5, 2, 2, 5, Hence, these terms are 1, or, 1, ·, 5, 2, 2, 5, & r, , 493
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494, , Senior Secondary School Mathematics for Class 11, , SUMMARY OF FACTS AND FORMULAE, 1., , (i) A sequence a1, a 2, a 3, …, an, … is called a geometrical progression, ak 1, (abbreviated as GP), if each term is nonzero and a r (constant) for, k, all k $ 1., The constant ratio r is called the common ratio., (ii) In a GP, we denote the first term by a, common ratio by r and nth term, by Tn ., , (iii) We have, Tn ar n 1., , (iv) nth term from the end of a GP with first term a, common ratio r and the, l, last term l is (n 1) ·, r, 2. For solving problems on GP, it is convenient to take, a, (i) 3 numbers in GP as r , a, ar., a a, (ii) 4 numbers in GP as 3 , r , ar, ar 3., r, a a, (iii) 5 numbers in GP as 2 , r , a, ar, ar 2., r, (iv) a, ar, ar 2, ar 3, … when their product is not given., 3. Sum of n terms of a GP with first term a and common ratio r is given by, Z]na, when r 1, ]], ]], n, ] a (1 r ), , when r 1, Sn ]], [] (1 r), ]] n, ]] a (r 1), ]], ,, ] (r 1) when r 1., \, 4. The sum of an infinite GP with first term a and common ratio r, where, a ·, |r| 1 is given by S , (1 r), 5. Properties of GP, Let a, ar, ar 2, ar 3, … be a GP. Then,, 1 1 1, 1, 1, 1, (i) a , ar , 2 , 3 , … is a GP with first term a and common ratio r ·, ar ar, (ii) ka, k (ar), k (ar 2), k (ar 3), … is a GP (where k ! 0) with first term ka and, common ratio r., k k k, k, k, (iii) a , ar , 2 , 3 , … is a GP (where k ! 0) with first term a and common, ar ar, 1, ratio r ·, (iv) a k, (ar) k, (ar 2) k, … is a GP (where k ! 0) with first term a k and common, ratio r k.,
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Some Special Series, , 495, , 13, , Some Special Series, SUM OF FIRST n NATURAL NUMBERS, THEOREM 1, , Prove that (1 2 3 … n) , n, , i.e., / k , k1, , PROOF, , 1, n ( n 1), 2, , 1, n (n 1) ., 2, , Consider the series (1 2 3 … n) ., This is an arithmetic series in which a 1, d 1 and l n., n, 1, 1, ;a Sn (a l )E ·, Sn (1 n) & Sn n (n 1), 2, 2, 2, n, 1, , , Hence, / k, n (n 1) ., 2, k1, , SUM OF THE SQUARES OF FIRST n NATURAL NUMBERS, THEOREM 2, , Prove that (1 2 2 2 3 2 … n 2) , n, , i.e., / k 2 , k1, , PROOF, , 1, n (n 1) (2n 1), 6, , 1, n (n 1)(2n 1) ., 6, , Let Sn 1 2 2 2 3 2 … n 2., Consider the identity: k 3 (k 1) 3 3k 2 3k 1., Putting k 1, 2, 3, …, (n 1), n successively in (i), we get, , ... (i), , 1 3 0 3 3 · 1 2 3 · 1 1;, 2 3 1 3 3 · 2 2 3 · 2 1;, 3 3 2 3 3 · 3 2 3 · 3 1;, … … … … …, … … … … …, (n 1) 3 (n 2) 3 3 · (n 1) 2 3 · (n 1) 1;, n 3 (n 1) 3 3 · n 2 3 · n 1., Adding columnwise, we get, , &, &, , (n 3 0 3) 3(1 2 2 2 3 2 … n 2) 3(1 2 3 … n) n, n, 1, 1, n 3 3 · / k 2 3 · n (n 1) n ;a (1 2 3 … n) n (n 1)E, 2, 2, k1, n, 3, 1, 3 d / k 2n n 3 n (n 1) n (2n 3 3n 2 n), 2, 2, k1, 495
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496, , Senior Secondary School Mathematics for Class 11, n, 1, 1, 1, 2, 3, 2, 2, < / k F 6 (2n 3n n) 6 n (2n 3n 1) 6 n (n 1) (2n 1) ., , k 1, n, 1, Hence, / k 2 n (n 1) (2n 1) ., 6, k1, , &, , SUM OF THE CUBES OF FIRST n NATURAL NUMBERS, THEOREM 3, , PROOF, , 2, 1, Prove that (1 3 2 3 3 3 … n 3) ' n (n 1)1, 2, 2, n, 1, 3, i.e., / k ' n (n 1)1 ·, 2, k1, , Let Sn 1 3 2 3 3 3 … n 3., Consider the identity: k 4 (k 1) 4 4k 3 6k 2 4k 1., Putting k 1, 2, 3, …, (n 1), n successively in (i), we get, , … (i), , 14 04 4 · 13 6 · 12 4 · 1 1, 24 14 4 · 23 6 · 22 4 · 2 1, 34 24 4 · 33 6 · 32 4 · 3 1, …, …, …, …, … …, …, …, …, …, … …, ( n 1) 4 ( n 2 ) 4 4 · ( n 1 ) 3 6 · ( n 1 ) 2 4 · ( n 1 ) 1, n 4 (n 1) 4 4 · n 3 6 · n 2 4 · n 1., Adding columnwise, we get, (n 4 0 4) 4 · {1 3 2 3 3 3 … n 3} 6 · {1 2 2 2 3 2 … n 2), 4 · {1 2 3 … n} (1 1 1 … to n terms), n, , n, , n, , k1, , k1, , &, , n 4 4 · d / k 3n 6 · d / k 2n 4 · d / kn n, , &, , 1, 1, n 4 · d / k n 6 · n (n 1) (2n 1) 4 · n (n 1) n, 6, 2, k1, , k1, n, , 4, , 3, , n, n, 1, 1, 2, (a / k n (n 1) (2n 1) and / k n (n 1)2, 6, 2, k1, k1, n, , &, , 4 · d / k 3n n 4 n (n 1) (2n 1) 2n (n 1) n, k1, , n, , &, , , d/ k n, 3, , k1, n, , 2, 1 · 4 3 2 1 2 2 1, (n 2n n ), n (n 2n 1) ' n (n 1)1, 2, 4, 4, , 2, 1, 3, d / k n ' n (n 1)1, 2, k1, , 2, 1, i.e., (1 3 2 3 3 3 … n 3) ' n (n 1)1 ·, 2, , SUM OF THE CUBES OF FIRST n ODD NATURAL NUMBERS, THEOREM 4, , Prove that [1 3 3 3 5 3 … (2n 1) 3] n 2 (2n 2 1), n, , i.e., / (2k 1) 3 n 2 (2n 2 1) ., k1
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Some Special Series, , 497, , Let Sn 1 3 3 3 5 3 … (2n 1) 3., , PROOF, , n, , Then, Sn / (2k 1) 3, k1, n, , / (8k 3 12k 2 6k 1), k1, , n, , n, , n, , k1, , k1, , k1, , 8 d / k 3n 12 d / k 2n 6 d / kn n [a 1 1 … to n terms = n], 8 · 1 n 2 (n 1) 2 12 · 1 n (n 1) (2n 1) 6 · 1 n (n 1) n, 6, 2, 4, 2n 2 (n 1) 2 2n (n 1) (2n 1) 3n (n 1) n, n (n 1) · {2n (n 1) 2 (2n 1) 3} n, n (n 1) (2n 2 2n 1) n n · {(n 1) (2n 2 2n 1) 1}, n (2n 3 n) n 2 (2n 2 1) ., , , {1 3 3 5 3 … (2n 1) 3} n 2 (2n 2 1) ., 3, , SUMMARY, , (i) (1 2 3 … n) , , 1, n (n 1), 2, , n, 1, i.e., d / kn n (n 1)., 2, k1, 2, 2, 2, 2, (ii) (1 2 3 … n ) , , 1, n (n 1) (2n 1), 6, , n, 1, i.e., d / k 2n n (n 1) (2n 1)., 6, k1, 2, 1, 3, 3, 3, 3, (iii) (1 2 3 … n ) ' n (n 1)1, 2, 2, n, 1, i.e., d / k 3n ' n (n 1)1 ·, 2, k1, , (iv) {1 3 3 3 5 3 … (2n 1) 3} n 2 (2n 2 1), n, , i.e., / (2k 1) 3 n 2 (2n 2 1)., k1, , SOLVED EXAMLES, EXAMPLE 1, SOLUTION, , If S1, S 2 and S 3 are the sums of first n natural numbers, their squares and, their cubes respectively then show that 9S 22 S 3 (1 8S1) ., We have, 1, S1 (1 2 3 … n) & S1 n (n 1);, 2, 1, 2, 2, 2, 2, , , , , , S 2 (1 2 3 … n ) & S 2 n (n 1) (2n 1);, 6, 1, 2, 3, 3, 3, and S 3 (1 2 3 … n ) & S 3 n 2 (n 1) 2., 4
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498, , Senior Secondary School Mathematics for Class 11, , 1 · 2 2, 1, n (n 1) (2n 1) 2 n 2 (n 1) 2 (2n 1) 2., 36, 4, 1 2 2· ·1, , , And, S 3 (1 8S1), n (n 1) '1 8 n (n 1)1, 2, 4, 1 n 2 (n 1) 2 (4n 2 4n 1) 1 n 2 (n 1) 2 (2n 1) 2., 4, 4, Hence, 9S 22 S 3 (1 8S1) ., , , EXAMPLE 2, SOLUTION, , 9S 22 9 #, , Find the sum to n terms of the series whose nth term is n (n 3) ., We have, Tk k (k 3) (k 2 3k) ., , sum to n terms is given by, n, , Sn / Tk, k1, n, , n, , n, , k1, , k1, , k1, , / (k 2 3k) / k 2 3 / k, 1 n (n 1)(2n 1) 3 · 1 n (n 1) 1 {n (n 1)(2n 1) 9n (n 1)}, 6, 2, 6, 1, 1, n (n 1) (2n 1 9) n (n 1) · 2(n 5) 1 n (n 1) (n 5) ., 6, 6, 3, 1, Hence, the required sum is n (n 1) (n 5) ., 3, EXAMPLE 3, SOLUTION, , Find the sum to n terms of the series whose nth term is (2n 1) 2., We have, Tk (2k 1) 2 (4k 2 4k 1) ., , sum to n terms is given by, n, , Sn / Tk, k1, n, , n, , n, , k1, , k1, , k1, , / (4k 2 4k 1) 4 · / k 2 4 · / k n, [a 1 1 … n times n], 1, 1, 4 · n (n 1) (2n 1) 4 · n (n 1) n, 6, 2, 2, n (n 1) (2n 1) 2n (n 1) n, 3, 1, · {2n (n 1) (2n 1) 6n (n 1) 3n}, 3, 1, [n (n 1) {2 (2n 1) 6} 3n] 1 · [4n (n 1) (n 1) 3n], 3, 3, 1, 1, 3, 2, (4n n) n (4n 1) ., 3, 3, 1, Hence, the required sum is n (4n 2 1) ., 3, EXAMPLE 4, SOLUTION, , Find the sum to n terms of the series whose nth term is (n 2 2 n) ., We have, Tk (k 2 2 k ) ., , , n, , Sn / Tk, k1, n, , n, , n, , k1, , k1, , k1, , / (k 2 2 k ) / k 2 / 2 k
Page 510 : Some Special Series, , 499, , 1 n (n 1) (2n 1) (2 2 2 2 3 … 2 n), 6, n, 1, 2, <a / k 6 n (n 1) (2n 1)F, k1, 2 ( 2 n 1), 1 n (n 1) (2n 1) , [ 2 2 2 … 2 n is a GP], 6, ( 2 1), 1 n (n 1) (2n 1) 2 (2 n 1) ., 6, 1, Hence, the required sum is n (n 1) (2n 1) 2 (2 n 1) ., 6, EXAMPLE 5, SOLUTION, , Sum the series 3 · 8 6 · 11 9 · 14 … to n terms., We have, Tk (kth term of 3, 6, 9, …) # (kth term of 8, 11, 14, …), {3 (k 1)# 3}#{8 (k 1)# 3} 3k (3k 5), (9k 2 15k) ., , , n, , Sn / Tk, k1, n, , n, , n, , k1, , k1, , k1, , / (9k 2 15k) 9 d / k 2n 15 d / kn, 9 · '1 n (n 1) (2n 1)1 15 · '1 n (n 1)1, 6, 2, 3, n (n 1) {(2n 1) 5} 3n (n 1) (n 3) ., 2, Hence, the required sum is 3n (n 1) (n 3) ., EXAMPLE 6, SOLUTION, , Sum the series 1 · 2 · 3 2 · 3 · 4 3 · 4 · 5 … to n terms., We have, Tk (kth term of 1, 2, 3, …) # (kth term of 2, 3, 4, …), # (kth term of 3, 4, 5, …), { 1 ( k 1) # 1} { 2 ( k 1 ) # 1 } # {3 (k 1 ) # 1 }, k (k 1) (k 2) (k 3 3k 2 2k)., , , n, , Sn / Tk, k1, n, , n, , n, , n, , k1, , k1, , k1, , k1, , / (k 3 3k 2 2k) / k 3 3 / k 2 2 / k, 1 n 2 (n 1) 2 3 · 1 n (n 1) (2n 1) 2 · 1 n (n 1), 6, 2, 4, 2 n, n, 1, 1, 3, 2, )a / k ' n (n 1)1 , / k n (n 1) (2n 1),, 2, 6, , , k 1, k 1, n, , /k, , k1, , 1 n (n 1) {n (n 1) 2 (2n 1) 4}, 4, , 1, n (n 1)2, 2
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500, , Senior Secondary School Mathematics for Class 11, , 1 n (n 1) (n 2 5n 6) 1 n (n 1) (n 2) (n 3) ., 4, 4, 1, Hence, the required sum is n (n 1) (n 2) (n 3) ., 4, EXAMPLE 7, SOLUTION, , Find the sum of the series 1 2 3 2 5 2 … to n terms., We have, 2, 2, 2, Tk kth term of (1 3 5 …), {1 (k 1)# 2} 2 (2k 1) 2 (4k 2 4k 1)., , , n, , Sn / Tk, k1, n, , / (4k 2 4k 1), k1, n, , n, , 4 / k 2 4 / k (1 1 … n times), k1, , k1, , 4 · 1 n (n 1) (2n 1) 4 · 1 n (n 1) n, 6, 2, n · {2 (n 1) (2n 1) 6 (n 1) 3} n (4n 2 1) ., 3, 3, n, 2, Hence, the required sum is (4n 1) ., 3, EXAMPLE 8, SOLUTION, , Find the sum 1 2 (1 2 2 2) (1 2 2 2 3 2) … to n terms., We have, 1, 1, Tk (1 2 2 2 3 2 … k 2) ' k (k 1) (2k 1)1 (2k 3 3k 2 k) ., 6, 6, , , n, , Sn / Tk, k1, , n, n, n, n, 1 · / (2k 3 3k 2 k) 1 · 2 · / k 3 1 · 3 · / k 2 1 · / k, 6 k1, 6, 6, 6 k1, k1, k1, n, n, n, 1 · / k3 1 · / k2 1 · / k, 3 k1, 2 k1, 6 k1, 2, 1 · '1 n (n 1)1 1 · 1 n (n 1) (2n 1) 1 · 1 n (n 1), 3 2, 2 6, 6 2, 1, 1, 1, 2, 2, , n (n 1) n (n 1) (2n 1) n (n 1), 12, 12, 12, 1 n (n 1) · {n (n 1) (2n 1) 1} 1 n (n 1) 2 (n 2) ., 12, 12, 1, 2, Hence, the required sum is n (n 1) (n 2) ., 12, , 1# 2 2 2 # 3 2 … n #(n 1) 2, , , 3n 5 ·, 3n 1, , EXAMPLE 9, , Prove that, , SOLUTION, , We have, kth term of the numerator k (k 1) 2 (k 3 2k 2 k) ., , 1 # 2 2 2 # 3 … n 2 #(n 1), 2, , kth term of the denominator k 2 (k 1) (k 3 k 2) .
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Some Special Series, n, , LHS , , k1, , n, , / ( k k 2), 3, , n, , n, , d / k n 2 · d / k n d / kn, , n, , / (k 3 2k 2 k), , , , 501, , , , 3, , k1, , k1, , 2, , k1, , n, , k1, , n, , d / k nd / k n, k1, , 3, , 2, , k1, , 1 2 2 · 1, 1, n (n 1) 2 n (n 1) (2n 1) n (n 1), 6, 2, 4, 1 2 21, n ( n 1), n (n 1) (2n 1), 6, 4, n, n, 1, 1, 2, )a d / kn 2 n (n 1); d / k n 6 n (n 1) (2n 1), k1, k1, n, 1 2, 3, 2, and d / k n n (n 1) 1, 4, k1, , n (n 1) (n 2) (3n 5), 12, #, 12, n (n 1) (n 2) (3n 1), [on simplifying], (3n 5), , RHS., (3n 1), , , EXAMPLE 10, , Find the sum of n terms of the series, 1 1 , 1, … ., (2 # 5) (5 # 8) (8 #11), , SOLUTION, , We have, 1, (kth term of 2, 5, 8, …)#(kth term of 5, 8, 11, …), 1, , { 2 ( k 1) # 3} # { 5 ( k 1 ) # 3 }, 1, 1, 1( 1, , , 2·, (3k 1) (3k 2) 3 (3k 1) (3k 2), 1, 1, 1, , , Tk (, 2·, 3 (3k 1) (3k 2), Putting k 1, 2, 3, …, n successively in (i), we get, Tk , , 1 1 1, T1 b l, 3 2 5, 1 1 1, T2 b l, 3 5 8, 1, 1 1, T3 b l, 3 8 11, … … … …, … … … …, 1, 1, 1, , (, 2·, 3 (3n 1) (3n 2), Adding columnwise, we get, Sn (T1 T2 T3 … Tn), Tn , , n, 1 b1 1 l , ·, 3 2 3n 2, 2 (3n 2), , … (i)
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502, EXAMPLE 11, SOLUTION, , Senior Secondary School Mathematics for Class 11, , Find the sum of n terms of the series 5 11 19 29 41 … ., Let Sn 5 11 19 29 … Tn 1 Tn, Sn , 5 11 19 … Tn 2 Tn 1 Tn, On subtraction, we get, 0 5 [6 8 10 12 … to (n 1) terms] Tn, (n 1)#[2 # 6 (n 2)# 2], &, Tn 5 , 2, &, Tn 5 (n 1) (n 4), &, , Tn n 2 3n 1., , , , Sn / Tk / (k 2 3k 1) / k 2 3 / k n, , n, , n, , n, , n, , k1, , k1, , k1, , k1, , 1 n (n 1) (2n 1) 3 # 1 n (n 1) n, 6, 2, 1, n (n 2) (n 4) ., 3, EXAMPLE 12, SOLUTION, , Find the sum of the first n terms of the series 3 7 13 21 31 … ., Let Sn 3 7 13 21 31 … Tn 1 Tn, Sn , 3 7 13 21 … Tn 2 Tn 1 Tn, On subtraction, we get, 0 3 [4 6 8 10 … to (n 1) terms] Tn, &, Tn 3 [4 6 8 … to (n 1) terms], (n 1), 3, #[2 # 4 (n 2)# 2], 2, 3 (n 1) (n 2) (n 2 n 1)., , , n, , n, , k1, n, , k1, n, , k1, , k1, , Sn / Tk / (k 2 k 1), / k2 / k n, , [a 1 1 1 … to n terms n], , 1 n (n 1) (2n 1) 1 n (n 1) n 1 n (n 2 3n 5) ., 6, 2, 3, EXAMPLE 13, SOLUTION, , Find the 50th term of the series 2 3 6 11 18 … ., Let, Sn 2 3 6 11 18 … Tn 1 Tn ., Again Sn , 2 3 6 11 … Tn 1 Tn ., On subtracting, we get, 0 2 [1 3 5 7 … to (n 1) terms] Tn, &, Tn 2 [1 3 5 7 … to (n 1) terms], (n 1), · [2 #1 (n 2)# 2] (n 1) 2 2., 2, T50 {(50 1) 2 2} {(49) 2 2} (2401 2) 2403., 2, , &
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Some Special Series, EXAMPLE 14, , SOLUTION, , 503, , Find the sum of the series, (3 3 2 3) (5 3 4 3) (7 3 6 3) … to (i) n terms, We have, Tk {(2k 1) 3 (2k) 3}, , (ii) 10 terms., , {8k 3 1 6k (2k 1) 8k 3} (12k 2 6k 1)., (i) sum to n terms is given by, n, , n, , k1, , k1, , Sn 12 / k 2 6 / k n, 12 # 1 n (n 1) (2n 1) 6 # 1 n (n 1) n, 6, 2, 2n (n 1) (2n 1) 3n (n 1) n, n (n 1) [4n 2 3] n n [(n 1) (4n 5) 1], n (4n 2 9n 6) (4n 3 9n 2 6n) ., (ii) Sum to 10 terms is given by, S10 10 #[4 #10 2 9 #10 6] 4960., , EXERCISE 13A, Find the sum of the series whose nth term is given by:, 1. (3n 2 2n), , 2. n (n 1) (n 4), , 3. (4n 6n 2n), , 4. (3n 2 3n 2), , 5. (2n 2 3n 5), , 6. (n 3 3 n), , 3, , 2, , Find the sum of the series:, 7. (2 2 4 2 6 2 8 2 … to n terms), 8. (2 3 4 3 6 3 8 3 … to n terms), 9. (5 2 6 2 7 2 … 20 2), 10. 1# 2 2 # 3 3 # 4 4 # 5 … to n terms, 11. 3 # 8 6 #11 9 #14 … to n terms, 12. 1# 2 2 2 # 3 2 3 # 4 2 … to n terms, 13. 1# 2 2 3 # 3 2 5 # 4 2 … to n terms, 14. 3 #1 2 5 # 2 2 7 # 3 2 … to n terms, 15. (1# 2 # 3) (2 # 3 # 4) (3 # 4 # 5) … to n terms, 16. (1# 2 # 4) (2 # 3 #7) (3 # 4 #10) … to n terms, 17., , 1 1 1 , … to n terms, (1# 2) (2 # 3) (3 # 4), , 18., , 1 1 1 , 1, …, (1# 3) (3 # 5) (5 #7), ( 2n 1) ( 2n 1), , 19., , 1 , 1, 1, 1, , …, (1# 6) (6 #11) (11#16), (5n 4) (5n 1)
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504, , 20., 21., 22., 23., 24., , Senior Secondary School Mathematics for Class 11, 3, 3, 3, 13 13 23 1 2 3 , … to n terms, , , , 1, 1 3, 1 3 5, 3 15 35 63 … to n terms, 1 5 12 22 35 … to n terms, 5 7 13 31 85 … to n terms, (1 2 3 … k ), , prove that, If S k , k, 1, (S 12 S 22 S 32 … S n2) , n (2n 2 9n 13)., 24, , 25. If Sn denotes the sum of the cubes of the first n natural numbers and sn, n Sk, denotes the sum of the first n natural numbers then find the value of / s ·, k1 k, ANSWERS (EXERCISE 13A), , 1., 4., 7., 10., 13., 16., , 1, n (n 1) (2n 3), 2, , 1, 3. n (n 1) 2 (n 2), n (n 1) (n 2) (3n 17), 12, 3, n, 1, 5. (4n 2 3n 23), 6. n 2 (n 1) 2 (3 n 1), 2n (n 2 1), 6, 2, 4, 2, 8. 2n 2 (n 1) 2, 9. 2840, n (n 1) (2n 1), 3, n (n 1) (n 2) (3n 5), 1, 11. 3n (n 1) (n 3), 12., n (n 1) (n 2), 3, 12, n (n 1) (3n 2 5n 1), n (n 1) (n 2) (n 3), n 3 2 , 15., (n 4n 4n 1) 14., 6, 2, 4, 2, n (n 1) (3n 19n 14), n, n, n, 17., 18., 19., 12, (n 1), (2n 1), (5n 1), 2., , 1, n (4n 2 6n 1), 3, 1, 25. n (n 1) (n 2), 6, , 1, n (2n 2 9n 13), 24, 1, 23. (3 n 8n 1), 2, 20., , 21., , 22., , 1 2 , n (n 1), 2, , HINTS TO SOME SELECTED QUESTIONS, 4. Tn (3n 3n 2)., 2, , , , n, , n, , k1, , k1, , Sn 3 d / k 2n 3 d / kn 2n, , [a 2 2 2 … n times 2n]., , 6. Tn (n 3 3 n) ., n, , n, , k1, , k1, , 1 2 2 2 n, n (n 1) {3 3 … 3 }, 4, _, n, ]Z], 1) bb, 1 2, 2 3 (3, `b ·, & Sn []] n (n 1) , 4, (3 1) b, \, a, 7. Tn {2 (n 1)# 2} 2 (2n) 2 4n 2., n, n, 1, Sn / Tk 4 d / k 2n 4 # n (n 1) (2n 1) ., 6, , , k 1, k 1, , , Sn d / k 3n d / 3 kn
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Some Special Series, 8. Tn {2 (n 1)# 2} 3 (2n) 3 8n 3., , , n, 1, Sn 8 ( / k 32 8 # n 2 (n 1) 2 2n 2 (n 1) 2., 4, k1, , 9. Given series (1 2 2 2 3 2 4 2 5 2 6 2 … 20 2) (1 2 2 2 3 2 4 2), '1 n (n 1) (2n 1), when n 201 30, 6, c 1 # 20 # 21# 41m 30 2840., 6, 10. Tn n (n 1) n 2 n., , , n, , n, , k1, , k1, , 2, Sn d / k n d / kn ·, , 11. Tn (nth term of 3, 6, 9, …)#(nth term of 8, 11, 14, …), {3 (n 1)# 3}#{8 (n 1)# 3} 3n (3n 5) 9n 2 15n., , , n, , n, , k1, , k1, , 2, Sn 9 d / k n 15 d / kn ·, , 12. Tn (nth term of 1, 2, 3, …)#(nth term of 2 2, 3 2, 4 2, …), n#(n 1) 2 n#(n 2 2n 1) n 3 2n 2 n., , , n, , n, , n, , k1, , k1, , k1, , 3, 2, Sn d / k n 2 d / k n d / kn ·, , 13. Tn (nth term of 1, 3, 5, …)#(nth term of 2 2, 3 2, 4 2, …), {1 (n 1)# 2}#(n 1) 2 (2n 1) (n 2 2n 1) (2n 3 3n 2 1) ., , , n, , n, , k1, , k1, , Sn 2 d / k 3n 3 d / k 2n n., , 14. Tn (nth term of 3, 5, 7, …)#(nth term of 1 2, 2 2, 3 2, …), {3 (n 1)# 2}# n 2 (2n 1) n 2 (2n 3 n 2) ., , , n, , n, , k1, , k1, 3, , 3, 2, Sn 2 d / k n d / k n ·, , 15. Tn n (n 1) (n 2) (n 3n 2 2n) ., , , n, , n, , n, , k1, , k1, , k1, , 3, 2, Sn d / k n 3 d / k n 2 d / kn ·, , 16. Tn (nth term of 1, 2, 3, …)#(nth term of 2, 3, 4, …)#(nth term of 4, 7, 10, …), n (n 1)#[4 (n 1)# 3] n (n 1) (3n 1) (3n 3 4n 2 n) ., A (n 1) Bn (A B) n A, 1, A B , , ·, 17. Tn , n (n 1), n (n 1), n (n 1) n n 1, A B 0 and A 1 & A 1 and B 1, 1, 1, Tn c n m, n 1, 1 1, T1 c m, 1 2, 1 1, , T2 c m, 2 3, 1 1, T3 c m, 3 4, … … … …, … … … …, , 505
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506, , Senior Secondary School Mathematics for Class 11, 1, 1, Tn 1 c n m, n 1, 1, 1, Tn c n m ·, n 1, Adding columnwise, we get, , 1 (n 1) 1 n ·, m, n1, n1, n1, A (2n 1) B (2n 1) 2 (A B) n (A B), 1, A, B, , , , , ·, 18. Tn , (2n 1) (2n 1) 2n 1 2n 1, (2n 1) (2n 1), (2n 1) (2n 1), 1, 1, ·, 2 (A B) 0 and A B 1 & A B 0, A B 1 & A , B , 2, 2, 1, 1, 1, 1·, Tn # ' , 2, 2n 1 2n 1, 1, 1 1, T1 #c m, 2, 1 3, 1, 1 1, T2 #c m, 2, 3 5, 1, 1, 1, T3 #c m, 2, 5 7, … … … … …, … … … … …, 1, 1, 1, 1, Tn 1 # ' , 2, 2n 3 2n 1, 1, 1, 1, 1·, Tn # ' , 2, 2n 1 2n 1, Adding columnwise, we get, n, 2n 1 1 , 1, 1, ·, m, Sn (T1 T2 T3 … Tn) c1 , 2, 2n 1, 2 (2n 1) (2n 1), Sn (T1 T2 T3 … Tn) c1 , , 19. Tn , , , 1, 1, 1, 1·, #' , 5, 5n 4 5n 1, 1, 1 1, T1 # ' 1, 5, 1 6, 1, 1 1, , #' 1, T2, 5, 6 11, 1, 1, 1, T3 # ' 1, 5, 11 16, … … … … …, … … … … …, , 1, 1, 1, 1, #' , 5, 5n 9 5n 4, 1, 1, 1, 1·, Tn # ' , 5, 5n 4 5n 1, Adding columnwise, we get, (5n 1 1), n, 1, 1, , ·, m, Sn (T1 T2 … Tn) c1 , 5, 5n 1, 5 (5n 1), (5n 1), 2, 1, 2, ' n (n 1)1, 3, (n 1), 2, 23 33 … n3, 1, , , 1 n2 1 n 1 ·, 20. Tn , 2, 4, 4, 4, 1 3 5 … (2n 1) n {1 (2n 1)}, 2, Tn 1 , , , , n, , Sn / Tk, k1, , n, n, 1 d / k 2n 1 d / kn 1 n., 2 k1, 4 k1, 4
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Some Special Series, , 507, , 21. Sn 3 15 35 63 … Tn 1 Tn, Sn , 3 15 35 … Tn 2 Tn 1 Tn ., On subtracting (ii) from (i), we get, 0 3 {12 20 28 … to (n 1) terms} Tn, 1, 2, & Tn 3 (n 1) {2 #12 (n 2)# 8} (4n 1), 2, n 2, 1, 1, 2, & Sn 4 d / k n n '4 # n (n 1) (2n 1)1 n n (4n 6n 1) ., 6, 3, k1, 24. S k , , , … (i), … (ii), , (1 2 3 … k) k (k 1) k 1, , , ·, 2, k, 2k, n, n, n, 1, 1, 1 n, 1, / S k2 / (k 1) 2 d / k 2n d / kn (1 1 … n times), 2 k1, 4, 4, 4, , , , k 1, k 1, k 1, 1 # 1 n (n 1) (2n 1) 1 n (n 1) 1 n, 4 6, 4, 4, n, 2, , [on simplification]., (2n 9n 13), 24, , 2, k, k, 1, 1, 25. S k ) / r 33 ' k (k 1)1 and s k ) / r3 k (k 1) ., 2, 2, r1, r1, , , , &, , 2, 1, , S k ' 2 k (k 1)1, , 1 k (k 1) 1 k 2 1 k, sk, 2, 2, 2, 1, , k (k 1), 2, n Sk, 1 n 2 1 n, / s ( / k 2 ( / k2, 2 k1, 2 k1, k, k1, , 1 # 1 # n (n 1) (2n 1) 1 # 1 n (n 1), 2 6, 2 2, 1 {n (n 1) (2n 1) 3n (n 1)} 1 n (n 1) (n 2) ., 6, 12, , EXERCISE 13B, Very-Short-Answer Questions, 1. Find the sum (2 4 6 8 … 100) ., 2. Find the sum (41 42 43 … 100) ., 3. Find the sum {(11) 2 (12) 2 (13) 2 … (20) 2} ., 4. Find the sum {(6) 3 (7) 3 (8) 3 (9) 3 (10) 3} ., n, , n, , k1, , k1, , 5. If / k 210, find the value of / k 2., n, , n, , k1, , k1, , 6. If / k 45, find the value of / k 3., 7. Find the sum of the series {2 2 4 2 6 2 … (2n) 2} ., 8. Find the sum of 10 terms of the geomeric series 2 6 18 … ., 9. Find the sum of n terms of the series whose rth term is (r 2 r ) .
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508, , Senior Secondary School Mathematics for Class 11, ANSWERS (EXERCISE 13B), , 1. 2550, , 2. 4230, , 3. 2485, , 4. 2800, , 5. 2870, , 6. 2025, , 1, , 8. 121 ( 6 2 ) 9. n (n 1) 2 n 1 2, 2, , 2, 7. n (n 1) (2n 1), 3, , HINTS TO SOME SELECTED QUESTIONS, 1., , 2., , 3., , 4., , 1, We know that (1 2 3 … n) n (n 1) ., 2, 1, , , , , , , (2 4 6 8 … 100) 2 #(1 2 3 … 50) c2 # # 50 # 51m 2550., 2, Given sum (1 2 3 … 40 41 … 100) (1 2 3 … 40), c 1 #100 #101m c 1 # 40 # 41m (5050 820) 4230., 2, 2, 1, We know that (1 2 2 2 3 2 … n 2) n (n 1) (2n 1) ., 6, given sum {1 2 2 2 … (10) 2 (11) 2 … (20) 2} {1 2 2 2 … (10) 2}, c 1 # 20 # 21# 41m c 1 #10 #11# 21m (2870 385) 2485., 6, 6, 2, 1, 3, 3, 3, 3, We know that (1 2 3 … n ) ' n (n 1)1 ·, 2, given sum {1 3 2 3 3 3 4 3 5 3 6 3 … (10) 3} {1 3 2 3 3 3 4 3 5 3}, , 2, 2, c 1 #10 #11m c 1 # 5 # 6m (55) 2 (15) 2 (3025 225) 2800., 2, 2, 1, 5. (1 2 3 … n) 210 & n (n 1) 210 & n 2 n 420 0, 2, & (n 21) (n 20) 0 & n 20., 1, (1 2 2 2 3 2 … n 2) n (n 1) (2n 1), where n 20, 6, c 1 # 20 # 21# 41m 2870., 6, 1, , , , , , 6. Given (1 2 3 … n) 45 & n (n 1) 45., 2, 2, n, 1, d / k 3n ' n (n 1)1 (45) 2 2025., 2, k1, , 7. Tk (2k) 2 4k 2., , , n, n, 1, 2, Sn / Tk 4 d / k 2n '4 # n (n 1) (2n 1)1 n (n 1) (2n 1) ., 6, 3, k1, k1, , 8. Given series is a geometric series in which a 2 , r 3 and n 10., a (r n 1), , (r 1), , 2 {( 3 ) 10 1}, , , , S, , &, , S 121 ( 6 2 ) ., , ( 3 1), , n, , n, , r1, n, , r1, n r, , #, , ( 3 1), ( 3 1), , , , 2 (243 1) ( 3 1), 2, , 9. Required sum / Tr / (r 2 r), / r / 2 1 n (n 1) (2 2 2 2 3 … 2 n), 2, r1, r1, 2 (2 n 1) 1, 1, n (n 1) 2 n 1 2., n (n 1) , 2, 2, (2 1), ,
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14, Measurement of Angles, TRIGONOMETRY, , The literal meaning of the word trigonometry is the science of measuring the, sides and the angles of triangles. However, in modern times it is used in a, wider context., , Te, rm, , in, , al, s, , id, , e, , We define it as the branch of mathematics which deals with the measurements of, angles of triangles and the problems related to these angles., →, B, ANGLE When a ray OA starting from its initial, position OA rotates about its end point O and, takes the final position OB, we say that angle AOB, (written as ∠AOB) has been formed., °, , Here, OA and OB are respectively known as, the initial side and the terminal side of ∠AOB, and, the point O is called its vertex., , O, (Vertex), , Initial side, , A, , The amount of rotation from the initial side to the terminal side is called, the measure of the angle., POSITIVE AND NEGATIVE ANGLES An angle formed by a rotating ray is said to be, positive or negative depending on whether it moves in an anticlockwise or a, clockwise direction respectively., , B, rm, Te, °, , O, , id, ls, ina, , O, , e, , Initial side, , Te, rm, , A, , ina, ls, , ide, , B, , A, , (Positive angle), MEASURING ANGLES, , Initial side, °, , (Negative angle), , Mainly, there are two systems of measuring angles., , (i) Sexagesimal System (Degree Measure), , The angle traced by a moving line about a point from its initial position to the, terminating position in making 1 360 of the complete revolution of a circle is said to have, a measure of 1 degree, written as 1°., 509
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510, , Senior Secondary School Mathematics for Class 11, , 1°, , is called a minute, written as 1′., 60, 1′, The one-sixtieth part of a minute, i.e.,, , is called a second, written as 1′′., 60, Also, we define 1 right angle = 90°., An angle measuring 180° is called a straight angle., The measures of the angles given below in (i), (ii), (iii), (iv), (v), (vi) are, respectively 40° , 90° , 230°, 420° , −30° and −230°., The one-sixtieth part of a degree, i.e.,, , B, , 230° O, , B, , 40°, , A, , 90°, , O, , O, , A, (i), , B, , A, , ( iii ), , ( ii ), , B, O, , 420°, O, , 30°, , A, , ( iv ), , B, , A, , B, , A, , O, 230°, ( vi ), , (v), , B, , (ii) Circular System (Radian Measure), , In this system, an angle is measured in radians., A radian is an angle subtended at the centre of a, circle by an arc whose length is equal to the radius of, the circle., We denote 1 radian by 1c ., , r, O, , r, , 1c, , A, , r, , A radian is a constant angle., Let us consider a circle with centre O and radius r., Let AB be an arc of length r. Join OA and, OB., , THEOREM 1, , PROOF, , B, r, , Then, ∠AOB = 1c ., Produce AO to meet the circle at C., Then, ∠AOC = a straight angle = 2 rt. O., Since the angles at the centre of a circle, are proportional to the lengths of the arcs, subtending them, we have, , C, , 1c, O, , r, , A
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Measurement of Angles, , 511, , ∠AOB, arc AB, r, 1, =, =, =, ∠AOC arc ABC 1 ⋅ ( 2πr) π, 2, ∠AOC 2rtO, =, = constant, ⇒ ∠AOB =, π, π, ⇒ 1c = constant., Hence, a radian is a constant angle., In a circle of radius r, if an arc of length l subtends an angle θ c at the, l, centre then θ = ⋅, r, Let us consider a circle with centre O and radius r., , THEOREM 2, , PROOF, , Let ∠AOB = θ c and let arc AB = l., Let C be a point on AB such that arc AC = r., , B, C, , Then, ∠AOC = 1 ., c, , r, , We know that the angles at the centre of a, circle are proportional to the lengths of the, arcs subtending them., ∠AOB arc AB l, ∴, =, =, ∠AOC arc AC r, θ, l, =, ⇒, r, 1, l, ⇒θ= ⋅, r, , O, , r, , A, , RELATION BETWEEN RADIAN AND DEGREE We know that a complete circle, subtends at its centre an angle whose measure is 2π radians as well as 360°., , ∴ ( 2π ) c = 360°., Hence, π c = 180°., TWO IMPORTANT CONVERSIONS, , ⎛ 180⎞ ° ⎛ 180 × 7 ⎞ °, (i) π c = 180° ⇒ 1c = ⎜, ⎟ =⎜, ⎟ = 57 °16′ 21′′ (approx.), ⎝ π ⎠, ⎝ 22 ⎠, c, , c, , 1 ⎞, ⎛ π ⎞, ⎛ 22, c, (ii) 180° = π c ⇒ 1° = ⎜, ×, ⎟ =⎜, ⎟ = ( 0.01746) ., ⎝ 180⎠, ⎝7, 180⎠, Degrees, , 30°, , 45°, , 60°, , 90°, , 180°, , 270°, , 360°, , Radians, , π/6, , π/4, , π/3, , π/2, , π, , 3 π/2, , 2π, , REMARK, , We express θ c by the symbol θ only.
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512, , Senior Secondary School Mathematics for Class 11, SUMMARY, , (i) In the sexagesimal system, we measure angles in, degrees, minutes and seconds., 1 right angle = 90° , 1° = 60 ′ and 1′ = 60 ′′ ., , (ii) In the circular measure, we measure angles in radians., Thus,, π c = 180° ., , (iii) If an arc of length l makes an angle θ c at the centre of a, circle of radius r , we have, θ=, , l, ⋅, r, , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Find the degree measure corresponding to each of the following radian, measures:, c, c, ⎛ 7π ⎞, ⎛ 3⎞, (ii) ⎜ ⎟, (iii) ( −2) c, (i) ⎜ ⎟, ⎝ 12 ⎠, ⎝ 4⎠, ⎛ 180⎞ °, (i) π c = 180° ⇒ 1c = ⎜, ⎟, ⎝ π ⎠, c, ⎛ 7π ⎞, ⎛ 180 7 π ⎞ °, ⇒ ⎜ ⎟ =⎜, ×, ⎟ = 105 °., ⎝ 12 ⎠, ⎝ π, 12 ⎠, c, , ⎛ 7π ⎞, Hence, ⎜ ⎟ = 105 °., ⎝ 12 ⎠, ⎛ 180⎞ °, c, (ii) π = 180° ⇒ 1c = ⎜, ⎟, ⎝ π ⎠, , c, , c, 7, 3⎞ °, ⎛ 3⎞, ⎛ 180 3 ⎞ ° ⎛, ⇒ ⎜ ⎟ =⎜, × ⎟ = ⎜180 ×, × ⎟, ⎝ 4⎠, ⎝ π, ⎠, ⎝, 22 4 ⎠, 4, = 42° 57′ 16′′ ., , ⎛ 3⎞, ∴ ⎜ ⎟ = 42° 57′ 16′′ ., ⎝ 4⎠, , ⎛ 180⎞ °, (iii) π c = 180° ⇒ 1c = ⎜, ⎟, ⎝ π ⎠, , 7, ⎧180, ⎫° ⎧, ⎫°, ⇒ ( −2) c = ⎨, × ( −2) ⎬ = ⎨180 ×, × ( − 2) ⎬, 22, ⎭, ⎭, ⎩ π, ⎩, = − (114° 32′ 44′′ )., ∴ ( −2) c = − (114° 32′ 44′′ )., , EXAMPLE 2, , Find the radian measure corresponding to each of the following degree, measures:, (i) 15°, (ii) 240°, (iii) −37 ° 30′
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Measurement of Angles, , 513, , c, , ⎛ π ⎞, (i) 180° = π c ⇒ 1° = ⎜, ⎟, ⎝ 180⎠, c, c, ⎛ π, ⎞, ⎛ π⎞, ⇒ 15 ° = ⎜, × 15⎟ = ⎜ ⎟ ⋅, ⎝ 180, ⎠, ⎝ 12⎠, c, ⎛ π⎞, Hence, 15 ° = ⎜ ⎟ ., ⎝ 12⎠, , SOLUTION, , ⎛ π ⎞, (ii) 180° = π c ⇒ 1° = ⎜, ⎟, ⎝ 180⎠, , c, , c, , c, , ⎛ π, ⎞, ⎛ 4π ⎞, ⇒ 240° = ⎜, × 240⎟ = ⎜ ⎟ ., ⎝ 180, ⎠, ⎝ 3⎠, c, , ⎛ 4π ⎞, Hence, 240° = ⎜ ⎟ ., ⎝ 3⎠, ⎛ π ⎞, (iii) 180° = π c ⇒ 1° = ⎜, ⎟, ⎝ 180⎠, , c, , c, , ⎧ π, ⎛ −75 ⎞ ⎫, ⇒ − 37 ° 30′ = ⎨, ×⎜, ⎟, ⎝ 2 ⎠ ⎬⎭, 180, ⎩, ⎡, ⎛, ⎢−37 ° 30′ = − ⎜ 37, ⎝, ⎢⎣, c, ⎛ −5 π ⎞, =⎜, ⎟ ., ⎝ 24 ⎠, , 1⎞°, ⎛ 75 ⎞ ° ⎤, ⎟ =−⎜ ⎟ ⎥, ⎝ 2⎠ ⎥, ⎠, 2, ⎦, , c, , ⎛ −5 π ⎞, Hence, ( −37 ° 30′ ) = ⎜, ⎟ ., ⎝ 24 ⎠, EXAMPLE 3, , SOLUTION, , O, c, , 25, cm, , Here, r = 25 cm and l = 11 cm., Let the measure of the required angle be θ c ., c, c, ⎛ l⎞, ⎛ 11 ⎞, Then, θ c = ⎜ ⎟ = ⎜ ⎟, ⎝r⎠, ⎝ 25 ⎠, ⎛ 11 180⎞ °, c, =⎜, ×, ⎟ [B π = 180°], ⎝ 25, π ⎠, 7, ⎛ 11, ⎞°, =⎜, ×, × 180⎟, ⎝ 25 22, ⎠, ⎛ 126⎞ °, =⎜, ⎟ = 25 ° 12′ ., ⎝ 5 ⎠, Hence, the required angle is 25 ° 12′., , r=, , EXAMPLE 4, , Find in degrees the angle subtended at the centre of a circle of diameter, 50 cm by an arc of length 11 cm., , A, , B, = 11 cm, , Find the radius of a circle in which a central angle of 72° intercepts an, arc of length 22 cm.( Use π = 22 7 .)
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514, , Senior Secondary School Mathematics for Class 11, , Let the required radius be r cm. Then,, c, c, π ⎞, ⎛, ⎛ 2π ⎞, l = 22 cm and θ = 72° = ⎜72 ×, ⎟ =⎜ ⎟ ., ⎝, ⎝ 5 ⎠, 180⎠, , SOLUTION, , l, l ⎛, 5 ⎞, ⇒ r = = ⎜ 22 ×, ⎟ cm, r, θ ⎝, 2π ⎠, 5, 7⎞, ⎛, = ⎜ 22 × × ⎟ cm, ⎝, 2 22⎠, 35, =, cm = 17.5 cm., 2, Hence, the radius of the circle is 17.5 cm., , Now, θ =, , EXAMPLE 5, , O, r cm, , 72°, , B, , A, = 22 cm, , The minute hand of a watch is 1.4 cm long. How far does its tip move in, 45 minutes? ( Use π = 22 7 .), In 60 minutes, the minute hand moves, through ( 2π ) c ., , SOLUTION, , 1.4 cm, , In 45 minutes, the minute hand moves, c, c, ⎛ 2π, ⎞, ⎛ 3π ⎞, through ⎜, × 45⎟ = ⎜, ⎟ ., ⎝ 60, ⎠, ⎝ 2 ⎠, , –, , c, , ∴, , ⎛ 3π ⎞, r = 1.4 cm and θ = ⎜, ⎟ ., ⎝ 2 ⎠, , c, , Distance moved by the tip of the minute, hand in 45 minutes is given by, 3π ⎞, 3 22⎞, ⎛, ⎛, l = r θ = ⎜1.4 ×, ⎟ cm = ⎜1.4 × × ⎟ cm = 6.6 cm., ⎝, ⎝, 2 ⎠, 2, 7⎠, EXAMPLE 6, , SOLUTION, , If the arcs of the same length in two circles subtend angles of 60° and, 75° at their respective centres, find the ratio of their radii., Let r1 and r2 be the radii of the two circles. Then,, c, c, π ⎞, ⎛, ⎛ π⎞, θ1 = 60° = ⎜ 60 ×, ⎟ =⎜ ⎟ ,, ⎝, ⎝ 3⎠, 180⎠, c, , c, , π ⎞, ⎛, ⎛ 5π ⎞, and θ 2 = 75 ° = ⎜75 ×, ⎟ =⎜ ⎟ ., ⎝, ⎝ 12 ⎠, 180⎠, Let the length of each arc be l cm. Then,, l = r1θ1 = r2θ 2, π⎞ ⎛, 5π ⎞, ⎛, ⇒ ⎜r1 × ⎟ = ⎜r2 ×, ⎟, ⎝, 3⎠ ⎝, 12 ⎠, r1 5, = ⋅, r2 4, Hence, r1 : r2 = 5 : 4., , O, , ⇒, , 60°, , O, , r1, , 75°, A, , B, , C, , r2, D
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Measurement of Angles, , The angles of a triangle are in AP and the ratio of the number of degrees, in the least to the number of radians in the greatest is 60 : π . Find the, angles in degrees and radians., , EXAMPLE 7, , SOLUTION, , 515, , Let the angles of the triangle be ( a − d) ° , a ° and ( a + d) °. Then,, ( a − d) + a + ( a + d) = 180 ⇒ 3 a = 180 ⇒ a = 60., Thus, the angles are ( 60 − d) ° , 60° and ( 60 + d) °., Number of degrees in the least angle = ( 60 − d)., π ⎫, ⎧, Number of radians in the greatest angle = ⎨( 60 + d) ×, ⎬⋅, 180 ⎭, ⎩, 60, π, ( 60 − d), =, ⇔ ( 60 − d) × π = 60 × ( 60 + d) ×, π, π, 180, ( 60 + d) ×, 180, ⇔ 3( 60 − d) = ( 60 + d), ⇔ 4d = 120 ⇔ d = 30., ∴ the required angles are ( 60 − 30) ° , 60° and ( 60 + 30) ° , i.e., 30°,, 60° and 90°., These angles in radians are, c, , c, , c, , c, , c, , c, , π ⎞ ⎛, π ⎞ ⎛, π ⎞, ⎛ π⎞ ⎛ π⎞ ⎛ π⎞, ⎛, ⎜ 30 ×, ⎟ , ⎜ 60 ×, ⎟ , ⎜ 90 ×, ⎟ , i.e., ⎜ ⎟ , ⎜ ⎟ , ⎜ ⎟ ., ⎝ 6⎠ ⎝ 3⎠ ⎝ 2⎠, ⎝, 180⎠ ⎝, 180⎠ ⎝, 180⎠, In a right-angled triangle, the difference between the two acute angles is, ( π 15) c . Find the angles in degrees., , EXAMPLE 8, , SOLUTION, , Clearly, the sum of the two acute angles of a right triangle is 90°., c, 180⎞ °, ⎛π⎞, ⎛π, Difference between the acute angles = ⎜ ⎟ = ⎜, ×, ⎟ = 12°., ⎝ 15 ⎠, ⎝ 15, π ⎠, Let the two acute angles be x° and y°. Then,, x + y = 90, x − y = 12, , … (i), … (ii), , Solving (i) and (ii), we get x = 51 and y = 39., Hence, the angles of the triangle are 51° , 39° and 90°., A wheel makes 360 revolutions in 1 minute. Through how many radians, does it turn in 1 second?, , EXAMPLE 9, , SOLUTION, , Number of revolutions made in 60 seconds = 360., 360, Number of revolutions made in 1 second =, = 6., 60, c, Angle moved in 1 revolution = ( 2π ) ., Angle moved in 6 revolutions = ( 2π × 6) c = (12π ) c ., , EXAMPLE 10, , Find the angle between the minute hand and the hour hand of a clock, when the time is 7.20.
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516, , Senior Secondary School Mathematics for Class 11, , SOLUTION, , Angle traced by the hour hand in 12 hours = 360°., 22, ⎛ 360 22⎞ °, hours = ⎜, × ⎟ = 220°., ⎝ 12, 3⎠, 3, Angle traced by the minute hand in 60 min = 360°., ⎛ 360, ⎞°, Angle traced by the minute hand in 20 min ⎜, × 20⎟ = 120°., ⎝ 60, ⎠, Angle traced by it in 7 h 20 min, i.e., in, , Hence, the required angle between the two hands, = ( 220° − 120°) = 100°., EXAMPLE 11, , SOLUTION, , A horse is tied to a post by a rope. If the horse moves along a circular, path, always keeping the rope tight, and describes 88 metres when it, traces 72° at the centre, find the length of the rope., , Let us denote the post by P and let PA be the length of the rope in the, tightest position. Suppose that the horse moves along the arc AB so, that ∠APB = 72° and arc AB = 88 m., B, Let the length of the rope PA be r metres., c, c, π ⎞, ⎛, ⎛ 2π ⎞, 88 m, Then, θ = 72° = ⎜72 ×, ⎟ = ⎜ ⎟ and l = 88 m., ⎝, ⎝ 5 ⎠, 180⎠, l, 5, 7⎞, 88, ⎛, m = ⎜ 88 × × ⎟ m = 70 m., ∴ r= =, 72°, ⎝, 2 22⎠, θ ( 2π /5), Hence, the length of the rope is 70 m., , P, , rm, , A, , EXERCISE 14, 1. Using a protractor, draw each of the following angles., (i) 60°, , (ii) 130°, , (iii) 300°, , (iv) 430°, , (v) −40°, , (vi) −220°, , (vii) −310°, , (viii) −400°, , 2. Express each of the following angles in radians., (i) 36°, , (ii) 120°, , (iii) 225°, , (v) 400°, , (vi) 7 ° 30′, , (vii) −270°, , (iv) 330°, (viii) − ( 22° 30′ ), , 3. Express each of the following angles in degrees., c, c, c, ⎛ 5π ⎞, ⎛ 18π ⎞, ⎛ 5⎞, (ii) − ⎜, (iii) ⎜ ⎟, (iv) −4 c, (i) ⎜ ⎟, ⎟, ⎝ 12 ⎠, ⎝ 5 ⎠, ⎝ 6⎠, 4. The angles of a triangle are in AP and the greatest angle is double the least., Find all the angles in degrees and radians., 5. The difference between the two acute angles of a right triangle is ( π 5 ) c ., Find these angles in radians and degrees., 6. Find the radius of a circle in which a central angle of 45° intercepts an arc of, length 33 cm. (Take π = 22 7 .), 7. Find the length of an arc of a circle of radius 14 cm which subtends an angle, of 36° at the centre.
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Measurement of Angles, , 517, , 8. If the arcs of the same length in two circles subtend angles 75° and 120° at, the centre, find the ratio of their radii., 9. Find the degree measure of the angle subtended at the centre of a circle of, diameter 60 cm by an arc of length 16.5 cm., 10. In a circle of diameter 30 cm, the length of a chord is 15 cm. Find the length, of the minor arc of the chord., 11. Find the angle in radians as well as in degrees through which a pendulum, swings if its length is 45 cm and its tip describes an arc of length 11 cm., 12. The large hand of a clock is 42 cm long. How many centimetres does its, extremity move in 20 minutes?, 13. A wheel makes 180 revolutions in 1 minute. Through how many radians, does it turn in 1 second?, 14. A train is moving on a circular curve of radius 1500 m at the rate of 66 km, per hour. Through what angle has it turned in 10 seconds?, 15. A wire of length 121 cm is bent so as to lie along the arc of a circle of radius, 180 cm. Find in degrees, the angle subtended at the centre by the arc., 16. The angles of a quadrilateral are in AP, and the greatest angle is double the, least. Express the least angle in radians., ANSWERS (EXERCISE 14), , ⎛ π⎞, 2. (i) ⎜ ⎟, ⎝5⎠, , c, , ⎛ 3π ⎞, (vii) − ⎜, ⎟, ⎝ 2 ⎠, , ⎛ 2π ⎞, (ii) ⎜ ⎟, ⎝ 3⎠, c, , c, , ⎛ 5π ⎞, (iii) ⎜ ⎟, ⎝ 4⎠, , ⎛ π⎞, (viii) − ⎜ ⎟, ⎝ 8⎠, , c, , ⎛ 11π ⎞, (iv) ⎜, ⎟, ⎝ 6 ⎠, , c, , ⎛ 20π ⎞, (v) ⎜, ⎟, ⎝ 9 ⎠, , 7. 8.8 cm, , c, , c, , 8. 8 : 5, , 9. 31° 30′, , c, c, ⎪⎧⎛ 7 π ⎞ ⎛ 3 π ⎞ ⎪⎫, ⎟ ⎬, ⎨⎜⎝ ⎟⎠ , ⎜⎝, 20 ⎠ ⎭⎪, ⎩⎪ 20, , 10. 15.71 cm, , c, , ⎛ 11 ⎞, 11. ⎜ ⎟ , 14° 12. 88 cm, ⎝ 45 ⎠, , c, , (iv) − ( 229° 5′ 27′′ ), , ⎪⎧⎛ 2π ⎞ ⎛ π ⎞ ⎛ 4π ⎞ ⎫⎪, 4. ( 40° , 60° , 80°), ⎨⎜ ⎟ , ⎜ ⎟ , ⎜ ⎟ ⎬ 5. ( 63 ° , 27 °),, ⎝ ⎠ ⎝ 3⎠ ⎝ 9 ⎠ ⎪, ⎭, ⎩⎪ 9, 6. 42 cm, , ⎛π⎞, (vi) ⎜ ⎟, ⎝ 24⎠, , c, , 3. (i) 75° (ii) −648° (iii) 47 ° 43′ 38′′, c, , c, , 13. ( 6π ) c, , 14. 7°, , 15. 38° 30′, , ⎛ π⎞, 16. ⎜ ⎟, ⎝ 3⎠, , c, , HINTS TO SOME SELECTED QUESTIONS, 4. Let the required angles be ( a − d )° , a° and ( a + d )° ., Then, the sum of the angles of a triangle being 180°, we have, ( a − d ) + a + ( a + d ) = 180 ⇔ 3 a = 180 ⇔ a = 60., Also, ( a + d ) = 2( a − d ) ⇔ a = 3 d, ⇔ 3 d = 60 [B a = 60], ⇔ d = 20.
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518, , Senior Secondary School Mathematics for Class 11, Hence, the angles in degrees are 40°, 60°, 80°., c, c, c, π ⎞ ⎛, π ⎞, π ⎞, ⎛, These angles in radians are ⎛⎜ 40 ×, ⎟ and ⎜ 80 ×, ⎟ , ⎜ 60 ×, ⎟ , i.e.,, ⎝, ⎝, 180 ⎠ ⎝, 180 ⎠, 180 ⎠, c, , c, , c, , ⎛ 2 π ⎞ , ⎛ π ⎞ and ⎛ 4 π ⎞ ., ⎟ ⎜ ⎟, ⎟, ⎜, ⎜, ⎝ 9 ⎠ ⎝ 3⎠, ⎝ 9 ⎠, 5. Let the acute angles of the given triangle be x c and y c. Then,, π, x+ y=, 2, π, x−y=, 5, 7π, 3π, and y =, Solving (i) and (ii), we get x =, ⋅, 20, 20, c, c, 7π⎞, ⎛ 3π ⎞ ., Thus, the required angles are ⎛⎜, ⎟ and ⎜, ⎟, ⎝ 20 ⎠, ⎝ 20 ⎠, , … (i), … (ii), , 7 π 180 ⎞ °, ⎛ 3 π × 180 ⎞ ° , i.e., 63° and 27 °., These angles in degrees are ⎛⎜, ×, ⎟ and ⎜, ⎟, ⎝ 20, ⎝ 20, π ⎠, π ⎠, 10. LetO be the centre of the given circle of radius 15 cm and let AB be the chord of length, 15 cm. Then, OA = OB = AB = 15 cm., ∴ CAOB is an equilateral triangle., And, therefore, θ = ∠AOB = 60 °, c, c, B, π ⎞, ⎛ π⎞, = ⎛⎜ 60 ×, ⎟ =⎜ ⎟ ., ⎝ 3⎠, ⎝, 180 ⎠, Let the minor arc AB = l cm. Then,, c, π, θ = ⎛⎜ ⎞⎟ and r = 15 cm., ⎝ 3⎠, l, π, ∴ θ = ⇒ l = rθ = ⎛⎜ 15 × ⎞⎟ cm, ⎝, r, 3⎠, 22, 110, = ⎛⎜ 5 × ⎞⎟ cm =, cm = 15.71 cm., ⎝, 7 ⎠, 7, , O, , 60°, , A, , 11. Here, r = 45 cm and l = 11cm., ∴, , Let the angle be θ c. Then,, c, l, 11, θ = = ⎛⎜ ⎞⎟, ⎠, ⎝, r, 45, , O, c, , 45 cm, , c, , Now, π = 180 °, 11, 180 11 ⎞ °, × ⎟, ⇒ ⎛⎜ ⎞⎟ = ⎛⎜, ⎝ 45 ⎠, ⎝ π, 45 ⎠, c, , 7, 11 °, = ⎛⎜ 180 ×, × ⎞⎟ = 14 ° ., ⎝, 22 45 ⎠, , q, , A, , 11 cm, , B
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15, Trigonometric, or Circular, Functions, Trigonometric, or Circular, Functions, , TRIGONOMETRIC (OR CIRCULAR) FUNCTIONS, , Let X′ OX and YOY′ be the coordinate axes., Taking O as the centre and a unit radius,, draw a circle, cutting the coordinate axes, at A , B , A′ and B′ , as shown in the figure., Suppose that a moving point starts, from A, and moves along the X, circumference of the circle in an, anticlockwise direction. Let it cover an arc, length θ and take the final position P( x , y)., Join OP. Then,, ∠AOP = θ, ⎡, θ, arc AP, l⎤, c, ⎢B ∠AOP = radius OP = 1 = θ , using θ = r ⎥ ⋅, ⎣, ⎦, , Y, B, , P(x, y), , 1, A, , O, , x, , y, M, , A, X, , B, Y, , Now, the six circular functions may be defined as under:, (i) cos θ = x, (ii) sin θ = y, 1, 1, (iii) sec θ = , x ≠ 0, (iv) cosec θ = , y ≠ 0, x, y, y, x, (vi) cot θ = , y ≠ 0, (v) tan θ = , x ≠ 0, x, y, REMARK, , Clearly, we have, 1, (i) sec θ =, cos θ, sin θ, (iii) tan θ =, cos θ, , 1, sin θ, 1, cos θ, (iv) cot θ =, =, tan θ sin θ, (ii) cosec θ =, , For any acute angle, say θ, we have, (i) cos2 θ + sin 2 θ = 1, (ii) 1 + tan 2 θ = sec2 θ, 2, 2, (iii) 1 + cot θ = cosec θ, Let X′ OX and YOY′ be the coordinate axes. Taking O as the centre and a, unit radius, draw a circle, meeting OX at A., Let P( x , y) be a point on the circle with ∠AOP = θ. Join OP., Draw PM ⊥ OA., , THEOREM 1, , PROOF, , 519
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Trigonometric, or Circular, Functions, , 521, , Signs of Trigonometric Functions in Various Quadrants, Quadrant, , Y, , Signs of various T-functions, , I, , All T-functions are positive., , II, , sin θ and cosec θ are positive., All others are negative., , III, , tan θ and cot θ are positive., All others are negative., , IV, , cos θ and sec θ are positive., All others are negative., , sine All, is +ve are +ve, I, II, tan O cos, is +ve is +ve, III, IV, , X, , X, , Y, , Remember, I, , II, , III, , IV, , All are +ve, , sin is +ve, , tan is +ve, , cos is +ve, , Values of T-functions of Some Particular Angles, , θ→, , 0, , sin θ, , 0, , cos θ, , 1, , tan θ, , 0, , π, 6, 1, 2, , π, 4, 1, 2, , π, 3, , 3, 2, 1, 3, , 1, 2, , 3, 2, 1, 2, , 1, , 3, , π, 2, , π, , 3π, 2, , 2π, , 1, , 0, , –1, , 0, , 0, , –1, , 0, , 1, , not, defined, , 0, , not, defined, , 0, , Domain and Range of Trigonometric Functions, , The domain and range of each one of the six trigonometric functions is given, below., T-function, , Domain, , Range, , sin x, , R, , [−1, 1], , cos x, , R, , [−1, 1], , tan x, cot x, cosec x, sec x, , {x ∈ R : x ≠ ( 2n + 1), , π, , 2,, , n ∈I}, , {x ∈ R : x ≠ nπ , n ∈ I }, {x ∈ R : x ≠ nπ , n ∈ I }, {x ∈ R : x ≠ ( 2n + 1), , π, , 2 , n ∈I}, , R, R, R−] − 1, 1[, R−] − 1, 1[, , PERIODIC FUNCTIONS A function f ( x) is said to be periodic if there exists a constant, real number p such that f ( x + p) = f ( x) for all x., The least positive value of p for which f ( x + p) = f ( x) is called the period, of f (x).
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522, , Senior Secondary School Mathematics for Class 11, , If f ( x) is a periodic function with period p then f ( ax + b), with a > 0 is a periodic function with period ( b a)., , AN IMPORTANT RESULT, , THEOREM 3, PROOF, , sin (θ + 2π ) = sin θ and cos (θ + 2π ) = cos θ., , Let X′ OX and YOY′ be the coordinate axes. With O as the centre, draw a, circle of unit radius, meeting OX at A., The circumference of this circle of, Y, unit radius is 2π., Let P be a point on this circle such, P, that arc AP = θ., If we start from the point A , move, A, along the circle and after making a, O, X, X, complete revolution, reach P then, the length of the arc covered is, (θ + 2π )., Since the trigonometric functions, Y, are defined in terms of the, coordinates of P, we have, cos (θ + 2π ) = cos θ and sin (θ + 2π ) = sin θ., (i) Since 2π is the least angle for which cos (θ + 2π ) = cos θ and, sin (θ + 2π ) = sin θ , it follows that, cos θ and sin θ are perodic functions, each with period 2π., Similarly, sec θ and cosec θ are periodic, each with period 2π; and, tan θ and cot θ are periodic, each with period π., A constant function is periodic, having no period., (ii) For any real number θ and any integer n, we have:, , REMARKS, , sin (θ + 2nπ ) = sin θ , cos (θ + 2nπ ) = cos θ, tan (θ + nπ ) = tan θ , cot (θ + nπ ) = cot θ, sec (θ + 2nπ ) = sec θ , cosec (θ + 2nπ ) = cosec θ, SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , −13, and θ lies in the second quadrant, find the values of all, 12, the other five trigonometric functions., If sec θ =, , We know that in the second quadrant, sin θ and cosec θ are positive, and all the other trigonometric functions are negative., Now, sec θ =, ∴, , −13, 1, −12, ⇒ cos θ =, =, ⋅, 12, sec θ, 13, , ⎫⎪ ⎛ 169 ⎞, ⎧⎪⎛ −13 ⎞ 2, 25, − 1⎟ =, tan 2 θ = ( sec2 θ − 1) = ⎨⎜, ⎟ − 1⎬ = ⎜, ⎪⎭ ⎝ 144 ⎠ 144, ⎪⎩⎝ 12 ⎠
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Trigonometric, or Circular, Functions, , 527, , 4. If sec θ = 2 and θ lies in Quadrant IV, find the values of all the other, trigonometric functions., 5. If sin x =, , −2 6, and x lies in Quadrant III, find the values of cos x and cot x., 5, , 6. If cos x =, , π, − 15, and < x < π , find the value of sin x., 4, 2, , 7. If sec x = −2 and π < x <, , 3π, , find the values of all the other five, 2, , trigonometric functions., 8. Find the value of, ⎛ 31π ⎞, (i) sin ⎜, ⎟, ⎝ 3 ⎠, ⎛ 13 π ⎞, (iv) cot ⎜, ⎟, ⎝ 4 ⎠, , ⎛ 17 π ⎞, (ii) cos ⎜, ⎟, ⎝ 2 ⎠, , ⎛ −25 π ⎞, (iii) tan ⎜, ⎟, ⎝ 3 ⎠, , ⎛ −25 π ⎞, (v) sec ⎜, ⎟, ⎝ 3 ⎠, , ⎛ − 41π ⎞, (vi) cosec ⎜, ⎟, ⎝ 4 ⎠, , (ii) sec ( −1470°), , (iii) tan ( −300°), , 9. Find the value of, (i) sin 405°, (iv) cot (585°), , (v) cosec ( −750°) (vi) cos ( −2220°), , 10. Prove that, π, π, π, π, + 2 cos2 + 3 sec2 + 4 cos2 = 8, 3, 4, 6, 2, π, π, π, π, π 7, (ii) sin cos 0 + sin cos + sin cos =, 6, 4, 4, 3, 6 4, π, π, π, 2 π, 2 π, (iii) 4 sin sin, + 3 cos tan + cosec, =4, 6, 3, 3, 4, 2, (i) tan 2, , ANSWERS (EXERCISE 15A), , 1, 1, 1. sin θ = − , tan θ =, , cot θ =, 2, 3, 2. cos θ =, , 3 , cosec θ = −2, sec θ =, , −2, 3, , 3, −1, 2, , tan θ =, , cot θ = − 3 , sec θ =, , cosec θ = −2, 2, 3, 3, , 3. sin θ =, , 3, −4, −3, −4, −5, , cos θ =, , tan θ =, , cot θ =, , sec θ =, 5, 5, 4, 3, 4, , 4. sin θ =, , −1, 1, , cos θ =, , tan θ = −1, cot θ = −1, cosec θ = − 2, 2, 2
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528, , Senior Secondary School Mathematics for Class 11, , 5. cos x = −, 7. sin x =, 3, 2, 1, 9. (i), 2, , 1, 1, , cot x =, 5, 2 6, , 6., , − 3, 1, , cos x = − , tan x =, 2, 2, (ii) 0, , 8. (i), , (ii), , 2, 3, , (iii) − 3, (iii), , 3, , 1, 4, 3 , cot x =, , 1, −2, , cosec x =, 3, 3, , (iv) 1, , (v) 2, , (vi) − 2, , (iv) 1, , (v) –2, , (vi), , 1, 2, , TRIGONOMETRIC FUNCTIONS OF SUM AND DIFFERENCE OF NUMBERS, THEOREM 1, , PROOF, , (i) cos (θ + φ) = cos θ cos φ − sin θ sin φ, (ii) cos (θ − φ) = cos θ cos φ + sin θ sin φ, , Let X′ OX and YOY′ be the coordinate axes. With O as the centre and, taking a unit radius, draw a circle, cutting the axes at A , A′ , B and B′ as, shown in the figure given below., Let ∠AOP = θ and ∠POQ = φ. Then, ∠AOQ = (θ + φ)., Let ∠AOR = − φ. Join PR and AQ., Y, (cos , sin ), B(0, 1), P, , (1, 0), X, , A, , O, , , , (1, 0), A, X, , [cos ( + ), sin ( + )] Q, (cos , sin ) R, , B (0, 1), Y, , (i) The points on the circle are given by, P(cos θ , sin θ), Q[cos (θ + φ), sin (θ + φ)], R[cos ( −φ), sin ( −φ)],, i.e., R(cos φ , − sin φ), A(1, 0), B( 0, 1), A′ ( −1, 0) and B′( 0, 1)., In CPOR and CQOA , we have, PO = QO = 1 unit, OR = OA = 1 unit and, ∠POR = ∠QOA [each equal to ( φ + ∠QOR)]., ∴ CPOR 8 CQOA., Hence, PR = QA., ∴, , PR 2 = QA 2, , ⇒ (cos θ − cos φ) 2 + (sin θ + sin φ) 2, = [cos (θ + φ) − 1]2 + [sin (θ + φ) − 0]2
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530, , Senior Secondary School Mathematics for Class 11, , ⎛π, ⎞, (iv) cot ⎜ − x⎟ =, ⎝2, ⎠, ⎛π, ⎞, (v) sec ⎜ − x⎟ =, ⎝2, ⎠, , ⎛π, ⎞, cos ⎜ − x⎟, ⎝2, ⎠ sin x, =, = tan x, π, ⎛, ⎞ cos x, sin ⎜ − x⎟, ⎝2, ⎠, , [using (i) and (ii)]., , 1, 1, ⎡, ⎛π, ⎞, =, = cosec x ⎢B cos ⎜ − x⎟ = sin, ⎝2, ⎠, π, ⎛, ⎞ sin x, ⎣, cos ⎜ − x⎟, ⎝2, ⎠, , ⎤, x⎥, ⎦, , = cosec x., ⎛π, ⎞, (vi) cosec ⎜ − x⎟ =, ⎝2, ⎠, , 1, 1, =, ⎛π, ⎞ cos x, sin ⎜ − x⎟, ⎝2, ⎠, , ⎡, ⎤, ⎛π, ⎞, ⎢B sin ⎜⎝ 2 − x⎟⎠ = cos x ⎥, ⎣, ⎦, , = sec x., SUMMARY, , THEOREM 3, , ⎛π, ⎞, (i) cos ⎜ − x⎟ = sin x, ⎝2, ⎠, , ⎛π, ⎞, (ii) sin ⎜ − x⎟ = cos x, ⎝2, ⎠, , ⎛π, ⎞, (iii) tan ⎜ − x⎟ = cot x, ⎝2, ⎠, , ⎛π, ⎞, (iv) cot ⎜ − x⎟ = tan x, ⎝2, ⎠, , ⎛π, ⎞, (v) sec ⎜ − x⎟ = cosec x, ⎝2, ⎠, , ⎛π, ⎞, (vi) cosec ⎜ − x⎟ = sec x, ⎝2, ⎠, , ⎛π, ⎞, (i) cos ⎜ + x⎟ = − sin x, ⎝2, ⎠, , ⎛π, ⎞, (ii) sin ⎜ + x⎟ = cos x, ⎝2, ⎠, , ⎛π, ⎞, (iii) tan ⎜ + x⎟ = − cot x, ⎠, ⎝2, , ⎛π, ⎞, (iv) cot ⎜ + x⎟ = − tan x, ⎠, ⎝2, , ⎛π, ⎞, (v) sec ⎜ + x⎟ = − cosec x, ⎠, ⎝2, PROOF, , ⎛π, ⎞, (vi) cosec ⎜ + x⎟ = sec x, ⎠, ⎝2, , π, π, ⎛π, ⎞, (i) cos ⎜ + x⎟ = cos cos x − sin sin x, ⎝2, ⎠, 2, 2, π, π, ⎡, ⎤, = ( 0 × cos x) − (1 × sin x) ⎢B cos = 0 and sin = 1⎥, 2, 2, ⎦, ⎣, = − sin x., ⎛π, ⎡π, ⎤, ⎞, (ii) sin ⎜ + x⎟ = sin ⎢ − ( − x) ⎥, ⎝2, ⎠, ⎣2, ⎦, = cos ( − x), = cos x, , ⎡, ⎤, ⎛π, ⎞, ⎢B sin ⎜⎝ 2 − θ⎟⎠ = cos θ ⎥, ⎣, ⎦, [B cos ( − x) = cos x ].
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Trigonometric, or Circular, Functions, , ⎛π, ⎞, (iii) tan ⎜ + x⎟ =, ⎝2, ⎠, =, , 531, , ⎛π, ⎞, sin ⎜ + x⎟, ⎝2, ⎠, ⎛π, ⎞, cos ⎜ + x⎟, ⎝2, ⎠, cos x, − sin x, ⎡, ⎤, ⎛π, ⎛π, ⎞, ⎞, ⎢B sin ⎜⎝ 2 + x⎟⎠ = cos x and cos ⎜⎝ 2 + x⎟⎠ = − sin x ⎥, ⎣, ⎦, , = − cot x., ⎛π, ⎞, (iv) cot ⎜ + x⎟ =, ⎝2, ⎠, , 1, 1, =, ⎛π, ⎞ − cot x, tan ⎜ + x⎟, ⎝2, ⎠, , ⎡, ⎛π, ⎞, ⎢B tan ⎜⎝ 2 + x⎟⎠ = − cot, ⎣, , ⎤, x⎥, ⎦, , ⎡, ⎛π, ⎞, ⎢B cos ⎜⎝ 2 + x⎟⎠ = − sin, ⎣, , ⎤, x⎥, ⎦, , = − tan x., ⎛π, ⎞, (v) sec ⎜ + x⎟ =, ⎝2, ⎠, , 1, 1, =, ⎛π, ⎞ − sin x, cos ⎜ + x⎟, ⎝2, ⎠, , = − cosec x., ⎛π, ⎞, (vi) cosec ⎜ + x⎟ =, ⎝2, ⎠, , 1, 1, =, ⎛π, ⎞ cos x, sin ⎜ + x⎟, ⎝2, ⎠, , ⎡, ⎤, ⎛π, ⎞, ⎢B sin ⎜⎝ 2 + x⎟⎠ = cos x ⎥, ⎣, ⎦, , = sec x., SUMMARY, , ⎛π, ⎞, (i) cos ⎜ + x⎟ = − sin x, ⎝2, ⎠, ⎛π, ⎞, (iii) tan ⎜ + x⎟ = − cot x, ⎝2, ⎠, , THEOREM 4, , ⎛π, ⎞, (iv) cot ⎜ + x⎟ = − tan x, ⎝2, ⎠, , ⎛π, ⎞, (v) sec ⎜ + x⎟ = − cosec x, ⎝2, ⎠, , ⎛π, ⎞, (vi) cosec ⎜ + x⎟ = sec x, ⎝2, ⎠, , (i) cos ( π − x) = − cos x, , (ii) sin ( π − x) = sin x, , (iii) tan ( π − x) = − tan x, , (iv) cot ( π − x) = − cot x, , (v) sec ( π − x) = − sec x, PROOF, , ⎛π, ⎞, (ii) sin ⎜ + x⎟ = cos x, ⎝2, ⎠, , (vi) cosec ( π − x) = cosec x, , (i) cos ( π − x) = cos π cos x + sin π sin x, = ( −1 × cos x) + ( 0 × sin x), = − cos x., , [B cos π = −1 and sin π = 0 ]
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Trigonometric, or Circular, Functions, , 533, , sin ( π + x) − sin x, =, cos ( π + x) − cos x, = tan x. [B sin ( π + x) = − sin x , cos ( π + x) = − cos x ], 1, 1, (iv) cot ( π + x) =, =, [B tan ( π + x) = tan x ], tan ( π + x) tan x, = cot x., 1, 1, (v) sec ( π + x) =, =, [B cos ( π + x) = − cos x ], cos ( π + x) − cos x, = − sec x., 1, 1, (vi) cosec ( π + x) =, =, [B sin ( π + x) = − sin x ], sin ( π + x) − sin x, = − cosec x., , (iii) tan ( π + x) =, , SUMMARY, , (i) cos ( π + x) = − cos x, (iii) tan ( π + x) = tan x, (v) sec ( π + x) = − sec x, , (ii) sin ( π + x) = − sin x, (iv) cot ( π + x) = cot x, (vi) cosec ( π + x) = − cosec x, , (i) cos ( 2π − x) = cos x, (ii) sin ( 2π − x) = − sin x, (iii) tan ( 2π − x) = − tan x, (iv) cot ( 2π − x) = − cot x, (v) sec ( 2π − x) = sec x, (vi) cosec ( 2π − x) = − cosec x, (i) cos ( 2π − x) = cos 2π cos x + sin 2π sin x, = (1 × cos x) + ( 0 × sin x) [B cos 2π = 1 and sin 2π = 0], = cos x., (ii) sin ( 2π − x) = sin [π + ( π + x)], = − sin ( π + x), [B sin ( π + θ) = − sin θ ], = − [− sin x] = sin x., The remaining parts are left to the student as an exercise., , THEOREM 6, , PROOF, , SUMMARY, , (i) cos (2π − x) = cos x, , (ii) sin (2π − x) = − sin x, , (iii) tan (2π − x) = − tan x (iv) cot (2π − x) = − cot x, (v) sec (2π − x) = sec x, , (vi) cosec (2π − x) = − cosec x, , Further, it is easy to verify that, (i) cos (2π + x) = cos x, , (ii) sin (2π + x) = sin x, , (iii) tan (2π + x) = tan x, , (iv) cot (2π + x) = cot x, , (v) sec (2π + x) = sec x, , (vi) cosec (2π + x) = cosec x
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Trigonometric, or Circular, Functions, , 537, , (ii) cos ( x + y) cos ( x − y), = (cos x cos y − sin x sin y)(cos x cos y + sin x sin y), = cos2 x cos2 y − sin 2 x sin 2 y, = cos2 x(1 − sin 2 y) − (1 − cos2 x) sin 2 y, = cos2 x − cos2 x sin 2 y − sin 2 y + cos2 x sin 2 y, = cos2 x − sin 2 y., , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Find the value of, (i) cos 480°, (iv) cosec ( −1410°), , (ii) sin 1230°, (v) cos ( −870°), , (iii) cot ( −135 °), (vi) tan 330°, , (i) cos 480° = cos ( 360° + 120°), = cos 120°, [B cos ( 360° + θ) = cos θ ], = cos (180° − 60°) = − cos 60° [B cos (180° − θ) = − cos θ ], 1, =− ⋅, 2, (ii) sin 1230° = sin ( 3 × 360° + 150°), = sin 150°, [B sin ( 2nπ + θ) = sin θ ], = sin (180° − 30°) = sin 30° [B sin (180° − θ) = sin θ ], 1, = ⋅, 2, (iii) cot ( −135 °) = − cot 135 °, [B cot ( − θ) = − cot θ ], = − cot ( 90° + 45 °) = tan 45 °, [B cot ( 90° + θ) = − tan θ ], = 1., (iv) cosec ( −1410°) = − cosec 1410° = − cosec ( 4 × 360° − 30°), = cosec 30°, [B cosec ( 2nπ − θ) = − cosec θ ], = 2., (v) cos ( −870°) = cos 870°, [B cos ( − θ) = cos θ ], = cos ( 2 × 360° + 150°), = cos 150°, [B cos ( 2nπ + θ) = cos θ ], = cos (180° − 30°) = − cos 30°, [B cos (180° − θ) = − cos θ ], − 3, =, ⋅, 2, (vi) tan 330° = tan ( 360° − 30°) = − tan 30°, [B tan ( 360° − θ) = − tan θ ], 1, =−, ⋅, 3
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538, EXAMPLE 2, , SOLUTION, , EXAMPLE 3, SOLUTION, , Senior Secondary School Mathematics for Class 11, , Find the values of all trigonometric functions of (i) 120° (ii) 150°., 3, ;, 2, −1, cos 120° = cos (180° − 60°) = − cos 60° =, ;, 2, ⎫, sin 120° ⎧ 3, tan 120° =, =⎨, × ( −2) ⎬ = − 3 ;, cos 120° ⎩ 2, ⎭, 1, 2, cosec 120° =, =, ;, sin 120°, 3, 1, sec 120° =, = − 2;, cos 120°, 1, −1, cot 120° =, =, ⋅, tan 120°, 3, 1, (ii) sin 150° = sin (180° − 30°) = sin 30° = ;, 2, − 3, cos 150° = cos (180° − 30°) = − cos 30° =, ;, 2, sin 150° ⎛ 1, 2 ⎞, −1, =⎜ ×, tan 150° =, ;, ⎟ =, cos 150° ⎝ 2 − 3 ⎠, 3, 1, cot 150° =, = − 3;, tan 150°, 1, −2, sec 150° =, =, ;, cos 150°, 3, 1, cosec 150° =, = 2., sin 150°, (i) sin 120° = sin (180° − 60°) = sin 60° =, , Show that sin 105 ° + cos 105 ° =, , 1, ⋅, 2, , We have, LHS = sin 105 ° + cos 105 °, , = sin ( 60° + 45 °) + cos ( 60° + 45 °), = (sin 60° cos 45 ° + cos 60° sin 45 ° ), + (cos 60° cos 45 ° − sin 60° sin 45 °), ⎧⎛ 3, 1 ⎞ ⎛1, 1 ⎞ ⎫ ⎧⎛ 1, 1 ⎞ ⎛ 3, 1 ⎞⎫, ×, = ⎨⎜, ×, ⎟⎬, ⎟ +⎜ ×, ⎟ −⎜, ⎟ ⎬ + ⎨⎜ ×, 2⎠ ⎝ 2, 2 ⎠ ⎭ ⎩⎝ 2, 2⎠ ⎝ 2, 2⎠ ⎭, ⎩⎝ 2, ⎛ 3, 1, 1, 3⎞, 1, =⎜, +, +, −, = RHS., ⎟ =, 2, ⎝ 2 2 2 2 2 2 2 2⎠, EXAMPLE 4, , Calculate the value of, (i) sin 15°, (ii) cos 15°, (iv) sin 75°, (v) cos 75°, , (iii) tan 15°, (vi) tan 75°
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Trigonometric, or Circular, Functions, SOLUTION, , (i) sin 15 ° = sin ( 45 ° − 30°), = sin 45 ° cos 30° − cos 45 ° sin 30°, ⎛ 1, 3⎞ ⎛ 1, 1 ⎞ ( 3 − 1), =⎜, ×, × ⎟ =, ⋅, ⎟ −⎜, ⎝, 2 ⎠, 2 2⎠, 2 2, ⎝ 2, (ii) cos 15 ° = cos ( 45 ° − 30°), = cos 45 ° cos 30° + sin 45 ° sin 30°, ⎛ 1, 3⎞ ⎛ 1, 1 ⎞ ( 3 + 1), =⎜, ×, ⋅, × ⎟ =, ⎟ +⎜, 2 ⎠ ⎝ 2 2⎠, 2 2, ⎝ 2, (iii) tan 15 ° = tan ( 45 ° − 30°), tan 45 ° − tan 30°, =, =, 1 + tan 45 ° tan 30°, , 1 ⎞, ⎛, ⎜1 −, ⎟, ⎝, ( 3 − 1), 3⎠, =, ⋅, 1 ⎞ ( 3 + 1), ⎛, 1 + ⎜1 ×, ⎟, ⎝, 3⎠, , (iv) sin 75 ° = sin ( 45 ° + 30°), = sin 45 ° cos 30° + cos 45 ° sin 30°, ⎛ 1, 3⎞ ⎛ 1, 1 ⎞ ( 3 + 1), =⎜, ×, ⋅, × ⎟ =, ⎟ +⎜, 2 ⎠ ⎝ 2 2⎠, 2 2, ⎝ 2, (v) cos 75 ° = cos ( 45 ° + 30° ), = cos 45 ° cos 30° − sin 45 ° sin 30°, ⎛ 1, 3⎞ ⎛ 1, 1⎞, =⎜, ×, × ⎟, ⎟ −⎜, ⎝, 2, 2⎠, 2, 2, ⎝, ⎠, =, , ( 3 − 1), ⋅, 2 2, , (vi) tan 75 ° = tan ( 45 ° + 30°), 1 ⎞, ⎛, ⎜1 +, ⎟, ⎝, tan 45 ° + tan 30°, 3⎠, =, =, 1 ⎞, 1 − tan 45 ° tan 30° ⎛, ⎜1 −, ⎟, ⎝, 3⎠, ( 3 + 1), =, ⋅, ( 3 − 1), EXAMPLE 5, SOLUTION, , Evaluate tan, , 13 π, ⋅, 12, , We have, 13 π, π⎞, ⎛, tan, = tan ⎜ π + ⎟, ⎝, 12, 12⎠, π, = tan, [B tan ( π + θ) = tan θ ], 12, ⎛ π π⎞, = tan ⎜ − ⎟, ⎝ 3 4⎠, , 539
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540, , Senior Secondary School Mathematics for Class 11, , π, π, − tan, 3, 4 = 3 −1, =, π, π 1+ 3, 1 + tan tan, 3, 4, ( 3 − 1) ( 3 − 1) ( 3 − 1) 2, =, ×, =, ( 3 − 1), ( 3 + 1) ( 3 − 1), ( 3 + 1 − 2 3) (4 − 2 3), =, =, = (2 −, 2, 2, tan, , EXAMPLE 6, , 3 )., , Prove that, 3, 2, 1, (ii) cos 50° cos 10° − sin 50° sin 10° =, 2, 1, (iii) cos 80° cos 20° + sin 80° sin 20° =, 2, 1, (iv) sin 36° cos 9° + cos 36° sin 9° =, 2, (i) sin 70° cos 10° − cos 70° sin 10° =, , (i) We know that sin θ cos φ − cos θ sin φ = sin (θ − φ)., Putting θ = 70° and φ = 10° in the this identity, we get, , SOLUTION, , 3, ⋅, 2, , sin 70° cos 10° − cos 70° sin 10° = sin (70° − 10°) = sin 60° =, (ii) We know that cos θ cos φ − sin θ sin φ = cos (θ + φ)., Putting θ = 50° and φ = 10° in the this identity, we get, cos 50° cos 10° − sin 50° sin 10° = cos (50° + 10°) = cos 60° =, (iii) We know that cos θ cos φ + sin θ sin φ = cos (θ − φ)., Putting θ = 80° and φ = 20° in this identity, we get, , 1, ⋅, 2, , cos 80° cos 20° + sin 80° sin 20° = cos ( 80° − 20°) = cos 60° =, (iv) We know that sin θ cos φ + cos θ sin φ = sin (θ + φ)., Putting θ = 36° and φ = 9° in this identity, we get, sin 36° cos 9° + cos 36° sin 9° = sin ( 36° + 9°) = sin 45 ° =, EXAMPLE 7, , Prove that, sin ( 40° + θ) cos (10° + θ) − cos ( 40° + θ) sin (10° + θ) =, , SOLUTION, , 1, ⋅, 2, , We know that sin x cos y − cos x sin y = sin ( x − y)., Putting x = ( 40° + θ) and y = (10° + θ) in this identity, we get, sin ( 40° + θ) cos (10° + θ) − cos ( 40° + θ) sin (10° + θ), = sin [( 40° + θ) − (10° + θ)], 1, = sin 30° = ⋅, 2, , 1, ⋅, 2, , 1, ⋅, 2
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Trigonometric, or Circular, Functions, , (ii) sin, , EXAMPLE 13, , SOLUTION, , ⎛ π π⎞, ⎜ + ⎟, ⎝ 4 6⎠, π, π, = sin cos +, 4, 6, ⎛ 1, 3⎞, =⎜, ×, ⎟ +, 2 ⎠, ⎝ 2, , 543, , 5π, = sin, 12, , Prove that, , π, π, sin, 4, 6, 1 ⎞ ( 3 + 1), ⎛ 1, × ⎟ =, ⋅, ⎜, ⎝ 2 2⎠, 2 2, , cos, , sin ( x + y) tan x + tan y, =, ⋅, sin ( x − y) tan x − tan y, , We have, sin ( x + y), sin ( x − y), sin x cos y + cos x sin y, =, sin x cos y − cos x sin y, tan x + tan y, [dividing num. and denom. by cos x cos y], =, tan x − tan y, , LHS =, , = RHS., EXAMPLE 14, , SOLUTION, , 15, 12, and cos φ =, , where θ and φ both lie in the first quadrant,, 17, 13, find the values of, (i) sin (θ + φ),, (ii) cos (θ − φ),, (iii) tan (θ + φ)., If sin θ =, , 15, 12, and cos φ =, ⋅, 17, 13, Since θ lies in the first quadrant, we have, sin θ > 0, cos θ > 0 and tan θ > 0., Again, since φ lies in the first quadrant, we have, sin φ > 0, cos φ > 0 and tan φ > 0., 225 ⎞, 64, ⎛, Now, cos2 θ = (1 − sin 2 θ) = ⎜1 −, ⎟ =, ⎝, ⎠, 289, 289, 64, 8, =, ⋅, ⇒ cos θ = +, 289 17, 144⎞, 25, ⎛, And, sin 2 φ = (1 − cos2 φ) = ⎜1 −, ⎟ =, ⎝, 169⎠ 169, 25, 5, =, ⋅, ⇒ sin φ = +, 169 13, sin θ ⎛ 15 17 ⎞ 15, and, =⎜, × ⎟ =, ∴ tan θ =, 8⎠, 8, cos θ ⎝ 17, 13 ⎞ 5, sin φ ⎛ 5, =⎜, × ⎟ =, ⋅, tan φ =, ⎝, 13 12 ⎠ 12, cos φ, Given: sin θ =
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546, SOLUTION, , Senior Secondary School Mathematics for Class 11, , Here, α lies in Quadrant III and, therefore, tan α is positive., 1, Now, cot α =, ⇒ tan α = 2., 2, Again, β lies in Quadrant II and, therefore, sin β is positive and cos β, is negative., −5, −3, Now, sec β =, ⇒ cos β =, ⋅, 3, 5, 9, 16 4, And, sin β = + 1 − cos2 β = + 1 −, =+, = ⋅, 25, 25 5, sin β 4 ⎛ −5 ⎞ − 4, ∴ tan β =, = ×⎜ ⎟ =, ⋅, cos β 5 ⎝ 3 ⎠, 3, tan α + tan β, ∴ tan (α + β ) =, 1 − tan α tan β, ⎛ 2⎞, 4, ⎜ ⎟, 2−, ⎝ 3⎠ ⎛ 2, 2, 3⎞, 3, =, =, ⋅, =⎜ × ⎟ =, ⎛ − 4⎞ ⎛ 11⎞ ⎝ 3 11⎠ 11, 1−2× ⎜ ⎟ ⎜ ⎟, ⎝ 3 ⎠ ⎝ 3⎠, , EXAMPLE 18, SOLUTION, , Prove that tan 56° =, , cos 11° + sin 11°, ⋅, cos 11° − sin 11°, , We have, LHS = tan 56° = tan ( 45 ° + 11°), , tan 45 ° + tan 11°, 1 − tan 45 ° tan 11°, 1 + tan 11°, =, 1 − tan 11°, sin 11°, 1+, cos 11° cos 11° + sin 11°, = RHS., =, =, sin 11° cos 11° − sin 11°, 1−, cos 11°, =, , EXAMPLE 19, SOLUTION, , Prove that tan 70° = tan 20° + 2 tan 50°., , We have, tan 70° = tan ( 20° + 50°), tan 20° + tan 50°, ⇒ tan 70° =, 1 − tan 20° tan 50°, ⇒, ⇒, ⇒, ⇒, ⇒, , tan 70° − tan 70° tan 20° tan 50° = tan 20° + tan 50°, tan 70° − tan ( 90° − 20°) tan 20° tan 50° = tan 20° + tan 50°, tan 70° − cot 20° tan 20° tan 50° = tan 20° + tan 50°, tan 70° − tan 50° = tan 20° + tan 50°, tan 70° = tan 20° + 2 tan 50°.
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Trigonometric, or Circular, Functions, EXAMPLE 20, SOLUTION, , 547, , Show that cot 2x cot x − cot 3 x cot 2x − cot 3 x cot x = 1., , We have, cot 3 x = cot ( 2x + x), cot 2x cot x − 1, ⇔ cot 3 x =, cot 2x + cot x, ⇔ cot 3 x cot 2x + cot 3 x cot x = cot 2x cot x − 1, ⇔ cot 2x cot x − cot 3 x cot 2x − cot 3 x cot x = 1., , EXERCISE 15B, 1. Find the value of, (i) cos 840°, (iv) sec ( − 420°), , (iii) tan ( −120°), , (ii) sin 870°, , (v) cosec ( − 690°) (vi) tan ( 225°), , (vii) cot ( −315 °) (viii) sin ( −1230°), , (ix) cos ( 495°), , 2. Find the values of all trigonometric functions of 135°., 3. Prove that, (i) sin 80° cos 20° − cos 80° sin 20° =, , 3, 2, , (ii) cos 45 ° cos 15 ° − sin 45 ° sin 15 ° =, , 1, 2, , (iii) cos 75 ° cos 15 ° + sin 75 ° sin 15 ° =, , 1, 2, , (iv) sin 40° cos 20° + cos 40° sin 20° =, , 3, 2, , (v) cos 130° cos 40° + sin 130° sin 40° = 0, 4. Prove that, (i) sin (50° + θ) cos ( 20° + θ) − cos (50° + θ) sin ( 20° + θ) =, , 1, 2, , (ii) cos (70° + θ) cos (10° + θ) + sin (70° + θ) sin (10° + θ) =, , 1, 2, , 5. Prove that, (i) cos (n + 2) x cos (n + 1) x + sin (n + 2) x sin (n + 1) x = cos x, ⎛π, ⎛π, ⎞, ⎞, (ii) cos ⎜ − x⎟ cos ⎜ − y⎟ − sin, ⎝4, ⎠, ⎝4, ⎠, , ⎛π, ⎛π, ⎞, ⎞, ⎜ − x⎟ sin ⎜ − y⎟ = sin ( x + y), ⎝4, ⎠, ⎝4, ⎠
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548, , Senior Secondary School Mathematics for Class 11, , ⎛π, ⎞, 2, tan ⎜ + x⎟, ⎝4, ⎠ ⎛ 1 + tan x ⎞, =⎜, ⎟ ⋅, ⎛π, ⎞ ⎝ 1 − tan x ⎠, tan ⎜ − x⎟, ⎝4, ⎠, , 6. Prove that, , 7. Prove that, ( 6 + 2), 4, cos 135 ° − cos 120°, (ii), = ( 3 − 2 2), cos 135 ° + cos 120°, (i) sin 75 ° =, , (iii) tan 15 ° + cot 15 ° = 4, 8. Prove that, (i) cos 15 ° − sin 15 ° =, , 1, 2, , (ii) cot 105 ° − tan 105 ° = 2 3, (iii), , tan 69° + tan 66°, = −1, 1 − tan 69° tan 66°, , 9. Prove that, , cos 9° + sin 9°, = tan 54°., cos 9° − sin 9°, , 10. Prove that, , cos 8° − sin 8°, = tan 37 °., cos 8° + sin 8°, , cos ( π + θ) cos ( −θ), = − cot θ ., ⎛π, ⎞, cos ( π − θ) cos ⎜ + θ⎟, ⎝2, ⎠, cos θ, sin ( −θ), tan ( 90° + θ), 12. Prove that, +, −, = 3., sin ( 90° + θ) sin (180° + θ), cot θ, , 11. Prove that, , sin (180° + θ) cos ( 90° + θ) tan ( 270° − θ) cot ( 360° − θ), = 1., sin ( 360° − θ) cos ( 360° + θ) cosec ( − θ) sin ( 270° + θ), 8, 12, 14. If θ and φ lie in the first quadrant such that sin θ =, and cos φ =, , find, 17, 13, the values of, 13. Prove that, , (i) sin (θ − φ),, , (ii) cos (θ + φ),, , (iii) tan (θ − φ)., , 15. If x and y are acute angles such that sin x =, ( x + y) =, , π, ⋅, 4, , 1, 1, and sin y =, , prove that, 5, 10, , 16. If x and y are acute angles such that cos x =, ( x − y) = −, , π, ⋅, 3, , 13, 1, and cos y = , prove that, 14, 7
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Trigonometric, or Circular, Functions, , 17. If sin x =, , 549, , 12, 4, π, π, and sin y = , where < x < π and 0 < y < , find the values of, 13, 5, 2, 2, , (i) sin ( x + y),, (ii) cos ( x + y), (iii) tan ( x − y)., 3, −24, 3π, 3π, 18. If cos x = and cos y =, , where, , find the, < x < 2π and π < y <, 5, 25, 2, 2, values of, (i) sin ( x + y),, , (ii) cos ( x − y),, , (iii) tan ( x + y)., , 19. Prove that, Trigonometric, or Circular, Functions, , ⎛π, ⎞ 1, (i) cos ⎜ + x⎟ = (cos x − 3 sin x), ⎠ 2, ⎝3, ⎛π, ⎛π, ⎞, ⎞, (ii) sin ⎜ + x⎟ + sin ⎜ − x⎟ = 2 cos x, ⎠, ⎝4, ⎠, ⎝4, (iii), , 1, ⎛π, ⎞ 1, cos ⎜ + x⎟ = (cos x − sin x), ⎝4, ⎠ 2, 2, , ⎛ 2π, ⎛ 2π, ⎞, ⎞, (iv) cos x + cos ⎜, + x⎟ + cos ⎜, − x⎟ = 0, ⎝ 3, ⎠, ⎝ 3, ⎠, 20. Prove that, 5π, π 1, sin, =, 12, 12 2, 5π, π 1, (ii) 2 cos, cos, =, 12, 12 2, (i) 2 sin, , (iii) 2 sin, , 5π, π (2 + 3), cos, =, 12, 12, 2, ANSWERS (EXERCISE 15B), , 1. (i) −, , 1, 1, 1, 1, (ii), (ix) −, (iii) 3 (iv) 2 (v) 2 (vi) 1 (vii) 1 (viii) −, 2, 2, 2, 2, , 2. sin 135 ° =, , 1, −1, , cos 135 ° =, , tan 135 ° = − 1, cot 135 ° = − 1,, 2, 2, , sec 135 ° = − 2 , cosec 135 ° = 2, 14. (i), , 21, 140, (ii), 221, 221, , (iii), , 21, 220, , 17. (i), , 16, 65, , (ii), , − 63, 65, , (iii), , 56, 33, , 18. (i), , 3, 5, , (ii), , − 44, −3, (iii), 125, 4
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550, , Senior Secondary School Mathematics for Class 11, HINTS TO SOME SELECTED QUESTIONS, , 4. (i) LHS = sin x cos y − cos x sin y, = sin ( x − y ) = sin [(50 ° + θ) − ( 20 ° + θ)] = sin 30 ° ., (ii) LHS = cos x cos y + sin x sin y, = cos ( x − y ) = cos [(70 ° + θ) − ( 10 ° + θ)] = cos 60 ° ., 5. (i) LHS = cos A cos B + sin A sin B, = cos ( A − B) = cos [(n + 2 )x − (n + 1)x] = cos x., (ii) LHS = cos A cos B − sin A sin B, ⎡ π, ⎤, π, = cos ( A + B) = cos ⎢ ⎛⎜ − x ⎞⎟ + ⎛⎜ − y ⎞⎟ ⎥, ⎝, ⎠, ⎝, ⎠, 4, 4, ⎣, ⎦, ⎡π, ⎤, = cos ⎢ − ( x + y )⎥ = sin ( x + y )., ⎣2, ⎦, 7. (i) sin 75 ° = sin ( 45 ° + 30 ° ), (ii) cos 135 ° = cos ( 180 ° − 45 ° ) = − cos 45 ° ., cos 120 ° = cos ( 180 ° − 60 ° ) = − cos 60 ° ., tan 45 ° − tan 30 °, (iii) tan 15 ° = tan ( 45 ° − 30 ° ) =, ⋅, 1 + tan 45 ° tan 30 °, 8. (i) cos 15 ° = cos ( 45 ° − 30 ° ) and sin 15 ° = sin ( 45 ° − 30 ° )., (ii) tan 105 ° = tan ( 60 ° + 45 ° )., (iii) LHS = tan ( 69 ° + 66 ° ) = tan 135 °, = tan ( 180 ° − 45 ° ) = − tan 45 ° = −1., 9. Dividing num. and denom. by cos 9° , we get, LHS =, Trigonometry, , 1 + tan 9 °, tan 45 ° + tan 9 °, =, 1 − tan 9 ° 1 − tan 45 ° tan 9 °, , = tan ( 45 ° + 9 ° ) = tan 54 ° ., 17. Clearly, sin x > 0 , cos x < 0 and tan x < 0., Also, sin y > 0 , cos y > 0 and tan y > 0., 18. In the 4th quadrant, cos x > 0 , sin x < 0 and tan x < 0., In the 3rd quadrant, tan y > 0 , sin y < 0 and cos y < 0., 9, 16 −4, =−, =, ,, 25, 25, 5, −4 5 ⎞ −4, tan x = ⎛⎜, × ⎟=, ⎝ 5, 3⎠, 3, −24, 576, cos y =, , sin y = − 1 −, =−, 25, 625, Now, find the required values., , ∴, , sin x = − 1 −, , 49, −7, 7, and tan y =, =, ⋅, 625 25, 24
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Trigonometric, or Circular, Functions, , 551, , SOME MORE TRIGONOMETRIC FUNCTIONS, THEOREM 1, , PROOF, , For all x , y ∈ R , we have, (a) 2sin x cos y = sin ( x + y) + sin ( x − y), (b) 2cos x sin y = sin ( x + y) − sin ( x − y), (c) 2cos x cos y = cos ( x + y) + cos ( x − y), (d) 2sin x sin y = cos ( x − y) − cos ( x + y), , We know that, sin ( x + y) = sin x cos y + cos x sin y, sin ( x − y) = sin x cos y − cos x sin y, , … (i), … (ii), , (a) Adding (i) and (ii), we get, 2sin x cos y = sin ( x + y) + sin ( x − y)., (b) Subtracting (ii) from (i), we get, 2cos x sin y = sin ( x + y) − sin ( x − y)., Further, we know that, cos ( x + y) = cos x cos y − sin x sin y, cos ( x − y) = cos x cos y + sin x sin y, (c) Adding (iii) and (iv), we get, 2cos x cos y = cos ( x + y) + cos ( x − y)., (d) Subtracting (iii) from (iv), we get, 2sin x sin y = cos ( x − y) − cos ( x + y)., THEOREM 2, , PROOF, , … (iii), … (iv), , For all x , y ∈ R , we have, , ⎛ x + y⎞, ⎛ x − y⎞, (i) sin x + sin y = 2 sin ⎜, ⎟ cos ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, ⎛ x + y⎞, ⎛ x − y⎞, (ii) sin x − sin y = 2 cos ⎜, ⎟ sin ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, ⎛ x + y⎞, ⎛ x − y⎞, (iii) cos x + cos y = 2 cos ⎜, ⎟ cos ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, ⎛ x + y⎞, ⎛ x − y⎞, (iv) cos x − cos y = − 2 sin ⎜, ⎟ sin ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, , ⎛ x + y⎞, ⎛ x − y⎞, Putting x = ( a + b) and y = ( a − b), we get a = ⎜, ⎟ and b = ⎜, ⎟⋅, ⎝ 2 ⎠, ⎝ 2 ⎠, (i) sin x + sin y = sin ( a + b) + sin ( a − b), = 2sin a cos b, ⎛ x + y⎞, ⎛ x − y⎞, = 2 sin ⎜, ⎟ cos ⎜, ⎟⋅, ⎝ 2 ⎠, ⎝ 2 ⎠, (ii) sin x − sin y = sin ( a + b) − sin ( a − b), = 2cos a sin b, ⎛ x + y⎞, ⎛ x − y⎞, = 2 cos ⎜, ⎟ sin ⎜, ⎟⋅, ⎝ 2 ⎠, ⎝ 2 ⎠
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552, , Senior Secondary School Mathematics for Class 11, , (iii) cos x + cos y = cos ( a + b) + cos ( a − b), = 2cos a cos b, ⎛ x + y⎞, ⎛ x − y⎞, = 2 cos ⎜, ⎟ cos ⎜, ⎟⋅, ⎝ 2 ⎠, ⎝ 2 ⎠, (iv) cos x − cos y = cos ( a + b) − cos ( a − b), = − 2sin a sin b, ⎛ x + y⎞, ⎛ x − y⎞, = − 2 sin ⎜, ⎟ sin ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, These formulae may be listed as under:, C + D⎞, ⎛C − D⎞, I. sin C + sin D = 2sin ⎛⎜, ⎟, ⎟ cos ⎜, ⎝ 2 ⎠, ⎝ 2 ⎠, II. sin C − sin D = 2cos ⎛⎜, ⎝, , C + D⎞, ⎛C − D⎞, ⎟, ⎟ sin ⎜, ⎝ 2 ⎠, 2 ⎠, , III. cos C + cos D = 2cos ⎛⎜, ⎝, , C + D⎞, ⎛C − D⎞, ⎟, ⎟ cos ⎜, ⎝ 2 ⎠, 2 ⎠, , C + D⎞, ⎛C − D⎞, IV. cos C − cos D = −2sin ⎛⎜, ⎟, ⎟ sin ⎜, ⎝ 2 ⎠, ⎝ 2 ⎠, , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Prove that, ⎛π, ⎛π, ⎞, ⎞, (i) cos ⎜ + x⎟ + cos ⎜ − x⎟ = 2 cos x, ⎠, ⎝4, ⎝4, ⎠, ⎛ 3π, ⎛ 3π, ⎞, ⎞, (ii) cos ⎜, + x⎟ − cos ⎜, − x⎟ = − 2 sin x, ⎝ 4, ⎠, ⎝ 4, ⎠, ⎛π, ⎛π, ⎞, ⎞, (i) LHS = cos ⎜ + x⎟ + cos ⎜ − x⎟, ⎝4, ⎠, ⎝4, ⎠, π, = 2 cos cos x [B cos ( A + B) + cos ( A − B) = 2 cos A cos B ], 4, 1, ⎛, ⎞, = ⎜2 ×, cos x⎟ = 2 cos x = RHS., ⎝, ⎠, 2, ⎛ 3π, ⎛ 3π, ⎞, ⎞, (ii) LHS = cos ⎜, + x⎟ − cos ⎜, − x⎟, ⎝ 4, ⎠, ⎝ 4, ⎠, = −2 sin, , 3π, sin x, 4, [B cos ( A + B) − cos ( A − B) = − 2 sin A sin B ], , π⎞, ⎛, = −2 sin ⎜ π − ⎟ sin x, ⎝, 4⎠
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Trigonometric, or Circular, Functions, , π, sin x, 4, 1 ⎞, ⎛, = ⎜ −2 ×, ⎟ sin x, ⎝, 2⎠, = −2 sin, , EXAMPLE 2, , SOLUTION, , EXAMPLE 3, , SOLUTION, , 553, , π⎞, π⎤, ⎡, ⎛, ⎢B sin ⎜⎝ π − 4 ⎟⎠ = sin 4 ⎥, ⎣, ⎦, = − 2 sin x = RHS., , Express each of the following as an algebraic sum of sines or cosines:, (i) 2 sin 3 x cos 2x, (ii) 2 cos 4x sin 2x, (iii) 2 cos 6x cos 4x, (iv) 2 sin 3 x sin 5 x, (i) We know that 2sin A cos B = sin ( A + B) + sin ( A − B)., ∴ 2 sin 3 x cos 2x = sin ( 3 x + 2x) + sin ( 3 x − 2x), = sin 5x + sin x., (ii) We know that 2cos A sin B = sin ( A + B) − sin ( A − B)., ∴ 2 cos 4x sin 2x = sin ( 4x + 2x) − sin ( 4x − 2x), = sin 6x − sin 2x., (iii) We know that 2cos A cos B = cos ( A + B) + cos ( A − B)., ∴ 2 cos 6x cos 4x = cos ( 6x + 4x) + cos ( 6x − 4x), = cos 10x + cos 2x., (iv) We know that 2sin A sin B = cos ( A − B) − cos ( A + B)., ∴ 2 sin 3 x sin 5 x = 2 sin 5 x sin 3 x, = cos (5 x − 3 x) − cos (5 x + 3 x), = cos 2x − cos 8x., Express each of the following as a product of sines or cosines or sine and, cosine:, (i) cos 5 x + cos 3 x, (ii) cos 5 x − cos 7 x, (iii) sin 7 x + sin 3 x, (iv) sin 5 x − sin 3 x, ⎛ C + D⎞, ⎛ C − D⎞, (i) We know that cos C + cos D = 2 cos ⎜, ⎟ cos ⎜, ⎟⋅, ⎝ 2 ⎠, ⎝ 2 ⎠, ⎛ 5x + 3x⎞, ⎛ 5x − 3x⎞, ∴ cos 5 x + cos 3 x = 2 cos ⎜, ⎟ cos ⎜, ⎟, ⎝, ⎝, 2 ⎠, 2 ⎠, = 2 cos 4x cos x., ⎛ C + D⎞, ⎛ C − D⎞, (ii) We know that cos C − cos D = − 2 sin ⎜, ⎟ sin ⎜, ⎟⋅, ⎝ 2 ⎠, ⎝ 2 ⎠, ⎛ 5x + 7x⎞, ⎛ 5x − 7x⎞, ∴ cos 5 x − cos 7 x = − 2 sin ⎜, ⎟ sin ⎜, ⎟, ⎝, ⎝, 2 ⎠, 2 ⎠, = − 2 sin 6x sin ( − x) = 2 sin 6x sin x., ⎛ C + D⎞, ⎛ C − D⎞, (iii) We know that sin C + sin D = 2 sin ⎜, ⎟ cos ⎜, ⎟⋅, ⎝ 2 ⎠, ⎝ 2 ⎠, ⎛ 7x + 3x⎞, ⎛ 7x − 3x⎞, ∴ sin 7 x + sin 3 x = 2 sin ⎜, ⎟ cos ⎜, ⎟, ⎝, ⎠, ⎝, 2, 2 ⎠, = 2 sin 5 x cos 2x.
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554, , Senior Secondary School Mathematics for Class 11, , ⎛ C + D⎞, ⎛ C − D⎞, (iv) We know that sin C − sin D = 2 cos ⎜, ⎟ sin ⎜, ⎟⋅, ⎝ 2 ⎠, ⎝ 2 ⎠, ⎛ 5x + 3x⎞, ⎛ 5x − 3x⎞, ∴ sin 5 x − sin 3 x = 2 cos ⎜, ⎟ sin ⎜, ⎟, ⎝, ⎝, 2 ⎠, 2 ⎠, = 2 cos 4x sin x., EXAMPLE 4, , SOLUTION, , Prove that, We have, LHS =, , cos 6x + cos 4x, = cot x., sin 6x − sin 4x, cos 6x + cos 4x, sin 6x − sin 4x, , ⎛ 6x, 2 cos ⎜, ⎝, =, ⎛ 6x, 2 cos ⎜, ⎝, , + 4x ⎞, ⎛ 6x, ⎟ cos ⎜, ⎝, 2 ⎠, + 4x ⎞, ⎛ 6x, ⎟ sin ⎜, ⎝, 2 ⎠, , − 4x ⎞, ⎟, 2 ⎠, − 4x ⎞, ⎟, 2 ⎠, , ⎧, ⎛C, ⎪Q cos C + cos D = 2 cos ⎜⎝, ⎪, ⎨, ⎪ sin C − sin D = 2 cos ⎛⎜ C, ⎪⎩, ⎝, 2 cos 5 x cos x cos x, =, =, 2 cos 5 x sin x sin x, , + D⎞, ⎛ C − D⎞ ⎫, ⎟ cos ⎜, ⎟, ⎝ 2 ⎠ ⎪⎪, 2 ⎠, ⎬, + D⎞, ⎛ C − D⎞ ⎪, ⎟ sin ⎜, ⎟, ⎝ 2 ⎠ ⎪⎭, 2 ⎠, , = cot x = RHS., EXAMPLE 5, , SOLUTION, , Prove that, We have, LHS =, , sin 3 x − sin x, = cot 2x., cos x − cos 3 x, sin 3 x − sin x, cos x − cos 3 x, , ⎛ 3x + x⎞, ⎛ 3x − x⎞, 2 cos ⎜, ⎟ sin ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, =, ⎛ x − 3x⎞, ⎛ x + 3x⎞, −2 sin ⎜, ⎟, ⎟ sin ⎜, ⎝ 2 ⎠, ⎝ 2 ⎠, ⎧, ⎛ C + D⎞, ⎛ C − D⎞ ⎫, ⎪Q sin C − sin D = 2 cos ⎜⎝ 2 ⎟⎠ sin ⎜⎝ 2 ⎟⎠ ⎪, ⎪, ⎪, ⎬, ⎨, C, +, D, C, D, −, ⎛, ⎛, ⎞, ⎞, ⎪ cos C − cos D = − 2sin ⎜, ⎟ sin ⎜, ⎟⎪, ⎪⎩, ⎝ 2 ⎠, ⎝ 2 ⎠ ⎪⎭, =, , cos 2x sin x, cos 2x sin x, =, − sin 2x sin ( − x) sin 2x sin x, , = cot 2x = RHS.
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Trigonometric, or Circular, Functions, , EXAMPLE 6, , SOLUTION, , Prove that, We have, LHS =, , 555, , sin 3 x − sin x, = 2 sin x., cos 2x, sin 3 x − sin x, cos 2x, , ⎛ 3x + x⎞, ⎛ 3x − x⎞, 2 cos ⎜, ⎟ sin ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, =, cos 2x, ⎡, ⎛ C + D⎞, ⎛ C − D⎞ ⎤, ⎢Q sin C − sin D = 2 cos ⎜⎝ 2 ⎟⎠ sin ⎜⎝ 2 ⎟⎠ ⎥, ⎣, ⎦, 2 cos 2x sin x, =, = 2 sin x = RHS., cos 2x, EXAMPLE 7, , SOLUTION, , Prove that, We have, LHS =, , =, , sin 5 x − 2 sin 3 x + sin x, = cosec 2x − cot 2x., cos 5 x − cos x, (sin 5 x + sin x) − 2 sin 3 x, cos 5 x − cos x, ⎛ 5x +, 2 sin ⎜, ⎝ 2, , x⎞, ⎛ 5x − x⎞, ⎟ cos ⎜, ⎟ − 2 sin 3 x, ⎠, ⎝ 2 ⎠, ⎛ 5x − x⎞, ⎛ 5x + x⎞, −2 sin ⎜, ⎟ sin ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, ⎡, ⎛ C + D⎞, ⎛ C − D⎞, ⎢B sin C + sin D = 2 sin ⎜⎝ 2 ⎟⎠ cos ⎜⎝ 2 ⎟⎠ and, ⎣, ⎛ C + D⎞, ⎛ C − D⎞ ⎤, cos C − cos D = − 2 sin ⎜, ⎟ sin ⎜, ⎟, ⎝ 2 ⎠ ⎥⎦, ⎝ 2 ⎠, , 2 sin 3 x cos 2x − 2 sin 3 x 2 sin 3 x(cos 2x − 1), =, −2 sin 3 x sin 2x, −2 sin 3 x sin 2x, (1 − cos 2x), 1, cos 2x, =, =, −, sin 2x, sin 2x sin 2x, =, , = cosec 2x − cot 2x = RHS., EXAMPLE 8, , SOLUTION, , If, , cos ( A + B) sin (C + D), =, , prove that tan A tan B tan C + tan D = 0., cos ( A − B) sin (C − D), , We have, , ⇒, , cos ( A, cos ( A, cos ( A, cos ( A, , + B) sin (C + D), =, − B) sin (C − D), + B) + cos ( A − B) sin (C + D) + sin (C − D), =, + B) − cos ( A − B) sin (C + D) − sin (C − D)
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556, , Senior Secondary School Mathematics for Class 11, , 2 cos A cos B, 2 sin C cos D, =, −2 sin A sin B 2 cos C sin D, ⇒ − cot A cot B = tan C cot D, ⇒ tan A tan B tan C = − tan D, ⇒ tan A tan B tan C + tan D = 0., Hence, tan A tan B tan C + tan D = 0., , ⇒, , Prove that, , EXAMPLE 9, , SOLUTION, , We have, , cos 4x + cos 3 x + cos 2x, = cot 3 x., sin 4x + sin 3 x + sin 2x, , (cos 4x + cos 2x) + cos 3 x, (sin 4x + sin 2x) + sin 3 x, ⎛ 4x + 2x ⎞, ⎛ 4x − 2x ⎞, 2 cos ⎜, ⎟ cos ⎜, ⎟ + cos 3 x, ⎝, ⎝, 2 ⎠, 2 ⎠, =, ⎛ 4x − 2x ⎞, ⎛ 4x + 2x ⎞, 2 sin ⎜, ⎟ cos ⎜, ⎟ + sin 3 x, ⎝, ⎝, 2 ⎠, 2 ⎠, [using formulae for (cos C + cos D) and (sin C + sin D)], , LHS =, , 2 cos 3 x cos x + cos 3 x cos 3 x( 2 cos x + 1), =, 2 sin 3 x cos x + sin 3 x sin 3 x( 2 cos x + 1), cos 3 x, =, = cot 3 x = RHS., sin 3 x, =, , EXAMPLE 10, SOLUTION, , Prove that sin x + sin 3 x + sin 5 x + sin 7 x = 4 sin 4x cos 2x cos x., , We have, LHS = (sin 7 x + sin x) + (sin 5 x + sin 3 x), , ⎛ 7x + x⎞, ⎛ 5x − 3x⎞, ⎛ 7x − x⎞, ⎛ 5x + 3x⎞, = 2 sin ⎜, ⎟ cos ⎜, ⎟, ⎟ + 2 sin ⎜, ⎟ cos ⎜, ⎝ 2 ⎠, ⎝, ⎝ 2 ⎠, ⎝, 2 ⎠, 2 ⎠, = 2 sin 4x cos 3 x + 2 sin 4x cos x, = 2 sin 4x (cos 3 x + cos x), ⎛ 3x + x⎞, ⎛ 3x − x⎞, = ( 2 sin 4x) × 2 cos ⎜, ⎟ cos ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, = ( 2 sin 4x) × 2 cos 2x cos x, = 4 sin 4x cos 2x cos x = RHS., EXAMPLE 11, , SOLUTION, , Prove that cos 2x cos, , x, 9x, ⎛ 5x⎞, − cos 3 x cos, = − sin 5 x sin ⎜ ⎟ ⋅, ⎝ 2⎠, 2, 2, , We have, 1⎡, x, 9x ⎤, LHS =, 2 cos 2x cos − 2 cos 3 x cos ⎥, 2 ⎢⎣, 2, 2⎦, 1⎡, x⎞, x⎞ ⎤ 1 ⎡, ⎛, ⎛, ⎛ 9x, ⎛ 9x, ⎞, ⎞⎤, + 3 x⎟ + cos ⎜, = ⎢cos ⎜ 2x + ⎟ + cos ⎜ 2x − ⎟ ⎥ − ⎢cos ⎜, − 3 x⎟ ⎥, ⎝, ⎝, ⎝ 2, ⎠, ⎝ 2, ⎠⎦, 2⎣, 2⎠, 2⎠ ⎦ 2 ⎣
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Trigonometric, or Circular, Functions, , 1, 2, 1, =, 2, =, , 557, , 5x, 3x⎞ 1 ⎛, 15 x, 3x⎞, ⎛, + cos, + cos, ⎜ cos, ⎟ − ⎜ cos, ⎟, ⎝, 2, 2⎠ 2⎝, 2, 2⎠, 5x, 15 x ⎞, ⎛, − cos, ⎜ cos, ⎟, ⎝, 2, 2 ⎠, , ⎡, ⎛ 5 x 15 x ⎞, ⎛ 15 x 5 x ⎞ ⎤, +, − ⎟⎥, ⎜, ⎜, ⎟, ⎝ 2, ⎝ 2, 1⎢, 2 ⎠, 2⎠, = ⎢−2 sin, sin, ⎥, 2⎢, 2, 2, ⎥, ⎢⎣, ⎥⎦, 5, x, ⎛ ⎞, = − sin 5 x sin ⎜ ⎟ = RHS., ⎝ 2⎠, EXAMPLE 12, SOLUTION, , Prove that, , We have, , sin 8x cos x − sin 6x cos 3 x, cos 2x cos x − sin 4x sin 3 x, , = tan 2x., , 2 sin 8x cos x − 2 sin 6x cos 3 x, 2 cos 2x cos x − 2 sin 4x sin 3 x, [sin ( 8x + x) + sin ( 8x − x)] − [sin ( 6x + 3 x) + sin ( 6x − 3 x)], =, [cos ( 2x + x) + cos ( 2x − x)] − [cos ( 4x − 3 x) − cos ( 4x + 3 x)], [B 2 sin A cos B = sin ( A + B) + sin ( A − B), etc. ], (sin 9x + sin 7 x) − (sin 9x + sin 3 x), =, (cos 3 x + cos x) − (cos x − cos 7x), (sin 7 x − sin 3 x), =, (cos 3 x + cos 7 x), , LHS =, , ⎛ 7x, 2 cos ⎜, ⎝, =, ⎛ 7x, 2 cos ⎜, ⎝, , + 3x⎞, ⎛ 7x, ⎟ sin ⎜, ⎝, 2 ⎠, + 3x⎞, ⎛ 7x, ⎟ cos ⎜, ⎝, 2 ⎠, , − 3x⎞, ⎟, 2 ⎠, − 3x⎞, ⎟, 2 ⎠, , (C + D ), (C − D ) ⎤, ⎡, sin, ⎢⎣Q sin C − sin D = 2 cos, 2, 2 ⎥⎦, cos 5 x sin 2x sin 2x, =, =, cos 5 x cos 2x cos 2x, = tan 2x = RHS., (sin x − sin y), ⎛ x − y⎞, EXAMPLE 13 Prove that, = tan ⎜, ⎟⋅, ⎝ 2 ⎠, (cos x + cos y), SOLUTION We have, sin x − sin y, LHS =, cos x + cos y, ⎛x + y ⎞, ⎛ x − y⎞, 2 cos ⎜, ⎟ sin ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, ⎛ x − y⎞, =, = tan ⎜, ⎟ = RHS., ⎝ 2 ⎠, ⎛ x + y⎞, ⎛ x − y⎞, 2 cos ⎜, cos, ⎜, ⎟, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠
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Trigonometric, or Circular, Functions, , EXAMPLE 17, SOLUTION, , Prove that, , cos 20° cos 40° cos 60° cos 80° =, , 559, , 1, ⋅, 16, , We have, cos 20° cos 40° cos 60° cos 80°, 1, = cos 60° cos 40°( 2 cos 80° cos 20°), 2, 1 1, = × cos 40° ⋅ {cos ( 80° + 20°) + cos ( 80° − 20°)}, 2 2, 1, = cos 40°(cos 100° + cos 60°), 4, 1⎞, 1, ⎛, = cos 40° ⎜ cos 100° + ⎟, ⎝, 2⎠, 4, 1, 1, ( 2 cos 100° cos 40°) + cos 40°, 8, 8, 1, 1, = [cos (100° + 40°) + cos (100° − 40°)] + cos 40°, 8, 8, 1, 1, = (cos 140° + cos 60°) + cos 40°, 8, 8, 1, 1, 1, = cos 140° +, + cos 40°, 8, 16 8, 1, 1, 1, = cos (180° − 40°) +, + cos 40°, 8, 16 8, 1, 1, 1, = − cos 40° +, + cos 40°, 8, 16 8, 1, =, = RHS., 16, =, , EXAMPLE 18, SOLUTION, , Prove that, , sin 10° sin 50° sin 60° sin 70° =, , 3, ⋅, 16, , We have, sin 10° sin 50° sin 60° sin 70°, 1, = sin 60° sin 10°( 2 sin 70° sin 50°), 2, 1, 3, = ×, × sin 10° ⋅ {cos (70° − 50°) − cos (70° + 50°)}, 2, 2, 3, sin 10° ⋅ {cos 20° − cos 120°}, 4, 3, 1⎞, ⎛, =, sin 10° ⎜ cos 20° + ⎟, ⎝, 4, 2⎠, =, , =, , 3, 3, ( 2 cos 20° sin 10°) +, sin 10°, 8, 8, , ⎡, ⎢⎣Q cos 120° = −, , 1⎤, 2 ⎥⎦
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560, , Senior Secondary School Mathematics for Class 11, , 3, 3, [sin ( 20° + 10°) − sin ( 20° − 10°)] +, sin 10°, 8, 8, ⎛ 3 1⎞, 3, 3, =, × ⎟ =, sin 30° = ⎜, ⋅, 8, 8, 2, 16, ⎝, ⎠, =, , EXERCISE 15C, Prove that, , 1. sin (150° + x) + sin (150° − x) = cos x, 2. cos x + cos (120° − x) + cos (120° + x) = 0, π⎞, π⎞, ⎛, ⎛, 3. sin ⎜ x − ⎟ + cos ⎜ x − ⎟ = 3 sin x, ⎝, ⎝, 6⎠, 3⎠, ⎛π, ⎞ 1 + tan x, 4. tan ⎜ + x⎟ =, ⎝4, ⎠ 1 − tan x, ⎛π, ⎞ 1 − tan x, 5. tan ⎜ − x⎟ =, ⎝4, ⎠ 1 + tan x, 6. Express each of the following as a product:, (i) sin 10x + sin 6x, (iii) cos 7 x + cos 5 x, , (ii) sin 7 x − sin 3 x, (iv) cos 2x − cos 4x, , 7. Express each of the following as an algebraic sum of sines or cosines:, (i) 2 sin 6x cos 4x, , (ii) 2 cos 5 x sin 3 x, , (iii) 2 cos 7 x cos 3 x, , (iv) 2 sin 8x sin 2x, , Prove that, , sin x + sin 3 x, = cot x, cos x − cos 3 x, , 9., , 10., , sin 5 x + sin 3 x, = tan 4x, cos 5 x + cos 3 x, , 11., , 12., , sin x + sin 3 x + sin 5 x, = tan 3 x, cos x + cos 3 x + cos 5 x, , 13., , (sin 7 x + sin 5 x) + (sin 9x + sin 3 x), = tan 6x, (cos 7 x + cos 5 x) + (cos 9x + cos 3 x), , 8., , sin 7 x − sin 5 x, = tan x, cos 7 x + cos 5 x, cos 9x − cos 5 x, − sin 2x, =, sin 17 x − sin 3 x cos 10x, , 14. cot 4x(sin 5 x + sin 3 x) = cot x(sin 5 x − sin 3 x), 15. (sin 3 x + sin x) sin x + (cos 3 x − cos x) cos x = 0, ⎛ x − y⎞, 16. (cos x − cos y) 2 + (sin x − sin y) 2 = 4 sin 2 ⎜, ⎟, ⎝ 2 ⎠
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Trigonometric, or Circular, Functions, , 17., , sin 2x − sin 2y, = cot ( x + y), cos 2y − cos 2x, , 18., , cos x + cos y, ⎛ x + y⎞, ⎛ x − y⎞, = cot ⎜, ⎟ cot ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, cos y − cos x, , 19., , sin x + sin y, ⎛ x + y⎞, ⎛ x − y⎞, = tan ⎜, ⎟ cot ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, sin x − sin y, , 20. sin 3 x + sin 2x − sin x = 4 sin x cos, , 561, , x, 3x, cos, 2, 2, , 21., , cos 4x sin 3 x − cos 2x sin x, = tan 2x, sin 4x sin x + cos 6x cos x, , 22., , cos 2x sin x + cos 6x sin 3 x, = cot 5 x, sin 2x sin x + sin 6x sin 3 x, , 1, 16, 3, 24. sin 20° sin 40° sin 60° sin 80° =, 16, 3, 25. cos 10° cos 30° cos 50° cos 70° =, 16, 1, 1, ⎛ x + y⎞ 3, 26. If cos x + cos y = and sin x + sin y = , prove that tan ⎜, ⎟ = ⋅, ⎝ 2 ⎠ 4, 3, 4, 23. sin 10° sin 30° sin 50° sin 70° =, , 27. Prove that, ( 3 + 1), 2, 1, (iii) cos 15 ° − sin 15 =, 2, (i) 2 cos 45 ° cos 15 ° =, , (ii) 2 sin 75 ° sin 15 ° =, , 1, 2, , ANSWERS (EXERCISE 15C), , 6. (i) 2 sin 8x cos 2x (ii) 2 cos 5 x sin 2x (iii) 2 cos 6x cos x (iv) 2 sin 3 x sin x, 7. (i) sin 10x + sin 2x (ii) sin 8x − sin 2x (iii) cos 10x + cos 4x, (iv) cos 6x − cos 10x, HINTS TO SOME SELECTED QUESTIONS, 2, , x + y⎞, ⎧, ⎛ x − y⎞ ⎫ + ⎧, ⎛ x + y⎞, ⎛ x − y⎞ ⎫, 16. LHS = ⎨ −2 sin ⎛⎜, ⎟⎬, ⎟ sin ⎜, ⎟ sin ⎜, ⎟⎬, ⎨ cos ⎜, ⎝, ⎠, ⎝, ⎠, ⎝, ⎠, ⎝ 2 ⎠⎭, 2, 2, 2, ⎩, ⎭, ⎩, = 4 sin 2 ⎛⎜, ⎝, , 2, , x − y⎞ ⎧ 2 ⎛ x + y⎞, 2 ⎛ x + y⎞ ⎫, 2 ⎛ x − y⎞, ⎟, ⎟ ⋅ ⎨ sin ⎜, ⎟ + cos ⎜, ⎟ ⎬ = 4 sin ⎜, ⎝ 2 ⎠, ⎝ 2 ⎠, ⎝ 2 ⎠⎭, 2 ⎠ ⎩
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562, , Senior Secondary School Mathematics for Class 11, 2 cos 4 x sin 3 x − 2 cos 2 x sin x, 2 sin 4 x sin x + 2 cos 6 x cos x, {sin ( 4 x + 3 x ) − sin ( 4 x − 3 x )} − {sin ( 2 x + x ) − sin ( 2 x − x )}, =, {cos ( 4 x − x ) − cos ( 4 x + x )} + {cos ( 6 x + x ) + cos ( 6 x − x )}, , 21. LHS =, , =, , (sin 7 x − sin 3 x ), (sin 7 x − sin x ) − (sin 3 x − sin x ), =, (cos 3 x − cos 5 x ) + (cos 7 x + cos 5 x ) (cos 7 x + cos 3 x ), , ⎛ 7 x + 3x ⎞, ⎛ 7 x − 3x ⎞, 2 cos ⎜, ⎟ sin ⎜, ⎟, ⎝, ⎠, ⎝, ⎠ sin 2 x, 2, 2, =, =, = tan 2 x = RHS., ⎛ 7 x + 3 x ⎞ s ⎛ 7 x − 3 x ⎞ cos 2 x, 2 cos ⎜, ⎟, ⎟ co ⎜, ⎝, ⎠, ⎝, ⎠, 2, 2, x, +, y, 1, ⎞ cos ⎛ x − y ⎞ = 1, 26. cos x + cos y = ⇒ 2 cos ⎛⎜, ⎜, ⎟, ⎟, ⎝ 2 ⎠ 3, ⎝ 2 ⎠, 3, x + y⎞, 1, ⎛ x − y⎞ = 1, sin x + sin y = ⇒ 2 sin ⎛⎜, ⎟, ⎟ cos ⎜, ⎝ 2 ⎠ 4, ⎝ 2 ⎠, 4, x + y⎞ 3, Dividing (ii) by (i), we get tan ⎛⎜, ⎟= ⋅, ⎝ 2 ⎠ 4, 27., , … (i), … (ii), , (i) Use 2 cos x cos y = cos ( x + y ) + cos ( x − y )., (ii) Use 2 sin x sin y = cos ( x − y ) − cos ( x + y )., (iii) LHS = cos 15 ° − cos ( 90 ° − 15 ° ) = cos 15 ° − cos 75 ° ., C + D⎞, ⎛D−C⎞, Now, use cos C − cos D = 2 sin ⎛⎜, ⎟, ⎟ sin ⎜, ⎝ 2 ⎠, ⎝ 2 ⎠, , TRIGONOMETRIC FUNCTIONS OF MULTIPLES OF ANGLES, THEOREM 1, , PROOF, , (i) sin 2x = 2 sin x cos x, (ii) cos 2x = (cos2 x − sin 2 x) = ( 2 cos2 x − 1) = (1 − 2 sin 2 x), 2 tan x, (iii) tan 2x =, 1 − tan 2 x, , (i) sin 2x = sin ( x + x), = sin x cos x + cos x sin x, = 2sin x cos x., (ii) cos 2x = cos ( x + x), = cos x cos x − sin x sin x, = cos2 x − sin 2 x., Now, cos 2x = cos2 x − sin 2 x, = (1 − sin 2 x − sin 2 x) = (1 − 2 sin 2 x), [B cos2 x = (1 − sin 2 x)]., 2, 2, Also, cos 2x = cos x − sin x, = cos2 x − (1 − cos2 x), [B sin 2 x = 1 − cos2 x ], = ( 2 cos2 x − 1)., ∴, , cos 2x = (cos2 x − sin 2 x) = (1 − 2 sin 2 x) = ( 2 cos2 x − 1).
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Trigonometric, or Circular, Functions, , 563, , (iii) tan 2x = tan ( x + x), tan x + tan x, =, 1 − (tan x × tan x), 2 tan x, =, ⋅, 1 − tan 2 x, 2 tan x, ∴, tan 2x =, ⋅, 1 − tan 2 x, REMARK, , The above results may be expressed as, (i) (1 − cos 2x) = 2 sin 2 x, (ii) (1 + cos 2x) = 2 cos2 x, (1 − cos 2x), (iii), = tan 2 x, (1 + cos 2x), , π, then prove that, 2, 2 tan x, 1 − tan 2 x, (ii) cos 2x =, (i) sin 2x =, 2, 1 + tan x, 1 + tan 2 x, π, Since x is not an odd multiple of , we have cos x ≠ 0., 2, (i) sin 2x = 2 sin x cos x, 2 sin x cos x, =, [B 1 = cos2 x + sin 2 x ], cos2 x + sin 2 x, ⎛ 2 sin x ⎞, ⎜, ⎟, ⎝ cos x ⎠, [dividing num. and denom. by cos2 x ], =, 1 + tan 2 x, 2 tan x, =, ⋅, 1 + tan 2 x, 2 tan x, ⋅, ∴ sin 2x =, 1 + tan 2 x, , THEOREM 2, , PROOF, , If x is not an odd multiple of, , (ii) cos 2x = cos2 x − sin 2 x, cos2 x − sin 2 x, =, cos2 x + sin 2 x, sin 2 x, 1−, cos2 x, =, sin 2 x, 1+, cos2 x, 1 − tan 2 x, =, ⋅, 1 + tan 2 x, 1 − tan 2 x, ⋅, ∴ cos 2x =, 1 + tan 2 x, , [B 1 = cos2 x + sin 2 x ], , [dividing num. and denom. by cos2 x ]
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564, , Senior Secondary School Mathematics for Class 11, , SUMMARY, , (i) sin 2x = 2sin x cos x, (ii) cos 2x = cos2 x − sin 2 x = 2cos2 x − 1 = 1 − 2sin 2 x, (iii) tan 2x =, , 2tan x, 1 − tan 2 x, , (iv) sin 2x =, , 2tan x, 1 + tan 2 x, , (v) cos 2x =, , 1 − tan 2 x, 1 + tan 2 x, , (i) sin 3 x = 3 sin x − 4 sin 3 x (ii) cos 3 x = 4 cos3 x − 3 cos x, 3 tan x − tan 3 x, (iii) tan 3 x =, 1 − 3 tan 2 x, (i) sin 3 x = sin ( 2x + x), = sin 2x cos x + cos 2x sin x, [B sin 2x = 2 sin x cos x ], = 2 sin x cos2 x + cos 2x sin x, = 2 sin x(1 − sin 2 x) + (1 − 2 sin 2 x) sin x, [B cos2 x = (1 − sin 2 x) and cos 2x = (1 − 2 sin 2 x)], = 3 sin x − 4 sin 3 x., (ii) cos 3 x = cos ( 2x + x), = cos 2x cos x − sin 2x sin x, = ( 2 cos2 x − 1) cos x − 2 sin 2 x cos x, [B cos 2x = ( 2 cos2 x − 1) and sin 2x = 2 sin x cos x ], 2, = ( 2 cos x − 1) cos x + 2(cos2 x − 1) cos x, [B sin 2 x = (1 − cos2 x)], = 4 cos3 x − 3 cos x., tan 2x + tan x, (iii) tan 3 x = tan ( 2x + x) =, 1 − tan 2x tan x, 2 tan x, + tan x, 3 tan x − tan 3 x, (1 − tan 2 x), =, ⋅, =, ⎛ 2 tan x ⎞, 1 − 3 tan 2 x, x, tan, 1−⎜, ⋅, ⎟, ⎝ 1 − tan 2 x ⎠, , THEOREM 3, , PROOF, , SUMMARY, , (i) sin 3 x = 3sin x − 4sin 3 x, (ii) cos 3 x = 4cos3 x − 3cos x, (iii) tan 3 x =, , 3tan x − tan 3 x, 1 − 3tan 2 x
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Trigonometric, or Circular, Functions, , 565, , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , 1, 3π, and π < x <, , find the values of, 2, 2, (i) sin 2x , (ii) cos 2x , (iii) tan 2x., , If sin x = −, , Since x lies in Quadrant III, we have sin x < 0, cos x < 0and tan x > 0., 1, Now, sin x = − (given)., 2, 1⎞, 3 − 3, ⎛, =, ∴ cos x = − 1 − sin 2 x = − ⎜1 − ⎟ = −, ⎠, ⎝, 4, 4, 2, and tan x =, , sin x ⎛ 1, 2 ⎞, 1, = ⎜− ×, ⋅, ⎟ =, cos x ⎝ 2 − 3 ⎠, 3, , (i) sin 2x = 2 sin x cos x, 3⎞, 3, ⎛ 1⎞ ⎛, = 2 × ⎜− ⎟ × ⎜−, ⋅, ⎟ =, ⎝ 2⎠ ⎝ 2 ⎠, 2, (ii) cos 2x = ( 2 cos2 x − 1), 3⎞, ⎛, ⎛3, ⎞ 1, = ⎜ 2 × ⎟ − 1 = ⎜ − 1⎟ = ⋅, ⎝, ⎝2, ⎠ 2, 4⎠, (iii) tan 2x =, EXAMPLE 2, , SOLUTION, , sin 2x ⎛ 3 2⎞, =⎜, × ⎟ =, cos 2x ⎝ 2, 1⎠, , 3., , 13, π, and < x < π , find the values of, 12, 2, (i) sin 2x , (ii) cos 2x , (iii) tan 2x., , If sec x = −, , Since x lies in Quadrant II, we have sin x > 0, cos x < 0and tan x < 0., −13, 1, −12, Now, sec x =, ⇒ cos x =, =, ⋅, 12, sec x, 13, 144, =+, 169, sin x 5, ⎛ −13 ⎞ −5, And, tan x =, =, ×⎜, ⋅, ⎟ =, cos x 13 ⎝ 12 ⎠ 12, , ∴, , sin x = + 1 − cos2 x = + 1 −, , 25, 5, =, ⋅, 169 13, , (i) sin 2x = 2 sin x cos x, 5, ( −12) ⎫ −120, ⎧, = ⎨2 ×, ×, ⋅, ⎬=, 13, 13 ⎭ 169, ⎩, (ii) cos 2x = 2 cos2 x − 1, 144 ⎞ 119, ⎛, = ⎜2 ×, − 1⎟ =, ⋅, ⎝, 169 ⎠ 169, (iii) tan 2x =, , sin 2x ⎛ −120 169⎞ −120, =⎜, ×, ⋅, ⎟ =, 119⎠, 119, cos 2x ⎝ 169
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566, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 3, , −3, 3π, and, < x < 2π , find the values of, 4, 2, (i) sin 2x , (ii) cos 2x , (iii) tan 2x., , If tan x =, , Since x lies in Quadrant IV, we have, cos x > 0, sin x < 0 and tan x < 0., , SOLUTION, , ∴, , 9⎞, ⎛, sec x = 1 + tan 2 x = ⎜1 + ⎟ =, ⎝, 16⎠, , 25 5, 4, =, ⇒ cos x = ⋅, 16 4, 5, , 16 ⎞, ⎛, And, sin x = − 1 − cos2 x = − ⎜1 − ⎟ = −, ⎝, 25 ⎠, sin x ⎛ −3 5 ⎞ −3, ∴ tan x =, =⎜, × ⎟ =, ⋅, cos x ⎝ 5, 4⎠, 4, (i) sin 2x = 2 sin x cos x, ( −3) 4 ⎫ −24, ⎧, = ⎨2 ×, × ⎬=, ⋅, 5, 5 ⎭ 25, ⎩, 2, (ii) cos 2x = ( 2 cos x − 1), 16, 7, ⎛, ⎞, = ⎜2 ×, − 1⎟ =, ⋅, ⎝, ⎠, 25, 25, sin 2x ⎛ −24 25 ⎞ −24, (iii) tan 2x =, =⎜, ×, ⋅, ⎟ =, 7⎠, 7, cos 2x ⎝ 25, EXAMPLE 4, , 1, , find the value of sin 3x., 3, 1, (ii) If cos x = , find the value of cos 3x., 2, (i) If sin x =, , (i) sin 3 x = ( 3 sin x − 4 sin 3 x), 3, ⎧⎪⎛, 4 ⎞ 23, 1⎞, ⎛ 1 ⎞ ⎫⎪ ⎛, = ⎨⎜ 3 × ⎟ − 4 × ⎜ ⎟ ⎬ = ⎜ 1 − ⎟ =, ⋅, ⎝ 3⎠ ⎪ ⎝, ⎠ 27, 27, 3⎠, ⎪⎩⎝, ⎭, (ii) cos 3 x = ( 4 cos3 x − 3 cos x), 1⎞ ⎛, 1⎞ ⎫ ⎛ 1 3 ⎞, ⎧⎛, = ⎨⎜ 4 × ⎟ − ⎜ 3 × ⎟ ⎬ = ⎜ − ⎟ = − 1., 8⎠ ⎝, 2⎠ ⎭ ⎝ 2 2 ⎠, ⎩⎝, , SOLUTION, , 4, and x is acute, find the value of tan, 5, 4, 5, cos x =, ⇒ sec x =, 5, 4, 25, ⇒ tan x = sec2 x − 1 =, −1 =, 16, 3⎞, ⎛, ⎜2 × ⎟, ⎝, 2 tan x, 4 ⎠ ⎛ 3 16⎞, =, ∴ tan 2x =, =⎜ × ⎟, 9⎞ ⎝ 2, 7⎠, 1 − tan 2 x ⎛, ⎜1 − ⎟, ⎝, 16⎠, , EXAMPLE 5, , SOLUTION, , 9, −3, =, ⋅, 25, 5, , If cos x =, , 2x., , 9, 3, = ⋅, 16 4, =, , 24, ⋅, 7
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Trigonometric, or Circular, Functions, , EXAMPLE 6, , If tan x =, , 567, , 1, 1, and tan y = , show that cos 2x = sin 4y., 7, 3, , We have, , SOLUTION, , 1⎞, ⎛, ⎜1 − ⎟, (1 − tan 2 x) ⎝, 49⎠ ⎛ 48 49⎞ 24, × ⎟ =, cos 2x =, =, ⋅, =⎜, 1 ⎞ ⎝ 49 50 ⎠ 25, (1 + tan 2 x) ⎛, ⎜1 + ⎟, ⎝, 49⎠, sin 4y = 2 sin 2y cos 2y, ( 2 tan y), (1 − tan 2 y), =2×, ×, 2, (1 + tan y) (1 + tan 2 y), , ∴, EXAMPLE 7, , SOLUTION, , EXAMPLE 8, , SOLUTION, , 1⎞, ⎛, ⎛, 2 × ⎜2 × ⎟, ⎜1 −, ⎝, ⎝, 3⎠, =, ×, 1⎞, ⎛, ⎛, ⎜1 + ⎟, ⎜1 +, ⎝, ⎝, 9⎠, cos 2x = sin 4y., , Prove that, π, π, 3, (i) sin cos =, 6, 6, 4, , 1⎞, ⎟, 9, 8, 9 ⎞ 24, 9⎠ ⎛ 4, × × ⎟ =, ⋅, =⎜ ×, 1 ⎞ ⎝ 3 10 9 10⎠ 25, ⎟, 9⎠, , (ii) cos2, , π, π, 3, − sin 2, =, 12, 12, 2, , π, π 1⎛, π, π⎞, cos = ⎜ 2 sin cos ⎟, 6, 6 2⎝, 6, 6⎠, 1, π⎞, ⎛, = × sin ⎜ 2 × ⎟, [B 2 sin x cos x = sin 2x ], ⎝, 2, 6⎠, 1, π ⎛1, 3⎞, 3, = sin = ⎜ ×, ⋅, ⎟ =, 2, 3 ⎝2, 2 ⎠, 4, π, π, π⎞, ⎛, (ii) cos2, − sin 2, = cos ⎜ 2 × ⎟, [B cos2 x − sin 2 x = cos 2x ], ⎝, 12, 12, 12⎠, π 3, = cos = ⋅, 6 2, (i) sin, , Prove that, sin 2x, (i), = cot x, 1 − cos 2x, , (ii), , 1 − cos 2x, = tan 2 x, 1 + cos 2x, , sin 2x, 1 − cos 2x, 2 sin x cos x, =, 2 sin 2 x, [B sin 2x = 2 sin x cos x , (1 − cos 2x) = 2 sin 2 x ], = cot x = RHS., sin 2x, Hence,, = cot x., 1 − cos 2x, , (i) LHS =
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Trigonometric, or Circular, Functions, , 571, , = ( 8 cos5 x − 10 cos3 x + 3 cos x) − 6(1 − cos2 x) cos x, + 8(1 − cos2 x) 2 ⋅ cos x, 5, 3, = 16 cos x − 20 cos x + 5 cos x., Hence, cos 5 x = 16 cos5 x − 20 cos3 x + 5 cos x., EXAMPLE 18, SOLUTION, , Prove that cos 6x = 32 cos6x − 48 cos4 x + 18 cos2 x − 1., , We have, cos 6x = cos 2( 3 x), = cos 2θ , where 3x = θ, = 2 cos2θ − 1 = 2 cos2 3 x − 1, = 2( 4 cos3 x − 3 cos x) 2 − 1 [B cos 3 x = ( 4 cos3 x − 3 cos x)], = 2( 9 cos2 x + 16 cos6x − 24 cos4 x) − 1, = 32 cos6x − 48 cos4 x + 18 cos2 x − 1., Hence, cos 6x = 32 cos6x − 48 cos4 x + 18 cos2 x − 1., Prove that cot 2x cot x − cot 3 x cot 2x − cot 3 x cot x = 1., We have, cot 3 x = cot ( 2x + x), cot 2x cot x − 1, ⇔ cot 3 x =, cot 2x + cot x, ⇔ cot 3 x cot 2x + cot 3 x cot x = cot 2x cot x − 1, ⇔ cot 2x cot x − cot 3 x cot 2x − cot 3 x cot x = 1., Hence, cot 2x cot x − cot 3 x cot 2x − cot 3 x cot x = 1., , EXAMPLE 19, SOLUTION, , EXAMPLE 20, , SOLUTION, , Find the value of, (i) sin 18°, (ii) cos 18°, (iii) cos 36°, (iv) sin 36°, (v) sin 72°, (vi) cos 72°, (vii) sin 54°, (viii) cos 54°, (i) Let θ = 18°. Then,, θ = 18° ⇒ 5θ = 90°, ⇒ 2θ = ( 90° − 3θ), ⇒ sin 2θ = sin ( 90° − 3θ) = cos 3θ, ⇒ 2 sin θ cos θ = 4 cos3 θ − 3 cos θ, ⇒ 2 sin θ cos θ − 4 cos3 θ + 3 cos θ = 0, ⇒ cos θ( 2 sin θ − 4 cos2 θ + 3) = 0, ⇒ 2 sin θ − 4 cos2 θ + 3 = 0 [B cos θ = cos 18° ≠ 0], ⇒ 2 sin θ − 4(1 − sin 2 θ) + 3 = 0, ⇒ 4 sin 2 θ + 2 sin θ − 1 = 0, −2 ± 4 + 16 ( −1 ± 5 ), ⇒ sin θ =, =, 8, 4, ( 5 − 1), ⇒ sin θ =, [B sin θ = sin 18° > 0]., 4, ( 5 − 1), ∴ sin 18° =, ⋅, 4
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572, , Senior Secondary School Mathematics for Class 11, , (ii) cos218° = (1 − sin 218°), ⎧ ( 5 − 1) 2 ⎫ ⎧ ( 6 − 2 5 ) ⎫ 10 + 2 5, = ⎨1 −, ⎬ = ⎨1 −, ⎬=, 16, 16, 16, ⎩, ⎭ ⎩, ⎭, 10 + 2 5, [B cos 18° > 0]., ⇔ cos 18° =, 4, (iii) cos 36° = (1 − 2 sin 218°), ⎧, ( 5 − 1) 2 ⎫ ⎧ ( 6 − 2 5 ) ⎫, = ⎨1 − 2 ⋅, ⎬ = ⎨1 −, ⎬, 16, 8, ⎩, ⎭ ⎩, ⎭, 5 +1, ⋅, 4, , =, , 1, , ⎧ ( 5 + 1) 2 ⎫, (iv) sin 36° = 1 − cos 36° = ⎨1 −, ⎬, 16, ⎩, ⎭, , 2, , 2, , 1, , ⎧10 − 2 5 ) ⎫ 2, 10 − 2 5, =⎨, ⋅, ⎬ =, 16, 4, ⎩, ⎭, 10 + 2 5, (v) sin 72° = sin ( 90° − 18°) = cos 18° =, ⋅, 4, ( 5 − 1), (vi) cos 72° = cos ( 90° − 18°) = sin 18° =, ⋅, 4, (vii) sin 54° = sin ( 90° − 36°) = cos 36° =, (viii) cos 54° = cos ( 90° − 36°) = sin 36° =, , ( 5 + 1), ⋅, 4, 10 − 2 5, ⋅, 4, , SUMMARY, , (i) sin 18 ° =, , ( 5 − 1), = cos 72°, 4, , (10 + 2 5), = sin 72°, 4, ( 5 + 1), (iii) cos 36° =, = sin 54 °, 4, (ii) cos 18 ° =, , (iv) sin 36° =, , EXAMPLE 21, SOLUTION, , Prove that sin, , We have, LHS = sin, , (10 − 2 5), = cos 54 °, 4, , π, 13 π −1, + sin, =, ⋅, 10, 10, 2, π, 13 π, + sin, 10, 10
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Trigonometric, or Circular, Functions, , 573, , π, 3π ⎞, π, 3π, ⎛, + sin ⎜ π +, [B sin ( π + θ) = − sin θ ], − sin, ⎟ = sin, ⎝, 10, 10 ⎠, 10, 10, = sin 18° − sin 54° = sin 18° − sin ( 90° − 36°), ⎧( 5 − 1) ( 5 + 1) ⎫ −1, = (sin 18° − cos 36°) = ⎨, −, ⎬=, 4, 4, ⎩, ⎭ 2, = sin, , ⎡, ( 5 − 1), ( 5 + 1) ⎤, and cos 36° =, ⎢B sin 18° =, ⎥, 4, 4, ⎣, ⎦, , EXAMPLE 22, SOLUTION, , Prove that (cos2 48° − sin 212°) =, , ( 5 + 1), ⋅, 8, , We have, LHS = (cos2 48° − sin 212°), , 1, ( 2 cos2 48° − 2 sin 212°), 2, 1, = {(1 + cos 96°) − (1 − cos 24°)}, 2, [B 2 cos2 θ = (1 + cos 2θ) and 2 sin 2 θ = (1 − cos 2θ)], 1, = (cos 96° + cos 24°), 2, 1⎡, ⎛ 96° + 24°⎞, ⎛ 96° − 24°⎞ ⎤, = ⎢2 cos ⎜, ⎟ cos ⎜, ⎟⎥, ⎝, ⎠⎦, ⎝, ⎠, 2⎣, 2, 2, =, , EXAMPLE 23, SOLUTION, , =, , 1 ( 5 + 1), 1, × 2 cos 60° cos 36° = ×, 2, 4, 2, , =, , ( 5 + 1), = RHS., 8, , Prove that (sin 2 72° − sin 2 60°) =, , ⎡, ( 5 + 1) ⎤, ⎢B cos 36° =, ⎥, 4 ⎦, ⎣, , ( 5 − 1), ⋅, 8, , We have, 2, 2, LHS = (sin 72° − sin 60°), 1, = ( 2 sin 2 72° − 2 sin 2 60°), 2, 1, = {(1 − cos 144°) − (1 − cos 120°)} [B 2 sin 2 θ = (1 − cos 2θ)], 2, 1, = (cos 120° − cos 144°), 2, 1 1, 1 1, 1⎤, ⎡, = − − cos 144° = − − cos (180° − 36°) ⎢B cos 120° = − ⎥, 4 2, 4 2, 2⎦, ⎣, 1 ( 5 + 1), 1 1, + cos 36° = − +, 4, 8, 4 2, ( 5 − 1), =, = RHS., 8, =−, , ⎡, ( 5 + 1) ⎤, ⎢B cos 36° =, ⎥, 4, ⎣, ⎦
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574, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 24, SOLUTION, , Prove that cos 6° cos 42° cos 66° cos 78° =, , 1, ⋅, 16, , We have, LHS = cos 6° cos 42° cos 66° cos 78°, , =, =, =, =, , =, , =, =, EXAMPLE 25, SOLUTION, , 1, ( 2 cos 66° cos 6°)( 2 cos 78° cos 42°), 4, 1, [cos ( 66° + 6°) + cos ( 66° − 6°)] ×, 4, [cos (78° + 42°) + cos (78° − 42°)], 1, (cos 72° + cos 60°)(cos 120° + cos 36°), 4, 1⎞ ⎛ 1, 1⎛, ⎞, ⎜ sin 18° + ⎟ ⎜ − + cos 36°⎟, ⎠, 2⎠ ⎝ 2, 4⎝, [B cos 72° = cos ( 90° − 18°) = sin 18° ], 1 ⎡( 5 − 1) 1 ⎤ ⎡ 1 ( 5 + 1) ⎤, + ⎥ ⎢− +, ⎢, ⎥, 4⎣, 4, 2⎦ ⎣ 2, 4, ⎦, ⎡, ( 5 − 1), ( 5 + 1) ⎤, and cos 36° =, ⎢B sin 18° =, ⎥, 4, 4, ⎣, ⎦, 1 ( 5 + 1) ( 5 − 1) (5 − 1), ⋅, ⋅, =, 4, 4, 4, 64, 4, 1, =, = RHS., 64 16, , Prove that sin, , We have, , π, 2π, 3π, 4π 5, sin, sin, sin, =, ⋅, 5, 5, 5, 5, 16, , π, 2π, 3π, 4π, sin, sin, sin, 5, 5, 5, 5, π, 2π, 2π ⎞, π⎞, ⎛, ⎛, = sin sin, sin ⎜ π −, ⎟ sin ⎜ π − ⎟, ⎝, ⎝, 5, 5, 5 ⎠, 5⎠, π, 2π, [B sin ( π − θ) = sin θ ], = sin 2 sin 2, 5, 5, 2, = (sin 36°) × (sin 72°) 2, = (sin 36°) 2 × (cos 18°) 2, [B sin 72° = sin ( 90° − 18°) = cos 18°], , LHS = sin, , =, , =, , (10 − 2 5 ) (10 + 2 5 ) (100 − 20), ×, =, 16, 16, (16 × 16), ⎡, 10 + 2 5 ⎤, 10 − 2 5, ⎢B sin 36° =, ⎥, and cos 18° =, 4, 4, ⎢⎣, ⎥⎦, 80, 5, =, = RHS., (16 × 16) 16
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Trigonometric, or Circular, Functions, EXAMPLE 26, , SOLUTION, , Find the value of, (i) sin 22° 30′, (i) sin 2 θ =, , (ii) cos 22° 30′, , 575, , (iii) tan 22° 30′, , (1 − cos 2θ), 2, , 1 ⎞, ⎛, ⎜1 −, ⎟, ⎝, 2 −1, 1, −, 45, °, (, cos, ), 2⎠, ⇒ sin 2 ( 22° 30′ ) =, =, =, 2, 2, 2 2, ⇒ sin ( 22° 30′ ) =, , (ii) cos2 θ =, , ( 2 − 1), ⋅, 2 2, , (1 + cos 2θ), 2, , 1 ⎞, ⎛, ⎜1 +, ⎟, ⎝, (1 + cos 45 ° ), 2 ⎠ ( 2 + 1), ⇒ cos2 ( 22° 30′ ) =, =, =, 2, 2, 2 2, ⇒ cos ( 22° 30′ ) =, (iii) tan 2 ( 22° 30′ ) =, =, , ( 2 + 1), ⋅, 2 2, , sin 2 ( 22° 30′ ) ( 2 − 1), ( 2 2), =, ×, cos2 ( 22° 30′ ), ( 2 2), ( 2 + 1), 2 − 1 ( 2 − 1) ( 2 − 1), =, ×, 2 + 1 ( 2 + 1) ( 2 − 1), , = ( 2 − 1) 2, ⇒ tan ( 22° 30′ ) = ( 2 − 1)., , EXERCISE 15D, 1. If sin x =, , π, 5, and 0 < x < , find the values of, 3, 2, , (i) sin 2x ,, (ii) cos 2x ,, (iii) tan 2x., −3, 3π, 2. If cos x =, and π < x <, , find the values of, 5, 2, (i) sin 2x ,, (ii) cos 2x ,, (iii) tan 2x., −5, π, 3. If tan x =, and < x < π , find the values of, 12, 2, (i) sin 2x ,, , (ii) cos 2x ,, (iii) tan 2x., 1, 4. (i) If sin x = , find the value of sin 3x., 6, −1, (ii) If cos x =, , find the value of cos 3x., 2
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576, , Senior Secondary School Mathematics for Class 11, , Prove that, , 5., , cos 2x, = cos x + sin x, cos x − sin x, , 6., , sin 2x, = tan x, 1 + cos 2x, , 7., , sin 2x, = cot x, 1 − cos 2x, , 8., , tan 2x, = tan x, 1 + sec 2x, , 9. sin 2x(tan x + cot x) = 2, , 10. cosec 2x + cot 2x = cot x, , 11. cos 2x + 2 sin 2 x = 1, , 12. (sin x − cos x) 2 = 1 − sin 2x, , 13. cot x − 2 cot 2x = tan x, , 14. (cos4 x + sin 4 x) =, , 15., , cos3 x − sin 3 x 1, = ( 2 + sin 2x), cos x − sin x 2, , 17. cos x cos 2x cos 4x cos 8x =, 18. Prove that, (i) 2 sin 22, , 16., , 1, ( 2 − sin 2 2x), 2, , 1 − cos 2x + sin x, = tan x, sin 2x + cos x, , sin 16x, 16 sin x, , 1°, 1°, 1, cos 22 =, 2, 2, 2, , (ii) 2 cos2 15 ° − 1 =, , (iii) 8 cos 3 20° − 6 cos 20° = 1, , 3, 2, , (iv) 3 sin 40° − 4 sin 3 40° =, , 3, 2, , 19. Prove that, ( 5 − 1), ( 5 − 1), (ii) sin 2 72° − cos2 30° =, 8, 8, 20. Prove that tan 6° tan 42° tan 66° tan 78° = 1., a, 21. If tan θ = , prove that a sin 2θ + b cos 2θ = b., b, (i) sin 2 24° − sin 2 6° =, , ANSWERS (EXERCISE 15D), , 4 5, 1. (i), 9, −120, 3. (i), 169, , −1, (ii), 9, 119, (ii), 169, , (iii) −4 5, (iii), , −120, 119, , 24, 2. (i), 25, 13, 4. (i), 27, , (ii), , −7, 25, , (iii), , −24, 7, , (ii) 1, , HINTS TO SOME SELECTED QUESTIONS, 3. Use the relations sin 2 x =, , 2 tan x, 2, , 1 + tan x, , and cos 2 x =, , cos 2 x − sin 2 x, = cos x + sin x = RHS., cos x − sin x, 2 sin x cos x, 6. LHS =, = tan x = RHS., 2 cos 2 x, 2 sin x cos x, 7. LHS =, = cot x = RHS., 2 sin 2 x, 5. LHS =, , 1 − tan 2 x, 1 + tan 2 x, , ⋅
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Trigonometric, or Circular, Functions, 8. LHS =, 10. LHS =, , 2 sin x cos x, sin 2 x, =, = tan x = RHS., ( 1 + cos 2 x ), 2 cos 2 x, cos 2 x 1 + cos 2 x, 2 cos 2 x, 1, +, =, =, = cot x = RHS., sin 2 x sin 2 x, sin 2 x, 2 sin x cos x, , ⎛ 1, 2( 1 − tan 2 x ) ⎫⎪ tan 2 x, 2 ⎞ ⎧⎪ 1, 13. LHS = ⎜, = tan x = RHS., −, −, ⎟=⎨, ⎬=, 2 tan x ⎪⎭ tan x, 2, tan, x, tan, x, tan, x, ⎝, ⎠ ⎩⎪, 14. LHS = (cos 2 x + sin 2 x ) 2 − 2 cos 2 x sin 2 x, 1, 1, = 1 − 2 cos 2 x sin 2 x = 1 − (sin 2 2 x ) = ( 2 − sin 2 2 x ) = RHS., 2, 2, 2 sin 2 x + sin x, sin x( 2 sin x + 1) sin x, 16. LHS =, =, =, = tan x = RHS., 2 sin x cos x + cos x cos x( 2 sin x + 1) cos x, 1, 17. LHS =, ( 2 sin x cos x ) cos 2 x cos 4 x cos 8 x, 2 sin x, 1, =, (sin 2 x cos 2 x ) cos 4 x cos 8 x, 2 sin x, 1, =, ( 2 sin 2 x cos 2 x ) cos 4 x cos 8 x, 4 sin x, 1, =, (sin 4 x cos 4 x ) cos 8 x, 4 sin x, 1, =, ( 2 sin 4 x cos 4 x ) cos 8 x, 8 sin x, 1, =, (sin 8 x cos 8 x ), 8 sin x, 1, =, ( 2 sin 8 x cos 8 x ), 16 sin x, sin 16 x, 1, (sin 16 x ) =, =, = RHS., 16 sin x, 16 sin x, 18., , ⎛, 1° ⎞, 1, (i) LHS = sin ⎜ 2 × 22 ⎟ = sin 45 ° =, ⋅, 2 ⎠, 2, ⎝, (ii) LHS = ( 2 cos 2θ − 1) = cos 2θ, where θ = 15 °, 3, ⋅, 2, (iii) LHS = 2( 4 cos 3 θ − 3 cos θ), where θ = 20 °, 1, = 2 × cos 3θ = 2 × cos 60 ° = ⎛⎜ 2 × ⎞⎟ = 1., ⎝, 2⎠, = cos 30 ° =, , (iv) LHS = ( 3 sin θ − 4 sin 3 θ), where θ = 40 °, = sin 3θ = sin 120 ° = sin 60 ° =, 20. LHS =, , sin 6 ° sin 42 ° sin 66 ° sin 78 °, ⋅, cos 6 ° cos 42 ° cos 66 ° cos 78 °, , 21. Use the relations, sin 2 θ =, , 2 tan θ, 2, , ( 1 + tan θ), , and cos 2 θ =, , 3, ⋅, 2, , 1 − tan 2 θ, 1 + tan 2 θ, , ⋅, , 577
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578, , Senior Secondary School Mathematics for Class 11, , TRIGONOMETRICAL FUNCTIONS OF HALF ANGLES, , For all x ∈ R , we have, x, x, (i) sin x = 2 sin cos, 2, 2, 2 x, (iii) 1 + cos x = 2 cos, 2, ⎛ x x⎞, (i) sin x = sin ⎜ + ⎟, ⎝ 2 2⎠, , THEOREM 1, , PROOF, , (ii) 1 − cos x = 2 sin 2, , x, x, x, x, cos + cos sin, 2, 2, 2, 2, x, x, = 2 sin cos, 2, 2, x, x, ∴ sin x = 2 sin cos ⋅, 2, 2, x, x, ⎛, ⎞, (ii) cos x = cos ⎜ + ⎟, ⎝ 2 2⎠, x, x, x, x, = cos cos − sin sin, 2, 2, 2, 2, 2 x, 2 x, = cos, − sin, 2, 2, x, ⎛, 2 x⎞, = ⎜1 − sin ⎟ − sin 2, ⎝, 2⎠, 2, x, ⎛, ⎞, = ⎜1 − 2 sin 2 ⎟ ⋅, ⎝, 2⎠, x, ∴ (1 − cos x) = 2 sin 2 ⋅, 2, ⎛ x x⎞, (iii) cos x = cos ⎜ + ⎟, ⎝ 2 2⎠, x, x, x, x, = cos cos − sin sin, 2, 2, 2, 2, 2 x, 2 x, = cos, − sin, 2, 2, x⎞, ⎛, 2 x, = cos, − ⎜1 − cos2 ⎟, 2 ⎝, 2⎠, = sin, , x, ⎛, ⎞, = ⎜ 2 cos2 − 1⎟ ⋅, ⎝, ⎠, 2, ∴ (1 + cos x) = 2 cos2, THEOREM 2, , x, ⋅, 2, , For all x ∈ R , we have, 2 tan ( x 2), (i) sin x =, 1 + tan 2 ( x 2), , (ii) cos x =, , 1 − tan 2 ( x 2), 1 + tan 2 ( x 2), , x, 2
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Trigonometric, or Circular, Functions, , PROOF, , 579, , ⎛ x⎞, ⎛ x⎞, (i) sin x = 2 sin ⎜ ⎟ cos ⎜ ⎟, ⎝ 2⎠, ⎝ 2⎠, x, ⎛ ⎞, ⎛ x⎞, 2 sin ⎜ ⎟ cos ⎜ ⎟, ⎝ 2⎠, ⎝ 2⎠, =, x, ⎛ ⎞, ⎛ x⎞, cos2 ⎜ ⎟ + sin 2 ⎜ ⎟, ⎝ 2⎠, ⎝ 2⎠, , =, , ⎧, ⎛ x⎞ ⎫, ⎪⎪ 2 sin ⎜⎝ 2 ⎟⎠ ⎪⎪, ⎨, ⎬, ⎪ cos ⎛⎜ x ⎞⎟ ⎪, ⎪⎩, ⎝ 2 ⎠ ⎪⎭, , ⎧ 2 ⎛ x⎞ ⎫, ⎪⎪ sin ⎜⎝ 2 ⎟⎠ ⎪⎪, 1+ ⎨, ⎬, ⎪ cos2 ⎛⎜ x ⎞⎟ ⎪, ⎪⎩, ⎝ 2 ⎠ ⎪⎭, , ⎡, 2 ⎛ x⎞ ⎤, ⎢dividing num. and denom. by cos ⎜⎝ 2 ⎟⎠ ⎥, ⎣, ⎦, , ⎛ x⎞, 2 tan ⎜ ⎟, ⎝ 2⎠, =, ⋅, ⎧, 2 ⎛ x⎞ ⎫, ⎨1 + tan ⎜⎝ ⎟⎠ ⎬, 2 ⎭, ⎩, ⎛ x⎞, 2 tan ⎜ ⎟, ⎝ 2⎠, ∴ sin x =, ⋅, ⎧, 2 ⎛ x⎞ ⎫, tan, +, 1, ⎜ ⎟⎬, ⎨, ⎝ 2⎠ ⎭, ⎩, ⎛ x⎞, ⎛ x⎞, (ii) cos x = cos2 ⎜ ⎟ − sin 2 ⎜ ⎟, ⎝ 2⎠, ⎝ 2⎠, ⎛ x⎞, ⎛ x⎞, cos2 ⎜ ⎟ − sin 2 ⎜ ⎟, ⎝ 2⎠, ⎝ 2⎠, =, 2 ⎛ x⎞, 2 ⎛ x⎞, cos ⎜ ⎟ + sin ⎜ ⎟, ⎝ 2⎠, ⎝ 2⎠, ⎛ x⎞, 1 − tan 2 ⎜ ⎟, ⎝ 2⎠, =, ⎛ x⎞, 1 + tan 2 ⎜ ⎟, ⎝ 2⎠, , ⎡, 2 ⎛ x⎞, 2 ⎛ x⎞ ⎤, ⎢B 1 = cos ⎜⎝ 2 ⎟⎠ + sin ⎜⎝ 2 ⎟⎠ ⎥, ⎣, ⎦, , ⎡, 2 ⎛ x⎞ ⎤, ⎢dividing num. and denom. by cos ⎜⎝ 2 ⎟⎠ ⎥ ⋅, ⎣, ⎦, , ⎛ x⎞, 1 − tan 2 ⎜ ⎟, ⎝ 2⎠, ∴ cos x =, ⋅, 2 ⎛ x⎞, 1 + tan ⎜ ⎟, ⎝ 2⎠
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580, , Senior Secondary School Mathematics for Class 11, , SUMMARY, , ⎛ x⎞, ⎛ x⎞, (i) sin x = 2sin ⎜ ⎟ cos ⎜ ⎟, ⎝ 2⎠, ⎝ 2⎠, ⎛ x⎞, (ii) (1 − cos x) = 2sin 2 ⎜ ⎟, ⎝ 2⎠, ⎛ x⎞, (iii) (1 + cos x) = 2cos2 ⎜ ⎟, ⎝ 2⎠, ⎛ x⎞, 2tan ⎜ ⎟, ⎝ 2⎠, (iv) sin x =, ⎛ x⎞, 1 + tan 2 ⎜ ⎟, ⎝ 2⎠, ⎛ x⎞, 1 − tan 2 ⎜ ⎟, ⎝ 2⎠, (v) cos x =, ⎛ x⎞, 1 + tan 2 ⎜ ⎟, ⎝ 2⎠, SOLVED EXAMPLES, , −4, π, and < x < π , find the values of, 3, 2, x, x, x, (iii) tan ⋅, (i) sin , (ii) cos ,, 2, 2, 2, Since x lies in Quadrant II, we have cos x < 0., −4, 16⎞ 25, ⎛, ⇒ sec2 x = (1 + tan 2 x) = ⎜1 + ⎟ =, B tan x =, ⎝, 3, 9⎠, 9, 1, 9, 2, ⇒ cos x =, =, sec2 x 25, 9, 3, ⇒ cos x = −, =− ⋅, 25, 5, π, π x π, Also,, <x<π ⇒, < <, 2, 4 2 2, x, ⇒ lies in Quadrant I, 2, x, x, ⇒ sin > 0 and cos > 0., 2, 2, 3⎞ 8, ⎛, 2 x, (i) 2 sin, = (1 − cos x) = ⎜1 + ⎟ =, ⎝, 2, 5⎠ 5, 8 4, 2 x, ⇒ sin, =, =, 2 10 5, x, x, 4, 2 ⎡, ⎤, B sin > 0⎥ ⋅, ⇒ sin = +, =, 2, 2, 5, 5 ⎢⎣, ⎦, , EXAMPLE 1, , SOLUTION, , If tan x =
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Trigonometric, or Circular, Functions, , x, 3⎞ 2, ⎛, = (1 + cos x) = ⎜1 − ⎟ =, ⎝, 5⎠ 5, 2, x ⎛ 2 ⎞ 1, ⇒ cos2 = ⎜, ⎟ =, 2 ⎝ 5 × 2⎠ 5, , (ii) 2 cos2, , x, x, 1, 1, ⎡, ⎤, =+, =, ⎢⎣B cos 2 > 0⎥⎦ ⋅, 2, 5, 5, ⎛ x⎞, sin ⎜ ⎟, ⎝ 2⎠ ⎛ 2, x, ⎞, (iii) tan =, =⎜, × 5 ⎟ = 2., ⎠, 2, ⎛ x⎞ ⎝ 5, cos ⎜ ⎟, ⎝ 2⎠, ⇒ cos, , EXAMPLE 2, , SOLUTION, , 3, 3π, and π < x <, , find the values of, 4, 2, x, x, x, (iii) tan ⋅, (i) sin , (ii) cos ,, 2, 2, 2, , If tan x =, , Since x lies in Quadrant III, we have cos x < 0., Now, tan x =, , 9 ⎞ 25, 3, ⎛, ⇒ sec2 x = (1 + tan 2 x) = ⎜1 + ⎟ =, ⎝, 16⎠ 16, 4, 1, 16, ⇒ cos2 x =, =, 2, sec x 25, , 16 −4, =, ⋅, 25, 5, 3π, π x 3π, Also, π < x <, ⇒, < <, 2, 2 2, 4, x, ⇒ lies in Quadrant II, 2, x, x, ⇒ sin > 0 and cos < 0., 2, 2, x, 4, 9, ⎛, ⎞, (i) 2 sin 2 = (1 − cos x) = ⎜1 + ⎟ =, ⎝, 2, 5⎠ 5, x, 9, ⇒ sin 2 =, 2 10, x, x, 9, 3, ⎡, ⎤, B sin > 0⎥ ⋅, ⇒ sin = +, =, ⎢, 2, 2, 10, 10, ⎦, ⎣, ⇒ cos x = −, , x, 4⎞ 1, ⎛, = (1 + cos x) = ⎜1 − ⎟ =, ⎝, 5⎠ 5, 2, x 1, ⇒ cos2 =, 2 10, x, x, 1, −1, ⎡, ⎤, ⇒ cos = −, =, ⎢⎣B cos 2 < 0⎥⎦ ⋅, 2, 10, 10, , (ii) 2 cos2, , 581
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582, , Senior Secondary School Mathematics for Class 11, , ⎛ x⎞, sin ⎜ ⎟, ⎝ 2⎠ ⎛ 3, 10 ⎞, =⎜, ×, ⎟ = −3., ⎛ x ⎞ ⎝ 10, −1 ⎠, cos ⎜ ⎟, ⎝ 2⎠, , x, (iii) tan =, 2, , EXAMPLE 3, , SOLUTION, , EXAMPLE 4, SOLUTION, , 1, and x lies in Quadrant III, find the values of, 3, x, x, x, (i) sin , (ii) cos ,, (iii) tan ⋅, 2, 2, 2, , If cos x = −, , Since x lies in Quadrant III, we have, 3π, π x 3π, π<x<, ⇒, < <, 2, 2 2, 4, x, ⇒ lies in Quadrant II, 2, x, x, ⇒ sin > 0 and cos < 0., 2, 2, 1⎞ 4, ⎛, 2 x, (i) 2 sin, = (1 − cos x) = ⎜1 + ⎟ =, ⎝, 2, 3⎠ 3, x, 2, ⇒ sin 2 =, 2 3, x, x, 2, 2, 3, 6 ⎡, ⎤, B sin > 0⎥ ⋅, ⇒ sin = +, =, ×, =, ⎢, 2, 2, 3, 3 ⎣, 3, 3, ⎦, 1⎞ 2, ⎛, 2 x, (ii) 2 cos, = (1 + cos x) = ⎜1 − ⎟ =, ⎝, 3⎠ 3, 2, 1, 2 x, ⇒ cos, =, 2 3, x, 1, −1, 3, 3, ⇒ cos = −, =, ×, =−, ⋅, 2, 3, 3, 3, 3, ⎛ x⎞, sin ⎜ ⎟, ⎝ 2⎠ ⎛ 6, x, 3 ⎞, (iii) tan =, =⎜, ×, ⎟ = − 2., 2, ⎛ x⎞ ⎝ 3, − 3⎠, cos ⎜ ⎟, ⎝ 2⎠, Prove that, We have, LHS =, , 1 + cos x, = ( cosec x + cot x) 2 ., 1 − cos x, 1 + cos x, 1 − cos x, , ⎛ x⎞, 2 cos2 ⎜ ⎟, ⎝ 2⎠, x, =, = cot 2, x, ⎛ ⎞, 2, 2 sin 2 ⎜ ⎟, ⎝ 2⎠, , ⎡, 2 ⎛ x⎞ ⎤, ⎢Q 1 + cos x = 2 cos ⎜⎝ 2 ⎟⎠ , ⎥, ⎥, ⎢, ⎢, 2 ⎛ x⎞ ⎥, cos, x, sin, 1, −, =, 2, ⎜, ⎟, ⎢, ⎝ 2 ⎠ ⎦⎥, ⎣
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Trigonometric, or Circular, Functions, , 583, , RHS = ( cosec x + cot x) 2, 2, , ⎛ 1, ⎛ 1 + cos x ⎞, cos x ⎞, =⎜, +, ⎟ =⎜, ⎟, ⎝ sin x sin x ⎠, ⎝ sin x ⎠, 2, , ∴, EXAMPLE 5, SOLUTION, , 2, , 2, , ⎧, ⎫ ⎧, ⎛ x⎞ ⎫, ⎛ x⎞, 2 cos2 ⎜ ⎟, ⎪⎪, ⎪⎪ ⎪⎪ cos ⎜⎝ 2 ⎟⎠ ⎪⎪, ⎝ 2⎠, 2 x, =⎨, ⋅, ⎬ = cot, ⎬ =⎨, x, x, x, 2, ⎪ 2 sin ⎛⎜ ⎞⎟ cos ⎛⎜ ⎞⎟ ⎪ ⎪ sin ⎛⎜ ⎞⎟ ⎪, ⎪⎩, ⎝ 2 ⎠ ⎪⎭, ⎝ 2⎠, ⎝ 2 ⎠ ⎪⎭ ⎪⎩, LHS = RHS., , Prove that, , cos x, = tan, (1 − sin x), , ⎛ π x⎞, ⎜ + ⎟⋅, ⎝ 4 2⎠, , We have, cos x, (1 − sin x), x⎞, ⎛ 2x, − sin 2 ⎟, ⎜ cos, ⎝, 2, 2⎠, =, x, x, x, x⎞, ⎛ 2, + sin 2 − 2 sin cos ⎟, ⎜ cos, ⎝, 2, 2, 2, 2⎠, ⎤, ⎡, ⎛ 2x, 2 x⎞, 2 x, 2 x, ⎢Q cos x = ⎜⎝ cos 2 − sin 2 ⎟⎠ , cos 2 + sin 2 = 1⎥, ⎥, ⎢, x, x, ⎥, ⎢, 2sin, and, sin, x, =, cos, ⎥⎦, ⎢⎣, 2, 2, x, x⎞ ⎛, x, x⎞, ⎛, ⎜ cos − sin ⎟ ⎜ cos + sin ⎟, ⎝, 2, 2⎠ ⎝, 2, 2⎠, =, 2, x, x⎞, ⎛, ⎜ cos − sin ⎟, ⎝, 2, 2⎠, x⎞, x, x⎞ ⎛, x⎞, π, ⎛, ⎛, ⎜ cos + sin ⎟ ⎜1 + tan ⎟, ⎜ tan + tan ⎟, ⎝, ⎝, 4, 2⎠, 2, 2⎠ ⎝, 2⎠, =, =, =, x⎞ ⎛, x⎞, x, x⎞, π, ⎛, ⎛, ⎜1 − tan ⎟ ⎜1 − tan ⋅ tan ⎟, ⎜ cos − sin ⎟, ⎝, ⎝, 2⎠ ⎝, 4, 2⎠, 2, 2⎠, x⎤, ⎡, ⎢⎣dividing num. and denom. by cos 2 ⎥⎦, ⎛ π x⎞, = tan ⎜ + ⎟ = RHS., ⎝ 4 2⎠, , LHS =, , EXAMPLE 6, , SOLUTION, , If α and β be two distinct real numbers such that (α − β ) ≠ 2nπ for any, intetger n, satisfying the equation a cos θ + b sin θ = c then prove that, a 2 − b2, 2ab, (i) cos (α + β ) = 2, (ii) sin (α + β ) = 2, ⋅, a + b2, a + b2, Since α and β satisfy the equation a cos θ + b sin θ = c, we have, … (i), a cos α + b sin α = c,, … (ii), a cos β + b sin β = c.
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Trigonometric, or Circular, Functions, , 585, , −1, and x lies in Quadrant IV, find the values of, 2, x, x, x, (i) sin ,, (ii) cos ,, (iii) tan ⋅, 2, 2, 2, x 12, and x lies in Quadrant I, find the values of, 4. If cos =, 2 13, 3. If sin x =, , (i) sin x ,, 5. If sin x =, , (ii) cos x ,, , (iii) cot x., , 3, π, x, and 0 < x < , find the value of tan ⋅, 5, 2, 2, , Prove that, , 6. cot, 8., 10., , ⎛ π x⎞, 7. tan ⎜ + ⎟ = tan x + sec x, ⎝ 4 2⎠, , x, x, − tan = 2 cot x, 2, 2, , 1 + sin x, = tan, 1 − sin x, , ⎛ π x⎞, 9. tan ⎜ + ⎟ + tan, ⎝ 4 2⎠, , ⎛ π x⎞, ⎜ + ⎟, ⎝ 4 2⎠, , ⎛ π x⎞, ⎜ − ⎟ = 2sec x, ⎝ 4 2⎠, , sin x, x, = tan, 1 + cos x, 2, , ANSWERS (EXERCISE 15E), , 1. (i), , 30, 6, (ii), (iii) 5, 6, 6, , 3. (i) sin, 4. (i), , 2. (i), , 1, 2 5, (ii), (iii) 2, 5, 5, , (2 − 3), (2 + 3), x, x, x, (ii) cos = −, (iii) tan = − ( 2 −, =, 2, 2, 2, 2, 2, , 120, 119, 119, (ii), (iii), 169, 169, 120, , 5., , 3), , 1, 3, , HINTS TO SOME SELECTED QUESTIONS, , 5. tan, , x, =, 2, , 1 − cos x, 1 + cos x, , , where cos x = 1 −, , 9, 4, = ⋅, 25 5, , x, x, cos ⎛⎜ ⎞⎟ sin ⎛⎜ ⎞⎟, ⎝ 2⎠, ⎝ 2⎠, 6. LHS =, −, =, x, x, sin ⎛⎜ ⎞⎟ cos ⎛⎜ ⎞⎟, ⎝ 2⎠, ⎝ 2⎠, , ⎧ 2 ⎛ x⎞, 2 ⎛ x⎞ ⎫, ⎨ cos ⎜ ⎟ − sin ⎜ ⎟ ⎬, ⎝ 2⎠ ⎭, ⎝, ⎠, 2, ⎩, x, x, sin cos, 2, 2, , =, , 2 cos x, = 2 cot x = RHS., sin x
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586, , Senior Secondary School Mathematics for Class 11, , ⎛ 1, sin x ⎞ ( 1 + sin x ), 7. RHS = ⎜, +, =, ⎟=, cos, x, cos, x⎠, cos x, ⎝, , π, 1 − cos ⎛⎜ + x ⎞⎟, ⎝2, ⎠, π, ⎛, ⎞, sin ⎜ + x ⎟, ⎝2, ⎠, , π x, 2 sin 2 ⎛⎜ + ⎞⎟, ⎝ 4 2⎠, π x, =, = tan ⎛⎜ + ⎞⎟ = LHS., ⎝ 4 2⎠, π x, π x, 2 sin ⎛⎜ + ⎞⎟ cos ⎛⎜ + ⎞⎟, ⎝ 4 2⎠, ⎝ 4 2⎠, π, 1 − cos ⎛⎜ + x ⎞⎟, 1 + sin x, 1 + sin x ( 1 + sin x ), ⎝2, ⎠, 8. LHS =, ×, =, =, π, cos x, ⎛, ⎞, 1 − sin x, 1 + sin x, sin ⎜ + x ⎟, ⎠, ⎝2, x⎞, 2 ⎛π, 2 sin ⎜ + ⎟, ⎝ 4 2⎠, π x, =, = tan ⎛⎜ + ⎞⎟ = RHS., ⎝ 4 2⎠, π x⎞, π x⎞, ⎛, ⎛, 2 sin ⎜ + ⎟ cos ⎜ + ⎟, ⎝ 4 2⎠, ⎝ 4 2⎠, sin x, 10. LHS =, =, ( 1 + cos x ), , x, x, 2 sin ⎛⎜ ⎞⎟ cos ⎛⎜ ⎞⎟, ⎝ 2⎠, ⎝ 2⎠, x, = tan = RHS., 2, 2 ⎛ x⎞, 2 cos ⎜ ⎟, ⎝ 2⎠, , q
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16, Conditional Identities Involving the, Angles of a Triangle, Before we begin, let us first recall the following formulae:, ⎛ x + y⎞, ⎛ x − y⎞, (i) sin x + sin y = 2sin ⎜, ⎟ cos ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, ⎛ x + y⎞, ⎛ x − y⎞, (ii) sin x − sin y = 2cos ⎜, ⎟ sin ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, ⎛ x + y⎞, ⎛ x − y⎞, (iii) cos x + cos y = 2cos ⎜, ⎟ cos ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, ⎛ x + y⎞, ⎛ x − y⎞, (iv) cos x − cos y = − 2sin ⎜, ⎟ sin ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, Conditional Identities Involving the Angles of a Triangle, , TYPE 1 IDENTITIES INVOLVING SINES AND COSINES, METHOD, , 1. Express the sum of the first two terms as a product., 2. In this product, express the sum of two angles in terms of the third angle,, using A + B + C = π., 3. Expand the third term by using one of the following relations:, sin 2θ = 2 sin θ cos θ and cos 2θ = ( 2 cos2θ − 1) = (1 − 2 sin 2θ)., 4. Take the common factor outside., 5. Express the T-ratio of a single angle into a sum of two angles, and use the, necessary formula from the ones given in the box above., SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , If A + B + C = π , prove that, sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C., We have, LHS = (sin 2A + sin 2B) + sin 2C, , [by formula], = 2 sin ( A + B) cos ( A − B) + 2 sin C cos C, = 2 sin ( π − C) cos ( A − B) + 2 sin C cos C, = 2 sin C cos ( A − B) + 2 sin C cos C [B sin ( π − C) = sin C ], 587
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588, , Senior Secondary School Mathematics for Class 11, , = 2sin C [cos ( A − B) + cos C ], = 2sin C [cos ( A − B) + cos {π − ( A + B)}], = 2sin C [cos ( A − B) − cos ( A + B)], , [B C = π − ( A + B)], , [B cos ( π − θ) = cos θ], , = 2 sin C [2 sin A sin B], = 4sin A sin B sin C = RHS ., Trigonometry, , ∴, , EXAMPLE 2, , sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C., , If A + B + C = π , prove that, sin 2A − sin 2B + sin 2C = 4 cos A sin B cos C., We have, , SOLUTION, , LHS = (sin 2A − sin 2B) + sin 2C, , ∴, EXAMPLE 3, , [by formula], = 2 cos ( A + B) sin ( A − B) + 2 sin C cos C, = 2 cos ( π − C) sin ( A − B) + 2 sin C cos C, = −2 cos C sin ( A − B) + 2 sin C cos C [B cos ( π − C) = − cos C], = −2cos C [sin ( A − B) − sin C ], = −2cos C [sin ( A − B) − sin {π − ( A + B)}] [B C = π − ( A + B)], = −2cos C [sin ( A − B) − sin ( A + B)], = −2 cos C [−2 cos A sin B], = 4cos A sin B cos C = RHS ., sin 2A − sin 2B + sin 2C = 4 cos A sin B cos C., , If A + B + C = π , prove that, cos 2A + cos 2B + cos 2C = − 1 − 4 cos A cos B cos C., We have, , SOLUTION, , LHS = cos 2A + cos 2B + cos 2C, , ∴, EXAMPLE 4, , SOLUTION, , = 2 cos ( A + B) cos ( A − B) + 2 cos2C − 1, = 2 cos ( π − C) cos ( A − B) + 2 cos2C − 1, = − 2 cos C cos ( A − B) + 2 cos2C − 1, = − 2 cos C [cos ( A − B) − cos C] − 1, = − 1 − 2cos C [cos ( A − B) − cos {π − ( A + B)}], = − 1 − 2cos C [cos ( A − B) + cos ( A + B)], = − 1 − 2 cos C [2 cos A cos B], = − 1 − 4cos A cos B cos C = RHS ., cos 2A + cos 2B + cos 2C = − 1 − 4 cos A cos B cos C., , If A + B + C = π , prove that, cos 2A + cos 2B − cos 2C = 1 − 4 sin A sin B cos C., We have, LHS = (cos 2A + cos 2B) − cos 2C, , = 2 cos ( A + B) cos ( A − B) − 2 cos2C + 1
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590, , Senior Secondary School Mathematics for Class 11, , C ⎡, ⎧ π ⎛ A + B ⎞ ⎫⎤, ⎛ A − B⎞, cos ⎜, ⎟ − sin ⎨ − ⎜, ⎟ ⎬⎥, ⎝ 2 ⎠, 2 ⎢⎣, ⎩ 2 ⎝ 2 ⎠ ⎭⎦, C ⎡, ⎛ A − B⎞, ⎛ A + B⎞ ⎤, = 2 cos ⎢cos ⎜, ⎟ − cos ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠ ⎥⎦, 2 ⎣, C ⎡, A, B⎤, = 2 cos ⎢2 sin, sin ⎥, 2 ⎣, 2, 2⎦, A, B, C, = 4 sin, sin cos = RHS ., 2, 2, 2, A, B, C, sin A + sin B − sin C = 4 sin, sin cos ⋅, 2, 2, 2, = 2 cos, , ∴, EXAMPLE 7, , If A + B + C = π , prove that, , A, B, C⎞, ⎛, cos A + cos B − cos C = ⎜ 4 cos cos sin ⎟ − 1., ⎝, 2, 2, 2⎠, , We have, , SOLUTION, , LHS = (cos A + cos B) − cos C, , ⎛ A + B⎞, ⎛ A − B⎞ ⎛, 2 C⎞, = 2 cos ⎜, ⎟ cos ⎜, ⎟ − ⎜1 − 2 sin ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠ ⎝, 2⎠, ⎛ π C⎞, ⎛ A − B⎞, 2C, = 2 cos ⎜ − ⎟ cos ⎜, ⎟ − 1 + 2 sin, ⎝ 2 2⎠, ⎝ 2 ⎠, 2, C ⎡, C⎤, ⎛ A − B⎞, = 2 sin ⎢cos ⎜, ⎟ + sin ⎥ − 1, ⎝ 2 ⎠, 2 ⎣, 2⎦, , C ⎡, ⎛ A − B⎞, ⎧ π ( A + B) ⎫⎤, cos ⎜, ⎟ + sin ⎨ −, ⎬ −1, ⎢, ⎝, ⎠, 2 ⎣, 2, 2 ⎭⎥⎦, ⎩2, C ⎡, ⎛ A − B⎞, ⎛ A + B⎞ ⎤, = 2 sin ⎢cos ⎜, ⎟ + cos ⎜, ⎟ −1, ⎝ 2 ⎠, ⎝ 2 ⎠ ⎥⎦, 2 ⎣, C ⎧, A, B⎫, = 2 sin ⎨2 cos cos ⎬ − 1, 2 ⎩, 2, 2⎭, A, B, C⎫, ⎧, = ⎨4 cos cos sin ⎬ − 1 = RHS ., 2, 2, 2⎭, ⎩, A, B, C⎞, ⎛, cos A + cos B − cos C = ⎜ 4 cos cos sin ⎟ − 1., ⎝, 2, 2, 2⎠, = 2 sin, , ∴, , TYPE 2 IDENTITIES INVOLVING SQUARES OF SINES AND COSINES, METHOD Change the squares of sines and cosines into cosines of double the, angles by using the formulae, 1, 1, cos2 θ = (1 + cos 2θ) and sin 2 θ = (1 − cos 2θ)., 2, 2, Now, proceed as above., EXAMPLE 8, , If A + B + C = π , prove that, sin 2 A + sin 2 B + sin 2C = 2(1 + cos A cos B cos C ).
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Conditional Identities Involving the Angles of a Triangle, , 591, , We have, , SOLUTION, , LHS = sin 2 A + sin 2 B + sin 2C, , 1, 1, 1, (1 − cos 2A) + (1 − cos 2B) + (1 − cos 2C ), 2, 2, 2, 3 1, = − (cos 2A + cos 2B + cos 2C ), 2 2, 3 1, = − ( −1 − 4 cos A cos B cos C ) [see Example 3], 2 2, = 2 + 2 cos A cos B cos C = 2(1 + cos A cos B cos C ) = RHS ., sin 2 A + sin 2 B + sin 2C = 2(1 + cos A cos B cos C )., =, , ∴, , If A + B + C = π , prove that, cos2 A + cos2 B − cos2C = 1 − 2 sin A sin B cos C., , EXAMPLE 9, , SOLUTION, , We have, LHS = cos2 A + cos2 B − cos2C, , 1, 1, 1, (1 + cos 2A) + (1 + cos 2B) − (1 + cos 2C ), 2, 2, 2, 1 1, = + (cos 2A + cos 2B − cos 2C ), 2 2, 1 1, = + (1 − 4 sin A sin B cos C ) [see Example 4], 2 2, = 1 − 2sin A sin B cos C = RHS ., cos2 A + cos2 B − cos2C = 1 − 2 sin A sin B cos C., =, , ∴, EXAMPLE 10, , SOLUTION, , In a CABC, prove that, π⎞, π⎞ 3, ⎛, ⎛, cos2 A + cos2 ⎜ A + ⎟ + cos2 ⎜ A − ⎟ = ⋅, ⎝, ⎝, 3⎠, 3⎠ 2, , We have, 2⎛, , LHS = cos A + cos ⎜ A +, 2, , ⎝, , π⎞, π⎞, 2⎛, ⎟ + cos ⎜ A − ⎟, ⎝, 3⎠, 3⎠, , 2π ⎞ ⎫, 2π ⎞ ⎫ 1 ⎧, 1⎧, 1, ⎛, ⎛, = (1 + cos 2A) + ⎨1 + cos ⎜ 2A +, ⎟ ⎬ + ⎨1 + cos ⎜ 2A − ⎟ ⎬, ⎝, ⎝, 3 ⎠⎭ 2⎩, 2⎩, 3 ⎠⎭, 2, 2π ⎞ ⎫, 2π ⎞, 1⎧, 3 1, ⎛, ⎛, = + cos 2A + ⎨cos ⎜ 2A +, ⎟ + cos ⎜ 2A −, ⎟⎬, ⎝, ⎝, 3 ⎠⎭, 3⎠, 2⎩, 2 2, 3 1, 2π, = + cos 2A + cos 2A cos, 2 2, 3, [B cos ( A + B) + cos ( A − B) = 2 cos A cos B], 3 1, 1, + cos 2A − cos 2A, 2 2, 2, 3, = = RHS., 2, =, , 2π, 1⎤, ⎡, ⎢⎣B cos 3 = − 2 ⎥⎦
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592, , Senior Secondary School Mathematics for Class 11, , If A + B + C = π , prove that, A, B, C, A, B, C, sin 2 + sin 2 − sin 2 = 1 − 2 cos cos sin ⋅, 2, 2, 2, 2, 2, 2, , EXAMPLE 11, , SOLUTION, , We have, A, B, C, + sin 2 − sin 2, 2, 2, 2, 1, 1, 1, = (1 − cos A) + (1 − cos B) − (1 − cos C ), 2, 2, 2, 1 1, = − (cos A + cos B − cos C ), 2 2, 1 1 ⎡⎛, A, B, C⎞, ⎤, = −, ⎜ 4 cos cos sin ⎟ − 1⎥ [see Example 7], 2 2 ⎢⎣⎝, 2, 2, 2⎠, ⎦, A, B, C, = 1 − 2 cos cos sin = RHS ., 2, 2, 2, A, B, C, A, B, C, sin 2 + sin 2 − sin 2 = 1 − 2 cos cos, sin ⋅, 2, 2, 2, 2, 2, 2, , LHS = sin 2, , ∴, EXAMPLE 12, , SOLUTION, , If A + B + C = π , prove that, A, B, C, A, B, C, cos2 + cos2 − cos2 = 2 cos cos sin ⋅, 2, 2, 2, 2, 2, 2, , We have, A, B, C, + cos2 − cos2, 2, 2, 2, 1, 1, 1, = (1 + cos A) + (1 + cos B) − (1 + cos C ), 2, 2, 2, 1 1, = + (cos A + cos B − cos C ), 2 2, A, B, C⎞, 1 1 ⎡⎛, ⎤, = + ⎢⎜ 4 cos cos sin ⎟ − 1⎥, 2 2 ⎣⎝, 2, 2, 2⎠, ⎦, A, B, C, = 2 cos cos sin = RHS ., 2, 2, 2, A, B, C, A, B, C, cos2 + cos2 − cos2 = 2 cos cos sin ⋅, 2, 2, 2, 2, 2, 2, , LHS = cos, , 2, , ∴, EXAMPLE 13, , SOLUTION, , If A + B + C = π , prove that, A, B, C, A, B, C⎞, ⎛, cos2 + cos2 + cos2 = 2 ⎜1 + sin, sin sin ⎟ ⋅, ⎝, 2, 2, 2, 2, 2, 2⎠, , We have, A, B, C, + cos2 + cos2, 2, 2, 2, 1, 1, 1, = (1 + cos A) + (1 + cos B) + (1 + cos C ), 2, 2, 2, 3 1, = + (cos A + cos B + cos C ), 2 2, , LHS = cos, , 2
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Conditional Identities Involving the Angles of a Triangle, , 593, , 3 1⎡, ⎤, ⎛ A + B⎞, ⎛ A − B⎞, +, 2 cos ⎜, ⎟ cos ⎜, ⎟ + cos C ⎥, ⎝ 2 ⎠, ⎝ 2 ⎠, 2 2 ⎢⎣, ⎦, 3, ⎛ π C⎞, ⎛ A − B⎞ 1 ⎛, 2 C⎞, = + cos ⎜ − ⎟ cos ⎜, ⎟ + ⎜1 − 2 sin ⎟, ⎝ 2 2⎠, ⎝ 2 ⎠ 2⎝, 2, 2⎠, =, , ⎡ ⎛ A + B⎞ ⎛ π C ⎞, 2 C⎤, ⎢B ⎜⎝ 2 ⎟⎠ = ⎜⎝ 2 − 2 ⎟⎠ and cos C = 1 − 2 sin 2 ⎥, ⎣, ⎦, C, C, ⎛ A − B⎞, 2, = 2 + sin cos ⎜, ⎟ − sin, ⎝ 2 ⎠, 2, 2, C ⎡, C⎤, ⎛ A − B⎞, = 2 + sin ⎢cos ⎜, ⎟ − sin ⎥, ⎝ 2 ⎠, 2 ⎣, 2⎦, ⎡, ⎧ π ⎛ A + B ⎞ ⎫⎤, ⎛ A − B⎞, ⎢cos ⎜⎝ 2 ⎟⎠ − sin ⎨ 2 − ⎜⎝ 2 ⎟⎠ ⎬⎥, ⎭⎦, ⎩, ⎣, C ⎡, ⎛ A − B⎞, ⎛ A + B⎞ ⎤, = 2 + sin ⎢cos ⎜, ⎟ − cos ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠ ⎥⎦, 2 ⎣, C ⎛, A, B⎞, = 2 + sin ⎜ 2 sin, sin ⎟, 2 ⎝, 2, 2⎠, = 2 + sin, , ∴, , C, 2, , A, B, C⎞, ⎛, = 2 ⎜1 + sin, sin sin ⎟ = RHS., ⎝, 2, 2, 2⎠, A, B, C⎞, ⎛, 2 A, 2 B, 2C, cos, sin sin ⎟ ., + cos, + cos, = 2 ⎜1 + sin, ⎝, 2, 2, 2, 2, 2, 2⎠, , TYPE 3 IDENTITIES INVOLVING TANGENTS, METHOD, , 1. Using A + B + C = π , express the sum of two angles in terms of the third., 2. Take tangents on both sides and expand the LHS., 3. Cross multiply and transpose., Similarly, we proceed for identities involving cotangents., EXAMPLE 14, , SOLUTION, , If A + B + C = π , prove that, tan A + tan B + tan C = tan A tan B tan C., , Here, A + B + C = π, ⇒ ( A + B) = ( π − C ), ⇒ tan ( A + B) = tan ( π − C ), tan A + tan B, = − tan C, ⇒, 1 − tan A tan B, ⇒ tan A + tan B = − tan C + tan A tan B tan C, ⇒ tan A + tan B + tan C = tan A tan B tan C., , EXAMPLE 15, , If A + B + C = π , prove that, A, B, B, C, C, A, tan, tan + tan tan + tan tan, = 1., 2, 2, 2, 2, 2, 2
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594, SOLUTION, , Senior Secondary School Mathematics for Class 11, , Here, A + B + C = π, ⎛ A B⎞ ⎛ π C ⎞, ⇒ ⎜ + ⎟ =⎜ − ⎟, ⎝2, 2⎠ ⎝ 2 2⎠, ⇒, , ⇒, , ⇒, ⇒, EXAMPLE 16, , SOLUTION, , If A + B + C = π , prove that, cot B cot C + cot C cot A + cot A cot B = 1., , Here, A + B + C = π, ⇒ ( A + B) = ( π − C ), ⇒ cot ( A + B) = cot ( π − C ) = − cot C, cot A cot B − 1, ⇒, = − cot C, cot A + cot B, ⇒, ⇒, , EXAMPLE 17, , SOLUTION, , C, ⎛ A B⎞, ⎛ π C⎞, tan ⎜ + ⎟ = tan ⎜ − ⎟ = cot, ⎝2, ⎝ 2 2⎠, 2⎠, 2, A, B, tan, + tan, 2, 2 = 1, A, B, C, 1 − tan, tan, tan, 2, 2, 2, C, A, B, C, A, B, tan tan, + tan tan = 1 − tan, tan, 2, 2, 2, 2, 2, 2, A, B, B, C, C, A, tan, tan + tan tan + tan tan, = 1., 2, 2, 2, 2, 2, 2, , cot A cot B − 1 = − cot C cot A − cot B cot C, cot B cot C + cot C cot A + cot A cot B = 1., , If A + B + C = π , prove that, A, B, C, A, B, C, cot, + cot + cot = cot, cot cot ⋅, 2, 2, 2, 2, 2, 2, , Here, A + B + C = π, ⎛ A B⎞ ⎛ π C ⎞, ⇒ ⎜ + ⎟ =⎜ − ⎟, ⎝2, 2⎠ ⎝ 2 2⎠, ⇒, , ⇒, , ⇒, ⇒, , C, ⎛ A B⎞, ⎛ π C⎞, cot ⎜ + ⎟ = cot ⎜ − ⎟ = tan, ⎝2, ⎝ 2 2⎠, 2⎠, 2, A, B, cot, cot − 1, 1, 2, 2, =, A, B, C, cot, + cot, cot, 2, 2, 2, A, B, A, B, C, C, cot, + cot = cot, cot cot − cot, 2, 2, 2, 2, 2, 2, A, B, C, A, B, C, cot, + cot + cot = cot, cot cot ⋅, 2, 2, 2, 2, 2, 2
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Conditional Identities Involving the Angles of a Triangle, , 595, , EXERCISE 16, If A + B + C = π , prove that, , 1. sin 2A + sin 2B − sin 2C = 4 cos A cos B sin C, 2. cos 2A − cos 2B − cos 2C = − 1 + 4 cos A sin B sin C, 3. cos 2A − cos 2B + cos 2C = 1 − 4 sin A cos B sin C, A, B, C, 4. sin A + sin B + sin C = 4 cos cos cos, 2, 2, 2, A, B, C, 5. cos A + cos B + cos C = 1 + 4 sin, sin sin, 2, 2, 2, sin 2A + sin 2B + sin 2C, A, B, C, 6., = 8 sin, sin sin, sin A + sin B + sin C, 2, 2, 2, 7. sin ( B + C − A) + sin (C + A − B) − sin ( A + B − C) = 4 cos A cos B sin C, cos A, cos B, cos C, 8., +, +, =2, sin B sin C sin C sin A sin A sin B, 9. cos2 A + cos2 B + cos2C = 1 − 2 cos A cos B cos C, 10. sin 2 A − sin 2 B + sin 2C = 2 sin A cos B sin C, A, B, C, A, B, C, 11. sin 2 + sin 2 + sin 2 = 1 − 2 sin, sin sin, 2, 2, 2, 2, 2, 2, 12. tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C, HINTS TO SOME SELECTED QUESTIONS, 7. LHS = sin ( A + B + C − 2 A ) + sin ( A + B + C − 2 B) − sin ( A + B + C − 2 C ), = sin ( π − 2 A ) + sin ( π − 2 B) − sin ( π − 2C ), = sin 2 A + sin 2 B − sin 2 C ., 8. Multiply both sides by 2 sin A sin B sin C ., 12. A + B + C = π ⇒ 2 A + 2 B = 2 π − 2 C, ⇒ tan ( 2 A + 2 B) = tan ( 2 π − 2 C ) = − tan 2 C ., , q
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17, Trigonometric Equations, TRIGONOMETRIC EQUATIONS, , Equations which involve trigonometric functions of unknown angles are known as, trigonometric equations., Trigonometric Equations, , For example, sin x =, , 1, , tan x = − 3 , cos x = sin 2x , etc., 2, , SOLUTION OF A TRIGONOMETRIC EQUATION A solution of a trigonometric equation is, the value of the unknown angle that satisfies the equation., , A trigonometric equation may have an infinite number of solutions., PRINCIPAL SOLUTIONS The solutions of a trigonometric equation for which 0 ≤ x < 2π, are called its principal solutions., GENERAL SOLUTION A solution of a trigonometric equation, generalised by means of, periodicity, is known as the general solution., , REMARK, , We shall use ‘I’ to denote the set of all integers., , GENERAL SOLUTIONS OF TRIGONOMETRIC EQUATIONS, THEOREM 1, , PROOF, , (i) sin θ = 0 ⇔ θ = nπ , n ∈ I, π, (ii) cos θ = 0 ⇔ θ = ( 2n + 1) , n ∈ I, 2, (iii) tan θ = 0 ⇔ θ = nπ , n ∈ I, , (i) sin θ = 0 ⇔ θ = 0, ± π , ± 2π , ± 3 π , …, ⇔ θ = nπ , n ∈ I., π, 3π, 5π, (ii) cos θ = 0 ⇔ θ = 0, ± , ±, ,±, ,…, 2, 2, 2, π, ⇔ θ = ( 2n + 1) , n ∈ I., 2, sin θ, (iii) tan θ = 0 ⇔, =0, cos θ, ⇔ sin θ = 0, ⇔ θ = 0, ± π , ± 2π , ± 3 π , …, ⇔ θ = nπ , n ∈ I., 596
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600, , Senior Secondary School Mathematics for Class 11, SOLVED EXAMPLES, , EXAMPLE 1, , SOLUTION, , EXAMPLE 2, , SOLUTION, , Find the principal solutions of each of the following equations:, 1, 1, 1, (ii) cos x =, (iii) tan x =, (i) sin x =, 2, 2, 3, 1, (i) The given equation is sin x = ⋅, 2, π 1, π⎞ 1, ⎛, We know that sin = and sin ⎜ π − ⎟ = ⋅, ⎝, 6 2, 6⎠ 2, π, 1, 5π, 1, and sin, =, =, ⋅, ∴ sin, 6, 2, 6, 2, π, 5π, Hence, the principal solutions are x = and x =, ⋅, 6, 6, 1, (ii) The given equation is cos x =, ⋅, 2, π, 1, π⎞, 1, ⎛, and cos ⎜ 2π − ⎟ =, We know that cos =, ⋅, ⎝, 4, 4⎠, 2, 2, π, 1, 7π, 1, and cos, =, ⋅, ∴ cos =, 4, 4, 2, 2, π, 7π, Hence, the principal solutions are x = and x =, ⋅, 4, 4, 1, (iii) The given equation is tan x =, ⋅, 3, π, 1, π⎞, 1, ⎛, and tan ⎜ π + ⎟ =, We know that tan =, ⋅, ⎝, ⎠, 6, 6, 3, 3, π, 1, 7π, 1, and tan, =, ⋅, ∴ tan =, 6, 6, 3, 3, π, 7π, Hence, the principal solutions are x = and x =, ⋅, 6, 6, Find the principal solutions of each of the following equations:, −1, − 3, (ii) cos x =, (i) sin x =, (iii) cot x = − 3, 2, 2, − 3, (i) The given equation is sin x =, ⋅, 2, π, 3, We know that sin =, ⋅, 3, 2, π⎞, π − 3, π⎞, π − 3, ⎛, ⎛, ∴ sin ⎜ π + ⎟ = − sin =, , and ⎜ 2π − ⎟ = − sin =, ⋅, ⎝, ⎝, 3⎠, 3, 2, 3⎠, 3, 2, 4π − 3, 5π − 3, and sin, =, =, ⋅, ∴ sin, 3, 2, 3, 2, 4π, 5π, and x =, Hence, the principal solutions are x =, ⋅, 3, 3
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Trigonometric Equations, , (ii) The given equation is cos x =, , 601, , −1, ⋅, 2, , π 1, = ⋅, 3 2, π⎞, π, π⎞, π, 1, 1, ⎛, ⎛, ∴ cos ⎜ π − ⎟ = − cos = − , and cos ⎜ π + ⎟ = − cos = − ⋅, ⎝, ⎝, 3⎠, 3, 2, 3⎠, 3, 2, We know that cos, , ∴ cos, , 2π, 1, 4π, 1, = − and cos, =− ⋅, 3, 2, 3, 2, , 2π, 4π, and x =, ⋅, 3, 3, 1, (iii) The given equation is cot x = − 3 ⇔ tan x = −, ⋅, 3, π, 1, We know that tan =, ⋅, 6, 3, π⎞, π, π⎞, π, −1, −1, ⎛, ⎛, ∴ tan ⎜ π − ⎟ = − tan =, , and tan ⎜ 2 π − ⎟ = − tan =, ⋅, ⎝, ⎝, 6⎠, 6, 6⎠, 6, 3, 3, Hence, the principal solutions are x =, , ∴ tan, , 5π, −1, 11π, −1, and tan, =, =, ⋅, 6, 6, 3, 3, , Hence, the principal solutions are x =, , 5π, 11π, and x =, ⋅, 6, 6, , EXAMPLE 3, , In each of the following, find the general value of x satisfying the, equation:, 1, 1, 1, (iii) tan x =, (i) sin x =, (ii) cos x =, 2, 2, 3, , SOLUTION, , (i) Given: sin x =, , 1, ⋅, 2, , The least value of x in [0, 2π [ for which sin x =, , 1, π, is x = ⋅, 4, 2, , π, π, ⇒ x = nπ + ( −1) n ⋅ , where n ∈ I., 4, 4, π, Hence, the general solution is x = nπ + ( −1) n ⋅ , where n ∈ I., 4, 1, (ii) Given: cos x = ⋅, 2, 1, π, The least value of x in [0, 2π [ for which cos x = is x = ⋅, 2, 3, π, π⎞, ⎛, ⇒ x = ⎜ 2nπ ± ⎟ , where n ∈ I., ∴ cos x = cos, ⎝, 3, 3⎠, ∴ sin x = sin, , π⎞, ⎛, Hence, the general solution is x = ⎜ 2nπ ± ⎟ , where n ∈ I., ⎝, 3⎠
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602, , Senior Secondary School Mathematics for Class 11, , (iii) Given: tan x =, , 1, ⋅, 3, , The least value of x in [0, 2π [ for which tan x =, , 1, π, is ⋅, 3 6, , π, π⎞, ⎛, ⇒ x = ⎜nπ + ⎟ , where n ∈ I., ⎝, 6, 6⎠, π⎞, ⎛, Hence, the general solution is x = ⎜nπ + ⎟ , where n ∈ I., ⎝, 6⎠, , ∴ tan x = tan, , EXAMPLE 4, , SOLUTION, , 3 cosec x = 2., , 2, 3, ⇒ sin x =, ⋅, 2, 3, 3 π, is ⋅, The least value of x in [0, 2π [ for which sin x =, 2, 3, π, ⎧, n π⎫, ⇒ x = ⎨nπ + ( −1) ⋅ ⎬, where n ∈ I., ∴ sin x = sin, 3, 3⎭, ⎩, π⎫, ⎧, Hence, the general solution is x = ⎨nπ + ( −1) n ⋅ ⎬, where n ∈ I., 3⎭, ⎩, , EXAMPLE 5, , SOLUTION, , Find the general value of x for which, Given:, , 3 cosec x = 2 ⇒ cosec x =, , In each of the following, find the general value of x satisfying the, equation:, −1, − 3, (i) sin x =, (iii) cot x = − 3, (ii) cos x =, 2, 2, − 3, π, π⎞, 4π, ⎛, = − sin = sin ⎜ π + ⎟ = sin, ⋅, ⎝, 2, 3, 3⎠, 3, 4π ⎞, 4π, ⎧, ⇒ x = ⎨nπ + ( −1) n ⋅, ∴ sin x = sin, ⎟ , where n ∈ I., 3⎠, 3, ⎩, 4π ⎞, ⎧, Hence, the general solution is x = ⎨nπ + ( −1) n ⋅, ⎟ , where n ∈ I., 3⎠, ⎩, −1, π, π⎞, 2π, ⎛, (ii) cos x =, = − cos = cos ⎜ π − ⎟ = cos, ⋅, ⎝, 2, 3, 3⎠, 3, 2π ⎞, 2π, ⎛, ⇒ x = ⎜ 2nπ ±, ∴ cos x = cos, ⎟ , where n ∈ I., ⎝, 3⎠, 3, 2π ⎞, ⎛, Hence, the general solution is x = ⎜ 2nπ ±, ⎟ , where n ∈ I., ⎝, 3⎠, (iii) cot x = − 3, −1, π, π⎞, 5π, ⎛, = − tan = tan ⎜ π − ⎟ = tan, ⇒ tan x =, ⎝, 6, 6⎠, 6, 3, 5π, ⇒ tan x = tan, 6, 5π ⎞, ⎛, ⇒ x = ⎜ nπ +, ⎟ , where n ∈ I., ⎝, 6⎠, (i) sin x =
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Trigonometric Equations, , 603, , 5π ⎞, ⎛, Hence, the general solution is x = ⎜nπ +, ⎟ , where n ∈ I., ⎝, 6⎠, EXAMPLE 6, , Find the general solution of each equation:, (i) 3 cot x + 1 = 0, (ii) cosec x + 2 = 0, , SOLUTION, , (i), , 3 cot x + 1 = 0, −1, ⇒ cot x =, 3, ⇒ tan x = − 3 = − tan, , π, = tan, 3, , 2π, 3, 2π ⎞, ⎛, ⇒ x = ⎜ nπ +, ⎟ , where n ∈ I., ⎝, 3⎠, , π⎞, 2π, ⎛, ⎜ π − ⎟ = tan, ⎝, 3⎠, 3, , ⇒ tan x = tan, , 2π ⎞, ⎛, Hence, the general solution is x = ⎜nπ +, ⎟ , where n ∈ I., ⎝, 3⎠, (ii) cosec x + 2 = 0, 1, π, π⎞, 5π, ⎛, = − sin = sin ⎜ π + ⎟ = sin, ⇒ sin x = −, ⎝, 4, 4⎠, 4, 2, 5π, ⇒ sin x = sin, 4, 5π ⎫, ⎧, ⇒ x = ⎨nπ + ( −1) n ⋅, ⎬, where n ∈ I., 4 ⎭, ⎩, 5π ⎫, ⎧, Hence, the general solution is x = ⎨nπ + ( −1) n ⋅, ⎬, where n ∈ I., 4 ⎭, ⎩, EXAMPLE 7, , SOLUTION, , Find the general solution of each of the equations:, 1, (i) sin 2x = −, (ii) tan 3 x = − 1, 2, 1, 7π, π, π⎞, ⎛, (i) sin 2x = − = − sin = sin ⎜ π + ⎟ = sin, ⎝, ⎠, 2, 6, 6, 6, 7π, ⇒ sin 2x = sin, 6, 7π ⎫, ⎧, ⇒ 2x = ⎨nπ + ( −1) n ⋅, ⎬, where n ∈ I, 6 ⎭, ⎩, 7π ⎫, ⎧nπ, + ( −1) n ⋅, ⇒ x=⎨, ⎬, where n ∈ I., 12 ⎭, ⎩ 2, 7π ⎫, ⎧nπ, Hence, the general solution is x = ⎨, + ( −1) n ⋅, ⎬, where n ∈ I., 12 ⎭, ⎩ 2, 3π, π, π⎞, ⎛, (ii) tan 3 x = −1 = − tan = tan ⎜ π − ⎟ = tan, ⎝, 4, 4⎠, 4, 3π, ⇒ tan 3 x = tan, 4
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604, , Senior Secondary School Mathematics for Class 11, , 3π ⎞, ⎛, ⇒ 3 x = ⎜ nπ +, ⎟ , where n ∈ I., ⎝, 4 ⎠, ⎛ nπ π ⎞, + ⎟ , where n ∈ I., ⇒ x=⎜, ⎝ 3, 4⎠, ⎛ nπ π ⎞, Hence, the general solution is x = ⎜, + ⎟ , where n ∈ I., ⎝ 3, 4⎠, EXAMPLE 8, , Find the general solution of each of the equations:, (ii) 2 cos2 x = 1, (iii) cot 2 x = 3, (i) 4 sin 2 x = 1, , SOLUTION, , (i) 4 sin 2 x = 1 ⇒ sin 2 x =, , 2, , 1 ⎛ 1⎞, π, = ⎜ ⎟ = sin 2, 4 ⎝ 2⎠, 6, 2, 2 π, ⇒ sin x = sin, 6, π⎫, ⎧, ⇒ x = ⎨nπ ± ⎬, where n ∈ I., 6⎭, ⎩, π⎞, ⎛, Hence, the general solution is x = ⎜nπ ± ⎟ , n ∈ I., ⎝, 6⎠, 2, , π, 1 ⎛ 1 ⎞, = ⎜ ⎟ = cos2, 2 ⎝ 2⎠, 4, π, ⇒ cos2 x = cos2, 4, π, ⎛, ⎞, ⇒ x = ⎜nπ ± ⎟ , where n ∈ I., ⎝, 4⎠, π⎞, ⎛, Hence, the general solution is x = ⎜nπ ± ⎟ , n ∈ I., ⎝, 4⎠, , (ii) 2 cos2 x = 1 ⇒ cos2 x =, , 2, , 1 ⎛ 1 ⎞, 2 π, =⎜, ⎟ = tan, 3 ⎝ 3⎠, 6, π, ⇒ tan 2 x = tan 2, 6, π, ⎛, ⎞, ⇒ x = ⎜nπ ± ⎟ , where n ∈ I., ⎝, 6⎠, π⎞, ⎛, Hence, the general solution is ⎜nπ ± ⎟ , where n ∈ I., ⎝, 6⎠, , (iii) cot 2 x = 3 ⇒ tan 2 x =, , EXAMPLE 9, SOLUTION, , Find the general solution of the equation sin 2x + sin 4x + sin 6x = 0., The given equation may be written as (sin 6x + sin 2x) + sin 4x = 0, ⇒, , 2 sin, , ( 6x + 2x), ( 6x − 2x), cos, + sin 4x = 0, 2, 2, (C + D), (C − D) ⎤, ⎡, cos, ⎢⎣B sin C + sin D = 2 sin, 2, 2 ⎥⎦
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Trigonometric Equations, , ⇒, ⇒, ⇒, ⇒, ⇒, ⇒, ⇒, , 2 sin 4x cos 2x + sin 4x = 0, sin 4x( 2 cos 2x + 1) = 0, sin 4x = 0 or 2 cos 2x + 1 = 0, 1, 2π, π, π⎞, ⎛, sin 4x = 0 or cos 2x = − = − cos = cos ⎜ π − ⎟ = cos, ⎝, ⎠, 2, 3, 3, 3, 2π, sin 4x = 0 or cos 2x = cos, 3, 2π ⎞, ⎛, 4x = nπ or 2x = ⎜ 2mπ ±, ⎟ , where m , n ∈ I, ⎝, 3⎠, x=, , nπ, π⎞, ⎛, or x = ⎜mπ ± ⎟ , where m , n ∈ I., ⎝, 4, 3⎠, , Hence, the general solution is given by x =, where m , n ∈ I., EXAMPLE 10, SOLUTION, , 605, , nπ, π⎞, ⎛, or x = ⎜mπ ± ⎟ ,, ⎝, 4, 3⎠, , Solve: 2 cos2 x + 3 sin x = 0., , We have, 2 cos2 x + 3 sin x = 0, ⇒, , 2(1 − sin 2 x) + 3 sin x = 0, , ⇒, , 2 sin 2 x − 3 sin x − 2 = 0, , ⇒, , 2 sin 2 x − 4 sin x + sin x − 2 = 0, , ⇒ 2 sin x(sin x − 2) + (sin x − 2) = 0, ⇒ (sin x − 2)( 2 sin x + 1) = 0, ⇒ (sin x − 2) = 0 or ( 2 sin x + 1) = 0, ⇒ sin x = 2 or ( 2 sin x + 1) = 0, ⇒ 2 sin x + 1 = 0 [B sin x = 2 is not possible ], 1, π, π⎞, 7π, ⎛, ⇒ sin x = − = − sin = sin ⎜ π + ⎟ = sin, ⎝, ⎠, 2, 6, 6, 6, 7π, ⇒ sin x = sin, 6, 7π ⎫, ⎧, ⇒ x = ⎨nπ + ( −1) n ⋅, ⎬, where n ∈ I., 6 ⎭, ⎩, 7π ⎫, ⎧, Hence, the general solution is given by x = ⎨nπ + ( −1) n ⋅, ⎬, where, 6 ⎭, ⎩, n ∈ I., EXAMPLE 11, , Find the general solution for each of the following equations:, (i) cos 4x = cos 2x, (ii) cos 3 x = sin 2x, (iii) sin 3 x + cos 2x = 0 (iv) sin mx + sin nx = 0, , SOLUTION, , (i) cos 4x = cos 2x, ⇒ cos 4x − cos 2x = 0
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Trigonometric Equations, , 607, , π⎞, ⎛, ⎛ 2nπ π ⎞, Hence, the general solution is x = ⎜ 2nπ − ⎟ or x = ⎜, − ⎟,, ⎝, ⎝ 5, 10⎠, 2⎠, where n ∈ I., π⎞, π⎞, ⎛, ⎛, NOTE ⎜ −2nπ − ⎟ and ⎜ 2nπ − ⎟ give the same result as n ∈ I., ⎝, ⎝, 2⎠, 2⎠, (iv) sin mx + sin nx = 0, (m + n) x, (m − n) x, cos, =0, ⇒ 2 sin, 2, 2, (m + n) x, (m − n) x, or sin, = 0 or cos, =0, 2, 2, (m + n) x, (m − n) x, π, = pπ or, = ( 2 q + 1) , where p , q ∈ I, ⇒, 2, 2, 2, 2 pπ, ( 2 q + 1) π, or x =, , where p , q ∈ I., ⇒ x=, (m + n), (m − n), 2 pπ, ( 2 q + 1) π, or x =, Hence, the general solution is x =, ,, (m + n), (m − n), where p , q ∈ I., Solve:, , 3 cos x − sin x = 1., , Given:, , 3 cos x − sin x = 1., , EXAMPLE 12, SOLUTION, , … (i), , Dividing both sides of (i) by ( 3 ) + ( −1) , i.e., by 2, we get, 2, , 2, , 3, 1, 1, cos x − sin x =, 2, 2, 2, π, π 1, ⇒ cos x cos − sin x sin =, 6, 6 2, π⎤, π⎞, π ⎡ 1, ⎛, Q = cos ⎥, ⇒ cos ⎜ x + ⎟ = cos, ⎝, 3⎦, 6⎠, 3 ⎢⎣ 2, π⎞, π, ⎛, ⇒ ⎜ x + ⎟ = 2nπ ± , where n ∈ I [B cos θ = cos α ⇒ θ = 2nπ ± α ], ⎝, 6⎠, 3, π⎞ ⎛, π⎞, π⎞ ⎛, π⎞, ⎛, ⎛, ⇒ ⎜ x + ⎟ = ⎜ 2nπ + ⎟ or ⎜ x + ⎟ = ⎜ 2nπ − ⎟ ,, ⎝, ⎝, 6⎠ ⎝, 3⎠, 6⎠ ⎝, 3⎠, π⎞, π⎞, ⎛, ⎛, x = ⎜ 2nπ + ⎟ or x = ⎜ 2nπ − ⎟ , where n ∈ I., ⎝, ⎝, 6⎠, 2⎠, π⎞, π⎞, ⎛, ⎛, Hence, the general solution is x = ⎜ 2nπ + ⎟ or x = ⎜ 2nπ − ⎟ ,, ⎝, ⎝, 6⎠, 2⎠, where n ∈ I., EXAMPLE 13, SOLUTION, , Solve: sec x − tan x =, , 3., , We have, sec x − tan x =, , 1, sin x, −, = 3 ⇒ (1 − sin x) =, cos x cos x, ⇒ 3 cos x + sin x = 1., , 3 ⇒, , 3 cos x
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608, , Senior Secondary School Mathematics for Class 11, , Thus, the given equation becomes, 3 cos x + sin x = 1., , … (i), , Dividing both sides of (i) by ( 3 ) + 1 , i.e., by 2, we get, 3, 1, 1, cos x + sin x =, 2, 2, 2, π, π 1, ⇒ cos x cos + sin x sin =, 6, 6 2, π⎞, π, ⎛, ⇒ cos ⎜ x − ⎟ = cos, ⎝, 6⎠, 3, π⎞, π, ⎛, ⇒ ⎜ x − ⎟ = 2nπ ± , where n ∈ I [B cos θ = cos α ⇒ θ = 2nπ ± α ], ⎝, 6⎠, 3, π⎞ ⎛, π⎞, π⎞ ⎛, π⎞, ⎛, ⎛, ⇒ ⎜ x − ⎟ = ⎜ 2nπ + ⎟ or ⎜ x − ⎟ = ⎜ 2nπ − ⎟, ⎝, ⎝, 6⎠ ⎝, 3⎠, 6⎠ ⎝, 3⎠, ⎛ π π⎞, ⎛ π π⎞, ⇒ x = 2nπ + ⎜ + + ⎟ or x = 2nπ + ⎜ − + ⎟ , where n ∈ I, ⎝ 3 6⎠, ⎝ 3 6⎠, π⎞, π⎞, ⎛, ⎛, ⇒ x = ⎜ 2nπ + ⎟ or x = ⎜ 2nπ − ⎟ , where n ∈ I, ⎝, ⎝, 2⎠, 6⎠, π⎞, ⎛, ⇒ x = ⎜ 2nπ − ⎟ , where n ∈ I, ⎝, 6⎠, π⎞ ⎤, ⎡, ⎛, ⎢Q sec x is not defined when x = ⎜⎝ 2nπ + 2 ⎟⎠ ⎥ ⋅, ⎣, ⎦, π⎞, ⎛, Hence, the general solution is x = ⎜ 2nπ − ⎟ , where n ∈ I., ⎝, 6⎠, 2, , 2, , EXERCISE 17, 1. Find the principal solutions of each of the following equations:, 3, 2, (iv) cot x = 3, (i) sin x =, , 1, 2, (v) cosec x = 2, (ii) cos x =, , (iii) tan x =, , 3, , 2, 3, 2. Find the principal solutions of each of the following equations:, −1, (ii) 2 cos x + 1 = 0, (i) sin x =, 2, (iii) tan x = − 1, (iv) 3 cosec x + 2 = 0, (v) tan x = − 3, , (vi) sec x =, , (vi), , 3 sec x + 2 = 0, , Find the general solution of each of the following equations:, , 3., , (i) sin 3 x = 0, (iv) cos 2x = 0, , 3x, =0, 2, 5x, (v) cos, =0, 2, (ii) sin, , π⎞, ⎛, (iii) sin ⎜ x + ⎟ = 0, ⎝, 5⎠, π⎞, ⎛, (vi) cos ⎜ x + ⎟ = 0, ⎝, 10⎠
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Trigonometric Equations, , (vii) tan 2x = 0, 4. (i) sin x =, , 3, 2, , 5. (i) cos x =, , −1, 2, , 6. (i) sin 2x =, , 1, 2, , 609, , π⎞, π⎞, ⎛, ⎛, (viii) tan ⎜ 3 x + ⎟ = 0 (ix) tan ⎜ 2x − ⎟ = 0, ⎝, ⎝, 6⎠, 4⎠, (ii) cos x = 1, , (iii) sec x = 2, , (ii) cosec x = − 2, , (iii) tan x = − 1, , (ii) cos 3 x =, , 1, 2, , (iii) tan, , 2x, =, 3, , 3, −2, 3, , 7. (i) sec 3 x = − 2, , (ii) cot 4x = − 1, , (iii) cosec 3 x =, , 8. (i) 4 cos2 x = 1, , (ii) 4 sin 2 x − 3 = 0, , (iii) tan 2 x = 1, , 9. (i) cos 3 x = cos 2x, , (ii) cos 5 x = sin 3 x, , (iii) cos mx = sin nx, , 10. sin x = tan x, , 11. 4 sin x cos x + 2 sin x + 2 cos x + 1 = 0, , 12. sec 2x = 1 − tan 2x, , 13. tan 3 x − 3 tan x = 0, , 14. sin x + sin 3 x + sin 5 x = 0, , 15. sin x tan x − 1 = tan x − sin x, , 16. cos x + sin x = 1, , 17. cos x − sin x = − 1, , 2, , 18., , 3 cos x + sin x = 1, , 19. 2 tan x − cot x + 1 = 0, , 20. sin x tan x − 1 = tan x − sin x, , 21. cot x + tan x = 2cosec x, ANSWERS (EXERCISE 17), , 1., , 2., , 3., , π, ,, 3, π, (iv) ,, 6, (i), , 2π, 3, 7π, 6, , π, ,, 3, π, (v) ,, 6, (ii), , 7 π 11π, ,, 6, 6, 4π 5 π, (iv), ,, 3 3, , 3π 5π, ,, 4, 4, 2π 5 π, (v), ,, 3 3, , nπ, , n ∈I, 3, , (iv) ( 2n + 1), (vii), , π, ,, 3, π, (vi) ,, 6, (iii), , π, , n ∈I, 4, , nπ, , n ∈I, 2, , (ii), , (iii), , 2nπ, , n ∈I, 3, , (v) ( 2n + 1), , 4π, 3, 11π, 6, , 3π 7π, ,, 4, 4, 5π 7π, (vi), ,, 6 6, , (ii), , (i), , (i), , 5π, 3, 5π, 6, , π⎞, ⎛, (iii) ⎜nπ − ⎟ , n ∈ I, ⎝, 5⎠, , π, 2π ⎞, ⎛, , n ∈ I (vi) ⎜nπ +, ⎟ , n ∈I, ⎝, 5, 5 ⎠, , ⎛ nπ π ⎞, ⎛ nπ π ⎞, (viii) ⎜, − ⎟ , n ∈ I (ix) ⎜, + ⎟ , n ∈I, ⎝ 3 18⎠, ⎝ 2, 8⎠
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Trigonometric Equations, , 611, , π⎞, ⎛, 16. x = ⎜ 2nπ + ⎟ or x = 2nπ , where n ∈ I, ⎝, 2⎠, π⎞, ⎛, 17. x = ⎜ 2nπ + ⎟ or x = ( 2n − 1) π , where n ∈ I, ⎝, 2⎠, π⎞, π⎞, ⎛, ⎛, 18. x = ⎜ 2nπ + ⎟ or x = ⎜ 2nπ − ⎟ , where n ∈ I, ⎝, ⎝, 2⎠, 6⎠, 19. x = nπ +, , 3π, 1, or x = mπ + tan −1 , where m , n ∈ I, 4, 2, , π, π⎞, ⎛, 20. x = ⎜mπ − ⎟ or x = nπ + ( −1) n, , where m , n ∈ I, ⎝, 4⎠, 2, 21. x = 2nπ ±, , π, , where n ∈ I, 3, , HINTS TO SOME SELECTED QUESTIONS, sin x, cos x, sin x cos x − sin x = 0, sin x (cos x − 1) = 0, sin x = 0 or cos x = 1, x = nπ or x = 2 mπ, where m , n ∈ I., , 10. sin x = tan x ⇒ sin x =, ⇒, ⇒, ⇒, ⇒, , 11. 4 sin x cos x + 2 sin x + 2 cos x + 1 = 0, ⇒ 2 sin x ( 2 cos x + 1) + ( 2 cos x + 1) = 0, ⇒ ( 2 cos x + 1)( 2 sin x + 1) = 0, 1, 1, ⇒ cos x = − or sin x = −, 2, 2, 1, π, π, 2π, ⇒ cos x = − = − cos = cos ⎛⎜ π − ⎞⎟ = cos, ⎝, 2, 3, 3⎠, 3, 1, π, π⎞, 7π, ⎛, or sin x = − = − sin = sin ⎜ π + ⎟ = sin, ⎝, 2, 6, 6⎠, 6, 2π ⎞, m 7π, or, where, ⇒ x = ⎛⎜ 2 nπ ±, x, =, m, π, +, (, −, 1, ), ⋅, ,, m , n ∈ I., ⎟, ⎝, 3⎠, 6, 12. sec2 2 x = 1 − tan 2 x, ⇒ 1 + tan 2 2 x = 1 − tan 2 x, ⇒ tan 2 2 x + tan 2 x = 0, ⇒ tan 2 x (tan 2 x + 1) = 0, ⇒ tan 2 x = 0 or tan 2 x = − 1, ⇒ 2x = nπ or tan 2 x = − tan, ⇒ x=, , π, π, 3π, = tan ⎛⎜ π − ⎞⎟ = tan, ⎝, 4, 4⎠, 4, , nπ, 3π, or 2 x = mπ +, , where m , n ∈ I, 2, 4
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612, , Senior Secondary School Mathematics for Class 11, ⇒, , x=, , nπ, mπ 3 π, or x =, +, , where m , n ∈ I., 2, 2, 8, , 13. tan 3 x − 3 tan x = 0 ⇒ tan x (tan 2 x − 3 ) = 0, ⇒ tan x = 0 or tan x =, , 3 or tan x = −, , ⇒ x = nπ or tan x = tan, , π, π, π, or tan x = − tan = tan ⎛⎜ π − ⎞⎟, ⎝, 3, 3, 3⎠, , ⇒ x = nπ or x = mπ +, , π, 2π, or x = pπ +, , where m , n, p ∈ I., 3, 3, , 15. The given equation becomes, (sin x − 1)(sin x + cos x ) = 0, ⇒, ⇒, ⇒, , sin x = 1 or tan x = − 1, π, 3π, sin x = sin or tan x = tan, 2, 4, 3π, n π, or x = mπ +, x = nπ + ( −1), , where m , n ∈ I., 2, 4, , 16. cos x + sin x = 1 ⇒, , 1, 1, 1, cos x +, sin x =, 2, 2, 2, , π, π, 1, + sin x sin =, 4, 4, 2, π⎞, π, ⎛, ⇒ cos ⎜ x − ⎟ = cos, ⎝, 4⎠, 4, ⇒ cos x cos, , π, π, = 2 nπ ±, 4, 4, π, π, π, π, or x − = 2 nπ −, ⇒ x − = 2 nπ +, 4, 4, 4, 4, π, or x = 2 nπ, where n ∈ I., ⇒ x = 2 nπ +, 2, ⇒ x−, , 17. cos x − sin x = − 1, ⇒, ⇒, , 1, 1, −1, cos x −, sin x =, 2, 2, 2, π, π, π, π, cos cos x − sin sin x = − cos = cos ⎛⎜ π − ⎞⎟, ⎝, 4, 4, 4, 4⎠, , ⇒, , π, 3π, cos ⎛⎜ x + ⎞⎟ = cos, ⎝, 4⎠, 4, , ⇒, , x+, , ⇒, ⇒, , 3, , π, 3π, = 2 nπ ±, 4, 4, π, 3π, π, 3π, or x + = 2 nπ −, x + = 2 nπ +, 4, 4, 4, 4, π, or x = ( 2 n − 1)π , where n ∈ I., x = 2 nπ +, 2
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Trigonometric Equations, 19. The given equation is, 2 tan 2 x + tan x − 1 = 0, ⇒, , (tan x + 1)( 2 tan x − 1) = 0, 1, ⇒ tan x = −1 or tan x =, 2, 1, 3π, or tan x = tan ⎛⎜ tan −1 ⎞⎟, ⇒ tan x = tan, ⎝, 2⎠, 4, 1, 3π, or x = mπ + tan −1 , where m , n ∈ I., ⇒ x = nπ +, 2, 4, 20. The given equation is, ⇒, ⇒, ⇒, ⇒, , sin x tan x + sin x − 1 − tan x = 0, sin x (tan x + 1) − (tan x + 1) = 0, (tan x + 1)(sin x − 1) = 0, π, π, tan x = − 1 = tan ⎛⎜ − ⎞⎟ or sin x = 1 = sin, ⎝ 4⎠, 2, π, π, x = ⎛⎜ mπ − ⎞⎟ or x = nπ + ( −1) n ⋅, ⎝, 4⎠, 2, , 21. The given equation is, cos x sin x, 2, +, =, ⇒ cos 2 x + sin 2 x = 2 cos x, sin x cos x sin x, 1, π, ⇒ cos x = = cos, 2, 3, π, ⇒ x = 2 nπ ± , where n ∈ I., 3, , q, , 613
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18, Solution of Triangles, The three sides and three angles of a triangle are called the parts, , INTRODUCTION, , of the triangle., When at least three parts of a triangle are known (i.e., any three parts), we, can compute the measures of the remaining three parts., The process of computing the measures of unknown parts of a triangle from those, of the given parts is known as solving the triangle., A, THE OBLIQUE TRIANGLE A triangle which does not, contain a right angle is called an oblique triangle., , In a CABC, we denote the lengths of the sides as, BC = a , CA = b and AB = c., THEOREM 1, , PROOF, , c, , b, , a, , B, , C, , (Sine Formula or Sine Rule) In any triangle, the sides are proportional to, the sines of the opposite angles., That is, in a CABC,, a, b, c, =, =, ⋅, sin A sin B sin C, , Let CABC be any triangle., We know that the sum of all the angles of a triangle is 180°. So, each, angle of a triangle cannot be obtuse., Let us assume here that ∠B is acute., Now, we consider the following cases:, (i) ∠C is acute, , (ii) ∠C is obtuse, , (iii) ∠C is a right angle, A, A, , A, c, , B, , h, , a, , D, , c, b, , c, b, , b = h, , h, C, , C, B, , C, a, , ( i ), , ( ii ), , Draw AD ⊥ BC or BC produced., 614, , D, B, , a, ( iii ), , C (D )
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Solution of Triangles, , 615, , Let AD = h., Then, in each figure, we have:, AD, h, … (1), = sin B ⇒, = sin B ⇒ h = c sin B, AB, c, Also, we have, AD, h, In Fig. (i),, = sin C ⇒, = sin C ⇒ h = b sin C., AC, b, AD, h, In Fig. (ii),, = sin ( π − C) = sin C ⇒, = sin C ⇒ h = b sin C., AC, b, AD, π, h, In Fig. (iii),, = 1 = sin = sin C ⇒, = sin C ⇒ b sin C., AC, 2, b, Thus, in each case, we have: h = b sin C, … (2), From (1) and (2), we get:, b, c, c sin B = b sin C ⇒, =, ⋅, sin B sin C, b, a, Similarly,, =, ⋅, sin B, sin A, a, b, c, Hence,, =, =, ⋅, sin A, sin B, sin C, (Cosine Formulae) Let a, b and c be the lengths of sides of C ABC,, opposite to ∠A , ∠B and ∠C respectively. Then,, (i) a 2 = b 2 + c2 − 2bc cos A, (ii) b 2 = c2 + a 2 − 2ca cos B, (iii) c2 = a 2 + b 2 − 2 ab cos C, Let CABC be given, as shown in Fig. (i) and Fig. (ii)., , THEOREM 2, , PROOF, , B, B, , A, , c, , a, , c, , b, , D, , (i), , C, , A, , a, , C, , b, , D, , ( ii ), , Solution of Triangles, , From Fig. (i), we get, BC 2 = BD 2 + DC 2, = BD 2 + ( AC − AD) 2, = BD 2 + AD 2 + AC 2 − 2AC ⋅ AD, = AB 2 + AC 2 − 2AC ⋅ AB cos A, [B BD 2 + AD 2 = AB 2 and AD = AB cos A], 2, 2, = AC + AB − 2AC ⋅ AB cos A, ⇒, a 2 = b 2 + c2 − 2bc cos A.
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616, , Senior Secondary School Mathematics for Class 11, , From Fig. (ii), we get, BC 2 = BD 2 + CD 2, = BD 2 + ( AD − AC) 2, = BD 2 + AD 2 + AC 2 − 2AC ⋅ AD, = AB 2 + AC 2 − 2AC ⋅ AB cos A, [B BD 2 + AD 2 = AB 2 and AD = AB cos A], 2, 2, 2, ⇒ a = b + c − 2bc cos A., Hence, from both the cases, we get, a 2 = b 2 + c2 − 2bc cos A., Similarly, b 2 = c2 + a 2 − 2ca cos B and c2 = a 2 + b 2 − 2ab cos C., REMARKS, , A convenient method writing cosine formulae is given below:, ( b 2 + c2 − a 2 ), (i) cos A =, 2bc, ( c2 + a 2 − b 2 ), (ii) cos B =, 2ca, ( a 2 + b 2 − c2 ), (iii) cos C =, 2ab, SUMMARY, , 1. Sine Formula, a, b, c, =, =, sin A sin B sin C, 2. Cosine Formulae, (i) a2 = b 2 + c 2 − 2bc cos A, (ii) b 2 = c 2 + a2 − 2ca cos B, (iii) c 2 = a2 + b 2 − 2ab cos C, THEOREM 3, , (Napier’s Analogies) In any CABC, prove that, , ( B − C) ( b − c), A, =, cot, 2, ( b + c), 2, (C − A) ( c − a), B, (ii) tan, =, cot, 2, ( c + a), 2, ( A − B) ( a − b), C, (iii) tan, =, cot, 2, ( a + b), 2, (i) tan, , PROOF, , By the sine formula, we have, a, b, c, =, =, = k (say)., sin A sin B sin C, Then, a = k sin A , b = k sin B and c = k sin C., ( b − c) k sin B − k sin C, (i)∴, =, ( b + c) k sin B + k sin C
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Solution of Triangles, , ⇒, , 617, , ( B + C), ( B − C), 2 cos, sin, (sin B − sin C), 2, 2, =, =, (sin B + sin C) 2 sin ( B + C) cos ( B − C), 2, 2, ( B + C), ( B − C), = cot, tan, 2, 2, ( B − C), ⎡ A ( B + C) π ⎤, ⎛ π A⎞, = ⎥, = cot ⎜ − ⎟ tan, ⎢⎣Q 2 +, ⎝ 2 2⎠, 2, 2⎦, 2, A, ( B − C), = tan, tan, 2, 2, ( B − C) ( b − c), A, tan, =, cot ⋅, 2, ( b + c), 2, , Similarly, (ii) and (iii) may be proved., , SOLVED EXAMPLES, EXAMPLE 1, , In any CABC, prove that, a (sin B − sin C) + b (sin C − sin A) + c(sin A − sin B) = 0., By the sine rule, we have, a, b, c, =, =, = k (say), sin A sin B sin C, , SOLUTION, , ⇒, ∴, , EXAMPLE 2, , SOLUTION, , a = k sin A , b = k sin B and c = k sin C., LHS = a (sin B − sin C) + b (sin C − sin A) + c(sin A − sin B), , = k sin A(sin B − sin C) + k sin B(sin C − sin A), + k sin C(sin A − sin B), = k[(sin A sin B − sin A sin C) + (sin B sin C − sin A sin B), + (sin A sin C − sin B sin C)], = ( k × 0) = 0 = RHS., , In any CABC, prove that, a sin ( B − C) + b sin (C − A) + c sin ( A − B) = 0., By the sine rule, we have, a, b, c, =, =, sin A sin B sin C, ⇒, ⇒, , sin A sin B sin C, =, =, = k (say), a, b, c, sin A = ka , sin B = kb and sin C = kc.
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618, , Senior Secondary School Mathematics for Class 11, , Substituting the values of sin B and sin C and using cosine formulae,, we have, a sin ( B − C) = a(sin B cos C − cos B sin C), ⎡, ( a 2 + b 2 − c2 ), ( c2 + a 2 − b 2 ) ⎤, = a ⎢kb ⋅, − kc ⋅, ⎥, 2ab, 2ac, ⎣, ⎦, k, = ⋅ [( a 2 + b 2 − c2 ) − ( c2 + a 2 − b 2 )] = k( b 2 − c2 )]., 2, Similarly, b sin(C − A) = b(sin C cos A − cos C sin A) = k( c2 − a 2 )., And, c sin( A − B) = c(sin A cos B − cos A sin B) = k( a 2 − b 2 )., ∴ a sin( B − C) + b sin (C − A) + c sin ( A − B), = k( b 2 − c2 ) + k( c2 − a 2 ) + k( a 2 − b 2 ), = k( b 2 − c2 + c2 − a 2 + a 2 − b 2 ) = k × 0 = 0., In any CABC, prove that, sin ( B − C) ( b 2 − c2 ), =, ⋅, sin ( B + C), a2, , EXAMPLE 3, , By the sine rule, we have, a, b, c, =, =, = k (say), sin A sin B sin C, ⇒ a = k sin A , b = k sin B and c = k sin C., , SOLUTION, , ∴, , RHS =, , ( b 2 − c2 ) k 2 sin 2 B − k 2 sin 2C, =, a2, k 2 sin 2 A, , (sin 2 B − sin 2C), sin ( B + C) ⋅ sin ( B − C), =, 2, sin A, sin 2 ( B + C), ⎡B ( A + B + C) = π ⇒ A = π − ( B + C) ⎤, ⎢∴ sin A = sin{π − ( B + C)} = sin( B + C) ⎥, ⎣, ⎦, sin ( B − C), =, = LHS., sin ( B + C), , =, , ∴, , RHS = LHS., , Hence,, EXAMPLE 4, , SOLUTION, , sin ( B − C) ( b 2 − c2 ), =, ⋅, sin ( B + C), a2, , In any C ABC, prove that, ( a − b), C, ( A − B), cos = sin, ⋅, c, 2, 2, By the sine rule, we have, a, b, c, =, =, = k (say), sin A, sin B, sin C, ⇒ a = k sin A , b = k sin B and c = k sin C.
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Solution of Triangles, , 619, , ( a − b), C, cos, c, 2, ( k sin A − k sin B), C k(sin A − sin B), C, =, cos =, ⋅ cos, k sin C, 2, k sin C, 2, ( A + B), ( A − B), 2 cos, sin, C, (sin A − sin B), C, 2, 2, ⋅ cos, =, ⋅ cos =, C, C, 2, 2, sin C, 2 sin cos, 2, 2, (C + D), (C − D) ⎤, ⎡, sin, ⎢⎣B (sin C − sin D) = 2 cos, 2, 2 ⎥⎦, , ∴, , LHS =, , ( A − B), ⎛ π C⎞, cos ⎜ − ⎟ ⋅ sin, ⎝ 2 2⎠, ⎡ A + B ⎛ π C⎞ ⎤, 2, =, = ⎜ − ⎟⎥, ⎢B, C, ⎝ 2 2⎠ ⎦, 2, ⎣, sin, 2, C, ( A − B), sin ⋅ sin, ( A − B), 2, 2, =, = sin, = RHS., C, 2, sin, 2, ( a − b), C, ( A − B), Hence,, cos = sin, ⋅, c, 2, 2, EXAMPLE 5, , SOLUTION, , In any CABC, prove that, ( b 2 − c2 ), ( c2 − a 2 ), ( a 2 − b2 ), sin 2A +, sin 2B +, sin 2C = 0., 2, 2, a, b, c2, By the sine rule, we have, a, b, c, =, =, = k (say), sin A sin B sin C, ⇒, , a = k sin A , b = k sin B and c = k sin C., , Applying sine rule and cosine formula, we get:, ( b 2 − c2 ), ( b 2 − c2 ), 2, =, sin, A, a2, a2, 2, ( b − c2 ), =, a2, , ⋅ ( 2 sin A cos A), 2, 2, 2, ⎛ 2a ⎞ ( b + c − a ), ⎜ ⎟⋅, ⎝ k⎠, 2bc, , ⎡, a, ( b 2 + c2 − a 2 ) ⎤, ⎢B sin A = and cos A =, ⎥, k, 2bc, ⎣, ⎦, 1, ⋅ ( b 2 − c2 )( b 2 + c2 − a 2 ), ( kabc), 1, ( c2 − a 2 ), Similarly,, ⋅ ( c2 − a 2 ) ( c2 + a 2 − b 2 ), sin 2B =, ( kabc), b2, =, , … (i), … (ii)
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620, , Senior Secondary School Mathematics for Class 11, , 1, ( a 2 − b2 ), ⋅ ( a 2 − b 2 ) ( a 2 + b 2 − c 2 )., sin 2C =, ( kabc), c2, From (i), (ii) and (iii), we get, And,, , … (iii), , ( b 2 − c2 ), ( c2 − a 2 ), ( a 2 − b2 ), sin 2A +, sin 2B +, sin 2C, 2, a, b2, c2, 1, =, ⋅ [( b 2 − c2 )( b 2 + c2 − a 2 ) + ( c2 − a 2 )( c2 + a 2 − b 2 ), ( kabc), + ( a 2 − b 2 )( a 2 + b 2 − c2 )], 1, =, × 0 = 0 = RHS., ( kabc), ( b 2 − c2 ), ( c2 − a 2 ), ( a 2 − b2 ), Hence,, sin 2A +, sin 2B +, sin 2C = 0., 2, 2, a, b, c2, LHS =, , EXAMPLE 6, , SOLUTION, , In any CABC, prove that, , 1, tan ( B − C ), ( b − c), 2, ⋅, =, ( b + c) tan 1 ( B + C ), 2, By the sine rule, we have, a, b, c, =, =, = k (say), sin A sin B sin C, ⇒ a = k sin A , b = k sin B and c = k sin C., ( b − c) k sin B − k sin C k(sin B − sin C), =, =, ∴ LHS =, ( b + c) k sin B + k sin C k(sin B + sin C), ( B + C), ( B − C), 2 cos, sin, (sin B − sin C), 2, 2, =, =, (sin B + sin C) 2 sin ( B + C) cos ( B − C), 2, 2, 1, tan ( B − C), 2, =, = RHS., 1, tan ( B + C), 2, 1, tan ( B − C), ( b − c), 2, Hence,, =, ⋅, ( b + c) tan 1 ( B + C), 2, , EXAMPLE 7, , In any C ABC, prove that, a(cos C − cos B) = 2( b − c) cos2, , SOLUTION, , A, ⋅, 2, , Applying cosine formulae, we have, LHS = a(cos C − cos B), , ⎡( a 2 + b 2 − c2 ) ( a 2 + c2 − b 2 ) ⎤ ( a 2 c + b 2 c − c3 − a 2 b − bc2 + b 3 ), = a⎢, −, ⎥=, 2ab, 2ac, 2bc, ⎣, ⎦
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Solution of Triangles, , 621, , ( b 3 − c3 ) + ( b 2 c − bc2 ) − ( a 2 b − a 2 c) ( b 3 − c3 ) + bc( b − c) − a 2 ( b − c), =, 2bc, 2bc, ⎧( b 2 + c2 − a 2 ) 2bc ⎫, [( b 2 + c2 + bc) + bc − a 2 ], = ( b − c), = ( b − c) ⋅ ⎨, +, ⎬, 2bc ⎭, 2bc, 2bc, ⎩, =, , ⎧( b 2 + c2 − a 2 ), ⎫, = ( b − c) ⎨, + 1⎬ = ( b − c)(1 + cos A), 2, bc, ⎭, ⎩, 2 A, = 2( b − c) cos, = RHS., 2, A, ∴ a(cos C − cos B ) = 2( b − c) cos2 ⋅, 2, EXAMPLE 8, SOLUTION, , In any CABC, prove that, a cos A + b cos B + c cos C = 2a sin B sin C., By the sine rule, we have, a, b, c, =, =, = k (say), sin A, sin B, sin C, ⇒ a = k sin A , b = k sin B and c = k sin C., ∴ LHS = a cos A + b cos B + c cos C, = k sin A cos A + k sin B cos B + k sin C cos C, 1, = k(sin 2A + sin 2B + sin 2C), 2, 1, = k[2 sin ( A + B) cos ( A − B) + 2 sin C cos C], 2, = k[sin ( A + B) cos ( A − B) + sin C cos C], = k[sin ( π − C) cos ( A − B) + sin C cos C], = k sin C[cos ( A − B) + cos C], = k sin C[cos ( A − B) + cos {π − ( A + B)}], = k sin C[cos ( A − B) − cos ( A + B)], = k sin C × 2 sin A sin B, = 2( k sin A) sin B sin C, = 2a sin B sin C = RHS., Hence, a cos A + b cos B + c cos C = 2a sin B sin C., , EXAMPLE 9, , SOLUTION, , In any CABC, prove that, ( b 2 − c2 ) cot A + ( c2 − a 2 ) cot B + ( a 2 − b 2 ) cot C = 0., , By the sine rule, we have, a, b, c, =, =, sin A sin B sin C, sin A sin B sin C, ⇒, =, =, = k (say), a, b, c, ⇒ sin A = ka , sin B = kb and sin C = kc., ( b 2 − c2 ) ⋅ cos A ( b 2 − c2 ) ( b 2 + c2 − a 2 ), ∴ ( b 2 − c2 ) cot A =, =, ⋅, sin A, ka, 2bc
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622, , Senior Secondary School Mathematics for Class 11, , =, , ( b 4 − c4 − a 2 b 2 + a 2 c2 ), 1, ⋅, ⋅ ( b 2 − c2 )( b 2 + c2 − a 2 ) =, ( 2kabc), 2kabc, , ( c4 − a 4 − b 2 c2 + a 2 b 2 ), ⋅, 2kabc, ( a 4 − b 4 − a 2 c2 + b 2 c2 ), And, ( a 2 − b 2 ) cot C =, ⋅, 2kabc, ∴ LHS = ( b 2 − c2 ) cot A + ( c2 − a 2 ) cot B + ( a 2 − b 2 ) cot C, 1, =, ⋅ [( b 4 − c4 − a 2 b 2 + a 2 c2 ) + ( c4 − a 4 − b 2 c2 + a 2 b 2 ), 2kabc, + ( a 4 − b 4 − a 2 c2 + b 2 c2 )], 1, =, × 0 = 0 = RHS., 2kabc, Hence, ( b 2 − c2 ) cot A + ( c2 − a 2 ) cot B + ( a 2 − b 2 ) cot C = 0., Similarly, ( c2 − a 2 ) cot B =, , EXAMPLE 10, , SOLUTION, , In any CABC, prove that, a 3 sin( B − C) + b 3 sin(C − A) + c3 sin( A − B) = 0., , By the sine rule, we have, a, b, c, =, =, = k (say), sin A sin B sin C, ⇒, ∴, , a = k sin A , b = k sin B and c = k sin C., a 3 sin( B − C) = k 3 sin 3 A ⋅ sin( B − C), = k 3 sin 2 A ⋅ sin A ⋅ sin( B − C), = k 3 sin 2 A[sin{π − ( B + C)} sin( B − C)], [B A + ( B + C) = π ], = k 3 sin 2 A[sin( B + C) sin( B − C)], = k 3 sin 2 A ⋅ (sin 2 B − sin 2C)., , Similarly, b sin(C − A) = k 3 sin 2 B(sin 2C − sin 2 A)., 3, , And, c3 sin( A − B) = k 3 sin 2C(sin 2 A − sin 2 B)., ∴, , a 3 sin( B − C) + b 3 sin(C − A) + c3 sin( A − B), = k 3 sin 2 A(sin 2 B − sin 2C) + k 3 sin 2 B(sin 2C − sin 2 A), + k 3 sin 2C(sin 2 A − sin 2 B), = 0., , EXAMPLE 11, , SOLUTION, , In any C ABC, prove that, C, C, ( a − b) 2 cos2 + ( a + b) 2 sin 2 = c2 ., 2, 2, , We have, LHS = ( a − b) cos, 2, , 2, , C, C, + ( a + b) 2 sin 2, 2, 2
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Solution of Triangles, , 623, , C, C⎞, C, C⎞, ⎛, ⎛, = a 2 ⎜ cos2 + sin 2 ⎟ + b 2 ⎜ cos2 + sin 2 ⎟, ⎝, ⎝, 2, 2⎠, 2, 2⎠, − 2ab cos2, , C, C, + 2ab sin 2, 2, 2, , C, C⎞, ⎛, = a 2 + b 2 − 2ab ⎜ cos2 − sin 2 ⎟, ⎝, 2, 2⎠, 2, 2, 2, [by cosine formula], = a + b − 2ab cos C = c = RHS., 2, 2C, 2, 2C, 2, + ( a + b) sin, =c ., ∴ ( a − b) cos, 2, 2, EXAMPLE 12 In a CABC, prove that, A, B, C, ( b − c) cot, + ( c − a) cot + ( a − b) cot = 0., 2, 2, 2, SOLUTION By the sine rule, we have, a, b, c, =, =, = k (say), sin A sin B sin C, ⇒ a = k sin A , b = k sin B and c = k sin C., A, A, = k(sin B − sin C) cot, ∴ ( b − c) cot, 2, 2, ⎛ A⎞, cos ⎜ ⎟, ⎝ 2⎠, ( B + C), ( B − C), = 2k cos, ⋅, sin, 2, 2, ⎛ A⎞, sin ⎜ ⎟, ⎝ 2⎠, ⎛ A⎞, cos ⎜ ⎟, ⎝ 2⎠, −, A, (, B, C, ), π, ⎛, ⎞, = 2k cos ⎜ − ⎟ sin, ⋅, ⎝ 2 2⎠, 2, ⎛ A⎞, sin ⎜ ⎟, ⎝ 2⎠, A, ⎛ ⎞, cos ⎜ ⎟, ⎝ 2⎠, A, ( B − C), = 2k sin, ⋅, sin, 2, 2, ⎛ A⎞, sin ⎜ ⎟, ⎝ 2⎠, ( B − C), A, = 2k sin, ⋅ cos, 2, 2, ( B − C), ⎧ π ( B + C) ⎫, = 2k sin, ⋅ cos ⎨ −, ⎬, 2, 2 ⎭, ⎩2, ( B − C), ( B + C), = 2k sin, ⋅ sin, 2, 2, = k(cos C − cos B)., Similarly, we have, B, C, ( c − a) cot = k(cos A − cos C) and ( a − b) cot = k(cos B − cos A)., 2, 2, A, B, C, ∴ ( b − c) cot, + ( c − a) cot + ( a − b) cot, 2, 2, 2, = k ⋅ [(cos C − cos B) + (cos A − cos C) + (cos B − cos A)] = 0.
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624, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 13, , SOLUTION, , In a CABC, if cos A =, , sin B, , show that the triangle is isosceles., 2 sin C, , By the sine rule, we have, a, b, c, =, =, sin A, sin B, sin C, sin A sin B sin C, =, =, = k (say), ⇒, a, b, c, ⇒, ∴, , sin A = ka , sin B = kb and sin C = kc., sin B, ( b 2 + c2 − a 2 ), kb, ⇒, =, cos A =, 2 sin C, 2bc, 2kc, ⇒ ( b 2 + c2 − a 2 ) = b 2, ⇒ c2 = a 2 ⇒ |c| = |a|, , ⇒ AB = BC., Hence, CABC is isosceles., EXAMPLE 14, , SOLUTION, , In a CABC, if a cos A = b cos B , show that the triangle is either isosceles, or right-angled., , By the sine rule, we have, a, b, =, = k (say), sin A sin B, ⇒, ∴, ⇒, , a = k sin A and b = k sin B., a cos A = b cos B, k sin A cos A = k sin B cos B, 1, 1, ⇒, (sin 2A) = (sin 2B), 2, 2, ⇒ sin 2A = sin 2B, ⇒ sin 2A − sin 2B = 0, ⇒ 2 cos ( A + B) sin ( A − B) = 0, ⇒ cos ( A + B) = 0 or sin ( A − B) = 0, π, or ( A − B) = 0, ⇒ ( A + B) =, 2, π, or A = B., ⇒ ∠C =, 2, Hence, CABC is right-angled or isosceles., , EXAMPLE 15, , SOLUTION, , In a CABC, if a = 18, b = 24, c = 30; find, (i) sin A , sin B , sin C, (ii) cos A , cos B , cos C, , Here, a 2 + b 2 = (18) 2 + ( 24) 2 = ( 324 + 576) = 900 = ( 30) 2 = c2 ., ∴ CABC is right-angled at C. Thus, ∠C = 90°.
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Solution of Triangles, , 625, , (i) By sine rule, we have, a, b, c, =, =, sin A sin B sin C, sin A sin B sin C, =, =, = k (say), ⇒, a, b, c, sin A sin B sin C, =, =, =k, ⇒, 18, 24, 30, ∴ sin A = 18k , sin B = 24k and sin C = 30k., , 1, ⋅, 30, 1⎞ 3, 1⎞ 4, ⎛, ⎛, ∴ sin A = 18k = ⎜18 ×, ⎟ = ; sin B = 24k = ⎜ 24 ×, ⎟ =, ⎝, ⎝, 30⎠ 5, 30⎠ 5, 1⎞, ⎛, and sin C = 30k = ⎜ 30 ×, ⎟ = 1., ⎝, 30⎠, 3, 4, Hence, sin A = , sin B = and sin C = 1., 5, 5, (ii) Using cosine formulae, we get, b 2 + c2 − a 2 ( 24) 2 + ( 30) 2 − (18) 2, cos A =, =, 2bc, 2 × 24 × 30, 576 + 900 − 324 1152 4, =, =, = ⋅, 1440, 1440 5, c2 + a 2 − b 2 ( 30) 2 + (18) 2 − ( 24) 2, cos B =, =, 2ca, 2 × 30 × 18, ( 900 + 324 − 576), 648, 3, =, =, = ⋅, 1080, 1080 5, cos C = cos 90° = 0., 4, 3, Hence, cos A = , cos B = and cos C = 0., 5, 5, But, ∠C = 90° ⇒ sin C = sin 90° = 1 ⇒ 30k = 1 ⇒ k =, , EXAMPLE 16, , SOLUTION, , By the sine formula, we have, a : b : c = sin A : sin B : sin C, = sin 45 ° : sin 60°: sin 75 °, 1, 3 ( 3 + 1), =, :, :, = 2 : 6 : ( 3 + 1)., 2, 2, 2 2, ⎛ 2⎞, In a CABC, if a = 2, b = 3 and sin A = ⎜ ⎟ , find ∠B., ⎝ 3⎠, a, b, 2, 3, =, ⇒, =, sin B, ( 2/ 3) sin B, sin A, 2 1⎞, ⎛, ⇒ sin B = ⎜ 3 × × ⎟ = 1, ⎝, 3 2⎠, ⇒ ∠B = 90°., , EXAMPLE 17, SOLUTION, , In a CABC, if ∠A = 45 ° , ∠B = 60° and ∠C = 75 ° , find the ratio of its, sides.
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626, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 18, , The angles of a triangle ABC are in AP and if it is being given that, b:c=, , SOLUTION, , 3 : 2 , find ∠A , ∠B and ∠C., , ∠A , ∠B , ∠C are in AP, ⇒ 2∠B = ∠A + ∠C, ⇒ 3 ∠B = ∠A + ∠B + ∠C = 180°, ⇒ ∠B = 60°., b, c, b sin B, Now,, =, ⇒, =, sin B sin C, c sin C, 3 sin 60°, ⇒, =, sin C, 2, ⎛ 3, ⇒ sin C = ⎜, ×, ⎝ 2, , 2×, , 1 ⎞, 1, ⎟ =, 3⎠, 2, , ⇒ ∠C = 45 °., ∴ ∠A = 180° − ( ∠B + ∠C) = [180° − ( 60° + 45 °)] = 75 °., Hence, ∠A = 75 ° , ∠B = 60° and ∠C = 45 °., EXAMPLE 19, SOLUTION, , 3 , find ∠B., , Let b = 2k and c = 3 k. Then,, b 2 + c2 − a 2, cos A =, 2bc, 4k 2 + 3 k 2 − a 2, ⇒ cos 30° =, 4 3 k2, 3, ⇒, × 4 3 k2 = 7 k2 − a2, 2, ⇒ k 2 = a 2 ⇒ k = a., Thus, b = 2a and c = 3 a., a, b, a, 2a, Now,, =, ⇒, =, sin A sin B, sin 30° sin B, 1 1⎞, ⎛, ⇒ sin B = ⎜ 2a × × ⎟ = 1, ⎝, 2 a⎠, ⇒ B = 90°., , EXAMPLE 20, SOLUTION, , In a CABC, if ∠A = 30° and b : c = 2 :, , Solve the triangle in which a = ( 3 + 1), b = ( 3 − 1) and ∠C = 60°., , ( A − B) ( a − c), C, =, cot , we get, 2, ( a + b), 2, ( A − B), 2, ⎛ 1, ⎞, tan, =, cot 30° = ⎜, × 3⎟ = 1, ⎝ 3, ⎠, 2, 2 3, A−B, ⇒, = 45 ° ⇒ A − B = 90°, 2, Also, A + B + C = 180° ⇒ A + B = 120°, On solving (i) and (ii), we get A = 105 ° and B = 15 °., , Using tan, , … (i), … (ii)
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Solution of Triangles, , 627, , a 2 + b 2 − c2, 2ab, ( 3 + 1) 2 + ( 3 − 1) 2 − c2 8 − c2, =, ⇒ cos 60° =, 4, 2( 3 + 1)( 3 − 1), 8 − c2 1, ∴, =, ⇒ 8 − c2 = 2 ⇒ c2 = 6 ⇒ c = 6., 4, 2, Hence, c = 6 cm , ∠A = 105 ° and ∠B = 15 °., Also, cos C =, , EXERCISE 18A, In any CABC, prove that, , 1. a( b cos C − c cos B) = ( b 2 − c2 ), 3., , 2. ac cos B − bc cos A = ( a 2 − b 2 ), , cos A cos B cos C ( a 2 + b 2 + c2 ), c − b cos A cos B, 4., +, +, =, =, 2abc, a, b, c, b − c cos A cos C, , 5. 2( bc cos A + ca cos B + ab cos C) = ( a 2 + b 2 + c2 ), A, B, C⎞, ⎛, 6. 4 ⎜ bc cos2 + ca cos2 + ab cos2 ⎟ = ( a + b + c) 2, ⎝, 2, 2, 2⎠, 7. a sin A − b sin B = c sin ( A − B), 9., 11., , sin( A − B) ( a 2 − b 2 ), =, sin( A + B), c2, ( a + b), C, ( A − B), sin, = cos, c, 2, 2, , 8. a 2 sin ( B − C) = ( b 2 − c2 ) sin A, 10., , ( b − c), A, ( B − C), cos = sin, a, 2, 2, , 12., , ( b + c), ( B + C), ( B − C), ⋅ cos, = cos, a, 2, 2, , 13. a 2 (cos2 B − cos2C) + b 2 (cos2C − cos2 A) + c2 (cos2 A − cos2 B) = 0, 14., , (cos2 B − cos2C) (cos2C − cos2 A) (cos2 A − cos2 B), +, +, =0, b+c, c+ a, a+b, , 15., , cos 2A cos 2B ⎛ 1, 1⎞, −, = ⎜ 2 − 2⎟, ⎝a, a2, b2, b ⎠, , 16. ( c2 − a 2 + b 2 ) tan A = ( a 2 − b 2 + c2 ) tan B = ( b 2 − c2 + a 2 ) tan C, 17. If in a CABC, ∠C = 90°, then prove that sin ( A − B) =, 18. In a CABC, if, , ( a 2 − b2 ), ⋅, ( a 2 + b2 ), , cos A cos B, =, , show that the triangle is isosceles., a, b, , 19. In aCABC, if sin 2 A + sin 2 B = sin 2C , show that the triangle is right-angled., 20. Solve the triangle in which a = 2 cm, b = 1 cm and c =, , 3 cm., , 21. In a CABC, if a = 3 cm , b = 5 cm and c = 7 cm , find cos A , cos B , cos C
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628, , Senior Secondary School Mathematics for Class 11, , 22. If the angles of a triangle are in the ratio 1 : 2 : 3, prove that its, corresponding sides are in the ratio 1 : 3 : 2., ANSWERS (EXERCISE 18A), , 20. ∠A = 90° , ∠B = 30° and ∠C = 60°, 13, 11, −1, and cos C =, 21. cos A =, , cos B =, 14, 14, 2, HINTS TO SOME SELECTED QUESTIONS, 6. LHS = 2 [bc( 1 + cos A ) + ca( 1 + cos B) + ab( 1 + cos C )], = 2[( bc + ca + ab ) + bc ⋅ cos A + ca ⋅ cos B + ab ⋅ cos C ], ⎡, ( b 2 + c 2 − a2 ), ( c 2 + a2 − b 2 ), ( a2 + b 2 − c 2 ) ⎤, = 2 ⎢( ab + bc + ca) + bc ⋅, + ca ⋅, + ab ⋅, ⎥, 2 ab, 2 bc, 2 ca, ⎢⎣, ⎥⎦, = a2 + b 2 + c 2 + 2( ab + bc + ca) = ( a + b + c ) 2 = RHS., a, b, c, 7., =, =, = k (say), sin A, sin B, sin C, ⇒ a = k sin A , b = k sin B and c = k sin C ., ∴ LHS = k sin 2 A − k sin 2 B, = k(sin 2 A − sin 2 B ) = k sin ( A + B ) ⋅ sin ( A − B ), = k sin ( π − C ) sin ( A − B) = k sin C sin ( A − B), = c sin ( A − B) = RHS., 8. LHS = k 2 sin 2 A[sin ( B − C )], = k 2 sin A sin [π − ( B + C )] sin ( B − C ), = k 2 sin A sin ( B + C ) sin ( B − C ), = k 2 sin A ⋅ (sin 2 B − sin 2C ) = RHS., 9. See Example 3., For questions 10, 11, 12 see Example 4., 13. a2 (cos 2 B − cos 2C ) = a2 [( 1 − sin 2 B) − ( 1 − sin 2C )] = a2 (sin 2C − sin 2 B), ⎛ c2 b 2 ⎞, 1 2 2, 2, = a2 ⎜, −, ⎟ = 2 ⋅ a ( c − b )., 2, 2, k ⎠ k, ⎝k, 14., , cos 2 B − cos 2C ( 1 − sin 2 B) − ( 1 − sin 2C ), =, ( b + c), ( b + c), =, , 15. LHS =, , (sin 2C − sin 2 B ) ( c 2 − b 2 ), 1, =, =, ( c − b )., ( b + c), k 2 ( b + c) k 2, , ( 1 − 2 sin 2 A ), 2, , −, , ( 1 − 2 sin 2 B), 2, , 1, 1 ⎞ 2 sin 2 A 2 sin 2 B, +, = ⎛⎜, −, ⎟−, 2, ⎝a, b2 ⎠, a2, b2, 2, 2 sin B, , a, b, 2 sin 2 A, 1, 1⎞, ⎛, +, =⎜, −, ⎟−, ⎝ a2 b 2 ⎠ k 2 sin 2 A k 2 sin 2 B, 1, 1⎞, 2, 2 ⎛ 1, 1⎞, = ⎛⎜, −, +, =⎜, −, ⎟−, ⎟⋅, ⎝ a2 b 2 ⎠ k 2 k 2 ⎝ a2 b 2 ⎠
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Solution of Triangles, 17. RHS =, =, , 629, , ( a2 − b 2 ), ( a2 + b 2 ), k 2 sin 2 A − k 2 sin 2 B, 2, , 2, , 2, , 2, , k sin A + k sin B, , =, , (sin 2 A − sin 2 B ), 2, , sin A + sin 2 ( 90 ° − A ), , = (sin 2 A − sin 2 B ), , = sin ( A + B) sin ( A − B) = (sin 90 ° ) sin ( A − B) = sin ( A − B)., cos A cos B, ( b 2 + c 2 − a2 ) a2 + c 2 − b 2, 18., =, ⇒, =, a, b, 2 abc, 2 abc, ⇒ b 2 + c 2 − a2 = a2 + c 2 − b 2 ⇒ 2 a2 = 2 b 2, ⇒ a2 = b 2 ⇒ a = b., ⎡, ⎤, a, b, c, =, =, = k (say)⎥, ⎢B, sin, A, sin, B, sin, C, ⎣, ⎦, ⇒ a2 + b 2 = c 2 ⇒ CABC is right-angled., , 19. sin 2 A + sin 2 B = sin 2C ⇒, , a2, , k, , 2, , +, , b2, k, , 2, , =, , c2, , k2, , 22. Let ∠A = x ° , ∠B = ( 2 x )° and ∠C = ( 3 x )° . Then ,, A + B + C = 180 ° ⇒ x + 2 x + 3 x = 180 ⇒ 6 x = 180 ⇒ x = 30., ∴ ∠A = 30 ° , ∠B = 60 ° and ∠C = 90 ° ., Ratio of sides = sin A : sin B : sin C = sin 30 ° : sin 60 ° : sin 90 °, 1, 3, = :, : 1 = 1: 3 : 2., 2 2, , PROBLEMS BASED ON SINE AND COSINE FORMULAE, EXAMPLE 1, , SOLUTION, , Two ships leave a port at the same time. One goes 24 km per hour in the, direction N 45° E and the other travels 32 km per hour in the direction, S 75° E. Find the distance between the ships at the end of 3 hours., , Let A and B be the positions of the two ships at the end of 3 hours., Then, OA = ( 24 × 3) km = 72 km ., And, OB = ( 32 × 3) km = 96 km ., Let, AB = x km., We have: ∠NOA = 45 ° and ∠SOB = 75 °, ∴ ∠AOB = 180° − ( 45 ° + 75 °) = 60°., Applying cosine formula on CAOB, we get, (72) 2 + ( 96) 2 − x 2, cos 60° =, 2 × 72 × 96, 1, 2, 2, ⇒ (72) + ( 96) − x 2 = × 2 × 72 × 96, 2, ⇒ 5184 + 9216 − x 2 = 6912, ⇒ x 2 = 14400 − 6912 = 7488, ⇒ x = 7488 = 86.53 km., Hence, the distance between the ships at the end of 3 hours is 86.53 km.
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630, EXAMPLE 2, , SOLUTION, , Senior Secondary School Mathematics for Class 11, , Two trees A and B are on the same side of a river. From a point C in the, river, the distances of trees A and B are 250 m and 300 m respectively. If, ∠C = 45 °, find the distance between the trees. (Use 2 = 1.44.), Let A and B be the trees and C be a point in the river., Then, CA = 250 m, CB = 300 m and ∠ACB = 45 °., Let AB = x metres., Applying cosine formula on CACB, we get, a 2 + b 2 − c2, cos C =, 2ab, ( 300) 2 + ( 250) 2 − x 2, ⇒ cos 45 ° =, 2 × 300 × 250, 1, 90000 + 62500 − x 2, =, ⇒, 150000, 2, 152500 − x 2, 1, 2, 2, ⇒, =, ×, =, 150000, 2, 2, 2, ⇒ 152500 − x 2 = 75000 × 2 = 75000 × 1.44 = 108000, ⇒, , x 2 = 152500 − 108000 = 44500, , ⇒, , x = 44500 = 210.95 m., , EXAMPLE 3, , The angle of elevation of the top point P of the, vertical tower PQ of height h from a point A on, the ground is 45° and from a point B, the angle, of elevation is 60° , where B is a point at a, distance d from the point A, measured along the, line AB, which makes an angle 30° with AQ., Prove that d = h( 3 − 1)., , SOLUTION, , In CPAQ, we have, AQ, AQ, = cot 45 ° = 1 ⇒, = 1 ⇒ AQ = h., PQ, h, From right CAQP, we get, AP 2 = AQ 2 + PQ 2 = ( h2 + h2 ) = 2h2 ⇒ AP = 2h., In CPBH , ∠BPH = 180° − ( 60° + 90°) = 30°., In CPAQ , ∠APQ = 180° − ( 45 ° + 90°) = 45 °., ∴ ∠APB = ( ∠APQ − ∠BPH) = ( 45 ° − 30°) = 15 °., In CAPB, we have, ∠PAB = 15 ° , ∠APB = 15 ° and ∠ABP = 180° − (15 ° + 15 °) = 150°., Using sine formula on CABP, we have, AB, AP, =, sin 15 ° sin 15 °, d, 2h, =, ⇒, sin 15 ° sin 30°
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Solution of Triangles, , ⇒, , d = 2 2h ⋅ sin 15 °, ( 3 − 1), ⇒ d = 2 2h ⋅, 2 2, = ( 3 − 1) h., Hence, d = ( 3 − 1) h., EXAMPLE 4, , SOLUTION, , EXAMPLE 5, , SOLUTION, , 631, , ⎤, ⎡, ⎥, ⎢Q sin 15 ° = (sin 45 ° − 30°), ⎢, = sin 45 ° cos 30° − cos 45 ° sin 30°⎥, ⎥, ⎢, 3, 1, 1 ( 3 − 1) ⎥, 1, ⎢, −, × =, ⋅, =, ×, ⎥, ⎢, 2, 2 2, 2 2, 2, ⎦, ⎣, , A lamp post is situated at the middle point M of the side AC of a, triangular plot ABC with BC = 7 m , CA = 8 m and AB = 9 m. The, lamp post subtends an angle tan −1 3 at the point B. Determine the, height of the lamp post., , We have:, a = BC = 7 m , b = AC = 8 m and C = AB = 9 m., Let M be the midpoint of AC and let MP be the, lamp post of height h metres., 1, Then, ∠MBP = θ = tan −1 ⋅, 3, Applying cosine formula on CABC, we get, ( a 2 + b 2 − c2 ) ( 49 + 64 − 81), 32 2, … (i), cos C =, =, =, =, 2ab, 2×7 × 8, 112 7, In CBMC, we have, 1, BC = 7 m , CM = CA = 4 m., 2, ∴ BM 2 = BC 2 + CM 2 − 2BC × CM × cos C, 2⎞, ⎛, [Using (i)], = ⎜ 49 + 16 − 2 × 7 × 4 × ⎟ = 49, ⎝, 7⎠, ⇒ BM = 7 m., From right CBMP, we have, PM h, h, tan θ =, =, ⇒, = tan(tan −1 3) = 3 ⇒ h = 21 m ., BM 7, 7, Hence, the height of the lamp post is 21 m., A tree stands vertically on a hill side which makes an angle of 15° with, the horizontal. From a point on the ground 35 m down the hill from the, base of the tree, the angle of elevation of the top of tree is 60°. Find the, height of the tree., Let PAQ be the hill, AB be the tree and PCX, be the horizontal. Let P be the point of, observation. Produce BA to meet PX at C., Let AB = H metres., Then, ∠XPA = 15 ° , PA = 35 m ,, ∠CPB = 60° and ∠PCA = 90°., ∴, ∠APB = ( 60° − 15 °) = 45 °., In CPAC , ∠PAC = 180° − (15 ° + 90°) = 75 °
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632, , Senior Secondary School Mathematics for Class 11, , ∴ ∠PAB = (180° − 75) = 105 °., And, ∠PBA = 180° − ( 45 ° + 105 °) = 30°., Applying sine rule on C PAB, we get, PA, AB, 35, H, =, ⇒, =, sin ∠PBA sin ∠APB, sin 30° sin 45 °, ∴ 35 × 2 = H × 2 ⇒ H = 35 2 m., Hence, the height of the tree is 35 2 m., , EXERCISE 18B, 1. Two boats leave a port at the same time. One travels 60 km in the direction, N 50° E while the other travels 50 km in the direction S 70° E. What is the, distance between the boats?, 2. A town B is 12 km south and 18 km west of a town A. Show that the bearing, of B from A is S 56° 20′ W. Also, find the distance of B from A., [Given: tan 56°20′ = 1.5.], 3. At the foot of a mountain, the angle of, elevation of its summit is 45°. After ascending, 1 km towards the mountain up an incline of, 30°, the elevation changes to 60° (as shown in, the given figure). Find the height of the, mountain. [Given: 3 = 1.73.], , ANSWERS (EXERCISE 18B), , 1. 55.68 km, , 2. 21.63 km, , 3. 1.365 km, , HINTS TO SOME SELECTED QUESTIONS, 1. Let the original position of each boat be at O. Let A and B be their respective final, positions, as shown in the figure., Then, ∠NOA = 50 ° , ∠SOB = 70 ° ,OA = 60 km and OB = 50 km, ∴ ∠AOB = 180 ° − (50 ° + 70 ° ) = 60 ° . Let AB = x km., By cosine formula on CAOB, we have, OA 2 + OB 2 − AB 2, cos ∠AOB =, 2 × OA × OB, ⇒, ⇒, , ( 60 ) 2 + (50 ) 2 − x 2, 2 × 60 × 50, 1, 3600 + 2500 − x 2 = × 2 × 60 × 50 ⇒ x 2 = 3100 ⇒ x =, 2, , cos 60 ° =, , Hence, the required distance between the boats is 55.68 km., , 3100 = 55.68 km.
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Solution of Triangles, , 633, , 2. Let B be the position of the town relative to A , as shown in the figure., From B, draw BC ⊥ AS., Then, AC = 12 km and BC = 18 km., Let ∠BAC = θ. Then,, BC 18 3, tan θ =, =, = ⇒ θ = 56 ° 20 ′ ., AC 12 2, ∴ bearing of B from A is S 56°20′ W., Also, AB = BC 2 + AC 2 = ( 18 ) 2 + ( 12 ) 2 =, , 324 + 144, , = 468 = 21.63 km., Hence, the distance of B from A is 21.63 km., SD, 3., = tan 45 ° = 1 ⇒ SD = FD = h km, FD, ⇒ ∠FSD = ∠SFD = 45 ° ., ∠ASC = 180 ° − ( 60 ° + 90 ° ) = 30 °, ⇒ ∠FSA = ( 45 ° − 30 ° ) = 15 ° and ∠SFA = 15 ° ., ∴ ∠SAF = 180 ° − ( 15 ° + 15 ° ) = 150 ° ., By the sine formula on CSAF, we have, 2h, 2h, AF, FS, 1, 2 2, =, ⇒, =, =, ⇒, = 2 2h, sin ∠FSA sin ∠SAF, sin 15 ° sin 150 ° sin 30 °, ( 3 − 1), ⎡, ( 3 − 1) ⎤, 2, 2, ⋅⎥, ⎢B FS = h + h = 2 h and sin 15 ° =, 2 2 ⎦, ⎣, ⇒, , h=, , ( 3 + 1) ( 3 + 1) (1.73 + 1) 2.73, 1, ×, =, =, =, = 1.365 km., 2, 2, 2, ( 3 − 1) ( 3 + 1), , q
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19, Graphs of Trigonometric Functions, GRAPH OF y = sin x, Graphs of Trigonometric Functions, , We know that sin x is periodic with period 2π., So, we draw the graph of sin x in the interval [0, 2π ]., We can then extend it easily by repeating it over intervals of length 2π., For this, we first construct the table given below., x, sin x, , 0, , π, , 0, , 1, , π, , 6, , 3, , 2, , π, , 3, , 2π, , 2, , 3, , 1, , 2, , 3, , = 0.5 = 0.87, , 2, , 5π, 1, , 6, , 2, , π, , 7π, , 0, , −1, , = 0.87 = 0.5, , 4π, , 6, , 3, , − 3, , 2, , 2, , 3π, , 2, , –1, , = −0.5 = − 0.87, , 5π, , 11π, , 3, , − 3, , 2, , −1, , 6, , 2π, , 2, , 0, , = − 0.87 = −0.5, , π, , On a graph paper we take 1 cm as 6 units on the x-axis, and 5 cm as 1 unit, on the y-axis. Taking the values of x along the x-axis and the values of sin x, along the y-axis, we plot the points ( 0, 0), ( π 6 , 0.5), ( π 3 , 0.87), ( π 2 ,1),, ( 2π 3 , 0.87), (5π 6 , 0.5), ( π , 0), ( 7π 6 , − 0.5), ( 4π 3 , − 0.87), ( 3π 2 , − 1), (5π 3 , − 0.87),, (11π 6 , − 0.5) and ( 2π , 0)., We join these points by a freehand curve and obtain the graph of, y = sin x , as shown below., Y 1, , O, X, , Y, , G r a p h o f y = s in x, , 0, 6, , 3, , 2, , 2, 3, , 5, 6, , 7, 6, , 1, , 634, , 4, 3, , 3, 2, , 5, 3, , 1 1, 6, , 2, , X
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Graphs of Trigonometric Functions, , 635, , GRAPH OF y = cos x, , We know that cos x is periodic with period 2π., So, we draw the graph of cos x in the interval [0, 2π ]., We can then extend it easily by repeating it over intervals of length 2π., For this, we first construct the table given below., Graphs of Trigonometric Functions, , x, , π, , 0, , cos x, , 3, , 1, , π, , 6, 2, , 1, , 3, 2, , π, , 2, , 2π, , 0, , −1, , = 0.87 = 0.5, , 3, 2, , 5π, , − 3, , 7π, , π, , 6, , 4π, , 6, , − 3, 2 –1, 2, , = −0.5 = −0.87, , On a graph paper, we take 1 cm as, unit on the y-axis., , −1, , 3π, , 3, , 5π, , 2, , 0, , 1, , 2, , = −0.87 = −0.5, π, , 3, , 2, , 11π, 3, , 6, , 2π, , 2, , 1, , = 0.5 = 0.87, , units on the x-axis, and 5 cm as 1, , 6, , Taking the values of x along the x-axis and the values of cos x along the, y-axis, we plot the points ( 0, 1), ( π 6 , 0.87), ( π 3 , 0.5), ( π 2 , 0), ( 2π 3 , − 0.5),, (5π 6 , − 0.87), ( π , − 1), ( 7π 6 , − 0.87), ( 4π 3 , − 0.5), ( 3π 2 , 0), (5π 3 , 0.5), (11π 6 , 0.87), and ( 2π , 1) on the graph paper and, joining these points successively by a, freehand curve, we obtain the graph of cos x , as shown below., Y, 1, , O, X, , Y, , G ra p h o f y = c o s x, , 0, 6, , 3, , 2, , 2, 3, , 5, 6, , 7, 6, , 4, 3, , 3, 2, , 5, 3, , 1 1, 6, , 2, , X, , 1, , GRAPH OF y = tan x, , We know that y = tan x is defined for all real values of x, except when x is an, odd multiple of π 2., Also, tan x is periodic with period π.
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636, , Senior Secondary School Mathematics for Class 11, , So, we draw the graph of y = tan x in [0, π ] and then extend it over other, intervals., As x increases from 0 to π 2 , tan x increases from 0 to ∞., As x crosses the value of π 2 , tan x becomes negative and is arbitrarily, large in magnitude., As x approaches π 2 from π , tan x comes to 0., Now, we can construct the table, given below., x, , 0, , tan x, , 0, , π, 6, 1, 3, =, 3, 3, 1.73, =, 3, = 0.58, , π, 4, , π, 3, , π, 2, , 2π, 3, , 3π, 4, , 1, , 3, , ∞, , − 3, , –1, , = 1.73, , = −1.73, , 5π, 6, −, , 1, − 3, =, 3, 3, −1.73, =, 3, = −0.58, , π, 0, , On a graph paper, we take 1 cm as π 6 units on the x-axis, and 2 cm as 1, unit on the y-axis., Taking the values of x on the x-axis and the values of tan x along the, y-axis, we plot the points (0, 0), ( π 6 , 0.58), ( π 4 , 1), ( π 3 , 1.73), ( π 2 , ∞),, ( 2π 3 , − 1.73), ( 3π 4 , −1), ( 5π 6 , − 0.58) and ( π , 0)., We join these points by a freehand curve and obtain the graph of tan x as, shown below.
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Graphs of Trigonometric Functions, , 637, , GRAPH OF y = a sin bx, , We know that −1 ≤ sin bx ≤ 1., ∴, − a ≤ a sin bx ≤ a., Also, sin x is periodic with period 2π., 2π, ⋅, ∴ sin bx is periodic with period, b, Hence, a sin bx is periodic with period, Draw the graph of y = 3 sin 2x., , EXAMPLE, SOLUTION, , 2π, ⋅, b, , −1 ≤ sin 2x ≤ 1 ⇒ − 3 ≤ 3 sin 2x ≤ 3., 2π, , i.e., π., 2, So, we draw the graph of y = 3 sin 2x in the interval [0, π ]., We thus prepare the table given below., , Also, 3 sin 2x is periodic with period, , x, , 0, , π, 6, , π, 4, , π, 3, , π, 2, , 2π, 3, , 3π, 4, , 5π, 6, , π, , 2x, , 0, , π, 3, , π, 2, , 2π, 3, , π, , 4π, 3, , 3π, 2, , 5π, 3, , 2π, , sin 2x, , 0, , –1, , –0.87, , 0, , 3 sin 2 x, , 0, , –3, , –2.61, , 0, , 3, 2, = 0.87, , 3, 2, = 0.87, , 1, , 2.61, , 3, , 2.61, , − 3, 2, = −0.87, , 0, , 0, , –2.61, , On a graph paper, we take 4 cm = π 3 units on the x-axis and 1 cm = 1 unit, on the y-axis., Now, we plot the points ( 0, 0), (π 6 , 2.61), ( π 4 , 3), ( π 3 , 2.61), ( π 2 , 0),, 2, π, ( 3 , − 2.61), ( 3π 4 , − 3), (5π 6 , − 2.61), and ( π , 0)., Join these points with a freehand curve to obtain the graph of y = 3 sin 2x ,, given below., Y, Graph of y = 3 sin 2x in [ 0, ], , 3, 2, 1, 0, 1, 2, 3, Y, , 6, , 4, , 3, , 2, , 2, 3, , 3, 4, , 5, 6, , X
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638, , Senior Secondary School Mathematics for Class 11, , GRAPH OF y = a cos bx, , We know that −1 ≤ cos bx ≤ 1., ∴, − a ≤ a cos bx ≤ a., Also, cos x is periodic with period 2π., 2π, ∴ cos bx is periodic with period, ⋅, b, Hence, a cos bx is periodic with period, Draw the graph of y = 3 cos 2x., , EXAMPLE 1, SOLUTION, , 2π, ⋅, b, , −1 ≤ cos 2x ≤ 1 ⇒ − 3 ≤ 3 cos 2x ≤ 3., , 2π, , i.e., π., 2, So, we draw the graph of y = 3 cos 2x in the interval [0, π ]., We thus prepare the table given below., , Also, 3 cos 2x is periodic with period, , x, , 0, , π, 6, , π, 4, , π, 3, , π, 2, , 2π, 3, , 3π, 4, , 5π, 6, , π, , 2x, , 0, , π, 3, , π, 2, , 2π, 3, , π, , 4π, 3, , 3π, 2, , 5π, 3, , 2π, , cos 2x, , 1, , 3 cos 2 x, , 3, , 1, 2, = 0.5, , 1, 2, = −0.5, , 0, , 1.5, , 0, , −, , –1, , –1.5, , –3, , 1, 2, = −0.5, −, , 0, , –1.5, , 0, , 1, 2, = 0.5, 1.5, , 1, , 3, , On a graph paper, we take 4 cm as π 3 units on the x-axis, and 1 cm = 1 unit, on the y-axis., Now, plot the points (0, 3), (π 6 , 1.5), ( π 4 , 0), ( π 3 , − 15, . ), ( π 2 , − 3),, 3, π, 5, π, 3 , −1.5), (, 4 , 0), ( 6 , 1.5) and ( π , 3)., Join them with a freehand curve to obtain the required graph, given, below., , ( 2π, , Y, 3, 2, 1, 0, 1, 2, 3, Y, , 6, , 4, , 3, , 2, , 2, 3, , 3 5, 4 6, , X
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Graphs of Trigonometric Functions, EXAMPLE 2, , SOLUTION, , 639, , 1, Draw the graphs of y = cos x and y = cos x in [0, 2π ] on the same, 2, scale., We know that, y = cos x is a periodic function with period 2π and amplitude 1, 1, y = cos x is a periodic function with period 4π and amplitude 1., 2, We may prepare the table as under:, , x, , 0, , cos x, 1, x, 2, 1, cos x, 2, , 1, 0, 1, , π, 6, 0.87, π, 12, , π, 3, 0.5, π, 6, , 0.97 0.87, , π, 2, 0, π, 4, 0.7, , 2π 5π, 3, 6, –0.5 –0.87, π, 5π, 3, 12, , –1, π, 2, , 7π, 4π, 6, 3, –0.87 –0.5, 7π, 2π, 12, 3, , 0.5, , 0, , –0.26 –0.5 –0.7 –0.87 –0.97, , 0.26, , π, , 3π, 2, 0, 3π, 4, , 5π, 3, 0.5, 5π, 6, , 11π, 6, 0.87, 11π, 12, , 2π, 1, π, –1, , On a graph paper, we take 1 cm = π 6 units on the x-axis, and 0.5 cm = 0.1, unit on the y-axis., For y = cos x , we plot the points (0, 1), ( π 6 , 0.87), ( π 3 , 0.5), ( π 2 , 0),, 5π, 7π, 4 π , −0.5), ( 3 π , 0), (5 π , 0.5),, 3 , −0.5), ( 6 , − 0.87), ( π , − 1), ( 6 , − 0.87), (, 3, 2, 3, 11, π, ( 6 , 0.87) and ( 2π , 1)., We join these points by a freehand curve to obtain the curve y = cos x., 1, For y = cos x , we plot the points (0, 1), ( π 6 , 0.97), ( π 3 , 0.87), ( π 2 , 0.7),, 2, ( 2π 3 , 0.5), (5π 6 , 0.26), ( π , 0), ( 7π 6 , − 0.26), ( 4π 3 , − 0.5), ( 3π 2 , − 0.7), (5π 3 , − 0.87),, ( 2π, , (11π 6 , − 0.97) and ( 2π , − 1)., 1, We join these points by a freehand curve to obtain the curve y = cos x., 2
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640, , Senior Secondary School Mathematics for Class 11, , EXERCISE 19, Draw the graph of each of the following functions:, , 1. sin 3x, , 2. 3 sin x, 3. 2 sin 3 x, x, 4. 2 cos 3 x, 5. sin, 2, 6. Draw the graphs of y = sin x and y = cos x in [0, 2π ] on the same axes., 7. Draw the graphs of y = cos x and y = cos 2x in [0, 2π ] on the same axes., q
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20, Straight Lines, INTRODUCTION, , A systematic study of geometry by using algebra was first carried out by, French mathematician R Descartes in 1637. This subject is now known as, Coordinate Geometry., In earlier classes, we have studied about coordinate axis, coordinate, plane, points in a plane, distance between two points and section formulae., Here, we give, in brief, the results derived so far., 1. Distance between the points A( x1 , y1 ) and B( x 2 , y 2 ) is given by, AB = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 ., 2. Area of a ABC whose vertices are A( x1 , y1 ), B( x 2 , y 2 ) and C( x 3 , y 3 ) is, given by, 1, a = ⋅|x1 ( y 2 − y 3 ) + x 2 ( y 3 − y1 ) + x 3 ( y1 − y 2 )| sq units., 2, REMARK The points A, B, C are collinear ⇔ area of a ABC = 0., 3. (i) If the point P( x , y) divides the join of A( x1 , y1 ) and B( x 2 , y 2 ) in the, ratio m : n then, mx 2 + nx1, my 2 + ny1, and y =, x=, ⋅, m+n, m+n, (ii) If P( x , y) is the midpoint of the line segment joining A( x1 , y1 ) and, B( x 2 , y 2 ) then, 1, 1, x = ( x1 + x 2 ) and y = ( y1 + y 2 )., 2, 2, SOLVED EXAMPLES, EXAMPLE 1, SOLUTION, , Find the distance between the points ( 2, − 3) and ( −6, 3)., Let A( 2, − 3) and B( −6, 3) be the given points. Then,, AB = ( −6 − 2) 2 + {3 − ( −3)}2 = ( −8) 2 + ( 3 + 3) 2 = ( −8) 2 + 62, = 64 + 36 = 100 = 10 units., 641
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642, EXAMPLE 2, , SOLUTION, , Senior Secondary School Mathematics for Class 11, , Using the distance formula, prove that the points A( −2, 3), B(1, 2) and, C(7 , 0) are collinear., We have, AB = (1 + 2) 2 + ( 2 − 3) 2 =, , 3 2 + ( −1) 2 = 10 units;, , BC = (7 − 1) 2 + ( 0 − 2) 2 = 62 + ( −2) 2 = 40 = 2 10 units;, AC = (7 + 2) 2 + ( 0 − 3) 2 = 92 + ( −3) 2 = 90 = 3 10 units., ∴, , AB + BC = ( 10 + 2 10) units = 3 10 units = AC., , Thus, AB + BC = AC , showing that the points A, B, C are collinear., SOME IMPORTANT RESULTS, , A quadrilateral is a, (i) rectangle, if its opposite sides are equal and the diagonals are equal;, (ii) square, if its all sides are equal and the diagonals are equal;, (iii) a parallelogram but not a rectangle, if its opposite sides are equal but the, diagonals are not equal;, (iv) a rhombus but not a square, if its all sides are equal but the diagonals are, not equal., EXAMPLE 3, , SOLUTION, , Prove that the points ( 0, 5), ( −2, − 2), (5 , 0) and (7 , 7) are the vertices of a, rhombus., Let the given points be A( 0, 5), B( −2, − 2), C(5 , 0) and D(7 , 7). Then,, ABCD is a quadrilateral in which:, AB = ( −2 − 0) 2 + ( −2 − 5) 2 = ( −2) 2 + ( −7) 2 = 4 + 49 = 53 units;, BC = (5 + 2) 2 + ( 0 + 2) 2 = 7 2 + 2 2 = 49 + 4 = 53 units;, CD = (7 − 5) 2 + (7 − 0) 2 = 2 2 + 7 2 = 4 + 49 = 53 units;, DA = (7 − 0) 2 + (7 − 5) 2 = 7 2 + 2 2 = 49 + 4 = 53 units., Thus, AB = BC = CD = DA = 53 units., Also, AC = (5 − 0) 2 + ( 0 − 5) 2 = 5 2 + ( −5) 2 = 50 = 5 2 units, And, BD = (7 + 2) 2 + (7 + 2) 2 = 9 2 + 9 2 = 81 + 81, = 162 = 9 2 units., ∴ AC ≠ BD., Thus, ABCD is a quadrilateral all of whose sides are equal but its, diagonals are not equal., Hence, ABCD is a rhombus., , EXAMPLE 4, , SOLUTION, , Find the area of the triangle whose vertices are A( 4, 4), B( 3 , − 16) and, C( 3 , − 2)., Let ( x1 = 4, y1 = 4); ( x 2 = 3 , y 2 = −16) and ( x 3 = 3 , y 3 = − 2).
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Straight Lines, , 643, , Then, area of a ABC is given by, 1, a = ⋅|x1 ( y 2 − y 3 ) + x 2 ( y 3 − y1 ) + x 3 ( y1 − y 2 )|, 2, 1, = ⋅|4( −16 + 2) + 3( −2 − 4) + 3( 4 + 16)|, 2, 1, 1, 1, = ⋅|− 56 − 18 + 60| = ⋅|− 14| = ⎛⎜ × 14⎞⎟ = 7 sq units., ⎝, ⎠, 2, 2, 2, Hence, the area of the given triangle is 7 sq units., Find the coordinates of the point which divides the line segment joining, the points A(5 , − 2) and B( 9, 6) in the ratio 3 : 1., , EXAMPLE 5, , Here ( x1 = 5 , y1 = − 2) and ( x 2 = 9, y 2 = 6). Also, m = 3 and n = 1., , SOLUTION, , Let the required point be P( x , y). Then,, (mx 2, (m, (my 2, y=, (m, x=, , + nx1 ) ( 3 × 9) + (1 × 5) 32, =, =, = 8,, + n), ( 3 + 1), 4, + ny1 ) ( 3 × 6) + {1 × ( −2)} (18 − 2) 16, =, =, =, = 4., + n), ( 3 + 1), 4, 4, , Hence, the required point is (8, 4)., EXAMPLE 6, , SOLUTION, , Find the coordinates of the midpoint of the line segment joining the, points A( −2, − 5) and B( 3 , − 1)., Let the required point be P( x , y). Then,, ( −2 + 3) 1, −5 + ( −1), = ,y=, = − 3., 2, 2, 2, 1, Hence, the required point is ⎛⎜ , − 3⎞⎟ ⋅, ⎝2, ⎠, x=, , EXAMPLE 7, , SOLUTION, , In what ratio is the line joining A( −1, 1) and B(5 , 7) divided by the line, x + y = 4?, Let the given line divide AB in the ratio m : 1., ⎛ 5m − 1 7m + 1⎞, Then, the point of division is C ⎜, ,, ⎟⋅, ⎝m+1 m+1⎠, This point C must lie on the line x + y = 4., 5m − 1 7m + 1, +, = 4 ⇔ (5m − 1) + (7m + 1) = 4(m + 1), m+1, m+1, 1, ⇔ 8m = 4 ⇔ m = ⋅, 2, 1, So, the required ratio is : 1, i.e., 1 : 2., 2, , ∴
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644, , Senior Secondary School Mathematics for Class 11, , EXERCISE 20A, 1. Find the distance between the points:, (i) A( 2, − 3) and B( −6, 3), (iii) P( −8, − 3) and Q( −2, − 5), , (ii) C( −1, − 1) and D( 8, 11), (iv) R( a + b , a − b) and S( a − b , a + b), , 2. Find the distance of the point P( 6, − 6) from the origin., 3. If a point P( x , y) is equidistant from the points A( 6, − 1) and B( 2, 3), find the, relation between x and y., 4. Find a point on the x-axis which is equidistant from the points A(7 , 6) and, B( −3 , 4)., 5. Find the distance between the points A( x1 , y1 ) and B( x 2 , y 2 ), when (i) AB is, parallel to the x-axis (ii) AB is parallel to the y-axis., 6. A is a point on the x-axis with abscissa −8 and B is a point on the y-axis with, ordinate 15. Find the distance AB., 7. Find a point on the y-axis which is equidistant from A( −4, 3) and B(5 , 2)., 8. Using the distance formula, show that the points A( 3 , − 2), B(5 , 2) and, C( 8, 8) are collinear., 9. Show that the points A(7 , 10), B( −2, 5) and C( 3 , − 4) are the vertices of an, isosceles right-angled triangle., 10. Show that the points A(1, 1), B( −1, −1) and C( − 3 , 3 ) are the vertices of, an equilateral triangle each of whose sides is 2 2 units., 11. Show that the points A( 2, − 2), B( 8, 4), C(5 , 7) and D( −1, 1) are the angular, points of a rectangle., 12. Show that A( 3 , 2), B( 0, 5), C( −3 , 2) and D( 0, −1) are the vertices of a square., 13. Show that A(1, − 2), B( 3 , 6), C(5 , 10) and D( 3 , 2) are the vertices of a, parallelogram., 14. Show that the points A( 2, − 1), B( 3 , 4), C( −2, 3) and D( −3 , − 2) are the, vertices of a rhombus., 15. If the points A( −2, − 1), B(1, 0), C( x , 3) and D(1, y) are the vertices of a, parallelogram, find the values of x and y., 16. Find the area ofa ABC whose vertices are A( −3 , − 5), B(5 , 2) andC( −9, − 3)., 17. Show that the points A( −5 , 1), B(5 , 5) and C(10, 7) are collinear., 18. Find the value of k for which the points A( −2, 3), B(1, 2) and C( k , 0) are, collinear., 19. Find the area of the quadrilateral whose vertices are A( −4, 5), B( 0, 7),, C(5 , − 5) and D( −4, − 2)., 20. Find the area of a ABC, the midpoints of whose sides AB , BC and CA are, D( 3 , − 1), E(5 , 3) and F(1, − 3) respectively., 21. Find the coordinates of the point which divides the join of A( −5 , 11) and, B( 4, − 7) in the ratio 2 : 7.
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Straight Lines, , 645, , 22. Find the ratio in which the x-axis cuts the join of the points A( 4, 5) and, B( −10, − 2). Also, find the point of intersection., 23. In what ratio is the line segment joining the points A( −4, 2) and B( 8, 3), divided by the y-axis? Also, find the point of intersection., ANSWERS (EXERCISE 20A), , 1. (i) 10 units, , (ii) 15 units, , (iii) 2 10 units, , (iv) ( 2 2b) units, , 2. 6 2 units, , 4. P( 3 , 0), 5. (i) |x 2 − x1| (ii)|y 2 − y1|, 7. P( 0, − 2), 15. x = 4, y = 2, 16. 29 sq units, 19. 60.5 sq units 20. 8 sq units, 21. P( −3 , 7), 7⎞, ⎛, 22. (5 : 2), P( −6, 0) 23. (1 : 2), P ⎜ 0, ⎟, ⎝ 3⎠, 3. x − y = 3, 6. 17 units, 18. k = 7, , HINTS TO SOME SELECTED QUESTIONS, 4. Let the required point be P( x , 0 )., Then, AP = BP ⇔ AP 2 = BP 2 ., 5. (i) When AB is parallel to the x-axis, then A and B have the same ordinate., Then, A( x1 , y ) and B( x2 , y ) are the given points. So, AB =|x2 − x1|., (ii) When AB is parallel to the y-axis, then A and B have the same abscissa., Then, A( x , y1 ) and B( x , y2 ) are the given points. So, AB =|y2 − y1|., 6. Given points are A( −8 , 0 ) and B( 0 , 15 )., 7. Let the required point be P( 0 , y )., 8. Show that AB + BC = AC ., 11. Show that AB = DC , BC = AD and diag. AC = diag. BD., 12. Show that AB = BC = CD = DA and diag. AC = diag. BD., 13. Show that AB = CD and BC = AD., 15. Midpoint of AC = Midpoint of BD., 17. Show that area of a ABC is zero., 19. Area of quad. ABCD = area (a ABC ) + area (a ACD )., 20. Let a ABC be given and let D, E, F be the midpoints of AB, BC and CA respectively., Then, ar (a ABC ) = 4 × ar (a DEF )., 22. Let the required ratio be k : 1 and let P( x , 0 ) be the point of division. Then,, k × ( −2 ) + 1 × 5, 5, =0 ⇔ k= ⋅, 2, k+1, So, the required ratio = 5 : 2., 23. Let the ratio be k : 1and let P( 0 , y ) be the point of division.
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646, , Senior Secondary School Mathematics for Class 11, , SLOPE OF A LINE, INCLINATION OF A LINE, , The angle of inclination or simply the inclination of a line is the angle θ which the part, of the line above the x-axis makes with the positive direction of the x-axis, measured in, anticlockwise direction., , Clearly, 0° ≤ θ < 180°., (i) The inclination of a line parallel to the x-axis or the x-axis itself, is 0°., (ii) The inclination of a line parallel to the y-axis or the y-axis itself, is 90°., HORIZONTAL LINE Any line parallel to the x-axis or the x-axis itself is called a, horizontal line., VERTICAL LINE Any line parallel to the y-axis or the y-axis itself is called a vertical, line., OBLIQUE LINE A line which is neither horizontal nor vertical is called a oblique line., REMARKS, , SLOPE OR GRADIENT OF A LINE, , If θ is the inclination of a non-vertical line, then m = tan θ is called the slope of the, line., EXAMPLE 1, , Find the slope of a line whose inclination is, (i) 45°, , SOLUTION, , (ii) 60°, , (iii) 150°, , Let m be the slope of the line. Then,, (i) m = tan 45 ° = 1., (ii) m = tan 60° =, , 3., , (iii) m = tan 150° = tan (180° − 30°) = − tan 30° =, REMARK 1, , −1, ⋅, 3, , The slope of a horizontal line is 0., We know that the inclination of a horizontal line is 0°., So, slope of a horizontal line is m = tan 0° = 0., , REMARK 2, , The slope of a vertical line is not defined., We know that the inclination of vertical line is 90°., So, slope of a vertical line is m = tan 90° , which is not defined.
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Straight Lines, , What is the inclination of a line whose slope is, , EXAMPLE 2, , (i) zero?, SOLUTION, , 647, , (ii) positive?, , (iii) negative?, , (iv) not defined?, , Let θ be the inclination of the given line. Then, m = tan θ., (i) m = 0 ⇒ tan θ = 0, [Q 0° ≤ θ < 180°], ⇒ θ = 0°, (ii) m > 0 ⇒ tan θ > 0, ⇒ θ lies between 0° and 90°, ⇒ θ is acute., (iii) m < 0 ⇒ tan θ < 0, ⇒ θ lies between 90° and 180°, ⇒ θ is obtuse., (iv) We know that a vertical line is the only line whose slope is not, defined. And, the inclination of a vertical line is 90°., Hence, the inclination of a line whose slope is not defined, is 90°., , SLOPE OF A LINE PASSING THROUGH TWO GIVEN POINTS, THEOREM 1, , PROOF, , Prove that the slope of a non-vertical line passing through the points, ( y − y1 ), ⋅, A( x1 , y1 ) and B( x 2 , y 2 ) is given by m = 2, ( x 2 − x1 ), , Let X′ OX and YOY′ be the coordinate axes. Let us consider a, non-vertical line CAB, passing through the points A( x1 , y1 ) and, B( x 2 , y 2 ). Let θ be its inclination., , Draw AL ⊥ OX, BM ⊥ OX and AN ⊥ BM., Then,, , ∠BAN = ∠ACX = θ, , (corr. A )., , From right triangle ANB, we have:, tan θ =, ⇒ m=, , BN ( BM − NM) ( BM − AL) ( y 2 − y1 ), =, =, =, AN, LM, (OM − OL) ( x 2 − x1 ), , ( y 2 − y1 ), ( x 2 − x1 ), , [Q tan θ = m].
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648, , Senior Secondary School Mathematics for Class 11, , Hence, the slope of the line joining the points A( x1 , y1 ) and B( x 2 , y 2 ) is, ( y − y1 ), given by, m = 2, ⋅, ( x 2 − x1 ), EXAMPLE 3, , SOLUTION, , Find the slope of the line passing through the points, (i) ( −2, 3) and ( 8, −5) (ii) ( 4, − 3) and ( 6, − 3) (iii) ( 3 , − 1) and ( 3 , 2), (i) Let A( −2, 3) and B( 8, − 5) be the given points. Then,, ⎡, ( y 2 − y1 ) ⎤, ⎢Q m = ( x − x ) ⎥, 2, 1 ⎦, ⎣, (ii) Let C( 4, − 3) and D( 6, − 3) be the given points. Then,, −3 − ( −3) −3 + 3, 0, slope of CD =, =, =, = 0., 6−4, 2, 2, slope of AB =, , ALITER, , −5 − 3, −8 −4, =, =, ⋅, 8 − ( −2) 10, 5, , The pointsC( 4, − 3) and D( 6, − 3) have the same y-coordinate., So, CD is a line parallel to the x-axis. Hence, its slope is 0., , (iii) Let P( 3 , −1) and Q( 3 , 2) be the given points. Then,, 2 − ( −1) 3, slope of PQ =, = , which is not defined., 3−3, 0, ALITER The points P( 3 , − 1) and Q( 3 , 2) have the same x-coordinate., So, PQ is a vertical line., Hence, its slope is not defined., EXAMPLE 4, , SOLUTION, , If the slope of the line passing through the points (2, 5) and ( x , 3) is 2,, find the value of x., Let A( 2, 5) and B( x , 3) be the given points. Then,, 3 −5, −2, slope of AB =, =, ⋅, x − 2 ( x − 2), −2, = 2 ⇔ 2x − 4 = − 2, ∴, ( x − 2), ⇔ 2x = 2 ⇔ x = 1., Hence, x = 1., , EXAMPLE 5, , SOLUTION, , Find the value of x so that the inclination of the line joining the points, ( x , − 3) and ( 2, 5) is 135°., Let A( x , − 3) and B( 2, 5) be the given points. Then,, 5 − ( −3), 8, slope of AB =, =, ⋅, 2−x, ( 2 − x), But, slope of AB = tan 135 ° = tan (180° − 45 °) = − tan 45 ° = − 1., 8, = −1 ⇔ x − 2 = 8 ⇔ x = 10., ∴, ( 2 − x), Hence, x = 10.
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Straight Lines, , 649, , Find the angle between the x-axis and the line joining the points ( 3 , − 1), and ( 4, − 2)., , EXAMPLE 6, , Let A( 3 , − 1) and B( 4, − 2) be the given points and let m be the slope of, the line AB. Then,, , SOLUTION, , ⎡, ( y 2 − y1 ) ⎤, ⎢Q m = ( x − x ) ⎥, 2, 1 ⎦, ⎣, Let θ be the angle between the x-axis and the line AB. Then,, m=, , −2 − ( −1) −2 + 1, =, = − 1., 4− 3, 1, , tan θ = m = −1 = − tan 45 ° = tan (180° − 45 °) = tan 135 °., ∴, , θ = 135 °., , Hence, the required angle is135°., SLOPES OF PARALLEL LINES, THEOREM 2, , PROOF, , Prove that two non-vertical lines are parallel, if and only if their slopes, are equal., , Let L1 and L2 be two non-vertical lines with slopes m1 and m2, respectively. Let their inclinations be θ1 and θ 2 respectively., First we assume that L1 and L2 are parallel., Then, L1 and L2 are parallel ⇒ θ1 = θ 2, ⇒ tan θ1 = tan θ 2, ⇒ m1 − m2 ., , Thus, non-vertical parallel lines have equal slopes., Conversely, let m1 = m2 . Then,, m1 = m2 ⇒ tan θ1 = tan θ 2, ⇒ θ1 = θ 2, , [Q 0° ≤ θ < 180°], , ⇒ L1 and L2 are parallel., Thus, non-vertical lines having equal slopes, are parallel. Hence, two, non-vertical lines are parallel if and only if their slopes are equal.
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Straight Lines, EXAMPLE 7, , SOLUTION, , Show that the line joining the points ( 2, − 3) and ( −5 , 1) is parallel to the, line joining the points (7 , − 1) and (0, 3)., Let A( 2, − 3), B( −5 , 1), C(7 , −1) and D( 0, 3) be the given points., 1 − ( −3), 4, 4, Then, slope of AB =, =, =−, −5 − 2, −7, 7, 3 − ( −1), 4, 4, And, slope of CD =, =, =−, ⋅, 0−7, −7, 7, ∴ slope of AB = slope of CD., Hence, AB || CD., Show that the line joining the points ( 2, − 5) and ( −2, 5) is perpendicular, to the line joining the points ( 6, 3) and (1, 1)., , EXAMPLE 8, , SOLUTION, , 651, , Let A( 2, − 5), B( −2, 5), C( 6, 3) and D(1, 1) be the given points., Let m1 and m2 be the slopes of AB and CD respectively. Then,, 5 − ( −5) 10, 5, =, =−, ⋅, m1 = slope of AB =, −2 − 2, −4, 2, 1 − 3 −2 2, =, =, ⋅, m2 = slope of CD =, 1 − 6 −5 5, −5, 2, ∴ m1m2 = ⎛⎜ ⎞⎟ × = −1., ⎝ 2⎠ 5, Hence, AB ⊥ CD., , EXAMPLE 9, , SOLUTION, , If the line through the points ( −2, 6) and (4, 8) is perpendicular to the line, through the points (8, 12) and ( x , 24), find the value of x., Let A( −2, 6), B( 4, 8), C( 8, 12) and D( x , 24) be the given points., Let m1 and m2 be the slopes of AB and CD respectively. Then,, ( 8 − 6), 2, 1, m1 = slope of AB =, =, =, ⋅, 4 − ( −2) 6, 3, ( 24 − 12), 12, =, ⋅, m2 = slope of CD =, ( x − 8), ( x − 8), Now, AB ⊥ CD ⇔ m1m2 = −1, 1, 12, ⇔, ×, = −1, 3 ( x − 8), ⇔ − x + 8 = 4 ⇔ x = 4., Hence, the required value of x is 4., Without using Pythagoras’s theorem show that the points (4, 4), (3, 5), and ( −1, −1) are the vertices of a right-angled triangle., Let A( 4, 4), B( 3 , 5) and C( −1, −1) be the vertices of a ABC., Let m1 and m2 be the slopes of AB and AC respectively. Then,, , EXAMPLE 10, , SOLUTION, , m1 = slope of AB =, , (5 − 4), = − 1., ( 3 − 4)
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652, , Senior Secondary School Mathematics for Class 11, , m2 = slope of AC =, ∴, , ( −1 − 4) −5, =, = 1., ( −1 − 4) −5, , m1m2 = − 1., , So, AB ⊥ AC and therefore, ∠CAB = 90°., Hence, the given points are the vertices of a right triangle., EXAMPLE 11, SOLUTION, , Using slopes, show that the points (5 , 1), (1, − 1) and (11, 4) are collinear., , Let A(5 , 1), B(1, − 1) and C(11, 4) be the given points. Then,, ( −1 − 1) −2 1, slope of AB =, =, =, (1 − 5), −4 2, 4 − ( −1), 5, 1, and slope of BC =, =, =, ⋅, 11 − 1, 10, 2, ∴ slope of AB = slope of BC, ⇒ AB || BC and have a point B in common, ⇒ A, B, C are collinear., Hence, the given points are collinear., , EXAMPLE 12, , SOLUTION, , Find the value of x for which the points ( x , − 1), ( 2, 1) and ( 4, 5) are, collinear., , Let A( x , − 1), B( 2, 1) and C( 4, 5) be the given collinear points., Then, by collinearity of A, B, C we have: slope of AB = slope of BC, 1 − ( −1) (5 − 1), 2, =, ⇒, = 2 ⇒ 2 − x = 1 ⇒ x = 1., 2−x, ( 4 − 2), 2−x, Hence, x = 1., , EXAMPLE 13, SOLUTION, , If the points ( h, 0), ( a , b) and ( 0, k) lie on a line, show that, , a, b, +, = 1., h, k, , Let A( h, 0), B( a , b) and C( 0, k) be the given collinear points., Since the given points A, B, C are collinear, we have, ∴, , slope of AB = slope of BC, b−0 k−b, b, ( b − k), =, ⇔, =, a−h 0−a, ( a − h), a, ⇔ ab = ( a − h)( b − k)
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Straight Lines, , ⇔ ak + hb = hk, a b, ⇔, + =1, h k, Hence,, EXAMPLE 14, , 653, , [on dividing both sides by hk], , a, b, +, = 1., h, k, , Using slopes, show that the vertices ( −2, − 1), ( 4, 0), ( 3 , 3) and ( −3 , 2), are the vertices of a parallelogram., , SOLUTION, , Let A( −2, − 1), B( 4, 0), C( 3 , 3) and D( −3 , 2) be the vertices of the, given quadrilateral ABCD. Then,, 0 − ( −1) 1, slope of AB =, = ;, 4 − ( −2) 6, 3−2, 1, slope of DC =, = ;, 3 − ( −3) 6, 3−0, 3, slope of BC =, =, = − 3;, 3 − 4 −1, 2 − ( −1), 3, slope of AD =, =, = − 3;, −3 − ( −2) −1, ∴ slope of AB = slope of DC ⇒ AB || DC ;, slope of BC = slope of AD ⇒ BC || AD., Hence, ABCD is a parallelogram., , EXAMPLE, , 15, , SOLUTION, , Clearly, the points P, Q, R, S are given by, 11, 17, 5 −5 ⎞, P( −1, 4), Q ⎛⎜5 , ⎞⎟ , R ⎛⎜ , −1⎞⎟ and S ⎛⎜ ,, ⎟⋅, ⎝ 2⎠, ⎝ 2, ⎠, ⎝2 2 ⎠, ⎛ 11 − 4⎞, ⎜, ⎟, ⎝ 2, ⎠, 3 1, =, = ;, ∴ slope of PQ =, (5 + 1), 12 4, ⎛ −1 + 5 ⎞, ⎜, ⎟, ⎝, 3 1, 2⎠, slope of SR =, =, = ;, ⎛ 17 − 5 ⎞ 12 4, ⎜, ⎟, ⎝ 2 2⎠, ⎛ −1 − 11⎞, ⎜, ⎟, ⎝, 2 ⎠ ⎛ −13 2⎞ −13, slope of QR =, =⎜, × ⎟ =, ;, ⎝ 2, 7⎠, 7, ⎛ 17 5⎞, − ⎟, ⎜, ⎝ 2, ⎠, ⎛ 5 4⎞, ⎜ − ⎟, ⎝2, ⎠ ⎛ −13 2⎞ −13, slope of PS =, =⎜, × ⎟ =, ⋅, 7⎠, 7, ⎛ 5 + 1⎞ ⎝ 2, ⎜, ⎟, ⎝2, ⎠, , A( −4, 2), B( 2, 6), C( 8, 5) and D( 9, − 7) are the vertices of a, quadrilateral ABCD. If P, Q, R, S are the midpoints of AB, BC, CD and, DA respectively, using slopes, show that PQRS is a parallelogram.
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Straight Lines, , EXAMPLE 16, SOLUTION, , 1, and 3., 2, , Let θ be the angle between the given lines., 1, Let m1 = and m2 = 3. Then,, 2, 1⎞ ⏐, ⏐ ⎛, ⎜3 − ⎟, m2 − m1 ⏐ ⏐ ⎝, (5/2⏐, ⏐, 2⎠ ⏐ ⏐, ⏐ =⏐, tan θ = ⏐, ⏐ = ⏐(5/2)⏐ = |1| = 1, 1, +, m, m, 1, ⎛, ⎞, ⏐, ⏐, 1 2⏐ ⏐⎜ 1 + 3 ⋅ ⎟ ⏐ ⏐, 2⎠⏐, ⏐⎝, ⇒ θ = 45 °., Hence, the angle between the given lines is 45°., , EXAMPLE 17, , SOLUTION, , Find the angle between the lines whose slopes are, , 655, , If A( −2, 1), B( 2, 3) and C( −2, − 4) be the vertices of a a ABC, show that, 2, tan B = ⋅, 3, , Clearly, B is the angle between BA and BC., 3 −1 1, Now, m1 = slope of BA =, =, 2+2 2, −4 − 3 7, and m2 = slope of BC =, =, ⋅, −2 − 2 4, ⏐ 7 −1 ⏐, m2 − m1 ⏐ ⏐ 4 2 ⏐ 2, ⏐, ⏐=, ∴ tan B = ⏐, = ⋅, ⏐1 + 7 ⋅ 1⏐, ⏐ 3, ⏐1 + m1m2⏐ ⏐, ⏐, 4 2⏐, 2, Hence, tan B = ⋅, 3, , EXAMPLE 18, , If the angle between two lines is, , π, 1, and the slope of one of the lines is ,, 4, 2, , find the slope of the other line., SOLUTION, , We know that the acute angle θ between two lines having slopes m1, and m2 is given by, ⏐m − m1 ⏐, ⏐⋅, tan θ = ⏐ 2, ⏐1 + m1m2⏐, π, 1, , m2 = m and θ = ⋅, 4, 2, Putting these values in (i), we get, ⏐m − 1 ⏐, 2m − 1⏐, ⏐, 2 ⏐ = tan π ⇔ ⏐, ⏐, ⏐= 1, ⏐, 4, ⏐1 + 1 m⏐, ⏐, ⏐2 + m⏐, ⏐, 2 ⏐, Let m1 =, , ... (i)
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656, , Senior Secondary School Mathematics for Class 11, , ⎛ 2m − 1⎞, ⇔ ⎜, ⎟ = ±1, ⎝ 2+ m⎠, 2m − 1, 2m − 1, ⇔, = 1 or, = −1, 2+m, 2+m, ⇔ ( 2m − 1) = ( 2 + m) or ( 2m − 1) = ( −2 − m), −1, ⋅, ⇔ m = 3 or m =, 3, −1, Hence, the slope of the other line is 3 or, ⋅, 3, , EXERCISE 20B, 1. Find the slope of a line whose inclination is, (i) 30°, (ii) 120°, (iii) 135°, 2. Find the inclination of a line whose slope is, 1, (i) 3, (ii), (iii) 1, 3, (v) − 3, , (iv) 90°, (iv) −1, , 3. Find the slope of a line which passes through the points, (i) (0, 0) and ( 4, − 2), (iii) (2, 5) and ( −4, − 4), , (ii) ( 0, − 3) and (2, 1), (iv) ( −2, 3) and ( 4, − 6), , −5, 4. If the slope of the line joining the points A( x , 2) and B( 6, − 8) is, , find the, 4, value of x., 5. Show that the line through the points (5, 6) and (2, 3) is parallel to the line, through the points ( 9, − 2) and ( 6, −5)., 6. Find the value of x so that the line through ( 3 , x) and (2, 7) is parallel to the, line through ( −1, 4) and (0, 6)., 7. Show that the line through the points ( −2, 6) and (4, 8) is perpendicular to, the line through the points ( 3 , − 3) and (5 , − 9)., 8. If A( 2, − 5), B( −2, 5), C( x , 3) and D(1, 1) be four points such that AB and CD, are perpendicular to each other, find the value of x., 9. Without using Pythagoras’s theorem, show that the points A(1, 2), B( 4, 5), and C( 6, 3) are the vertices of a right-angled triangle., 10. Using slopes, show that the points A( 6, − 1), B(5 , 0) and C( 2, 3) are, collinear., 11. Using slopes, find the value of x for which the points A(5 , 1), B(1, − 1) and, C( x , 4) are collinear., 12. Using slopes, show that the points A( −4, − 1), B( −2, − 4), C( 4, 0) and D( 2, 3), taken in order, are the vertices of a rectangle., 13. Using slopes, prove that the points A( −2, − 1), B(1, 0), C( 4, 3) and D(1, 2), are the vertices of a parallelogram.
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Straight Lines, , 657, , 14. If the three points A( h, k), B( x1 , y1 ) andC( x 2 , y 2 ) lie on a line then show that, ( h − x1 )( y 2 − y1 ) = ( k − y1 )( x 2 − x)., 15. If the points A( a , 0), B( 0, b) and P( x , y) are collinear, using slopes, prove that, x y, + = 1., a b, 16. A line passes through the points A( 4, − 6) and B( −2, −5). Show that the line, AB makes an obtuse angle with the x-axis., 17. The vertices of a quadrilateral are A( −4, 2), B( 2, 6), C( 8, 5) and D( 9, − 7)., Using slopes, show that the midpoints of the sides of the quad. ABCD form, a parallelogram., 18. Find the slope of the line which makes an angle of 30° with the positive, direction of the y-axis, measured anticlockwise., 1, 19. Find the angle between the lines whose slopes are 3 and, ⋅, 3, 20. Find the angle between the lines whose slopes are ( 2 − 3 ) and ( 2 + 3 )., 21. If A(1, 2), B( −3 , 2) and C( 3 , − 2) be the vertices of a a ABC, show that, 2, 4, (iii) tan C = ⋅, (i) tan A = 2, (ii) tan B =, 3, 7, 22. If θ is the angle between the lines joining the points A( 0, 0) and B( 2, 3), and, 11, the points C( 2, − 2) and D( 3 , 5), show that tan θ =, ⋅, 23, 23. If θ is the angle between the diagonals of a parallelogram ABCD whose, vertices are A( 0, 2), B( 2, − 1), C( 4, 0) and D( 2, 3). Show that tan θ = 2., 24. Show that the points A( 0, 6), B( 2, 1) and C(7 , 3) are three corners of a, square ABCD. Find (i) the slope of the diagonal BD and (ii) the coordinates, of the fourth vertex D., 25. A(1, 1), B(7 , 3) and C( 3 , 6) are the vertices of a a ABC. If D is the midpoint, of BC and AL ⊥ BC, find the slopes of (i) AD and (ii) AL., ANSWERS (EXERCISE 20B), , 1, 1. (i), 3, , (ii) − 3, , 2. (i) 60°, 1, 3. (i) −, 2, , (ii) 30°, (ii) 2, , (iii) −1, , (iv) not defined, , (iii) 45°, (iv) 135°, (v) 120°, 3, 3, (iii), 4. x = − 2, 6. x = 9, (iv) −, 2, 2, 18. − 3, , 8. x = 6, , 11. x = 11, , 20. 60°, , 7, 7, 4, 24. (i) (ii) (5, 8) 25. (i), (ii), 3, 8, 3, , 19. 30°
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658, , Senior Secondary School Mathematics for Class 11, HINTS TO SOME SELECTED QUESTIONS, , 11. For collinearity, we must have, slope of AB = slope of BC., 12. Let m1 , m 2 , m 3 , m 4 be the slopes of AB, BC , CD and DA respectively. Then, show that, m1 = m 3 , m 2 = m 4 and m1m 2 = − 1., 14. Slope of AB = slope of BC, ( y1 − k ) ( y2 − y1 ), ( k − y1 ) ( y2 − y1 ), =, ⇒, =, ⇒, ( x1 − h ) ( x2 − x1 ), ( h − x1 ) ( x2 − x1 ), ⇒ ( h − x1 )( y2 − y1 ) = ( k − y1 )( x2 − x1 )., 15. Since the points A, B, P are collinear, we have, b−0 y−b, (slope of AB = slope of BP) ⇒, =, 0−a x−0, ⇒ bx + ay = ab ⇒, , x y, + = 1., a b, , 16. Slope of AB is negative., 11, 17, 5 −5, 17. Midpoints of AB, BC, CD and DA are P( −1, 4 ), Q ⎛⎜ 5 , ⎞⎟ , R ⎛⎜, , − 1⎞⎟ and S ⎛⎜ , ⎞⎟ ⋅, ⎝ 2⎠ ⎝ 2, ⎠, ⎝2 2 ⎠, 1, −13, Now, show that (slope of PQ = slope of RS = ) and (slope ofQR = slope of PS =, )., 4, 7, 18. The given line makes an angle of ( 90 ° + 30 ° ) = 120 ° with the positive direction of the, x-axis. Hence, m = tan 120 ° = − 3 ., 24. BD ⊥ AC ., , VARIOUS FORMS OF EQUATIONS OF A LINE, EQUATION OF A LINE, , The equation of a straight line is the linear relation between two variables x and y,, which is satisfied by the coordinates of each and every point on the line and not by, those of any other point in the cartesian plane., SOME ELEMENTARY RESULTS, , (i), , EQUATION OF X-AXIS:, , We know that the ordinate of each point on the x-axis is 0. If P( x , y) is any, point on the x-axis, then y = 0., Hence, the equation of x-axis is y = 0., (ii), , EQUATION OF Y-AXIS:, , We know that the abscissa of each point on the y-axis is 0. If P( x , y) is any, point on the y-axis, then x = 0., Hence, the equation of y-axis is x = 0., (iii), , EQUATION OF A LINE PARALLEL TO Y-AXIS:, , Let AB be a straight line parallel to the y-axis lying on its right-hand side, at a distance a from it.
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Straight Lines, , 659, , Then, the abscissa of each, point on AB is a., If P( x , y) is any point on AB, then x = a., Thus, the equation of a vertical, line at a distance a from the, y-axis, lying on its right-hand, side is x = a., Similarly, the equation of a, vertical line at a distance a from, the y-axis, lying on its left-hand side is x = − a., (iv), , EQUATION OF A LINE PARALLEL TO X-AXIS:, , Let AB be a straight line, parallel to the x-axis, lying, above it at a distance b from it., Then, the ordinate of each, point on AB is b., If P( x , y) is any point on AB,, then y = b., Thus, the equation of a, horizontal line at a distance b, from the x-axis, lying above the, x-axis is y = b., Similarly, the equation of a horizontal line at a distance b from the x-axis, lying, below the x-axis is y = − b., SUMMARY, , (i) Equation of x-axis is y = 0., (ii) Equation of y-axis is x = 0., (iii) Equation of a vertical line on RHS of the y-axis at a distance, a from it is x = a.., (iv) Equation of a vertical line on LHS of the x-axis at a distance, a from it is x = −a.., (v) Equation of a horizontal line lying above the x-axis at a, distance b from it is y = b., (vi) Equation of a horizontal line lying below the x-axis at a, distance b from it is y = −b., EXAMPLE 1, SOLUTION, , Write down the equation of (i) x-axis (ii) y-axis., We know that, (i) the equation of x-axis is y = 0., (ii) the equation of y-axis is x = 0.
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660, EXAMPLE 2, , Senior Secondary School Mathematics for Class 11, , Write down the equation of a line parallel to the x-axis, (i) at a distance of 5 units above the x-axis,, (ii) at a distance of 4 units below the x-axis., , SOLUTION, , (i) The equation of a line parallel to the x-axis at a distance of 5 units, above x-axis is y = 5., (ii) The equation of a line parallel to the x-axis at a distance of 4 units, below the x-axis is y = − 4., , EXAMPLE 3, , Write down the equation of a line parallel to the y-axis, (i) at a distance of 7 units on left-hand side of the y-axis,, (ii) at a distance of 3 units on right-hand side of the y-axis., , SOLUTION, , (i) The equation of a line parallel to the y-axis at a distance of 7 units, on its left is x = − 7., (ii) The equation of a line parallel to the y-axis at a distance of 3 units, on its right is x = 3., , EXAMPLE 4, , SOLUTION, , EXAMPLE 5, , Find the equations of the lines parallel to the axes and passing through, the point ( −3 , 5)., We know that, (i) the equation of a line, parallel to the x-axis and, passing through ( −3 , 5) is, y = 5., (ii) the equation of a line, parallel to the y-axis and, passing through ( −3 , 5) is, x = − 3., Find the values of k for which the line, ( k − 3) x + ( k 2 − 4) y + ( k − 1)( k − 6) = 0, (i) is parallel to the x-axis,, (ii) is parallel to the y-axis,, (iii) passes through the origin., , SOLUTION, , Given equation of the line is ( k − 3) x + ( k 2 − 4) y + ( k − 1)( k − 6) = 0., (i) The given line will be parallel to the x-axis, when the coefficient, of y is zero,, i.e., when k 2 − 4 = 0., This happens when k = ± 2., (ii) The given line will be parallel to the y-axis, when the coefficient, of x is zero,, i.e., when k − 3 = 0 or k = 3.
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Straight Lines, , 661, , (iii) The given line will pass through the origin, if (0, 0) satisfies the, given equation,, i.e., when ( k − 1)( k − 6) = 0., This happens when k = 1 or k = 6., EXAMPLE 6, , SOLUTION, , Find the equation of a line which is equidistant from the lines x = − 3 and, x = 5., Let X′ OX and YOY′ be the coordinate axes, , Let L1 be the line x = − 3. Then, it is parallel to the y-axis., It meets the x-axis at A( −3 , 0)., Let L2 be the line x = 5. Then, it is parallel to the y-axis., It meets the x-axis at B(5 , 0)., −3 + 5 0 + 0⎞, Midpoint of AB is M ⎛⎜, ,, ⎟ , i.e., M(1, 0)., ⎝ 2, 2 ⎠, So, the line x = 1 is equidistant from L1 and L2 ., EQUATION OF A LINE IN POINT–SLOPE FORM, THEOREM 1, , SOLUTION, , Prove that the equation of a non-vertical line with slope m and passing, through the point ( x1 , y1 ) is ( y − y1 ) = m( x − x1 )., Let L be a non-vertical line with slope m and passing through a point, A( x1 , y1 ).
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662, , Senior Secondary School Mathematics for Class 11, , Let P( x , y) be an arbitrary point on L., ( y − y1 ), Then, slope of line L =, ⋅, ( x − x1 ), ( y − y1 ), ∴, = m ⇔ ( y − y1 ) = m( x − x1 )., ( x − x1 ), Hence, the required equation is, ( y − y1 ) = m( x − x1 )., EXAMPLE 1, SOLUTION, , EXAMPLE 2, SOLUTION, , EXAMPLE 3, SOLUTION, , Find the equation of a line passing through the point (4, 3) and having, slope 2., We know that the equation of a line with slope m and passing, through the point ( x1 , y1 ) is given by, ( y − y1 ) = m( x − x1 )., Here, m = 2, x1 = 4 and y1 = 3., Hence, the required equation is, ( y − 3) = 2( x − 4), i.e., 2x − y − 5 = 0., Find the equation of a line which makes an angle of 135° with the x-axis, and passes through the point (3, 5)., Here, m = tan 135 ° = tan (180° − 45 °) = − tan 45 ° = −1., Hence, the required equation is, y −5, = − 1 ⇔ ( y − 5) = 3 − x ⇔ x + y − 8 = 0., x−3, Find the equation of a line passing through the point ( 3 , − 4) and parallel, to the x-axis., Since the given line is parallel to the x-axis, we have m = 0., So, the required equation is, y+4, = 0 ⇔ y + 4 = 0., y−3, , EQUATION OF A LINE IN TWO-POINT FORM, THEOREM 2, , PROOF, , Prove that the equation of a non-vertical line passing through two given, points A( x1 , y1 ) and B( x 2 , y 2 ) is given by, ( y − y1 ) ( y 2 − y1 ), =, ⋅, ( x − x1 ) ( x 2 − x1 ), , Let L be the line passing through the, points A( x1 , y1 ) and B( x 2 , y 2 ) and, let P( x , y) be an arbitrary point on L., Then, clearly we have, slope of AP = slope of AB, ( y − y1 ) ( y 2 − y1 ), ⇒, =, ,, ( x − x1 ) ( x 2 − x1 ), which is the required equation.
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Straight Lines, EXAMPLE 4, SOLUTION, , 663, , Find the equation of a line passing through the points ( −1, 1) and ( 2, − 4)., Let the given points be A( −1, 1) and B( 2, − 4)., We know that the equation of a line passing through the points, ( y − y1 ) ( y 2 − y1 ), =, ⋅, ( x1 , y1 ) and ( x 2 , y 2 ) is given by, ( x − x1 ) ( x 2 − x1 ), Here, x1 = −1, y1 = 1, x 2 = 2 and y 2 = − 4., So, the equation of a line passing through the given points is, ( y − 1), ( −4 − 1), ( y − 1) −5, =, ⇔, =, {x − ( −1)} 2 − ( −1), ( x + 1), 3, ⇔ 3( y − 1) = − 5( x + 1), ⇔ 5 x + 3 y + 2 = 0., Hence, the required equation is 5 x + 3 y + 2 = 0., , EXAMPLE 5, , SOLUTION, , EXAMPLE 6, , SOLUTION, , Show that the three points ( 3 , 0), ( −2, − 2) and (8, 2) are collinear. Also,, find the equation of the straight line on which these points lie., Let A( 3 , 0), B( −2, − 2) and C( 8, 2) be the given points., Then, the equation of line AB is given by, ⎡, y − 0 −2 − 0, y − y1 y 2 − y1 ⎤, =, ⎢using x − x = x − x ⎥, x − 3 −2 − 3, 1, 2, 1⎦, ⎣, y, 2, ⇒, =, x−3 5, ... (i), ⇒ 5 y = 2x − 6, ⇒ 2x − 5 y − 6 = 0, which is the required equation., Putting x = 8 and y = 2 in (i), we get, LHS = ( 2 × 8) − (5 × 2) − 6 = 0 = RHS., Thus, the point C( 8, 2) also lies on (i)., Hence, the given points lie on the same straight line, whose equation, is 2x − 5 y − 6 = 0., Show that the points ( a , 0), ( 0, b) and ( 3 a , − 2b) are collinear. Also find, the equation of the line containing them., Let A( a , 0), B( 0, b) and C( 3 a , − 2b) be the given points., Then, the equation of line AB is given by, y−0 b−0, =, ⇒ − ay = bx − ab, x −a 0−a, ⇒ bx + ay − ab = 0., Thus, the equation of line AB is bx + ay − ab = 0., Putting x = 3 a and y = − 2b in (i), we get, LHS = b ⋅ 3 a + a( −2b) − ab = 3 ab − 2ab − ab = 0 = RHS., Thus, the point C( 3 a , − 2b) also lies on AB., , ... (i)
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664, , Senior Secondary School Mathematics for Class 11, , Hence, the given points are collinear and the equation of the line, containing them is bx + ay − ab = 0., EXAMPLE 7, , SOLUTION, , Find the equations of the sides of a triangle whose vertices are, A( −1, 8), B( 4, − 2) and C( −5 , − 3)., Here, we make use of the equation of a, line in two-point form., y − 8 −2 − 8, Equation of AB is, =, x+1, 4+1, ⇒ 5( y − 8) + 10( x + 1) = 0, ⇒ 10x + 5 y − 30 = 0, ⇒ 2x + y − 6 = 0;, y + 2 −3 + 2, equation of BC is, =, ⇒, x − 4 −5 − 4, ⇒, y − 8 −3 − 8, equation of AC is, =, ⇒, x + 1 −5 + 1, ⇒, , EXAMPLE 8, , SOLUTION, , − 9( y + 2) = − ( x − 4), x − 9y − 22 = 0;, − 4( y − 8) = −11( x + 1), 11x − 4y + 43 = 0., , Find the equations of the medians of a a ABC whose vertices are, A( 2, 5), B( −4, 9) and C( −2, − 1)., The vertices of a ABC are A( 2, 5), B( −4, 9) and C( −2, − 1)., Let D, E, F be the midpoints of BC , CA and AB, respectively., Then, these points are, −4 − 2 9 − 1⎞ ⎛ −2 + 2 −1 + 5 ⎞, ,, ,, ⎟ , E⎜, ⎟, 2, 2 ⎠ ⎝ 2, 2 ⎠, 2 − 4 5 + 9⎞, and F ⎛⎜, ,, ⎟, ⎝ 2, 2 ⎠, D ⎛⎜, ⎝, , i.e., D( −3 , 4), E( 0, 2) and F( −1, 7)., Thus, AD , BE and CF are the medians of a ABC., y −5, 4−5, y −5 1, Equation of median AD is, =, ⇒, =, x − 2 −3 − 2, x−2 5, ⇒ 5( y − 5) = x − 2, ⇒ x − 5 y + 23 = 0., Equation of median BE is, , y −9 2−9, y − 9 −7, =, ⇒, =, x + 4 0+ 4, x+4, 4, ⇒ 4( y − 9) + 7( x + 4) = 0, ⇒ 7 x + 4y − 8 = 0.
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Straight Lines, , Equation of median CF is, , y+1, 7+1, =, ⇒ ( y + 1) = 8( x + 2), x + 2 −1 + 2, ⇒ 8x − y + 15 = 0., , Find the equation of the perpendicular bisector of the line joining the, points A( 2, 3) and B( 6, − 5)., , EXAMPLE 9, , SOLUTION, , 665, , Here, A( 2, 3) and B( 6, − 5) are, the end points of the given line, segment., The midpoint of AB is, 2 + 6 3 − 5⎞, ,, ⎟ , i.e., M( 4, − 1)., 2, 2 ⎠, −5 − 3, Slope of AB =, = − 2., 6−2, M ⎛⎜, ⎝, , Let LM ⊥ AB and let the slope of LM be m., 1, [Q LM ⊥ AB]., Then, m × ( −2) = − 1 ⇒ m =, 2, Clearly, LM is the perpendicular bisector of AB., 1, and passing through M( 4, − 1)., Now, LM is a line with slope =, 2, So, the required equation is, y − ( −1) 1, =, ⇒ 2( y + 1) = ( x − 4), x−4, 2, ⇒ x − 2y − 6 = 0., EXAMPLE 10, , If A( 2, 1), B( −2, 3) and C( 4, 5) are the vertices of a a ABC then find the, equation of, (i) the median through C, (ii) the altitude through B, (iii) the right bisector of side BC., , SOLUTION, , (i) Let F be the midpoint of side AB., Then, CF is the median through C., , 2 + ( −2) 1 + 3 ⎞, Coordinates of F are F ⎛⎜, ,, ⎟ , i.e., F( 0, 2)., ⎝, 2, 2 ⎠
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666, , Senior Secondary School Mathematics for Class 11, , So, the equation of median CF is given by, y −5 2−5, y −5, 3, =, ⇔, =, x −4 0−4, x−4 4, ⇔ 4( y − 5) = 3( x − 4) ⇔ 3 x − 4y + 8 = 0., Hence, the equation of median CF is 3 x − 4y + 8 = 0., (ii) Draw BL ⊥ AC. Then, BL is the altitude through B., 5 −1, Slope of AC =, = 2., 4−2, Let the slope BL be m., Since BL ⊥ AC, we have 2m = − 1 and therefore, m =, , −1, ⋅, 2, , −1, ⋅, 2, So, the equation of BL is given by, y − 3 −1, =, ⇔ 2( y − 3) = − ( x + 2) ⇔ x + 2y − 4 = 0., x+2, 2, Hence, the equation of altitude BL is x + 2y − 4 = 0., , Thus, the slope of BL is, , (iii) Let D be the midpoint of BC., , −2 + 4 3 + 5 ⎞, Then, the coordinates of D are D ⎛⎜, ,, ⎟ , i.e., D(1, 4)., ⎝ 2, 2 ⎠, Through D, draw DP ⊥ BC., 5−3 2 1, Slope of BC =, = = ⋅, 4+ 2 6 3, Let the slope of PD be m., 1, Since PD ⊥ BC , we have m × = −1 ⇒ m = − 3., 3, So, the slope of PD is −3., So, the equation of PD is given by, y−4, = − 3 ⇔ − 3( x − 1) = ( y − 4) ⇔ 3 x + y − 7 = 0., x −1, Hence, the equation of the right bisector of BC is 3 x + y − 7 = 0., , EXAMPLE 11, , SOLUTION, , Find the equation of the bisector of ∠A of a ABC, whose vertices are, A( −2, 4), B(5 , 5) and C( 4, − 2)., , Let AD be the bisector of ∠A., Then, BD : DC = AB : AC., Now,|AB| = (5 − 4) 2 + (5 + 2) 2, = 50 = 5 2., and|AC| = ( −2 − 4) 2 + ( 4 + 2) 2, = 72 = 6 2.
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Straight Lines, , 667, , BD AB 5 2 5, =, =, =, ⇒ BD : DC = 5 : 6., DC AC 6 2 6, Thus, D divides BC in the ratio 5 : 6., ⎛ 5 × 4 + 6 × 5 5 × ( −2) + 6 × 5 ⎞, So, the coordinates of D are ⎜, ,, ⎟,, 5+6, 5+6, ⎝, ⎠, , ∴, , 50 20, i.e., D ⎛⎜ , ⎞⎟, ⎝ 11 11 ⎠, ∴, , equation of line AD is given by, 20, −4, y − 4 11, y − 4 −24 −1, =, =, =, ⇔, 3, x + 2 50 + 2, x + 2 72, 11, ⇔ 3( y − 4) = − ( x + 2) ⇔ x + 3 y − 10 = 0., Hence, the equation of the bisector of ∠A is x + 3 y − 10 = 0., , EXERCISE 20C, 1. Find the equation of a line parallel to the x-axis at a distance of, (i) 4 units above it, (ii) 5 units below it., 2. Find the equation of a line parallel to the y-axis at a distance of, (i) 6 units to its right (ii) 3 units to its left., 3. Find the equation of a line parallel to the x-axis and having intercept −3 on, the y-axis., 4. Find the equation of a horizontal line passing through the point ( 4, − 2)., 5. Find the equation of a vertical line passing through the point ( −5 , 6)., 6. Find the equation of a line which is equidistant from the lines x = − 2 and, x = 6., 7. Find the equation of a line which is equidistant from the lines y = 8 and, y = −2., 8. Find the equation of a line, (i) whose slope is 4 and which passes through the point (5 , − 7);, (ii) whose slope is −3 and which passes through the point ( −2, 3);, 2π, (iii) which makes an angle of, with the positive direction of the x-axis, 3, and passes through the point (0, 2)., 9. Find the equation of a line whose inclination with the x-axis is 30° and, which passes through the point (0, 5)., 10. Find the equation of a line whose inclination with the x-axis is 150° and, which passes through the point ( 3 , − 5)., 11. Find the equation of a line passing through the origin and making an angle, of 120° with the positive direction of the x-axis.
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668, , Senior Secondary School Mathematics for Class 11, , 12. Find the equation of a line which cuts off intercept 5 on the x-axis and, makes an angle of 60° with the positive direction of the x-axis., 13. Find the equation of the line passing through the point P( 4, − 5) and parallel, to the line joining the points A( 3 , 7) and B( −2, 4)., 14. Find the equation of the line passing through the point P( −3 , 5) and, perpendicular to the line passing through the points A( 2, 5) and B( −3 , 6)., 15. Find the slope and the equation of the line passing through the points:, (i) ( 3 , − 2) and ( −5 , − 7), (ii) ( −1, 1) and ( 2, − 4), (iii) (5, 3) and ( −5 , − 3), (iv) ( a , b) and ( −a , b), 16. Find the angle which the line joining the points (1, 3 ) and ( 2 ,, with the x-axis., , 6) makes, , 17. Prove that the points A(1, 4), B( 3 , − 2) and C( 4, − 5) are collinear. Also find, the equation of the line on which these points lie., 18. If A( 0, 0), B( 2, 4) and C( 6, 4) are the vertices of a a ABC, find the equations, of its sides., 19. If A( −1, 6), B( −3 , − 9) and C(5 , − 8) are the vertices of a a ABC, find the, equations of its medians., 20. Find the equation of the perpendicular bisector of the line segment whose, end points are A(10, 4) and B( −4, 9)., 21. Find the equations of the altitudes of a a ABC, whose vertices are, A( 2, − 2), B(1, 1) and C( −1, 0)., 22. If A( 4, 3), B( 0, 0) and C( 2, 3) are the vertices of a a ABC, find the equation, of the bisector of ∠A., 23. The midpoints of the sides BC, CA and AB of a a ABC are D( 2, 1), E( −5 , 7), and F( −5 , − 5) respectively. Find the equations of the sides of a ABC., 24. If A(1, 4), B( 2, − 3) and C( −1, − 2) are the vertices of a a ABC, find the, equation of, (i) the median through A, (ii) the altitude through A, (iii) the perpendicular bisector of BC., ANSWERS (EXERCISE 20C), , 1. (i) y − 4 = 0 (ii) y + 5 = 0, 3. y + 3 = 0, 4. y + 2 = 0, 7. y = 3, 9. x −, 11., , 2. (i) x − 6 = 0 (ii) x + 3 = 0, 5. x + 5 = 0, 6. x = 2, , 8. (i) 4x − y − 27 = 0 (ii) 3 x + y + 3 = 0 (iii) 3 x + y − 2 = 0, 3y + 5 3 = 0, , 3 x + y = 0 12., , 10. x +, , 3 y + ( −3 + 5 3 ) = 0, , 3 x − y − 5 3 = 0 13. 3 x − 5 y − 37 = 0 14. 5 x − y + 20 = 0, , 5, −5, 3, 15. (i) , 5 x − 8y − 31 = 0 (ii), , 5 x + 3 y + 2 = 0 (iii) , 3 x − 5 y = 0 (iv) 0, y = b, 8, 3, 5, 16. 60°, , 17. 3 x + y − 7 = 0 18. y = 4, 2x − 3 y = 0, 2x − y = 0
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Straight Lines, , 669, , 19. 29x + 4y + 5 = 0, 8x − 5 y − 21 = 0, 13 x + 14y + 47 = 0 20. 28x − 10y − 19 = 0, 21. 2x + y − 2 = 0, 3 x − 2y − 1 = 0, x − 3 y + 1 = 0, 22. x − 3 y + 5 = 0, 23. x − 2 = 0, 6x − 7 y + 79 = 0, 6x + 7 y + 65 = 0, 24. (i) 13 x − y − 9 = 0 (ii) 3 x − y + 1 = 0 (iii) 3 x − y − 4 = 0, HINTS TO SOME SELECTED QUESTIONS, −2 + 6 ⎞, 6. Midpoint of A( −2 , 0 ) and B( 6 , 0 ) is M ⎛⎜, , 0 ⎟ , i.e., M( 2 , 0 )., ⎝ 2, ⎠, So, the required line is x = 2., 8 − 2⎞, 7. Midpoint of A( 0 , 8 ) and B( 0 , − 2 ) is M ⎛⎜ 0 ,, ⎟ , i.e., M( 0 , 3 )., ⎝, 2 ⎠, So, the required line is y = 3., 12. The given line passes through the point (5, 0) and m = tan 60 ° = 3. So, the required, y−0, equation is, = 3 ⇔ y = 3 ( x − 5 )., x−5, 4−7, 3, 3, 13. Slope of the given line = slope of AB =, = ⋅ Thus, m = ⋅, −2 − 3 5, 5, y+5 3, =, ⇔ 3 x − 5 y − 37 = 0., ∴ required equation is, x−4 5, 6−5, −1, 14. Slope of AB = m1 =, =, ⋅, −3 − 2 5, −1, Now, m1m 2 = − 1 ⇔, × m 2 = − 1 ⇔ m 2 = 5., 5, y−5, = 5 ⇔ 5 x − y + 20 = 0., ∴ required equation is, x+ 3, ⎛ 6 − 3⎞, 16. m = tan θ = ⎜, ⎟=, ⎝ 2−1 ⎠, , 3 ⇔ θ = 60 ° .
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670, , Senior Secondary School Mathematics for Class 11, , EQUATION OF A LINE IN SLOPE–INTERCEPT FORM, , Prove that the equation of a line with slope m and making an intercept c, on the y-axis, is given by y = mx + c., PROOF Let L be a given line with slope m,, making an intercept c on the y-axis., Then, L cuts the y-axis at a point C, such that OC = c., Clearly, the coordinates of C are, (0, c)., Thus, the line L has slope m and, passes through the point C( 0, c)., So, by the point–slope form, the, equation of L is, ( y − c) = m( x − 0) ⇒ y = mx + c., Hence, y = mx + c is the required equation., REMARK The equation of a line with slope m and y-intercept c is given by, y = mx + c., THEOREM 1, , Prove that the equation of a line with slope m and making an intercept d, on the x-axis, is given by y = m( x − d)., PROOF Let L be a given line with slope m,, making an intercept d on the x-axis., Then, L cuts the x-axis at a point D, such that OD = d., Clearly, the coordinates of D are, ( d , 0)., Thus, the line L has slope m and, passes through the point D( d , 0)., So, by the point–slope form, the, equation of L is, ( y − 0), = m ⇔ y = m( x − d)., ( x − d), Hence, y = m( x − d) is the required equation., REMARK The equation of a line with slope m and x-intercept d is given by, y = m( x − d)., THEOREM 2, , SUMMARY, , (i) The equation of a line with slope m and y-intercept c is, given by, y = mx + c., (ii) The equation of a line with slope m and x-intercept d is, given by, y = m( x − d).
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Straight Lines, , 671, , SOLVED EXAMPLES, EXAMPLE 1, SOLUTION, , Find the equation of a line whose slope is, , We know that the equation of a line with slope m and y-intercept c is, given by y = mx + c., 1, −5, Here, m = and c =, ⋅, 2, 4, Hence, the required equation of the line is, y=, , EXAMPLE 2, , SOLUTION, , EXAMPLE 3, , SOLUTION, , 1, −5, and y-intercept equal to, ⋅, 2, 4, , 1, 5, x−, ⇒ 4y = 2x − 5, 2, 4, ⇒ 2x − 4y − 5 = 0., , Find the equation of a line which, intersects the y-axis at a distance of, 3 units above the origin and makes, an angle of 30° with the positive, direction of the x-axis., 1, Here, m = tan 30° =, and c = 3., 3, Hence, the required equation of, the line is, 1, [Q y = mx + c], y=, x + 3 ⇒ 3y = x + 3 3, 3, ⇒ x − 3 y + 3 3 = 0., Find the equation of a line which cuts off an intercept of 4 units on, negative direction of the y-axis and makes an angle of 120° with the, positive direction of the x-axis., We know that the equation of a line with slope m and y-intercept c is, given by, y = mx + c.
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672, , Senior Secondary School Mathematics for Class 11, , Here, m = tan 120° = tan (180° − 60°) = − tan 60° = − 3 and c = −4., Hence, the required equation of the line is, y = − 3x − 4 ⇒, EXAMPLE 4, , 3x + y + 4 = 0, , [Q y = mx + c]., , Find the equation of a line for which tan θ =, , 1, and x-intercept equal to, 3, , 5 units., SOLUTION, , EXAMPLE 5, , SOLUTION, , We know that the equation of a line with slope m and x-intercept d is, given by y = m( x − d)., 1, and d = 5., Here, m = tan θ =, 3, Hence, the required equation of the line is, 1, y = ( x − 5) ⇒ 3 y = x − 5, 3, ⇒ x − 3 y − 5 = 0., Find the equation of a line which cuts the x-axis at a distance of 3 units to, the left of the origin and has a slope equal to −2., We know that the equation of a line with slope m and x-intercept d is, given by y = m( x − d)., Here, m = −2 and d = − 3., Hence, the required equation of the line is, y = − 2( x + 3) ⇒ 2x + y + 6 = 0., , EXAMPLE 6, , SOLUTION, , EXAMPLE 7, SOLUTION, , Reduce the equation 6x + 3 y − 5 = 0 to the slope–intercept form and find, its slope and y-intercept., We know that the equation of a line with slope m and y-intercept c is, given by y = mx + c., Now, 6x + 3 y − 5 = 0 ⇒ 3 y = − 6x + 5, 5, ⇒ y = − 2x + ⋅, 3, 5, ∴ m = − 2 and c = ⋅, 3, 5, Hence, slope = −2 and y-intercept = ⋅, 3, Prove that the lines x + 2y − 9 = 0 and 2x + 4y + 5 = 0 are parallel., Let the slopes of the given lines be m1 and m2 respectively., Then, x + 2y − 9 = 0 ⇒ 2y = − x + 9, 1, 9, ⇒ y=− x+ ⋅, 2, 2, And, 2x + 4y + 5 = 0 ⇒ 4y = − 2x − 5, 1, 5, ⇒ y=− x− ⋅, 2, 4
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Straight Lines, , 673, , 1, 1, and m2 = − ⋅, 2, 2, Thus, m1 = m2 ., , ∴, , m1 = −, , Hence, the given lines are parallel., Show that the lines 27 x − 18y + 25 = 0 and 2x + 3 y + 7 = 0 are, perpendicular to each other., , EXAMPLE 8, , SOLUTION, , Let the slopes of the given lines be m1 and m2 respectively., Then, 27 x − 18y + 25 = 0 ⇒ 18y = 27 x + 25, 3, 25, ⇒ y= x+, 2, 18, And, 2x + 3 y + 7 = 0 ⇒ 3 y = − 2x − 7, −2, 7, ⇒ y=, x− ⋅, 3, 3, 3, −2, and m2 =, ∴ m1 =, 2, 3, ⎛ 3 ⎞ ⎛ −2⎞, ⇒ m1m2 = ⎜ ⎟ × ⎜ ⎟ = − 1, ⎝ 2⎠ ⎝ 3 ⎠, Hence, the given lines are perpendicular to each other., Find the angle made by the line x +, direction of the x-axis., , EXAMPLE 9, , SOLUTION, , 3 y − 6 = 0 with the positive, , Let the required angle be θ., Now, x +, , 3y − 6 = 0 ⇒ y = −, , 1, 6, x+, ⋅, 3, 3, , 1, 1, ⇔ tan θ = −, = − tan 30° = tan (180° − 30°) = tan 150°., 3, 3, ∴ θ = 150°., Hence, the given line makes an angle of 150° with the positive, direction of the x-axis., , ∴ m=−, , EXAMPLE 10, , SOLUTION, , Find the angle made by the line x cos 30° + y sin 30° + sin 120° = 0, with the positive direction of the x-axis., , Let the required angle be θ., Now, x cos 30° + y sin 30° + sin 120° = 0, ⇒, ⇒, , ⎛ 3⎞, 3, ⎛ 1⎞, =0, ⎜, ⎟ x+⎜ ⎟ y+, ⎝ 2⎠, 2, ⎝ 2 ⎠, 3x + y +, , 3 =0, , ⇒ y = (− 3)x + (− 3), ⇒ y = mx + c, where m = − 3 and c = − 3., Now, m = − 3, ⇔ tan θ = − 3 = − tan 60° = tan (180° − 60°) = tan 120°.
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Straight Lines, SOLUTION, , 675, , We have, 2, 7, x− ⋅, 3, 3, 2, slope of the given line =, 3, 2x − 3 y = 7 ⇒ y =, , ∴, , and slope of a line parallel to the given line =, Now, the equation of a line with slope, point ( 2, − 5) is given by, y − ( −5) 2, y+5 2, =, ⇒, =, x−2, 3, x−2 3, , 2, ⋅, 3, , 2, and passing through the, 3, ⎤, ⎡ ( y − y1 ), ⎢Q ( x − x ) = m ⎥, 1, ⎦, ⎣, , ⇒ 3( y + 5) = 2( x − 2), ⇒ 2x − 3 y − 19 = 0, which is the required equation., EXAMPLE 14, , SOLUTION, , Find the equation of the line passing through the point ( −2, − 4) and, perpendicular to the line 3 x − y + 5 = 0., , We have, 3x − y + 5 = 0 ⇒ y = 3x + 5, ∴, , [Q y = mx + c], , slope of the given line = 3, , and slope of a line perpendicular to the given line = −, , 1, 3, , ⎡, 1 ⎤, ⎢Q m2 = − m ⎥ ⋅, 1⎦, ⎣, Now, the equation of a line with slope −, point ( −2, − 4) is given by, , 1, and passing through the, 3, , ⎡ ( y − y1 ), ⎤, ⎢Q ( x − x ) = m ⎥, 1, ⎣, ⎦, ⇒ 3( y + 4) = − ( x + 2), ⇒ x + 3 y + 14 = 0, which is the required equation., , y − ( −4), 1, y + 4 −1, =−, ⇒, =, x − ( −2), 3, x+2, 3, , Find the equation of the line whose y-intercept is −3 and which is, perpendicular to the line 3 x − 2y + 5 = 0., 3, 5, 3 x − 2y + 5 = 0 ⇒ y = x +, 2, 2, 3, ∴ slope of the given line =, ⋅, 2, −2 ⎡, −1 ⎤, ∴ slope of a line perpendicular to the given line =, Q m2 =, ⋅, 3 ⎢⎣, m1 ⎥⎦, , EXAMPLE 15, , SOLUTION
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676, , Senior Secondary School Mathematics for Class 11, , Now, the equation of a line with slope, , −2, and y-intercept −3 is, 3, , given by, −2, x−3, [Q y = mx + c], 3, 2x + 3 y + 9 = 0, which is the required equation., y=, , ⇒, , Find the equation of the line perpendicular to the line x − 7 y + 5 = 0, and having x-intercept 3., 1, 5, x − 7y + 5 = 0 ⇒ y = x + ⋅, 7, 7, 1, ⋅, ∴ slope of the given line =, 7, ⎡, −1 ⎤, ∴ slope of a line perpendicular to the given line = − 7 ⎢Q m2 =, ⎥⋅, m, 1⎦, ⎣, Thus, m = − 7 and d = 3, where d = x-intercept., Now, the equation of a line with slope −7 and x-intercept 3 is given by, , EXAMPLE 16, , SOLUTION, , y = − 7( x − 3), , [Q y = m( x − d)], , ⇒ 7 x + y − 21 = 0, which is the required equation., EXAMPLE 17, , SOLUTION, , Find the equations of the lines through the point (3, 2), which make an, angle of 45° with the line x − 2y = 3., , Let the slope of the required line be m., y−2, Then, its equation is, = m., x−3, 1, 3, Given line is x − 2y = 3 ⇒ y = x − ⋅, 2, 2, 1, Clearly, the slope of this line is ⋅, 2, It is given that the angle between (i) and (ii) is 45°., ∴, , ⇔, ∴, , ⏐m − 1 ⏐, ⏐, 2 ⏐ = tan 45 °, 1, ⏐, ⏐1 + m⏐, ⏐, ⏐, 2 ⏐, , ⎡, ⎢using tan θ, ⎣, , ... (i), ... (ii), , ⎤, ⏐ m − m1 ⏐, ⏐⎥, =⏐ 2, 1, +, m, m, ⏐, 1 2⏐⎦, , −1, 2m − 1, ⋅, = − 1 ⇔ m = 3 or m =, 2+m, 3, y−2, y − 2 −1, the required equation is, = 3 or, =, x−3, x−3 3, 2m − 1, = 1 or, 2+m, , i.e., 3 x − y − 7 = 0 or x + 3 y − 9 = 0.
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Straight Lines, , 677, , EXERCISE 20D, 1. Find the equation of the line whose, , 2., , 3., 4., 5., 6., 7., 8., 9., 10., , (i) slope = 3 and y-intercept = 5, (ii) slope = − 1 and y-intercept = 4, 2, (iii) slope = − and y-intercept = − 3, 5, Find the equation of the line which makes an angle of 30° with the, positive direction of the x-axis and cuts off an intercept of 4 units with the, negative direction of the y-axis., 5π, and which makes an, Find the equation of the line whose inclination is, 6, intercept of 6 units on the negative direction of the y-axis., Find the equation of the line cutting off an intercept −2 from the y-axis and, equally inclined to the axes., Find the equation of the bisectors of the angles between the coordinate, axes., Find the equation of the line through the point ( −1, 5) and making an, intercept of −2 on the y-axis., Find the equation of the line which is parallel to the line 2x − 3 y = 8 and, whose y-intercept is 5 units., Find the equation of the line passing through the point (0, 3) and, perpendicular to the line x − 2y + 5 = 0., Find the equation of the line passing through the point (2, 3) and, perpendicular to the line 4x + 3 y = 10., Find the equation of the line passing through the point (2, 4) and, perpendicular to the x-axis., , 11. Find the equation of the line that has x-intercept −3 and which is, perpendicular to the line 3 x + 5 y = 4., 12. Find the equation of the line which is perpendicular to the line 3 x + 2y = 8, and passes through the midpoint of the line joining the points (6, 4) and, ( 4, − 2)., 13. Find the equation of the line whose y-intercept is −3 and which is, perpendicular to the line joining the points ( −2, 3) and ( 4, −5)., 14. Find the equation of the line passing through ( −3 , 5) and perpendicular to, the line through the points (2, 5) and ( −3 , 6)., 15. A line perpendicular to the line segment joining the points (1, 0) and (2, 3), divides it in the ratio 1 : 2. Find the equation of the line., ANSWERS (EXERCISE 20D), , 1. (i) 3 x − y + 5 = 0 (ii) x + y − 4 = 0 (iii) 2x + 5 y + 15 = 0, 2. x − 3 y − 4 3 = 0, 4. x − y − 2 = 0 or x + y + 2 = 0, , 3. x + 3 y + 6 3 = 0, 5. x − y = 0 or x + y = 0
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678, , Senior Secondary School Mathematics for Class 11, , 6. 7 x + y + 2 = 0 7. 2x − 3 y + 15 = 0, , 8. 2x + y − 3 = 0, , 9. 3 x − 4y + 6 = 0, , 10. x = 2, 11. 5 x − 3 y + 15 = 0 12. 2x − 3 y − 7 = 0 13. 3 x − 4y − 12 = 0, 14. 5 x − y + 20 = 0, 15. 3 x + 9y − 13 = 0, HINTS TO SOME SELECTED QUESTIONS, 2. m = tan 30 ° =, 3. m = tan, , 1, and c = − 4., 3, , 5π, π, π −1, = tan ⎛⎜ π − ⎞⎟ = − tan =, ⋅ And, c = − 6., ⎝, 6, 6⎠, 6, 3, , 4. m = tan 45 ° = 1or m = tan 135 ° = tan ( 180 ° − 45 ° ) = − tan 45 ° = − 1. Also, c = − 2., 5. m = tan 45 ° = 1or m = tan 135 ° = − 1. And, c = 0., 6. Let the required equation be y = mx − 2., Since it passes through ( −1, 5 ), we have m × ( −1) − 2 = 5 ⇒ m = − 7 ., 2, 8, 2, x − ⋅ Its slope = ⋅, 3, 3, 3, 2, ∴ slope of the required line = and c = 5., 3, 1, 1, 5, 8. Given line is y = x + ⋅ Its slope is ⋅, 2, 2, 2, 1, m × = − 1 ⇒ m = − 2., 2, Let y = − 2 x + c be the required line., 7. Given line is y =, , Since it passes through (0, 3), we have 3 = ( −2 ) × 0 + c ⇒ c = 3., ∴, , m = − 2 and c = 3., , 10. Required line is parallel to the y-axis. So, its equation is x = 2 ., 11. Required line passes through ( −3 , 0 )., 5+ 3 ⎞, 12. Required line passes through the point ⎛⎜, , 0 ⎟ , i.e., (4, 0)., ⎝ 2, ⎠, 13. Let A( −2 , 3 ) and B( 4 , − 5 ) be the given points., Slope of AB =, ∴, , y=, , −5 − 3 −4, −4, 3, =, ⋅ So, m × ⎛⎜ ⎞⎟ = − 1 ⇒ m = ⋅, ⎝ 3⎠, 4+ 2, 3, 4, , 3, x− 3, 4, , [Q c = − 3]., , 14. Given points are A( 2 , 5 ) and B( −3 , 6 )., Slope of AB =, ∴, , 6−5, −1, −1, =, ⋅ So, m × ⎛⎜ ⎞⎟ = − 1 ⇒ m = 5., ⎝ 5 ⎠, −3 − 2 5, , required equation is, , y−5, = 5 ⇔ 5 x − y + 20 = 0., x+ 3
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Straight Lines, , 679, , 15. Let M divide the join of A( 1, 0 ) and B( 2 , 3 ) in the ratio 1 : 2., ⎛ 1× 2 + 2 × 1 1× 3 + 2 × 0 ⎞, ⎛4 ⎞, Then, the point M is ⎜, ,, ⎟ , i.e., M ⎜ , 1⎟ ⋅, ⎝3 ⎠, 1+ 2, 1+ 2, ⎝, ⎠, 3−0, Slope of AB =, = 3., 2−1, 1, ⎛ 4 ⎞, Let L be the required line. Then, slope of L = − and it passes through M ⎜ 1⎟ ⋅, ⎝ 3, ⎠, 3, Let L be the required line. Then, L ⊥ AB., 1, 4, slope of L = − and it passes through M ⎛⎜ , 1⎞⎟ ⋅, ∴, ⎝3 ⎠, 3, , EQUATION OF A LINE IN INTERCEPTS FORM, , Prove that the equation of a line making intercepts a and b on the x-axis, and the y-axis respectively, is, x y, + = 1., a b, Let X′ OX and YOY′ be the, coordinate axes., Let AB be the line cutting the, x-axis and the y-axis at A and B, respectively such that OA = a and, OB = b., Then, these points are A( a , 0) and, B( 0, b)., So, the equation of line AB is given by, y−0 b−0, [two-point form], =, x −a 0−a, y, b, =, ⇔ b( x − a) = − ay, ⇔, x − a −a, ⇔ bx + ay = ab, x y, [on dividing both sides by ab]., ⇔, + =1, a b, Thus, the line making intercepts a and b on the x- and y-axes, x y, respectively is + = 1., a b, , THEOREM 1, , PROOF, , SOLVED EXAMPLES, EXAMPLE 1, SOLUTION, , Find the equation of the line which makes intercepts 2 and −3 on the, x-axis and the y-axis respectively., We know that the equation of a line making intercepts a and b on the, x y, x-axis and y-axis respectively, is + = 1., a b
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680, , Senior Secondary School Mathematics for Class 11, , Here a = 2 and b = − 3., Hence, the required equation is, x, y, x y, +, =1 ⇔, − = 1 ⇔ 3 x − 2y − 6 = 0., 2 −3, 2 3, EXAMPLE 2, SOLUTION, , EXAMPLE 3, , SOLUTION, , EXAMPLE 4, , Find the equation of the line which passes through the point (3, 4) and the, sum of whose intercepts on the axes is 14., Let the intercepts made by the line on the x-axis and y-axis be a and, (14 − a) respectively., x, y, [intercept form]., Then, its equation is +, =1, a (14 − a), Since it passes through the point (3, 4), we have, 3, 4, +, = 1 ⇔ 3(14 − a) + 4 a = a(14 − a), a (14 − a), ⇔ a 2 − 13 a + 42 = 0 ⇔ ( a − 6)( a − 7) = 0, ⇔ a = 6 or a = 7., Now, a = b ⇔ b = 14 − 6 = 8., And, a = 7 ⇔ b = 14 − 7 = 7., So, the required equation is, x y, x y, + = 1 or, + =1, 6 8, 7 7, i.e., 4x + 3 y − 24 = 0 or x + y − 7 = 0., Find the equations of the lines which cut off intercepts on the axes whose, sum and product are 1 and −6 respectively., x y, Let the required equation be + = 1., a b, Then x-intercept = a and y-intercept = b., ... (i), ∴ a + b =1, and ab = −6, ... (ii), −6, from (ii) in (i), we get, Putting b =, a, 6, a − = 1 ⇔ a 2 − a − 6 = 0 ⇔ ( a − 3)( a + 2) = 0, a, ⇔ a = 3 or a = −2., Now, a = 3 ⇔ b = (1 − a) = (1 − 3) = − 2., And, a = − 2 ⇔ b = (1 − a) = (1 + 2) = 3., ∴ the required equation is, x, y, x, y, +, = 1 or, +, =1, 3 −2, −2 3, i.e., 2x − 3 y − 6 = 0 or 3 x − 2y + 6 = 0., If M( a , b) is the midpoint of a line segment intercepted between the axes,, x y, show that the equation of the line is + = 2., a b
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Straight Lines, SOLUTION, , Let the required equation of the line be, x, y, +, = 1., c d, Then, it cuts the x-axis and y-axis, at the points A( c, 0) and B( 0, d)., Let M( a , b) be the midpoint of AB., , 681, , ... (i), , c+ 0, 0+ d, = a and, =b, 2, 2, ⇔ c = 2a and d = 2b., So, the required equation is, x, y, x y, +, =1 ⇔, + = 2., 2a 2b, a b, Then,, , Hence, the required equation is, EXAMPLE 5, , x y, + = 2., a b, , Find the equation of a line which passes through the point ( −3 , 7) and, makes intercepts on the axes, equal in magnitude but opposite in sign., , SOLUTION, , Let the required line make intercepts a and −a on the x-axis and y-axis, respectively., x, y, ... (i), Then, its equation is +, = 1 ⇔ x − y = a., a −a, Since this line passes through the point ( −3 , 7), we have, a = ( −3 − 7) = − 10., So, the required equation of the line is, x, y, +, = 1 ⇔ − x + y = 10 ⇔ x − y + 10 = 0., −10 10, Hence, the required equation is x − y + 10 = 0., , EXAMPLE 6, , Find the intercepts cut off by the line 2x − y + 16 = 0 on the coordinate, axes., We have, x, y, 2x − y + 16 = 0 ⇔ 2x − y = −16 ⇔, +, =1, −8 16, [on dividing both sides by −16], , SOLUTION, , ∴, EXAMPLE 7, , SOLUTION, , the x-intercept = − 8 and the y-intercept = 16., , Find the equation of the line so that the line segment intercepted between, the axes is bisected at the point (2, 3)., Let AB be the given line segment making intercepts a and b on the, x-axis and y-axis respectively., x y, Then, the equation of line AB is + = 1., a b
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682, , Senior Secondary School Mathematics for Class 11, , Moreover, these points are A( a , 0), and B( 0, b). Let M( 2, 3) be the, midpoint of AB. Then,, a+0, 0+ b, = 2 and, =3, 2, 2, ⇒ a = 4 and b = 6., Hence, the required equation of, the given line is, x y, + = 1 ⇔ 3 x + 2y − 12 = 0., 4 6, Hence, the required equation is 3 x + 2y − 12 = 0., EXAMPLE 8, , SOLUTION, , Find the equation of the line so that, the line segment intercepted, between the axes is divided by the, point P( −5 , 4) in the ratio 1 : 2., Let AB be the given line segment, making intercepts a and b on the, x-axis and y-axis respectively., Then, the equation of line AB is, x y, + = 1., a b, So, these points are A( a , 0) and B( 0, b)., Now, P( −5 , 4) divides AB in the ratio 1 : 2., (1 × 0 + 2 × a), (1 × b + 2 × 0), −15, and b = 12., = − 5 and, =4 ⇔ a=, ∴, (1 + 2), (1 + 2), 2, So, the required equation of the line is, x, y, −2x, y, +, =1 ⇔, +, =1, −, 15, 15, 12, ⎛, ⎞ 12, ⎜, ⎟, ⎝ 2 ⎠, ⇔ 8x − 5 y + 60 = 0., Hence, the required equation of the line is 8x − 5 y + 60 = 0., , EXAMPLE 9, , SOLUTION, , x y, Find the equation of a line drawn perpendicular to the line + = 1, 4 6, through the point, where it meets the y-axis., x, y, The given line is, +, = 1., 4, 6, It meets the y-axis at B( 0, 6)., x y, Also, + = 1 ⇔ 3 x + 2y = 12, 4 6, −3, ⇔ y=, x + 6., 2, −3, Slope of the given line is, ⋅, 2
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Straight Lines, , 683, , Let the slope of the line perpendicular to the given line be m. Then,, 2, ⎛ 3⎞, m × ⎜− ⎟ = −1 ⇒ m =, ⎝ 2⎠, 3, , [Q m1m2 = −1]., , 2, and passes through the point, Now, the required line has slope, 3, B( 0, 6). So, its equation is, y−6 2, =, ⇔ 2x = 3 y − 18 ⇔ 2x − 3 y + 18 = 0., x−0 3, Hence, the required equation is 2x − 3 y + 18 = 0., Find the equation of the line passing through the point of intersection, of the lines 4x + 7 y − 3 = 0 and 2x − 3 y + 1 = 0, which has equal, intercepts on the axes., AN IMPORTANT NOTE The point of intersection of two lines is obtained by, solving their equations simultaneously., x y, SOLUTION Let the equation of the required line be, + = 1,, a a, i.e., x + y = a, ... (i), The equations of the given lines are:, EXAMPLE 10, , 4x + 7 y = 3, 2x − 3 y = − 1, On solving (ii) and (iii), we get x =, , ... (ii), ... (iii), 1, 5, ,y=, ⋅, 13, 13, , ⎛ 1 5⎞, So, the point of intersection of the given lines is P ⎜ , ⎟ ⋅, ⎝ 13 13 ⎠, ⎛ 1 5⎞, Now, line (ii) passes through the point P ⎜ , ⎟ ⋅, ⎝ 13 13 ⎠, 1, 5, Putting x =, and y =, in (i), we get, 13, 13, 5⎞, 6, ⎛ 1, a=⎜, + ⎟ =, ⋅, ⎝ 13 13 ⎠ 13, 6, Hence, the required equation is x + y =, ,, 13, i.e., 13 x + 13 y − 6 = 0., EXAMPLE 11, , SOLUTION, , Find the area of the triangle formed by the coordinate axes and the line, ax + by = 2ab., , Let X′ OX and YOY′ be the coordinate axes., x, y, Now, ax + by = 2ab ⇒, +, = 1., 2b 2a, Suppose that this line meets the x-axis and y-axis at A and B, respectively., Then, OA = 2b and OB = 2a.
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684, , Senior Secondary School Mathematics for Class 11, , ∴, , EXAMPLE 12, , SOLUTION, , ⎛1, ⎞, area of a BOA = ⎜ × OA × OB⎟ sq units, ⎝2, ⎠, ⎛1, ⎞, = ⎜ × 2b × 2a⎟ = 2ab sq units., ⎝2, ⎠, , The area of the triangle formed by the coordinate axes and a line is 6, square units and the length of its hypotenuse is 5 units. Find the equation, of the line., , Let X′ OX and YOY′ be the, coordinate axes, and let AB be the, part of the given line intercepted by, the axes., Let OA = a and OB = b., 1, Then, ab = 6 ⇔ ab = 12, 2, , ... (i), , ... (ii), and, a 2 + b 2 = 25, 12, Putting b =, from (i) in (ii), we get, a, 144, a 2 + 2 = 25 ⇒ a 4 − 25 a 2 + 144 = 0, a, ⇒ ( a 2 − 16)( a 2 − 9) = 0 ⇒ a 2 = 16 or a 2 = 9, ⇒ a = 4 or a = 3 [Q length is always positive]., Now, ( a = 4 ⇒ b = 3) and ( a = 3 ⇒ b = 4)., Hence, the required equation of the line is, x, y, x, y, +, = 1 or, +, =1, 4, 3, 3, 4, i.e., 3 x + 4y − 12 = 0 or 4x + 3 y − 12 = 0., , EXERCISE 20E, 1. Find the equation of the line which cuts off intercepts −3 and 5 on the x-axis, and y-axis respectively., 2. Find the equation of the line which cuts off intercepts 4 and −6 on the x-axis, and y-axis respectively.
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Straight Lines, , 685, , 3. Find the equation of the line that cuts off equal intercepts on the coordinate, axes and passes through the point (4, 7)., 4. Find the equation of the line which passes through the point ( 3 , − 5), and cuts off intercepts on the axes which are equal in magnitude but, opposite in sign., 5. Find the equation of the line passing through the point (2, 2) and cutting off, intercepts on the axes, whose sum is 9., 6. Find the equation of the line which passes through the point ( 22, − 6) and, whose intercept on the x-axis exceeds the intercept on the y-axis by 5., 7. Find the equation of the line whose portion intercepted between the axes is, bisected at the point ( 3 , − 2)., 8. Find the equation of the line whose portion intercepted between the, coordinate axes is divided at the point (5, 6) in the ratio 3 : 1., 9. A straight line passes through the point ( −5 , 2) and the portion of the line, intercepted between the axes is divided at this point in the ratio 2 : 3. Find, the equation of the line., x y, 10. If the straight line + = 1 passes through the points ( 8, − 9) and (12, −15),, a b, find the values of a and b., ANSWERS (EXERCISE 20E), , 1. 5 x − 3 y + 15 = 0 2. 3 x − 2y − 12 = 0 3. x + y − 11 = 0, 4. x − y − 8 = 0, 5. x + 2y − 6 = 0 or 2x + y − 6 = 0, 6. 6x + 11y − 66 = 0 or x + 2y − 10 = 0, 7. 2x − 3 y − 12 = 0 8. 2x + 5 y − 40 = 0 9. 3 x − 5 y + 25 = 0 10. a = 2, b = 3, HINTS TO SOME SELECTED QUESTIONS, 6. Let the required equation be, , y, x, + = 1., a+ 5 a, , Since it passes through ( 22 , − 6 ), we have, , 22, 6, − = 1., a+ b a, , This gives a2 − 11a + 30 = 0 ⇔ ( a − 6 )( a − 5 ) = 0 ⇔ a = 6 or, y, y, x, x, So, the required equations are, + = 1 or, + = 1., 11 6, 10 5, x y, 7. Let the required equation be + = 1., a b, Then, it cuts the axes at A( a, 0 ) and B( 0 , b )., ⎛ a + 0 = 3 and 0 + b = − 2 ⎞ ⇔ a = 6 , b = − 4., ∴, ⎜, ⎟, ⎝ 2, ⎠, 2, x y, + = 1., a b, Then, it cuts the axes at A( a, 0 ) and B( 0 , b )., Let P(5 , 6 ) divide AB in the ratio 3 : 1., , 8. Let the required equation be, , a = 5.
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686, , Senior Secondary School Mathematics for Class 11, , Then,, , 3 × 0 + 1× a, 3 × b + 1× 0, = 5 and, = 6 ⇔ a = 20 and b = 8., 3+ 1, 3+ 1, , x y, + = 1passes through the points A( 8 , − 9 ) and B( 12 , − 15 ),, a b, 8 9, 12 15, we have − = 1 and, −, = 1., a b, a, b, 3, 8, On solving, we get = 1 and = 4. This gives a = 2 , b = 3., b, a, , 10. Since, , EQUATION OF A LINE IN NORMAL FORM, , Let p be the length of perpendicular (or normal) from the origin to a given, non-vertical line L, and let α be the angle between the normal and the, positive direction of the x-axis. Then, prove that the equation of the line L, is given by, x cos α + y sin α = p., Let X′ OX and YOY′ be the coordinate axes and let L be the given line., Draw ON perpendicular to L., Let ON = p, and let the angle between the positive direction of the, x-axis and ON be α., Draw perpendicular NM on the x-axis., All possible positions of the line L are shown below., In each case, we have, , THEOREM 1, , PROOF
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Straight Lines, , 687, , OM, NM, = cos α and, = sin α, ON, ON, ⇔ OM = p cos α and NM = p sin α., Thus, the coordinates of N are ( p cos α , p sin α )., Moreover, line L is perpendicular to ON., −1, −1, − cos α, =, =, ⋅, ∴ slope of the line L =, slope of ON tan α, sin α, Thus, the line L passes through the point N ( p cos α , p sin α ) and has a, − cos α, slope m =, ⋅, sin α, So, the equation of the line L is given by, y − p sin α − cos α, [point–slope form], =, x − p cos α, sin α, ⇔ ( y − p sin α ) sin α = − cos α ( x − p cos α ), ⇔ x cos α + y sin α = p(cos2 α + sin 2 α ), ⇔ x cos α + y sin α = p , which is the required equation., SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Find the equation of a line whose perpendicular distance from the origin, is 5 units and the angle between the positive direction of the x-axis and, the perpendicular is 30°., Here p = 5 units and α = 30°., So, the equation of the given line in normal form is, x cos α + y sin α = p , where α = 30° and p = 5 units, , ⇔ x cos 30° + y sin 30° = 5
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688, , Senior Secondary School Mathematics for Class 11, , ⎛ 3⎞, ⎛ 1⎞, ⇔ x⎜, ⎟ + y⎜ ⎟ = 5, ⎝ 2⎠, 2, ⎝, ⎠, ⇔, , 3 x + y − 10 = 0,, which is the required equation., , EXAMPLE 2, , SOLUTION, , Find the equation of the line whose, perpendicular distance from the, origin is 3 units and the angle, between the positive direction of, x-axis and the perpendicular is 15°., Here p = 3 units and α = 15 °., So, the equation of the given line in, normal form is, x cos α + y sin α = p , whereα = 15 °, and p = 3 units, ⇔ x cos 15 ° + y sin 15 ° = 3, x( 3 + 1) y( 3 − 1), +, =3, 2 2, 2 2, [Q cos 15 ° = cos ( 45 ° − 30°) = cos 45 ° cos 30° + sin 45 ° sin 30°, sin 15 ° = sin ( 45 ° − 30°) = sin 45 ° cos 30° − cos 45 ° sin 30°], ⇔ ( 3 + 1) x + ( 3 − 1) y − 6 2 = 0, which is the required equation., ⇔, , EXAMPLE 3, , SOLUTION, , Find the equation of a line whose perpendicular distance from the origin, is 8 units and the angle between the positive direction of the x-axis and, the perpendicular is 135°., Here p = 8 units and α = 135 °., So, the equation of the given line in, normal form is, x cos α + y sin α = p ,, where α = 135 ° and p = 8 units, ⇔ x cos 135 ° + y sin 135 ° = 8, ⎛ −1 ⎞, ⎛ 1 ⎞, ⇔ x ⎜ ⎟ + y ⎜ ⎟ = 18, ⎝ 2⎠, ⎝ 2⎠, [Q cos 135 ° = cos (180° − 45 °) = − cos 45 °, sin 135 ° = sin (180° − 45 °) = sin 45 °], ⇔ − x + y = 4 ⇔ x − y + 4 = 0, which is the required equation., , EXAMPLE 4, , Find the equation of a line whose perpendicular distance from the origin, is 2 units and the angle between the perpendicular segment and the, positive direction of the x-axis is 240°.
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Straight Lines, SOLUTION, , 689, , Here, p = 2 units and α = 240°., , So, the equation of the given line in normal form is, x cos α + y sin α = p , where α = 240° and p = 2 units, ⇔ x cos 240° + y sin 240° = 2, ⎛− 3⎞, ⎛ −1⎞, ⇔ x⎜ ⎟ + y⎜, ⎟ = 2 [Q cos 240° = cos (180° + 60°) = − cos 60°, ⎝ 2⎠, ⎝ 2 ⎠, sin 240° = sin (180° + 60°) = − sin 60°], ⇔ −x − 3 y = 4 ⇔ x +, , 3 y + 4 = 0, which is the required equation., , EXERCISE 20F, 1. Find the equation of the line for which, (i) p = 3 and α = 45 °, (ii) p = 5 and α = 135 °, (iii) p = 8 and α = 150°, (iv) p = 3 and α = 225 °, (v) p = 2 and α = 300°, (vi) p = 4 and α = 180°, 2. The length of the perpendicular segment from the origin to a line is 2 units, 1, and the inclination of this perpendicular is α such that sin α = and α is, 3, acute. Find the equation of the line., 3. Find the equation of the line which is at a distance of 3 units from the origin, 5, such that tan α = , where α is the acute angle which this perpendicular, 12, makes with the positive direction of the x-axis., ANSWERS (EXERCISE 20F), , 1. (i) x + y − 3 2 = 0 (ii) x − y + 5 2 = 0 (iii) 3 x − y + 16 = 0, (iv) x + y + 3 2 = 0 (v) x −, 2. 2 2x + y − 6 = 0, , 3 y − 4 = 0 (vi) x + 4 = 0, 3. 12x + 5 y − 39 = 0
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690, , Senior Secondary School Mathematics for Class 11, , GENERAL EQUATION OF A LINE, , Prove that the equation Ax + By + C = 0 always represents a line,, provided A and B are not simultaneously zero., Let Ax + By + C = 0, where A ≠ 0 or B ≠ 0., CASE 1 When A = 0 and B ≠ 0, In this case, the given equation becomes, ⎛ −C ⎞, By + C = 0 ⇒ y = ⎜ ⎟ ,, ⎝ B⎠, which represents a line parallel to the x-axis., CASE 2 When A ≠ 0 and B = 0, In this case, the given equation becomes, ⎛ −C ⎞, Ax + C = 0 ⇒ x = ⎜ ⎟ ,, ⎝A⎠, which represents a line parallel to the y-axis., CASE 3 When A ≠ 0 and B ≠ 0, In this case, we have, Ax + By + C = 0 ⇒ By = − Ax + ( −C), ⎛ −A ⎞, ⎛ −C ⎞, ⇒ y=⎜, ⎟ x+⎜ ⎟⋅, ⎝ B ⎠, ⎝ B⎠, This is clearly an equation of a line in slope–intercept form,, −C, ⎛ −A ⎞, ⋅, y = mx + c, where m = ⎜, ⎟ and c =, ⎝ B ⎠, B, Thus, Ax + By + C = 0, when A ≠ 0 or B ≠ 0, always represents, a line., , THEOREM 1, , PROOF, , Prove that every line has an equation of the form Ax + By + c = 0, where, A , B , C are constants., For a given line, there are two possibilities., CASE 1 The given line is parallel to the y-axis., In this case, the equation is of the form, x = c ., Now, x = c ⇔ 1 ⋅ x + 0 ⋅ y − c = 0, ⇔ Ax + By + C = 0, where C = − c ., Thus, it is of the form Ax + By + C = 0., CASE 2 The given line is not parallel to the y-axis., In this case, the equation is of the form, y = mx + c ⇒ mx − y + c = 0, ⇒ Ax + By + C = 0,, where A = m , B = − 1 and C = c ., Thus, it takes the form Ax + By + C = 0., , THEOREM 2, , PROOF, , REDUCTION OF GENERAL FORM TO STANDARD FORM, , Let the given equation of the line be Ax + By + C = 0.
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Straight Lines, , 691, , I. SLOPE–INTERCEPT FORM:, , Ax + By + C = 0 ⇒ By = − Ax − C, ⎛ −A ⎞, ⎛ −C ⎞, ⇒ y=⎜, ⎟x + ⎜ ⎟ ⋅, ⎝ B ⎠, ⎝ B⎠, , −A, −C, and c =, ⋅, B, B, This is clearly the equation of a line in slope–intercept form, y = mx + c,, −A, −C, where m =, and c =, ⋅, B, B, ⇒ y = mx + c, where m =, , II. INTERCEPTS FORM:, , Ax + By + C = 0 ⇒ Ax + By = − C, ⎛A⎞, ⎛ B⎞, ⇒ ⎜ ⎟ x + ⎜ ⎟ y = 1 [on dividing both sides by −C], ⎝ −C ⎠, ⎝ −C ⎠, x, y, ⇒, +, =1, ⎛ −C ⎞, ⎛ −C ⎞, ⎜ ⎟, ⎜ ⎟, ⎝A⎠, ⎝ B⎠, x y, −C, −C, and b =, ⇒, + = 1, where a =, ⋅, a b, A, B, x, y, This is clearly the equation of a line in intercept form, +, = 1,, a b, −C, −C, where a =, and b =, ⋅, A, B, III. NORMAL FORM:, , Ax + By + C = 0 ⇒, , A, A +B, 2, , 2, , x+, , B, A +B, 2, , 2, , ⋅y +, , C, A + B2, 2, , =0, , [on dividing throughout by A 2 + B 2 ], −A, −B, C, ⇒, x+, y=, 2, 2, 2, 2, 2, A +B, A +B, A + B2, ⇒ x cos α + y sin α = p ,, −A, −B, C, where cos α =, and p =, , sin α =, ⋅, A 2 + B2, A 2 + B2, A 2 + B2, This is the equation of a line in normal form, x cos α + ysin α = p,, −A, −B, C, and p =, where cos α =, , sin α =, ⋅, A 2 + B2, A 2 + B2, A 2 + B2, SOLVED EXAMPLES, EXAMPLE 1, , Reduce the equation, , 3 x + y + 2 = 0 to, , (i) slope–intercept form and find the slope and y-intercept., (ii) intercepts form and find the intercepts on the axes.
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692, SOLUTION, , Senior Secondary School Mathematics for Class 11, , We have, (i) 3 x + y + 2 = 0 ⇒ y = − 3 x − 2., This is of the form y = mx + c, where m = − 3 and c = −2., ∴, , y = − 3 x − 2 is in slope–intercept form., , Here, slope = − 3 and y-intercept = −2., (ii), , 3x + y + 2 = 0 ⇒, , 3x + y = −2, , ⎛ 3⎞, ⎛ 1⎞, ⇒ ⎜, ⎟ x + ⎜ ⎟ y =1, ⎝ −2⎠, −, 2, ⎝, ⎠, x, y, ⇒, +, = 1., −2, ⎛ −2 ⎞, ⎜, ⎟, ⎝ 3⎠, −2, x, y, This is of the form +, and b = −2., = 1, where a =, a b, 3, x, y, Thus,, +, = 1 is in intercepts form., −2, ⎛ −2 ⎞, ⎜, ⎟, ⎝ 3⎠, −2, and y-intercept = −2., Here, x-intercept =, 3, EXAMPLE 2, , SOLUTION, , Reduce the equation 3 x − 2y + 4 = 0 to intercepts form and find the, length of the segment intercepted between the axes., We have, 3 x − 2y + 4 = 0 ⇒ 3 x − 2y = − 4, 3x y, + = 1 [on dividing both sides by −4], −4 2, x, y, ⇒, + =1, −, 4, 2, ⎛ ⎞, ⎜ ⎟, ⎝ 3⎠, x y, −4, and b = 1., ⇒, + = 1, where a =, a b, 3, ⇒
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Straight Lines, , 693, , x, y, + = 1 is the required equation in intercepts form., 2, ⎛ −4⎞, ⎜ ⎟, ⎝ 3⎠, −4, x-intercept =, and y-intercept = 2., 3, If AB is the part of the line intercepted between the axes then the end, ⎛ −4 ⎞, points of this line segment are A ⎜ , 0⎟ and B( 0, 2)., ⎝ 3 ⎠, , ∴, , 2, , Length AB =, EXAMPLE 3, , 16, 2, ⎛ −4, ⎞, − 0⎟ + ( 0 − 2) 2 =, +4=, 13 units., ⎜, ⎝ 3, ⎠, 9, 3, , Reduce the equation, , 3 x + y + 2 = 0 to the normal form, , x cos α + y sin α = p, and hence find the values of α and p., SOLUTION, , We have, 3x + y + 2 = 0 ⇒ − 3x − y = 2, , [keeping constant +ve], , ⎛− 3⎞, ⎛ −1⎞, ⇒ ⎜, ⎟x + ⎜ ⎟y =1, ⎝ 2⎠, ⎝ 2 ⎠, [on dividing throughout by ( 3 ) 2 + 12 ], ⇒ x cos α + y sin α = p ,, − 3, −1, and p = 1., , sin α =, 2, 2, Since cos α < 0 and sin α < 0, α lies in the third quadrant., sin α ⎛ −1⎞ ⎛ −2 ⎞, 1, Now, tan α =, = ⎜ ⎟ ×⎜, = tan (180° + 30°) = tan 210°, ⎟=, cos α ⎝ 2 ⎠ ⎝ 3 ⎠, 3, ⇒ α = 210°., where cos α =, , Thus, α = 210° and p = 1., Hence, the given equation in normal form is given by, x cos 210° + y sin 210° = 1., EXAMPLE 4, , Reduce the equation x +, , 3 y + 5 = 0 to the normal form, , x cos α + y sin α = p, and hence find the values of α and p., SOLUTION, , We have, x + 3y + 5 = 0 ⇒ −x − 3y = 5, , [keeping constant +ve], , ⎛− 3⎞, 5, ⎛ −1⎞, ⇒ ⎜ ⎟x + ⎜, ⎟y =, ⎝ 2⎠, 2, 2, ⎝, ⎠, [on dividing throughout by ( −1) 2 + ( − 3 ) 2 ], ⇒ x cos α + y sin α = p ,, 5, −1, − 3, and p = ⋅, where cos α =, , sin α =, 2, 2, 2
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694, , Senior Secondary School Mathematics for Class 11, , Since cos α < 0 and sin α < 0, α lies in third quadrant., Now, tan α =, ∴, , sin α ⎛ − 3 ⎞, =⎜, ⎟ × ( −2) =, cos α ⎝ 2 ⎠, , 3 = tan (180° + 60°) = tan 240°, , α = 240°., , 5, ⋅, 2, Hence, the given equation in normal form is, 5, x cos 240° + y sin 240° = ⋅, 2, , Thus, α = 240° and p =, , EXAMPLE 5, , SOLUTION, , Reduce the equation y + 4 = 0 to the normal form x cos α + y sin α = p,, and hence find the values of α and p., We have, [keeping constant positive], y + 4= 0 ⇒ −y = 4, ⇒ x cos α + y sin α = p ,, where cos α = 0, sin α = − 1 and p = 4., Now, (cos α = 0 and sin α = − 1) ⇒ α = 270°., Hence, the required normal form of the given equation is, x cos 270° + y sin 270° = 4., Clearly, α = 270° and p = 4., , EXERCISE 20G, 1. Reduce the equation 2x − 3 y − 5 = 0 to slope–intercept form, and find from, it the slope and y-intercept., 2. Reduce the equation 5 x + 7 y − 35 = 0 to slope–intercept form, and hence, find the slope and the y-intercept of the line., 3. Reduce the equation y + 5 = 0 to slope–intercept form, and hence find the, slope and the y-intercept of the line., 4. Reduce the equation 3 x − 4y + 12 = 0 to intercepts form. Hence, find the, length of the portion of the line intercepted between the axes., 5. Reduce the equation 5 x − 12y = 60 to intercepts form. Hence, find the, length of the portion of the line intercepted between the axes., 6. Find the inclination of the line:, (iii) 3 x − y − 4 = 0, (i) x + 3 y + 6 = 0 (ii) 3 x + 3 y + 8 = 0, 7. Reduce the equation x + y − 2 = 0 to the normal form x cos α + y sin α = p,, and hence find the values of α and p., 8. Reduce the equation x + 3 y − 4 = 0 to the normal form x cos α + y sin α = p,, and hence find the values of α and p., 9. Reduce each of the following equations to normal form:, (i) x + y − 2 = 0, (iv) 2y − 3 = 0, , (ii) x + y + 2 = 0, (v) 4x + 3 y − 9 = 0, , (iii) x + 5 = 0
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Straight Lines, , 695, ANSWERS (EXERCISE 20G), , 2, 5, 2, −5, 1. y = x − , m = and c =, 3, 3, 3, 3, , −5, −5, 2. y =, and c = 5, x + 5, m =, 7, 7, x, y, 4., + = 1, 5 units, −4 3, , 3. y = 0 ⋅ x − 5 , m = 0 and c = − 5, 5., , x, y, +, = 1, 13 units, 12 −5, , 6. (i) 150° (ii) 135° (iii) 60°, , 7. x cos 45 ° + y sin 45 ° = 1; α = 45 ° , p = 1, 8. x cos 60° + y sin 60° = 2; α = 60° , p = 2, 9. (i) x cos 45 ° + y sin 45 ° = 2, (iii) x cos 180° + y sin 180° = 5, , (ii) x cos 225 ° + y sin 225 ° = 1, (iv) x cos 90° + y sin 90° =, , (v) x cos α + y sin α = p , where cos α =, , 3, 2, , 4, 3, 9, , sin α = and p =, 5, 5, 5, , DISTANCE OF A POINT FROM A LINE, THEOREM 3, , Prove that the length of perpendicular from a given point P( x1 , y1 ) on a, line Ax + By + C = 0, is given by, d=, , PROOF, , |Ax1 + By1 + C|, A 2 + B2, , ⋅, , Let L be the line represented by the equation Ax + By + C = 0 and let, P( x , y) be a given point outside it., Let this line L intersect the x-axis and, y-axis at the points Q and R, respectively, and let PM ⊥ QR., Now,, Ax + By + C = 0 ⇒ Ax + By = −C, Ax By, +, =1, −C −C, x, y, ⇒, +, = 1., −, C, −, ⎛, ⎞ ⎛ C⎞, ⎜, ⎟ ⎜, ⎟, ⎝ A⎠ ⎝ B⎠, ⇒, , Thus, the given line intersects the x-axis and the y-axis at the points, ⎛ −C ⎞, ⎛ −C ⎞, Q⎜, , 0⎟ and R ⎜ 0,, ⎟⋅, ⎝, ⎝ A, ⎠, B⎠, Draw PM ⊥ QR.
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696, , Senior Secondary School Mathematics for Class 11, , 1 ⏐, C ⎞⏐, ⎛ −C, ⎞, ⎛ −C ⎞ ⎛, ⋅⏐x1 ⎜, − 0⎟ + 0 ⋅ ( 0 − y1 ) + ⎜, ⎟ ⋅ ⎜ y + ⎟⏐, ⎠, ⎝ A ⎠ ⎝ 1 B ⎠⏐, 2 ⏐ ⎝ B, y, C ⎞⏐, 1 ⏐ ⎛x, 1 C, = ⋅⏐− C ⎜ 1 + 1 +, ⎟⏐ = ⋅⏐ ( Ax1 + By1 + C)⏐, A, AB ⎠⏐ 2 ⏐AB, 2 ⏐ ⎝B, ⏐, C, ... (i), =, ( Ax1 + By1 + C), 2AB, , Area of a PQR =, , Also, area of a PQR =, =, , 1, 1, × QR × PM = × PM × (OQ) 2 + (OR) 2, 2, 2, 1, × PM ×, 2, , 2, , ⎛ −C ⎞, ⎛ −C ⎞, ⎜, ⎟ +⎜, ⎟, ⎝ A⎠, ⎝ B⎠, , 2, , −C, −C ⎤, ⎡, ⎢⎣Q OQ = A and OR = B ⎥⎦, 1, × PM ×, 2, From (i) and (ii), we get, =, , C2 C2, +, A 2 B2, , ... (ii), , C2 C2 ⏐ C, 1, × PM ×, +, =, ⋅ ( Ax1 + By1 + C )⏐, 2, A 2 B 2 ⏐2AB, ⏐, C, C, 1, ⇒, ×, × A 2 + B 2 × PM =, ⋅|Ax1 + By1 + C|, 2 AB, 2AB, ⇒ PM =, , |Ax1 + By1 + C|, , Hence, d =, REMARK, , A 2 + B2, , ⋅, , |Ax1 + By1 + C|, , where PM = d., A 2 + B2, , Clearly, the length of perpendicular from the origin on the line, |C|, ⋅, Ax + By + C = 0 is, A 2 + B2, , DISTANCE BETWEEN TWO PARALLEL LINES, THEOREM 4, , SOLUTION, , Prove that the distance between two parallel lines Ax + By + C1 = 0 and, Ax + By + C 2 = 0 is given by, |C 2 − C1|, d=, ⋅, A 2 + B2, Let L1 and L2 be two given parallel lines represented by, Ax + By + C1 = 0, , ... (i), , and Ax + By + C 2 = 0., , ... (ii), , ⎛ C ⎞, Putting y = 0 in (i), we get x = ⎜ − 1 ⎟ ⋅, ⎝ A⎠
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Straight Lines, , 697, , ⎛ C, ⎞, Thus, L1 intersects the x-axis at Q = ⎜ − 1 , 0⎟ ⋅, ⎝ A ⎠, Let QR ⊥ L2 and let d =|QR|. Then,, ⎛ C, ⎞, d = length of perpendicular from Q ⎜ − 1 , 0⎟ on Ax + By + C 2 = 0, ⎝ A ⎠, ⏐A × ( − C1 ) + B × 0 + C ⏐, 2, A, ⏐ |C 2 − C1|, ⏐, =, ⋅, ⇒ d=, 2, 2, A +B, A 2 + B2, Distance between two parallel lines y = mx + C1 and y = mx + C 2 is, given by, |C − C1|, d= 2, ⋅, 1 + m2, , REMARK, , SUMMARY, , (i) Length of perpendicular from a point P( x1 , y1 ) on the, |Ax1 + By1 + C|, line Ax + By + C = 0 is given by, d =, ⋅, A 2 + B2, (ii) Distance between two parallel lines Ax + By + C1 = 0, |C 2 − C1|, and Ax + By + C 2 = 0 is given by, d =, ⋅, A 2 + B2, (iii) Distance between y = mx + C1 and y = mx + C 2 is, |C − C1|, given by, d = 2, ⋅, 1 + m2, SOLVED EXAMPLES, EXAMPLE 1, SOLUTION, , Find the distance of the point (4, 1) from the line 3 x − 4y + 12 = 0., Clearly, the required distance is the length of perpendicular from the, point P( 4, 1) on the line 3 x − 4y + 12 = 0.
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698, , Senior Secondary School Mathematics for Class 11, , Let the required distance be d. Then,, d=, , |3 × 4 − 4 × 1 + 12|, 3 + ( −4), 2, , 2, , =, , 20, units = 4 units., 5, , EXAMPLE 2, , Find the distance of the point ( −1, 1) from the line 12( x + 6) = 5( y − 2)., , SOLUTION, , The given point is P( −1, 1) and the given line is 12x − 5 y + 82 = 0., Let the required distance be d. Then,, d = length of perpendicular from P( −1, 1) on the line12x − 5 y + 82 = 0, =, , EXAMPLE 3, , SOLUTION, , EXAMPLE 4, SOLUTION, , |12 × ( −1) − 5 × 1 + 82|, (12) + ( −5), 2, , =, , 65, 65, =, = 5 units., 169 13, , Find the length of perpendicular from the point ( a , b) to the line, x y, + = 1., a b, The given point is P( a , b) and the given line is bx + ay − ab = 0., Let d be the length of perpendicular from P( a , b) to the line, bx + ay − ab = 0., |b × a + a × b − ab|, |ab|, units., Then, d =, =, 2, 2, b +a, a 2 + b2, Find the length of perpendicular from the origin to the line, 4x + 3 y − 2 = 0., The given point is P( 0, 0) and the given line is 4x + 3 y − 2 = 0., Led d be the length of perpendicular from P( 0, 0) to the line, 4x + 3 y − 2 = 0., Then, d =, , EXAMPLE 5, , 2, , |4 × 0 + 3 × 0 − 2|, 42 + 3 2, , =, , 2, unit., 5, , If p is the length of perpendicular from the origin to the line whose, intercepts on the axes are a and b then show that, 1, 1, 1, =, +, ⋅, p 2 a 2 b2, , SOLUTION, , The equation of the line making intercepts a and b on the axes is, given by, x, y, x y, ... (i), +, = 1, i.e., + − 1 = 0., a b, a b, Since p is the length of perpendicular fromO(0, 0) to line (i), we have, ⏐1 × 0 + 1 × 0 − 1⏐, b, ⏐a, ⏐, =, p=, 1, 1, +, a 2 b2, , |−1|, =, 1, 1, +, a 2 b2, , 1, 1, 1, +, a 2 b2
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700, EXAMPLE 7, , SOLUTION, , Senior Secondary School Mathematics for Class 11, , If p1 and p 2 are the lengths of perpendiculars from the origin to the line, xsec θ + ycosec θ = a and x cos θ − y sin θ = a cos 2θ respectively then, prove that 4p12 + p 22 = a 2 ., We have, xsec θ + ycosec θ = a ⇒, , x, y, +, − a = 0., cos θ sin θ, , ... (i), , Now, p1 is the length of perpendicular from the origin to line (i), so, we have, ⏐, ⏐ 1 × 0 + 1 × 0 − a⏐, ⏐, sin θ, |a|, ⏐cos θ, ⏐, p1 =, sin 2θ, = |a|sin θ cos θ =, 2, 1, 1, +, cos2 θ sin 2 θ, ⇒ 2 p1 =|a|sin 2θ, ⇒, , 4 p12 = a 2 sin 2 2θ., , ... (ii), , The other line is x cos θ − y sin θ − a cos 2θ = 0., ... (iii), Now, p 2 is the length of perpendicular from the origin to line (ii), so, we have, |0 × cos θ − 0 × sin θ − a cos 2θ|, p2 =, = |a cos 2θ|, cos 2 θ + sin 2 θ, ⇒, , p 22 = a 2 cos2 2θ., , ... (iv), , Adding (ii) and (iv), we get, 4p12 + p 22 = a 2 (sin 2 2θ + cos2 2θ), ⇒, EXAMPLE 8, , SOLUTION, , 4p12 + p 22 = a 2 ., , What are the points on the y-axis whose perpendicular distance from the, x y, line − = 1 is 3 units?, 3 4, x y, ... (i), − = 1 ⇒ 4x − 3 y − 12 = 0., 3 4, Let P( 0, y) be a point on the y-axis whose perpendicular distance, from the line 4x − 3 y − 12 = 0 is 3 units. Then,, |4 × 0 − 3 y − 12|, = 3 ⇒ |−3 y − 12| = 15 ⇒ |−3( y + 4)| = 15, 42 + ( −3) 2, ⇒ |− ( y + 4)| = 5 ⇒ |y + 4| = 5, ⇒ ( y + 4 = 5) or ( y + 4 = − 5), ⇒ y = 1 or y = − 9., Hence, the required points are (0, 1) and ( 0, − 9)., , EXAMPLE 9, , Find the distance between the parallel lines 15 x + 8y − 34 = 0 and, 15 x + 8y + 31 = 0.
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Straight Lines, SOLUTION, , 701, , Converting each of the given equations to the form y = mx + C , we get, −15, 17, ... (i), x+, 8, 4, −15, 31, ... (ii), 15 x + 8y + 31 = 0 ⇒ y =, x−, 8, 8, Clearly, the slopes of the given lines are equal and so they are parallel., 15 x + 8y − 34 = 0 ⇒ y =, , The given lines are of the form y = mx + C1 and y = mx + C 2 , where, −15, 17, −31, and C 2 =, m=, , C1 =, ⋅, 4, 8, 8, ∴, , distance between the given lines, =, , =, , |C 2 − C1|, 1+m, , 2, , , where m =, , −31 17, −, 8, 4, ⎛ −15 ⎞, 1+ ⎜, ⎟, ⎝ 8 ⎠, , 2, , −15, 17, −31, and C 2 =, , C1 =, 8, 4, 8, , ⎛ 65 ⎞, ⏐− 65⏐, ⎜ ⎟, ⎝ 8 ⎠ ⎛ 65, 8 ⎞ 65, ⏐ 8⏐, units., =, =, =⎜, × ⎟ =, 17 ⎠ 17, 225, 289 ⎝ 8, 1+, 64, 64, , Hence, the distance between the given lines is, EXAMPLE 10, SOLUTION, , 65, units., 17, , Find the distance between the lines 3 x − 4y + 9 = 0 and 6x − 8y − 17 = 0., , Converting each of the given equations to the form y = mx + C , we get, 3, 9, ... (i), 3 x − 4y + 9 = 0 ⇒ y = x +, 4, 4, 3, 17, ... (ii), x−, 4, 8, Clearly, the slopes of the given lines are equal and so they are parallel., 6x − 8y − 17 = 0 ⇒ y =, , Let the given lines be y = mx + C1 and y = mx + C 2 ,, 3, 9, −17, m = , C1 = and C 2 =, ⋅, 4, 4, 8, |C − C1|, ∴ distance between given lines = 2, 1 + m2, ⏐−17 − 9⏐ ⏐−35⏐, 4⏐ ⏐ 8 ⏐ ⎛ 35, ⏐ 8, =, =, =⎜, ×, ⎝ 8, ⎛ 5⎞, 9, ⎜ ⎟, 1+, ⎝ 4⎠, 16, =, , 7, units., 2, , Hence, the distance between the given lines is, , 7, units., 2, , where, , 4⎞, ⎟, 5⎠
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702, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 11, , SOLUTION, , Prove that the line 5 x − 2y − 1 = 0 is mid-parallel to the lines, 5 x − 2y − 9 = 0 and 5 x − 2y + 7 = 0., , Converting each of the given equations to the form y = mx + C ,, we get, 5, 1, ... (i), 5 x − 2y − 1 = 0 ⇒ y = x −, 2, 2, 5, 9, ... (ii), 5 x − 2y − 9 = 0 ⇒ y = x −, 2, 2, 5, 7, ... (iii), 5 x − 2y + 7 = 0 ⇒ y = x +, 2, 2, Clearly, the slope of (i) is equal to the slope of each of (ii) and (iii)., So, the line (i) is parallel to each of the given lines (ii) and (iii)., Let the given lines be y = mx + C , y = mx + C1 and y = mx + C 2, 5, 1, 9, 7, respectively. Then, m = , C = − , C1 = − and C 2 = ⋅, 2, 2, 2, 2, Let d1 and d 2 be the distances of (i) from (ii) and (iii) respectively., ⏐− 9 + 1 ⏐, ( 4 × 2), 2, 8, ⏐ 2 2⏐, Then, d1 =, units, = |− 4|⋅, =, =, =, 2, 29, 29, 29, 25, 1+m, 1+, 4, ⏐7 + 1 ⏐, |C 2 − C| ⏐ 2 2⏐ ⎛, 2 ⎞, 8, units., and d 2 =, =, = ⎜4 ×, ⎟ =, 2, ⎠, ⎝, 25, 29, 29, 1+m, 1+, 4, Thus, d1 = d 2 ., This shows that (i) is equidistant from (ii) and (iii)., Hence, 5 x − 2y − 1 = 0 is mid-parallel to the lines 5 x − 2y − 9 = 0 and, 5 x − 2y + 7 = 0., |C1 − C|, , EXAMPLE 12, , SOLUTION, , Find the equation of the line midway between the parallel lines, 9x + 6y − 7 = 0 and 3 x + 2y + 6 = 0., , Converting each of the given equations to the form y = mx + C ,, we get, −3, 7, ... (i), 9x + 6y − 7 = 0 ⇒ y =, x+, 2, 6, −3, ... (ii), 3 x + 2y + 6 = 0 ⇒ y =, x−3, 2, −3, Clearly, the slope of each one of the given lines is, ⋅, 2, Let the given lines be y = mx + C1 and y = mx + C 2 ., −3, 7, Then, m =, , C1 = and C 2 = −3., 2, 6
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Straight Lines, , 703, , Let L be the required line. Then, L is parallel to each one of (i) and (ii),, and equidistant from each one of them., −3, ⋅, ∴ slope of L =, 2, −3, ... (iii), Let the equation of L be y =, x+C, 2, Then, distance between (i) and (iii) must be equal to the distance, between (ii) and (iii)., |C1 − C| |C 2 − C|, ∴, =, ⇒ |C1 − C| = |C 2 − C|, 1+m2, 1 + m2, 7, 7, ⇒ ⏐ − C⏐ = |− 3 − C| ⇒ ⏐ − C⏐ = |3 + C|, ⏐6, ⏐, ⏐6, ⏐, 7, −11, −11, ⇒ − C = 3 + C ⇒ 2C =, ⇒C=, ⋅, 6, 6, 12, ∴, , equation of L is y =, , −3, 11, x − , i.e., 18x + 12y + 11 = 0., 2, 12, , Hence, the line 18x + 12y + 11 = 0 is midway between the parallel, lines 9x + 6y − 7 = 0 and 3 x + 2y + 6 = 0., EXAMPLE 13, , SOLUTION, , Find the coordinates of a point on the line x + y + 3 = 0, whose distance, from the line x + 2y + 2 = 0 is 5., , Let the required point be P( a , b). Then,, a + b + 3 = 0., Also,, , |a + 2b + 2|, 12 + 22, , ... (i), , = 5 ⇒ |a + 2b + 2| = 5, , ⇒ ( a + 2b + 2 = 5) or ( a + 2b + 2 = − 5), ⇒ a + 2b = 3 or a + 2b = − 7., Thus, a + 2b = 3, ... (ii), and, ... (iii), a + 2b = − 7, On solving (i) and (ii), we get a = −9 and b = 6., On solving (i) and (iii), we get a = 1 and b = −4., Hence, the required points are A( −9, 6) and B(1, − 4)., , EXERCISE 20H, 1., 2., 3., 4., , Find the distance of the point ( 3 , − 5) from the line 3 x − 4y = 27., Find the distance of the point ( −2, 3) from the line 12x = 5 y + 13., Find the distance of the point ( −4, 3) from the line 4( x + 5) = 3( y − 6)., Find the distance of the point (2, 3) from the line y = 4., , 5. Find the distance of the point (4, 2) from the line joining the points (4, 1) and, (2, 3).
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704, , Senior Secondary School Mathematics for Class 11, , 6. Find the length of perpendicular from the origin to each of the following, lines:, (i) 7 x + 24y = 50 (ii) 4x + 3 y = 9 (iii) x = 4, 7. Prove that the product of the lengths of perpendiculars drawn from, x, y, the points A( a 2 − b 2 , 0) and B( − a 2 − b 2 , 0) to the line cos θ + sin θ = 1,, a, b, is b 2 ., 8. Find the values of k for which the length of perpendicular from the point, (4, 1) on the line 3 x − 4y + k = 0 is 2 units., 9. Show that the length of perpendicular from the point (7 , 0) to the line, 5 x + 12y − 9 = 0 is double the length of perpendicular to it from the point, (2, 1)., 10. The points A( 2, 3), B( 4, − 1) and C( −1, 2) are the vertices of a ABC. Find the, length of perpendicular from C on AB and hence find the area of a ABC., 11. What are the points on the x-axis whose perpendicular distance from the, x y, line + = 1 is 4 units?, 3 4, 12. Find all the points on the line x + y = 4 that lie at a unit distance from the, line 4x + 3 y = 10., 13. A vertex of a square is at the origin and its one side lies along the line, 3 x − 4y − 10 = 0. Find the area of the square., 14. Find the distance between the parallel lines 4x − 3 y + 5 = 0 and, 4x − 3 y + 7 = 0., 15. Find the distance between the parallel lines 8x + 15 y − 36 = 0 and, 8x + 15 y + 32 = 0., 16. Find the distance between the parallel lines y = mx + c and y = mx + d., 17. Find the distance between the parallel lines p( x + y) + q = 0 and, p( x + y) − r = 0., 18. Prove that the line 12x − 5 y − 3 = 0 is mid-parallel to the lines, 12x − 5 y + 7 = 0 and 12x − 5 y − 13 = 0., 19. The perpendicular distance of a line from the origin is 5 units and its slope, is −1. Find the equation of the line., ANSWERS (EXERCISE 20H), , 2, 1. unit, 5, , 2. 4 units, , 2, units, 6. (i) 2 units, 2, 7, units, 7 sq units, 10., 5, 5., , 13, 3., units, 5, (ii), , 9, units, 5, , 4. 1 unit, , (iii) 4 units 8. k = 2 or k = − 18, , 11. (8, 0) and ( −2, 0)
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Straight Lines, , 12. (3, 1) and ( −7 , 11), , 13. 4 sq units, , 15. 4 units, , 16., , 705, , 2, units, 5, |q + r|, 17., 2p, , 14., , |d − c|, 1 + m2, , 19. x + y + 5 2 = 0 or x + y − 5 2 = 0, HINTS TO SOME SELECTED QUESTIONS, 1. The given line is 0 ⋅ x + 1 ⋅ y − 4 = 0., 5. Equation of the line joining the points A( 4 , 1) and B( 2 , 3 ) is, y−1 3−1, =, , i.e., x + y − 5 = 0., x−4 2−4, 6. (iii) The given line is x + 0 ⋅ y − 4 = 0., 7. Put a2 − b 2 = c. Then,, ⏐c cos θ − 1⏐, ⏐⏐, ⏐c cos θ + 1⏐, ⏐c cos θ + 0 − 1⏐, ⏐ ⏐, ⏐− c cos θ + 0 − 1⏐, ⏐ ⏐, ⏐, ⏐, a, a, a, ⏐, ⏐, ⏐ a, ⏐, ⏐, ⏐, ⏐, ⏐, ×, =, ( d1 × d2 ) =, ⎛ cos 2 θ sin 2 θ ⎞, cos 2 θ sin 2 θ, cos 2 θ sin 2 θ, ⎜ 2 +, ⎟, +, +, 2, 2, 2, 2, b2 ⎠, ⎝ a, a, b, a, b, ⏐c 2 cos 2 θ, ⏐, − 1⏐, ⏐, 2, a, |( a2 − b 2 ) cos 2 θ − a2| 2, ⏐, ⏐, =, ×b, × a2 b 2 =, 2, 2, 2, 2, ( b 2 cos 2 θ + a2 sin 2 θ), ( b cos θ + a sin θ), =, , |a2 ( 1 − sin 2 θ) − b 2 cos 2 θ − a2|, 2, , 2, , 2, , 2, , ( b cos + a sin θ), , × b2 =, , |− ( a2 sin 2 θ + b 2 cos 2 θ)|, ( b 2 cos 2 θ + a2 sin 2 θ), , × b2, , = b2., 8., , |3 × 4 − 4 × 1 + k|, 3 2 + ( −4 ) 2, , 10. Equation of AB is, , = 2 ⇔ |8 + k| = 10 ⇔ ( 8 + k ) = 10 or ( 8 + k ) = − 10 ., y − 3 −4, =, ⇒ 2 x + y − 7 = 0., 2, x−2, , Length of perpendicular, |2 × ( −1) + 2 − 7| 7, units., =, 5, 2 2 + 12, ar(a ABC ) =, , from, , C( −1, 2 ), , to, , 2x + y − 7 = 0, , is, , 7 ⎞, 1, 7, 1, × AB ×, = ⎛⎜ × 20 ×, ⎟ sq units = 7 sq units., 2, 5⎠, 5 ⎝2, , 11. Let the required point be P( x , 0 )., Given line is 4 x + 3 y − 12 = 0., Now, , |4 x + 3 × 0 − 12|, 42 + 32, , = 4 ⇔ |x − 3| = 5 ⇒ x − 3 = 5 or, ⇒ x = 8 or, , x− 3= −5, , x = − 2., , equal, , to
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706, , Senior Secondary School Mathematics for Class 11, , 12. Let x = a. Then, x + y = 4 ⇒ y = ( 4 − a)., Let the required point be P( a, 4 − a). Then,, |4 a + 3( 4 − a) − 10|, 42 + 32, ∴, , = 1 ⇒ |a + 2| = 5 ⇒ a + 2 = 5 or, , ⇒ a = 3 or, required points are (3, 1) and ( −7 , 11)., , a+ 2= −5, , a = − 7., , 13. Side of the square = length of perp. from (0, 0) on 3 x − 4 y − 10 = 0, =, , |3 × 0 − 4 × 0 − 10|, 3 2 + ( −4 ) 2, , = 2 units., , Area of the square = ( 2 × 2 ) sq units = 4 sq units., 16. Putting x = 0 in y = mx + c , we get y = c., Thus, P( 0 , c ) is a point on y = mx + c., Required distance = length of perp. from P( 0 , c ) on y = mx + d, =, , |m × 0 − c + d|, 1 + m2, , =, , |d − c|, 1 + m2, , ⋅, , 17. Putting x = 0 in px + py + q = 0, we get y = ( − q/ p )., ⎛ − q⎞, Required distance = length of perp. from P ⎜ 0, ⎟ on px + py − r = 0, p⎠, ⎝, ⏐, ⏐, ⎛ − q⎞, ⏐p × 0 + p × ⎜ ⎟ − r ⏐, p, ⎝, ⎠, ⏐ = |− q − r| = |q + r| ⋅, =⏐, 2, 2, 2p, 2p, p +p, 19. Let the required equation be y = − x + c., |0 + 0 − c|, ∴, = 5 ⇒ |c| = 5 2 ⇒ c = 5 2, 12 + 12, , or, , c = − 5 2., , MISCELLANEOUS PROBLEMS, It may be mentioned here that when the equations of two lines are given,, then on solving them simultaneously, we get the point of intersection of the, given lines., EXAMPLE 1, SOLUTION, , Find the point of intersection of the lines 5 x + 7 y = 3 and 2x − 3 y = 7., The given equations are:, 5x + 7y − 3 = 0, 2x − 3 y − 7 = 0, On solving (i) and (ii), by cross multiplication, we get, x, y, 1, =, =, ( −49 − 9) ( −6 + 35) ( −15 − 14), , ... (i), ... (ii)
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Straight Lines, , 707, , ⎛ −58⎞, ⎛ 29 ⎞, x=⎜, ⎟ = 2 and y = ⎜, ⎟ = −1., ⎝ −29⎠, ⎝ −29⎠, Hence, the point of intersection of the given lines is P( 2, −1)., ⇒, , EXAMPLE 2, SOLUTION, , EXAMPLE 3, SOLUTION, , EXAMPLE 4, SOLUTION, , Find the equation of the line parallel to the y-axis and drawn through the, point of intersection of the lines x − 7 y + 15 = 0 and 2x + y = 0., The given equations are:, ... (i), x − 7 y + 15 = 0, ... (ii), 2x + y = 0, On solving (i) and (ii), we get x = − 1 and y = 2., Thus, the given lines intersect at the point P( −1, 2)., The line parallel to the y-axis and drawn through P( −1, 2) is given by, x = − 1, i.e., x + 1 = 0., Hence, the required equation is x + 1 = 0., Find the equation of the line passing through the intersection of the lines, x + 2y + 3 = 0 and 3 x + 4y + 7 = 0, and parallel to the line y − x = 8., The given intersecting lines are:, ... (i), x + 2y + 3 = 0, ... (ii), 3 x + 4y + 7 = 0, By cross multiplication, we get, x, y, 1, x y, 1, =, =, ⇒, = =, (14 − 12) ( 9 − 7) ( 4 − 6), 2 2 −2, 2, 2, = −1 and y =, = − 1., ⇒ x=, −2, −2, Thus, the lines (i) and (ii) intersect at the point P( −1, − 1)., Now, y − x = 8 ⇒ y = x + 8., ... (iii), Slope of line (iii) is 1., Slope of the line parallel to this line is 1., y+1, So, the equation of the required line is, = 1 ⇒ x − y = 0., x+1, Find the value of k for which the lines 3 x + y = 2, kx + 2y = 3 and, 2x − y = 3 may intersect at a point., The given lines are:, ... (i), 3x + y − 2 = 0, ... (ii), kx + 2y − 3 = 0, ... (iii), 2x − y − 3 = 0, On solving (i) and (iii) by cross multiplication, we get, x, y, 1, x, y, 1, =, =, ⇒, = =, ( −3 − 2) ( −4 + 9) ( −3 − 2), −5 5 −5, ⎛ −5 ⎞, ⎛ 5⎞, ⇒ x = ⎜ ⎟ = 1 and y = ⎜ ⎟ = −1., ⎝ −5 ⎠, ⎝ −5 ⎠, Thus, the point of intersection of (i) and (iii) is P(1, − 1).
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708, , Senior Secondary School Mathematics for Class 11, , For the given lines to intersect at a point, x = 1 and y = − 1 must satisfy, (ii) also., ∴ ( k × 1) + 2 × ( −1) − 3 = 0 ⇒ k − 5 = 0 ⇒ k = 5., Hence, k = 5., EXAMPLE 5, SOLUTION, , Show that the lines x − y = 6, 4x − 3 y = 20 and 6x + 5 y + 8 = 0 are, concurrent. Also find their point of intersection., The given equations are:, ... (i), x −y −6=0, ... (ii), 4x − 3 y − 20 = 0, ... (iii), 6x + 5 y + 8 = 0, On solving (i) and (ii) by cross multiplication, we get, x, y, 1, x, y 1, =, =, ⇒, =, = ⇒ x = 2, y = − 4., ( 20 − 18) ( −24 + 20) ( −3 + 4), 2 −4 1, Thus, the lines (i) and (ii) intersect at the point P( 2, − 4)., Putting x = 2 and y = − 4 in (iii), we get, LHS = 6 × 2 + 5 × ( −4) + 8 = 0 = RHS., This shows that the point P( 2, − 4) also lies on (iii)., Thus, all the given three lines intersect at the same point., Hence, the given lines are concurrent and their point of intersection, is P( 2, − 4)., , EXAMPLE 6, , SOLUTION, , If the three lines y = m 1 x + c1 , y = m2 x + c2 and y = m3 x + c3 are, concurrent then show that, m 1 ( c2 − c3 ) + m2 ( c3 − c1 ) + m3 ( c1 − c2 ) = 0., The given lines are:, ... (i), m1 x − y + c1 = 0, ... (ii), m2 x − y + c2 = 0, ... (iii), m3 x − y + c3 = 0, On solving (i) and (ii) by cross multiplication, we get, x, y, 1, =, =, ( − c2 + c1 ) (m2 c1 − m 1 c2 ) ( −m 1 + m2 ), (m2 c1 − m 1 c2 ), ( c1 − c2 ), and y =, ⋅, ⇒ x=, (m2 − m 1 ), (m2 − m 1 ), ⎛ c1 − c2 m2 c1 − m 1 c2 ⎞, ⎟⋅, Thus, the point of intersection of (i) and (ii) is P ⎜⎜, ,, m2 − m 1 ⎟⎠, ⎝ m2 − m 1, Since the given three lines meet at a point, the point P must therefore, lie on (iii) also., ( c1 − c2 ), (m2 c1 − m 1 c2 ), + c3 = 0, −, ∴ m3 ⋅, (m2 − m 1 ), (m2 − m1 ), ⇒ m3 ( c1 − c2 ) − (m2 c1 − m 1 c2 ) + c3 (m2 − m 1 ) = 0, ⇒ m 1 ( c2 − c3 ) + m2 ( c3 − c1 ) + m3 ( c1 − c2 ) = 0.
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Straight Lines, EXAMPLE 7, , SOLUTION, , 709, , Find the area of the triangle formed by the lines y − x = 0, x + y = 0 and, x − c = 0., Let the sides AC , AB and BC of a ABC be, represented by y − x = 0, x + y = 0 and, x − c = 0 respectively., On solving these equations pairwise, we, get the points A( 0, 0), B( c, − c) andC( c, c)., ∴ area of a ABC is given by, 1, a = |x1 ( y 2 − y 3 ) + x 2 ( y 3 − y1 ) + x 3 ( y1 − y 2 )|, 2, 1, = ⋅|0 ⋅ ( −c − c) + c( c − 0) + c( 0 + c)|sq units, 2, ⎛1, ⎞, = ⎜ × 2c2 ⎟ sq units = c2 sq units., ⎝2, ⎠, Hence, the required area is c2 sq units., , EXAMPLE 8, , SOLUTION, , Find the area of the triangle formed by the lines y = m 1 x + c1 ,, y = m2 x + c2 and x = 0., Let the sides AC, AB and BC of a ABC, be represented by the equations:, ... (i), m 1 x − y + c1 = 0, ... (ii), m2 x − y + c2 = 0, ... (iii), x=0, On solving (i) and (ii) by cross, multiplication, we have, x, y, 1, =, =, ( − c2 + c1 ) (m2 c1 − m 1 c2 ) ( − m 1 + m2 ), ⇒ x=, , ( c1 − c2 ), (m2 − m 1 ), , and, , y=, , (m2 c1 − m 1 c2 ), (m2 − m 1 ), , ⋅, , ⎛ c1 − c2 m2 c1 − m 1 c2 ⎞, ⎟⋅, Thus, the lines AC and AB intersect at A ⎜⎜, ,, m2 − m 1 ⎟⎠, ⎝ m2 − m 1, On solving (ii) and (iii), we get B( 0, c2 )., On solving (i) and (iii), we get C( 0, c1 )., ∴ area of a ABC is given by, 1, ⋅|x1 ( y 2 − y 3 ) + x 2 ( y 3 − y1 ) + x 3 ( y1 − y 2 )|, 2, ⏐ 1 ( c1 − c2 ) 2, 1 ⏐ ( c1 − c2 ), ⋅ ( c2 − c1 ) + 0 + 0⏐= ⋅, ⋅, = ⋅⏐, 2 ⏐(m2 − m 1 ), ⏐ 2 |m 1 − m2|, , a=
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710, , Senior Secondary School Mathematics for Class 11, 2, 1 ( c1 − c2 ), Hence, the area of the triangle formed by the given lines is ⋅, ⋅, 2 |m 1 − m2|, , EXAMPLE 9, , SOLUTION, , Find the image of the point P( 3 , 8) with respect to the line x + 3 y = 7,, assuming the given line to be a plane mirror., Let AB be the given line whose equation is x + 3 y − 7 = 0, 1, 7, Now, x + 3 y = 7 ⇒ y = − x + ⋅, 3, 3, 1, Slope of AB = − ⋅, 3, , .... (i), , Let P( 3 , 8) be the given point and, let Q( x1 , y1 ) be its image in AB., Join PQ, intersecting AB at N., Then, PN ⊥ AB and PN = QN ., Let the slope of PQ be m. Then,, ⎛ 1⎞, m × ⎜ − ⎟ = −1 ⇒ m = 3, ⎝ 3⎠, , [Q AB ⊥ PQ]., , Equation of line PQ is given by, y−8, = 3 ⇒ y − 8 = 3 x − 9 ⇒ 3 x − y − 1 = 0., x−3, , ... (ii), , On solving (i) and (ii), we get, x, y, 1, x, y, 1, =, =, ⇒, =, =, ( −3 − 7) ( −21 + 1) ( −1 − 9), −10 −20 −10, ⇒ x = 1 and y = 2., Thus, AB and PQ intersect at the point N(1, 2)., Also, N bisects PQ., 8 + y1, ⎛ 3 + x1, ⎞, = 1,, = 2⎟ ⇒ x1 = −1 and y1 = − 4 ., ∴ ⎜, ⎠, ⎝ 2, 2, Hence, the image of the given point P( 3 , 8) is Q( −1, − 4)., , EXERCISE 20I, 1. Find the points of intersection of the lines, 4x + 3 y = 5 and x = 2y − 7., 2. Show that the lines x + 7 y = 23 and5 x + 2y = 16 intersect at the point (2, 3)., 3. Show that the lines 3 x − 4y + 5 = 0, 7 x − 8y + 5 = 0 and 4x + 5 y = 45 are, concurrent. Also find their point of intersection., 4. Find the value of k so that the lines 3 x − y − 2 = 0, 5 x + ky − 3 = 0 and, 2x + y − 3 = 0 are concurrent.
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Straight Lines, , 711, , 5. Find the image of the point P(1, 2) in the line x − 3 y + 4 = 0., 6. Find the area of the triangle formed by the lines x + y = 6, x − 3 y = 2 and, 5 x − 3 y + 2 = 0., 7. Find the area of the triangle formed by the lines x = 0, y = 1 and 2x + y = 2., 8. Find the area of the triangle, the equations of whose sides are y = x , y = 2x, and y − 3 x = 4., 9. Find the equation of the perpendicular drawn from the origin to the line, 4x − 3 y + 5 = 0. Also, find the coordinates of the foot of the perpendicular., 10. Find the equation of the perpendicular drawn from the point P( −2, 3) to the, line x − 4y + 7 = 0. Also, find the coordinates of the foot of the, perpendicular., 11. Find the equations of the medians of a triangle whose sides are given by the, equations 3 x + 2y + 6 = 0, 2x − 5 y + 4 = 0 and x − 3 y − 6 = 0., ANSWERS (EXERCISE 20I), , 1. ( −1, 3), , 3. (5, 5), , 6. 12 sq units, , 7. 1 sq unit, ⎛ 4 3⎞, 9. 3 x + 4y = 0, ⎜ − , ⎟, ⎝ 5 5⎠, , 4. k = −2, , ⎛ 6 7⎞, 5. ⎜ , ⎟, ⎝ 5 5⎠, , 8. 4 sq units, ⎛ −27 23 ⎞, 10. 4x + y + 5 = 0, ⎜, , ⎟, ⎝ 17 17 ⎠, , 11. 41x − 112y − 70 = 0, 16x − 59y − 120 = 0 and 25 x − 53 y + 50 = 0, , SHIFTING OF ORIGIN, TRANSLATION OF AXES, , Sometimes it is desirable to shift the origin to a new point and the new axes are, transformed parallel to the original axes., A transformation of this kind is known as translation of axes., Let P( x , y) be a point with reference to, O as origin and OX and OY as the axes., Draw PM ⊥ OX, so that OM = x and PM = y., Let the new origin be O′( h, k) referred, to the old axes, i.e., OL = h and O′ L = k., Draw O′ X′ || OX, meeting PM at M′., Also, draw O′ Y′ || OY , meeting OM at L., Now, we have the new origin at O′ and, the new axes as O′ X′ and O′ Y′., Let O′ M′ = x′ and M′ P = y′ with reference to the new axes. Then, P( x′ , y′ ), are the new coordinates of P.
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712, , Senior Secondary School Mathematics for Class 11, , Now, OM = OL + LM ⇒ x = h + x′ ⇒ x = x′ + h ⇒ x′ = x − h., And, MP = MM′ + PM′ ⇒ y = k + y′ ⇒ y = y′ + k ⇒ y′ = y − k., Hence, x′ = x − h and y′ = y − k are the relations between the new and old, coordinates., SOLVED EXAMPLES, EXAMPLE 1, SOLUTION, , EXAMPLE 2, SOLUTION, , EXAMPLE 3, SOLUTION, , EXAMPLE 4, , SOLUTION, , If the origin be shifted to the point ( 2, − 3) by a translation of coordinate, axes, find the new coordiantes of the point ( 4, 7)., Let the given point be P( x , y). Then, x = 4 and y = 7., Let the origin O be shifted to the point O′( h, k), where h = 2, k = − 3., Then, the new coordinates of P are ( x′ , y′ ), where, x′ = x − h = ( 4 − 2) = 2, and y′ = y − k = (7 + 3) = 10., Hence, the coordinates of (4, 7) in the new system are (2, 10)., If the origin is shifted to the point ( 2, 3), the coordinates of a point, become (5 , − 4). Find the original coordinates, when the axes are parallel., Let the original coordinates be P( x , y)., Let the origin O be shifted to the new point O′( h, k), where h = 2 and, k = 3., Then, the new coordinates of P are ( x′ , y′ ), where x′ = 5 and y′ = − 4., Now, x′ = x − h ⇒ 5 = x − 2 ⇒ x = 7., And, y′ = y − k ⇒ − 4 = y − 3 ⇒ y = − 1., Hence, the original coordinates are (7, –1)., At what point the origin be shifted, if the coordinates of a point ( 4, 5), become ( 3 , 7)?, Let O′( h, k) be the point to which the origin O be shifted., Then, ( x = 4, y = 5) and ( x′ = 3 , y′ = 7)., Now, x′ = x − h ⇒ h = x − x′ = ( 4 − 3) = 1., And, y′ = y − k ⇒ k = y − y′ = (5 − 7) = − 2., Hence, the origin O must be shifted to the point O′ (1, − 2)., If the origin be shifted to the point ( 3 , − 1), find the new equation of the, line 2x − 3 y + 5 = 0., Let the origin O be shifted to the point O′( h, k), where h = 3 and, k = − 1., Let the new coordinates of P( x , y) be P( x′ , y′ )., Then, x′ = x − h ⇒ x′ = x − 3 ⇒ x = x′ + 3., And, y′ = y − k ⇒ y′ = y + 1 ⇒ y = y′ − 1., So, the new equation becomes:, 2( x′ + 3) − 3( y′ − 1) + 5 = 0 ⇒ 2x′ − 3 y′ + 14 = 0.
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Straight Lines, , 713, , Hence, the equation of the straight line in the new system is, 2x − 3 y + 14 = 0., EXAMPLE 5 Find the point to which the origin be shifted after a translation, so that, the equation x 2 + y 2 − 4x − 8y + 3 = 0 will have no first degree terms., SOLUTION Let the origin O be shifted to a point O′( h, k)., Let the new coordinates of P( x , y) be P( x′ , y′ )., Then, x′ = x − h ⇒ x = x′ + h., And, y′ = y − k ⇒ y = y′ + k ., So, the new equation becomes:, ( x′ + h) 2 + ( y′ + k) 2 − 4( x′ + h) − 8( y′ + k) + 3 = 0, ⇒ ( x′ 2 + h2 + 2x′ h) + ( y′ 2 + k 2 + 2y′ k) − 4( x′ + h) − 8( y′ + k) + 3 = 0, ⇒ x′ 2 + y′ 2 + ( 2h − 4) x′ + ( 2k − 8) y′ + ( h2 + k 2 − 4h − 8k + 3) = 0., Since we are required to get an equation free from first degree terms,, so we have:, (2h − 4 = 0 and 2k − 8 = 0), ⇒ (2h = 4 and 2k = 8) ⇒ ( h = 2 and k = 4)., Hence, the origin O should be shifted to the point O′( 2, 4)., , EXERCISE 20J, 1. If the origin is shifted to the point (1, 2) by a translation of the axes, find the, new coordinates of the point (3, –4)., 2. If the origin is shifted to the point (–3, –2) by a translation of the axes, find, the new coordinates of the point (3, –5)., 3. If the origin is shifted to the point (0, –2) by a translation of the axes, the, coordinates of a point become (3, 2). Find the original coordinates of the, point., 4. If the origin is shifted to the point (2, –1) by a translation of the axes, the, coordinates of a point become (–3, 5). Find the original coordinates of the, point., 5. At what point must the origin be shifted, if the coordinates of a point (–4, 2), become (3, –2)?, Find what the given equation becomes when the origin is shifted to the point (1, 1)., 7. xy − y 2 − x + y = 0, 6. x 2 + xy − 3 x − y + 2 = 0, 9. xy − x − y + 1 = 0, 8. x 2 − y 2 − 2x + 2y = 0, 10. Transform the equation 2x 2 + y 2 − 4x + 4y = 0 to parallel axes when the, origin is shifted to the point (1, –2)., ANSWERS (EXERCISE 20J), , 1. ( 2, − 6), 5. ( −7 , 4), , 2. ( 6, − 3), 6. x 2 + xy = 0, , 3. (3, 0), 7. y 2 − xy = 0, , 4. ( −1, 4), 8. x 2 − y 2 = 0
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714, , 9. xy = 0, , Senior Secondary School Mathematics for Class 11, , 10. 2x 2 + y 2 − 6 = 0, , EQUATION OF FAMILY OF LINES PASSING THROUGH THE POINT OF INTERSECTION, OF TWO LINES, , Let the two intersecting lines L1 and L2 be given by, … (i), A1 x + B1 y + C1 = 0, and, … (ii), A 2 x + B2 y + C 2 = 0, From (i) and (ii), we can form an equation, … (iii), ( A1 x + B1 y + C1 ) + k( A 2 x + B2 y + C 2 ) = 0,, where k is an arbitrary constant, called parameter., Clearly, for any value of k, the equation (iii) is of first degree in x and y., Hence, it represents a family of lines., Putting some value of k, we get a particular member of this family., SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , EXAMPLE 2, SOLUTION, , Find the equation of the line drawn through the point of intersection of, the lines 4x − 3 y + 7 = 0 and 2x + 3 y + 5 = 0 and passing through the, point ( −4, 5)., The equation of any line through the point of intersection of the given, lines is of the form, ( 4x − 3 y + 7) + k( 2x + 3 y + 5) = 0, … (i), ⇒ ( 4 + 2k) x + ( 3 k − 3) y + (5 k + 7) = 0, If it passes through the point (–4, 5), we have, ( 4 + 2k)( −4) + ( 3 k − 3) ⋅ 5 + (5 k + 7) = 0, ⇒ −16 − 8k + 15 k − 15 + 5 k + 7 = 0 ⇒ 12k = 24 ⇒ k = 2., Substituting k = 2 in (i), we get, 8x + 3 y + 17 = 0, which is the required equation., Find the equation of the line through the intersection of lines 3 x + 4y = 7, and x − y + 2 = 0 and whose slope is 5., The given lines are 3 x + 4y − 7 = 0 and x − y + 2 = 0., The equation of any line through the point of intersection of the given, lines is of the form., … (i), ( 3 x + 4y − 7) + k( x − y + 2) = 0, ⇒ ( 3 + k) x + ( 4 − k) y + ( 2k − 7) = 0, ⇒ ( 4 − k) y = − ( 3 + k) x + (7 − 2k), −( 3 + k), (7 − 2k), x+, ⇒ y=, 4−k, ( 4 − k), ( k + 3), (7 − 2k), x+, ⋅, ⇒ y=, ( k − 4), ( 4 − 2k)
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Straight Lines, , Slope of this line is, , 715, , ( k + 3), ⋅, ( k − 4), , k+ 3, 23, = 5 ⇒ k + 3 = 5 k − 20 ⇒ 4k = 23 ⇒, ⋅, k−4, 4, 23, in (i), we get, Substituting k =, 4, 23, ( 3 x + 4y − 7) +, ( x − y + 2) = 0, 4, ⇒ 4( 3 x + 4y − 7) + 23( x − y + 2) = 0, ⇒ 35 x − 7 y + 18 = 0, which is the required equation., , ∴, , EXAMPLE 3, , SOLUTION, , EXAMPLE 4, , Find the equation of the line through the intersection of lines, x + 2y − 3 = 0 and 4x − y + 7 = 0 and which is parallel to the line, 5 x + 4y − 20 = 0., −5, 5 x + 4y − 20 = 0 ⇒ y =, x+5, 4, −5, ∴ slope of the given line =, 4, −5, and slope of the required line =, ⋅, 4, Now, the equation of any line through the intersection of the given, lines is of the form, … (i), ( x + 2y − 3) + k( 4x − y + 7) = 0, ⇒ (1 + 4k) x + ( 2 − k) y + (7 k − 3) = 0, ⇒ ( 2 − k) y = − (1 + 4k) x + ( 3 − 7 k), − (1 + 4k), ( 3 − 7 k), x+, ⇒ y=, ( 2 − k), ( 2 − k), (1 + 4k), ( 3 − 7 k), x+, ⋅, ⇒ y=, ( k − 2), ( 2 − k), (1 + 4k), Slope of this line =, ( k − 2), (1 + 4k) −5, ∴, =, ⇒ 4 + 16k = −5 k + 10, ( k − 2), 4, 6 2, ⇒ 21k = 6 ⇒ k =, = ⋅, 21 7, 2, Substituting k = in (i), we get, 7, 2, ( x + 2y − 3) + ( 4x − y + 7) = 0, 7, ⇒ (7 x + 14y − 21) + ( 8x − 2y + 14) = 0, ⇒ 15 x + 12y − 7 = 0, which is the required equation., Find the equation of the line through the intersection of the lines, 3 x + y − 9 = 0 and 4x + 3 y − 7 = 0 and which is perpendicular to the, line 5 x − 4y + 1 = 0.
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716, , SOLUTION, , Senior Secondary School Mathematics for Class 11, , 5, 1, x+ ⋅, 4, 4, 5, Slope of the given line is ⋅, 4, Now, the equation of any line through the intersection of the given, lines is of the form, … (i), ( 3 x + y − 9) + k( 4x + 3 y − 7) = 0, ⇒ ( 3 + 4k) x + (1 + 3 k) y − ( 9 + 7 k) = 0, ⇒ (1 + 3 k) y = − ( 3 + 4k) x + ( 9 + 7 k), ( 3 + 4k), ( 9 + 7 k), … (ii), x+, ⇒ y=−, (1 + 3 k), (1 + 3 k), , 5 x − 4y + 1 = 0 ⇒ y =, , Let m be the slope of the line perpendicular to the required line., 5, −4, Then, m × = −1 ⇒ m =, ⋅, 4, 5, −( 3 + 4k) −4, ∴ we must have, =, (1 + 3 k), 5, −11, ⋅, ⇒ 15 + 20k = 4 + 12k ⇒ 8k = −11 ⇒ k =, 8, −11, in (i), we get, Substituting k =, 8, 11, ( 3 x + y − 9) −, ( 4 + 3 y − 7) = 0, 8, ⇒ ( 24x + 8y − 72) − 44x − 33 y + 77 = 0, ⇒ 20x + 25 y − 5 = 0 ⇒ 4x + 5 y − 1 = 0., EXAMPLE 5, SOLUTION, , EXAMPLE 6, SOLUTION, , Find the equation of the x line parallel to the y-axis and drawn through, the point of intersection of the lines x − 7 y + 5 = 0 and 3 x + y − 7 = 0., The equation of any line through the point of intersection of the given, lines is of the form, ( x − 7 y + 5) + k( 3 x + y − 7) = 0, … (i), ⇒ (1 + 3 k) x + ( k − 7) y + (5 − 7 k) = 0, If the line is parallel to y-axis then coefficient of y should be 0, i.e.,, k − 7 = 0, which gives k = 7., Substituting k = 7 in (i), we get: 22x − 44 = 0, ∴ x − 2 = 0, which is the required equation., Find the equation of the line through the intersection of the lines, 2x + 3 y = 4 and x − 5 y + 7 = 0 that has its x-intercept equal to − 4., The given lines are 2x + 3 y − 4 = 0 and x − 5 y + 7 = 0., The equation of any line through the intersection of the given lines is, of the form, ( 2x + 3 y − 4) + k( x − 5 y + 7) = 0, … (i), ⇒ ( 2 + k) x + ( 3 − 5 k) y + (7 k − 4) = 0
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Straight Lines, , 717, , If this line has x-intercept –4, then the point (–4, 0) lies on (i)., ∴ ( 2 + k)( −4) + (7 k − 4) = 0 ⇒ − 8 − 4k + 7 k − 4 = 0, ⇒ 3 k = 12 ⇒ k = 4., Substituting k = 4 in (i), we get, 6x − 17 y + 24 = 0, which is the required equation., , EXERCISE 20K, 1. Find the equation of the line drawn through the point of intersection of, the lines x − 2y + 3 = 0 and 2x − 3 y + 4 = 0 and passing through the point, (4, –5)., 2. Find the equation of the line drawn through the point of intersection of the, lines x − y = 7 and 2x + y = 2 and passing through the origin., 3. Find the equation of the line drawn through the point of intersection of the, −2, lines x + y = 9 and 2x − 3 y + 7 = 0 and whose slope is, ⋅, 3, 4. Find the equation of the line drawn through the point of intersection of the, lines x − y = 1 and 2x − 3 y + 1 = 0 and which is parallel to the line, 3 x + 4y = 12., 5. Find the equation of the line through the intersection of the lines 5 x − 3 y = 1, and 2x + 3 y = 23 and which is perpendicular to the line 5 x − 3 y = 1., 6. Find the equation of the line through the intersection of the lines, 2x − 3 y = 0 and 4x − 5 y = 2 and which is perpendicular to the line, x + 2y + 1 = 0., 7. Find the equation of the line through the intersection of the lines, x − 7 y + 5 = 0 and 3 x + y − 7 = 0 and which is parallel to x-axis., 8. Find the equation of the line through the intersection of the lines, 2x − 3 y + 1 = 0 and x + y − 2 = 0 and drawn parallel to y-axis., 9. Find the equation of the line through the intersection of the lines, 2x + 3 y − 2 = 0 and x − 2y + 1 = 0 and having x-intercept equal to 3., 10. Find the equation of the line passing through the intersection of the lines, 3 x − 4y + 1 = 0 and 5 x + y − 1 = 0 and which cuts off equal intercepts from, the axes., ANSWERS (EXERCISE 20K), , 1. 7 x + 3 y − 13 = 0, 3. 2x + 3 y − 23 = 0, 5. 63 x + 105 y − 781 = 0, 7. y = 1, 9. x + 5 y − 3 = 0, , 2. 4x + 3 y = 0, 4. 3 x − 4y + 24 = 0, 6. 2x − y − 4 = 0, 8. x = 1, 10. 23 x + 23 y = 11
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718, , Senior Secondary School Mathematics for Class 11, HINTS TO SOME SELECTED QUESTIONS, , 7. When a line is parallel to x-axis, then coefficient of x should be 0., 10. Let the required line be, ( 3 x − 4 y + 1) + k(5 x + y − 1) = 0, ⇒ ( 3 + 5k ) x + ( k − 4) y + ( 1 − k ) = 0, ⇒ ( 3 + 5 k ) x + ( k − 4 ) y = ( k − 1), ( 3 + 5k ), ( k − 4), y, x, ⇒, =1, x+, y=0 ⇒, +, ( k − 1), ( k − 1), ( k − 1), ( k − 1), ( 3 + 5k ) ( k − 4), k−1, k−1, −7, ⇒ k − 4 = 3 + 5k ⇒ 4k = − 7 ⇒ k =, ⋅, =, ∴, 4, 3 + 5k k − 4, , q
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21, Circle, Coordinate Geometry, , CIRCLE A circle is the set of all points in a plane which are at a constant distance from, a fixed point in the plane., The fixed point is called the centre and the constant distance is called the, radius of the circle., , Equation of a Circle in Standard Form, , The equation of a circle with centre ( h, k) and radius r is given by, ( x − h) 2 + ( y − k) 2 = r 2 ., Let us consider a circle with centre C( h, k) and, radius r., Let P( x , y) be an arbitrary point on the circle., , THEOREM 1, , PROOF, , Then,|CP| = r ⇒, , ( x − h) 2 + ( y − k) 2 = r, , ⇒ ( x − h) 2 + ( y − k) 2 = r 2 ,, which is the required equation of the given, circle., THEOREM 2, , PROOF, , The equation of a circle with centre at the origin and radius r is given by, x2 + y2 = r 2., , The equation of a circle with centre C( 0, 0) and radius r is given by, ( x − 0) 2 + ( y − 0) 2 = r 2 ⇒ x 2 + y 2 = r 2 ., , Equation of a Circle, the End Points of whose Diameter are given, THEOREM 3, , PROOF, , The equation of a circle described on the line joining A( x1 , y1 ) and, B( x 2 , y 2 ) as a diameter is ( x − x1 )( x − x 2 ) + ( y − y1 )( y − y 2 ) = 0., , Let A( x1 , y1 ) and B( x 2 , y 2 ) be the end points of a diameter of the given, circle, and let P( x , y) be any point on the circle., Join AP and BP., Then, the angle in a semicircle being a, right angle,, we have ∠APB = 90°., ( y − y1 ), Now, slope of AP =, ⋅, ( x − x1 ), ( y − y2 ), And, slope of BP =, ⋅, ( x − x2 ), 719
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720, , Circle, , Senior Secondary School Mathematics for Class 11, , Since AP ⊥ BP , we have, ( y − y1 ) ( y − y 2 ), ⋅, = − 1 ⇒ ( x − x1 )( x − x 2 ) + ( y − y1 )( y − y 2 ) = 0,, ( x − x1 ) ( x − x 2 ), which is the required equation of the circle., SUMMARY, , (i), , The equation of a circle with centre C( h, k) and, radius r is given by ( x − h) 2 + (y − k) 2 = r 2 ., , (ii) The equation of a circle with centre C( 0, 0) and, radius r is given by x2 + y2 = r 2 ., (iii) The equation of a circle with A( x1 , y1 ) and B( x 2 , y 2 ), as the end points of a diameter is given by, ( x − x1 )( x − x2 ) + (y − y1 )(y − y2 ) = 0., SOLVED EXAMPLES, , Find the equation of a circle with centre ( 3 , − 2) and radius 5., We know that the equation of a circle with centre, C( h, k) and radius r is given by, ( x − h) 2 + ( y − k) 2 = r 2 ., Here, h = 3 , k = − 2 and r = 5., ∴ the required equation of the circle is, ( x − 3) 2 + ( y + 2) 2 = 5 2, ⇒ x 2 + y 2 − 6x + 4y − 12 = 0., , EXAMPLE 1, SOLUTION, , EXAMPLE 2, , SOLUTION, , Find the equation of a circle whose centre is ( 2, − 1) and which passes, through the point ( 3 , 6)., Let C( 2, − 1) be the centre of the given, circle and let it pass through the point, P( 3 , 6). Then, radius of the circle, = |CP| = ( 3 − 2) 2 + ( 6 + 1) 2 = 50., ∴ the required equation of the circle is, ( x − 2) 2 + ( y + 1) 2 = ( 50) 2, ⇒ x 2 + y 2 − 4x + 2y − 45 = 0., , EXAMPLE 3, , SOLUTION, , Find the equation of the circle passing through the point ( 2, 4) and, having its centre at the intersection of the lines x − y = 4 and, 2x + 3 y + 7 = 0., The equations of the given lines are:, x−y=4, 2x + 3 y = − 7, , … (i), … (ii)
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Circle, , 721, , Solving (i) and (ii) simultaneously, we get x = 1 and y = − 3., So, the point of intersection of the given lines is C(1, − 3)., ∴ centre of the given circle is C(1, − 3)., Also, the circle passes through the point P( 2, 4)., ∴ radius of the circle, =|CP|= (1 − 2) 2 + ( −3 − 4) 2 = 50., ∴ the required equation of the circle is, ( x − 1) 2 + ( y + 3) 2 = ( 50) 2, ⇒ x 2 + y 2 − 2x + 6y − 40 = 0., EXAMPLE 4, SOLUTION, , Find the equation of a circle of radius 5 units, whose centre lies on the, x-axis and which passes through the point ( 2, 3)., It is given that the centre of the circle lies on the x-axis., So, let C( k , 0) be the centre of the circle., Also, it is given that it passes through the point P( 2, 3)., ∴ radius of the circle =|CP|, = ( k − 2) 2 + ( 0 − 3) 2, = k 2 − 4k + 13, ⇒, , k 2 − 4k + 13 = 5, , [B radius = 5 units], , ⇒ k − 4k + 13 = 25 ⇒ k 2 − 4k − 12 = 0, ⇒ ( k − 6)( k + 2) = 0 ⇒ k = 6 or k = − 2., ∴ centre of the circle is ( 6, 0) or ( −2, 0)., Hence, the required equation of the circle is, 2, , ( x − 6) 2 + ( y − 0) 2 = 5 2 or ( x + 2) 2 + ( y − 0) 2 = 5 2, ⇒ x 2 + y 2 − 12x + 11 = 0 or x 2 + y 2 + 4x − 21 = 0., Thus, there are two circles satisfying the given conditions., EXAMPLE 5, SOLUTION, , Find the equation of a circle, the end points of one of whose diameters are, A( 2, − 3) and B( −3 , 5)., We know that the equation of a circle, the end points of one of whose, diameters are ( x1 , y1 ) and ( x 2 , y 2 ), is given by, ( x − x1 )( x − x 2 ) + ( y − y1 )( y − y 2 ) = 0., Here, x1 = 2, y1 = −3 and x 2 = −3 , y 2 = 5 ., ∴, , the required equation of the circle is, ( x − 2)( x + 3) + ( y + 3)( y − 5) = 0., ⇒ x 2 + y 2 + x − 2y − 21 = 0., , EXAMPLE 6, SOLUTION, , Find the equation of the circle drawn on the diagonal of the rectangle as, its diameter, whose sides are x = 6, x = − 3 , y = 3 and y = −1., Let ABCD be the given rectangle and let its sides AD , BC , AB and CD, be represented by the equations x = − 3 , x = 6, y = −1 and y = 3, respectively.
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722, , Senior Secondary School Mathematics for Class 11, , Then, the coordinates of A and C are, A( −3 , − 1) and C( 6, 3)., So, the equation of the circle with AC, as diameter is given by, ( x + 3)( x − 6) + ( y + 1)( y − 3) = 0, ⇒ x 2 + y 2 − 3 x − 2y − 21 = 0., EXAMPLE 7, SOLUTION, , If y = 2x is a chord of the circle x 2 + y 2 − 10x = 0, find the equation of, the circle with this chord as a diameter., The points of intersection of the given chord and the given circle are, obtained by simultaneously solving, y = 2x and x 2 + y 2 − 10x = 0., Putting y = 2x in x 2 + y 2 − 10x = 0, we get, 5 x 2 − 10x = 0 ⇔ 5 x( x − 2) = 0 ⇔ x = 0 or x = 2., Now, x = 0 ⇒ y = 0, and x = 2 ⇒ y = 4., ∴ the points of intersection of the, given chord and the given circle, are A( 0, 0) and B( 2, 4)., ∴ the required equation of the, circle with AB as diameter is, ( x − 0)( x − 2) + ( y − 0)( y − 4) = 0, ⇒ x 2 + y 2 − 2x − 4y = 0., , EXAMPLE 8, SOLUTION, , Find the equation of a circle with centre ( h, k) and touching the x-axis., Clearly, the radius of the circle = k., So, the required equation is, ( x − h) 2 + ( y − k) 2 = k 2, ⇒ x 2 + y 2 − 2hx − 2ky + h2 = 0., , EXAMPLE 9, , SOLUTION, , Find the equation of a circle with centre ( h, k), and touching the y-axis., Clearly, the radius of the circle = h., So, the required equation is, ( x − h) 2 + ( y − k) 2 = h2, ⇒ x 2 + y 2 − 2hx − 2ky + k 2 = 0.
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Circle, EXAMPLE 10, SOLUTION, , 723, , Find the equation of a circle with centre ( h, k) and touching both the axes., , Clearly, radius = h = k = c (say)., ∴ the equation of the circle is, ( x − c) 2 + ( y − c) 2 = c2, ⇒ x 2 + y 2 − 2c( x + y) + c2 = 0,, where c = h = k., , EXERCISE 21A, Find the equation of a circle with, , 1. centre (2, 4) and radius 5, 2. centre (–3, –2) and radius 6, 3. centre ( a , a) and radius 2, 4. centre ( a cos α , a sin α ) and radius a, 5. centre ( − a , − b) and radius a 2 − b 2, 6. centre at the origin and radius 4, 7. Find the centre and radius of each of the following circles:, 2, , (i) ( x − 3) 2 + ( y − 1) 2 = 9, (iii) ( x + 5) 2 + ( y − 3) 2 = 20, , 2, , 1, 1, 1, (ii) ⎛⎜ x − ⎞⎟ + ⎛⎜ y + ⎞⎟ =, ⎝, ⎝, 2⎠, 3⎠, 16, (iv) x 2 + ( y − 1) 2 = 2, , 8. Find the equation of the circle whose centre is ( 2, − 5) and which passes, through the point (3, 2)., 9. Find the equation of the circle of radius 5 cm, whose centre lies on the y-axis, and which passes through the point (3, 2)., 10. Find the equation of the circle whose centre is (2, –3) and which passes, through the intersection of the lines 3 x + 2y = 11 and 2x + 3 y = 4., 11. Find the equation of the circle passing through the point (–1, 3) and having, its centre at the point of intersection of the lines x − 2y = 4 and, 2x + 5 y + 1 = 0., 12. If two diameters of a circle lie along the lines x − y = 9 and x − 2y = 7, and, the area of the circle is 38.5 sq cm, find the equation of the circle., 13. Find the equation of the circle, the coordinates of the end points of one of, whose diameters are, (i) A( 3 , 2) and B( 2, 5), (ii) A(5 , − 3) and B( 2, − 4), (iii) A( −2, − 3) and B( −3 , 5), (iv) A( p , q) and B(r , s), 14. The sides of a rectangle are given by the equations x = −2, x = 4, y = −2 and, y = 5. Find the equation of the circle drawn on the diagonal of this rectangle, as its diameter.
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724, , Senior Secondary School Mathematics for Class 11, ANSWERS (EXERCISE 21A), , 1. x + y − 4x − 8y − 5 = 0, , 2. x + y + 6x + 4y − 23 = 0, , 3. x + y − 2ax − 2ay = 0, , 4. x 2 + y 2 − 2ax cos α − 2ay sin α = 0, , 5. x 2 + y 2 + 2ax + 2by + 2b 2 = 0, , 6. x 2 + y 2 − 16 = 0, , 2, , 2, , 2, , 2, , 7. (i) Centre (3, 1), radius = 3, , 2, , 2, , 1, 1 −1, (ii) Centre ⎛⎜ , ⎞⎟ , radius =, ⎝2 3 ⎠, 4, , (iii) Centre (–5, 3), radius = 2 5 (iv) Centre (0, 1), radius = 2, 8. x 2 + y 2 − 4x + 10y − 21 = 0, 9. ( x 2 + y 2 − 12y + 11 = 0) or ( x 2 + y 2 + 4y − 21 = 0), 10. x 2 + y 2 − 4x + 6y + 3 = 0, , 11. x 2 + y 2 − 4x + 2y − 20 = 0, , 12. 4x 2 + 4y 2 − 88x − 16y + 451 = 0, 13., , (i) x 2 + y 2 − 5 x − 7 y + 16 = 0, (iii) x 2 + y 2 + 5 x − 2y − 9 = 0, , (ii) x 2 + y 2 − 7 x + 7 y + 22 = 0, (iv) ( x − p)( x − r) + ( y − q)( y − s) = 0, , 14. x + y − 2x − 3 y − 18 = 0, 2, , 2, , HINTS TO SOME SELECTED QUESTIONS, 9. Let the centre of the circle be C ( 0 , k )., It is given that the circle passes through P( 3 , 2 )., Then,|CP|2 = 5 2 ⇔ ( 0 − 3 ) 2 + ( 2 − k ) 2 = 5 2, ⇔ ( 2 − k ) = ± 4 ⇔ k = 6 or k = − 2., ∴ the centre of the circle is (0, 6) or (0, –2)., Hence, the equation of the circle is, [( x − 0 ) 2 + ( y − 6 ) 2 = 5 2 ] or [( x − 0 ) 2 + ( y + 2 ) 2 = 5 2 ]., 12. We know that the point of intersection of two diameters of a circle is the centre of, the circle., On solving x − y = 9 and x − 2 y = 7 , we get the centre as C( 11, 2 )., 77, 77 7, 49, 7, Now, πr 2 =, ⇔ |r 2| = ⎛⎜ × ⎞⎟ =, ⇔ r= ⋅, ⎝ 2 22 ⎠, 2, 4, 2, 49, ∴ the required equation is ( x − 11) 2 + ( y − 2 ) 2 =, ⋅, 4, , GENERAL EQUATION OF A CIRCLE, , The general equation of a circle is of the form x 2 + y 2 + 2gx + 2 fy + c = 0., Also, every equation of the form x 2 + y 2 + 2gx + 2 fy + c = 0 represents, a circle if ( g 2 + f 2 − c) > 0., We know that the equation of a circle with centre ( h, k) and radius r is, given by, ( x − h) 2 + ( y − k) 2 = r 2, , THEOREM, , PROOF
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Circle, , 725, , ⇒ x 2 + y 2 − 2hx − 2ky + ( h2 + k 2 − r 2 ) = 0, ⇒ x 2 + y 2 + 2gx + 2 fy + c = 0, where − h = g , − k = f and ( h2 + k 2 − r 2 ) = c., Hence, the general equation of a circle is of the form, x 2 + y 2 + 2gx + 2 fy + c = 0., Conversely, let x 2 + y 2 + 2gx + 2 fy + c = 0 be the given equation. Then,, x 2 + y 2 + 2gx + 2 fy + c = 0, ⇒ ( x 2 + 2gx + g 2 ) + ( y 2 + 2 fy + f 2 ) = ( g 2 + f 2 − c), ⇒ ( x + g) 2 + ( y + f ) 2 = ( g 2 + f 2 − c) 2, ⇒ {x − ( − g)} 2 + {y − ( − f )} 2 = ( g 2 + f 2 − c) 2, ⇒ ( x − h) 2 + ( y − k) 2 = r 2 , where h = − g , k = − f and r = g 2 + f 2 − c ., Clearly, this equation will represent a circle if ( g 2 + f 2 − c) > 0., The centre of this circle is ( − g , − f ) and its radius is g 2 + f 2 − c ., NOTE, , (i) If ( g 2 + f 2 − c) = 0 then the above equation represents a point circle., (ii) If ( g 2 + f 2 − c) < 0 then the above equation does not represent a circle., , REMARK, , The equation of a circle has the following properties:, (i) It is a second-degree equation in x and y., (ii) It contains no term of xy., (iii) Coefficient of x 2 = coefficient of y 2 ., , Thus, the equation ax 2 + by 2 + 2hxy + 2gx + 2 fy + c = 0 will represent a, circle only when a = b and h = 0., SUMMARY, , The general equation of a circle is, x2 + y2 + 2gx + 2fy + c = 0., Its centre is ( − g , − f ) and radius = g 2 + f 2 − c., , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Show that the equation x 2 + y 2 − 6x + 4y − 36 = 0 represents a circle., Also, find its centre and radius., The given equation is x 2 + y 2 − 6x + 4y − 36 = 0., This is of the form x 2 + y 2 + 2gx + 2 fy + c = 0,, where 2g = − 6, 2 f = 4 and c = − 36., ∴ g = − 3 , f = 2 and c = − 36., Hence, the given equation represents a circle., Centre of the circle = ( − g , − f ) = ( 3 , − 2)., Radius of the circle = g 2 + f 2 − c =, , 3 2 + ( −2) 2 + 36 = 49 = 7 units.
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726, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 2, , Show that the equation 3 x 2 + 3 y 2 + 12x − 18y − 11 = 0 represents a, circle. Also, find its centre and radius., 3 x 2 + 3 y 2 + 12x − 18y − 11 = 0, 11, = 0., ⇒ x 2 + y 2 + 4x − 6y −, 3, 2, 2, This is of the form x + y + 2gx + 2 fy + c = 0,, , SOLUTION, , where 2g = 4, 2 f = − 6 and c = −, 11, ⋅, 3, Hence, the given equation represents a circle., Centre of the circle = ( − g , − f ) = ( −2, 3)., , ∴, , … (i), 11, ⋅, 3, , g = 2, f = −3 and c = −, , Radius of the circle = g 2 + f 2 − c, = ( −2) 2 + 3 2 +, EXAMPLE 3, , SOLUTION, , 11, 50, units., =, 3, 3, , Find the equation of a circle passing through the points (5 , 7), ( 6, 6) and, ( 2, − 2). Find its centre and radius., Let the required equation of the circle be, x 2 + y 2 + 2gx + 2 fy + c = 0., , … (i), , Since it passes through each of the points (5, 7), (6, 6) and ( 2, − 2), each, one of these points must satisfy (i)., … (ii), ∴ 25 + 49 + 10g + 14 f + c = 0 ⇒ 10g + 14 f + c + 74 = 0, … (iii), 36 + 36 + 12g + 12 f + c = 0 ⇒ 12g + 12 f + c + 72 = 0, … (iv), 4 + 4 + 4g − 4 f + c = 0, ⇒ 4g − 4 f + c + 8 = 0, Subtracting (ii) from (iii), we get, … (v), 2g − 2 f − 2 = 0 ⇒ g − f = 1., Subtracting (iv) from (iii), we get, … (vi), 8g + 16 f + 64 = 0 ⇒ g + 2 f = − 8., Solving (v) and (vi), we get g = − 2 and f = − 3., Putting g = − 2 and f = − 3 in (ii), we get c = − 12., Putting g = − 2, f = − 3 and c = − 12 in (i), we get, x 2 + y 2 − 4x − 6y − 12 = 0,, which is the required equation of the circle., Centre of this circle = ( − g , − f ) = ( 2, 3)., And, its radius = g 2 + f 2 − c = 4 + 9 + 12 = 25 = 5 units., EXAMPLE 4, , SOLUTION, , Find the equation of the circle passing through the vertices of a triangle, whose sides are represented by the equations x + y = 2, 3 x − 4y = 6 and, x − y = 0., Let the sides AB , BC and CA of CABC be represented by the, equations x + y = 2, 3 x − 4y = 6 and x − y = 0 respectively.
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Circle, , 727, , Solving x + y = 2 and 3 x − 4y = 6, we get, B( 2, 0)., Solving 3 x − 4y = 6 and x − y = 0, we get, C( −6, − 6)., Solving x + y = 2 and x − y = 0, we get, A(1, 1)., Let the required equation of the circle be, x 2 + y 2 + 2gx + 2 fy + c = 0., , … (i), , Since it passes through A(1, 1), B( 2, 0) and C( −6, − 6), each of these, points must satisfy (i)., 12 + 12 + 2g + 2 f + c = 0 ⇒ 2g + 2 f + c + 2 = 0, 22 + 02 + 4g + c = 0 ⇒ 4g + c + 4 = 0, ( −6) 2 + ( −6) 2 − 12g − 12 f + c = 0 ⇒ −12g − 12 f + c + 72 = 0, Subtracting (ii) from (iii), we get, 2g − 2 f + 2 = 0 ⇔ g − f = − 1., Subtracting (iv) from (iii), we get, 16g + 12 f − 68 = 0 ⇔ 4g + 3 f = 17., Solving (v) and (vi), we get g = 2 and f = 3., Putting g = 2 in (iii), we get c = − 12., Hence, the required equation is, x 2 + y 2 + 4x + 6y − 12 = 0., , ∴, , EXAMPLE 5, , SOLUTION, , … (ii), … (iii), … (iv), … (v), … (vi), , Show that the points (5 , 5), ( 6, 4), ( −2, 4) and (7 , 1) are concyclic, i.e., all, lie on the same circle. Find the equation, centre and radius of this circle., Let the equation of the circle passing through the points (5, 5), (6, 4), and (7, 1) be given by, x 2 + y 2 + 2gx + 2 fy + c = 0., Then, each of these points must satisfy (i)., 25 + 25 + 10g + 10 f + c = 0 ⇒ 10g + 10 f + c = − 50., 36 + 16 + 12g + 8 f + c = 0 ⇒ 12g + 8 f + c = − 52., 49 + 1 + 14g + 2 f + c = 0 ⇒ 14g + 2 f + c = − 50., Subtracting (ii) from (iv), we get, 4g − 8 f = 0 ⇒ g − 2 f = 0., Subtracting (iii) from (iv), we get, 2g − 6 f = 2 ⇒ g − 3 f = 1., Solving (v) and (vi), we get g = − 2 and f = − 1., Putting these values in (ii), we get c = − 20., Putting g = − 2, f = − 1 and c = − 20 in (i), we get, x 2 + y 2 − 4x − 2y − 20 = 0., , … (i), … (ii), … (iii), … (iv), … (v), … (vi), , … (vii), , This is the equation of the circle passing through the points (5 , 5),, ( 6, 4) and (7, 1).
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728, , Senior Secondary School Mathematics for Class 11, , Putting x = − 2 and y = 4 in (vii), we get, LHS = 4 + 16 + 8 − 8 − 20 = 0 = RHS., This shows that the point (–2, 4) also lies on (vii)., Hence, the points (5, 5), (6, 4), ( −2, 4) and (7 , 1) all lie on the same, circle, given by (vii)., Centre of this circle = ( − g , − f ) = ( 2, 1)., Radius of this circle = g 2 + f 2 − c = ( −2) 2 + ( −1) 2 + 20, = 25 = 5 units., EXAMPLE 6, , SOLUTION, , Find the equation of the circle which passes through the centre of the, circle x 2 + y 2 + 8x + 10y − 7 = 0 and is concentric with the circle, 2x 2 + 2y 2 − 8x − 12y − 9 = 0., , We have 2x 2 + 2y 2 − 8x − 12y − 9 = 0, 9, ⇒ x 2 + y 2 − 4x − 6y − = 0 ⇒ x 2 + y 2 + 2gx + 2 fy + c = 0,, 2, −9, where g = − 2, f = − 3 and c =, ⋅, 2, Centre of this circle = ( − g , − f ) = ( 2, 3)., ∴ the centre of the required circle = C( 2, 3)., Again, x 2 + y 2 + 8x + 10y − 7 = 0, ⇒ x 2 + y 2 + 2gx + 2 fy + c = 0, where g = 4, f = 5 and c = − 7., Centre of this circle = ( − g , − f ) = ( −4, − 5)., So, the required circle passes through the point P( −4, − 5)., Radius of the required circle = CP = ( 2 + 4) 2 + ( 3 + 5) 2, = 36 + 64 = 100 = 10., Hence, the required equation of the circle is, ( x − 2) 2 + ( y − 3) 2 = (10) 2 ⇒ x 2 + y 2 − 4x − 6y − 87 = 0., , EXAMPLE 7, , SOLUTION, , Find the equation of the circle whose centre lies on the line x − 4y = 1, and which passes through the points ( 3 , 7) and (5 , 5)., Let the required equation of the circle be, , x 2 + y 2 + 2gx + 2 fy + c = 0., Its centre is ( − g , − f ), which lies on the line x − 4y = 1., ∴ − g + 4 f = 1., , … (i), … (ii), , Also, it is given that (i) passes through the points (3, 7) and (5, 5)., 3 2 + 7 2 + 6g + 14 f + c = 0 ⇒ 6g + 14 f + c = −58., 5 2 + 5 2 + 10g + 10 f + c = 0 ⇒ 10g + 10 f + c = −50., Subtracting (iii) from (iv), we get, 4g − 4 f = 8 ⇔ g − f = 2 ., Solving (ii) and (v), we get g = 3 and f = 1., Putting g = 3 and f = 1 in (iii), we get c = −90., , ∴, , … (iii), … (iv), … (v)
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Circle, , 729, , ∴ g = 3 , f = 1 and c = −90., Putting these values in (i), we get the required equation as, x 2 + y 2 + 6x + 2y − 90 = 0., EXAMPLE 8, , SOLUTION, , Find the equation of a circle concentric with the circle, 2x 2 + 2y 2 − 6x + 8y + 1 = 0, and of double its area., The given equation of the circle is, 2x 2 + 2y 2 − 6x + 8y + 1 = 0, 1, ⇒ x 2 + y 2 − 3 x + 4y + = 0 ⇒ x 2 + y 2 + 2gx + 2 fy + c = 0,, 2, −3, 1, where g =, , f = 2 and c = ⋅, 2, 2, 3, Centre of this circle = ( − g , − f ) = ⎛⎜ , − 2⎞⎟ ., ⎝2, ⎠, Radius of this circle = g 2 + f 2 − c, , 23, ⋅, 2, ⎛3, ⎞, ∴ the centre of the required circle = ⎜ , − 2⎟ ⋅, ⎝2, ⎠, Let r1 be the radius of the required circle. Then,, 2, ⎧⎪, ⎛ 23 ⎞ ⎫⎪, 23, 2, πr12 = 2 ⎨π × ⎜, ⋅, ⎟ ⎬ ⇒ r1 =, 2, ⎝ 2 ⎠ ⎪, ⎪⎩, ⎭, =, , 9, 1, + 4− =, 4, 2, , Hence, the equation of the required circle is, 2, , ⇒, ⇒, EXAMPLE 9, , SOLUTION, , ⎛ x − 3 ⎞ + ( y + 2) 2 = 23, ⎜, ⎟, ⎝, 2, 2⎠, 21, 2, 2, x + y − 3 x + 4y −, =0, 4, 2, 2, 4x + 4y − 12x + 16y − 21 = 0., , Find the equation of a circle which passes through the origin and cuts off, intercepts −2 and 3 from the x-axis and the y-axis respectively., Let the required equation of the circle be, x 2 + y 2 + 2gx + 2 fy + c = 0., , … (i), , Clearly, the circle passes through the, points O( 0, 0), A( −2, 0) and B( 0, 3)., Putting x = 0 and y = 0 in (i), we get c = 0., Thus, (i) becomes, … (ii), x 2 + y 2 + 2gx + 2 fy = 0., Putting x = − 2 and y = 0 in (ii), we get, 4g = 4 ⇔ g = 1.
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730, , Senior Secondary School Mathematics for Class 11, , Putting x = 0 and y = 3 in (ii), we get, −3, 6f = −9 ⇔ f =, ⋅, 2, −3, Putting g = 1 and f =, in (ii), we get x 2 + y 2 + 2x − 3 y = 0,, 2, which is the required equation of the circle., , EXERCISE 21B, 1. Show that the equation x + y − 4x + 6y − 5 = 0 represents a circle. Find, its centre and radius., 2. Show that the equation x 2 + y 2 + x − y = 0 represents a circle. Find, its centre and radius., 2, , 2, , 3. Show that the equation 3 x 2 + 3 y 2 + 6x − 4y − 1 = 0 represents a circle., Find its centre and radius., 4. Show that the equation x 2 + y 2 + 2x + 10y + 26 = 0 represents a point, circle. Also, find its centre., 5. Show that the equation x 2 + y 2 − 3 x + 3 y + 10 = 0 does not represent a, circle., 6. Find the equation of the circle passing through the points, (i) ( 0, 0), (5 , 0) and (3, 3), (ii) (1, 2) , (3, –4) and (5, –6), (iii) (20, 3), (19, 8) and (2, –9), Also, find the centre and radius in each case., 7. Find the equation of the circle which is circumscribed about the triangle, whose vertices are A( −2, 3), B(5 , 2) and C( 6, − 1). Find the centre and, radius of this circle., 8. Find the equation of the circle concentric with the circle, x 2 + y 2 + 4x + 6y + 11 = 0 and passing through the point P(5 , 4)., 9. Show that the points A(1, 0), B( 2, − 7), C( 8, 1) and D( 9, − 6) all lie on the, same circle. Find the equation of this circle, its centre and radius., 10. Find the equation of the circle which passes through the points (1, 3) and, ( 2, − 1), and has its centre on the line 2x + y − 4 = 0., 11. Find the equation of the circle concentric with the circle, x 2 + y 2 − 4x − 6y − 3 = 0 and which touches the y-axis., 12. Find the equation of the circle concentric with the circle, x 2 + y 2 − 6x + 12y + 15 = 0 and of double its area., 13. Prove that the centres of the three circles x 2 + y 2 − 4x − 6y − 12 = 0,, x 2 + y 2 + 2x + 4y − 5 = 0 and x 2 + y 2 − 10x − 16y + 7 = 0 are collinear., 14. Find the equation of the circle which passes through the points A(1, 1) and, B( 2, 2) and whose radius is 1. Show that there are two such circles., 15. Find the equation of a circle passing through the origin and intercepting, lengths a and b on the axes.
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Circle, , 731, , 16. Find the equation of the circle circumscribing the triangle formed by the, lines x + y = 6, 2x + y = 4 and x + 2y = 5., 17. Show that the quadrilateral formed by the straight lines x − y = 0,, 3 x + 2y = 5 , x − y = 10 and 2x + 3 y = 0 is cyclic and hence find the equation, of the circle., 18. If (–1, 3) and (α , β ) are the extremities of the diameter of the circle, x 2 + y 2 − 6x + 5 y − 7 = 0, find the coordinates (α , β )., ANSWERS (EXERCISE 21B), , 1. Centre (2, –3), radius = 3 2, , 1, 2. Centre ⎛⎜ − ,, ⎝ 2, , 1⎞, 1, ⎟ , radius =, ⎠, 2, 2, , 2, 4, 3. Centre ⎛⎜ −1, ⎞⎟ , radius =, ⎝, ⎠, 3, 3, , 4. Centre ( −1, − 5), , 5 1, 26, (i) x 2 + y 2 − 5 x − y = 0, centre ⎛⎜ , ⎞⎟ and radius =, ⎝ 2 2⎠, 2, , 6., , (ii) x 2 + y 2 − 22x − 4y + 25 = 0, centre (11, 2) and radius = 10, (iii) x 2 + y 2 − 14x − 6y − 111 = 0, centre (7, 3) and radius = 13, 7. x 2 +, 8. x 2 +, 9. x 2 +, 10. x 2 +, 12. x 2 +, 14. x 2 +, 15. x 2 +, 17. x 2 +, , y 2 − 2x + 2y − 23 = 0, centre (1, –1) and radius = 5, y 2 + 4x + 6y − 85 = 0, y 2 − 10x + 6y + 9 = 0, centre C(5 , − 3) and radius = 5, 11. x 2 + y 2 − 4x − 6y + 9 = 0, y 2 − 3 x − 2y − 1 = 0, 2, y − 6x + 12y − 15 = 0, y 2 − 4x − 2y + 4 = 0, x 2 + y 2 − 2x − 4y + 4 = 0, 16. x 2 + y 2 − 17 x − 19y + 50 = 0, y 2 − ax − by = 0, 18. B(7 , − 8), y 2 − 6x + 4y = 0, HINTS TO SOME SELECTED QUESTIONS, , 4. Show that the radius of the circle is 0., , −3, 3, , f = and c = 10., 2, 2, 7. Find the equation of the circle passing through the points A( −2 , 3 ), B(5 , 2 ) and, C( 6 , − 1)., 8. Centre of the required circle is C( −2 , − 3 )., 5. Show that ( g 2 + f 2 − c ) < 0 , where g =, , The circle passes through the point P(5 , 4 )., Radius =|CP| = (5 + 2 ) 2 + ( 4 + 3 ) 2 = 7 2 ., ∴, , the required equation is ( x + 2 ) 2 + ( y + 3 ) 2 = 98., , 11. Centre of the given circle is (2, 3)., Centre of the required circle is (2, 3)., The required circle touches the y-axis, so its radius = 2., ∴ the required equation is ( x − 2 ) 2 + ( y − 3 ) 2 = 2 2 .
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732, , Senior Secondary School Mathematics for Class 11, , 13. Centres of the given circles are A( 2 , 3 ), B( −1, − 2 ) and C(5 , 8 ) respectively. Now, show, that BA + AC = BC ., Hence B, A , C are collinear., 14. Let the centre of the circle be C ( h , k )., Then,|CA| = 1and|CB| = 1, ⇒, , ( h − 1) 2 + ( k − 1) 2 = 12 and ( h − 2 ) 2 + ( k − 2 ) 2 = 12, , ⇒, , h 2 + k 2 − 2h − 2k + 1 = 0, 2, , … (i), , 2, , and h + k − 4 h − 4 k + 7 = 0 ., , … (ii), , Subtracting (ii) from (i), we get, 2 h + 2 k − 6 = 0 ⇒ h + k = 3 ⇒ k = ( 3 − h )., Putting this value in (i), we get, h 2 + ( 3 − h ) 2 − 2 h − 2( 3 − h ) + 1 = 0, ⇒ ( h − 2 )( h − 1) = 0 ⇒ h = 2 or h = 1., ∴ k = 1 or k = 2., Hence, the centres are (2, 1) and (1, 2)., 15. The circle passes through the points O ( 0 , 0 ), A( a, 0 ) and B( 0 , b )., 5, 18. Centre of the circle is C ⎛⎜ 3 , − ⎞⎟ ⋅, ⎝, 2⎠, One end of the diameter AB is A( −1, 3 )., Let the other end of the diameter be B(α , β )., Then, C is the midpoint of AB., −1 + α, 3 + β −5 ⎞, = 3 and, =, ∴ ⎛⎜, ⎟ ⇒ (α = 7 and β = −8)., ⎝ 2, 2, 2 ⎠, , q
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22, Parabola, CONIC SECTIONS Under this heading, we shall, study about the curves, namely parabola, ellipse and, hyperbola. These curves can be obtained as, intersections of a plane with a double-napped right, circular cone., NAPPES Let L be a fixed vertical line and let M be, another line intersecting it at a fixed point A and, inclined to it at an angle α., When M is rotated around the line L such that, α remains constant, then the surface generated is a, double-napped right circular hollow cone., The line L is called the axis of the cone, A is called its vertex and the line M, is called the generator., The two parts, separated by the point A are called nappes., , CONIC SECTIONS When a right circular cone is intersected by a plane, the curves, obtained are known as conic sections., , INTERSECTION OF NAPPES OF A CONE AND A PLANE, , Suppose that a plane cuts double-napped cones each with semi-vertical, angle α., Let β be the angle made by the plane with the axis of the cone., CASE 1 When the plane cuts the nappes, but not at the vertex, In this case, the following situations arise., (i) When α < β = 90°, the section is a circle. [See Fig. (i)], (ii) When α < β < 90° , the section is an ellipse. [See Fig. (ii)], 733
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734, , Senior Secondary School Mathematics for Class 11, , (iii) When β = α , the section is a parabola. [See Fig. (iii)], (iv) When 0 ≤ β < α , the plane cuts both the nappes and the curves of, intersection is a hyperbola. [See Fig. (iv)], , CASE 2, , When the plane cuts the nappe at the vertex, In this case, the following situations arise., (i) When α < β ≤ 90° , then the section is a point. [Fig. (v)], (ii) When β = α , then the section is a line. [Fig. (vi)], (iii) When 0 ≤ β < α , then the section is a pair of straight lines., [Fig. (vii)], , PARABOLA, PARABOLA It is the path traced by a point which moves in a plane in such a way that, its distance from a fixed point is always equal to its distance from a fixed line, both, lying in the same plane, whereas the given fixed point does not lie on the given line., The fixed point is called the focus of the parabola and the fixed line is, called its directrix.
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Parabola, , 735, , A line through the focus and perpendicular to the directrix is called the, axis of the parabola., The point of intersection of the, parabola with its axis is called the vertex of, the parabola., In the adjoining figure, C is a parabola, with focus F and the line DD′ as its, directrix., Clearly, x-axis is the axis of the, parabola and O(0, 0) is its vertex., If we take an arbitrary point P on the, parabola and draw PM ⊥ DD′ then by the definition of a parabola, we have, PF = PM., RIGHT-HANDED PARABOLA (or First Standard Form):, , y 2 = 4ax , a > 0, , Let X′ OX and YOY′ be the coordinate axes, and let a > 0 be given., Let us consider a parabola whose focus, is F( a , 0) and the directrix is the line DD′,, whose equation is x + a = 0., Let P( x , y) be an arbitrary point on the, parabola. Let PM ⊥ DD′., Then, by the definition of a parabola,, we have PF = PM., Now, PF = PM ⇒ PF2 = PM 2, ⇒ ( x − a) 2 + y 2 = ( x + a) 2 [Q length of perp. from P( x , y), on the line ( x + a) = 0 is ( x + a)], ⇒ y 2 = ( x + a) 2 − ( x − a) 2, ⇒ y 2 = 4ax ( a > 0)., Thus, every point on the parabola satisfies the equation, y 2 = 4ax., Conversely, let P( x , y) be a point satisfying the equation, y 2 = 4ax., Let F( a , 0) be a given point and ( x + a) = 0 be a given line DD′, and let, PM ⊥ DD′. Then,, PF = ( x − a) 2 + y 2, = ( x − a) 2 + 4ax, , [Q y 2 = 4ax], , = ( x + a) 2, = ( x + a) = PM, [Q PM = length of perp. from P( x , y) on the line ( x + a) = 0]., This shows that P( x , y) lies on the parabola, y 2 = 4ax.
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736, , Senior Secondary School Mathematics for Class 11, , Hence, the equation of a parabola with vertex at the origin, focus at the, point F( a , 0) and directrix x = − a is y 2 = 4ax., This equation is known as the first standard form of a parabola or a righthanded parabola., FOCAL CHORD Any chord of a parabola passing through its focus is called a focal, chord., FOCAL DISTANCE OF A POINT The distance of any point of a parabola from its focus is, called the focal distance of that point., LATUS RECTUM OF A PARABOLA A chord of a parabola, passing through its focus and, perpendicular to its axis, is called the latus rectum of the parabola., In the figure given below, LFL′ is the latus rectum of the given parabola, with focus F., LENGTH OF LATUS RECTUM OF A PARABOLA, , Let LFL′ be the latus rectum of a, parabola y 2 = 4ax and let|FL| = l., Then, clearly the coordinates of L and L′, are L( a , l) and L′( a , − l )., Since the point L( a , l ) lies on the, parabola y 2 = 4ax , we have, l 2 = 4a 2 ⇒ l = 2a ., ∴ |LL′| = 2 l = 4a., Hence, the length of the latus, rectum is 4a., The end points of the latus rectum are L( a , 2a) and L′( a , − 2a)., The equation of the latus rectum is x = a., SUMMARY, , y 2 = 4ax , a > 0 is a parabola whose, (i) focus is F( a , 0), (ii) vertex is O( 0, 0), (iii) directrix is the line x + a = 0, (iv) axis is the line y = 0, (v) length of the latus rectum is 4a, (vi) latus rectum is the line x = a., LEFT-HANDED PARABOLA (or 2nd Standard Form):, , y2 = − 4ax, a > 0, , Proceeding as above, we can easily show that y 2 = − 4ax is the equation of a, parabola whose focus is F( − a , 0) and directrix is x − a = 0., We summarise the properties of the parabola, y 2 = − 4ax as given below.
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Parabola, , 737, , SUMMARY, , y 2 = − 4ax , a > 0 is a parabola whose, (i) focus is F( − a , 0), (ii) vertex is O( 0, 0), (iii) directrix is the line x − a = 0, (iv) axis is the line y = 0, (v) length of the latus rectum is 4a, (vi) latus rectum is the line x = − a., UPWARD PARABOLA (or 3rd Standard Form):, , x2 = 4ay, a > 0, , This is easy to show that x 2 = 4ay is the equation of a parabola whose focus is, F( 0, a) and directrix is y + a = 0., We summarise the properties of the, parabola, x 2 = 4ay as given below., SUMMARY, , x 2 = 4ay , a > 0 is a parabola whose, (i) focus is F( 0, a ), (ii) vertex is O( 0, 0), (iii) directrix is the line y + a = 0, (iv) axis is the line x = 0, (v) length of latus rectum is 4a, (vi) latus rectum is the line y − a = 0., DOWNWARD PARABOLA (or 4th Standard Form):, , x2 = − 4ay, a > 0, , We can easily prove that x 2 = − 4ay is the equation of a parabola whose focus is, F( 0, − a) and directrix is y − a = 0., We summarise the properties of the parabola, x 2 = − 4ay as given below., SUMMARY, , x 2 = − 4ay , a > 0 is a parabola whose, (i) focus is F( 0, − a), (ii) vertex is O( 0, 0), (iii) directrix is the line y − a = 0, (iv) axis is the line x = 0, (v) length of latus rectum is 4a, (vi) latus rectum is the line y + a = 0.
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738, , Senior Secondary School Mathematics for Class 11, , MAIN FACTS ABOUT ALL TYPES OF PARABOLAS, Focus Vertex, , Equation, Equation, of, of axis, directrix, , Length, of latus, rectum, , Equation, of latus, rectum, , ( a , 0), , (0, 0), , x+a=0, , y=0, , 4a, , x−a=0, , (ii) y 2 = − 4ax , a > 0 ( −a , 0) (0, 0), , x−a=0, , y=0, , 4a, , x+a=0, , (0, 0), , y+a=0, , x=0, , 4a, , y−a=0, , (iv) x 2 = − 4ay , a > 0 ( 0, − a ) (0, 0), , y−a=0, , x=0, , 4a, , y+a=0, , Parabola, (Equation), (i) y 2 = 4ax , a > 0, , (Right-handed), (Left-handed), (iii) x 2 = 4ay , a > 0, , ( 0, a ), , (Upward), (Downward), SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , EXAMPLE 2, , SOLUTION, , Find the coordinates of the focus and the vertex, the equations of the, directrix and the axis, and length of the latus rectum of the parabola, y 2 = 8x., The given equation is of the form, y 2 = 4ax , where 4a = 8, i.e., a = 2., This is a right-handed parabola., Its focus is F( a , 0), i.e., F( 2, 0)., Its vertex is O( 0, 0)., The equation of the directrix is, x = − a , i.e., x = −2., Its axis is x-axis, whose equation, is y = 0., Length of latus rectum = 4a = ( 4 × 2) units = 8 units., Find the coordinates of the focus and the vertex, the equations of the, directrix and the axis, and length of the latus rectum of the parabola, y 2 = −12x., The given equation is of the form, y 2 = − 4ax , where 4a = 12, i.e., a = 3., This is a left-handed parabola., Its focus is F( − a , 0), i.e., F( −3 , 0)., Its vertex is O( 0, 0)., The equation of the directrix is x = a ,, i.e., x = 3., Its axis is x-axis, whose equation is, y = 0., Length of latus rectum = 4a = ( 4 × 3) units = 12 units.
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Parabola, EXAMPLE 3, , SOLUTION, , 739, , Find the coordinates of the focus and the vertex, the equations of the, directrix and the axis, and length of the latus rectum of the parabola, x 2 = 6y., 3, The given equation is of the form x 2 = 4ay , where 4a = 6, i.e., a = ⋅, 2, So, this is a case of upward parabola., 3, Its focus is F( 0, a), i.e., F⎛⎜ 0, ⎞⎟ ⋅, ⎝ 2⎠, Its vertex is O( 0, 0)., The equation of the directrix is, −3, y = − a , i.e., y =, , i.e., 2y + 3 = 0., 2, Its axis is y-axis, whose equation, is x = 0., 3, Length of latus rectum = 4a = ⎛⎜ 4 × ⎞⎟ units = 6 units., ⎝, 2⎠, , EXAMPLE 4, , Find the coordinates of the focus and the vertex, the equations of the, directrix and the axis, and length of latus rectum of the parabola, x 2 = −16y., , SOLUTION, , The given equation is of the form, x 2 = − 4ay , where 4a = 16, i.e.,, a = 4., So, it is a case of downward, parabola., Its focus is F( 0, − a), i.e., F( 0, − 4)., Its vertex is O( 0, 0)., The equation of the directrix is, y = a , i.e., y = 4., Its axis is y-axis, whose equation, is x = 0., Length of latus rectum = 4a = ( 4 × 4) units = 16 units., , EXAMPLE 5, , Find the equation of the parabola with focus at F( 3 , 0) and directrix, x = −3., Since the focus lies on the x-axis,, so x-axis is the axis of the, parabola., Focus F( 3 , 0) lies to the righthand side of the origin., So, it is a right-handed parabola., Let the required equation be, y 2 = 4ax., Then, focus is F( a , 0). So, a = 3., Hence, the required equation is y 2 = 12x., , SOLUTION
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740, EXAMPLE 6, , Senior Secondary School Mathematics for Class 11, , Find the equation of the parabola with vertex at the origin and y + 3 = 0, as its directrix. Also, find its focus., , SOLUTION, , Let the vertex of the parabola, be O( 0, 0)., Now, y + 3 = 0 ⇒ y = −3., Thus, the directrix is a line, parallel to the x-axis at a, distance of 3 units below the, x-axis., So, the focus of the parabola lies, on the y-axis., Consequently, the focus is, F( 0, 3)., Hence, the equation of the, parabola is, x 2 = 4ay , where a = 3,, i.e., x 2 = 12y., , EXAMPLE 7, , Find the equation of the parabola with vertex at (0, 0) and focus at (0, 2)., Also, find the equation of its directrix., Let O( 0, 0) be the vertex and F( 0, 2) be the, focus of the given parabola., Since focus lies on the y-axis, so y-axis is the, axis of the parabola., Also, the focus lies above the x-axis., So, it is a case of upward parabola., Let the equation of the parabola be x 2 = 4ay., Then, its focus is F( 0, a) and so a = 2., Hence, the required equation is x 2 = 8y., Also, the equation of the directrix is y = −2., , SOLUTION, , EXAMPLE 8, SOLUTION, , Find the equation of the parabola with vertex at the origin, passing, through the point P( 3 , − 4) and symmetric about the y-axis., It is given that the vertex of the parabola is, O( 0, 0) and it is symmetric about the, y-axis., So, its equation is x 2 = 4ay or x 2 = − 4ay., Since the parabola passes through the, point P( 3 , − 4), so it lies in the 4th, quadrant., ∴ it is a downward parabola., Let its equation be x 2 = − 4ay., Since it passes through the point P( 3 , − 4),, we have
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Parabola, , 3 2 = − 4 × a × ( − 4) ⇒ a =, , 741, , 9, ⋅, 16, , So, the required equation is, 9, −9, x2 = − 4 ×, y ⇒ x2 =, y ⇒ 4x 2 + 9y = 0., 16, 4, EXAMPLE 9, , SOLUTION, , Find the equation of the parabola with vertex at the origin, the axis along, the x-axis and passing through the point P( 2, 3)., It is given that the vertex of the, parabola is O( 0, 0) and its axis lies, along the x-axis., So, its equation is y 2 = 4ax or y 2 = − 4ax., Since it passes through the point, P( 2, 3), so it lies in the first quadrant., ∴ it is a right-handed parabola., So, its equation is y 2 = 4ax., Now, P( 2, 3 ) lies on it, so, 9, 3 2 = 4a × 2 ⇒ a = ⋅, 8, Hence, the required equation is y 2 = 4 ×, , 9, 9, x ⇒ y 2 = x., 8, 2, , EXERCISE 22, 1. Find the coordinates of the focus and the vertex, the equations of the, directrix and the axis, and length of the latus rectum of the parabola:, (i) y 2 = 12x, (ii) y 2 = 10x, (iii) 3 y 2 = 8x, 2. Find the coordinates of the focus and the vertex, the equations of the, directrix and the axis, and length of the latus rectum of the parabola:, (ii) y 2 = − 6x, (iii) 5 y 2 = −16x, (i) y 2 = −8x, 3. Find the coordinates of the focus and the vertex, the equations of the, directrix and the axis, and length of the latus rectum of the parabola:, (ii) x 2 = 10y, (iii) 3 x 2 = 8y, (i) x 2 = 16y, 4. Find the coordinates of the focus and the vertex, the equations of the, directrix and the axis, and length of the latus rectum of the parabola:, (i) x 2 = −8y, (ii) x 2 = −18y, (iii) 3 x 2 = −16y, 5. Find the equation of the parabola with vertex at the origin and focus at, F( −2, 0)., 6. Find the equation of the parabola with focus F( 4, 0) and directrix x = − 4., 7. Find the equation of the parabola with focus F( 0, − 3) and directrix y = 3., 8. Find the equation of the parabola with vertex at the origin and focus F( 0, 5).
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742, , Senior Secondary School Mathematics for Class 11, , 9. Find the equation of the parabola with vertex at the origin, passing through, the point P(5 , 2) and symmetric with respect to the y-axis., 10. Find the equation of the parabola which is symmetric about the y-axis and, passes through the point P( 2, − 3)., ANSWERS (EXERCISE 22), , 1., , (i) F( 3 , 0), O( 0, 0), x + 3 = 0, y = 0, 12 units, 5, (ii) F ⎛⎜ , 0⎞⎟ , O( 0, 0), 2x + 5 = 0, y = 0, 10 units, ⎝2 ⎠, 2, 8, (iii) F ⎛⎜ , 0⎞⎟ , O( 0, 0), 3 x + 2 = 0, y = 0, units, ⎝3 ⎠, 3, , 2., , (i) F( −2, 0), O( 0, 0), x = 2, y = 0, 8 units, −3 ⎞, (ii) F ⎛⎜, , 0⎟ , O( 0, 0), 2x − 3 = 0, y = 0, 6 units, ⎝ 2 ⎠, 16, −4, units, (iii) F ⎛⎜ , 0⎞⎟ , O( 0, 0), 5 x − 4 = 0, y = 0,, ⎝5, ⎠, 5, , 3., , (i) F( 0, 4), O( 0, 0), y + 4 = 0, x = 0, 16 units, 5, (ii) F ⎛⎜ 0, ⎞⎟ , O( 0, 0), 2y + 5 = 0, x = 0, 10 units, ⎝ 2⎠, 2, 8, (iii) F ⎛⎜ 0, ⎞⎟ , O( 0, 0), 3 y + 2 = 0, x = 0,, units, ⎝ 3⎠, 3, , 4., , (i) F( 0, − 2), O( 0, 0), y = 2, x = 0, 8 units, −9, (ii) F ⎛⎜ 0, ⎞⎟ , O( 0, 0), 2y − 9 = 0, x = 0, 18 units, ⎝, 2⎠, −4, 16, units, (iii) F ⎛⎜ 0, ⎞⎟ , O( 0, 0), 3 y − 4 = 0, x = 0,, ⎝, 3, 3⎠, , 5. y 2 = −8x, , 6. y 2 = 16x, , 7. x 2 = −12y, , 8. x 2 = 20y, , 9. 2x 2 = 25 y, , 10. 3 x 2 = − 4y, q
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23, Ellipse, ELLIPSE It is the path traced by a point which moves in a plane in such a way that the, sum of its distance from two fixed points in the plane is a constant., The two fixed points are called the foci of the ellipse., NOTE, , The plural of focus is foci., , In the given figure, F1 and F2 are two, fixed points and P is a point which moves, in such a way that PF1 + PF2 = constant., The path traced by the point P is called an, ellipse, and the points F1 and F2 are called, its foci., SOME MORE TERMS RELATED TO AN ELLIPSE, (I) CENTRE OF THE ELLIPSE, , The midpoint of the line segment joining the, foci, is called the centre of the ellipse., In the given figure, F1 and F2 are the, foci of the ellipse and O is its centre,, where OF1 = OF2 ., (II) AXES OF THE ELLIPSE, , The line segment through the foci of the ellipse with its end points on, the ellipse, is called its major axis., In the given figure, AB is the major axis of the ellipse., , MAJOR AXIS:, , MINOR AXIS: The line segment through the centre and perpendicular to the major axis, with its end points on the ellipse, is called its minor axis., In the given figure, CD is the minor axis of the ellipse., (III) VERTICES OF AN ELLIPSE, , The end points of the major axis of an ellipse are called its vertices., In the given figure, A and B are the vertices of the ellipse., AN IMPORTANT NOTE, , In an ellipse, we take:, Length of the major axis = AB = 2a., Length of the minor axis = CD = 2b., 743
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744, , Senior Secondary School Mathematics for Class 11, , Distance between the foci = F1 F2 = 2c., Length of the semi-major axis = a., Length of the semi-minor axis = b., (IV) ECCENTRICITY OF AN ELLIPSE, , c, is always constant and it is denoted by e, called the eccentricity of, a, the ellipse., c, For an ellipse, we have 0 < e < 1 ⎡Q c < a ⇔ e = < 1⎤ ⋅, ⎢⎣, ⎥⎦, a, , The ratio, , AN IMPORTANT RESULT, , In the given ellipse, it is being given that AB = 2a , CD = 2b and F1 F2 = 2c., Prove that: ( a 2 − c2 ) = b 2 ., , PROOF, , Clearly, COD is the perpendicular bisector of F1 F2 ., Join F1C and F2C., Now, F1O = c, OC = b ⇒ F1C = c2 + b 2 ., And, OF2 = c, OC = b ⇒ F2C = c2 + b 2 ., Since B and C both lie on the given ellipse, we have, F1 B + F2 B = F1C + F2C = constant [by the definition of an ellipse], ⇒ ( F1O + OB) + (OB − OF2 ) = F1C + F2C, ⇒ ( c + a) + ( a − c) = 2 c2 + b 2, [Q F1O = c, OF2 = c, OB = a and F1C = F2C = c2 + b 2 ], ⇒ a = c2 + b 2 ⇒ a 2 = ( c2 + b 2 ) ⇒ ( a 2 − c2 ) = b 2 ., Hence, ( a 2 − c2 ) = b 2 ., , STANDARD EQUATION OF AN ELLIPSE, THEOREM, , Prove that the standard equation of an ellipse is, x2 y2, + 2 = 1,, a2, b, where a and b are the lengths of the semi-major axis and the semi-minor, axis respectively and a > b.
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Ellipse, PROOF, , 745, , Let X′ OX and YOY′ be the coordinate axes., Let us consider an ellipse in which a and b are the lengths of the, semi-major axis and semi-minor axis respectively., , Choose a real number c such that c2 = ( a 2 − b 2 )., Let F1 ( − c, 0) and F2 ( c, 0) be the two fixed points and let P( x , y) be an, arbitrary point on the given ellipse, moving on it in such a way that, PF1 + PF2 = 2a. Then,, PF1 + PF2 = 2a, ⇒, , ( x + c) 2 + y 2 + ( x − c) 2 + y 2 = 2a, , ⇒, , ( x + c) 2 + y 2 = 2a − ( x − c) 2 + y 2, , ⇒ ( x + c) 2 + y 2 = 4a 2 + ( x − c) 2 + y 2 − 4a ( x − c) 2 + y 2, [on squaring both sides], ⇒ ( x + c) 2 − ( x − c) 2 − 4a 2 = −4a ( x − c) 2 + y 2, ⇒ 4cx − 4a 2 = −4a ( x − c) 2 + y 2, cx, ⇒ ( x − c) 2 + y 2 = a −, a, c2 x 2, 2, 2, 2, ⇒ ( x − c) + y = a − 2cx + 2, a, c2 x 2, 2, 2, 2, 2, ⇒ x − 2 + y = (a − c ), a, c2 ⎞, 2⎛, ⇒ x ⎜ 1 − 2 ⎟ + y 2 = ( a 2 − c2 ), a ⎠, ⎝, x 2 ( a 2 − c2 ), + y 2 = ( a 2 − c2 ), a2, x2, y2, =1, ⇒ 2 + 2, ( a − c2 ), a, x2 y2, ⇒ 2 + 2 =1, a, b, , [on dividing both sides by −4a], [on squaring both sides], , ⇒, , [on dividing both sides by ( a 2 − c2 )], , Thus, every point on the ellipse satisfies the equation, , [Q ( a 2 − c2 ) = b 2 ]., x2 y2, + 2 = 1., a2, b
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746, , Senior Secondary School Mathematics for Class 11, , Conversely, let the equation of the given curve be, , x2 y2, + 2 = 1 and let, a2, b, , P( x , y) be an arbitrary point on this curve., Then,, , ⎛, x2 y2, x2 ⎞, + 2 = 1 ⇒ y 2 = b2 ⎜1 − 2 ⎟, 2, a, b, a ⎠, ⎝, ⇒ y2 =, , b2 ( a 2 − x 2 ), ⋅, a2, , … (i), , Also, let ( a 2 − b 2 ) = c2 ., , … (ii), , Let F1 ( − c, 0) and F2 ( c, 0) be two fixed points on the x-axis. Then,, PF1 = ( x + c) 2 + y 2, = ( x + c) 2 +, , b2 ( a 2 − x 2 ), a2, , [using (i)], , = ( x + c) 2 +, , ( a 2 − c2 )( a 2 − x 2 ), a2, , [using (ii)], , c2 x 2, cx, = ⎛⎜ a + ⎞⎟, ⎝, a⎠, a2, cx, Similarly, PF2 = ⎛⎜ a − ⎞⎟ ⋅, ⎝, a⎠, = a 2 + 2cx +, , 2, , cx, = ⎛⎜ a + ⎞⎟, ⎝, a⎠, , cx, cx, PF1 + PF2 = ⎛⎜ a +, + a − ⎞⎟, ⎝, a, a⎠, ⇔ PF1 + PF2 = 2a., This shows that the given curve is an ellipse., , ∴, , x2 y2, + 2 = 1., a2, b, The foci of an ellipse always lie on the major axis., , Hence, the equation of the ellipse is, AN IMPORTANT NOTE, , HORIZONTAL ELLIPSE, , In the given equation of an ellipse, if the coefficient of x 2 has the larger denominator, then its major axis lies along the x-axis., Such an ellipse is called a horizontal ellipse., x2 y2, Thus, 2 + 2 = 1 is an horizontal ellipse, if a 2 > b 2 ., a, b, x2 y2, EXAMPLE, +, = 1 is an horizontal ellipse., 16, 9, LATUS RECTUM OF A HORIZONTAL ELLIPSE, , The latus rectum of an ellipse is a line segment perpendicular to the major axis,, passing through any of the foci with end points lying on the ellipse.
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Ellipse, , 747, , To Find the Length of Latus Rectum of the Ellipse, , x2 y 2, +, = 1, a > b., a2 b 2, , Let X′ OX and YOY′ be the coordinate axes., Let F1 ( − c, 0) and F2 ( c, 0) be the foci of the given ellipse, where, c2 = ( a 2 − b 2 )., Let LF1 M and PF2Q be the latus rectums., Let PF2 = l. Then, the coordinates of P are P( c, l)., , x2 y2, + 2 = 1, we have, a2, b, 2, 2, 2, 2, ⎛, c, l, l, c ⎞, +, = 1 ⇒ 2 = ⎜1 − 2 ⎟, a 2 b2, b, a ⎠, ⎝, , Since P( c, l) lies on the ellipse, , ⎛, ⎡, ( a 2 − b2 ) ⎤, c2 ⎞, ⇒ l 2 = b 2 ⎜1 − 2 ⎟ = b 2 ⎢1 −, ⎥, a ⎠, a2, ⎝, ⎣, ⎦, b4, b2, ⇒ l=, ⋅, 2, a, a, 2, 2, b, 2b, ∴ PF2 = l =, ⇔ PF2Q = 2l =, ⋅, a, a, 2b 2, Similarly, LF1 M = 2l =, ⋅, a, ⇒ l2 =, , Hence, the length of the latus rectum is, , 2b 2, ⋅, a, , RESULTS ON HORIZONTAL ELLIPSE, , (i) The standard form of equation of a horizontal ellipse is, x2 y2, + 2 = 1, 0 < b < a., a2, b, (ii) Its centre is O( 0, 0)., (iii) Its vertices are A( − a , 0) and B( a , 0)., (iv) Its foci are F1 ( − c, 0) and F2 ( c, 0), where c2 = ( a 2 − b 2 )., , [Q c2 = ( a 2 − b 2 )]
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748, , Senior Secondary School Mathematics for Class 11, , (v) Length of the major axis, AB = 2a and length of the minor axis, CD = 2b., (vi) Equation of the major axis is y = 0 and that of the minor axis is x = 0., (vii) Length of the latus rectum =, (viii) Eccentricity, e =, , c, =, a, , 2b 2, ⋅, a, , a 2 − b2, ⋅, a, , VERTICAL ELLIPSE, , In the given equation of an ellipse, if the coefficient of x 2 has the smaller denominator,, then its major axis lies along the y-axis., Such an ellipse is called a vertical ellipse., , RESULTS ON VERTICAL ELLIPSE, , x2 y2, (i) The standard form of the equation of a vertical ellipse is 2 + 2 = 1,, b, a, where 0 < b < a., (ii) Its centre is O( 0, 0)., (iii) Its vertices are A( 0, − a) and B( 0, a)., (iv) Its foci are F1 ( 0, − c) and F2 ( 0, c), where a 2 − b 2 = c2 ,, i.e., F1 ( 0, − ae) and F2 ( 0, ae)., (v) Length of the major axis = 2a and length of the minor axis = 2b.
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Ellipse, , 749, , (vi) Equation of the major axis is x = 0 and that of the minor axis is y = 0., 2b 2, (vii) Length of latus rectum =, ⋅, a, (viii) Eccentricity, e =, , c, =, a, , a 2 − b2, ⋅, a, , SUMMARY, , Horizontal Ellipse, x2 y 2, +, =1, a2 b 2, , Properties, , Vertical Ellipse, x2 y 2, +, =1, b 2 a2, , 0 < b < a and c 2 = (a2 − b 2 ) 0 < b < a and c 2 = (a2 − b 2 ), (i) Centre, (ii) Vertices, , (0, 0), A( − a , 0), B( a , 0), , (0, 0), A( 0, − a) and B( 0, a ), , (iii) Foci, , F1 ( − c, 0) and F2 ( c, 0) or F1 ( 0, − c) and F2 ( 0, c) or, ( −ae, 0) and ( ae, 0), ( 0, − ae) and ( 0, ae), the, , 2a, , 2a, , (v) Length of the, minor axis, (vi) Equation of the, major axis, , 2b, , 2b, , y=0, , x=0, , x=0, , y=0, , 2b 2, a, , 2b 2, a, , (iv) Length of, major axis, , (vii) Equation of the, minor axis, (viii) Length of the, latus rectum, (ix) Eccentricity, , e=, , c, =, a, , a 2 − b2, a, , e=, , c, =, a, , a 2 − b2, a, , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Find the lengths of the major and minor axes; coordinates of the vertices, and the foci, the eccentricity and length of the latus rectum of the ellipse:, x2 y2, +, = 1., 16, 9, 2, 2, x, y, Given equation is, +, = 1., 16, 9, x2 y2, This is of the form 2 + 2 = 1, where a 2 > b 2 ., a, b, So, it is an equation of a horizontal ellipse., Now, (a 2 = 16 and b 2 = 9 ⇒ (a = 4 and b = 3)., ∴, , c = a 2 − b 2 = 16 − 9 = 7.
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750, , Senior Secondary School Mathematics for Class 11, , Thus, a = 4, b = 3 and c = 7., (i) Length of the major axis = 2a = ( 2 × 4) units = 8 units., Length of the minor axis = 2b = ( 2 × 3) units = 6 units., (ii) Coordinates of the vertices are A( − a , 0) and B( a , 0), i.e., A( −4, 0), and B( 4, 0)., (iii) Coordinates of the foci are F1 ( − c, 0) and F2 ( c, 0), i.e., F1 ( − 7 , 0), and F2 ( 7 , 0)., 7, c, (iv) Eccentricity, e = =, ⋅, 4, a, 2b 2 ( 2 × 9), 9, (v) Length of the latus rectum =, units = units., =, 3, 2, a, EXAMPLE 2 Find the lengths of the major and minor axes; coordinates of the vertices, and the foci; the eccentricity and length of the latus rectum of the ellipse:, 4x 2 + 9y 2 = 144., SOLUTION, , The given equation may be written as, x2 y2, +, = 1., 36 16, 2, 2, x, y, This is of the form 2 + 2 = 1, where a 2 > b 2 ., a, b, So, it is an equation of a horizontal ellipse., Now, ( a 2 = 36 and b 2 = 16) ⇒ ( a = 6 and b = 4)., ∴, , c = a 2 − b2 =, , 36 − 16 = 20 = 2 5., , Thus, a = 6, b = 4 and c = 2 5., (i) Length of the major axis = 2a = ( 2 × 6) units = 12 units., Length of the minor axis = 2b = ( 2 × 4) units = 8 units., (ii) Coordinates of the vertices are A( − a , 0) and B( a , 0), i.e., A( −6, 0), and B( 6, 0)., (iii) Coordinates of the foci are F1 ( − c, 0) and F2 ( c, 0), i.e., F1 ( −2 5 , 0), and F2 ( 2 5 , 0)., c 2 5, 5, (iv) Eccentricity, e = =, =, ⋅, a, 6, 3, 2b 2 ( 2 × 16), 16, units., units =, (v) Length of the latus rectum =, =, 6, a, 3, EXAMPLE 3 Find the lengths of the major and minor axes; coordinates of the vertices, and the foci; the eccentricity and length of the latus rectum of the ellipse:, x2 y2, +, = 1., 4, 36, 2, 2, x, y, SOLUTION Given equation is, +, = 1., 4, 36, x2 y2, This is of the form 2 + 2 = 1, where a 2 > b 2 ., b, a, So, it is an equation of a vertical ellipse., Now, ( b 2 = 4 and a 2 = 36) ⇒ ( b = 2 and a = 6).
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Ellipse, , ∴, , c = a 2 − b2 =, , 36 − 4 =, , 751, , 32 = 4 2., , Thus, a = 6, b = 2 and c = 4 2., (i) Length of the major axis = 2a = ( 2 × 6) units = 12 units., Length of the minor axis = 2b = ( 2 × 2) units = 4 units., (ii) Coordinates of its vertices are A( 0, − a) and B( 0, a), i.e., A( 0, − 6), and B( 0, 6)., (iii) Coordinates of its foci are F1 ( 0, − c) and F2 ( 0, c), i.e., F1 ( 0, − 4 2), and F2 ( 0, 4 2 )., c 4 2 2 2, (iv) Eccentricity, e = =, =, ⋅, a, 6, 3, 2, 2b, 4, ( 2 × 4), units = units., (v) Length of the latus rectum =, =, 6, 3, a, EXAMPLE 4 Find the lengths of the major and minor axes; coordinates of the vertices, and the foci; the eccentricity and length of the latus rectum of the ellipse:, 4x 2 + y 2 = 100., SOLUTION, , The given equation of the ellipse may be written as, x2, y2, +, = 1., 25 100, 2, 2, x, y, This is of the form 2 + 2 = 1, where a 2 > b 2 ., b, a, So, it is an equation of a vertical ellipse., Now, ( b 2 = 25 and a 2 = 100) ⇒ ( b = 5 and a = 10)., ∴, , c = a 2 − b 2 = 100 − 25 = 75 = 5 3., , Thus, a = 10, b = 5 and c = 5 3., (i) Length of the major axis = 2a = ( 2 × 10) units = 20 units., Length of the minor axis = 2b = ( 2 × 5) units = 10 units., (ii) Coordinates of the vertices are A( 0, − a) and B( 0, a), i.e.,, A( 0, − 10) and B( 0, 10)., (iii) Coordinates of the foci are F1 ( 0, − c) and F2 ( 0, c), i.e., F1 ( 0, −5 3 ), and F2 ( 0, 5 3 )., 3, c 5 3, (iv) Eccentricity, e = =, =, ⋅, 10, 2, a, 2b 2 ( 2 × 25), (v) Length of the latus rectum =, units = 5 units., =, 10, a, EXAMPLE 5 Find the equation of an ellipse whose vertices are at ( ± 5 , 0) and foci at, ( ± 4, 0)., SOLUTION, , Since the vertices of the given ellipse are on the x-axis, so it is a, horizontal ellipse., x2 y2, Let the equation of the ellipse be 2 + 2 = 1, where a 2 > b 2 ., a, b, Its vertices are ( ± a , 0) and therefore, a = 5., Its foci are ( ± c, 0) and therefore, c = 4.
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752, , Senior Secondary School Mathematics for Class 11, , ∴ c2 = ( a 2 − b 2 ) ⇒ b 2 = ( a 2 − c2 ) = ( 25 − 16) = 9., Thus, a 2 = 5 2 = 25 and b 2 = 9., x2 y2, Hence, the required equation is, +, = 1., 25, 9, EXAMPLE 6, SOLUTION, , Find the equation of an ellipse whose foci are ( ±4, 0) and the eccentricity is, , 1, ⋅, 3, , Since the foci of the given ellipse are on the x-axis, so it is a horizontal, ellipse., Let the required equation of the ellipse be, x2 y2, + 2 = 1, where a 2 > b 2 ., a2, b, Let its foci be ( ± c, 0). Then, c = 4., c, c, 4, Also, e =, ⇔ a= =, = 12., a, e (1/ 3), Now, c2 = ( a 2 − b 2 ) ⇒ b 2 = ( a 2 − c2 ) = (144 − 16) = 128., ∴ a 2 = (12) 2 = 144 and b 2 = 128., x2, y2, Hence, the required equation is, +, = 1., 144 128, , EXAMPLE 7, , SOLUTION, , Find the equation of an ellipse whose major axis lies on the x-axis and, which passes through the points (4, 3) and (6, 2)., Since the major axis of the ellipse lies on the x-axis, so it is a horizontal, ellipse., Let the required equation of the ellipse be, x2 y2, … (i), + 2 = 1 (where a 2 > b 2 )., a2, b, 16, 9, … (ii), Since (4, 3) lies on (i), we have 2 + 2 = 1., a, b, 36, 4, … (iii), Also, since (6, 2) lies on (i), we have 2 + 2 = 1., a, b, 1, 1, Putting 2 = u and 2 = v in (ii) and (iii), we get, a, b, … (iv), 16u + 9v = 1, … (v), 36u + 4v = 1, On multiplying (iv) by 9 and (v) by 4, and subtracting, we get, 1, 1, 1, 65v = 5 ⇔ v =, ⇔ 2 =, ⇔ b 2 = 13., 13, 13, b, 1, Putting v =, in (iv), we get, 13, 4, 4, 1, 1, 9⎞, ⎛, ⇔ u = ⎛⎜, × ⎞⎟ =, 16u = ⎜1 − ⎟ ⇔ 16u =, ⎝, ⎠, ⎝, ⎠, 13, 13 16, 52, 13, 1, 1, 2, ⇔ 2 =, ⇔ a = 52., 52, a
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Ellipse, , 753, , Thus, a 2 = 52 and b 2 = 13., Hence, the required equation is, EXAMPLE 8, , SOLUTION, , x2 y2, +, = 1., 52 13, , Find the equation of the ellipse, the ends of whose major axis are ( ±3 , 0), and the ends of whose minor axis are ( 0, ± 2)., Clearly, the major axis of the given ellipse lies on the x-axis. So, it is a, horizontal ellipse., x2 y2, + 2 = 1, where a 2 > b 2 ., a2, b, Its vertices are ( ± a , 0) and, therefore, a = 3, [Q ends of the major axis are the vertices]., Ends of the minor axis are C( 0, − 2) and D( 0, 2)., ∴ CD = 4, i.e., length of the minor axis = 4 units., ∴ 2b = 4 ⇔ b = 2., Now, ( a = 3 and b = 2) ⇒ ( a 2 = 9 and b 2 = 4)., x2 y2, Hence, the required equation is, +, = 1., 9, 4, Let the required equation be, , EXAMPLE 9, , SOLUTION, , Find the equation of the ellipse whose centre lies at the origin, major axis, 2, lies on the x-axis, the eccentricity is and the length of the latus rectum, 3, is 5 units., Since the major axis of the ellipse lies on the x-axis, it is a horizontal, ellipse., x2 y2, + 2 = 1, where a 2 > b 2 ., a2, b, 2b 2, Length of its latus rectum =, ⋅, a, 2, 2, 2, 2b, 2a (1 − e ), [Q b 2 = a 2 (1 − e2 )], =5 ⇔, =5, ∴, a, a, 5, 5, ⎛ 5 9⎞ 9, =⎜ × ⎟ = ⋅, =, ⇔ a=, 4⎞ ⎝ 2 5 ⎠ 2, 2(1 − e2 ), ⎛, 2⎜1 − ⎟, ⎝, 9⎠, 81 ⎛, 4, 81 5 ⎞ 45, 2, 2, 2, Also, b = a (1 − e ) =, × ⎜1 − ⎞⎟ = ⎛⎜, × ⎟ =, ⋅, 4 ⎝, 9⎠ ⎝ 4, 9⎠, 4, 2, 45, 81, ⎛ 9⎞, and b 2 =, ⋅, ∴ a2 = ⎜ ⎟ =, ⎝ 2⎠, 4, 4, Hence, the required equation is, Let the required equation be, , 4x 2 4y 2, x2, y2, +, =1 ⇔, +, = 1 ⇔ 20x 2 + 36y 2 = 405., ( 81/4) ( 45/4), 81, 45
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754, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 10, , SOLUTION, , Find the equation of an ellipse whose vertices are ( 0, ± 13) and the foci are, ( 0, ± 5)., , Since the vertices of the ellipse lie on the y-axis, it is a vertical ellipse., x2 y2, + 2 = 1, where a 2 > b 2 ., b2, a, Its vertices are ( 0, ± a) and therefore, a = 13., Let the required equation be, , Let its foci be ( 0, ± c). Then, c = 5., ∴, , b 2 = ( a 2 − c2 ) = (169 − 25) = 144., , Thus, b 2 = 144 and a 2 = (13) 2 = 169., x2, y2, Hence, the required equation is, +, = 1., 144 169, EXAMPLE 11, , SOLUTION, , Find the equation of the ellipse whose foci are ( 0, ± 6) and the length of, whose minor axis is 16., , Since the foci of the given ellipse lie on the y-axis, it is a vertical ellipse., Let the required equation be, , x2 y2, + 2 = 1, where a 2 > b 2 ., b2, a, , Let c2 = ( a 2 − b 2 )., Then, its foci are ( 0, ± c) and therefore, c = 6., 1, Also, b = length of the semi-minor axis = ⎛⎜ × 16⎞⎟ = 8., ⎝2, ⎠, ∴ a 2 = ( c2 + b 2 ) = ( 36 + 64) = 100., Thus, b 2 = 64 and a 2 = 100., Hence, the required equation is, EXAMPLE 12, , SOLUTION, , x2, y2, +, = 1., 64 100, , Find the equation of the ellipse whose foci are ( 0, ± 5) and the length of, whose major axis is 20., , Since the foci of the ellipse lie on the y-axis, it is a vertical ellipse., Let the required equation be, , x2 y2, + 2 = 1, where a 2 > b 2 ., b2, a, , Let c2 = ( a 2 − b 2 )., Its foci are ( 0, ± c) and therefore, c = 5., , 1, Also, a = length of the semi-major axis = ⎛⎜ × 20⎞⎟ = 10., ⎝2, ⎠, Now, c2 = ( a 2 − b 2 ) ⇔ b 2 = ( a 2 − c2 ) = (100 − 25) = 75., Thus, a 2 = (10) 2 = 100 and b 2 = 75., x2, y2, Hence, the required equation is, +, = 1., 75 100
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Ellipse, , EXAMPLE 13, , 755, , Find the equation of the ellipse for which e =, ( 0, ± 10)., , SOLUTION, , 4, and whose vertices are, 5, , Since the vertices of the ellipse lie on the y-axis, it is a vertical ellipse., Let the required equation be, , x2 y2, + 2 = 1, where a 2 > b 2 ., b2, a, , Its vertices are ( 0, ± a) and therefore, a = 10., Let c2 = ( a 2 − b 2 )., Then, e =, , 4, c, ⇒ c = ae = ⎛⎜10 × ⎞⎟ = 8., ⎝, 5⎠, a, , Now, c2 = ( a 2 − b 2 ) ⇔ b 2 = ( a 2 − c2 ) = (100 − 64) = 36., ∴, , a 2 = (10) 2 = 100 and b 2 = 36., , Hence, the required equation is, EXAMPLE 14, , SOLUTION, , x2, y2, +, = 1., 36 100, , Find the equation of the ellipse with centre at the origin, major-axis on, the y-axis and passing through the points (3, 2) and (1, 6)., , Since the major axis of the ellipse lies on the y-axis, it is a vertical, ellipse., x2 y2, + 2 = 1 (where a 2 > b 2 )., b2, a, 9, 4, Since (3, 2) lies on (i), we have 2 + 2 = 1., b, a, 1, 36, Also, since (1, 6) lies on (i), we have 2 + 2 = 1., b, a, 1, 1, Putting 2 = u and 2 = v , these equations become:, b, a, Let the required equation be, , … (i), … (ii), … (iii), , 9u + 4v = 1, , … (iv), , and u + 36v = 1, , … (v), , On multiplying (v) by 9 and subtracting (iv) from it, we get, 8, 1, 1, 1, 320v = 8 ⇔ v =, =, ⇔ 2 =, ⇔ a 2 = 40., 320 40, 40, a, 1, Putting v =, in (v), we get, 40, 1, 9, 1, 1, 1, u + ⎛⎜ 36 × ⎞⎟ = 1 ⇔ u = ⎛⎜1 − ⎞⎟ =, ⇔ 2 =, ⇔ b 2 = 10., ⎝, ⎝, 40⎠, 10⎠ 10, 10, b, Thus, b 2 = 10 and a 2 = 40., Hence, the required equation is, , x2 y2, +, = 1., 10 40
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756, , Senior Secondary School Mathematics for Class 11, , EXERCISE 23, Find the (i) lengths of major and minor axes, (ii) coordinates of the vertices,, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of, each of the following ellipses., 1., , x2 y2, +, =1, 25, 9, , 4. x 2 + 4y 2 = 100, 7., , x2 y2, +, =1, 4, 25, , 10. 9x 2 + y 2 = 36, , 2., , x2 y2, +, =1, 49 36, , 5. 9x 2 + 16y 2 = 144, 8., , x2 y2, +, =1, 9, 16, , 11. 16x 2 + y 2 = 16, , 3. 16x 2 + 25 y 2 = 400, 6. 4x 2 + 9y 2 = 1, 9. 3 x 2 + 2y 2 = 18, 12. 25 x 2 + 4y 2 = 100, , 13. Find the equation of the ellipse whose vertices are at ( ±6, 0) and foci at, ( ±4, 0)., 14. Find the equation of the ellipse whose vertices are at ( 0, ± 4) and foci at, ( 0, ± 7 )., 15. Find the equation of the ellipse the ends of whose major and minor axes are, ( ±4, 0) and ( 0, ± 3) respectively., 16. The length of the major axis of an ellipse is 20 units and its foci are( ±5 3 , 0)., Find the equation of the ellipse., 17. Find the equation of the ellipse whose foci are ( ±2, 0) and the eccentricity, 1, is ⋅, 2, 1, 18. Find the equation of the ellipse whose foci are at ( ±1, 0) and e = ⋅, 2, 4, 19. Find the equation of the ellipse whose foci are at ( 0, ± 4) and e = ⋅, 5, 20. Find the equation of the ellipse with centre at the origin, major axis on the, x-axis and passing through the points (4, 3) and ( −1, 4)., 3, 21. Find the equation of the ellipse with eccentricity , foci on the y-axis, centre, 4, at the origin and passing through the point (6, 4)., 22. Find the equation of the ellipse which passes through the point (4, 1) and, having its foci at ( ±3 , 0)., 23. Find the equation of an ellipse, the lengths of whose major and minor axes, are 10 and 8 units respectively., 2, 24. Find the equation of an ellipse whose eccentricity is , the latus rectum is, 3, 5 and the centre is at the origin.
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Ellipse, , 757, , 25. Find the eccentricity of an ellipse whose latus rectum is one half of its, minor axis., 26. Find the eccentricity of an ellipse whose latus rectum is one half of its, major axis., ANSWERS (EXERCISE 23), , 1. (i) 10 units, 6 units, 4, (iv) e =, 5, , (ii) A( −5 , 0) and B(5 , 0) (iii) F1 ( −4, 0) and F2 ( 4, 0), (v) 3.6 units, , 2. (i) 14 units, 12 units, , (ii) A( −7 , 0) and B(7 , 0), , (iii) F1 ( − 13 , 0) and F2 ( 13 , 0), 3. (i) 10 units, 8 units, (iv) e =, , 3, 5, , (ii) A( −10, 0) and B(10, 0), , (iii) F1 ( −5 3 , 0) and F2 (5 3 , 0), , 7, 4, , 6. (i) 1 unit,, , 3, 2, , (v) 5 units, , 9, units, 2, , 1, 1, (ii) A ⎛⎜ − , 0⎞⎟ and B ⎛⎜ , 0⎞⎟, ⎝2 ⎠, ⎝ 2 ⎠, , ⎛− 5 ⎞, ⎛ 5 ⎞, (iii) F1 ⎜, , 0⎟ and F2 ⎜, , 0⎟, ⎝ 6, ⎠, ⎝ 6, ⎠, , (iv) e =, , 5, 3, , (v), , 4, unit, 9, , (ii) A( 0, − 5 ) and B( 0, 5 ), , 7. (i) 10 units, 4 units, (iii) F1 ( 0, − 21 ) and F2 ( 0,, , 21 ), , (iv) e =, , 21, 5, , (v), , 8, units, 5, , (ii) A( 0, − 4) and B( 0, 4), , 8. (i) 8 units, 6 units, (iii) F1 ( 0, − 7 ) and F2 ( 0, 7 ), , (iv) e =, , 7, 4, , 1, (v) 4 units, 2, , (ii) A( 0, − 3) and B( 0, 3), , 9. (i) 6 units, 2 6 units, (iii) F1 ( 0, − 3 ) and F2 ( 0,, , (iv) e =, , (ii) A( −4, 0) and B( 4, 0) (iii) F1 ( − 7 , 0), F2 ( 7 , 0), (v), , 2, unit, 3, , 72, units, 7, , 32, units, 5, , 4. (i) 20 units, 10 units, , (iv) e =, , (v), , (ii) A( −5 , 0) and B(5 , 0) (iii) F1 ( −3 , 0) and F2 ( 3 , 0), (v), , 5. (i) 8 units, 6 units, , 13, 7, , (iv) e =, , 3), , (iv) e =, , 1, 3, , (v) 4 units
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758, , Senior Secondary School Mathematics for Class 11, , (ii) A( 0, − 6) and B( 0, 6), , 10. (i) 12 units, 4 units, (iii) F1 ( 0, − 4 2 ) and F2 ( 0, 4 2 ), , (iv) e =, , 2 2, 3, , (v) 1, , 1, units, 3, , (ii) A( 0, − 4) and B( 0, 4), , 11. (i) 8 units, 2 units, (iii) F1 ( 0, − 15 ) and F2 ( 0, 15 ), , (iv) e =, , 15, 4, , (v), , 1, unit, 2, , (ii) A( 0, − 5 ) and B( 0, 5 ), , 12. (i) 10 units, 4 units, (iii) F1 ( 0, − 21) and F2 ( 0,, , (iv) e =, , 21), , 21, 5, , (v) 1, , 3, units, 5, , 13., , x2 y2, +, =1, 36 20, , 14., , x2 y2, +, =1, 9, 16, , 15., , x2 y2, +, =1, 16, 9, , 16., , x2, y2, +, =1, 100 25, , 17., , x2 y2, +, =1, 16 12, , 18., , x2 y2, +, =1, 4, 3, , 19., , x2 y2, +, =1, 9, 25, , 20. 7 x 2 + 15 y 2 = 247, , 22., , x2 y2, +, =1, 18, 9, , 23., , 25. e =, , 3, 2, , x2 y2, +, =1, 25 16, , 26. e =, , 21. 16x 2 + 7 y 2 = 688, 24., , 4x 2 4y 2, +, =1, 81, 45, , 1, 2, HINTS TO SOME SELECTED QUESTIONS, , 2, , 6. Given equation is, ∴, , a2 =, , ∴, , a=, , 2, , y, x, +, = 1., ( 1/ 4 ) ( 1/ 9 ), , 1 2 1, 1 1, 5, , b = and c 2 = ( a2 − b 2 ) = ⎛⎜ − ⎞⎟ =, ⋅, ⎝ 4 9 ⎠ 36, 4, 9, , 1, 1, 5, , b = and c =, ⋅, 2, 3, 6, , 15. The ends of the major axis are ( ± a, 0 ) and the ends of the minor axis are ( 0 , ± b )., ∴ a = 4 and b = 3., x2, 2, , +, , y2, , = 1., a, b2, Length of the major axis = 20 ⇔ 2 a = 20 ⇔ a = 10., Foci are ( ± c , 0 ) ⇒ c = 5 3 ., , 16. Let the equation of the ellipse be, , Now, c 2 = ( a2 − b 2 ) ⇔ b 2 = ( a2 − c 2 ) = ( 100 − 75 ) = 25., ∴, , the required equation is, , y2, x2, +, = 1., 100 25, , 19. Clearly, the major axis lies along the y-axis., x2 y2, Let the required equation be, +, = 1., b2, a2
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Ellipse, , 759, , 4, c, ⋅ So, a = = 5., 5, e, b 2 = ( a2 − c 2 ) = ( 25 − 16 ) = 9., , Here, c = 4 and e =, ∴, , 21. Since the centre of the ellipse lies at the origin and its foci lie on the y-axis, it is a, vertical ellipse., x2 y2, Let its equation be, … (i), +, = 1., b2, a2, Let c 2 = a2 − b 2 ., 3, c 3, 3, Now, e =, ⇔ =, ⇔ c = a., 4, a 4, 4, 9 2 ⎞ 7 a2, a ⎟=, ∴ b 2 = ( a2 − c 2 ) = ⎛⎜ a2 −, ⋅, ⎝, 16 ⎠, 16, So, the equation of the ellipse is, y2, x2, +, = 1 ⇔ 16 x 2 + 7 y 2 = 7 a2 ., 2, 7a, a2, 16, Since it passes through (6, 4), putting x = 6 and y = 4 in (ii), we get, 7 a2 = 688., Hence, the required equation is 16 x 2 + 7 y 2 = 688., 23. 2 a = 10 and 2 b = 8., 2 a2 ( 1 − e 2 ), 2b 2, 9, 24., =5 ⇒, =5 ⇒ a= ⋅, a, a, 2, 5 9 ⎞ 45, 2 5a, ⎛, Also, b =, =⎜ × ⎟=, ⋅, 2 ⎝ 2 2⎠, 4, ⎧ 2b 2 1, a⎫, 25. ⎨, = × 2b ⇒ b = ⎬ ⋅, 2, 2⎭, ⎩ a, a2, So, b 2 = a2 ( 1 − e 2 ) ⇒, = a2 ( 1 − e 2 )., 4, 2b 2 1, b, 1, 26., = × 2 a ⇒ a2 = 2 b 2 ⇒ a = 2 b. So, e = =, ⋅, a, 2, a, 2, , q, , ... (ii)
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24, Hyperbola, HYPERBOLA It is the set of all points in a plane, the difference of whose distances, from two fixed points in the plane is a constant., , The two fixed points are called the foci of the hyperbola., SOME MORE TERMS RELATED TO A HYPERBOLA, (I) CENTRE OF THE HYPERBOLA, , The midpoint of the line segment joining the foci is called the centre of the hyperbola., In the given figure, F1 and F2 are the foci of the hyperbola and O is its, centre, where OF1 = OF2 ., , (II) AXES OF THE HYPERBOLA, TRANSVERSE AXIS:, , The line through the foci of the hyperbola is called its, , transverse axis., In the given figure, X′ OX is the transverse axis of the hyperbola., CONJUGATE AXIS: The line through the centre and perpendicular to the transverse, axis of the hyperbola is called its conjugate axis., , Thus, COD is the conjugate axis of the hyperbola in the given figure., (III) VERTICES OF THE HYPERBOLA, , The points at which the hyperbola intersects the transverse axis are called its vertices., In the given figure, A and B are the vertices of the hyperbola., (IV) LENGTH OF TRANSVERSE AXIS, , The distance between the two vertices of a hyperbola is called the length of its, transverse axis., In the given hyperbola, length of the transverse axis = AB., 760
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Hyperbola, AN IMPORTANT NOTE, , 761, , In a hyperbola, we take:, Length of the transverse axis = AB = 2a., Distance between the foci = F1 F2 = 2c, where c > a., Length of the conjugate axis = 2b, where b 2 = ( c2 − a 2 )., , (V) ECCENTRICITY OF A HYPERBOLA, , The ratio, , c, is always constant, called the eccentricity of the hyperbola and it is denoted, a, , by e., Here, c > a ⇔, , c, > 1 ⇔ e > 1., a, , STANDARD EQUATION OF A HYPERBOLA, , Prove that the standard equation of a hyperbola is, x2 y2, −, = 1., a 2 b2, Let X′ OX and YOY′ be the coordinate axes., Let us consider a hyperbola with centre at O( 0, 0) and its foci at, F1 ( − c, 0) and F2 ( c, 0)., , THEOREM, , PROOF, , Let P( x , y) be an arbitrary point on the hyperbola such that, PF1 − PF2 = 2a. Then,, PF1 − PF2 = 2a, ⇒, , ( x + c) 2 + y 2 − ( x − c) 2 + y 2 = 2a, , ⇒, , ( x + c) 2 + y 2 = 2a + ( x − c) 2 + y 2, , ⇒ ( x + c) 2 + y 2 = 4a 2 + ( x − c) 2 + y 2 + 4a ( x − c) 2 + y 2, ⇒ ( x + c) 2 − ( x − c) 2 − 4a 2 = 4a ( x − c) 2 + y 2, ⇒ 4cx − 4a 2 = 4a ( x − c) 2 + y 2, cx, ⇒ ⎛⎜, − a⎞⎟ = ( x − c) 2 + y 2, ⎝ a, ⎠, 2, , cx, ⇒ ⎛⎜, − a⎞⎟ = ( x − c) 2 + y 2, ⎝ a, ⎠
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Hyperbola, , 763, , x2 y2, −, = 1, lies on the hyperbola., a 2 b2, Hence, the equation of a hyperbola with origin (0, 0) and transverse, axis along the x-axis is given by, Thus, every point that satisfies, , REMARK, , x2 y2, −, = 1., a 2 b2, A hyperbola in which a = b is called an equilateral hyperbola., , LATUS RECTUM OF A HYPERBOLA, , The latus rectum of a hyperbola is a line segment perpendicular to the transverse axis,, through any of the foci with its end points lying on the hyperbola., x2 y2, TO FIND THE LENGTH OF LATUS RECTUM OF THE HYPERBOLA, −, =1, a2 b 2, Let X′OX and YOY ′ be the coordinate axes., Let F1 ( − c, 0) and F2 ( c, 0) be the foci of the given hyperbola, where, c2 = ( a 2 + b 2 )., Let LF1 M and PF2Q be the latus rectums., Let PF2 = l., Then, the coordinates of P are P( c, l )., Since P( c, l ) lies on the hyperbola, x2 y2, −, = 1, we have, a 2 b2, ⎞, c2, l2, l 2 ⎛ c2, − 2 = 1 ⇒ 2 = ⎜ 2 − 1⎟, 2, a, b, b, ⎠, ⎝a, ⎛ c2, ⎞, ⎫, ⎧( a 2 + b 2 ), ⇒ l 2 = b 2 ⎜ 2 − 1⎟ = b 2 ⋅ ⎨, − 1⎬ [Q c 2 = ( a 2 + b 2 )], 2, a, ⎠, ⎝a, ⎭, ⎩, b4, 2, ⇒ l = 2, a, b2, ⇒ l=, ⋅, a, b2, 2b 2, ∴ PF2 = l =, ⇔ PF2Q = 2l =, ⋅, a, a, 2, 2b, Similarly, LF1 M = 2l =, ⋅, a, 2b 2, Hence, the length of the latus rectum is, ⋅, a, SUMMARY OF THE RESULTS ON HORIZONTAL HYPERBOLA, , (i) The standard equation of a horizontal hyperbola is, (ii) Its centre is O( 0, 0)., , x2 y2, −, = 1., a 2 b2
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764, , Senior Secondary School Mathematics for Class 11, , (iii) X′ OX is the transverse axis and YOY′ is the conjugate axis., (iv) Its foci are F1 ( − c, 0) and F2 ( c, 0), i.e., F1 ( − ae, 0) and F2 ( ae, 0)., (v) Its vertices are A( − a , 0) and B( a , 0)., (vi) Its eccentricity is, e =, , c, =, a, , a 2 + b2, ⋅, a, , (vii) Length of the transverse axis = 2a and its equation is y = 0., (viii) Length of the conjugate axis = 2b and its equation is x = 0., 2b 2, ⋅, a, We have similar results for vertical hyperbola, given below., , (ix) Length of its latus rectum =, , SUMMARY OF THE RESULTS ON VERTICAL HYPERBOLA, , (i) The, , standard equation, y2 x2, hyperbola is 2 − 2 = 1., a, b, , of, , a, , vertical, , (ii) Its centre is O( 0, 0)., (iii) YOY′ is the transverse axis and X′ OX is the, conjugate axis., (iv) Its foci are F1 ( 0, − c) and F2 ( 0, c), i.e.,, F1 ( 0, − ae) and F2 ( 0, ae)., (v) Its vertices are A( 0, − a) and B( 0, a)., (vi) Its eccentricity is, e =, , c, =, a, , a 2 + b2, ⋅, a, , (vii) Length of the transverse axis = 2a and its, equation is x = 0., (viii) Length of the conjugate axis = 2b and its equation is y = 0., (ix) Length of its latus rectum =, , 2b 2, ⋅, a
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Hyperbola, , 765, , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Find the lengths of the axes; the coordinates of the vertices and the foci;, the eccentricity and length of the latus rectum of the hyperbola, x2 y2, −, = 1., 36 64, x2 y2, The equation of the given hyperbola is, −, = 1., 36 64, 2, 2, x, y, Comparing the given equation with 2 − 2 = 1, we get, a, b, a 2 = 36 and b 2 = 64., ∴ a = 6, b = 8 and c = a 2 + b 2 = 36 + 64 = 100 = 10., (i) Length of the transverse axis = 2a = ( 2 × 6) units = 12 units., Length of the conjugate axis = 2b = ( 2 × 8) units = 16 units., (ii) The coordinates of the vertices are A( − a , 0) and B( a , 0), i.e.,, A( −6, 0) and B( 6, 0)., (iii) The coordinates of the foci are F1 ( − c, 0) and F2 ( c, 0), i.e.,, F1 ( −10, 0) and F2 (10, 0)., c 10 5, (iv) Eccentricity, e = =, = ⋅, a, 6, 3, 2b 2 ⎛ 2 × 64⎞, 64, (v) Length of the latus rectum =, units., =⎜, ⎟ units =, ⎝ 6 ⎠, 3, a, , EXAMPLE 2, , SOLUTION, , Find the lengths of the axes; the coordinates of the vertices and the foci;, the eccentricity and length of the latus rectum of the hyperbola, 9x 2 − 16y 2 = 144., 9x 2 − 16y 2 = 144 ⇒, , x2 y2, −, = 1., 16, 9, , Thus, the equation of the given hyperbola is, Comparing the given equation with, a 2 = 16 and b 2 = 9., , x2 y2, −, = 1., 16, 9, , x2 y2, −, = 1, we get, a 2 b2, , ∴ a = 4, b = 3 and c = a 2 + b 2 = 16 + 9 = 25 = 5., (i) Length of the transverse axis = 2a = ( 2 × 4) units = 8 units., Length of the conjugate axis = 2b = ( 2 × 3 ) units = 6 units., (ii) The coordinates of the vertices are A( − a , 0) and B( a , 0), i.e.,, A( −4, 0) and B( 4, 0)., (iii) The coordinates of the foci are F1 ( − c, 0) and F2 ( c, 0), i.e.,, F1 ( −5 , 0) and F2 (5 , 0)., c 5, (iv) Eccentricity, e = = ⋅, a 4, 2b 2 ⎛ 2 × 9⎞, 9, (v) Length of the latus rectum =, =⎜, ⎟ units = units., ⎝, ⎠, 4, 2, a
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766, EXAMPLE 3, , SOLUTION, , Senior Secondary School Mathematics for Class 11, , Find the equation of the hyperbola whose vertices are ( ±2, 0) and the foci, are ( ±3 , 0)., Since the vertices of the given hyperbola are of the form ( ± a , 0), it is a, horizontal hyperbola., Let the required equation be, Then, its vertices are ( ± a , 0)., , x2 y2, −, = 1., a 2 b2, , But, it is given that the vertices are ( ±2, 0)., ∴, , a = 2., , Let its foci be ( ± c, 0)., But, it is given that the foci are ( ±3 , 0)., ∴, , c = 3., , Now, b 2 = ( c2 − a 2 ) = ( 3 2 − 22 ) = ( 9 − 4) = 5., Thus, a 2 = 22 = 4 and b 2 = 5., x2 y2, Hence, the required equation is, −, = 1., 4, 5, EXAMPLE 4, , SOLUTION, , Find the equation of the hyperbola whose foci are ( ±5 , 0) and the, transverse axis is of length 8., Since the foci of the given hyperbola are of the form ( ± c, 0), it is a, horizontal hyperbola., x2 y2, −, = 1., a 2 b2, Length of its transverse axis = 2a., Let the required equation be, , ∴ 2a = 8 ⇔ a = 4 ⇔ a 2 = 16., Let its foci be ( ± c, 0)., Then, c = 5, , [Q foci are ( ±5 , 0)]., , ∴ b = ( c − a ) = (5 2 − 42 ) = ( 25 − 16) = 9., Thus, a 2 = 16 and b 2 = 9., x2 y2, Hence, the required equation is, −, = 1., 16, 9, 2, , EXAMPLE 5, , SOLUTION, , 2, , 2, , Find the equation of the hyperbola whose foci are ( ±4, 0) and the length of, the latus rectum is 12 units., Since the foci of the given hyperbola are of the form ( ± c, 0), it is a, horizontal hyperbola., Let the required equation be, Let its foci be ( ± c, 0)., Then, c = 4, , x2 y2, −, = 1., a 2 b2, , [Q foci are ( ±4, 0)].
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Hyperbola, , 767, , Now, c = 4 ⇔ c2 = 16 ⇔ a 2 + b 2 = 16., 2b 2, Length of the latus rectum = 12 ⇔, = 12 ⇔ b 2 = 6a., a, From (i) and (ii), we get, a 2 + 6a = 16 ⇔ a 2 + 6a − 16 = 0, ⇔ ( a + 8)( a − 2) = 0, ⇔ a=2, , … (i), … (ii), , [Q a cannot be negative]., , Putting a = 2 in (ii), we get b 2 = ( 6 × 2) = 12., Thus, a 2 = 22 = 4 and b 2 = 12., Hence, the required equation is, EXAMPLE 6, , SOLUTION, , x2 y2, −, = 1., 4, 12, , Find the equation of the hyperbola whose vertices are ( ±7 , 0) and the, 4, eccentricity is ⋅, 3, Since the vertices of the given hyperbola are of the form ( ± a , 0), it is a, horizontal hyperbola., Let the required equation of the hyperbola be, Then, its vertices are ( ± a , 0)., , x2 y2, −, = 1., a 2 b2, , But, the vertices are ( ±7 , 0)., ∴, , a = 7 ⇔ a 2 = 49., c, 4, 28, Also, e =, ⇔ c = ae = ⎛⎜7 × ⎞⎟ =, ⋅, ⎝, ⎠, a, 3, 3, , ⎡ 28 2, ⎤ 343, Now, c2 = ( a 2 + b 2 ) ⇔ b 2 = ( c2 − a 2 ) = ⎢⎛⎜ ⎞⎟ − 49⎥ =, ⋅, ⎝ 3⎠, 9, ⎣, ⎦, 343, Thus, a 2 = 49 and b 2 =, ⋅, 9, ∴ the required equation is, x2, y2, x 2 9y 2, −, =1 ⇔, −, = 1 ⇔ 7 x 2 − 9y 2 = 343., 49 ( 343/9), 49 343, EXAMPLE 7, , Find the equation of the hyperbola whose eccentricity is, are ( ±2, 0)., , SOLUTION, , 3, and whose foci, 2, , Since the foci of the given hyperbola are of the form ( ± c, 0), it is a, horizontal hyperbola., Let the required equation be, Let the foci be ( ± c, 0)., Then, c = 2, , x2 y2, −, = 1., a 2 b2, , [Q foci are ( ±2, 0)].
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768, , Senior Secondary School Mathematics for Class 11, , ∴, , e=, , c, c, 2, 4, ⇔ a= =, = ⋅, a, e ( 3/2) 3, , 16, 20, And, c2 = ( a 2 + b 2 ) ⇔ b 2 = ( c2 − a 2 ) = ⎛⎜ 4 − ⎞⎟ =, ⋅, ⎝, 9⎠, 9, 16, 20, and b 2 =, Thus, a 2 =, ⋅, 9, 9, Hence, the required equation is, x2, y2, 9x 2 9y 2, =1 ⇔, −, = 1 ⇔ 45 x 2 − 36y 2 = 80., =, 16, 20, ⎛ 16⎞ ⎛ 20⎞, ⎜ ⎟ ⎜ ⎟, ⎝ 9⎠ ⎝ 9⎠, EXAMPLE 8, , SOLUTION, , Find the lengths of the axes; the coordinates of the vertices and the foci;, the eccentricity and length of the latus rectum of the hyperbola, y2 x2, −, = 1., 4, 9, y2 x2, The given equation is, −, = 1., 4, 9, Clearly, the given equation represents a vertical hyperbola., y2 x2, Comparing the given equation with 2 − 2 = 1, we get, a, b, (a 2 = 4 and b 2 = 9) ⇒ ( a = 2 and b = 3 )., ∴ c = a 2 + b 2 = 4 + 9 = 13., (i) Length of the transverse axis = 2a = ( 2 × 2) units = 4 units., Length of the conjugate axis = 2b = ( 2 × 3 ) units = 6 units., (ii) The coordinates of the vertices are A( 0, − a ) and B( 0, a ), i.e.,, A( 0, − 2) and B( 0, 2)., (iii) The coordinates of the foci are F1 ( 0, − c) and F2 ( 0, c), i.e.,, F1 ( 0, − 13 ) and F2 ( 0, 13 )., c, 13, =, ⋅, a, 2, 2b 2 ⎛ 2 × 9⎞, (v) Length of the latus rectum =, =⎜, ⎟ units = 9 units., ⎝ 2 ⎠, a, Find the lengths of the axes; the coordinates of the vertices and the foci;, the eccentricity and length of the latus rectum of the hyperbola, y 2 − 16x 2 = 16., , (iv) Eccentricity, e =, , EXAMPLE 9, , SOLUTION, , y2 x2, −, = 1., 16, 1, Clearly, the given equation represents a vertical hyperbola., y2 x2, Comparing the given equation with 2 − 2 = 1, we get, a, b, (a 2 = 16 and b 2 = 1) ⇒ ( a = 4 and b = 1)., y 2 − 16x 2 = 16 ⇒, , ∴, , c = a 2 + b 2 = 16 + 1 = 17.
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Hyperbola, , 769, , (i) Length of the transverse axis = 2a = ( 2 × 4) units = 8 units., Length of the conjugate axis = 2b = ( 2 × 1) units = 2 units., (ii) The coordinates of the vertices are A( 0, − a) and B( 0, a), i.e.,, A( 0, − 4) and B( 0, 4)., (iii) The coordinates of the foci are F1 ( 0, − c) and F2 ( 0, c), i.e.,, F1 ( 0, − 17 ) and F2 ( 0, 17 )., (iv) Eccentricity, e =, , c, 17, =, ⋅, a, 4, , (v) Length of the latus rectum =, EXAMPLE 10, , SOLUTION, , 2b 2 ⎛ 2 × 1⎞, 1, =⎜, ⎟ unit = unit., ⎝, ⎠, 4, a, 2, , Find the equation of the hyperbola whose vertices are ( 0, ± 3) and the foci, are ( 0, ± 5)., , Since the vertices of the given hyperbola are of the form ( 0, ±a ), it is a, vertical hyperbola., y2 x2, Let the required equation be 2 − 2 = 1., a, b, Then, its vertices are ( 0, ± a )., But, it is given that the vertices are ( 0, ± 3)., ∴ a = 3., Let its foci be ( 0, ± c)., But, it is given that the foci are ( 0, ± 5 )., ∴, , c = 5., , Now, b 2 = ( c2 − a 2 ) = (5 2 − 3 2 ) = ( 25 − 9) = 16., Thus, a 2 = 3 2 = 9 and b 2 = 16., y2 x2, Hence, the required equation is, −, = 1., 9, 16, EXAMPLE 11, , Find the equation of the hyperbola whose eccentricity is, vertices are ( 0, ± 6). Also, find the coordinates of its foci., , SOLUTION, , 5, and whose, 3, , Since the vertices of the given hyperbola are of the form ( 0, ± a), it is a, vertical hyperbola., y2 x2, Let the required equation be 2 − 2 = 1., a, b, Then, its vertices are ( 0, ± a )., But, it is given that the vertices are ( 0, ± 6)., 5, ∴ a = 6. Also, e = ⋅, 3, c, 5, Now, = e ⇒ c = ae = ⎛⎜ 6 × ⎞⎟ = 10., ⎝, a, 3⎠, 2, 2, 2, 2, And, c = ( a + b ) ⇔ b = ( c2 − a 2 ) = {(10) 2 − 62} = (100 − 36) = 64.
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770, , Senior Secondary School Mathematics for Class 11, , Thus, a 2 = 62 = 36 and b 2 = 64., y2 x2, −, = 1., 36 64, Also, the coordinates of its foci are ( 0, ±c), i.e., ( 0, ±10)., , Hence, the required equation is, , EXAMPLE 12, , SOLUTION, , Find the equation of the hyperbola whose foci are ( 0, ±12) and the length, of whose latus rectum is 36., , Since the coordinates of the foci of the given hyperbola are of the, form ( 0, ±c), it is a case of vertical hyperbola., Let the required equation be, Clearly, c = 12., , y2 x2, −, = 1., a 2 b2, , 2b 2, = 36 ⇔ b 2 = 18a., a, Now, c2 = ( a 2 + b 2 ) ⇔ a 2 = ( c2 − b 2 ) = (144 − 18a), Length of the latus rectum = 36 ⇔, , ⇔ a 2 + 18a − 144 = 0 ⇔ ( a + 24)( a − 6) = 0, ⇔ a=6, , [Q a is non-negative]., , Thus, a 2 = 62 = 36 and b 2 = 18a = (18 × 6) = 108., x2, y2, Hence, the required equation is, −, = 1., 36 108, EXAMPLE 13, , SOLUTION, , Find the equation of the hyperbola with centre at the origin, length of the, transverse axis 6 and one focus at (0, 4)., , Since the coordinates of the foci are of the form ( 0, ± c), it is a case of, vertical hyperbola., y2 x2, −, = 1., a 2 b2, Clearly, c = 4, [Q ( 0, c) = ( 0, 4)]., Length of the transverse axis = 6 ⇔ 2a = 6 ⇔ a = 3., Let its equation be, , Also, c2 = ( a 2 + b 2 ) ⇔ b 2 = ( c2 − a 2 ) = ( 42 − 3 2 ) = (16 − 9) = 7., Thus, a 2 = 3 2 = 9 and b 2 = 7., Hence, the required equation is, EXAMPLE 14, , SOLUTION, , y2 x2, −, = 1., 9, 7, , Find the equation of the hyperbola whose foci are at ( 0, ± 6) and the, length of whose conjugate axis is 2 11., , Since the foci of the given hyperbola are of the form ( 0, ± c), it is a case, of vertical hyperbola., y2 x2, −, = 1., a 2 b2, Let its foci be ( 0, ± c)., , Let its equation be
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Hyperbola, , 771, , But, the foci are at ( 0, ± 6)., ∴ c = 6., Length of the conjugate axis = 2 11, 2b = 2 11 ⇔ b = 11 ⇔ b 2 = 11., Also, c2 = ( a 2 + b 2 ) ⇔ a 2 = ( c2 − b 2 ) = ( 36 − 11) = 25., Thus, a 2 = 25 and b 2 = 11., Hence, the required equation is, EXAMPLE 15, , SOLUTION, , y2 x2, −, = 1., 25 11, , Find the equation of the hyperbola whose foci are at ( 0, ± 10 ) and which, passes through the point (2, 3)., , Since the foci of the given hyperbola are of the form ( 0, ±c), it is a case, of vertical hyperbola., y2 x2, −, = 1., a 2 b2, Let its foci be ( 0, ± c)., , ... (i), , Let its equation be, , But, the foci are ( 0, ± 10 )., ∴, , c = 10 ⇔ c2 = 10 ⇔ ( a 2 + b 2 ) = 10, , Since (i) passes through (2, 3), we have, , ... (ii), [Q c2 = ( a 2 + b 2 )]., , 9, 4, −, = 1., a 2 b2, , 9, 4, 9, 4, [using (ii)], − 2 =1 ⇔ 2 −, =1, 2, a, b, a, (10 − a 2 ), ⇔ 9(10 − a 2 ) − 4a 2 = a 2 (10 − a 2 ), ⇔ a 4 − 23 a 2 + 90 = 0, ⇔ ( a 2 − 18)( a 2 − 5 ) = 0, ⇔ a2 = 5, [Q a 2 = 18 ⇒ b 2 = − 8, which is not possible]., 2, 2, Thus, a = 5 and b = 5., y2 x2, Hence, the required equation is, −, = 1, i.e., y 2 − x 2 = 5., 5, 5, , Now,, , EXERCISE 24, Find the (i) lengths of the axes, (ii) coordinates of the vertices, (iii) coordinates, of the foci, (iv) eccentricity and (iv) length of the latus rectum of each of the, following the hyperbola:, 1., , x2 y2, −, =1, 9, 16, , 3. x 2 − y 2 = 1, , 2., , x2 y2, −, =1, 25, 4, , 4. 3 x 2 − 2y 2 = 6
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772, , Senior Secondary School Mathematics for Class 11, , 5. 25 x 2 − 9y 2 = 225, 8., , 6. 24x 2 − 25 y 2 = 600, , y2 x2, −, =1, 9, 27, , 9. 3 y 2 − x 2 = 108, , 7., , y2 x2, −, =1, 16 49, , 10. 5 y 2 − 9x 2 = 36, , 11. Find the equation of the hyperbola with vertices at( ±6, 0) and foci at( ±8, 0)., 12. Find the equation of the hyperbola with vertices at( 0, ± 5) and foci at( 0, ± 8)., 13. Find the equation of the hyperbola whose foci are ( ± 29 , 0) and the, transverse axis is of the length 10., 14. Find the equation of the hyperbola whose foci are ( ±5 , 0) and the conjugate, axis is of the length 8. Also, find its eccentricity., 15. Find the equation of the hyperbola whose foci are ( ±3 5 , 0) and the length, of the latus rectum is 8 units., 16. Find the equation of the hyperbola whose vertices are ( ± 2, 0) and the, eccentricity is 2., 17. Find the equation of the hyperbola whose foci are ( ± 5 , 0) and the, eccentricity is, , 5, ⋅, 3, , 18. Find the equation of the hyperbola, the length of whose latus rectum is 4, and the eccentricity is 3., 19. Find the equation of the hyperbola with eccentricity, , 2 and the distance, , between whose foci is 16., 20. Find the equation of the hyperbola whose vertices are ( 0, ± 3) and the, 4, eccentricity is ⋅ Also find the coordinates of its foci., 3, 21. Find the equation of the hyperbola whose foci are ( 0, ± 13) and the length of, whose conjugate axis is 24., 22. Find the equation of the hyperbola whose foci are ( 0, ± 10) and the length of, whose latus rectum is 9 units., 23. Find the equation of the hyperbola having its foci at ( 0, ± 14 ) and passing, through the point P( 3 , 4)., ANSWERS (EXERCISE 24), , 1. (i) 6 units, 8 units, 5, (iv) e =, 3, , (ii) A( −3 , 0), B( 3 , 0), 2, (v) 10 units, 3, , (iii) F1 ( −5 , 0), F2 (5 , 0)
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Hyperbola, , 2. (i) 10 units, 4 units, 29, 5, , (iv) e =, , 3. (i) 2 units, 2 units, (iv) e = 2, , (ii) A( −5 , 0), B(5 , 0), (v) 1, , 773, , (iii) F1 ( − 29 , 0), F2 ( 29 , 0), , 3, units, 5, , (ii) A( −1, 0), B(1, 0), , (iii) F1 ( − 2 , 0), F2 ( 2 , 0), , (v) 2 units, , 4. (i) 2 2 units, 2 3 units (ii) A( − 2 , 0), B( 2 , 0) (iii) F1 ( − 5 , 0), F2 ( 5 , 0), 5, 2, , (iv) e =, , 5. (i) 6 units, 10 units, 34, 3, , (iv) e =, , 6. (i) 10 units, 4 6 units, (iv) e = 1, , 2, 5, , 7. (i) 8 units, 14 units, (iv) e =, , 65, 4, , 8. (i) 6 units, 6 3 units, (iv) e = 2, , (v) 3 2 units, (ii) A( −3 , 0), B( 3 , 0), (v) 16, , 2, units, 3, , (ii) A( −5 , 0), B(5 , 0), (v) 9, , (iii) F1 ( −7 , 0), F2 (7 , 0), , 3, units, 5, , (ii) A( 0, − 4), B( 0, 4), , (iii) F1 ( 0, − 65 ), F2 ( 0, 65 ), , 1, (v) 24 units, 2, (ii) A( 0, − 3), B( 0, 3), , (iii) F1 ( 0, − 6), F2 ( 0, 6), , (v) 18 units, , 9. (i) 12 units, 12 3 units (ii) A( 0, − 6), B( 0, 6), (iv) e = 2, 12, units, 4 units, 10. (i), 5, , (iii) F1 ( − 34 , 0), F2 ( 34 , 0), , (iii) F1 ( 0, −12), F2 ( 0, 12), , (v) 36 units, 6⎞, 6⎞ ⎛, ⎛, (ii) A ⎜ 0, −, ⎟ , B ⎜ 0,, ⎟, ⎝, 5⎠, 5⎠ ⎝, , ⎛, ⎛, 14 ⎞, 14 ⎞, (iii) F1 ⎜ 0, − 2, ⎟ , F2 ⎜ 0, 2, ⎟, 5⎠, 5⎠, ⎝, ⎝, , (iv) e =, , 14, 3, , (v), , 4 5, units, 3, , 11., , x2 y2, −, =1, 36 28, , 12., , y2 x2, −, =1, 25 39, , 13., , x2 y2, −, =1, 25, 4, , 14., , 5, x2 y2, −, = 1, e =, 3, 9, 16, , 15., , x2 y2, −, =1, 25 20, , 16., , x2 y2, −, =1, 4, 12, , 17., , x2 y2, −, =1, 3, 2, , 18. 16x 2 − 2y 2 = 1, , 20., , y2 x2, −, = 1, ( 0, ± 4), 9, 7, , 21., , 22., , y2 x2, −, =1, 64 36, , 23. y 2 − x 2 = 7, , 19. x 2 − y 2 = 32, , y2, x2, −, =1, 25 144
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774, , Senior Secondary School Mathematics for Class 11, HINTS TO SOME SELECTED QUESTIONS, 2, , 10. Given equation is, , 2, , y, x, −, = 1., ( 36/5 ), 4, , 36 2, 36, 56, , b = 4 and c 2 = ( a2 + b 2 ) = ⎛⎜, + 4 ⎞⎟ =, ⋅, ⎝ 5, ⎠ 5, 5, c ⎛ 56, 5 ⎞ 2 14, 14, Also, e = = ⎜, ×, =, ⋅, ⎟=, a ⎝ 5, 6 ⎠, 6, 3, , ∴, , a2 =, , 18. e = 3 ⇔, , c, = 3 ⇔ c = 3 a ⇔ c 2 = 9 a2 ⇔ a2 + b 2 = 9 a2 ⇔ 8 a2 = b 2 ., a, , … (i), , 2, , 2b, = 4 ⇔ b 2 = 2 a., a, , … (ii), , ∴, , 8 a2 = 2 a ⇔ 2 a( 4 a − 1) = 0 ⇔ a =, , ∴, , a2 =, , 1, and b 2 = ⎛⎜ 2 ×, ⎝, 16, , 1, ⋅, 4, , 1⎞ 1, ⎟= ⋅, 4⎠ 2, , 19. 2 c = 16 ⇔ c = 8., Also, e = 2., c, c, 8, Now, e = ⇒ a = =, = 4 2., e, a, 2, ∴ b 2 = ( c 2 − a2 ) = ( 64 − 32 ) = 32., Thus, a2 = 32 and b 2 = 32 ., , q
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25, Applications of Conic Sections, Applications of Conic Sections, , Conic sections have their applications in various fields. Some of them are, listed below., 1. We know that when an object is thrown in space then the path traced by the, object (called a projectile) is a parabola. By using various properties of a, parabola, we can find the equation of the path traced by the projectile., Also, we can obtain other useful information such as the greatest height, attained by the projectile, its horizontal range, and so on., 2. In a parabolic reflector, light rays or sound waves directed parallel to its, axis converge at the focus and are then reflected by it, parallel to the axis., Because of this property, parabolic reflectors are used in cars, automobiles,, loudspeakers, solar cookers, telescopes, etc., 3. Cables shaped as parabolic arcs are used in the construction of suspension, bridges., 4. The planets of the solar system move in elliptical paths with the sun at one, of the foci. Artificial satellites are made to move in elliptical paths around, the earth., 5. Hyperbolas have their applications in the field of ballistics. Suppose a gun, is fired and its sound reaches two posts, situated at the two foci of a, hyperbola, at different times. From the time difference, the distance, between the two listening posts can be calculated., SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , The focus of a parabolic mirror is at a, distance of 5 cm from its vertex and the, mirror is 15 cm deep. Find the diameter of, the mirror., Let us take the vertex of the mirror at the, origin with its axis lying along the, positive direction of the x-axis., Let the equation of the parabolic shape of, the mirror be y 2 = 4ax., Since the distance between the focus and, the vertex is 5 cm, we have a = 5., 775
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776, , Senior Secondary School Mathematics for Class 11, , ∴, , the equation of the parabola is, y 2 = 4 × 5 × x, i.e., y 2 = 20x., , … (i), , The depth of the mirror, OM = 15 cm., ∴ the coordinates of one end of the diameter AB are (15 , p), where, AM = p cm., Clearly, A(15 , p) lies on the parabola y 2 = 20x., , Coordinate Geometry, , ∴ p 2 = ( 20 × 15), ⇒ p = 300 = 10 3, ⇒ AM = 10 3 cm, ⇒ AB = 2 × AM = 20 3 cm., Hence, the diameter of the mirror is 20 3 cm., , EXAMPLE 2, , A parabolic reflector is 9 cm deep and its diameter is 24 cm. How far is, its focus from the vertex?, Let us take the vertex of the reflector at, the origin, and let its axis lie along the, positive direction of the x-axis., Let the equation of the parabolic shape of, the reflector be, … (i), y 2 = 4ax., , SOLUTION, , As given, we have, OM = 9 cm and AB = 24 cm, 1, ⇒ OM = 9 cm and AM = AB = 12 cm., 2, ∴ the coordinates of A are ( 9, 12)., Since A lies on (i), we have, 12 × 12 = 4 × a × 9 ⇒ a = 4., ∴ the equation of the parabola is y 2 = 16x., Its focus is F( a , 0), i.e., F( 4, 0)., Hence, the focus is at a distance of 4 cm from the vertex., EXAMPLE 3, , SOLUTION, , An arch is in the form of a parabola with its axis vertical. The arch is, 10 m high and 5 m wide at the base. Find its width at a distance of 2 m, from its vertex., , Let O be the origin and OY be the, vertical axis of the parabolic, arch., Then, the equation of the arch is, … (i), x 2 = 4ay., ∴ BA = 5 m and OC = 10 m., 1, 1, Now, OL = ML = BA = 25, . m., 2, 2, And, AL = OC = 10 m.
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Applications of Conic Sections, , 777, , ∴ the coordinates of A are (2.5, 10). Since A( 25, . , 10) lies on (i), we have, ⎛ 2.5 × 2.5 ⎞, 5, 2, ⇒ a=⎜, ⋅, ( 2.5) = 4a × 10, ⎟ =, ⎝ 4 × 10 ⎠ 32, 5, 5, … (ii), ∴ x 2 = ⎛⎜ 4 ×, × y⎞⎟ ⇒ x 2 = y., ⎝, ⎠, 32, 8, Draw DEF|| ML such that OE = 2 m., Let the coordinates of F be F( x1 , 2)., Since F( x1 , 2) lies on (ii), we have, 5, 5, 5, ⋅, x12 = ⎛⎜ × 2⎞⎟ ⇒ x12 =, ⇒ x1 =, ⎝8, ⎠, 2, 4, ⎛, 5, 5⎞, ∴ EF =, m ⇒ DF = 2 × EF = ⎜ 2 ×, ⎟ m = 5 m., 2, 2⎠, ⎝, Hence, the required width of the arch at a distance of 2 m from its, vertex is 5 metres., An equilateral triangle is inscribed in the parabola y 2 = 4ax so that one, angular point of the triangle is at the vertex of the parabola. Find the, length of each side of the triangle., , EXAMPLE 4, , Let COAB be an equilateral triangle, inscribed in the parabola y 2 = 4ax., Let OA = OB = AB = l., , SOLUTION, , Let AB cut the x-axis at M., Then, ∠AOM = ∠BOM = 30º., l 3, OM, = cos 30º ⇒ OM = l cos 30º =, ,, 2, OA, AM, l, and, = sin 30º ⇒ AM = l sin 30º = ⋅, 2, OA, ⎛ l 3 l⎞, , ⎟⋅, ∴ the coordinates of A are ⎜, ⎝ 2 2⎠, , ∴, , Since A lies on the parabola y 2 = 4ax, we have, l2, l 3, = 4a ×, ⇒ l = 8a 3., 4, 2, Hence, the length of each side of the triangle is 8a 3 units., EXAMPLE 5, , SOLUTION, , Find the area of the triangle formed by the lines joining the vertex of the, parabola x 2 = 12y to the ends of its latus rectum., The vertex of the parabola x 2 = 12y is O( 0, 0)., Comparing x 2 = 12y with x 2 = 4ay, we get a = 3., The coordinates of its focus S are ( 0, 3).
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778, , Senior Secondary School Mathematics for Class 11, , Clearly, the ends of its latus rectum are, A( −2a , a) and B( 2a , a), i.e., A( −6, 3) and B( 6, 3)., ⏐ 0 0 1⏐, 1, ∴ area of COBA = ⋅⏐ 6 3 1⏐, ⏐, 2 ⏐, ⏐−6 3 1⏐, 1, = {1 × (18 + 18)}, 2, = 18 sq units., EXAMPLE 6, , SOLUTION, , A water jet coming out of the small opening O of a fountain reaches its, maximum height of 4 metres at a distance of 0.5 metre from the vertical., Find the height of the jet above the horizontal OX at a distance of, 0.75 metre from the point O., The path of a water jet is a parabola., Let the equation of this parabolic path be, … (i), y = ax 2 + bx + c., Let AB be the maximum height reached., The path is symmetrical about AB., Let the jet strike the x-axis at E., Then, AE = OA = 0.5 m, ⇒ OE = 2(OA) = ( 2 × 0.5 ) m = 1 m., Clearly, AB = 4 m., Thus, the points O , B and E are, O( 0, 0), B( 0.5 , 4) and E(1, 0)., , Since these points lie on (i), we have, 1, 1, c = 0, a + b = 4 and a + b = 0, 4, 2, ⇒ c = 0, a + 2b = 16 and a + b = 0, ⇒ a = −16, b = 16 and c = 0., ∴ the equation of the parabola is y = −16x 2 + 16x., Let P be a point on OE such that OP = 0.75 m., Draw PD ⊥ OX, meeting the parabola at D., Let PD = h metres., Then, the coordinates of D are ( 0.75 , h)., Since D( 0.75 , h) lies on (ii), we have, 2, 3, 3, h = −16 × ⎛⎜ ⎞⎟ + ⎛⎜16 × ⎞⎟ ⇒ h = ( −9 + 12) = 3., ⎝ 4⎠, ⎝, 4⎠, ∴ the required height, PD = h m = 3 m., , … (ii)
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Applications of Conic Sections, , 779, , The cable of a uniformly loaded suspension bridge hangs in the form of a, parabola. The roadway which is horizontal and 100 m long is supported, by vertical wires attached to the cable, the longest wire being 30 m and, the shortest being 6 m. Find the length of a supporting wire attached to, the roadway 18 m from the middle., , EXAMPLE 7, , Let X′OX be the x-axis,, taken along the roadway,, and let YOY′ be the y-axis., Let CAB be the bridge in the, form of a parabola having, its vertex at A( 0, 6) such, that A is the centre of the, bridge., Let BL and CM be the, longest vertical wires, each, of length 30 m, and let OA, be the shortest vertical, wire, 6 m in length., The equation of the parabolic form of the bridge is, , SOLUTION, , x 2 = 4a ( y − 6)., , … (i), , It passes through B(50, 30)., Putting x = 50 and y = 30 in (i), we get, 2500, ⋅, 2500 = 4a( 30 − 6) ⇒ a =, 96, 2500, in (i), we get, Putting a =, 96, 2500, … (ii), x2 = 4 ×, ( y − 6) ⇒ 6x 2 = 625 y − 3750 ., 96, Let l m be the length of the supporting wire 18 m from the middle., Then, P(18, l) must satisfy (ii)., ∴, , 6 × 18 × 18 = 625 l − 3750 ⇒ 625 l = 5694, ⇒ l = 9.11., Hence, the length of the supporting wire 18 m from the middle is, 9.11 m., , EXAMPLE 8, , SOLUTION, , A rod of length 12 cm moves with its ends always touching the, coordinate axes. Determine the equation of the path of a moving point P, on the rod which is 3 cm from the end in contact with the x-axis., Let AB be the rod and P( x , y) be a point on it such that, AP = 3 cm and PB = 9 cm., From P, draw PM ⊥ OX and PN ⊥ OY., Let AM = p and BN = q., Then CBNP and CPMA are similar.
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780, , Senior Secondary School Mathematics for Class 11, , BN, BP, q 9, =, ⇒ =, ⇒ q = 3 y., PM PA, y 3, MA PA, p 3, 1, And,, =, ⇒, =, ⇒ p = x., NP, BP, x 9, 3, 1, 4x, ,, ∴ OA = OM + MA = x + p = x + x =, 3, 3, and OB = ON + BN = y + q = y + 3 y = 4y., In CBOA, we have, ∴, , OA 2 + OB 2 = AB 2, ⎛ 4x ⎞, ⇒ ⎜ ⎟, ⎝ 3⎠, 16x 2, ⇒, 9, x2, +, ⇒, 81, , 2, , + ( 4y ) 2 = (12) 2, + 16y 2 = 144, y2, = 1., 9, , Hence, the path of P is an ellipse whose equation is, , A man running in a racecourse notes that the sum of the distances of the, two flag posts from him is always 10 m, and the distance between the flag, posts is 8 m. Find the equation of the path traced by the man., , EXAMPLE 9, , SOLUTION, , x2 y2, +, = 1., 81, 9, , We know that an ellipse is the locus of a point that moves in such a, way that the sum of its distances from two fixed points (called foci) is, constant., So, the path traced by the man is an ellipse., Let the equation of the ellipse be, x2 y2, + 2 = 1, where b 2 = a 2 (1 − e2 )., a2, b, 2ae = 8, 4, ⇒ a = 5 and e =, 5, 16 ⎞, ⎛, ⇒ b 2 = a 2 (1 − e2 ) = 25 ⎜1 − ⎟ = 9, ⎝, 25 ⎠, ⇒ b = 3., x2 y2, Hence, the equation of the path is, +, = 1., 25, 9, , Clearly,, , EXAMPLE 10, , SOLUTION, , 2a = 10 and, , An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the, centre. Find the height of the arch at a point 1.5 m from one end., , Let the equation of the ellipse of which the given arch is a part be, x2 y2, + 2 = 1., a2, b, , … (i)
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Applications of Conic Sections, , 781, , It is being given that, A′ A = 8 m and OB = 2 m, ⇒ 2a = 8 and b = 2, ⇒ a = 4 and b = 2., So, the equation of the arch is, x2 y2, +, = 1., 16, 4, Let P be a point on A′ OA such that AP = 1.5 m., , Draw PM ⊥ A′ A, meeting the ellipse at M., Then, AP = 1.5 m (given)., Now, OP = (OA − AP) = ( 4 − 1.5) m = 2.5 m. Let PM = p m., So, the coordinates of M are ( 2.5 , p)., Since M lies on the ellipse, we have, 625 ⎞, 39, p2 ⎛, (2.5) 2 p 2, +, =1 ⇒, = ⎜1 −, ⎟ ⇒ p2 =, 1600⎠, 16, 16, 4, 4 ⎝, 39 6.24, =, = 1.56., 4, 4, Hence, the height of the arch at a point 1.5 m from one end is 1.56 m., ⇒ p=, , EXERCISE 25, 1. The focus of a parabolic mirror is at a distance of 6 cm from its vertex. If the, mirror is 20 cm deep, find its diameter., 2. A parabolic reflector is 5 cm deep and its diameter is 20 cm. How far is its, focus from the vertex?, 3. The towers of a bridge, hung in the form of a parabola, have their tops 30 m, above the roadway, and are 200 m apart. If the cable is 5 m above the, roadway at the centre of the bridge, find the length of the vertical, supporting cable, 30 m from the centre.
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782, , Senior Secondary School Mathematics for Class 11, , 4. A rod of length 15 cm moves with its ends always touching the coordinate, axes. Find the equation of the locus of a point P on the rod, which is at a, distance of 3 cm from the end in contact with the x-axis., 5. A beam is supported at its ends by supports which are 12 m apart. Since the, load is concentrated at its centre, there is a deflection of 3 cm at the centre,, and the deflected beam is in the shape of a parabola. How far from the centre, is the deflection 1 cm?, ANSWERS (EXERCISE 25), , 1. 8 30 cm, , 2. 5 cm, , 4. 5 x 2 + 9y 2 = 81, , 3. 7.25 m, , 5. 2 6 m, , HINTS TO SOME SELECTED QUESTIONS, 5. The beam takes the shape of a parabola, whose equation is of the form x 2 = 4 ay. … (i), 3 ⎞, Since the point P ⎛⎜ 6 ,, ⎟ lies on it, we have, ⎝ 100 ⎠, ⎛ 4 × a × 3 ⎞ = 36 ⇔ a = 300 m., ⎜, ⎟, ⎝, 100 ⎠, Let AB be the deflection. Then,, 1, 2 ⎞, m and B is B ⎛⎜ x,, AB =, ⎟⋅, ⎝ 100 ⎠, 100, 2, ∴ x 2 = 4 × 300 ×, ⇔ x = 2 6 m., 100, , q
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26, Three-Dimensional Geometry, COORDINATES OF A POINT IN SPACE Let O be the origin, and let OX, OY and OZ, be three mutually perpendicular lines, taken as the x-axis, the y-axis and the, z-axis respectively in such a way that they form a right-handed system., The planes, YOZ, ZOX and XOY are, respectively known as yz-plane, zx-plane and, xy-plane., These planes, known as coordinate, planes, divide the space into eight parts,, called octants., Let P be a point in space. Through P,, draw planes parallel to coordinate planes,, and meeting the axes OX, OY and OZ in, points A, B and C respectively. Complete the, parallelepiped whose coterminous edges are, OA, OB and OC., , Let OA = x , OB = y and OC = z., Then, we say that the coordinates of P are (x, y, z)., REMARK, , In fact, we have,, , (i) x = distance of P from yz-plane;, (ii) y = distance of P from xz-plane;, (iii) z = distance of P from xy-plane., As a consequence of the above facts, we, find that, (i) Every point in yz-plane has, x-coordinate zero., (ii) Every point in xz-plane has y-coordinate zero., (iii) Every point in xy-plane has z-coordinate zero., (iv) Any point on the x-axis is of the form (x, 0, 0)., (v) Any point on the y-axis is of the form (0, y, 0)., (vi) Any point on the z-axis is of the form (0, 0, z)., 783
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784, , Senior Secondary School Mathematics for Class 11, , The following table shows the signs of coordinates in eight octants., I, , II, , III, , IV, , V, , VI, , VII, , VIII, , x, , +, , −, , −, , +, , +, , −, , −, , +, , y, , +, , +, , −, , −, , +, , +, , −, , −, , z, , +, , +, , +, , +, , −, , −, , −, , −, , EXAMPLE 1, , SOLUTION, , EXAMPLE 2, , SOLUTION, , EXAMPLE 3, SOLUTION, EXAMPLE 4, SOLUTION, , In which octant does the given point lie?, (i) ( −2, 4, 3), (ii) ( 3 , − 2, − 5), (iii) ( −6, 3 , − 4), (iv) ( −3 , − 1, 4), (v) (1, − 3 , 6), (vi) ( 4, 7 , − 2), It is clear from the above table that the point, (i) ( −2, 4, 3) lies in octant II;, (ii) ( 3 , − 2, − 5) lies in octant VIII;, (iii) ( −6, 3 , − 4) lies in octant VI;, (iv) ( −3 , − 1, 4) lies in octant III;, (v) (1, − 3 , 6) lies in octant IV;, (vi) ( 4, 7 , − 2) lies in octant V., If a point lies on the y-axis then what are its x-coordinate and, z-coordinate?, If a point lies on the y-axis then its x-coordinate is 0 and its, z-coordinate is 0., If a point lies in xy-plane then what is its z-coordinate?, If a point lies in xy-plane then its z-coordinate is 0., In which plane does the point ( 0, 5 , − 4) lie?, Clearly, the point ( 0, 5 , − 4) lies in the yz-plane., , EXERCISE 26A, 1. If a point lies on the z-axis then find its x-coordinate and y-coordinate., 2. If a point lies on yz-plane then what is its x-coordinate?, 3. In which plane does the point ( 4, − 3 , 0) lie?, 4. In which octant does each of the given points lie?, (i) ( −4, − 1, − 6), (iv) ( 4, − 3 , − 2), , (ii) ( 2, 3 , − 4), , (iii) ( −6, 5 , − 1), , (v) ( −1, − 6, 5), , (vi) (4, 6, 8), ANSWERS (EXERCISE 26A), , 1. x = 0, y = 0, 4. (i) VII, , 2. x = 0, , (ii) V (iii) VI, , (iv) VIII, , 3. xy-plane, (v) III, , (vi) I
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Three-Dimensional Geometry, , 785, , (Distance Formula) Prove that the distance between the points, , THEOREM 1, , P( x1 , y1 , z1 ) and Q( x 2 , y 2 , z2 ) is given by, PQ = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 + (z2 − z1 ) 2 ., PROOF, , Let O be the origin, and let P( x1 , y1 , z1 ) and Q( x 2 , y 2 , z2 ) be the given, point. From P and Q, draw perpendiculars PM and QN respectively on, the xy-plane., Then, the coordinates of M and N are, M( x1 , y1 , 0) and N ( x 2 , y 2 , 0)., The two-dimensional coordinates of M, and N referred to OX and OY are, M( x1 , y1 ) and N ( x 2 y 2 )., ∴, , MN 2 = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 ., , Now, from P draw PR ⊥ QN ., Then, PR is parallel and equal to MN., Now, in right triangle PRQ, we have, PQ 2 = PR 2 + RQ 2, = MN 2 + (QN − RN ) 2, = MN 2 + (QN − PM) 2, [Q PM = z1 and QN = z2 ]., , = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 + (z2 − z1 ) 2, ∴, , PQ = ( x 2 − x1 ) + ( y 2 − y1 ) + (z2 − z1 ) ., 2, , COROLLARY, , 2, , 2, , The distance of the point P( x , y , z) from the origin O( 0, 0, 0) is, OP = ( x − 0) 2 + ( y − 0) 2 + (z − 0) 2 = x 2 + y 2 + z 2 ., , EXAMPLE 1, SOLUTION, , Find the distance between the points A( −2, 1, − 3) and B( 4, 3 , − 6)., We have, AB = {[4 − ( −2)]2 + ( 3 − 1) 2 + [−6 − ( −3)]2 }, = {62 + 22 + ( −3) 2} = 49 = 7 units., , EXAMPLE 2, , SOLUTION, , Show that the points A( 0, 7 , 10), B( −1, 6, 6) and C( −4, 9, 6) form an, isosceles right-angled triangle., We have, AB = ( −1 − 0) 2 + ( 6 − 7) 2 + ( 6 − 10) 2 = 18 = 3 2 ,, BC = ( −4 + 1) 2 + ( 9 − 6) 2 + ( 6 − 6) 2 = 18 = 3 2 ,, and AC = ( −4 − 0) 2 + ( 9 − 7 ) 2 + ( 6 − 7 ) 2 =, , 36 = 6., , Clearly, AB = BC and AB 2 + BC 2 = (18 + 18) = 36 = AC 2 ., Hence, triangle ABC is an isosceles right-angled triangle.
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786, EXAMPLE 3, , SOLUTION, , Senior Secondary School Mathematics for Class 11, , Prove that the points A( 3 , − 2, 4), B(1, 1, 1) and C( −1, 4, − 2) are, collinear., We have, AB = (1 − 3) 2 + (1 + 2) 2 + (1 − 4) 2 = 4 + 9 + 9 = 22 ,, BC = ( −1 − 1) 2 + ( 4 − 1) 2 + ( −2 − 1) 2 = 4 + 9 + 9 = 22 ,, and AC = ( −1 − 3) 2 + ( 4 + 2) 2 + ( −2 − 4) 2 = 16 + 36 + 36, = 88 = 2 22., ∴ AB + BC = AC., This shows that the given points are collinear., , EXAMPLE 4, , SOLUTION, , Find the equation of the curve formed by the set of all points whose, distances from the points ( 3 , 4, −5 ) and ( −2, 1, 4) are equal., Let P( x , y , z) be any point on the given curve, and let A( 3 , 4, − 5 ) and, B( −2, 1, 4) be the given points., Then, PA = PB, ⇒, , PA 2 = PB 2, , ⇒ ( x − 3) 2 + ( y − 4) 2 + (z + 5 ) 2 = ( x + 2) 2 + ( y − 1) 2 + (z − 4) 2, ⇒ 10x + 6y − 18z − 29 = 0., Hence, the required curve is 10x + 6y − 18z − 29 = 0., EXAMPLE 5, , SOLUTION, , Find the equation of the curve formed by the set of all those points the, sum of whose distances from the points A( 4, 0, 0) and B( −4, 0, 0) is, 10 units., Let P( x , y , z) be an arbitrary point on the given curve. Then,, PA + PB = 10, ⇒, , ( x − 4) 2 + y 2 + z 2 + ( x + 4) 2 + y 2 + z 2 = 10, , ⇒, , ( x + 4) 2 + y 2 + z 2 = 10 − ( x − 4) 2 + y 2 + z 2, , ... (i), , ⇒ ( x + 4) + y + z = 100 + ( x − 4) + y + z − 20 ( x − 4) + y + z 2, [on squaring both sides of (i)], 2, , 2, , 2, , 2, , 2, , 2, , ⇒ 16x = 100 − 20 ( x − 4) 2 + y 2 + z 2, ⇒ 5 ( x − 4) 2 + y 2 + z 2 = ( 25 − 4x), ⇒ 25[( x − 4) 2 + y 2 + z 2 ] = 625 + 16x 2 − 200x, ⇒ 9x 2 + 25 y 2 + 25z 2 − 225 = 0., Hence, the required equation of the curve is, 9x 2 + 25 y 2 + 25z 2 − 225 = 0., , 2, , 2
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Three-Dimensional Geometry, , 787, , SOME USEFUL RESULTS, , A quadrilateral ABCD is a, (i) parallelogram, if AB = CD and BC = DA ;, (ii) rectangle, if AB = CD , BC = DA and AC = BD ;, (iii) rhombus, if AB = BC = CD = DA but AC ≠ BD ;, (iv) square, if AB = BC = CD = DA and AC = BD., , EXERCISE 26B, 1. Find the distance between the points:, (i) A(5 , 1, 2) and B( 4, 6, − 1), (iii) R(1, − 3 , 4) and S( 4, − 2, − 3), , (ii) P(1, − 1, 3) and Q( 2, 3 , − 5), (iv) C( 9, − 12, − 8) and the origin, , 2. Show that the points A(1, − 1, − 5 ), B( 3 , 1, 3) and C( 9, 1, − 3) are the vertices, of an equilateral triangle., 3. Show that the points A( 4, 6, − 5 ), B( 0, 2, 3) and C( −4, − 4, − 1) form the, vertices of an isosceles triangle., 4. Show that the points A( 0, 1, 2), B( 2, − 1, 3) and C(1, − 3 , 1) are the vertices of, an isosceles right-angled triangle., 5. Show that the points A(1, 1, 1), B( −2, 4, 1), C(1, − 5 , 5 ) and D( 2, 2, 5 ) are the, vertices of a square., 6. Show that the points A(1, 2, 3), B( −1, − 2, − 1), C( 2, 3 , 2) and D( 4, 7 , 6) are, the vertices of a parallelogram. Show that ABCD is not a rectangle., 7. Show that the points P( 2, 3 , 5 ), Q( −4, 7 , − 7), R( −2, 1, −10) and S( 4, − 3 , 2), are the vertices of a rectangle., 8. Show that the points P(1, 3 , 4), Q( −1, 6, 10), R( −7 , 4, 7) and S( −5 , 1, 1) are, the vertices of a rhombus., 9. Show that D( −1, 4, − 3) is the circumcentre of triangle ABC with vertices, A( 3 , 2, − 5 ), B( −3 , 8, − 5 ) and C( −3 , 2, 1)., 10. Show that the following points are collinear:, (i) A( −2, 3 , 5 ), B(1, 2, 3) and C(7 , 0, − 1), (ii) A( 3 , − 5 , 1), B( −1, 0, 8) and C(7 , − 10, − 6), (iii) P( 3 , −2, 4), Q(1, 1, 1) and R( −1, 4, 2), 11. Find the equation of the curve formed by the set of all points which are, equidistant from the points A( −1, 2, 3) and B( 3 , 2, 1)., 12. Find the point on the y-axis which is equidistant from the points A( 3 , 1, 2), and B(5 , 5 , 2).
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788, , Senior Secondary School Mathematics for Class 11, , 13. Find the point on the z-axis which is equidistant from the points A(1, 5 , 7), and B(5 , 1, − 4)., 14. Find the coordinates of the point which is equidistant from the points, A( a , 0, 0), B( 0, b , 0), C( 0, 0, c) and O( 0, 0, 0)., 15. Find the point in yz-plane which is equidistant from the points, A( 3 , 2, − 1), B(1, − 1, 0) and C( 2, 1, 2)., 16. Find the point in xy-plane which is equidistant from the points, A( 2, 0, 3), B( 0, 3 , 2) and C( 0, 0, 1)., ANSWERS (EXERCISE 26B), , 1. (i) 35 units, , (ii) 9 units, , 11. 2x − z = 0, , 12. (0, 3, 0), , 31 −3 ⎞, 15. ⎛⎜ 0,, ,, ⎟, ⎝ 16 16 ⎠, , 16. (3, 2, 0), , (iii) 59 units, , (iv) 17 units, , 3, 13. ⎛⎜ 0, 0, ⎞⎟, ⎝, 2⎠, , a b c, 14. ⎛⎜ , , ⎞⎟, ⎝ 2 2 2⎠, , HINTS TO SOME SELECTED QUESTIONS, 6. Show that AB = CD , BC = DA and AC ≠ BD., 7. Show that AB = CD , BC = DA and AC = BD., 8. Show that PQ = QR = RS = SP and PR ≠ QS., 9. Show that DA = DB = DC ., 11. Let P( x , y , z) be an arbitrary point on the given curve. Then,, PA = PB ⇒ PA 2 = PB 2 ., 12. Any point on the y-axis is of the form P( 0 , y , 0 )., 14. Let the required point be P( x , y , z). Then,, PA 2 = PB 2 = PC 2 = PO 2 ⇔ PA 2 − PO 2 = 0 , PB 2 − PO 2 = 0 and PC 2 − PO 2 = 0 ., 15. Any point in the yz-plane is of the form P( 0 , y , z)., |AP| =|BP| =|CP| ⇔ AP 2 = BP 2 and BP 2 = CP 2 ., ( 0 − 3 ) 2 + ( y − 2 ) 2 + (z + 1) 2 = ( 0 − 1) 2 + ( y + 1) 2 + (z − 0 ) 2, and ( 0 − 1) 2 + ( y + 1) 2 + (z − 0 ) 2 = ( 0 − 2 ) 2 + ( y − 1) 2 + (z − 2 ) 2, ⇒, , 3 y − z − 6 = 0 and 4 y + 4z − 7 = 0., 31, −3, and z =, ⋅, 16, 16, 31 −3 ⎞, Hence, the required point is ⎛⎜ 0 ,, , ⎟⋅, ⎝ 16 16 ⎠, , On solving, we get y =
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Three-Dimensional Geometry, , 789, , SECTION FORMULAE, THEOREM 2, , (Section Formula) Let P( x1 , y1 , z1 ) and Q( x 2 , y 2 , z2 ) be two points and, , let R( x , y , z) be a point on PQ dividing it in the ratio m : n. Prove that, x=, PROOF, , mx 2 + nx1, my 2 + ny1, mz2 + nz1, and z =, ,y=, ⋅, m+n, m+n, m+n, , From P, Q and R, draw perpendiculars PL, QM and RN on the xy-plane., Also, draw PS ⊥ QM, meeting QM and RN at S and T respectively., From similar triangles PRT and PQS, we have, RT, PR, =, QS, PQ, ⇒, , RN − TN, m, =, QM − SM m + n, , ⇒, , RN − PL, m, =, QM − PL m + n, , z − z1, m, =, z2 − z1 m + n, mz2 + nz1, ⇒ z=, ⋅, m+n, mx 2 + nx1, my 2 + ny1, Similarly, x =, and y =, ⋅, m+n, m+n, mx 2 + nx1, my 2 + ny1, mz2 + nz1, Hence, x =, ,y=, ,z=, ⋅, m+n, m+n, m+n, ⇒, , The coordinates of the midpoint of the join of P( x1 , y1 , z1 ) and, Q( x 2 , y 2 , z2 ) are, , COROLLARY 1, , ⎛ x1 + x 2 y1 + y 2 z1 + z2 ⎞, ,, ,, ⎜, ⎟⋅, ⎝ 2, 2, 2 ⎠, PROOF, , Let R be the midpoint of PQ., Putting m = n = 1 in the above result, we find that the coordinates of, R are, ⎛ x1 + x 2 , y1 + y 2 , z1 + z2 ⎞ ⋅, ⎜, ⎟, ⎝ 2, 2, 2 ⎠, The coordinates of a point R which divides the join of P( x1 , y1 , z1 ) and, Q( x 2 , y 2 , z2 ) externally in the ratio m : n are, , COROLLARY 2, , ⎛ mx 2 − nx1 my 2 − ny1 mz2 − nz1 ⎞, ,, ,, ⎜, ⎟⋅, m −n, m −n ⎠, ⎝ m −n
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790, PROOF, , Senior Secondary School Mathematics for Class 11, , Replacing n by −n in Theorem 2, we find that the coordinates of R are, ⎛ mx 2 − nx1 my 2 − ny1 mz2 − nz1 ⎞, ,, ,, ⎜, ⎟⋅, m −n, m −n ⎠, ⎝ m −n, , CENTROID OF A TRIANGLE, THEOREM 3, , Show that the centroid of the triangle with vertices A( x1 , y1 , z1 ),, B( x 2 , y 2 , z2 ) and C( x 3 , y 3 , z3 ) is, ⎛ x1 + x 2 + x 3 , y1 + y 2 + y 3 , z1 + z2 + z3 ⎞ ⋅, ⎜, ⎟, ⎝, ⎠, 3, 3, 3, , PROOF, , Let D be the midpoint of BC. Join, A and D., Then, the coordinates of D are, ⎛ x 2 + x 3 , y 2 + y 3 , z2 + z3 ⎞ ⋅, ⎜, ⎟, ⎝ 2, 2, 2 ⎠, Let G be the centroid of triangle, ABC. Then, G lies on AD and, divides it in the ratio 2 : 1., ∴ the coordinates of G are, ⎡ ⎛ x2 + x3 ⎞, ⎤, ⎛ y2 + y3 ⎞, ⎛ z2 + z3 ⎞, ⎢ 2 ⋅ ⎜⎝ 2 ⎟⎠ + 1 ⋅ x1 2 ⋅ ⎜⎝ 2 ⎟⎠ + 1 ⋅ y1 2 ⋅ ⎜⎝ 2 ⎟⎠ + 1 ⋅ z1 ⎥, ,, ,, ⎢, ⎥,, 2+1, 2+1, 2+1, ⎢, ⎥, ⎢⎣, ⎥⎦, x + x 2 + x 3 y1 + y 2 + y 3 z1 + z2 + z3 ⎞, i.e., ⎛⎜ 1, ,, ,, ⎟⋅, ⎝, ⎠, 3, 3, 3, , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Find the coordinates of the point which divides the join of the points, P(5 , 4, 2) and Q( −1, − 2, 4) in the ratio 2 : 3., Let R( x , y , z) be the required point. Then,, 2( −1) + 3 × 5, 2( −2) + 3 × 4, 2× 4+ 3 × 2, ,y=, ,z=, 2+ 3, 2+ 3, 2+ 3, 13, 8, 14, or x =, , y = and z =, ⋅, 5, 5, 5, 13 8 14, So, the required point is R ⎛⎜ , , ⎞⎟ ⋅, ⎝5 5 5⎠, x=, , EXAMPLE 2, , Find the coordinates of the point which divides the join of the points, A( 2, − 1, 3) and B( 4, 3 , 1) externally in the ratio 3 : 4.
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Three-Dimensional Geometry, SOLUTION, , 791, , Let C( x , y , z) be the required point. Then,, , or, , 3 × 4−4× 2, 3 × 3 − 4( −1) 3 × 1 − 4 × 3, ,y=, ,, 3−4, 3−4, 3−4, x = − 4, y = − 13 and z = 9., , ∴, , the required point is C( −4, − 13 , 9)., , x=, , EXAMPLE 3, , SOLUTION, , Find the ratio in which the join of the points P( 2, − 1, 3) and Q( 4, 3 , 1), 20 5 15, is divided by the point ⎛⎜ , , ⎞⎟ ⋅, ⎝7 7 7⎠, Let the required ratio be λ : 1. Then, the coordinates of R are, ⎛ 4λ + 2 3 λ − 1 λ + 3 ⎞, ,, ,, ⎜, ⎟⋅, ⎝ λ + 1 λ + 1 λ + 1⎠, 20 5 15, But, the coordinates of R are ⎛⎜ , , ⎞⎟, ⎝7 7 7⎠, 3, 4λ + 2 20, or λ = ⋅, ∴, =, 4, λ+1, 7, 3, So, the required ratio is : 1, i.e., 3 : 4., 4, , EXAMPLE 4, , SOLUTION, , Find the ratio in which the line segment, joining the points P( 2, 3 , 4), and Q( −3 , 5 , − 4) is divided by the yz-plane. Also, find the point of, intersection., Let PQ be the divided by the yz-plane at a point R in the ratio λ : 1., Then, the coordinates of R are, ⎛ −3 λ + 2 5 λ + 3 −4λ + 4⎞, ... (i), ,, ,, ⎜, ⎟⋅, λ+1, λ+1 ⎠, ⎝ λ+1, Since R lies on the yz-plane, the x-coordinate or R is therefore 0., 2, −3 λ + 2, = 0, or λ = ⋅, ∴, 3, λ+1, 2, So, the required ratio is : 1, i.e., 2 : 3., 3, 2, Putting λ = in (i), the point of intersection of the line segment PQ, 3, 19 4, and the yz-plane is ⎛⎜ 0, , ⎞⎟ ⋅, ⎝ 5 5⎠, , EXAMPLE 5, , SOLUTION, , Find the ratio in which the join of A( 2, 1, 5) and B( 3 , 4, 3) is divided by, the plane 2x + 2y − 2z = 1. Also, find the coordinates of the point of, division., Suppose the given plane intersects AB at a point C and let the, required ratio be λ : 1.
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792, , Senior Secondary School Mathematics for Class 11, , Then, the coordinates of C are, ⎛ 3 λ + 2 4λ + 1 3 λ + 5 ⎞, ,, ,, ⎜, ⎟⋅, ⎝ λ+1 λ+1 λ+1⎠, , … (1), , Since C lies on the plane 2x + 2y − 2z = 1, this point must satisfy the, equation of the plane., ⎛ 3 λ + 2⎞, ⎛ 4λ + 1⎞, ⎛ 3λ + 5⎞, 5, ⋅, 2⎜, ⎟ + 2⎜, ⎟ − 2⎜, ⎟ = 1 or λ =, ⎝ λ+1⎠, ⎝ λ +1⎠, ⎝ λ+1⎠, 7, 5, So, the required ratio is : 1, i.e., 5 : 7., 7, 5, 29 9 25, in (i), the required point of division is C ⎛⎜ , , ⎞⎟ ⋅, Putting λ =, ⎝ 12 4 6 ⎠, 7, , ∴, , EXAMPLE 6, , SOLUTION, , Three vertices of a parallelogram ABCD are A( 3 , − 1, 2), B(1, 2, − 4), and C( −1, 1, 2). Find the coordinates of the fourth vertex D., Let D( x , y , z) be the required point. Then, the midpoint of diagonal, BD is, ⎛ x + 1 , y + 2 , z − 4⎞ ⋅, ⎜, ⎟, ⎝ 2, 2, 2 ⎠, And, the midpoint of diagonal AC is, ⎛ 3 − 1 , −1 + 1 , 2 + 2⎞ , i.e., (1, 0, 2)., ⎜, ⎟, ⎝ 2, 2, 2 ⎠, But, the midpoints of the diagonals of a parallelogram always coincide., x+1, y+2, z−4, ∴, = 1,, = 0 and, = 2., 2, 2, 2, So, x = 1, y = − 2 and z = 8., Hence, the required point is D(1, − 2, 8)., , EXAMPLE 7, , SOLUTION, , The midpoints of the sides of a triangle are (1, 5 , − 1), ( 0, 4, − 2) and, ( 2, 3 , 4). Find its vertices., Let A( x1 , y1 , z1 ), B( x 2 , y 2 , z2 ) and C( x 3 , y 3 , z3 ) be the vertices of the, given triangle, and let D(1, 5 , − 1), E( 0, 4, − 2) and F( 2, 3 , 4) be the, midpoints of the sides BC, CA and AB respectively. Then,, x2 + x3, y + y3, z + z3, = 1; 2, = 5; 2, = 1;, 2, 2, 2, x 3 + x1, y + y1, z +z, = 0; 3, = 4; 3 1 = − 2;, 2, 2, 2, x1 + x 2, y1 + y 2, z +z, = 2;, = 3 and 1 2 = 4., 2, 2, 2, Thus, x 2 + x 3 = 2; x 3 + x1 = 0; x1 + x 2 = 4;, y 2 + y 3 = 10; y 3 + y1 = 8; y1 + y 2 = 6;, z2 + z3 = − 2; z3 + z1 = − 4; z1 + z2 = 8.
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Three-Dimensional Geometry, , 793, , Adding first three equations, we get, 2( x1 + x 2 + x 3 ) = 6 or x1 + x 2 + x 3 = 3., Thus, x1 = 1, x 2 = 3 and x 3 = −1., Adding next three equations, we get, 2( y1 + y 2 + y 3 ) = 24 or y1 + y 2 + y 3 = 12., ∴ y1 = 2; y 2 = 4 and y 3 = 6., Adding last three equations, we get, 2(z1 + z2 + z3 ) = 2 or z1 + z2 + z3 = 1., ∴ z1 = 3 , z2 = 5 and z3 = − 7., Hence, the vertices of the given triangle are, A(1, 2, 3); B( 3 , 4, 5) and C( −1, 6, − 7)., EXAMPLE 8, , SOLUTION, , Using the section formula, prove that the three points, A( −2, 3 , 5), B(1, 2, 3) and C(7 , 0, − 1) are collinear., Suppose C divides AB in the ratio λ : 1., λ−2, 2λ + 3, 3λ + 5, Then,, = 7,, = 0 and, = − 1., λ+1, λ+1, λ+1, −3, From each of these equations, we get λ =, ⋅, 2, This shows that C divides AB externally in the ratio 3 : 2., So, C lies on the line joining A and B., Hence, the given points A, B, C are collinear., , EXERCISE 26C, 1. Find the coordinates of the point which divides the join of A( 3 , 2, 5) and, B( −4, 2, − 2) in the ratio 4 : 3., 2. Let A( 2, 1, − 3) and B(5 , − 8, 3) be two given points. Find the coordinates of, the points of trisection of the line segment AB., 3. Find the coordinates of the point that divides the join of A( −2, 4, 7) and, B( 3 , − 5 , 8) extremally in the ratio 2 : 1., 4. Find the ratio in which the point R(5 , 4, − 6) divides the join of P( 3 , 2, − 4), and Q( 9, 8, − 10)., 5. Find the ratio in which the point C(5 , 9, − 14) divides the join of A( 2, − 3 , 4), and B( 3 , 1, − 2)., 6. Find the ratio in which the line segment having the end points A( −1, − 3 , 4), and B( 4, 2, −1) is divided by the xz-plane. Also, find the coordinates of the, point of division.
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794, , Senior Secondary School Mathematics for Class 11, , 7. Find the coordinates of the point where the line joining A( 3 , 4, 1) and, B(5 , 1, 6) crosses the xy-plane., 8. Find the ratio in which the plane x − 2y + 3z = 5 divides the join of, A( 3 , − 5 , 4) and B( 2, 3 , − 7). Find the coordinates of the point of intersection, of the line and the plane., 9. The vertices of a triangle ABC are A( 3 , 2, 0), B(5 , 3 , 2) and C( −9, 6, − 3)., The bisector AD of ∠A meets BC at D. Find the coordinates of D., 10. If the three consecutive vertices of a parallelogram be A( 3 , 4, − 1),, B(7 , 10, − 3) and C(5 , − 2, 7), find the fourth vertex D., 11. Two vertices of a triangle ABC are A( 2, − 4, 3) and B( 3 , − 1, − 2), and its, centroid is (1, 0, 3). Find its third vertex C., 12. If origin is the centroid of traingle ABC with vertices A( a , 1, 3), B( −2, b , − 5), and C( 4, 7 , c), find the values of a, b, c., 13. The midpoints of the sides of a triangle are (1, 5 , − 1), ( 0, 4, − 2) and (2, 3, 4)., Find its vertices., ANSWERS (EXERCISE 26C), , 1. ( −1, 2, 1), , 2. ( 3 , − 2, −1), ( 4, −5 , 1), , 3. ( 8, − 14, 9), , 4. 1 : 2, , 5. 3 : 2 (externally), , 6. 3 : 2, (2, 0, 1), , 13 23 ⎞, 7. ⎛⎜ ,, , 0⎟, ⎝5 5, ⎠, , 13 −9 −2⎞, 8. 2 : 3, ⎛⎜ ,, , ⎟, ⎝5 5 5⎠, , 19 57 17 ⎞, 9. ⎛⎜ ,, , ⎟, ⎝ 8 16 16 ⎠, , 10. (1, − 8, 9), , 11. ( −2, 5 , 8), , 12. a = − 2, b = − 8, c = 2, , 13. (1, 2, 3); ( 3 , 4, 5); ( −1, 6, − 7), HINTS TO SOME SELECTED QUESTIONS, 2. Let P and Q be the points of trisection of AB. Then, P divides AB in the ratio 1 : 2 and Q, divides AB in the ratio 2 : 1., , 7. Suppose the xy-plane divides AB in the ratio λ : 1 ., Then,, , 6λ + 1, 1, =0 ⇒ λ=− ⋅, λ+ 1, 6, , Divide AB in the ratio 1 : ( − 6 )., 8. D divides BC in the ratio AB : AC ., , q
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27, Limits, CONCEPT OF LIMIT, Consider the function f ( x) =, Clearly, f (1) =, , ( x 2 − 1), ., ( x − 1), , 0, , which is meaningless., 0, , Thus, f ( x) is not defined at x = 1., Now,, , f ( x) =, , ( x 2 − 1) ( x − 1) ( x + 1), =, = ( x + 1), only when x ≠ 1., ( x − 1), ( x − 1), , If we give to x, a value not exactly 1 but slightly more than 1 then clearly,, the value of f ( x) is slightly more than 2. Now, if we go on decreasing this value, and take it nearer to 1 then clearly the value of f ( x) will come nearer to 2, as, shown below., If x = 1.1, , then f ( x) = 2.1., , If x = 1.01, , then f ( x) = 2.01., , If x = 1.001, … … …, , then f ( x) = 2.001., …, …, …, , If x = 1.0000001, … … …, , then f ( x) = 2.0000001., …, …, …, , Thus, as the value of x approaches 1, the value of f ( x) approaches 2, and, we write:, as, , x → 1, f ( x) → 2., , (The symbol ‘→’ stands for ‘approaches to’.), Similarly, if we give to x, a value slightly less than 1 then the value of f ( x), is slightly less than 2. Now, if we go on increasing this value and take it nearer, to 1, the value of f ( x) will come nearer to 2, as shown below., If x = 0.9, , then f ( x) = 1.9., , If x = 0.99, , then f ( x) = 1.99., 795
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796, , Senior Secondary School Mathematics for Class 11, , If x = 0.999, … … …, If x = 0.999999, … … …, , then, …, then, …, , f ( x) = 1.999., … …, f ( x) = 1.999999., … …, , Thus, in this case also, as x → 1, f ( x) → 2., We express this fact as, , LIMIT, , ⎛ x 2 − 1⎞, lim ⎜, ⎟ = 2., x→ 1 ⎝ x − 1 ⎠, , We say that lim f ( x) = l if whenever x → a , f ( x) → l., x→ a, , Working rules for finding lim f ( x), x→a, , RULE I, , Put x = a in the given function. If f ( a) is a definite value then, lim f ( x) = f ( a)., , x→ a, , EXAMPLES, , 1. lim sin x = sin 0 = 0., x→ 0, , 2. lim ( x 2 + 5 x − 2) = (12 + 5 × 1 − 2) = 4., x→1, , If f ( a) is indeterminate, we adopt the rules given below., RULE II, , If f ( x) is a rational function then factorize the numerator and the, denominator. Cancel out the common factors and then put x = a., , EXAMPLE, , SOLUTION, , Evaluate, , x→1, , EXAMPLE, , SOLUTION, , (ii) lim, , x→1, , ( x 2 − 4x + 3), ⋅, ( x − 1), , ⎛ x 2 − 9⎞, ( x + 3) ( x − 3), (i) lim ⎜, = lim ( x + 3) = 6., ⎟ = lim, x→ 3 ⎝ x − 3 ⎠, x→ 3, x→ 3, ( x − 3), (ii) lim, , RULE III, , ⎛ x 2 − 9⎞, (i) lim ⎜, ⎟, x→ 3 ⎝ x − 3 ⎠, , ( x − 1) ( x − 3), ( x 2 − 4x + 3), = lim ( x − 3) = −2., = lim, x→1, x→1, ( x − 1), ( x − 1), , If the given function contains a surd then simplify it by using, conjugate surds. After simplification, put x = a ., Evaluate, , ⎧ 1+ x − 1−x⎫, lim ⎨, ⎬⋅, x, ⎭, ⎩, , x→ 0, , We have, ⎧ 1+ x − 1−x⎫, ⎧( 1 + x − 1 − x ( 1 + x + 1 − x ) ⎫, ⋅, lim ⎨, ⎬ = xlim, ⎬, ⎨, →0, x, x, ( 1 + x + 1 − x) ⎭, ⎭, ⎩, ⎩, , x→ 0
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Limits, , 797, , {(1 + x) − (1 − x)}, x( 1 + x + 1 − x ), 2x, = lim, x → 0 x( 1 + x + 1 − x ), 2, = lim, =1, x→ 0 ( 1 + x + 1 − x), = lim, , x→ 0, , RULE IV, , [putting x = 0]., , If the given function contains a series which is capable of being, expanded then after making proper expansion and simplifying,, cancel the common factors in the numerator and denominator, if any., Then, put x = a., , Some important expansions are given below:, (i) For|x| < 1, we have the binomial expansion, ⎧, ⎫, n(n − 1) 2 n(n − 1) (n − 2) 3, x +, x + …⎬, (1 + x) n = ⎨1 + nx +, 2!, 3!, ⎭, ⎩, ⎛ x n − an ⎞, n −1, (ii) ⎜, + ax n − 2 + a 2 x n − 3 + … + a n − 1 ), ⎟ = (x, ⎝ x−a ⎠, ⎫, ⎧, x2 x3, xn, (iii) ex = ⎨1 + x +, +, +…+, + …⎬, 2! 3 !, n!, ⎭, ⎩, 2, ⎫, ⎧, x, (iv) a x = ⎨1 + x(log a) +, (log a) 2 + …⎬, 2!, ⎭, ⎩, 2, 3, 4, ⎫, ⎧, x, x, x, (v) log (1 + x) = ⎨x −, +, −, + …⎬, 2, 3, 4, ⎭, ⎩, ⎫, ⎧, x 3 x5 x 7, (vi) sin x = ⎨x −, +, −, + …⎬, 3! 5! 7!, ⎭, ⎩, 2, 4, 6, ⎫, ⎧, x, x, x, (vii) cos x = ⎨1 −, +, −, + …⎬, 2, 4, 6, !, !, !, ⎭, ⎩, , ⎫, ⎧, 2 5, x3, (viii) tan x = ⎨x +, +, x + …⎬, 3, 15, ⎭, ⎩, , We shall also make use of the following theorems on limits. The proof of, these theorems is beyond the scope of this book., Fundamental Theorems on Limits (without proof), (i) lim { f ( x) + g( x)} = ⎧⎨lim f ( x) ⎫⎬ + ⎧⎨lim g( x) ⎫⎬, x→ a, ⎭ ⎩x → a, ⎭, ⎩x → a, (ii) lim { f ( x) − g( x)} = ⎧⎨lim f ( x) ⎫⎬ − ⎧⎨lim g( x) ⎫⎬, x→ a, ⎭ ⎩x → a, ⎭, ⎩x → a, ⎧, ⎫, (iii) lim {c ⋅ f ( x)} = c ⋅ ⎨lim f ( x) ⎬, where c is a constant, x→ a, ⎭, ⎩x → a, ⎧, (iv) lim { f ( x) ⋅ g( x)} = ⎨lim f ( x) ⎫⎬ ⋅ ⎧⎨lim g( x) ⎫⎬, x→ a, ⎭ ⎩x → a, ⎭, ⎩x → a
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798, , Senior Secondary School Mathematics for Class 11, , ⎧ lim f ( x) ⎫, ⎨ x→ a, ⎬, ⎭ , provided lim g( x) ≠ 0, ⎩, x→ a, ⎧ lim g( x) ⎫, ⎬, ⎨ x→ a, ⎭, ⎩, (vi) If f ( x) ≤ g( x) for all x then lim f ( x) ≤ lim g( x)., ⎧ f ( x) ⎫, (v) lim ⎨, ⎬=, x → a g( x), ⎭, ⎩, , x→ a, , x→ a, , Things to Remember, , C + D⎞, ⎛ C − D⎞, (i) sin C + sin D = 2 sin ⎛⎜, ⎟ cos ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, C + D⎞, ⎛ C − D⎞, (ii) sin C − sin D = 2 cos ⎛⎜, ⎟ sin ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, C + D⎞, ⎛ C − D⎞, (iii) cos C + cos D = 2 cos ⎛⎜, ⎟ cos ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, C + D⎞, ⎛ C − D⎞, (iv) cos C − cos D = −2 sin ⎛⎜, ⎟ sin ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, , Some more Important Theorems on Limits, THEOREM 1, , PROOF, , ⎛ x n − an ⎞, n −1, lim ⎜, ⎟ = na , where a > 0., x→ a ⎝ x − a ⎠, , We know that, ∴, , ⎛ x n − an ⎞, n −1, + ax n − 2 + a 2 x n − 3 + … + a n − 1 )., ⎜, ⎟ = (x, ⎝ x−a ⎠, , ⎛ x n − an ⎞, n −1, lim ⎜, + ax n − 2 + a 2 x n − 3 + … + a n −1 ), ⎟ = lim ( x, x→ a ⎝ x − a ⎠, x→ a, = na n − 1, , [putting x = a]., , ⎛ e − 1⎞, lim ⎜, ⎟ = 1., ⎝ x ⎠, x, , THEOREM 2, , PROOF, , x→ 0, , We have, ⎫, ⎧⎛, ⎞, x2 x3, +, + …⎟ − 1 ⎪, ⎪ ⎜1 + x +, 2! 3 !, ⎛, ⎞⎫, ⎛ e − 1⎞, ⎠, x2, ⎪⎝, ⎪⎧, lim ⎜, Q ex = ⎜1 + x +, + …⎟ ⎬, ⎟ = lim ⎨, ⎨, ⎬, x→ 0 ⎝ x ⎠, x→ 0, x, !, 2, ⎝, ⎠⎭, ⎪⎩, ⎪, ⎪⎭, ⎪⎩, 2, 3, ⎧⎛, ⎧ ⎛, ⎞⎫, ⎞⎫, x, x, x, x2, +, + …⎟ ⎪, +, + …⎟ ⎪, ⎪⎜ x +, ⎪ x ⎜1 +, !, !, 2, 3, !, !, 2, 3, ⎝, ⎠, ⎝, ⎠⎪, ⎪, ⎪, ⎪, = lim ⎨, = lim ⎨, ⎬, ⎬, x→ 0, x→ 0, x, x, ⎪, ⎪, ⎪, ⎪, x, , ⎪⎩, ⎛, ⎞, x, x2, = lim ⎜1 +, +, + …⎟, x→ 0 ⎝, 2! 3 !, ⎠, , =1, , [putting x = 0]., , ⎪⎭, , ⎪⎩, , ⎪⎭
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Limits, , THEOREM 3, , 799, , ⎛ a x − 1⎞, lim ⎜, ⎟ = log a., x→ 0 ⎝, x ⎠, , ⎧⎧, ⎫ ⎫, x2, (log, ), (log a) 2 + …⎬ − 1 ⎪, 1, x, a, +, +, ⎨, ⎪, x, 2!, ⎛ a − 1⎞, ⎪⎩, ⎭ ⎪ [on expanding a x], PROOF lim ⎜, ⎟ = lim ⎨, ⎬, x→ 0 ⎝, x, x ⎠ x→ 0 ⎪, ⎪, ⎪⎭, ⎪⎩, ⎧, ⎫, x, (log a) 2 + …⎬, x ⎨(log a) +, 2, !, ⎭, = lim ⎩, x→ 0, x, ⎫, ⎧, x, x2, = lim ⎨(log a) +, (log a) 2 +, (log a) 3 + …⎬, x→ 0, 2, 3, !, !, ⎭, ⎩, [putting x = 0]., = log a, THEOREM 4, , lim (1 + x)1/x = e., , x→ 0, , 1⎛1 ⎞, 1⎛1 ⎞⎛1 ⎞, ⎧, ⎫, ⎜ − 1⎟, ⎜ − 1⎟ ⎜ − 2⎟, ⎪, ⎪, ⎝, ⎠, 1, ⎪, x x, x⎝x ⎠⎝x ⎠ 3, 1/x, 2, PROOF lim (1 + x), = lim ⎨1 + ⋅ x +, ⋅ x + … ∞⎬, ⋅x +, x→0, x→0, 2!, 3!, ⎪ x, ⎪, ⎪⎩, ⎪⎭, 1/ x, [using the binomial expansion of (1 + x) ], ⎧, ⎫, (1 − x) (1 − x) (1 − 2x), = lim ⎨1 + 1 +, +, + …⎬, x→ 0, 2, !, 3, !, ⎭, ⎩, ⎛, ⎞, 1, 1, [putting x = 0], = ⎜1 + 1 +, +, + …⎟, 2! 3 !, ⎝, ⎠, = e., log (1 + x), = 1., x, ⎛, ⎞, x2 x3, +, − …⎟, ⎜x −, 2, 3, ⎝, ⎠, log (1 + x), PROOF lim, = lim, x→ 0, x→ 0, x, x, THEOREM 5, , lim, , x→ 0, , ⎛, ⎞, x x2, = lim ⎜1 − +, − …⎟ = 1, x→ 0 ⎝, 2, 3, ⎠, , [expanding log (1 + x)], [putting x = 0]., , SUMMARY, , ⎛ x n − an ⎞, n −1, 1. lim ⎜, ⎟ = na , where a > 0., x→ a ⎝ x − a ⎠, , ⎛ ex − 1⎞, 2. lim ⎜, ⎟ = 1., x→ 0 ⎝ x ⎠, , ⎛ a x − 1⎞, 3. lim ⎜, ⎟ = log e a., x→ 0 ⎝, x ⎠, , 5. lim, , 4. lim (1 + x)1/x = e., x→ 0, , x→ 0, , log (1 + x), = 1., x
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800, , Senior Secondary School Mathematics for Class 11, SOLVED EXAMPLES, , EXAMPLE 1, , SOLUTION, , EXAMPLE 2, , SOLUTION, , Evaluate, , ⎧ x12 − a12 ⎫, lim ⎨, ⎬⋅, x→ a, ⎩ x−a ⎭, ⎡, ⎛ x n − an ⎞, n −1 ⎤, ⎜, ⎟ = na ⎥ ⋅, ⎢Q xlim, →a ⎝ x − a ⎠, ⎦, ⎣, , ⎧ x12 − a12 ⎫, ( 12 − 1 ), = 12 a11, lim ⎨, ⎬ = 12 a, x→ a, ⎩ x−a ⎭, Evaluate, , ⎛ x5 − 32⎞, lim ⎜ 3, ⎟⋅, x→ 2 ⎝ x − 8 ⎠, , ⎛ x5 − 32⎞, ⎛ x5 − 25 ⎞, lim ⎜ 3, ⎟ = lim ⎜ 3, ⎟, x→ 2 ⎝ x − 8 ⎠, x → 2 ⎝ x − 23 ⎠, ⎧⎛ x5 − 25 ⎞ ⎛ x 3 − 23 ⎞ ⎫, = lim ⎨⎜, ⎟ ÷⎜, ⎟⎬, x→ 2 ⎝ x − 2 ⎠, ⎝ x − 2 ⎠⎭, ⎩, , ⎛ x5 − 25 ⎞, ⎛ x 3 − 23 ⎞, = lim ⎜, ⎟ ÷ lim ⎜, ⎟, x→ 2 ⎝ x − 2 ⎠, x→ 2 ⎝ x − 2 ⎠, , = [(5 × 24 ) ÷ ( 3 × 22 )], 80, 20, ⋅, = ⎛⎜ ⎞⎟ =, ⎝ 12 ⎠, 3, EXAMPLE 3, , Evaluate, lim, , x→ 1, , x→ 1, , x15 − 1, ⋅, x10 − 1, , ⎧⎛ x15 − 1⎞ ⎛ x10 − 1⎞ ⎫, ⎟ ÷⎜, ⎟⎬, ⎨⎜, ⎩⎝ x − 1 ⎠ ⎝ x − 1 ⎠ ⎭, ⎛ x10 − 1⎞, ⎛ x15 − 1⎞, = lim ⎜, ⎟ ÷ lim ⎜, ⎟, x→ 1 ⎝ x − 1 ⎠, x→ 1 ⎝ x − 1 ⎠, , x −1, = lim, x10 − 1 x → 1, 15, , SOLUTION, , lim, , = (15 × 114 ) ÷ (10 × 19 ) =, EXAMPLE 4, , SOLUTION, , ⎡, ⎛ x n − an ⎞, n −1 ⎤, ⎜, ⎟ = na ⎥, ⎢Q xlim, →a ⎝ x − a ⎠, ⎦, ⎣, , Evaluate, , 15 3, = ⋅, 10 2, , ⎛ x m − am ⎞, lim ⎜ n, ⎟⋅, x → a ⎝ x − an ⎠, , ⎧⎛ x m − a m ⎞ ⎛ x n − a n ⎞ ⎫, ⎛ x m − am ⎞, lim ⎜ n, ⎟ = lim ⎨⎜, ⎟⎬, ⎟ ÷⎜, n, x→ a ⎝ x − a ⎠, x→ a ⎝ x − a ⎠, ⎝ x − a ⎠⎭, ⎩, ⎛ x m − am ⎞, ⎛ x n − an ⎞, = lim ⎜, ⎟ ÷ lim ⎜, ⎟, x→ a ⎝ x − a ⎠, x→ a ⎝ x − a ⎠, = (ma m − 1 ) (na n − 1 ), =, , ma m − 1 m m − n, = a, ., n, na n − 1
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Limits, , EXAMPLE 5, , SOLUTION, , Evaluate, , ( x + 2) 3/2 − ( a + 2) 3/2, ⋅, x→ a, x−a, lim, , ( x + 2) 3/2 − ( a + 2) 3/2, x→ a, x−a, ( x + 2) 3/2 − ( a + 2) 3/2, =, lim, (x + 2 ) → (a + 2 ), ( x + 2) − ( a + 2), lim, , ⎛3, , =, , EXAMPLE 6, , SOLUTION, , 801, , ⎜, 3, ⋅ ( a + 2) ⎝ 2, 2, , ⎞, − 1⎟, ⎠, , =, , 3, ( a + 2)1/2, 2, , ⎡, ⎛ x n − an ⎞, n−1⎤, ⎜, ⎟ = na, ⎥⋅, ⎢Q xlim, →a ⎝ x − a ⎠, ⎦, ⎣, , (1 + x) n − 1, ⋅, x→ 0, x, , Evaluate lim, , Put (1 + x) = y, so that when x → 0 then y → 1., ⎛ y n − 1⎞, (1 + x) n − 1, = lim ⎜, ⎟, x→ 0, y→1 ⎝ y −1⎠, x, , ∴, , lim, , ⎛ y n − 1n ⎞, (n − 1 ), = lim ⎜, = (n × 1) = n., ⎟ =n × 1, y→1 ⎝ y −1 ⎠, EXAMPLE 7, , SOLUTION, , Evaluate, , ⎛ 1 − x −1/3 ⎞, lim ⎜, ⎟⋅, x → 1 ⎝ 1 − x −2/3 ⎠, , ⎡ (1 − x −1/3 ) ⎤, ⎛ 1 − x −1/3 ⎞, lim ⎜, lim, =, ⎟, ⎢, ⎥, x → 1 ⎝ 1 − x −2/3 ⎠, x → 1 (1) 2 − ( x −1/3 ) 2, ⎣, ⎦, (1 − x −1/3 ), x → 1 (1 − x −1/3 )(1 + x −1/3 ), 1, 1, 1, = lim, =, = ⋅, x → 1 (1 + x − 1 / 3 ), {1 + (1) − 1 / 3 } 2, = lim, , EXAMPLE 8, , SOLUTION, , Evaluate, , lim, , x→ 0, , lim, , x→ 0, , ( 1 + 3x − 1 − 3x), ⋅, x, , [CBSE 2000C], , ( 1 + 3x − 1 − 3x), x, ⎧( 1 + 3 x − 1 − 3 x ) ( 1 + 3 x + 1 − 3 x ) ⎫, = lim ⎨, ×, ⎬, x→ 0, x, ( 1 + 3x + 1 − 3x) ⎭, ⎩, {(1 + 3 x) − (1 − 3 x), 6x, = lim, = lim, x → 0 x( 1 + 3 x + 1 − 3 x ), x → 0 x( 1 + 3 x + 1 − 3x), 6, 6, 6, = lim, =, = = 3., x → 0 ( 1 + 3 x + 1 − 3 x), ( 1 + 1) 2
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802, , EXAMPLE 9, , SOLUTION, , Senior Secondary School Mathematics for Class 11, , ⎧, ( x 2 − 4), lim ⎨, ⎩ 3x − 2 − x +, 2, ⎫, ⎧, ( x − 4), lim ⎨, ⎬, x→ 2, 3, x, −, 2, −, x, +, 2, ⎭, ⎩, ⎧, ( x 2 − 4), = lim ⎨, x→ 2 ( 3x − 2 − x, ⎩, Evaluate, , x→ 2, , ⎫, ⎬⋅, 2⎭, , + 2), , ×, , ( 3x − 2 +, ( 3x − 2 +, , x − 2) ⎫, ⎬, x − 2) ⎭, , ⎧( x − 4)( 3 x − 2 + x − 2) ⎫, = lim ⎨, ⎬, x→ 2, ( 3 x − 2) − ( x + 2), ⎭, ⎩, 2, , ⎧( x 2 − 4)( 3 x − 2 +, = lim ⎨, x→ 2, 2( x − 2), ⎩, = lim, , x→ 2, , =8, EXAMPLE 10, , SOLUTION, , Evaluate, , ( x + 2)( 3 x − 2 +, 2, [putting x = 2]., , x − 2), , [cancelling ( x − 2)], , ⎛ e3 x − 1⎞, lim ⎜, ⎟⋅, x→ 0 ⎝, x ⎠, , ⎧⎛ e3 x − 1⎞, ⎫, ⎛ e3 x − 1⎞, lim ⎜, [Q ( x → 0) ⇒ ( 3 x → 0)], ⎟ = lim ⎨⎜, ⎟ × 3⎬, x→ 0 ⎝, x ⎠ 3 x → 0 ⎩⎝ 3 x ⎠, ⎭, ⎛ ey − 1⎞, = 3 × lim ⎜, ⎟ , where y = 3 x, y→ 0 ⎝, y ⎠, = ( 3 × 1) = 3, , EXAMPLE 11, , SOLUTION, , x − 2) ⎫, ⎬, ⎭, , Evaluate, , ⎤, ⎡, ⎛ ey − 1⎞, ⎜, ⎟ = 1⎥ ⋅, ⎢Q ylim, →0⎝, y ⎠, ⎦, ⎣, , ⎛ ex − e2 ⎞, lim ⎜, ⎟⋅, x→ 2 ⎝ x − 2 ⎠, , [CBSE 2003], , Put ( x − 2) = y, so that when x → 2 then y → 0., ⎛ ex − e2 ⎞, ⎛ ey + 2 − e2 ⎞, ∴ lim ⎜, ⎟ = lim ⎜, ⎟, x→2 ⎝ x − 2 ⎠, y→ 0 ⎝, y, ⎠, ⎧, ⎛ ey − 1⎞ ⎫, = lim ⎨e2 ⋅ ⎜, ⎟⎬, y→ 0, ⎝ y ⎠⎭, ⎩, ⎛ ey − 1⎞, 2, 2, = e2 ⋅ lim ⎜, ⎟ = ( e × 1) = e, y→ 0 ⎝, y ⎠, , EXAMPLE 12, , ⎤, ⎡, ⎛ ey − 1⎞, ⎜, ⎟ = 1⎥ ⋅, ⎢Q ylim, →0⎝, y, ⎠, ⎦, ⎣, , Evaluate:, ⎛ e− x − 1⎞, ⎛ ex − e− x ⎞, ⎛ ex + e− x − 2⎞, (i) lim ⎜, ⎟ (ii) lim ⎜, ⎟ (iii) lim ⎜, ⎟ [CBSE 2004], x→ 0 ⎝, x→ 0 ⎝, x→ 0 ⎝, x ⎠, x ⎠, x2, ⎠
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Limits, , SOLUTION, , 803, , ⎛ e− x − 1⎞, ⎛ ey − 1⎞, (i) lim ⎜, ⎟ = lim ⎜, ⎟ , where − x = y, x→ 0 ⎝, x ⎠ y → 0 ⎝ −y ⎠, ⎤, ⎡, ⎛ ey − 1⎞, ⎜, ⎟ = 1⎥ ⋅, ⎢Q ylim, →0 ⎝, y ⎠, ⎦, ⎣, , ⎛ ey − 1⎞, = − lim ⎜, ⎟ = −1, y→0 ⎝, y ⎠, ⎛ ex − e− x ⎞, ⎧( ex − 1) − ( e− x − 1) ⎫, (ii) lim ⎜, ⎟ = lim ⎨, ⎬, x→ 0 ⎝, x ⎠ x→ 0 ⎩, x, ⎭, ⎧⎛ ex − 1⎞ ⎛ e− x − 1⎞ ⎫, = lim ⎨⎜, ⎟ −⎜, ⎟⎬, x→ 0 ⎝ x ⎠, ⎝ x ⎠⎭, ⎩, ⎛ ex − 1⎞, ⎛ e− x − 1⎞, = lim ⎜, ⎟ − lim ⎜, ⎟, x→ 0 ⎝ x ⎠, x→ 0 ⎝, x ⎠, ⎛ ey − 1⎞, = 1 − lim ⎜, ⎟ ,, y → 0 ⎝ −y ⎠, , ⎤, ⎡, ⎛ ex − 1⎞, where − x = y ⎢Q lim ⎜, ⎟ = 1⎥, x→ 0 ⎝ x ⎠, ⎦, ⎣, ⎤, ⎡, ⎛ ey − 1⎞, ⎛ ey − 1⎞, = 1 + lim ⎜, ⎟ = 1⎥ ⋅, ⎟ = (1 + 1) = 2 ⎢Q lim ⎜, y→0 ⎝, y→0 ⎝, y ⎠, y, ⎠, ⎦, ⎣, ⎛ ex + e− x − 2⎞, ⎛ e2 x + 1 − 2ex ⎞, (iii) lim ⎜, lim, =, ⎟, ⎜, ⎟, x→ 0 ⎝, x2, x 2 ex, ⎠ x→ 0 ⎝, ⎠, ⎧⎪⎛ ex − 1⎞ 2 1 ⎫⎪, = lim ⎨⎜, ⎟ ⋅ x⎬, x→ 0 ⎝ x ⎠, e ⎪, ⎪⎩, ⎭, 2, x, ⎛ e − 1⎞, 1, = lim ⎜, ⎟ ⋅ lim x, x→ 0 ⎝ x ⎠, x→ 0 e, 1, = ⎛⎜12 × 0 ⎞⎟ = (1 × 1) = 1, ⎝, e ⎠, EXAMPLE 13, , SOLUTION, , Evaluate, , ⎤, ⎡, ⎛ ex − 1⎞, ⎜, ⎟ = 1⎥ ⋅, ⎢Q xlim, →0 ⎝ x ⎠, ⎦, ⎣, , ⎛ 3 x − 2x ⎞, lim ⎜, ⎟⋅, x→ 0 ⎝, x ⎠, , ⎛ 3 x − 2x ⎞, ⎧( 3 x − 1) − ( 2x − 1) ⎫, lim ⎜, ⎟ = lim ⎨, ⎬, x→ 0 ⎝, x ⎠ x→ 0 ⎩, x, ⎭, ⎛ 3 x − 1⎞, ⎛ 2x − 1⎞, = lim ⎜, ⎟ − lim ⎜, ⎟, x→ 0 ⎝, x ⎠ x→ 0 ⎝ x ⎠, = (log 3 − log 2) = log, , 3, 2, , ⎡, ⎛ a x − 1⎞, ⎜, ⎟ = log, ⎢Q xlim, →0 ⎝, x ⎠, ⎣, , ⎤, a⎥ ⋅, ⎦
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804, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 14, , SOLUTION, , Evaluate, , ⎛ 3 2 x − 23 x ⎞, lim ⎜, ⎟⋅, x→ 0 ⎝, x, ⎠, , ⎛ 3 2 x − 23 x ⎞, ⎧( 3 2 x − 1) − ( 23 x − 1) ⎫, lim ⎜, ⎟ = lim ⎨, ⎬, x→ 0 ⎝, x, x, ⎠ x→ 0 ⎩, ⎭, ⎛ 3 2 x − 1⎞, ⎛ 23 x − 1⎞, = lim ⎜, ⎟ − lim ⎜, ⎟, x→ 0 ⎝, x ⎠ x→ 0 ⎝ x ⎠, ⎧⎛ 3 2 x − 1⎞ ⎫, ⎧⎛ 23 x − 1⎞ ⎫, = lim ⎨⎜, ⎟ ⋅ 3⎬, ⎟ ⋅ 2⎬ − lim ⎨⎜, x→ 0 ⎝, x→ 0 ⎝ 3x ⎠, ⎭, ⎩ 2x ⎠ ⎭, ⎩, ⎛ 3 2 x − 1⎞, ⎛ 23 x − 1⎞, = 2 ⋅ lim ⎜, ⎟ − 3 ⋅ lim ⎜, ⎟, 2x→ 0 ⎝, 3x→ 0 ⎝ 3x ⎠, 2x ⎠, ⎡, ⎛ a y − 1⎞, ⎜, ⎟ = log, ⎢Q ylim, →0 ⎝, y ⎠, ⎣, 9, = [log ( 3) 2 − log ( 2) 3 ] = (log 9 − log 8) = log ⎛⎜ ⎞⎟ ⋅, ⎝ 8⎠, , = ( 2 log 3 − 3 log 2), , EXAMPLE 15, , Evaluate, , ⎤, a⎥, ⎦, , ⎛ 3 2 x − 1⎞, lim ⎜ 3 x ⎟ ⋅, x→ 0 ⎝ 2 − 1⎠, , ⎧⎛ 3 2 x − 1⎞, ⎫, ⎟ ⋅ 2x ⎬, ⎨⎜, ⎛ 3 − 1⎞, ⎭, ⎩⎝ 2x ⎠, lim ⎜, ⎟ = lim, x → 0 ⎝ 23 x − 1 ⎠, x → 0 ⎧⎛ 23 x − 1⎞, ⎫, ⎟ ⋅ 3x⎬, ⎨⎜, 3, x, ⎝, ⎠, ⎭, ⎩, 2x, , SOLUTION, , ⎛ 3 2 x − 1⎞, lim ⎜, ⎟, 2 2 x → 0 ⎝ 2x ⎠, = ⋅, 3, ⎛ 23 x − 1⎞, lim ⎜, ⎟, 3x→ 0 ⎝ 3x ⎠, , EXAMPLE 16, , SOLUTION, , Evaluate, , lim, , x→1, , =, , 2 log 3, ⋅, 3 log 2, , =, , 2 log 3 log ( 3 2 ) log 9, =, =, ⋅, 3 log 2 log ( 23 ) log 8, , lim, , x→1, , ⎤, ⎡, ⎛ a y − 1⎞, ⎜, ⎟ = log a ⎥, ⎢Q ylim, →0 ⎝, y ⎠, ⎦, ⎣, , x2 − x, ⋅, x −1, , ⎧( x 2 − x ( x + 1) ( x 2 + x ) ⎫, ×, × 2, ⎨, ⎬, ⎩ ( x − 1) ( x + 1) ( x + x ) ⎭, ( x 4 − x)( x + 1), x( x 3 − 1)( x + 1), = lim, =, lim, x → 1 ( x − 1)( x 2 +, x ) x → 1 ( x − 1)( x 2 + x ), , (x2 − x), = lim, ( x − 1) x → 1
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806, , Senior Secondary School Mathematics for Class 11, , ∴, , ⎛ y 4 − 1⎞, ⎛ y 4 − 14 ⎞, (1 + x) 4 − 1, = lim ⎜, ⎟ = lim ⎜, ⎟, x→ 0, y →1 ⎝ y − 1 ⎠, y →1 ⎝ y − 1 ⎠, x, lim, , ⎡, ⎛ x n − an ⎞, n −1 ⎤, ⎜, ⎟ = na ⎥, ⎢Q xlim, →a ⎝ x − a ⎠, ⎣, ⎦, , = 4 × 1( 4 − 1 ), = ( 4 × 13 ) = 4., Evaluate, , EXAMPLE 20, , SOLUTION, , lim, , x→ 0, , (1 + x) 6 − 1, ⋅, (1 + x)5 − 1, , Put (1 + x) = y, so that when x → 0 then y → 1., ∴, , (1 + x) 6 − 1, x → 0 (1 + x)5 − 1, lim, , ⎛ y 6 − 1⎞, = lim ⎜ 5 ⎟ =, y → 1 ⎝ y − 1⎠, , ⎛ y 6 − 1⎞, lim ⎜, ⎟, y →1 ⎝ y − 1⎠, ⎛ y − 1⎞, lim ⎜, ⎟, ⎝ y − 1⎠, 5, , y →1, , =, , =, , ⎛ y 6 − 16 ⎞, lim ⎜, ⎟, y →1 ⎝ y − 1 ⎠, ⎛y −1 ⎞, lim ⎜, ⎟, ⎝ y −1 ⎠, 5, , 5, , =, , 6 × 1( 6 − 1 ), 5 × 1(5 − 1 ), , y →1, , ⎡, ⎛ x n − an ⎞, n −1 ⎤, ⎜, ⎟ = na ⎥ ⋅, ⎢Q xlim, →a ⎝ x − a ⎠, ⎦, ⎣, , 6 × 15 6, =, 5 × 14 5, , EXERCISE 27A, Evaluate the following limits:, 1. lim (5 − x), , 2. lim ( 6x 2 − 4x + 3), , ⎛ x 2 + 9⎞, 3. lim ⎜, ⎟, x→ 3 ⎝ x + 3 ⎠, , ⎛ x 2 − 4x ⎞, 4. lim ⎜, ⎟, x→ 3 ⎝ x − 2 ⎠, , ⎛ x 2 − 25 ⎞, 5. lim ⎜, ⎟, x→5 ⎝ x − 5 ⎠, , ⎛ x 3 − 1⎞, 6. lim ⎜, ⎟, x→1 ⎝ x − 1 ⎠, , ⎛ x 3 + 8⎞, 7. lim ⎜, ⎟, x → −2 ⎝ x + 2 ⎠, , ⎛ x 4 − 81⎞, 8. lim ⎜, ⎟, x→ 3 ⎝ x − 3 ⎠, , ⎛ x 2 − 4x + 3 ⎞, 9. lim ⎜ 2, ⎟, x → 3 ⎝ x − 2x − 3 ⎠, , ⎛ 4x 2 − 1⎞, 10. lim ⎜, ⎟, 1, x → ⎝ 2x − 1 ⎠, , x→ 2, , x→1, , 2, , ⎛ x − 64⎞, 11. lim ⎜ 2, ⎟, x → 4 ⎝ x − 16 ⎠, , ⎛ x5 − 32⎞, 12. lim ⎜ 3, ⎟, x→ 2 ⎝ x − 8 ⎠, , ⎛ x5/2 − a5/2 ⎞, 13. lim ⎜, ⎟, x→ a ⎝, x−a ⎠, , ⎧( x + 2)5/3 − ( a + 2)5/3 ⎫, 14. lim ⎨, ⎬, x→ a, x−a, ⎭, ⎩, , 3
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Limits, , 807, , ⎛ x n − 1⎞, 15. lim ⎜, ⎟, x→1 ⎝ x − 1 ⎠, , ⎛ x − a⎞, 16. lim ⎜, ⎟⋅, x→ a ⎝ x − a ⎠, , ⎛ x + h − x⎞, 17. lim ⎜, ⎟, h→0 ⎝, h, ⎠, , 18. lim, , ⎛ 1 + x − 1⎞, 19. lim ⎜, ⎟, x→ 0 ⎝, x, ⎠, , ⎛ 2 − x − 2 + x⎞, 20. lim ⎜, ⎟, x→ 0 ⎝, x, ⎠, , ⎛ 1 + x + x 2 − 1⎞, ⎟, 21. lim ⎜, ⎟, x→ 0 ⎜, x, ⎠, ⎝, , ⎛ 3 − x − 1⎞, 22. lim ⎜, ⎟, x→ 0 ⎝, 2−x ⎠, , ⎛, ⎞, 2x, 23. lim ⎜, ⎟, x→ 0 ⎝ a + x − a − x ⎠, , ⎛ 3 + x − 5 − x⎞, 24. lim ⎜, ⎟, x→1 ⎝, x2 − 1, ⎠, , ⎛, ⎞, x2 − 4, 25. lim ⎜, ⎟, x→ 2 ⎝ x + 2 −, 3x − 2⎠, , ⎛ 3 − 5 + x⎞, 26. lim ⎜, ⎟, x→ 4 ⎝ 1 − 5 − x ⎠, , ⎛ a + x − a⎞, 27. lim ⎜, ⎟, x→ 0 ⎝ x, a( a + x) ⎠, , ⎛ 1 + x2 − 1 + x ⎞, ⎟, 28. lim ⎜, ⎟, 3, x→ 0 ⎜, ⎝ 1 + x − 1 + x⎠, , ⎛ x4 − 3x2 + 2 ⎞, 29. lim ⎜ 3, ⎟, x → 1 ⎝ x − 5 x 2 + 3 x + 1⎠, , ⎛ 3 x + 3 3 − x − 12⎞, 30. lim ⎜ 3 − x, ⎟, x→ 2 ⎝ 3, − 3 x/2 ⎠, , ⎛ e4 x − 1⎞, 31. lim ⎜, ⎟, x→ 0 ⎝, x ⎠, , ⎛ e2 + x − e2 ⎞, 32. lim ⎜, ⎟, x→ 0 ⎝, x, ⎠, , ⎛ ex − e4 ⎞, 33. lim ⎜, ⎟, x→ 4 ⎝ x − 4 ⎠, , ⎛ e3 x − e2 x ⎞, 34. lim ⎜, ⎟, x→ 0 ⎝, x ⎠, , ⎛ ex − x − 1⎞, 35. lim ⎜, ⎟, x→ 0 ⎝, x, ⎠, , ⎛ ebx − eax ⎞, 36. lim ⎜, ⎟, 0< a < b, x→ 0 ⎝, x ⎠, , ⎛ a x − bx ⎞, 37. lim ⎜, ⎟, x→ 0 ⎝, x ⎠, , ⎛ ax − a−x ⎞, 38. lim ⎜, ⎟, x→ 0 ⎝, x ⎠, , ⎛ 2x − 1⎞, 39. lim ⎜, ⎟, x→ 0 ⎝, x ⎠, , ⎛ 3 2 + x − 9⎞, 40. lim ⎜, ⎟, x→ 0 ⎝, x ⎠, , h→ 0, , 1 ⎧ 1, 1 ⎫, −, ⎬, ⎨, h ⎩ x+h, x⎭, , ANSWERS (EXERCISE 27B), , 1. 3, 1, 9., 2, 1, 16., 2 a, , 2. 5, , 3. 3, , 10. 2, , 11. 6, , 17., , 1, −1, 18. 3/2, 2x, 2 x, , 4. −3, 20, 12., 3, 1, 19., 2, , 5. 10, 6. 3, 7. 12, 5 3/2, 5, 14. ( a + 2) 2/3, 13. a, 2, 3, −1, 1, 1, 20., 21., 22., 2, 2, ( 3 + 1), , 8. 108, 15. n, 23. 2 a
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808, , 24., , Senior Secondary School Mathematics for Class 11, , 1, −1, 1, 25. −8 26., 27., 4, 3, 2a 3/2, , 34. 1, , 35. 0, , 36. ( b − a), , 1, −4, 30., 31. 4, 2, 3, , 28. 1 29., , 37. (log a − log b), , 32. e2, , 33. e4, , 38. 2log a 39. log 2 40. 9 log 3, , HINTS TO SOME SELECTED QUESTIONS, ⎧( x − x + h ) ( x + x + h ) ⎫, 18. Given limit = lim ⎨, ×, ⎬, h → 0 ⎪ h( x + h ) x, ( x + x + h ) ⎪⎭, ⎩, −h, = lim, h → 0 h( x + h ) x ( x + x + h ), = lim, , h→0, , −1, ( x + h) x ( x +, , x + h), , =, , −1, 2 x 3/2, , ⋅, , ⎧( 3 − 5 + x ) ( 1 + 5 − x ) ( 3 + 5 + x ) ⎫, 26. Given limit = lim ⎨, ×, ×, ⎬, x→ 4 ⎪ (1 − 5 − x ), ( 1 + 5 − x ) ( 3 + 5 + x ) ⎪⎭, ⎩, = lim, , x→ 4, , = lim, , x→ 4, , {9 − (5 + x )} ( 1 + 5 − x ), {1 − (5 − x )} ( 3 + 5 + x ), (1 + 5 − x), −( 3 + 5 + x ), , =, , (1 +, , = lim, , x→ 4, , 1), , −( 3 +, , =, , 9), , ( 4 − x )( 1 + 5 − x ), −( 4 − x )( 3 + 5 + x ), , 2, 1, =− ⋅, −6, 3, , ⎧( 1 + x 2 − 1 + x ) ( 1 + x 3 +, 28. Given limit = lim ⎪⎨, ×, 3, 3, x→ 0, ⎪⎩ ( 1 + x − 1 + x ) ( 1 + x +, = lim, , x→ 0, , = lim, , x→ 0, , = lim, , x→ 0, , 29. Given limit = lim, , x→1, , {( 1 + x 2 ) − ( 1 + x )} × ( 1 + x 3 +, 3, , 2, , {( 1 + x ) − ( 1 + x )} × ( 1 + x +, x( x − 1)( 1 + x 3 +, , 1+ x), , ( x + 1)( 1 + x 2 +, , 1+ x), , ( x − 1)( x + 1)( x 2 − 2 ), ( x − 1)( x 2 − 4 x − 1), , ( 1 + x2 +, ( 1 + x2 +, , 1+ x), 1+ x), , 1+ x), =, , = lim, , ⎫, ⎧ x 33, ⎪⎪ 3 + x − 12 ⎪⎪, 3, 30. Given limit = lim ⎨, ⎬ = tlim, 3, x→ 2, →3, ⎪ 3 − 3 x/2 ⎪, ⎪⎭, ⎪⎩ 3 x, , 1+ x), , ×, , 1+ x), , x( x − 1)( x + 1)( 1 + x 2 +, ( 1 + x3 +, , 1+ x), , x→1, , ( 1+, , 1), , 1⋅ ( 1 +, , 1), , = 1., , ( x + 1)( x 2 − 2 ), ( x 2 − 4 x − 1), , ⎛ t 2 + 27 − 12 ⎞, ⎜, ⎟, t2, ⎜, ⎟, 27, ⎜, ⎟, −t, ⎝, ⎠, t2, , =, , [put 3 x/ 2 = t ], , ⎛ t 4 − 12t 2 + 27 ⎞, (t 2 − 3 )(t + 3 )(t − 3), = lim ⎜, ⎟ = lim, 3, 2, t→ 3 ⎝, 27 − t, ⎠ t → 3 ( 3 − t )( 9 + 3t + t ), = lim, , t→ 3, , (t 2 − 3 )(t + 3 ), −( 9 + 3t + t 2 ), , =, , 1, ⋅, 2, , ( 9 − 3 )( 3 + 3 ) −36 −4, =, =, ⋅, −( 9 + 9 + 9 ), 27, 3, , 1 + x ) ⎫⎪, ⎬, 1 + x ) ⎪⎭
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Limits, , 809, , 33. Put ( x − 4 ) = y , so that when x → 4 then y → 0., ∴, , ⎛ ey+ 4 − e4 ⎞, ⎛ e y − 1⎞, 4, 4, 4, given limit = lim ⎜, ⎟ = e ⋅ lim ⎜, ⎟ = ( e × 1) = e ., y→0 ⎝, y, →, 0, y, ⎠, ⎝ y ⎠, , 34. Given limit = lim, , x→ 0, , = lim, , x→ 0, , ( e 3 x − 1) − ( e 2 x − 1), x, ⎛ e 2 x − 1⎞, ( e 3 x − 1), − lim ⎜, ⎟, x→ 0 ⎝, x, x ⎠, , ⎧ ⎛ e 3 x − 1⎞ ⎫, = lim ⎨ 3 ⎜, ⎟ ⎬ − lim, 3x→ 0 ⎪ ⎝, 3 x ⎠ ⎪⎭ 2 x → 0, ⎩, , ⎧ ⎛ e 2 x − 1⎞ ⎫, ⎟⎬, ⎨2 ⎜, ⎪⎩ ⎝ 2 x ⎠ ⎪⎭, , = ( 3 × 1) − ( 2 × 1) = ( 3 − 2 ) = 1., ⎧ ⎛ e x − 1⎞, ⎫, ⎛ e x − 1⎞, 35. Given limit = lim ⎨ ⎜, ⎟ − 1⎬ = lim ⎜, ⎟ − 1 = ( 1 − 1) = 0 ., x → 0 ⎪⎝, x, →, 0, x ⎠, ⎝ x ⎠, ⎪⎭, ⎩, ⎧ ⎛ e bx − 1 ⎞ ⎫, ⎧ ⎛ e ax − 1 ⎞ ⎫, ⎧ ( e bx − 1) − ( e ax − 1) ⎫, 36. Given limit = lim ⎨, = lim ⎨ ⎜, ⎟ ⋅ b ⎬ − lim ⎨ ⎜, ⎟ ⋅ a⎬, ⎬, x→ 0, bx → 0 ⎪ ⎝ bx ⎠, ax → 0 ⎪ ⎝ ax ⎠, x, ⎩, ⎭, ⎩, ⎭⎪, ⎩, ⎭⎪, = ( 1 × b ) − ( 1 × a) = ( b − a)., ⎧ ( ax − 1) − ( b x − 1) ⎫, ⎛ ax − 1⎞, ⎛ b x − 1⎞, 37. Given limit = lim ⎨, = lim ⎜, ⎟ − lim ⎜, ⎟ = (log a − log b )., ⎬, x→ 0, x→ 0 ⎝, x, x ⎠ x→ 0 ⎝ x ⎠, ⎩, ⎭, ⎧ ⎛ a2 x − 1⎞ 2 ⎫ ⎡, 2⎤, 38. Given limit = lim ⎨ ⎜, = 2 log a., ⎟ × x ⎬ = (log a) ×, ⎢⎣, x → 0 ⎪⎝ 2x ⎠, 1 ⎥⎦, a, ⎪, ⎩, ⎭, , TRIGONOMETRIC LIMITS, For finding the limits of trigonometric functions, we use trigonometric, transformations and simplify., The following theorems are quite important., THEOREM 1, , PROOF, , (i) lim sin θ = 0, θ→0, , (ii) lim cos θ = 1, θ→0, , Consider a triangle ABC in which ∠C is a right angle and ∠ABC = θ, radians., CA, BC, Then,, sin θ =, and cos θ =, ⋅, BA, BA, A, , If we keep BC fixed and go on decreasing θ, we find, that A keeps on coming nearer and nearer to C., ∴ when θ → 0 then A → C., ∴ θ → 0 ⇒ CA → 0 and BA → BC, , B, , C
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Limits, , ⎛ sin ax ⎞, (iii) lim ⎜, ⎟ = xlim, x → 0 ⎝ sin bx ⎠, →0, , 811, , sin ax ⎞, ax ⋅ ⎛⎜, ⎟, ⎝ ax ⎠, sin bx ⎞, bx ⋅ ⎛⎜, ⎟, ⎝ bx ⎠, , ⎛ sin ax ⎞, ⎜, ⎟, ⎝ ax ⎠, a, = ⋅ lim, b x → 0 ⎛ sin bx ⎞, ⎜, ⎟, ⎝ bx ⎠, sin ax, lim, a ax → 0 ax, = ⋅, [Q x → 0 ⇒ ax → 0 and bx → 0], b lim sin bx, bx → 0, bx, ⎡Q lim sin θ = 1⎤ ⋅, ⎢⎣ θ → 0 θ, ⎥⎦, , a 1, a, = ⎛⎜ × ⎞⎟ =, ⎝ b 1⎠ b, EXAMPLE 2, , SOLUTION, , Evaluate, , sin 5 x, ⋅, tan 3 x, , lim, , x→ 0, , sin 5 x, lim, = lim, x → 0 tan 3 x, x→ 0, , 5, = ⋅, 3, , EXAMPLE 3, , SOLUTION, , Evaluate, , lim, , x→ 0, , lim, , x→ 0, , ⎛ sin 5 x ⎞, 5x ⋅ ⎜, ⎟, ⎝ 5x ⎠, tan 3 x ⎞, 3 x ⋅ ⎜⎛, ⎟, ⎝ 3x ⎠, , sin 5 x ⎞, lim ⎛⎜, ⎟, ⎝ 5x ⎠, ⎛ 5 1⎞ 5, =⎜ × ⎟ = ⋅, 3, tan, x, ⎞ ⎝ 3 1⎠ 3, lim ⎛⎜, ⎟, 3x→ 0 ⎝, 3x ⎠, 5x→ 0, , (1 − cos x), ⋅, x2, , [CBSE 2000], , (1 − cos x), 2 sin 2 ( x/2), lim, =, 2, x→ 0, x2, ⎛ x⎞ × 4, ⎜ ⎟, ⎝ 2⎠, 2, , =, , EXAMPLE 4, , ⎧ sin ( x/2) ⎫ ⎛ 1, 1, 1, 2⎞, lim ⎨, ⎬ = ⎜ × 1 ⎟⎠ = ⋅, 2 x → 0 ⎩ ( x/2) ⎭ ⎝ 2, 2, , ⎛ 1 − cos 4x ⎞, Evaluate (i) lim ⎜, ⎟, x → 0 ⎝ 1 − cos 5 x ⎠, ⎛ 1 − cos mx ⎞, (ii) lim ⎜, ⎟⋅, x → 0 ⎝ 1 − cos nx ⎠, , [CBSE 2002C]
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Limits, , EXAMPLE 6, , SOLUTION, , Evaluate, , ⎛ 1 + cos x ⎞, lim ⎜, ⎟⋅, ⎝ tan 2 x ⎠, , x→π, , ⎛ 1 + cos x ⎞, cos2 x(1 + cos x), lim ⎜, ⎟ = lim, 2, x → π ⎝ tan x ⎠, x→ π, sin 2 x, = lim, , cos2 x(1 + cos x), (1 − cos2 x), , = lim, , cos2 x, 1, =, (1 − cos x) 2, , x→ π, , x→ π, , EXAMPLE 7, , SOLUTION, , Evaluate, , lim, , x→ 0, , lim, , x→ 0, , x tan x, x sin x, = lim, (1 − cos x) x → 0 cos x(1 − cos x), = lim, , x sin 2 x, sin x cos x(1 − cos x), , = lim, , x (1 − cos2 x), x (1 + cos x), = lim, sin x cos x (1 − cos x) x → 0 sin x cos x, , = lim, , (1 + cos x) ⎛, x, 2, ⋅ lim, = ⎜1 × ⎞⎟ = 2., ⎝, cos x, 1⎠, sin x x → 0, , x→ 0, , x→ 0, , Evaluate:, (i) lim ( sec x − tan x), x→, , SOLUTION, , [putting x = π]., , x tan x, ⋅, (1 − cos x), , x→ 0, , EXAMPLE 8, , 813, , (ii) lim, , π, 2, , x→ 0, , ( cosec x − cot x), x, , ⎛ 1, sin x ⎞, (i) lim ( sec x − tan x) = lim ⎜, −, ⎟, π, π ⎝ cos x, cos x ⎠, x→, x→, 2, , 2, , = lim, , π, x→, 2, , (1 − sin x), (1 − sin x) cos x, = lim, π, cos x, cos2 x, x→, 2, , = lim, , (1 − sin x) cos x, (1 − sin 2 x), , = lim, , cos x, =0, (1 + sin x), , π, x→, 2, , π, x→, 2, , ⎡putting x = π ⎤ ⋅, ⎢⎣, 2 ⎥⎦
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814, , Senior Secondary School Mathematics for Class 11, , (ii) lim, , x→ 0, , ( cosec x − cot x), = lim, x→ 0, x, = lim, , x→ 0, , = lim, , x→ 0, , = lim, , x→ 0, , Evaluate:, , EXAMPLE 9, , SOLUTION, , ⎛ 1, cos x ⎞, −, ⎜, ⎟, ⎝ sin x sin x ⎠, x, (1 − cos x), (1 − cos x) sin x, = lim, x, →, 0, x sin x, x sin 2 x, (1 − cos x) sin x, sin x, = lim, x → 0 x(1 + cos x), x (1 − cos2 x), 1, 1, sin x, 1, ⋅ lim, = ⎛⎜1 × ⎞⎟ = ⋅, ⎝, ⎠, x, →, 0, 2, 2, x, (1 + cos x), , ⎛ tan x − sin x ⎞, (i) lim ⎜, ⎟, x→ 0 ⎝, sin 3 x, ⎠, , (ii) lim, , x→ 0, , (tan x − sin x), x3, , ⎛ sin x − sin x cos x ⎞, (tan x − sin x), = lim ⎜, ⎟, 3, x, →, 0, sin x, cos x sin 3 x, ⎝, ⎠, , (i) lim, , x→ 0, , sin x(1 − cos x), (1 − cos x), = lim, 3, x, →, 0, cos x sin x, cos x sin 2 x, (1 − cos x), = lim, x → 0 cos x (1 − cos2 x), 1, 1, = lim, =, [putting x = 0]., x → 0 cos x (1 + cos x), 2, = lim, , x→ 0, , (ii) lim, , x→ 0, , ⎛ sin x − sin x cos x ⎞, (tan x − sin x), = lim ⎜, ⎟, x→ 0 ⎝, x3, x 3 cos x, ⎠, = lim, , x→ 0, , sin x (1 − cos x), = lim, x→ 0, x 3 cos x, , ⎧ sin x 2 sin 2 ( x/2), 1 ⎫, ⋅, ⋅, ⎬, ⎨, 2, cos x ⎭, x, ⎩ x, , ⎧ sin x 2 sin 2 ( x/2) 1, 1 ⎫, = lim ⎨, ⋅, ⋅ ⋅, ⎬, 2, x→ 0, 4 cos x ⎭, ( x/2), ⎩ x, 2, ⎧ sin ( x/2) ⎫, sin x, 1, 1, = ⋅ lim, ⋅ lim ⎨, ⎬ ⋅ xlim, x→ 0, → 0 cos x, (, /, ), 2 x→ 0 x, x, 2, ⎭, ⎩, 1, 1, ⎛, ⎞, 2, = ⎜ × 1 × 1 × 1⎟ = ⋅, ⎝2, ⎠ 2, ⎛ x 3 cot x ⎞, lim ⎜, ⎟⋅, x → 0 ⎝ 1 − cos x ⎠, ⎛ x 3 cot x ⎞, lim ⎜, ⎟, x → 0 ⎝ 1 − cos x ⎠, ⎧ x 3 cot x, (1 + cos x) ⎫, = lim ⎨, ×, ⎬, x → 0 (1 − cos x), (1 + cos x) ⎭, ⎩, , EXAMPLE 10, , SOLUTION, , Evaluate, , = lim, , x→ 0, , x 3 cos x (1 + cos x), x 3 cos x (1 + cos x), =, lim, x→ 0, sin x (1 − cos2 x), sin 3 x
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816, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 14, , SOLUTION, , Evaluate, , lim, , (1 − cos x, , lim, , x→ 0, , (1 − cos x, , x→ 0, , x, , x, , cos 2x ), 2, , ⋅, , [CBSE 2004C, ‘05], , cos 2x ), 2, , ⎧(1 − cos x cos 2x ) (1 + cos x cos 2x ) ⎫, = lim ⎨, ×, ⎬, x→ 0, x2, (1 + cos x cos 2x ) ⎭, ⎩, ⎫, ⎧(1 − cos2 x cos 2x), 1, = lim ⎨, ×, ⎬, 2, x→ 0, x, (, +, cos, cos, ), 1, x, 2, x, ⎭, ⎩, (1 − cos2 x cos 2x), 1, = lim, × lim, x→ 0, x → 0 (1 + cos x, x2, cos 2x ), ⎧1 − cos2 x (1 − 2 sin 2 x) ⎫, 1, = lim ⎨, ⎬×, 2, x→ 0, x, +, (, 1, 1 1), ⎭, ⎩, 2, 2, 2, (1 − cos x + 2 sin x cos x) 1, = lim, ×, x→ 0, 2, x2, (sin 2 x + 2 sin 2 x cos2 x) 1, sin 2 x(1 + 2 cos2 x), 1, = × lim, =, ×, lim, 2 x→ 0, 2 x→ 0, x2, x2, 2, , sin x ⎞, 1, × lim ⎛⎜, ⎟ × lim (1 + 2 cos2 x), x→ 0, 2 x→ 0 ⎝ x ⎠, 1, 3, = × 12 × (1 + 2 × 12 ) = ⋅, 2, 2, =, , EXAMPLE 15, , SOLUTION, , Evaluate, , lim, , π, x→, 4, , (sin x − cos x), ⋅, ⎛x − π⎞, ⎜, ⎟, ⎝, 4⎠, , [CBSE 2000, ’03], , π, π, Put ⎛⎜ x − ⎞⎟ = y so that when x → then y → 0., ⎝, ⎠, 4, 4, (sin x − cos x), ∴ lim, π, ⎛x − π⎞, x→, ⎜, ⎟, 4, ⎝, 4⎠, ⎧, ⎨sin, = lim ⎩, y→0, , ⎛ π + y⎞ − cos ⎛ π + y⎞ ⎫, ⎜, ⎟, ⎜, ⎟⎬, ⎝4, ⎠, ⎝4, ⎠⎭, y, , [putting ( x −, , π, ) = y], 4, , π, π, π, π, ⎧⎛, ⎞⎫, ⎞ ⎛, ⎨⎜⎝ sin cos y + cos sin y⎟⎠ − ⎜⎝ cos cos y − sin sin y⎟⎠ ⎬, 4, 4, 4, 4, ⎭, ⎩, = lim, y→0, y, =, , ⎛ sin y ⎞, 2, × lim ⎜, ⎟ = ( 2 × 1) = 2 ., 2 y→0 ⎝ y ⎠
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Limits, , ⎛π, ⎞, 45. lim ⎜ − x⎟ tan x, π ⎝ 2, ⎠, x→, , 819, , 46. lim, , x→ 0, , ( 1 + 2x − 1 − 2x ), [CBSE 1998C], sin x, , 2, , ( a + h) 2 sin ( a + h) − a 2 sin a, ( e3 + x − sin x − e3 ), 48. lim, h→0, x→ 0, h, x, tan x, tan x, (e, (e, − 1), − 1), 50. lim, 49. lim, x→ 0, x→ 0, tan x, x, ax + x cos x, sin ax + bx, 51. lim, 52. lim, , where a , b , a + b ≠ 0, x→ 0, x → 0 ax + sin bx, b sin x, sin ( π − x), tan 2x, 53. lim, 54. lim, π, π, x→ π, π ( π − x), x→, 2 x−, 2, cos 2x − 1, 55. lim, 56. lim ( cosec x − cot x), x → 0 cos x − 1, x→ 0, 1 − cos 2mx, 1 − cos mx, 57. lim, 58. lim, x → 0 1 − cos 2nx, x → 0 1 − cos nx, 47. lim, , 59. lim, , sin 2mx, sin 2nx, , 60. lim, , 61. lim, , sec 4x − sec 2x, sec 3 x − sec x, , 62. lim, , x→ 0, , x→ 0, , 63. lim, , x→ 0, , 65. lim, , x→ 0, , 67. lim, , π, x→, 4, , x→ 0, , x→ 0, , 1 + sin x − 1 − sin x, x, , 66. lim, , cos x − cos a, cot x − cot a, , 68. lim, x→, , π, 2, , 69. lim, , cot 2 x − 3, cosec x − 2, , 70. lim, , 71. lim, , 1 − tan x, 1 − 2 sin x, , 72. lim, , π, x→, 6, , π, x→, 4, , sin 2 x, 2 − 3 cos x − sin x, ( 6x − π ) 2, , x→ a, , tan 3 x − tan x, π⎞, ⎛, cos ⎜ x + ⎟, ⎝, 4⎠, , 2 − 1 + cos x, , 64. lim, , π, x→, 6, , cos ax − cos bx, cos cx − 1, , sin 2x + sin 3 x, 2x + sin 3 x, , x→ π, , π, x→, 6, , 2 − 1 + sin x, 2 cos2 x, 2 + cos x − 1, ( π − x) 2, 2 sin 2 x + sin x − 1, 2 sin 2 x − 3 sin x + 1, , ANSWERS (EXERCISE 27B), , 1., , 2, 3, , 9. −3, , 2., , 5, 8, , 10. 1, , 3, 5, 1, 11., 2, 3., , 4., , α, β, , 12. −2, , 5., , 4, 7, , 13. 2, , 3, 4, 1, 14., 2, 6., , 7., , m, n, , 15. 1, , 8. 0, 16., , 1, 2
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820, , 17., 25., , Senior Secondary School Mathematics for Class 11, , 1, 3, m2, n2, , 18., , 1, 4, , 1, 6, 1, 27., 2, 19., , 26. 1, , 1, 2, , 28. 4, 1, 2, , 9, 2, 1, 29., 2, , 21., , 22., , 1, 2, , 30. −1, , 37. − sin a, , 23., , 2, 3, , 31. 4, , 35. −1, , 36., , 39. 2 a cos a, , 40. 2, , 41. 8, , 45. 1, , 47. ( 2a sin a + a 2 cos a), 48. ( e3 − 1), 1, 52. 1, 53., 54. 2, 55. 4, 56. 0, π, 3, 1, 1, 60. 1, 61., 62., 63. 1, 64., 2, 36, 4 2, 1, 67. – 4 68., 66. sin 3 a, 69. 4, 8, , 33. 2, , 50., 58., 65., 71., , 34. −2, , 20., , 46. 2, a+1, 1, 51., b, m, m2, 59. 2, n, n, ( a 2 − b2 ), c2, 2, 72. –3, , 4, 9, , 32. 2, , 38. cos a, , 43. 2 cos 2, , 42. 4, , 24., , 1, 15, 49. 1, m2, 57. 2, n, 44., , 70., , 1, 4, , HINTS TO SOME SELECTED QUESTIONS, sin 5 x ⎞, sin x ⎞, sin 3 x ⎞, ⎛, ⎛, 8. Given limit = ⎛⎜ lim, ⎟, ⎟ − ( 2 × 3 ) ⋅ ⎜ lim, ⎟ + 5 ⋅ ⎜ lim, ⎝ x → 0 5x ⎠, ⎝ x→ 0 x ⎠, ⎝ x → 0 3x ⎠, = [1 − ( 6 × 1) + (5 × 1)] = 0 ., 9. Given limit = lim, x→, , π, 6, , ( 2 sin x − 1)(sin x + 1), (sin x + 1), = lim, ( 2 sin x − 1)(sin x − 1) x → π (sin x − 1), 6, , ⎛ sin π + 1⎞, ⎜, ⎟, ⎝, ⎠, 6, = −3 ., =, ⎛ sin π − 1⎞, ⎜, ⎟, ⎝, ⎠, 6, sin 2 x ⎞, 2 × ⎛⎜ lim, ⎟+ 3, ( 2 × 1) + 3 5, ⎝ x → 0 2x ⎠, 10. Given limit =, = = 1., =, sin 3 x ⎞ 2 + ( 3 × 1) 5, ⎛, 2 + ⎜ 3 × lim, ⎟, x→ 0, ⎝, 3x ⎠, tan 2 x ⎞, 2 × ⎛⎜ lim, ⎟−1, ⎝ x → 0 2x ⎠, ( 2 × 1) − 1 1, 11. Given limit =, = ⋅, =, x, tan, ( 3 − 1), 2, ⎞, 3 − lim ⎛⎜, ⎟, x→ 0 ⎝ x ⎠, ⎛ x − 2 × tan 2 x ⎞, ⎜, ⎟, ⎝, 2 x ⎠ ⎧ 0 − ( 2 × 1) ⎫, 12. Given limit = lim, =⎨, ⎬ = −2., x→ 0, 1, ⎩, ⎭, ⎛ tan x ⎞, ⎜, ⎟, ⎝ x ⎠, ⎛ cos x + sin x ⎞, ⎜, ⎟, ⎝, x ⎠ ( 1 + 1), 13. Given limit = lim, = 2., =, x→ 0, ( 0 + 1), ⎛ x + tan x ⎞, ⎜, ⎟, ⎝, x ⎠
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Limits, 14. Given limit = lim, , x→ 0, , = lim, , x→ 0, , 15. Given limit = lim, , x→ 0, , 16. Given limit =, , sin x ( 1 − cos x ), , = lim, , cos x sin 3 x, ( 1 − cos x ), , x→ 0, , cos x ( 1 − cos 2 x ), , = lim, , x→ 0, , 821, , ( 1 − cos x ), cos x sin 2 x, 1, 1, = ⋅, cos x ( 1 + cos x ) 2, , x, = 1., sin x, , 1, 2x, 1, 1, × lim, = ⎛⎜ × 1⎞⎟ = ⋅, ⎠ 2, 2 2 x → 0 tan 2 x ⎝ 2, , 20. Given limit = lim, , x→ 0, , 21. Given limit = lim, , x→ 0, , 24. Given limit = lim, , x→ 0, , ( 1 − cos x ), ( 1 − cos 2 x ), , = lim, , x→ 0, , 3x ⎞, 2 sin 2 ⎛⎜, ⎟, ⎝ 2 ⎠, , =, , 2, , ⎛ 3x ⎞ × 4, ⎜, ⎟, ⎝ 2 ⎠, 9, 2 sin 2 2 x, 2, , 2 sin 3 x, , 1, 1, = ⋅, ( 1 + cos x ) 2, , 9, ⋅, 2, , = lim, , x→ 0, , sin 2 2 x, sin 2 3 x, , 2, , 2, , ⎛ sin 2 x ⎞ × 4 x 2, ⎜, ⎟, ⎝ 2x ⎠, , sin 2 x ⎞, lim ⎛⎜, ⎟, 4 2 x → 0 ⎝ 2x ⎠, = lim, = ⋅, 2, x → 0 sin 3 x 2, ⎛, ⎞ × 9 x 2 9 lim ⎛ sin 3 x ⎞, ⎜, ⎟, ⎜, ⎟, 3 x → 0 ⎝ 3x ⎠, ⎝ 3x ⎠, 4, = ⎛⎜ ×, ⎝9, , 26. Given limit = lim, , x→ 0, , 1⎞ 4, ⎟= ⋅, 1⎠ 9, 2 sin x( 1 − cos x ), x3, , ⎡, ⎤, ⎢, ⎥, 2, sin, sin, (, /, x, 2, x, 2, ), ⎛, ⎞, ⎥ = 1., = lim ⎢ 2 ⎜, ⎟⋅, 2, x→ 0 ⎢ ⎝ x ⎠, ⎛ x⎞ × 4 ⎥, ⎢, ⎥, ⎜ ⎟, ⎝ 2⎠, ⎣, ⎦, , tan ( x/ 2 ) 1 1, ( 1 − cos x ), 2 sin 2 ( x/ 2 ), = lim, = lim, × = ⋅, x→ 0, x → 0 2 x sin ( x/ 2 ) cos ( x/ 2 ), x→ 0, 2 2, x sin x, ⎛ x⎞, ⎜ ⎟, ⎝ 2⎠, (cos 2 x − 1), −( 1 − cos 2 x ), 30. Given limit = lim, = lim, x→ 0, x→ 0, x sin 2 x, x sin 2 x, , 29. Given limit = lim, , = lim, , x→ 0, , 35. Given limit = lim, , tan x, −2 sin 2 x, = − lim, = −1., x→ 0, 2 x sin x cos x, x, ( 3 sin x − 4 sin 3 x − 3 sin x ), ( π − x)3, , x→ π, , = lim, , x→ π, , −4 sin 3 x, ( π − x), , 3, , = lim, , θ→0, , 3, , −4 sin 3 ( π + θ), ( −θ) 3, , sin θ ⎞, = − 4 lim ⎛⎜, ⎟ = ( −4 × 1) = −4 ., θ→0 ⎝ θ ⎠, , , where ( x − π ) = θ
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822, , Senior Secondary School Mathematics for Class 11, , 36. Putting ( π − 2x ) = t, we have, 1 + cos ( π − t ), 1 − cos t, given limit = lim, = lim, t→ 0, t→ 0, t2, t2, = lim, , 2 sin 2 (t / 2 ), , t→ 0, , 37. Given limit = lim, , −2 sin ⎛⎜, ⎝, , x→ a, , t2, , 2, , =, , 1, 2, ⎧ sin (t / 2 ) ⎫, 2⎞, ⎛1, ⋅ lim ⎨, ⎬ =⎜ ×1 ⎟= ⋅, ⎝2, ⎠ 2, 4 t → 0 ⎩ (t / 2 ) ⎭, , x + a⎞, ⎛ x − a⎞, ⎟ sin ⎜, ⎟, ⎝ 2 ⎠, 2 ⎠, = − lim, x→ a, ( x − a), , = − sin a × 1 = − sin a., x + a⎞, ⎛ x − a⎞, 2 cos ⎛⎜, ⎟ sin ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, 39. Given limit = lim, ×( x +, x→ a, ( x − a), , x − a⎞ ⎫, ⎧, sin ⎛⎜, ⎟, ⎪⎪, x + a⎞, ⎝ 2 ⎠ ⎪⎪, ⎛, ⎟⋅, ⎨ sin ⎜, ⎬, ⎝ 2 ⎠, ⎛ x − a⎞ ⎪, ⎪, ⎜, ⎟, ⎝ 2 ⎠ ⎪⎭, ⎪⎩, , a), , x − a⎞, ⎧, sin ⎛⎜, ⎟, ⎪⎪, ⎝ 2 ⎠, x + a⎞, ⎛, ×( x +, = lim ⎨ cos ⎜, ⎟⋅, x→ a, ⎝ 2 ⎠, ⎛ x − a⎞, ⎪, ⎜, ⎟, ⎝ 2 ⎠, ⎪⎩, , ⎫, ⎪⎪, a )⎬, ⎪, ⎪⎭, , = (cos a × 1 × 2 a ) = 2 a cos a., 41. Given limit = lim, , 2 sin 4 x sin x, , x2, 2 sin 4 x cos x, 42. Given limit = lim, x → 0 2 cos 5 x sin x, x→ 0, , sin 4 x sin x, = lim ⎛⎜, ×, × 8 ⎞⎟ = ( 8 × 1 × 1) = 8., x → 0 ⎝ 4x, ⎠, x, , ⎧ ⎛ sin 4 x × 4 ⎞ ⋅ cos x ⎫, ⎟, ⎪⎪ ( 1 × 4 ) × 1, ⎪⎪ ⎜⎝ 4 x, ⎠, = 4., = lim ⎨, ⎬=, x→ 0, x, sin, 1× 1, ⎛, ⎞, ⎪ ⎜, ⎟ ⋅ cos 5 x ⎪, ⎪⎭, ⎪⎩ ⎝ x ⎠, ( 2 + x + 2 − x) ⎫, ⎧( 2 + x − 2 + x) ⎫, 2 cos ⎧⎨, ⎬ sin ⎨, ⎬, 2, 2, ⎩, ⎭, ⎩, ⎭, 43. Given limit = lim, x→ 0, x, sin x, = ( 2 cos 2 ) ⋅ lim, = ( 2 cos 2 ) × 1 = ( 2 cos 2 )., x→ 0, x, 44. Given limit = lim, , x→ 0, , sin x ⋅ sin x, 2 sin 2 x, = lim, 2 sin 5 x sin 3 x x → 0 sin 5 x ⋅ sin 3x, , sin x ⎞, lim ⎛⎜, ⎟ ⋅ lim, ⎝ x ⎠ x→ 0, =, sin 5 x, × 5 ⎞⎟ × lim, lim ⎛⎜, x → 0 ⎝ 5x, ⎠ x→ 0, x→ 0, , =, , 1× 1, 1, =, ⋅, ( 1 × 5 ) × ( 1 × 3 ) 15, , ⎛ sin x ⎞, ⎜, ⎟, ⎝ x ⎠, ⎛ sin 3 x ×, ⎜, ⎝ 3x, , 3 ⎞⎟, ⎠
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Limits, , 45., , 823, , π, π, Put ⎛⎜ − x ⎞⎟ = y so that when x → then y → 0., ⎝2, ⎠, 2, ∴, , y, π, given limit = lim y tan ⎛⎜ − y ⎞⎟ = lim y cot y = lim, = 1., y→0, y → 0 tan y, ⎝2, ⎠ y→0, , 46. Given limit = lim, , x→ 0, , = lim, , x→ 0, , ( 1 + 2 x − 1 − 2 x )( 1 + 2 x +, sin x ( 1 + 2 x +, , 1 − 2x ), , 1 − 2x ), , {( 1 + 2 x ) − ( 1 − 2 x )}, sin x ( 1 + 2 x +, , 1 − 2x ), , x, 1, = 4 ⋅ lim, ⋅ lim, x → 0 sin x x → 0 ( 1 + 2 x +, , 1 − 2x ), , 1, = ⎛⎜ 4 × 1 × ⎞⎟ = 2 ., ⎝, 2⎠, ( a2 + h 2 + 2 ah ) sin ( a + h ) − a2 sin a, h, sin ( a + h ) − sin a, 2, = a ⋅ lim, + lim h sin ( a + h ) + lim 2 a sin ( a + h ), h→0, h→ 0, h→0, h, , 47. Given limit = lim, , h→0, , h, h, cos ⎛⎜ a + ⎞⎟ ⋅ sin ⎛⎜ ⎞⎟, ⎝, ⎝ 2⎠, 2⎠, = a ⋅ lim, + 0 + 2 a sin a, h→0, ( h/ 2 ), 2, , sin ( h/ 2 ), h, = a2 ⋅ lim cos ⎛⎜ a + ⎞⎟ ⋅ lim, + 2 a sin a, h→0, ⎝, 2 ⎠ h → 0 ( h/ 2 ), = ( a2 × cos a × 1) + 2 a sin a = ( a2 cos a + 2 a sin a)., ⎛ e x − 1⎞, sin x, 48. Given limit = e 3 ⋅ lim ⎜, = ( e 3 × 1 − 1) = ( e 3 − 1)., ⎟ − lim, x→ 0 ⎝, x ⎠ x→ 0 x, 50. Given limit = lim, , x→ 0, , ⎧ ( e tan x − 1) tan x ⎫, ( e tan x − 1), = lim ⎨, ×, ⎬, x→ 0, x, x ⎭, ⎩ tan x, =, , lim, , tan x → 0, , ax + x cos x, a + cos x, 51. lim, = lim, x→ 0, x → 0 ⎛ sin x ⎞, b sin x, b⎜, ⎟, ⎝ x ⎠, =, , tan x, ( e tan x − 1), × lim, = ( 1 × 1) = 1., x→ 0, tan x, x, [dividing num. and denom. by x], , ( a + 1) ( a + 1), =, ⋅, ( b × 1), b, , ⎧ sin ax + b ⎫, sin ax + bx, ⎪, ⎪, 52. lim, [dividing num. and denom. by x], = lim ⎨ x, sin bx ⎬, x → 0 ax + sin bx, x→ 0, ⎪a +, ⎪, ⎩, x ⎭, ⎧ a⎛ sin ax ⎞ + b ⎫, ⎟, ⎨ ⎜, ⎬, ⎝ ax ⎠, ⎭ = ( a × 1) + b = a + b = 1., = lim ⎩, x→ 0 ⎧, sin bx ⎞ ⎫ a + ( b × 1) a + b, ⎛, ⎟⎬, ⎨a + b ⎜, ⎝ bx ⎠ ⎭, ⎩
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Limits, , 825, , sin mx sin mx ⎞, sin mx, sin mx, lim ⎛⎜, ⋅, ⋅ lim, ⎟, 2 lim, ⎝ mx, mx ⎠ m x → 0 mx x → 0 mx, =, ⋅, =, ⋅, n2 lim sin nx ⋅ lim sin nx, n2 lim ⎛ sin nx ⋅ sin nx ⎞, ⎜, ⎟, x→ 0, x → 0 ⎝ nx, nx x → 0 nx, nx ⎠, m2, , x→ 0, , ⎛ m 2 1 × 1⎞ m 2, =⎜, ×, ⋅, ⎟=, 2, 1 × 1 ⎠ n2, ⎝n, ⎧ sin 2 x + sin 3 x ⎫, sin 2 x + sin 3 x, ⎪, x ⎪, 60. lim, [dividing num. and denom. by x], = lim ⎨ x, sin 3x ⎬, x→ 0, x→ 0, 2 x + sin 3 x, ⎪ 2+, ⎪, ⎩, ⎭, x, sin 2 x, sin 3 x, + 3×, 2×, 3x, 2x, = lim, sin 3 x, x→ 0, 2 +3×, 3x, sin 2 x, sin 3 x, + 3 × lim, 2 × lim, x→ 0, x→ 0, 2x, 3 x = ( 2 × 1 + 3 × 1) = 5 = 1., =, sin 3 x, 5, ( 2 + 3 × 1), 2 + 3 × lim, x→ 0, 3x, 1 ⎫, ⎧ 1, −, ⎬, ⎨, cos 4 x cos 2 x ⎭, sec 4 x − sec 2 x, ⎩, 61. lim, = lim, = lim, x → 0 sec 3 x − sec x, x→ 0 ⎧, 1, 1 ⎫ x→ 0, −, ⎨, ⎬, ⎩ cos 3 x cos x ⎭, , ⎧ cos 2 x − cos 4 x ⎫, ⎪⎪ cos 4 x cos 2 x ⎪⎪, ⎨ cos x − cos 3 x ⎬, ⎪, ⎪, ⎪⎩ cos 3 x cos x ⎪⎭, , ⎧ cos 2 x − cos 4 x cos 3 x cos x ⎫, = lim ⎨, ×, ⎬, x → 0 ⎩ cos x − cos 3 x, cos 4 x cos 2 x ⎭, ⎧ 2 sin 3 x sin x cos x cos 3 x ⎫, = lim ⎨, ×, ⎬, x → 0 ⎩ 2 sin 2 x sin x, cos 2 x cos 4 x ⎭, ⎧ sin 3 x × 3 ⎫, cos x cos 3x, ⎪, ⎪, = lim ⎨ 3 x, ⎬ × lim, sin 2 x, x→ 0, x → 0 cos 2 x cos 4 x, ⎪, ⎪, ×2, ⎩ 2x, ⎭, ⎛ 1 × 3 1⎞ 3, =⎜, × ⎟= ⋅, ⎝ 1 × 2 1⎠ 2, 62. lim, , x→ 0, , 2 − 1 + cos x, sin 2 x, , = lim, , x→ 0, , 2 − 2 cos 2 ( x/ 2 ), ( 1 − cos 2 x ), , = lim, , x→ 0, , x, 2 ⎛⎜ 1 − cos ⎞⎟, ⎝, 2⎠, ( 1 − cos x )( 1 + cos x ), , x, 2 ⎛⎜ 1 − cos ⎞⎟, ⎝, 1, 2⎠, = lim, =, lim, x→ 0 ⎛, 2 x→ 0, 2 x⎞, ⎜ 2 sin ⎟ ( 1 + cos x ), ⎝, 2⎠, 1, 1, 1, =, lim, =, 2 x → 0 ⎛ 1 + cos x ⎞ ( 1 + cos x ), 2, ⎜, ⎟, ⎝, 2⎠, , ⎛ 1 − cos x ⎞, ⎜, ⎟, ⎝, 2⎠, ⎛ 1 − cos 2 x ⎞ ( 1 + cos x ), ⎜, ⎟, ⎝, 2⎠, 1, 1, ×, =, ⋅, ( 1 + 1)( 1 + 1) 4 2
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826, , Senior Secondary School Mathematics for Class 11, , 63. Given limit = lim, , ( 1 + sin x − 1 − sin x ), , x→ 0, , = lim, , x→ 0, , x, x( 1 + sin x +, , x→ 0, , x→, , π, 6, , ( 1 + sin x +, , 1 − sin x ), , ( 1 + sin x +, , 1 − sin x ), , ( 1 + sin x ) − ( 1 − sin x ), , = 2 × lim, , 64. Given limit = lim, , ×, , 1 − sin x ), , sin x, 1, × lim, = ⎛⎜ 2 × 1 ×, x → 0 ( 1 + sin x + 1 − sin x ), ⎝, x, , 1⎞, ⎟ = 1., 2⎠, , ⎛ 3, ⎞, 1, 2 − 2⎜, cos x + sin x ⎟, 2, ⎝ 2, ⎠, ( 6x − π )2, , π, π, ⎤, ⎡, 2 ⎢ 1 − ⎛⎜ cos cos x + sin sin x ⎞⎟ ⎥, ⎝, ⎠⎦, 6, 6, ⎣, = lim, π, ( 6x − π )2, x→, 6, , π ⎤, ⎡, x π, 2 ⎢ 1 − cos ⎛⎜ x − ⎞⎟ ⎥, 2 × 2 sin 2 ⎛⎜ − ⎞⎟, ⎝, ⎝ 2 12 ⎠, 6⎠⎦, ⎣, = lim, = lim, 2, x, π, π, ( 6x − π )2, x→, →, x π, 144 ⎛⎜ − ⎞⎟, 6, 2, 12, ⎝ 2 12 ⎠, 2, , ⎧ sin ⎛ x − π ⎞ ⎫, ⎜, ⎟, ⎪⎪, ⎝ 2 12 ⎠ ⎪⎪, 4, ⎛ 1 × 12 ⎞ = 1 ⋅, =, × lim ⎨, ⎟, ⎬ =⎜, x, π, ⎝ 36, ⎠ 36, x, π, 144, ⎛ − ⎞ ⎪, →, ⎪, ⎜, ⎟, 2, 12, ⎪⎩ ⎝ 2 12 ⎠ ⎪⎭, a + b⎞, ⎛ a − b⎞x, −2 sin ⎛⎜, ⎟, ⎟ x ⋅ sin ⎜, ⎝ 2 ⎠, ⎝ 2 ⎠, 65. Given limit = lim, x→ 0, cx, −2 sin 2 ⎛⎜ ⎞⎟, ⎝ 2⎠, , = lim, , x→ 0, , a + b⎞, a − b⎞, sin ⎛⎜, sin ⎛⎜, ⎟x, ⎟x, a + b⎞, ⎝ 2 ⎠, ⎝ 2 ⎠, a − b⎞, ⎛, ×⎜, × ⎛⎜, ⎟x, ⎟x⋅, ⎝ 2 ⎠, ⎝ 2 ⎠, ⎛ a + b⎞x, ⎛ a − b⎞x, ⎜, ⎟, ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, ⎧ sin ⎛ cx ⎞ ⎫, ⎜ ⎟⎪, ⎪⎪, ⎝ 2 ⎠⎪, ⎬, ⎨, cx, ⎪ ⎛⎜ ⎞⎟ ⎪, ⎩⎪ ⎝ 2 ⎠ ⎭⎪, , a + b⎞⎛ a − b⎞ 4 ⎫, = ⎧⎨ ⎛⎜, ⎟ × 2 ⎬ ⋅ lim, ⎟⎜, ⎩⎝ 2 ⎠ ⎝ 2 ⎠ c ⎭ x → 0, , 2, , a + b⎞, a − b⎞, sin ⎛⎜, sin ⎛⎜, ⎟x, ⎟x, ⎝ 2 ⎠, ⎝ 2 ⎠, ⋅ lim, ⎛ a + b ⎞ x x→ 0 ⎛ a − b ⎞ x, ⎜, ⎟, ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, , ⎧ ( a2 − b 2 ), ⎫ ( a2 − b 2 ), ⋅, =⎨, × 1 × 1⎬ =, 2, c, c2, ⎩, ⎭, cos x − cos a, 66. Given limit = lim, ⋅ sin x sin a, x → a cos x sin a − sin x cos a
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Limits, , 827, , x + a⎞, ⎛ x − a⎞, −2 sin ⎛⎜, ⎟, ⎟ sin ⎜, ⎝ 2 ⎠, ⎝ 2 ⎠, ⋅ sin x sin a, = lim, x→ a, sin ( a − x ), x + a⎞, ⎛ x − a⎞, −2 sin ⎛⎜, ⎟, ⎟ sin ⎜, ⎝ 2 ⎠, ⎝ 2 ⎠, ⋅ sin x sin a, = lim, x→ 0, x − a⎞, ⎛ x − a⎞, −2 sin ⎛⎜, ⎟ cos ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠, x + a⎞, sin ⎛⎜, ⎟, ⎝ 2 ⎠, = lim, ⋅ sin x sin a = sin 3 a., x→ 0, x − a⎞, ⎛, cos ⎜, ⎟, ⎝ 2 ⎠, tan x(tan x + 1)(tan x − 1), 67. Given limit = lim, π, π, x→, cos ⎛⎜ x + ⎞⎟, 4, ⎝, 4⎠, tan x(tan x + 1)(sin x − cos x ), = lim, π, π, x→, cos x ⋅ cos ⎛⎜ x + ⎞⎟, 4, ⎝, 4⎠, tan x(tan x + 1)(cos x − sin x ), = − lim, π, π, x→, cos x ⋅ cos ⎛⎜ x + ⎞⎟, 4, ⎝, 4⎠, 1, 1, tan x(tan x + 1)⎛⎜, cos x −, sin x ⎞⎟, ⎝ 2, ⎠, 2, = − 2 lim, π, π, x→, cos x ⋅ cos ⎛⎜ x + ⎞⎟, 4, ⎝, 4⎠, π, tan x(tan x + 1) ⋅ cos ⎛⎜ x + ⎞⎟, ⎝, tan x(tan x + 1), 4⎠, = − 2 lim, = − 2 lim, π, π, π⎞, cos x, ⎛, x→, x, →, cos x ⋅ cos ⎜ x + ⎟, 4, 4, ⎝, 4⎠, 1 × ( 1 + 1), =− 2⋅, = − 4., ⎛ 1 ⎞, ⎜, ⎟, ⎝ 2⎠, 68. Given limit = lim, , ( 2 − 1 + sin x ), , π, x→, 2, , = lim, , π, x→, 2, , = lim, , π, x→, 2, , =, , 2 cos 2 x, 2 − ( 1 + sin x ), 2 ( 1 − sin 2 x ), , ( 2+, , 1 + sin x ), , ( 2+, , 1 + sin x ), , 1, 1 + sin x ), , ( 2+, , ( 1 − sin x ), 2 ( 1 − sin x )( 1 + sin x ), , 1, 1, ⋅ lim, 2 x → π ( 1 + sin x )( 2 +, 2, , =, , ×, , ×, , 1, 1, 1, ×, = ⋅, 2 2× 2 2 8, , ×, , ( 2+, , 1 + sin x ), , 1, 1 + sin x )
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828, , Senior Secondary School Mathematics for Class 11, , 69. Given limit = lim, x→, , π, 6, , 70. Given limit = lim, , x→ π, , = lim, , x→ π, , cosec2 x − 4, = lim ( cosec x + 2 ) = ( 2 + 2 ) = 4., cosec x − 2 x → π, 6, , ( 2 + cos x − 1)( 2 + cos x + 1), ( π − x ) 2 ( 2 + cos x + 1), ( 1 + cos x ), ( π − x ) 2 ( 2 + cos x + 1), , = lim, , h→0, , 1 + cos ( π − h ), h 2 { 2 + cos ( π − h ) + 1}, [putting π − x = h], , = lim, , h→0, , ( 1 − cos h ), h 2 { 2 − cos h + 1}, , 2, , 2 sin ( h/ 2 ), , = lim, , h→0, , 2, , ⎛ h ⎞ × 4 × { 2 − cos h + 1}, ⎜ ⎟, ⎝ 2⎠, , 2, , =, , 1, 1, 1, ⎧ sin ( h/ 2 ) ⎫, lim ⎨, = ⎛⎜ × 12 ×, ⎬ ⋅, 2 h → 0 ⎩ ( h/ 2 ) ⎭ lim { 2 − cos h + 1} ⎝ 2, h→0, , cos x − sin x, 71. Given limit = lim, = lim, π, ⎧ 1 − sin x ⎫ x → π, x→, ⎬, 4 cos x ⋅ 2 ⎨, 4, ⎩ 2, ⎭, , 1⎞ 1, ⎟= ⋅, 2⎠ 4, , 1, 1, sin x ⎫⎬, 2 ⎧⎨, cos x −, 2, ⎩ 2, ⎭, π, ⎛, ⎞, 2 cos x ⎜ sin − sin x ⎟, ⎝, ⎠, 4, , ⎧ sin π cos x − cos π sin x ⎫, ⎨, ⎬, 4, 4, ⎭, = lim ⎩, π, ⎛ sin π − sin x ⎞, x→, cos, x, ⎟, ⎜, 4, ⎠, ⎝, 4, π, sin ⎛⎜, ⎝4, = lim, π, π, x→, cos x ⋅ 2 cos ⎛⎜ +, 4, ⎝8, = lim, x→, , π, 4, , − x ⎞⎟, ⎠, x⎞, ⎛ π x⎞, ⎟ sin ⎜ − ⎟, ⎝ 8 2⎠, 2⎠, , π x, π x, π x, cos ⎛⎜ − ⎞⎟, 2 sin ⎛⎜ − ⎞⎟ cos ⎛⎜ − ⎞⎟, ⎝ 8 2⎠, ⎝ 8 2⎠, ⎝ 8 2⎠, = lim, π, π x, π x, π x, cos x ⋅ 2 cos ⎛⎜ + ⎞⎟ sin ⎛⎜ − ⎞⎟ x → 4 cos x ⋅ cos ⎛⎜ + ⎞⎟, ⎝ 8 2⎠, ⎝ 8 2⎠, ⎝ 8 2⎠, , π π, cos ⎛⎜ − ⎞⎟, ⎝ 8 8⎠, cos 0, 1, =, =, = 2., =, 2, 2, π, π π⎞, ⎛, π, 1, ⎛, ⎞, cos ⋅ cos ⎜ + ⎟ ⎛ cos ⎞, ⎟, ⎜, ⎟, ⎝ 8 8 ⎠ ⎜⎝, 4, ⎝ 2⎠, 4⎠, 72. Given limit = lim, , π, x→, 6, , = lim, x→, , π, 6, , = lim, , π, x→, 6, , 2 sin 2 x + 2 sin x − sin x − 1, 2 sin 2 x − 2 sin x − sin x + 1, 2 sin x(sin x + 1) − (sin x + 1), 2 sin x(sin x − 1) − (sin x − 1), (sin x + 1)( 2 sin x − 1), (sin x + 1), = lim, (sin x − 1)( 2 sin x − 1) x → π (sin x − 1), 6
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Limits, , 829, , ⎛ sin π + 1⎞ ⎛ 1 + 1⎞, ⎜, ⎟ ⎜, ⎟, ⎝, ⎠ ⎝2, ⎠ 3, 6, =, =, = × ( −2 ) = −3 ., 1, π, ⎛ sin − 1⎞ ⎛ − 1⎞ 2, ⎜, ⎟ ⎜, ⎟, ⎝, ⎠ ⎝2, ⎠, 6, , LEFT-HAND LIMIT, Let f ( x) be a given function and let x approach to a from the left. Then, the limit of, f ( x), as x approaches to a from the left, is denoted by lim− f ( x)., x→ a, , Method for Finding lim f (x ), x → a−, , Replace x by ( a − h) and take the limit as h → 0., RIGHT-HAND LIMIT, Let f ( x) be a given function and let x approach to a from the right. Then, the limit of, f ( x), as x approaches to a from the right, is denoted by lim+ f ( x)., x→ a, , Method for Finding lim f (x ), x → a+, , Replace x by ( a + h) and take the limit as h → 0., LIMIT OF A FUNCTION, , We say that lim f ( x) exists only when lim f ( x) = lim f ( x), −, +, x→ a, , x→ a, , x→ a, , and this common value is known as lim f ( x)., x→ a, , SOLVED EXAMPLES, EXAMPLE 1, , ⎧2x + 3 , x ≤ 0, Let f ( x) = ⎨, ⎩ 3( x + 1), x > 0., (ii) lim f ( x), Find (i) lim f ( x), x→ 0, , SOLUTION, , x→1, , (i) We have, lim+ f ( x) = lim 3( 0 + h + 1) = lim 3( h + 1) = 3., x→ 0, , h→0, , h→0, , lim− f ( x) = lim 3( 0 − h + 1) = lim 3( − h + 1) = 3., , x→ 0, , ∴, , h→0, , h→0, , lim+ ( x) = lim− f ( x) = 3., , x→ 0, , x→ 0, , Hence, lim f ( x) = 3., x→ 0, , (ii) We have, lim f ( x) = lim f (1 + h) = lim 3(1 + h + 1) = lim 3( 2 + h) = 6., +, x→1, , h→0, , h→0, , h→0, , lim− f ( x) = lim f (1 − h) = lim 3(1 − h + 1) = lim 3( 2 − h) = 6., , x→1, , h→0, , h→0, , h→0
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830, , Senior Secondary School Mathematics for Class 11, , ∴, , lim f ( x) = lim− f ( x) = 6., , x → 1+, , x→1, , Hence, lim f ( x) = 6., x→1, , EXAMPLE 2, , ⎧x 2 − 1, x ≤ 1, Let f ( x) = ⎨, 2, ⎩− x − 1, x > 1., Find lim f ( x)., x→1, , We have, , SOLUTION, , lim f ( x) = lim f (1 + h) = lim {−(1 + h) 2 − 1} = lim ( −2 − h2 − 2h) = −2., , x → 1+, , h→0, , h→0, , h→0, , lim f ( x) = lim f (1 − h) = lim {(1 − h) − 1} = lim (1 + h2 − 2h − 1), 2, , x → 1−, , h→0, , h→0, , h→0, , = lim ( h2 − 2h) = 0., h→0, , ∴, , lim+ f ( x) ≠ lim− f ( x)., , x→1, , x→1, , Hence, lim f ( x) does not exist., x→1, , EXAMPLE 3, , ⎧|x|, ⎪ ,x≠0, Let f ( x) = ⎨ x, ⎪⎩0, x = 0., Find lim f ( x)., x→ 0, , We have, , SOLUTION, , |h|, h, = lim = 1., h h→0 h, |− h|, h, = −1., lim f ( x) = lim f ( 0 − h) = lim f ( − h) = lim, = lim, h→0, h→0, h → 0 −h, h → 0 −h, x → 0−, Thus, lim f ( x) ≠ lim f ( x)., lim f ( x) = lim f ( 0 + h) = lim f ( h) = lim, , x→ 0+, , x→ 0+, , h→0, , h→0, , h→0, , x → 0−, , Hence, lim f ( x) does not exist., x→ 0, , Caution, , Here note that, , EXAMPLE 4, , h, = 1, since h ≠ 0., h, , ⎧a + bx , x < 1, ⎪, Let f ( x) = ⎨4,, x =1, ⎪b − ax , x > 1., ⎩, If lim f ( x) = f (1), find the values of a and b., x→1, , SOLUTION, , Clearly, f (1) = 4., lim+ f ( x) = lim f (1 + h) = lim {b − a(1 + h)} = ( b − a)., x→1, , h→0, , h→0, , lim f ( x) = lim f (1 − h) = lim {a + b(1 − h)} = ( a + b)., , x → 1−, , h→0, , h→0
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Limits, , 831, , Now, lim f ( x) = f (1) = 4, x→1, , ⇒, , lim+ f ( x) = lim− f ( x) = f (1) = 4, , x→1, , x→1, , ⇒ ( b − a) = ( a + b) = 4, ⇒ b − a = 4 and b + a = 4 ⇒ a = 0, b = 4., Hence, a = 0 and b = 4., EXAMPLE 5, , ⎧|x|+1, x < 0, ⎪, Let f ( x) = ⎨0,, x=0, ⎪|x|−1, x > 0, ⎩, For what value(s) of a does lim f ( x) exists?, x→ a, , SOLUTION Case 1., , When a < 0., In this case, we have, lim f ( x) = lim f ( a + h) = lim {|a + h|+ 1} = |a|+ 1., +, h→0, , x→ a, , h→0, , lim f ( x) = lim f ( a − h) = lim {|a − h|+ 1} = |a|+ 1., , x → a−, , ∴, , h→0, , h→0, , lim+ f ( x) = lim− f ( x) = |a|+ 1., , x→ a, , x→ a, , So, in this case, lim f ( x) exists., x→ a, , Case 2., , When a < 0., In this case, we have, lim+ f ( x) = lim f ( a + h) = lim {|a + h|− 1} = |a|− 1., h→0, , x→ a, , h→0, , lim− f ( x) = lim f ( a − h) = lim {|a − h| − 1} = |a|− 1., h→0, , x→ a, , ∴, , h→0, , lim+ f ( x) = lim− f ( x) = |a|− 1., , x→ a, , x→ a, , So, in this case, lim f ( x) exists., x→ a, , Case 3., , When a < 0., In this case, we have, lim f ( x) = lim f ( 0 + h) = lim {|h|− 1} = −1., +, h→0, , x→ 0, , h→0, , lim f ( x) = lim f ( 0 − h) = lim {|− h|+ 1} = 1., , x → 0−, , h→0, , h→0, , Thus, lim f ( x) ≠ lim f ( x)., +, −, x→ 0, , x→ 0, , Hence, lim f ( x) does not exist., x→ 0, , Thus, lim f ( x) exists only when a ≠ 0., x→ a, , EXAMPLE 6, , ⎧mx 2 + n, x < 0, ⎪, Let f ( x) = ⎨nx + m , 0 ≤ x ≤ 1, ⎪nx 3 + m , x > 1., ⎩, For what values of integers m and n, lim f ( x) and lim f ( x) both exist?, x→ 0, , x→1
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Limits, SOLUTION, , 833, , We have, lim f ( x) = lim f ( k + h) = lim [k + h] = lim k = k, , x → k+, , h→0, , h→0, , h→0, , {Q [k + h] = k}, , and lim f ( x) = lim f ( k − h) = lim [k − h] = lim ( k − 1) = ( k − 1), −, x→ k, , h→0, , h→0, , h→0, , Thus, lim f ( x) ≠ lim f ( x)., +, −, x→ k, , {Q [k − h] = k − 1}, , x→ k, , Hence, lim f ( x) does not exist., x→ k, , If f is an odd function and lim f ( x) exists then prove that this limit must, x→ 0, be 0., , EXAMPLE 9, , SOLUTION, , Let f be an odd function and let lim f ( x) exist. Then, x→ 0, lim f ( x) = lim f ( x), x→ 0+, , ⇒, ⇒, ⇒, , x → 0−, , lim f ( 0 + h) = lim f ( 0 − h), , h→0, , h→0, , lim f ( h) = lim f ( − h) = − lim f ( h) [Q f being odd, f ( − h) = − f ( h)], , h→0, , h→0, , h→0, , 2 lim f ( h) = 0 ⇒ lim f ( h) = 0 ⇒ f ( x) = 0., h→0, , Hence, lim f ( x) = 0., , h→0, , x→ 0, , EXAMPLE 10, , If f is an even function, prove that lim f ( x) = lim f ( x)., −, +, x→ 0, , SOLUTION, , x→ 0, , Let f be an even function. Then,, lim f ( x) = lim f ( 0 − h) = lim f ( − h), , x → 0−, , h→0, , h→0, , [Q f being even, f ( − h) = f ( h)], , = lim f ( h), h→0, , = lim f ( 0 + h) = lim+ f ( x)., h→0, , x→ 0, , Hence, lim f ( x) = lim f ( x)., −, +, x→ 0, , x→ 0, , ⎛ e − 1⎞, Show that lim ⎜ 1/x, ⎟ does not exist., x→ 0 ⎝ e, + 1⎠, 1/x, , EXAMPLE 11, , SOLUTION, , ⎛ e1/x − 1 ⎞, Let f ( x) = ⎜ 1/x, ⎟ ⋅ Then,, ⎝ e + 1⎠, lim+ f ( x) = lim f ( 0 + h) = lim f ( h), x→ 0, , h→0, , h→0, , 1 ⎞, ⎛, ⎜ 1 − 1/h ⎟ ⎛ 1 − 0 ⎞, ⎛ e1/h − 1 ⎞, e, = lim ⎜ 1/h, ⎟ =⎜, ⎟ = lim ⎜, ⎟ = 1., h→0 ⎝e, + 1⎠ h → 0 ⎜ 1 + 1 ⎟ ⎝ 1 + 0⎠, /, h, 1, ⎝, e ⎠, lim f ( x) = lim f ( 0 − h) = lim f ( − h), , x → 0−, , h→0, , h→0
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834, , Senior Secondary School Mathematics for Class 11, , ⎛ 1, ⎞, ⎜ 1/ h − 1 ⎟ ⎛ 0 − 1 ⎞, ⎛ e − 1/ h − 1 ⎞, e, = lim ⎜ −1/h, ⎟ =⎜, ⎟ = lim ⎜, ⎟ = −1., h→0 ⎝e, + 1⎠ h → 0 ⎜ 1 + 1⎟ ⎝ 0 + 1⎠, ⎝ e1/h, ⎠, Thus, lim f ( x) ≠ lim f ( x)., x→ 0+, , x → 0−, , Hence, lim f ( x) does not exist., x→ 0, , EXAMPLE 12, SOLUTION, , Show that lim, , x, , x → 0 |x|, , Let f ( x) =, , does not exist., , x, ⋅ Then,, |x|, , h→0, , ∴, , h, , h, = 1., h, x→ 0, −h, −h, lim f ( x) = lim f ( 0 − h) = lim f ( − h) = lim, = lim, = −1., −, h, →, 0, h, →, 0, h, →, 0, h, →, 0, |− h|, h, x→ 0, lim f ( x) ≠ lim f ( x)., lim+ f ( x) = lim f ( 0 + h) = lim f ( h) = lim, , x→ 0+, , h→0, , h → 0 |h|, , = lim, , h→0, , x → 0−, , Hence, lim f ( x) does not exist., x→ 0, , EXERCISE 27C, 1. If f ( x) = |x|− 3 , find lim f ( x)., x→ 3, , ⎧x, ⎪ ,x≠0, 2. Let f ( x) = ⎨|x|, ⎪0, x = 0., ⎩, Show that lim f ( x) does not exist., x→ 0, , ⎧|x − 3|, ,x≠ 3, ⎪, 3. Let f ( x) = ⎨( x − 3), ⎪0,, x = 3., ⎩, Show that lim f ( x) does not exist., x→ 3, , ⎧1 + x 2 , 0 ≤ x ≤ 1, 4. Let f ( x) = ⎨, ⎩2 − x , x > 1., Show that lim f ( x) does not exist., x→1, , ⎧ x − |x|, ,x≠0, ⎪, 5. Let f ( x) = ⎨ x, ⎪⎩2,, x = 0., Show that lim f ( x) does not exist., x→ 0
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Limits, , 835, , 0< x ≤1, ⎧5 x − 4,, 6. Let f ( x) = ⎨ 3, ⎩4x − 3 x , 1 < x < 2., Find lim f ( x)., x→1, , ⎧4x − 5 , x ≤ 2, 7. Let f ( x) = ⎨, ⎩ x − a , x > 2., If lim f ( x) exists then find the value of a., x→ 2, , ⎧ 3x, ,x≠0, ⎪, 8. Let f ( x) = ⎨|x|+ 2x, ⎪0,, x = 0., ⎩, Show that lim f ( x) does not exist., x→ 0, , ⎧cos x , x ≥ 0, 9. Let f ( x) = ⎨, ⎩x + k , x < 0., Find the value of k for which lim f ( x) exists., x→ 0, , 1, 10. Show that lim does not exist., x→ 0 x, 1, 11. Show that lim, = ∞., x → 0 |x|, 12. Show that lim e−1/x does not exist., x→ 0, , 13. Show that lim sin, x→ 0, , 14. Show that lim, , x→ 2, , 1, does not exist., x, , x, does not exist., [x], , π, ⎧ k cos x, ,x≠, ⎪ π − 2x, 2, 15. Let f ( x) = ⎨, π, ⎪3,, x= ⋅, ⎩, 2, ⎛ π⎞, If lim f ( x) = f ⎜ ⎟ , find the value of k., π, ⎝ 2⎠, x→, 2, , ANSWERS (EXERCISE 27C), , 1. 0, , 6. 1, , 7. a = −1, , 9. k = 1, , 10. k = 6, , HINTS TO SOME SELECTED QUESTIONS, 1., , lim f ( x ) = lim f ( 3 + h ) = lim {|3 + h| − 3} = lim ( 3 + h − 3 ) = lim h = 0 ., , x → 3+, , h→0, , h→0, , h→0, , h→0, , lim f ( x ) = lim f ( 3 − h ) = lim {|3 − h| − 3} = lim ( 3 − h − 3 ) = lim ( − h ) = 0 ., , x → 3−, , h→0, , h→0, , h→0, , h→0
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836, , Senior Secondary School Mathematics for Class 11, ∴, , lim f ( x ) = lim f ( x ) = 0 ⇒ lim f ( x ) = 0., , x → 3+, , x → 3−, , x→ 3, , h, , 2. lim f ( x ) = lim f ( 0 + h ) = lim f ( h ) = lim, x→ 0+, , h→0, , h→0, , h → 0|h|, , lim f ( x ) = lim f ( 0 − h ) = lim f ( − h ) = lim, , x → 0−, , ∴, , h→0, , = lim, , h→0, , −h, , h → 0|− h|, , h→0, , h, =1, h, , = lim, , h→0, , [Q h > 0 ], , −h, = −1., h, , lim f ( x ) ≠ lim f ( x ) and so lim f ( x ) does not exist., , x→ 0+, , x → 0−, , x→ 0, , |3 + h − 3|, |h|, h, 3. lim f ( x ) = lim f ( 3 + h ) = lim, = lim, = lim = 1., h→0 h, h→0, h → 0 ( 3 + h − 3), h→0 h, x → 3+, |3 − h − 3|, |− h|, h, = lim, = lim, = −1., lim f ( x ) = lim f ( 3 − h ) = lim, −, h, →, 0, h, →, 0, h, →, 0, (, 3, −, h, −, 3, ), −h h → 0 −h, x→ 3, ∴, , [Q h > 0 ], , lim f ( x ) ≠ lim f ( x ) and so lim f ( x ) does not exist., , x → 3+, , x → 3−, , x→ 3, , 4. lim f ( x ) = lim f ( 1 + h ) = lim {2 − ( 1 + h )} = lim ( 1 − h) = 1., x → 1+, , h→0, , h→0, , h→0, , lim f ( x ) = lim f ( 1 − h ) = lim {1 + ( 1 − h ) 2 } = lim ( 1 + 1 + h 2 − 2 h) = 2 ., , x → 1−, , ∴, , h→0, , h→0, , h→0, , lim f ( x ) ≠ lim f ( x ) and so lim f ( x ) does not exist., , x → 1+, , x → 1−, , x→1, , 6. lim f ( x ) = lim f ( 1 + h ) = lim {4( 1 + h ) 3 − 3( 1 + h )} = ( 4 × 13 − 3 ) = 1., x → 1+, , h→0, , h→0, , lim f ( x ) = lim f ( 1 − h ) = lim {5( 1 − h ) − 4} = (5 − 4 ) = 1., , x → 1−, , ∴, 7., , h→0, , h→0, , lim f ( x ) = lim f ( x ) = 1 and so lim f ( x ) = 1., , x → 1+, , x → 1−, , x→1, , lim f ( x ) = lim f ( 2 + h ) = lim ( 2 + h − a) = ( 2 − a)., , x → 2+, , h→0, , h→0, , lim f ( x ) = lim f ( 2 − h ) = lim {4( 2 − h ) − 5} = ( 8 − 5 ) = 3., , x→ 2−, , h→0, , h→0, , Since lim f ( x ) exists, we must have lim f ( x ) = lim f ( x )., x→ 2, , ∴, , 2 − a = 3 ⇒ a = ( 2 − 3 ) = −1., , x → 2+, , x→ 2−, , 9. lim f ( x ) = lim f ( 0 + h ) = lim cos h = 1., x→ 0+, , h→0, , h→0, , lim f ( x ) = lim f ( 0 − h ) = lim f ( − h ) = lim {( − h ) + k} = k ., , x → 0−, , h→0, , h→0, , h→0, , Since lim f ( x ) exists, we have lim f ( x ) = lim f ( x ) and hence k = 1., x→ 0+, , x→ 0, , x → 0−, , 1, 1, 10. Let f ( x ) = ⋅ Then, lim f ( x ) = lim f ( 0 + h ) = lim f ( h ) = lim, = ∞., h→0, h→0, h→0 h, x, x→ 0+, 1, And, lim f ( x ) = lim f ( 0 − h ) = lim f ( − h ) = lim, = −∞., h→0, h→0, h → 0 −h, x → 0−, ∴, , lim f ( x ) ≠ lim f ( x ) and hence lim f ( x ) does not exist., , x→ 0+, , 11. Let f ( x ) =, , x → 0−, , x→ 0, , 1, ⋅ Then,, |x|, , lim f ( x ) = lim f ( 0 + h ) = lim f ( h ) = lim, , x→ 0+, , h→0, , h→0, , 1, , h → 0 |h|, , = lim, , h→0, , 1, =∞, h, , [Q h > 0 ]
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Limits, , 837, , lim f ( x ) = lim f ( 0 − h ) = lim f ( − h ) = lim, , x → 0−, , ∴, , lim, , h→0, , 1, , x → 0 |x|, , 1, , h → 0|− h|, , h→0, , = lim, , h→0, , 1, = ∞., h, , = ∞., , 12. Let f ( x ) = e −1/x. Then,, lim f ( x ) = lim f ( 0 + h ) = lim f ( h ) = lim e −1/h = lim, , x→ 0+, , h→0, , h→0, , h→0, , h→0, , 1, 1/ h, , e, , lim f ( x ) = lim f ( 0 − h ) = lim f ( − h ) = lim e1/h = ∞., , x → 0−, , ∴, , h→0, , h→0, , h→0, , =, , 1, = 0., ∞, [Q e ∞ = ∞], , lim f ( x ) does not exist., , x→ 0, , 13. Let f ( x ) = sin, , 1, ⋅, x, , lim f ( x ) = lim f ( 0 + h ) = lim f ( h ) = lim sin, , x→ 0+, , h→0, , h→0, , h→0, , 1, ⋅, h, , 1, oscillates between –1 and 1 and, h, 1, thus it does not approach to a definite value. So, lim sin does not exist., h→0, h, 1, Hence, lim sin does not exist., x→ 0, x, x, 14. Let f ( x ) =, ⋅ Then,, [x], 2+ h, lim f ( x ) = lim f ( 2 + h ) = lim, =1, [Q h → 0 , [2 + h] = 2], h→0, h → 0 [ 2 + h], x → 2+, Clearly, when h is given different values, then sin, , lim f ( x ) = lim f ( 2 − h ) = lim, , x → 2−, , ∴, , 15., , h→0, , h→0, , lim f ( x ) ≠ lim f ( x ) and so lim f ( x ) does not exist., , x → 2+, , x → 2−, , π, lim f ( x ) = lim f ⎛⎜ + h ⎞⎟ = lim, h→0 ⎝ 2, ⎠ h→0, π+, , x→, , 2, , = lim, , h→0, , ∴, , 2−h, 2−h, = lim, =2, [2 − h] h → 0 1, , x→ 2, , π, k cos ⎛⎜ + h ⎞⎟, ⎝2, ⎠, − k sin h, = lim, h→0, π, −2 h, ⎛, ⎞, π − 2⎜ + h⎟, ⎝2, ⎠, , k ⎛ sin h ⎞ k, ⎜, ⎟= ⋅, 2⎝ h ⎠ 2, , k, = 3 ⇒ k = 6., 2, , q, , [Q [2 − h] = 1]
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28, Differentiation, DERIVATIVE OF A FUNCTION, Let y = f ( x) be a given continuous function. Then, the value of y depends upon the, value of x and it changes with a change in the value of x. We use the word, increment to denote a small change, i.e., an increase or decrease, in the values of x, and y., Let δy be an increment in y, corresponding to an increment δx in x. Then,, y = f ( x) and y + δy = f ( x + δx)., On subtraction, we get δy = f ( x + δx) − f ( x)., δy f ( x + δx) − f ( x), =, ⋅, ∴, δx, δx, δy, f ( x + δx) − f ( x), So,, lim, = lim, ⋅, δx → 0 δ x, δx → 0, δx, The above limit, if it exists finitely, is called the derivative or differential, coefficient of y = f ( x) with respect to x, and it is denoted by, dy, d, or, { f ( x)} or f ′( x)., dx, dx, The process of finding the derivative is known as differentiation., dy, δy, f ( x + δx) − f ( x), = lim, = lim, ⋅, δ, x, →, 0, δ, x, →, 0, dx, δx, δx, DERIVATIVE AT A POINT The value of f ′( x), obtained by putting x = a , is called the, dy, derivative of f ( x) at x = a , and it is denoted by f ′( a) or ⎧⎨ ⎫⎬, ⋅, ⎩ dx ⎭x = a, REMARK, , If f ′( a) exists, we say that f ( x) is differentiable at x = a., If f ′( x) exists for every value of x in the domain of the function, we say, that f ( x) is differentiable., Y, , GEOMETRICAL SIGNIFICANCE OF A DERIVATIVE, , Let y = f ( x) be a continuous function., Then, we draw its graph. Suppose it is a, curve. Let a be a point in the domain of the, given function. Let P[a , f ( a)] be a point on, this curve, and let Q[a + h, f ( a + h)] be, some neighbouring point on it., f ( a + h) − f ( a), Slope of chord PQ =, ( a + h − a), 838, , Q[(a+h), f(a+h)], , P[a, f(a)], X, , T, , O, , R, Y, , X
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Differentiation, , =, , 839, , f ( a + h) − f ( a), ⋅, h, , Let the point Q move along the curve such that Q → P., As Q comes closer and closer to P along the curve, the chord QP, approaches the tangent at P., This happens when h → 0., f ( a + h) − f ( a), lim, = lim (slope of chord PQ), ∴, h→0, Q→P, h, or, , f ′ ( a) = the slope of the tangent at P., , Hence, the derivative of f ( x) at x = a is the slope of the tangent to the curve, y = f ( x) at the point [a , f ( a)]., PHYSICAL SIGNIFICANCE OF A DERIVATIVE Let s = f (t) be a function representing, the distance travelled by a particle moving in a straight line in time t., , Let (s + δs) be the distance travelled by it in time (t + δt)., ∴, s + δs = f (t + δt)., On subtraction, we get δs = f (t + δt) − f (t)., δs f (t + δt) − f (t), average velocity =, =, ., ∴, δt, δt, Now, when δt → 0, the average velocity becomes the instantaneous, velocity., f (t + δt) − f (t), δs, = lim, = f ′ (t)., ∴ velocity at time t = lim, δt → 0 δ t, δt → 0, δt, DIFFERENTIATION FROM THE FIRST PRINCIPLE Obtaining the derivative of a given, function by using the definition is called differentiation from the first principle or, ab initio or by delta method., , Some Important Derivatives Using the First Principle, THEOREM 1, , PROOF, , From the first principle, we have, d, ( x n ) = nx n − 1 , where n is a fixed number, integer or rational., dx, , Let y = x n ., , … (i), , Let δy be an increment in y, corresponding to an increment δx in x., Then,, … (ii), y + δy = ( x + δx) n ., On subtracting (i) from (ii), we get δy = ( x + δx) n − x n, or, ∴, , δy ( x + δx) n − x n, =, ⋅, δx, δx, dy, δy, = lim, δ, x, →, 0, dx, δx
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844, , Senior Secondary School Mathematics for Class 11, , SUMMARY, , We may summarise the above results as given below:, d n, ( x ) = nxn − 1, dx, , (ii), , d x, ( e ) = ex, dx, , (iii), , d, ( sin x) = cos x, dx, , (iv), , d, (cos x) = − sin x, dx, , (v), , d, (tan x) = sec2 x, dx, , (vi), , d, (sec x) = sec x tan x, dx, , d, (cosec x) = − cosec x cot x, dx, , (viii), , d, (cot x) = − cosec2 x., dx, , (i), , (vii), , Some more Results on Differentiation, THEOREM 9, PROOF, , The derivative of a constant function is zero, i.e.,, , Let f ( x) = c be a constant function. Then, for any change δx in x, there, is no change in the value of the function., ∴ f ( x) = c and f ( x + δx) = c., So, from first principle, we have, f ( x + δx) − f ( x), c − c⎞, d, ( c) = f ′ ( x) = lim, = lim ⎛⎜, ⎟ = 0., δx → 0, δx → 0 ⎝ δ x ⎠, dx, δx, Hence,, , THEOREM 10, , PROOF, , d, ( c) = 0., dx, , d, ( c) = 0, where c is a constant., dx, , Let f ( x) be a differentiable function and let c be a fixed real number., d, d, Then, c ⋅ f ( x) is also differentiable and, {c ⋅ f ( x)} = c ⋅, { f ( x)}., dx, dx, , Let y = c ⋅ f ( x) and y + δy = c ⋅ f ( x + δx)., Then,, δy = c ⋅ f ( x + δx) − c ⋅ f ( x)., δy c ⋅ f ( x + δx) − c ⋅ f ( x), ∴, =, ., δx, δx, dy, δy, c ⋅ f ( x + δx) − c ⋅ f ( x), So,, = lim, = lim, dx δx → 0 δx δx → 0, δx, f ( x + δx) − f ( x), d, = c ⋅ lim, = c⋅, { f ( x)} ., δx → 0, dx, δx, dy, d, d, d, Thus,, = c⋅, { f ( x)} , i.e.,, {c ⋅ f ( x)} = c ⋅, { f ( x)}., dx, dx, dx, dx, d, Now, f ( x) being differentiable, it follows that, { f ( x)} exists and, dx, d, therefore, c ⋅, { f ( x)} exists., dx
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846, , Senior Secondary School Mathematics for Class 11, , We may extend the above result for a finite number of differentiable, , REMARK, , functions. Thus, we have, d, d, d, d, [ f1 ( x) + f 2 ( x) + … + f n ( x)] =, { f1 ( x)} +, { f 2 ( x)} + … +, { f n ( x)}., dx, dx, dx, dx, Find the derivative of ( x 3 + ex + 3 x + cot x) with respect to x., , EXAMPLE 2, SOLUTION, , We have, d, d, d x, d, d, ( x 3 + ex + 3 x + cot x) =, (x3) +, (e ) +, ( 3 x) +, (cot x), dx, dx, dx, dx, dx, = 3 x 2 + ex + 3 x(log 3) − cosec2 x., 3, Find the derivative of ⎛⎜ 9x 2 + + 5 sin x⎞⎟ with respect to x., ⎝, ⎠, x, , EXAMPLE 3, , SOLUTION, , We have, , d, 3, ( 9x 2 + + 5 sin x), dx, x, d, d, d, = 9⋅, (x2) + 3 ⋅, ( x −1 ) + 5 ⋅, (sin x), dx, dx, dx, = 9 × 2x + 3 ⋅ ( −1) x −2 + 5 cos x = 18x −, , THEOREM 12, , PROOF, , 3, + 5 cos x., x2, , If f ( x) and g( x) be differentiable functions then { f ( x) − g( x)} is also, differentiable, and, d, d, d, { f ( x) − g( x)} =, { f ( x)} −, {g( x)}., dx, dx, dx, , d, d, { f ( x) − g( x)} =, { f ( x) + ( −1) ⋅ g( x)}, dx, dx, d, d, =, { f ( x)} +, {( −1) ⋅ g( x)}, dx, dx, d, d, d, d, =, { f ( x)} + ( −1) ⋅, {g( x)} =, { f ( x)} −, {g( x)}., dx, dx, dx, dx, d, d, d, ∴, { f ( x) − g( x)} =, { f ( x)} −, {g( x)}., dx, dx, dx, d, Now, f ( x) and g( x) being differentiable, it follows that, { f ( x)} and, dx, d, {g( x)} both exist., dx, d, d, Consequently,, { f ( x)} −, {g( x)} exists., dx, dx, d, So, by the above result,, { f ( x) − g( x)} exists., dx, Hence, { f ( x) − g( x)} is differentiable., We have
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Differentiation, , 847, , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Differentiate the following functions with respect to x:, ⎛ x 2 + 4 − 2 tan x + 6e⎞, ⎜, ⎟, ⎝, ⎠, x2 3, d ⎛ 2, 4, 2, ⎞, ⎜ x + 2 − tan x + 6e⎟, ⎠, dx ⎝, 3, x, 2 d, d, d, d, (x2) + 4⋅, ( x −2 ) − ⋅, (tan x) + 6 ⋅, ( e), 3 dx, dx, dx, dx, 2, = 2x + 4 ⋅ ( −2) x −3 − sec2 x + 6 × 0, 3, 8, 2, = 2x − 3 − sec2 x., 3, x, =, , EXAMPLE 2, , SOLUTION, , Find the derivative of, , We have, , d, ⎡, ( e) = 0⎤, Q, ⎢⎣ dx, ⎥⎦, , ⎧ 3, ⎫, 5, 6, 2 tan x, −, +, −, + 7⎬⋅, ⎨3, cos, sin, sec, x, x, x, x, ⎭, ⎩, , ⎫, d ⎧ 3, 5, 6, 2 tan x, −, +, −, + 7⎬, ⎨, cos x sin x, sec x, dx ⎩ 3 x, ⎭, , d, { 3 x −1/3 − 5 sec x + 6cosec x − 2 sin x + 7}, dx, d, d, d, = 3⋅, ( x −1/3 ) − 5 ⋅, ( sec x) + 6 ⋅, ( cosec x), dx, dx, dx, =, , −2 ⋅, = 3 ⋅ ⎛⎜ −, ⎝, −1, ⎧, = ⎨ 4/3, ⎩x, EXAMPLE 3, , 1 ⎞ − 4/ 3, − 5 sec x tan x − 6cosec x cot x − 2 cos x, ⎟ x, 3⎠, ⎫, − 5 sec x tan x − 6cosec x cot x − 2 cos x ⎬, ⎭, , Differentiate the following functions:, (i) ( x 2 − 5 x + 6)( x − 3), , SOLUTION, , d, d, (sin x) +, (7), dx, dx, , (i), , 1 ⎞, ⎛, (ii) ⎜ x +, ⎟, ⎝, x⎠, , 2, , (iii), , d, {( x 2 − 5 x + 6) ( x − 3)}, dx, d, ( x 3 − 8x 2 + 21x − 18), dx, d, d, d, d, =, (x3) − 8⋅, ( x 2 ) + 21 ⋅, ( x) −, (18), dx, dx, dx, dx, = 3 x 2 − 8 × 2x + 21 × 1 − 0 = ( 3 x 2 − 16x + 21)., =, , 3 x 2 + 2x + 5, x
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848, , Senior Secondary School Mathematics for Class 11, , (ii), , (iii), , EXAMPLE 4, , SOLUTION, , If, y=, ∴, , EXAMPLE 5, , SOLUTION, , EXAMPLE 6, , SOLUTION, , d, dx, , 2, ⎧⎪⎛, 1 ⎞ ⎫ d ⎧⎛, 1, ⎞ ⎫, x, +, ⎜, ⎟, ⎨⎝, ⎬=, ⎨⎜ x + + 2⎟⎠ ⎬, x, x ⎠ ⎪⎭ dx ⎩⎝, ⎪⎩, ⎭, d, d, d, =, ( x) +, ( x −1 ) +, ( 2), dx, dx, dx, 1, = 1 + ( −1)x −2 + 0 = ⎛⎜1 − 2 ⎞⎟ ⋅, ⎝, x ⎠, , d ⎧ 3 x 2 + 2x + 5 ⎫ d, { 3 x 3/2 + 2x1/2 + 5 x −1/2}, ⎬=, ⎨, dx ⎩, dx, x, ⎭, [on dividing each term by x ], d, d 1/ 2, d, = 3⋅, ( x 3/ 2 ) + 2 ⋅, (x ) + 5 ⋅, ( x − 1/ 2 ), dx, dx, dx, 1, 1, 3, = 3 ⋅ x1/2 + 2 ⋅ x −1/2 + 5 ⋅ ⎛⎜ − ⎞⎟ x −3/2, ⎝ 2⎠, 2, 2, 1, 5, 9, =, x +, − x −3/2 ., 2, x 2, y=, , 1 − cos 2x, ,, 1 + cos 2x, , find, , dy, ., dx, , 1 − cos 2x, 2 sin 2 x, =, = tan x., 1 + cos 2x, 2 cos2 x, dy, d, =, (tan x) = sec2 x., dx, dx, , ⎛, ⎞, dy, x2 x3, If y = ⎜1 + x +, = y., +, + … ∞⎟ ⋅ show that, dx, 2!, 3!, ⎠, ⎝, We have, y = ex., dy, d x, ∴, =, ( e ) = ex = y., dx dx, If u = 3 t 4 − 5 t 3 + 2t 2 − 18t + 4, find, , du, at t = 1., dt, , du d, =, ( 3 t 4 − 5 t 3 + 2t 2 − 18t + 4), dt dt, d, d, d, d, d, = 3 ⋅ (t 4 ) − 5 ⋅ (t 3 ) + 2 ⋅ (t 2 ) − 18 ⋅ (t) +, ( 4), dt, dt, dt, dt, dt, = 3 × 4t 3 − 5 × 3 t 2 + 2 × 2t − 18 × 1 + 0, = 12t 3 − 15 t 2 + 4t − 18., ⎛ du⎞, = (12 × 13 − 15 × 12 + 4 × 1 − 18), ∴ ⎜ ⎟, ⎝ dt ⎠ t = 1, = (12 − 15 + 4 − 18) = − 17.
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Differentiation, , 849, , EXERCISE 28A, Differentiate the following functions:, 1, 1. (i) x −3, (ii) 3 x, 2. (i), x, 3. (i) 3 x −5, , 1, 5x, , (ii), , (ii), , 1, x, , (iii), , 3, , 1, x, , (iii) 6 ⋅ 3 x 2, 4, +, x, , (ii) 5 x − 3/2 +, , 4. (i) 6x5 + 4x 3 − 3 x 2 + 2x − 7, , x −, , 7, x, , (iii) ax 3 + bx 2 + cx + d , where a , b , c, d are constants, 5. (i) 4x 3 + 3 ⋅ 2x + 6 ⋅ 8 x −4 + 5 cot x, 1, 2, + x 2 − 2x + 6x −2/3 − x 6, 3, x, 1, 2, 3, 6 cot x, 6. (i) 4 cot x − cos x +, −, +, +9, 2, cos x sin x cosec x, (ii), , x 3, − +, 3 x, , x −, , (ii) −5 tan x + 4 tan x cos x − 3 cot x sec x + 2sec x − 13, 7. (i) ( 2x + 3) ( 3 x − 5), 1, (iv) ⎛⎜ x − ⎞⎟, ⎝, x⎠, 8. (i), (iv), , 2, , 1 ⎛, 1 ⎞, (iii) ⎛⎜ x + ⎞⎟ ⎜ x −, ⎟, ⎝, x⎠ ⎝, x⎠, , (ii) x(1 + x) 3, 1, (v) ⎛⎜ x 2 + 2 ⎞⎟, ⎝, x ⎠, , 3, , (vi) ( 2x 2 + 5 x − 1)( x − 3), , 3 x 2 + 4x − 5, x, , (ii), , ( x 3 + 1)( x − 2), x2, , (1 + x), 3, x, , (v), , ax 2 + bx + c, x, , x, , 9. (i) If y = 6x5 − 4x 4 − 2x 2 + 5 x − 9, find, , dy, dx, , (iii), , x−4, 2 x, , (vi), , a + b cos x, sin x, , at x = − 1., , π, at x = ⋅, dx, 3, dy, ( 2 − 3 cos x), π, at x = ⋅, (iii) If y =, , find, dx, sin x, 4, (ii) If y = (sin x + tan x), find, , dy, , dy, 1 ⎞, ⎛, 10. If y = ⎜ x +, + y = 2 x., ⎟ , show that 2x ⋅, ⎝, dx, x⎠, ⎛ x, 11. If y = ⎜, +, ⎝ a, 12. If y =, , a ⎞, ⎛ dy ⎞ ⎛ x a ⎞, ⎟ , prove that ( 2xy) ⎜ ⎟ = ⎜ − ⎟ ⋅, ⎝ dx ⎠ ⎝ a x ⎠, x ⎠, , dy, 1 + cos 2x, , find, ⋅, 1 − cos 2x, dx
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850, , Senior Secondary School Mathematics for Class 11, , 13. y =, , dy, 1 − tan 2 ( x/2), , find, ⋅, 2, dx, 1 + tan ( x/2), 1 − tan 2 ( x/ 2 ), , Hint, , 1 + tan 2 ( x/ 2 ), , = cos x., , ANSWERS (EXERCISE 28A), , 1. (i), , −3, x4, , 3. (i), , −15, x6, , 4., , (ii), , 1 − 2/ 3, x, 3, , (ii), , −1, 5x2, , 2. (i), , −1, x2, , 1 − 3/ 2, x, 2, , (ii) −, , (iii) −, , 1 −4/3, x, 3, , (iii) 4x −1/3, , (i) 30x 4 + 12x 2 − 6x + 2, , (ii), , −15 −5/2, 1, 7, x, − 2x −3/2 +, +, 2, 2 x x2, , (iii) 3 ax 2 + 2bx + c, 5., , (i) 12x 2 + ( 3 log 2) ⋅ 2x − 3 x −3/2 − 5 cosec2 x, (ii), , 6., , 1, 3, 1, 1, + 2 +, + x −3/2 + 2x − 2x(log 2) − 4x −5/3 − 4x5, 3 x, 2 x 2, , (i) −4cosec2 x +, , 1, sin x + 2sec x tan x + 3 cosec x cot x − 6 sin x, 2, , (ii) −5 sec2 x + 4 cos x + 3 cosec x cot x + 2sec x tan x, 7. (i) 12x − 1, 1, (iv) 2 ⎛⎜ x − 3 ⎞⎟, ⎝, x ⎠, 8., , (ii) 4x 3 + 9x 2 + 6x + 1, (v) 6x5 −, , (iii), , 6, 6, + 6x − 3, 7, x, x, , 5, (i) ⎛⎜ 3 + 2 ⎞⎟, ⎝, x ⎠, , (ii) 3 x 2 − 2 −, , 3 1/ 2, ( x + x −5/2 ), 2, , (vi) 6x 2 − 2x − 16, 1, 4, +, x2 x3, , (iii), , 1, 1, +, 4 x x x, , (iv), , (v), , b, c, 3, a x +, − x −3 / 2, 2, 2, 2 x, , (vi) − cosec x ( a cot x + bcosec x), , 9. (i) 55, , (ii) 9/ 2, , (iii) 2( 3 − 2), , 1 −5/6 7 1/6, x, + x, 6, 6, , 12. −cosec2 x, , 13. − sin x
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Differentiation, , (ii), , EXAMPLE 2, SOLUTION, , EXAMPLE 3, SOLUTION, , EXAMPLE 4, SOLUTION, , EXAMPLE 5, , SOLUTION, , 881, , d, d, ( x 2 ex sin x) =, [( x 2 ex) sin x], dx, dx, d, d, = ( x 2 ex) ⋅, (sin x) + sin x ⋅, ( x 2 ex), dx, dx, d, d x, ⎧, ⎫, = x 2 ex cos x + sin x ⋅ ⎨x 2 ⋅, ( e ) + ex ⋅, (x2)⎬, dx, dx, ⎩, ⎭, 2 x, 2 x, x, = x e cos x + sin x ⋅ [x e + 2xe ], = x 2 ex cos x + x 2 ex sin x + 2xex sin x, = xex[x cos x + x sin x + 2 sin x]., , Differentiate, , x 2 tan x., , d, ( x 2 tan x), dx, d, d, ⎧, ⎫, = ⎨x 2 ⋅, (tan x) + tan x ⋅, (x2)⎬, dx, dx, ⎩, ⎭, = ( x 2 sec2 x + 2x tan x)., , We have, , Differentiate ( ex sin x + x p cos x)., d x, ( e sin x + x p cos x), dx, d x, d, =, ( e sin x) +, ( x p cos x), dx, dx, d x ⎫ ⎧ p d, d, d, ⎧, ⎫, = ⎨e x ⋅, (sin x) + sin x ⋅, ( e ) ⎬ + ⎨x ⋅, (cos x) + cos x ⋅, (xp)⎬, dx, dx, dx, ⎩ dx, ⎭ ⎩, ⎭, = ( ex cos x + ex sin x) + [x p ( − sin x) + cos x ⋅ ( px p − 1 )], = ex(cos x + sin x) − x p sin x + px p − 1 cos x, = ex(cos x + sin x) − x p − 1 ( x sin x − p cos x)., , We have, , Differentiate, d x 3, [e ( x +, dx, , ex( x 3 +, , x )., , d, d x, (x3 + x) + (x3 + x) ⋅, (e ), dx, dx, 1, ⎧, ⎫, = ex ⎨ 3 x 2 + x −1/2 ⎬ + ( x 3 + x ) ⋅ ex, 2, ⎩, ⎭, 1, ⎫, x ⎧ 3, 2, = e ⎨x + 3 x +, + x⎬⋅, 2 x, ⎩, ⎭, , x )] = ex ⋅, , ⎛ ex cos x ⎞, Differentiate ⎜, ⎟ , using the product rule., ⎝ x3 ⎠, We have, , d ⎧ ex cos x ⎫ d, [( ex cos x) ⋅ x −3 ], ⎨, ⎬=, dx ⎩ x 3 ⎭ dx, d, d x, = ex cos x ⋅, ( x −3 ) + x −3 ⋅, ( e cos x), dx, dx
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Differentiation, , 883, , Derivative of the Quotient of two Functions, THEOREM, , (Quotient rule) If f(x) and g(x) are two differentiable functions and, , g( x) ≠ 0 then, , f ( x), g( x), , is also differentiable, and, , d, d, ⎧ f ( x) ⎫ g( x) ⋅ dx { f ( x)} − f ( x) ⋅ dx {g( x)}, ⋅, ⎨, ⎬=, [g( x)]2, ⎩ g( x) ⎭, f ( x), f ( x + δx), Let y =, and y + δy =, ., g( x), g( x + δx), d, dx, , PROOF, , Then,, ∴, , δy, 1 ⎧ f ( x + δx) f ( x) ⎫, =, ⋅⎨, −, ⎬⋅, δx δx ⎩ g( x + δx) g( x) ⎭, , dy, δy, 1 ⎧ f ( x + δx) f ( x) ⎫, −, = lim, = lim, ⎨, ⎬, dx δx → 0 δx δx → 0 δx ⎩ g( x + δx) g( x) ⎭, g( x) ⋅ f ( x + δx) − g( x + δx) ⋅ f ( x), = lim, δx → 0, δx ⋅ g( x + δx) ⋅ g( x), g( x) ⋅ f ( x + δx) − g( x) ⋅ f ( x) + g( x) ⋅ f ( x) − g( x + δx) ⋅ f ( x), δx ⋅ g( x + δx) ⋅ g( x), [on subtracting and adding g( x) ⋅ f ( x) in numerator], ⎡, ⎧ f ( x + δx) − f ( x) ⎫, ⎧ g( x + δx) − g( x) ⎫⎤, f ( x) ⋅ ⎨, = ⎢ lim g( x) ⋅ ⎨, ⎬ − δlim, ⎬⎥, δ, x, →, 0, x, →, 0, x, δ, δx, ⎩, ⎭, ⎩, ⎭⎦, ⎣, , = lim, , δx → 0, , d, ⎡, = ⎢g( x) ⋅, dx, ⎣, d, g( x) ⋅, dx, =, d, ∴, dx, , ⎡, ⎤, 1, × ⎢ lim, δx → 0 g( x + δx) ⋅ g( x) ⎥, ⎣, ⎦, 1, d, ⎤, { f ( x)} − f ( x) ⋅, {g( x)} ⎥ ×, 2, dx, ⎦ [g( x)], d, { f ( x)} − f ( x) ⋅, {g( x)}, dx, [g( x)]2, , d, d, ⎧ f ( x) ⎫ g( x) ⋅ dx { f ( x)} − f ( x) ⋅ dx {g( x)}, ⋅, ⎨, ⎬=, [g( x)]2, ⎩ g( x) ⎭, , Now, each of f ( x) and g( x) being differentiable, it follows that, and, , d, { f ( x)}, dx, , d, {g( x)} both exist., dx, , So, by the above result, it follows that, is differentiable., , d, dx, , ⎧ f ( x) ⎫, f ( x), ⎨, ⎬ exists, and hence, g, ( x), g, (, x, ), ⎩, ⎭
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884, , REMARK, , Senior Secondary School Mathematics for Class 11, , dv ⎫, ⎧ du, − u⋅ ⎬, ⎨v ⋅, d ⎛ u⎞ ⎩ dx, dx ⎭, The above result may be expressed as, , i.e.,, ⎜ ⎟ =, dx ⎝ v ⎠, v2, derivative of the quotient of two functions, =, , ( denom. × derivative of num.) − (num. × derivative of denom.), ( denominator) 2, , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Differentiate:, ex, (i), x, (i), , d, dx, , (ii), , d, dx, , ⎛ 2x + 3 ⎞, ex, (iii), (ii) ⎜ 2, ⎟, (1 + sin x), ⎝ x −5⎠, d x, x d, ⎛ ex ⎞ x ⋅ dx ( e ) − e ⋅ dx ( x) xex − ex ⋅ 1 ex( x − 1), =, =, ⋅, ⎜ ⎟ =, x2, x2, x2, ⎝ x⎠, d, d, 2, 2, ⎛ 2x + 3 ⎞ ( x − 5) ⋅ dx ( 2x + 3) − ( 2x + 3) ⋅ dx ( x − 5), =, ⎜ 2, ⎟, ( x 2 − 5) 2, ⎝ x −5⎠, , ( x 2 − 5) ⋅ 2 − ( 2x + 3) ⋅ 2x −2( x 2 + 3 x + 5), =, ⋅, ( x 2 − 5) 2, ( x 2 − 5) 2, d x, x d, ⎞ (1 + sin x) ⋅ dx ( e ) − e ⋅ dx (1 + sin x), d ⎛, ex, (iii), ⎟ =, ⎜, dx ⎝ 1 + sin x ⎠, (1 + sin x) 2, =, , =, , EXAMPLE 2, , SOLUTION, , Differentiate, , (1 + sin x) ⋅ ex − ex (cos x) (1 + sin x − cos x) ex, =, ⋅, (1 + sin x) 2, (1 + sin x) 2, , ⎛ x2 + 5x − 6⎞, ⎜ 2, ⎟⋅, ⎝ 4x − x + 3 ⎠, , Using the quotient rule, we have, ⎛ x2 + 5x − 6⎞, ⎜ 2, ⎟, ⎝ 4x − x + 3 ⎠, d, d, ( 4x 2 − x + 3) ⋅, ( x 2 + 5 x − 6) − ( x 2 + 5 x − 6) ⋅, ( 4x 2 − x + 3), dx, dx, =, ( 4x 2 − x + 3) 2, d, dx, , =, , EXAMPLE 3, , ( 4x 2 − x + 3)( 2x + 5) − ( x 2 + 5 x − 6)( 8x − 1) ( 9 + 54x − 21x 2 ), =, ⋅, ( 4x 2 − x + 3) 2, ( 4x 2 − x + 3) 2, , Differentiate, , ⎛ x 2 sin x ⎞, ⎜, ⎟⋅, ⎝ 1−x ⎠
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Differentiation, SOLUTION, , 885, , By the quotient rule, we have, d, d, (1 − x) ⋅, ( x 2 sin x) − x 2 sin x ⋅, (1 − x), d ⎛ x 2 sin x ⎞, dx, dx, ⎟ =, ⎜, dx ⎝ 1 − x ⎠, (1 − x) 2, d, d, ⎧, ⎫, (1 − x) ⎨x 2 ⋅, (sin x) + sin x ⋅, ( x 2 ) ⎬ − ( x 2 sin x)( −1), dx, dx, ⎩, ⎭, =, (1 − x) 2, =, =, , (1 − x) 2, , ⋅, , By the quotient rule, we have, d, d, (1 + tan x) ⋅, (1 − tan x) − (1 − tan x) ⋅, (1 + tan x), dy, dx, dx, =, dx, (1 + tan x) 2, (1 + tan x)( − sec2 x) − (1 − tan x)( sec2 x), =, (1 + tan x) 2, 2, −2sec x, −2, =, =, (1 + tan x) 2 (cos2 x)(1 + tan 2 x + 2 tan x), −2, −2, ⋅, =, =, 2, (, sin, 1, +, 2x), ⎧, ⎫, sin x 2 sin x, (cos2 x) ⎨1 +, +, ⎬, cos x ⎭, cos2 x, ⎩, Differentiate:, ⎛ sin x + cos x ⎞, (i) ⎜, ⎟, ⎝ sin x − cos x ⎠, , EXAMPLE 5, , SOLUTION, , x 2 (1 − x) cos x + ( x sin x)( 2 − x), , ⎧ 1 − tan x ⎫, dy, −2, If y = ⎨, =, ⋅, ⎬, show that, 1, tan, +, x, dx, (, 1, +, sin, 2x), ⎩, ⎭, , EXAMPLE 4, , SOLUTION, , (1 − x)[x 2 cos x + 2x sin x] + x 2 sin x, (1 − x) 2, , ⎛ sec x − 1 ⎞, (ii) ⎜, ⎟, ⎝ sec x + 1⎠, , d ⎛ sin x + cos x ⎞, ⎟, ⎜, dx ⎝ sin x − cos x ⎠, d, d, (sin x − cos x) ⋅ (sin x + cos x) − (sin x + cos x) ⋅ (sin x − cos x), dx, dx, =, (sin x − cos x) 2, (sin x − cos x)(cos x − sin x) − (sin x + cos x)(cos x + sin x), =, (sin x − cos x) 2, (i), , −[(sin x − cos x) 2 + (sin x + cos x) 2 ], (sin x − cos x) 2, −2(sin 2 x + cos2 x), −2, ⋅, =, =, 2, (sin x + cos2 x − 2 sin x cos x) (1 − sin 2x), =
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886, , Senior Secondary School Mathematics for Class 11, , (ii), , d ⎛ sec x − 1 ⎞, ⎟, ⎜, dx ⎝ sec x + 1⎠, d, d, ( sec x + 1) ⋅, ( sec x − 1) − ( sec x − 1) ⋅, ( sec x + 1), dx, dx, =, 2, ( sec x + 1), ( sec x + 1) sec x tan x − ( sec x − 1) sec x tan x 2sec x tan x, =, =, ⋅, ( sec x + 1) 2, ( sec x + 1) 2, , EXERCISE 28D, Differentiate:, 1., , 2x, x, , 2., , ⎛ 2x 2 − 4 ⎞, 5. ⎜ 2, ⎟, ⎝ 3x + 7⎠, 9., 13., , log x, x, , ⎛ x 2 + 3 x − 1⎞, 6. ⎜, ⎟, ⎝ x+2 ⎠, , x, (a + x2), , 10., , 2cot x, x, , 14., , 2, , ⎛ sin x − cos x ⎞, 17. ⎜, ⎟, ⎝ sin x + cos x ⎠, , x4, sin x, , 3., , ex, (1 + x), , 4., , 7., , ( x 2 − 1), ( x + 7 x + 1), , ⎛ 5 x 2 + 6x + 7 ⎞, 8. ⎜ 2, ⎟, ⎝ 2x + 3 x + 4⎠, , 11., , 2, , a+ x, a− x, , ⎛ 1 + sin x ⎞, 15. ⎜, ⎟, ⎝ 1 − sin x ⎠, , sin x, (1 + cos x), , ex sin x, sec x, , 21., , 23., , x tan x, ( sec x + tan x), , ⎛ ax 2 + bx + c ⎞, 24. ⎜ 2, ⎟, ⎝ px + qx + r ⎠, , ⎛ ex + sin x ⎞, 19. ⎜, ⎟, ⎝ 1 + log x ⎠, , 2x cot x, x, , 22., , (ii) sec x [Hint Write cot x =, , cos x, log x, , ⎛ 1 − cos x ⎞, 16. ⎜, ⎟, ⎝ 1 + cos x ⎠, , ⎛ sec x − tan x ⎞, 18. ⎜, ⎟, ⎝ sec x + tan x ⎠, , 20., , 26. (i) cot x, , 12., , ex, (1 + x 2 ), , ex ( x − 1), ( x + 1), , ⎛ sin x − x cos x ⎞, 25. ⎜, ⎟, ⎝ x sin x + cos x ⎠, , cos x, 1, and sec x =, .], sin x, cos x, ANSWERS (EXERCISE 28D), , 1., , 2 ( x log 2 − 1), x2, , 2., , (1 − log x), x2, , 3., , xe, (1 + x) 2, , 5., , 52x, ( 3 x 2 + 7) 2, , 6., , ( x 2 + 4x + 7), ( x + 2) 2, , 7., , (7 x 2 + 4x + 7), ( x 2 + 7 x + 1) 2, , 8., , 3( x 2 + 4x + 1), ( a2 − x 2 ), 9. 2, 2, 2, ( 2x + 3 x + 4), (a + x2)2, , x, , x, , 10., , 4., , ex(1 − x) 2, (1 + x 2 ) 2, , x 3 ( 4 sin x − x cos x), sin 2 x
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Differentiation, , 11., , a, x ⋅( a − x)2, , 887, , 12., , 13., , −( 2xcosec2 x + cot x), 1, 14., 3/2, (, 1, +, cos, x), x, , 15., , 16., , 2 sin x, (1 + cos x) 2, , 18., , 17., , 2, (1 + sin 2x), , − ( x sin x log x + cos x), x(log x) 2, 2 cos x, (1 − sin x) 2, 2sec x (tan x − sec x), ( sec x + tan x), , x( e + cos x)(1 + log x) − ( e + sin x), 20. ex[cos 2x + sin x cos x], x(1 + log x) 2, 1, 2x[− xcosec2 x + x(log 2) cot x − log x], ex( x 2 + 1), 2, 22., 21., ( x + 1) 2, x3/2, xsec x( sec x − tan x) + tan x, ( aq − bp) x 2 + 2( ar − pc) x + ( br − cq), 24., 23., ( sec x + tan x), ( px 2 + qx + r) 2, x, , x, , 19., , 25., , x2, ( x sin x + cos x) 2, , 26. (i) −cosec2 x, , (ii) sec x tan x, , Derivative of a Function of a Function, CHAIN RULE, , If y = f (t) and t = g( x) then, , dy ⎛ dy dt ⎞, =⎜, ×, ⎟⋅, dx ⎝ dt dx ⎠, , This rule may be extended further., If y = f (t), t = g(u) and u = h( x) then, , dy ⎛ dy dt du⎞, =⎜, ×, ×, ⎟⋅, dx ⎝ dt du dx ⎠, , SOLVED EXAMPLES, EXAMPLE 1, SOLUTION, , Differentiate, , (i) sin x 3, , (ii) sin 3 x, , (iii) esin x, , (i) Let y = sin x 3 ., Put x 3 = t , so that y = sin t and t = x 3 ., ∴, So,, , dy, , dt, = cos t and, = 3x2., dt, dx, dy ⎛ dy dt ⎞, =⎜, ×, ⎟, dx ⎝ dt dx ⎠, = 3 x 2 cos t = 3 x 2 cos x 3, , [Q t = x 3 ]., , d, (sin x 3 ) = 3 x 2 cos x 3 ., dx, (ii) Let y = sin 3 x = (sin x) 3 ., Put sin x = t , so that y = t 3 and t = sin x., Hence,
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890, SOLUTION, , Senior Secondary School Mathematics for Class 11, , 1 − tan x, ⋅, 1 + tan x, 1 − tan x, 1 − tan x, Putting, ⋅, = t , we get y = t and t =, 1 + tan x, 1 + tan x, Let y =, , ∴, , dy 1 −1/2, 1, ., = t, =, dt 2, 2 t, , dt, And,, =, dx, =, ∴, , EXAMPLE 7, SOLUTION, , EXAMPLE 8, SOLUTION, , (1 + tan x) ⋅, , d, d, (1 − tan x) − (1 − tan x) ⋅, (1 + tan x), dx, dx, 2, (1 + tan x), , (1 + tan x)( − sec2 x) − (1 − tan x)( sec2 x), −2 sec2 x, =, ⋅, (1 + tan x) 2, (1 + tan x) 2, , 1, −2 sec2 x, dy ⎛ dy dt ⎞, =⎜, ×, ×, ⎟ =, dx ⎝ dt dx ⎠ 2 t (1 + tan x) 2, 1 + tan x, , =, , − sec2 x, ×, (1 + tan x) 2, , =, , − sec x, ⋅, (1 + tan x) 3/2 (1 − tan x)1/2, , 1 − tan x, 2, , If y = sin ( sin x + cos x ), find, , dy, ⋅, dx, , Putting (sin x + cos x) = t and (sin x + cos x) = t = u, we get, y = sin u, u = t and t = (sin x + cos x)., dy, du 1 −1/2, 1, ∴, = cos u,, = t, =, du, dt 2, 2 t, dt, and, = (cos x − sin x)., dx, dy ⎛ dy du dt ⎞ cos u, So,, =⎜, ×, ×, ⋅ (cos x − sin x), ⎟ =, dx ⎝ du dt dx ⎠, 2 t, cos t, [Q u = t ], =, ⋅ (cos x − sin x), 2 t, cos( sin x + cos x )(cos x − sin x), [Q t = (sin x + cos x)]., =, 2 sin x + cos x, Differentiate, , sin 2x cos 3 x., , By using the product rule, we have, d, d, d, (sin 2x cos 3 x) = (sin 2x) ⋅, (cos 3 x) + (cos 3 x) ⋅, (sin 2x), dx, dx, dx, = (sin 2x) ⋅ ( − sin 3 x) ⋅ 3 + (cos 3 x) ⋅ (cos 2x) ⋅ 2, [using chain rule], = ( −3 sin 2x sin 3 x + 2 cos 3 x cos 2x).
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Differentiation, , SOLUTION, , Using the product rule, we have, d ax, {e cos ( bx + c)}, dx, d, d ax, = eax ⋅, {cos ( bx + c)} + cos ( bx + c) ⋅, (e ), dx, dx, d, d, = eax {− sin ( bx + c)} ⋅, ( bx + c) + cos ( bx + c) ⋅ eax ⋅, ( ax), dx, dx, [using the chain rule], = − beax sin ( bx + c) + aeax cos ( bx + c), = eax [a cos ( bx + c) − b sin ( bx + c)]., , EXAMPLE 10, , SOLUTION, , eax cos ( bx + c)., , Differentiate, , EXAMPLE 9, , 891, , Differentiate, (i) Let y =, , (i), , 1+ x, 1−x, , (ii), , 1+ x, ⋅, 1−x, , x, 1 − x2, , Put, , 1+ x, 1+ x, ⋅, = t, so that y = t and t =, 1−x, 1−x, , ∴, , dy 1 −1/2, 1, = t, =, ⋅, dt 2, 2 t, , dt, And,, =, dx, =, ∴, , (1 − x) ⋅, , d, d, (1 + x) − (1 + x) ⋅, (1 − x), dx, dx, 2, (1 − x), , (1 − x) ⋅ 1 − (1 + x) ( − 1), 2, =, 2, (1 − x), (1 − x) 2, , 1, 2, dy ⎛ dy dt ⎞, =⎜, ×, ×, ⎟ =, ⎝, ⎠, dx, dt dx, 2 t (1 − x) 2, 1−x, =, 1 + x (1 − x) 2 1 + x, (1 − x), 1−x, 1, =, ⋅, (1 − x) 3/2 (1 + x)1/2, =, , 1, , 2, , d, (ii), dx, , d, d, ( x) − x ⋅, { 1 − x2}, 1 − x2 ⋅, ⎛, ⎞, x, dx, dx, ⎜, ⎟ =, ⎜ 1 − x2 ⎟, (1 − x 2 ), ⎝, ⎠, 1, 1 − x 2 ⋅ 1 − x ⋅ (1 − x 2 ) −1/2 ( −2x), 2, =, (1 − x 2 ), x2, 1 − x2 +, 2, 2, 1, 1 − x 2 = (1 − x ) + x =, =, ⋅, 2, 2 3/ 2, (1 − x ), (1 − x ), (1 − x 2 ) 3/2
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892, , Senior Secondary School Mathematics for Class 11, , EXERCISE 28E, Differentiate the following with respect to x:, 1. sin 4x, 2. cos 5x, 3. tan 3x, 5. cot 2 x, , 6. tan 3 x, , 9. ecot x, , 10., , sin x, , 7. tan, , 18., , sin x 3, , 21. cos 3 x sin 5 x 22. sin x sin 2x, 2x, , 3x, , 24. e sin 3 x, , 25. e cos 2x, , 28. e(x sin x +, , 29., , 32., , cos x), , a2 − x 2, a +x, 2, , 2, , 33., , ex + e− x, ex − e− x, 1 + sin x, 1 − sin x, , 8. ex, , x, , 11. (5 + 7 x) 6, , 13. ( 3 x 2 − x + 1) 4 14. ( ax 2 + bx + c) n 15., 17. cos2 x 3, , 4. cos x 3, 12. ( 3 − 4x)5, , 1, ( x 2 − x + 3) 3, x sin x, , 19., , 23. cos (sin, 26. e, , −5 x, , 4, , 16. sin 2 ( 2x + 3), cot, , 20., , x, , ax + b ), 27. cos ( x 3 . ex), , cot 4x, , 30., , e 2 x + e −2 x, e 2 x − e −2 x, , 31., , 34., , 1 + ex, 1 − ex, , 35., , dy, , when, dx, 36. y = sin ( sin x + cos x ), , 37. y = ex log (sin 2x), , ⎛ 1 − x2 ⎞, 38. y = cos ⎜, ⎟, ⎝ 1 + x2 ⎠, , ⎛ 1 + x2 ⎞, 39. y = sin ⎜, ⎟, ⎝ 1 − x2 ⎠, , 1 − x2, 1 + x2, e2 x + x 3, cosec 2x, , Find, , 40. y =, , sin x + x 2, cot 2x, , 41. If y =, , cos x − sin x, dy, , show that, + y 2 + 1 = 0., cos x + sin x, dx, , 42. If y =, , cos x + sin x, dy, π⎞, ⎛, = sec2 ⎜ x + ⎟ ⋅, , show that, ⎝, dx, 4⎠, cos x − sin x, , 43. If y =, , 1−x, dy, , prove that (1 − x 2 ), + y = 0., 1+ x, dx, , 44. If y =, , sec x − tan x, dy, , show that, = sec x (tan x + sec x)., sec x + tan x, dx, , [CBSE 2004], , ANSWERS (EXERCISE 28E), , 1. 4 cos 4x, , 2. −5 sin 5 x, , 3. 3 sec2 3 x, , 4. −3 x 2 sin x 3
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Differentiation, , 5. −2 cot x cosec2 x, 9. −( cosec2 x)( ecot x), , 6. 3 tan 2 x sec2 x 7., 10., , cos x, 2 sin x, , −3( 2x − 1), ( x 2 − x + 3) 4, , 19., , ( x cos x + sin x), 2 x sin x, , sec2 x, 2 x, , 4, , 8. 4x 3 ( ex ), , 11. 42(5 + 7 x)5, , 12. −20( 3 − 4x) 4, , 14. n( 2ax + b)( ax 2 + bx + c) n − 1, , 13. 4( 6x − 1)( 3 x 2 − x + 1) 3, 15., , 893, , 16. 2 sin ( 4x + 6) 17. −3 x 2 sin ( 2x 3 ), −cosec2, , 20., , 4 x . cot, , 18., , 3 x 2 cos x 3, 2 sin x 3, , x, x, , 21. 5 cos 3 x cos 5 x − 3 sin 3 x sin 5 x, 22. 2 sin x cos 2x + cos x sin 2x, −a, 24. e2 x( 3 cos 3 x + 2 sin 3 x), 23., (cos ax + b ) . sin (sin ax + b ), 2 ax + b, 25. e3 x( 3 cos 2x − 2 sin 2x), , 26. − e−5 x( 4cosec2 4x + 5 cot 4x), , 27. − x 2 ex ( x + 3) ⋅ sin ( x 3 ex), −4, −8, 29. x − x 2, 30. 2 x −2 x 2, (e − e ), (e − e ), , 28. ( x cos x). e(x sin x + cos x), −2x, 31., (1 + x 2 ) 3 / 2 (1 − x 2 )1/2, , 32., , −2a 2 x, ( a + x ) ( a 2 − x 2 )1/2, , 34., , ex, (1 − e )(1 − e2 x)1 / 2, , 36., , 38., , 33., , 2 3/2, , 2, , 35. 2e2 x(cos 2x + sin 2x) + 2x 3 cos 2x + 3 x 2 sin 2x, , x, , (cos x − sin x) cos ( sin x + cos x ), , 37. ex[2 cot 2x + log (sin 2x)], , 2 sin x + cos x, 4x, ⋅ sin, (1 + x 2 ) 2, , ⎛ 1 − x2 ⎞, ⎟, ⎜, ⎝ 1 + x2 ⎠, , 1, or sec x ( sec x + tan x), (1 − sin x), , 2, , 39., , ⎛ 1 + x2 ⎞, 4x, ⋅ cos ⎜, ⎟, 2 2, (1 − x ), ⎝ 1 − x2 ⎠, , 40. 2 (sin x + x 2 ) sec2 2x + (cos x + 2x) tan 2x, , HINTS TO SOME SELECTED QUESTIONS, 35. Write, , 1, = sin 2 x., cosec 2 x, , 1, 1, cos x +, sin x sin x cos π + cos x sin π, 2, 2, 4, 4 = tan ⎛ x + π ⎞ ⋅, 42. y =, =, ⎜, ⎟, π, π, 1, 1, ⎝, 4⎠, cos x −, sin x cos x cos − sin x sin, 4, 4, 2, 2, , q
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29, , Mathematical Reasoning, Introduction, , Logic is the science of reasoning., By reasoning we draw valid inferences from some well-known facts., In mathematical language, there are two kinds of reasoning, namely, inductive and deductive., Mathematical induction is some sort of inductive reasoning, which we, have studied in an earlier chapter., Now, in this chapter we shall discuss deductive reasoning., SENTENCES, , A group of words which makes complete sense is called a sentence., , We communicate our ideas to others through sentences., Sentences are classified as declarative (or assertive); imperative; exclamatory, and interrogative., DECLARATIVE SENTENCES, , A sentence that makes an assertion is called a declarative or, , assertive sentence., Each of the following is a declarative sentence., (i) The earth revolves round the sun., (ii) The sum of the angles of a triangle is 180°., (iii) 2 is the only even prime number., IMPERATIVE SENTENCES A sentence that expresses a request or a command is called an, imperative sentence., EXAMPLES, , Each of the following is an imperative sentence., (i) Switch on the light., (ii) Do your homework., (iii) Give me a glass of water., (iv) Keep quiet., INTERROGATIVE SENTENCES A sentence that asks some question is called an, interrogative sentence., EXAMPLES, , EXAMPLES, , Each of the following is an interrogative sentence., (i) Where are you coming from?, (ii) How far is Mumbai from here?, (iii) Is every set finite?, 894
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Mathematical Reasoning, EXCLAMATORY SENTENCES, , 895, , A sentence that expresses some strong feeling is called an, , exclamatory sentence., Each of the following is an exclamatory sentence., (i) What a beautiful girl!, (ii) Listen to me, Tanvy!, (iii) May you live long!, (iv) Hurrah! We have won., Actually, in mathematical terms, we shall be interested only in particular, types of declarative sentences, called statements., EXAMPLES, , STATEMENTS A declarative sentence which can be judged to be true or false but not, both, is called a mathematically acceptable statement or simply a statement., EXAMPLE 1, , SOLUTION, , Check whether the following sentences are statements. Give reasons for, your answer., (i) The earth is a planet., (ii) There are 35 days in a month., (iii) Mathematics is difficult., (iv) Every square is a rectangle., (v) Every set is a finite set., (vi) Every real number is a complex number., (vii) 9 is less than 7., (viii) The square of a natural number is an even number., (i) It is a scientifically established fact that the earth is a planet., So, the given sentence is true., Hence, it is a statement., (ii) We know that the maximum number of days in a given month is, 31. So, the given sentence is false., Therefore, it is a statement., (iii) For some people mathematics can be easy and for some others it, can be difficult., So, we cannot judge whether the given sentence is true or false., Hence, it is not a statement., (iv) In a square, every angle measures 90° and its opposite sides are, equal. So, it is a rectangle., Thus, the given sentence is true., Hence, it is a statement., (v) Since there are sets which are not finite, so the given sentence is, false., Therefore, it is a statement., (vi) Every real number a can be written as a = ( a + 0i )., So, every real number is a complex number., Therefore, the given sentence is true., Hence, it is a statement.
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896, , Senior Secondary School Mathematics for Class 11, , (vii) Since 9 is greater than 7, so the given sentence is false., Hence, it is a statement., (viii) We know that 2 and 3 are natural numbers such that 22 = 4,, which is even and 3 2 = 9, which is odd., So, the given sentence is sometimes true and sometimes false., Therefore, it is not a statement., REMARK (I), , Sentences involving variable time such as ‘yesterday’, ‘today’ and, 'tomorrow' are not statements., , EXAMPLES, , None of the sentences given below is a statement., (i) Today is Sunday., (ii) Yesterday was a windy day., (iii) Tomorrow is a holiday., Sentences involving variable places such as 'here', 'there', etc., are not, statements., , REMARK (II), , EXAMPLES, , REMARK (III), , EXAMPLES, , EXAMPLE 2, , SOLUTION, , None of the sentences given below is a statemement., (i) Meerut is far from here., (ii) This place is very hot., (iii) Rohit will reach there in time., Sentences involving variable pronouns such as 'he', 'she', etc., are not, statements., None of the sentences given below is a statemement., (i) He is a doctor., (ii) She is a graduate., , Check whether the following sentences are statements., (i) Answer this question., (ii) Everyone in this room is rich., (iii) There is no rain without clouds., (iv) Mathematics is fun., (v) She is a commerce graduate., (vi) 2 is a rational number., (vii) Fire is always hot., (viii) The sides of a quadrilateral have equal length., (i) It is an order and therefore, it is not a statement., (ii) From the context it is not clear which room is referred here and, also the term rich is not well defined., So, the given sentence is not a statement., (iii) This sentence is always true and therefore, it is a statement., (iv) For those who like mathematics it may be a fun but for others it, may not be.
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Mathematical Reasoning, , (v), (vi), (vii), (viii), , 897, , So, the given sentence cannot be judged to be true or false., Hence, it is not a statement., It is not known who she is. So, the given sentence is not a, statement., Clearly, 2 is an irrational number. So, the given sentence is false., Therefore, it is a statement., Clearly, the given sentence is always true., Hence, it is a statement., All the sides of a square as well as that of a rhombus are equal., But, the sides of a trapezium as well as that of a rectangle are, inequal., So, the given sentence is sometimes true and sometimes false., Hence, it is not a statement., , NEGATION OF A STATEMENT, , The denial of an assertion contained in a statement is called its negation., ~ p denotes the negation or denial of p., The denial of a statement can be expressed in various ways., EXAMPLE 3, SOLUTION, , EXAMPLE 4, , SOLUTION, , Let p: Kolkata is a city. Express the denial of p in three different ways., This can be done as shown below., ~ p: Kolkata is not a city., or ~ p: It is false that Kolkata is a city., or ~ p: It is not the case that Kolkata is a city., Write the negation of each of the following statements in two different, ways., (i) Africa is a continent., (ii) 5 is rational., (iii) All integers are rational numbers., (iv) Some prime numbers are odd numbers., (v) Everyone in Germany speaks German., Given below are the required negations:, (i) Africa is not a continent., Or, It is false that Africa is a continent., (ii) 5 is not rational., Or, It is false that 5 is rational., (iii) There exists an integer which is not a rational number., Or, It is false that all integers are rational numbers., (iv) No prime number is an odd number., Or, It is false that some prime numbers are odd numbers., (v) At least one person in Germany does not speak German., Or, It is false that everyone in Germany speaks German.
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898, , Senior Secondary School Mathematics for Class 11, , EXAMPLE 5, , SOLUTION, , Write the negation of each of the following statements:, p: All birds have wings., q: For every real number x , x 2 > x., r: There exists a real number x such that 0 < x < 1., s: For every real number x, either x > 1 or x < 1., We have:, ~ p: There exists a bird which has no wings., ~ q: There exists a real number x such that x 2 is not greater than x., ~ r: There does not exist a real number x such that 0 < x < 1., ~ s: There exists a real number x such that neither x > 1 nor x < 1., , EXERCISE 29A, 1. Which of the following sentences are statements? In case of a statement, mention whether it is true or false., (i) The sun is a star., (ii) 7 is an irrational number., (iii) The sum of 5 and 6 is less than 10., (iv) Go to your class., (v) Ice is always cold., (vi) Have you ever seen the Red Fort?, (vii) Every relation is a function., (viii) The sum of any two sides of a triangle is always greater than the, third side., (ix) May God bless you!, 2. Which of the following sentences are statements? In case of a statement,, mention whether it is true or false., (i) Paris is in France., (ii) Each prime number has exactly two factors., (iii) The equation x 2 + 5|x|+ 6 = 0 has no real roots., (iv) ( 2 + 3 ) is a complex number., (v) Is 6 a positive integer?, (vi) The product of –3 and –2 is –6., (vii) The angles opposite to the equal sides of an isosceles triangle, are equal., (viii) Oh! it is too hot., (ix) Monika is a beautiful girl., (x) Every quadratic equation has at least one real root.
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Mathematical Reasoning, , 899, , 3. Which of the following statements are true and which are false? In each case, give a valid reason for your answer., (i) p: 11 is an irrational number., (ii) q: Circle is a particular case of an ellipse., (iii) r: Each radius of a circle is a chord of the circle., (iv) s: The centre of a circle bisects each chord of the circle., (v) t: If a and b are integers such that a < b , then − a > − b., (vi) u: The quadratic equation x 2 + x + 1 = 0 has no real roots., 4. Write the negation of each of the following statements:, (i) Every natural number is greater than 0., (ii) Both the diagonals of a rectangle are equal., (iii) The sum of 4 and 5 is 8., (iv) The number 6 is greater than 4., (v) Every natural number is an integer., (vi) The number –5 is a rational number., (vii) All cats scratch., (viii) There exists a rational number x such that x 2 = 3., (ix) All students study mathematics at the elementary level., (x) Every student has paid the fees., (xi) There is some integer k for which 2k = 6., (xii) None of the students of this class has passed., ANSWERS (EXERCISE 29A), , 1. (i) Statement, True, , (iii), (v), (vii), (ix), , (ii) Statement, True, Statement, False, (iv) Imperative sentence, not a statement, Statement, True, (vi) Interrogative sentence, not a statement, Statement, False, (viii) Statement, True, Exclamatory sentence, not a statement, , 2. (i) Statement, True, , (iii), (v), (vi), (viii), (ix), (x), , (ii) Statement, True, Statement, True, (iv) Statement, False, Interrogative sentence, not a statement, Statement, False, (vii) Statement, True, Exclamatory sentence, not a statement, Not a statement, the term ‘beautiful’ is vague., Statement, False, , 3. (i) True, as 11 being prime 11 is irrational., , x2 y2, +, = 1., a 2 b2, Putting b = a , it becomes x 2 + y 2 = a 2 , which is the equation of a circle., , (ii) True. The equation of an ellipse is
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900, , Senior Secondary School Mathematics for Class 11, , (iii), (iv), (v), (vi), , False, since the radius of a circle is not a chord., False, since there are chords which do not pass through the centre, True, since a < b ⇒ − a > − b is true., True. Since D = (12 − 4 × 1 × 1) − 3 < 0., , 4. (i) There exists a natural number which is not greater than 0., , (ii), (iii), (iv), (v), (vi), (vii), (viii), (ix), , It is false that both the diagonals of a rectangle are equal., It is false that the sum of 4 and 5 is 8., The number 6 is not greater than 4., There exists a natural number which is not an integer., The number –5 is not a rational number., There exists a cat which does not scratch., There does not exist a rational number x such that x 2 = 3., There exists a student who does not study mathematics at the, elementary level., (x) At least one student has not paid the fees., (xi) For all integers k, 2k ≠ 6., (xii) At least one student of this class has passed., , LOGICAL CONNECTIVES, SIMPLE STATEMENTS A statement which does not contain any other statement as its, component part is called a simple statement., EXAMPLES Each of the following is a simple statement., (i) Chennai is the capital of Tamil Nadu., (ii) The set of real numbers is an infinite set., (iii) The sum of 6 and 3 is greater than 10., COMPOUND STATEMENTS A compound statement is a combination of two or more, simple sentences., The method of combining two or more simple sentences is known as, compounding., LOGICAL CONNECTIVES The words or phrases which connect two or more simple, sentences are called logical connectives or simply connectives., These connectives are:, , (i) and, , (ii) or, , (iii) If .... then ...., , (iv) if and only if, , CONNECTIVES AND THEIR SYMBOLS, , Connective, , Symbol, , Nature of compound statement, , (i) and, , ∧, , Conjunction, , (ii) or, , ∨, , Disjunction, , (iii) If ... then ..., , ⇒, , Implication (or conditional), , (iv) if and only if, , ⇔, , Biconditional (or equivalence)
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Mathematical Reasoning, , 901, , Each of the following is a compound statement., (i) 2 is a rational number and 2 is an irrational number., (ii) The sum of 3 and 5 is 7 or 8., (iii) If it rains then the school may be closed., (iv) A triangle is equiangular if and only if it is equilateral., The connectives in (i), (ii), (iii) and (iv) are respectively and , or, If ... then, if, and only if., EXAMPLES, , COMPOUNDING OF SENTENCES, 1. CONJUNCTION, , Two simple sentences connected by the word ‘and’ are said to form a, , conjunction., We use the symbol ‘∧’ to denote ‘and’., TRUTH TABLE FOR p, , ∧ q Note that ‘p ∧ q’ is true only when each one of the statements, , p and q is true., Based on the above fact, we prepare the truth table for ‘p ∧ q’ as under., Truth table for p ∧ q, p, , q, , p∧q, , T, , T, , T, , T, , F, , F, , F, , T, , F, , F, , F, , F, , Remember that:, (i) The compound statement with ‘and’ is true if all its component, statements are true., (ii) The compound statement with ‘and’ is false if any of its, component statements is false or all of its component statements, are false., EXAMPLE 1, , SOLUTION, , Split each of the following compound statements into simple sentences, and determine whether it is true or false:, (i) The grass is green and the sky is blue., (ii) Agra is in Uttar Pradesh and Shimla is in Punjab., (iii) All rational numbers are real and all real numbers are not complex, numbers., (iv) x = 5 and x = 2 are the roots of the equation 3 x 2 − x − 10 = 0., (i) Let p: The grass is green., And q: The sky is blue., Clearly, each one of p and q is a true statement and therefore,, (p and q) is true., Hence, the given statement is true.
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902, , Senior Secondary School Mathematics for Class 11, , (ii) Let p: Agra is in Uttar Pradesh., And q: Shimla is in Punjab., Clearly, p is true and q is false., Therefore, (p and q) is false., Hence, the given statement is false., (iii) Let p: All rational numbers are real., And, q: All real numbers are not complex numbers., Clearly, p is true and q is false., ∴ (p and q) is false., Hence,the given statement is false., (iv) Let p: x = 5 is a root of the equation 3 x 2 − x − 10 = 0., And q: x = 2 is a root of the equation 3 x 2 − x − 10 = 0., Clearly, 3 × 5 2 − 5 − 10 = 60 ≠ 0. So, x = 5 is not a root of the, equation 3 x 2 − x − 10 = 0., And, 3 × 22 − 2 − 10 = 0. So, x = 2 is a root of 3 x 2 − x − 10 = 0., ∴ p is false and q is true., Consequently, (p and q) is false., Hence, the given statement is false., 2. DISJUNCTION, , Two simple sentences connected by the word ‘or’ are said to form a, , disjunction., We use the symbol ‘∨’ to denote ‘or’., Clearly, (p or q) is true when at least one of them is true., TRUTH TABLE FOR (p ∨ q), , Clearly, ( p ∨ q) is true only when (p is true or q is true), , or both are true., Truth table for ( p ∨ q), , p, , q, , p∨q, , T, T, F, F, , T, F, T, F, , T, T, T, F, , Remember that:, (i) A compound statement with ‘or’ is true when at least one of them is true., (ii) A compound statement with ‘or’ is false when each one of them is false., In (p or q) we say that exclusive ‘or’ is used when both p and q are not true, simultaneously., Otherwise, inclusive ‘or’ is used.
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Mathematical Reasoning, , 903, , For each of the following statements, determine whether an inclusive, ‘or’ or exclusive ‘or’ is used. Give reasons for your answer., (i) For identification you need a passport or an Adhar Card., (ii) The school is closed if it is a holiday or a Sunday., (iii) 3 is a rational number or an irrational number., (iv) Two lines intersect at a point or are parallel., (v) Students can take Sanskrit or French as their third language., (i) Here ‘or’ is inclusive since a person can have both a passport and, an Adhar Card for identification., (ii) Here ‘or’ is inclusive since the school is closed on holiday as well, as on Sunday., (iii) Let p : 3 is a rational number., And, q : 3 is an irrational number., Clearly, p is false and q is true., ∴ (p or q) is true and ‘or’ is exclusive., (iv) Here ‘or’ is exclusive because it is not possible for two lines to, intersect and be parallel together., (v) Here ‘or’ is exclusive because a student cannot take both Sanskrit, and French., , EXAMPLE 2, , SOLUTION, , OPEN SENTENCES AND THEIR TRUTH SETS, , Let p( x) be a declarative sentence involving a variable x and let A be a given set, such that for a ∈ A , p( a) is true or false., Then, p( x) is called an open sentence defined on A., The subset of A consisting of all those values of x which convert p( x) into a, true statement is called a Truth Set of the open sentence p( x)., Write down the truth set of each of the following open sentences:, (i) p( x): x + 5 < 9, x ∈ N ., (ii) p( x): x + 3 < 3, x ∈ N ., (iii) p( x): x + 5 > 7, x ∈ R., (iv) p( x): 2x 2 + 5 x − 3 = 0, x ∈ I., We have:, (i) Truth Set = {x ∈ N : x + 5 < 9}, = {x ∈ N : x < 4} = {1, 2, 3}., (ii) Truth Set = {x ∈ N : x + 3 < 3}, = {x ∈ N : x < 0} = φ., (iii) Truth Set = {x ∈ R : x + 5 > 7}, = {x ∈ R : x > 2} =]2, ∞[., 2, (iv) 2x + 5 x − 3 = 0 ⇒ 2x 2 + 6x − x − 3 = 0 ⇒ 2x( x + 3) − ( x + 3) = 0, 1, ⇒ ( x + 3)( 2x − 1) = 0 ⇒ x = −3 or x = ⋅, 2, , EXAMPLE 3, , SOLUTION
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904, , Senior Secondary School Mathematics for Class 11, , ∴ truth set = {x ∈ I : 2x 2 + 5 x − 3 = 0} = {−3}, , ⎡ 1, ⎤, ⎢⎣Q 2 ∉ I ⎥⎦ ⋅, , QUANTIFIERS AND QUANTIFIED STATEMENTS, , Two Important Symbols:, , (i) The symbol ‘∀’ stands ‘for all values of ’., This is known as universal quantifier., (ii) The symbol ‘∃’ stands for ‘there exists’., The symbol ∃ is known as existential quantifier., QUANTIFIED STATEMENT An open sentence with a quantifier becomes a, statement, called a quantified statement., EXAMPLE 4, , SOLUTION, , Use quantifiers to convert each of the following open sentences defined, on N, into a true statement:, (i) x + 5 = 8, (ii) x 2 > 0, (iii) x + 2 < 4, , (i) ∃ x ∈ N such that x + 5 = 8 is a true statement, since x = 3 ∈ N, satisfies x + 5 = 8., (ii) x 2 > 0 ∀ x ∈ N is a true statement, since the square of every, natural number is positive., (iii) ∃ x ∈ N such that x + 2 < 4 is a true statement, since x = 1 ∈ N, satisfies x + 2 < 4., Let A = {1, 2, 3 , 4}. Examine whether the statements given below are, true or false., (i) ∃ x ∈ A such that x + 3 = 8., (ii) ∀ x ∈ A , x + 2 < 7., (iii) ∃ x ↔ A such that x + 1 < 3., (iv) ∀ x ∈ A , x + 3 ≥ 5., (i) Clearly, no number in A satisfies x + 3 = 8., ∴ the given statement is false., (ii) Every number in A satisfies x + 2 < 7., So, the given statement is true., (iii) Clearly, x = 1 ↔ A satisfies x + 1 < 3., ∴ the given statement is true., (iv) Since x = 1 ∈ A does not satisfy x + 3 ≥ 5, the given statement is, false., , EXAMPLE 5, , SOLUTION, , EXERCISE 29B, 1. Split each of the following into simple sentences and determine whether it, is true or false., (i) A line is straight and extends indefinitely in both the directions., (ii) A point occupies a position and its location can be determined.
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Mathematical Reasoning, , 905, , (iii) The sand heats up quickly in the sun and does not cool down fast, at night., (iv) 32 is divisible by 8 and 12., (v) x = 1 and x = 2 are the roots of the equation x 2 − x − 2 = 0., (vi) 3 is rational and 3 is irrational., (vii) All integers are rational numbers and all rational numbers are not, real numbers., (viii) Lucknow is in Uttar Pradesh and Kanpur is in Uttarakhand., 2. Split each of the following into simple sentences and determine whether it is, true or false. Also, determine whether an ‘inclusive or’ or ‘exclusive or’ is, used., (i) The sum of 3 and 7 is 10 or 11., (ii) (1 + i) is a real or a complex number., (iii) Every quadratic equation has one or two real roots., (iv) You are wet when it rains or you are in a river., (v) 24 is a multiple of 5 or 8., (vi) Every integer is rational or irrational., (vii) For getting a driving licence you should have a ration card or a, passport., (viii) 100 is a multiple of 6 or 8., (ix) Square of an integer is positive or negative., (x) Sun rises or Moon sets., 3. Find the truth set in case of each of the following open sentences defined, on N:, (i) x + 2 < 10, , (ii) x + 5 < 4, , (iii) x + 3 > 2, , 4. Let A = {2, 3, 5, 7}. Examine whether the statements given below are true or, false., (i) ∃ x ∈ A such that x + 3 > 9., (ii) ∃ x ∈ A such that x is even., (iii) ∃ x ∈ A such that x + 2 = 6., (iv) V x ∈ A , x is prime., (v) V x ∈ A , x + 2 < 10, (vi) V x ∈ A , x + 4 ≥ 11, ANSWERS (EXERCISE 29B), , 1. (i) p: A line is straight., , q: A line extends indefinitely in both the directions., (p and q) is true., (ii) p: A point occupies a position., q: The location of a point can be determined., (p and q) is true.
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906, , Senior Secondary School Mathematics for Class 11, , (iii) p: The sand heats up quickly in the sun., q: The hot sand does not cool down fast at night., (p and q) is true., (iv) p: 32 is divisible by 8., q: 32 is divisible by 12., (p and q) is not true., (v) p: x = 1 is a root of the equation x 2 − x − 2 = 0., q: x = 2 is a root of the equation x 2 − x − 2 = 0., (p and q) is not true., (vi) p: 3 is rational., q: 3 is irrational., (p and q) is true., (vii) p: All integers are rational numbers., q: All rational numbers are not real numbers., (p and q) is false., (viii) p: Lucknow is in Uttar Pradesh., q: Kanpur is in Uttarakhand., (p and q) is false., 2. (i) p: The sum of 3 and 7 is 10., , q: The sum of 3 and 7 is 11., (p or q) is true. ‘or is exclusive’, (ii) p: (1 + i) is a real number., q: (1 + i) is a complex number., (p or q) is true. ‘or is exclusive’, (iii) p: Every quadratic equation has one real root., q: Every quadratic equation has two real roots., (p or q) is false., (iv) p: You are wet when it rains., q: You are wet when you are in a river., (p or q) is true. ‘or is inclusive’, (v) Let p: 24 is a multiple of 5., and q: 24 is a multiple of 8., (p or q) is true. ‘or is exclusive’, (vi) Let p: Every integer is rational., q: Every integer is irrational., (p or q) is true. ‘or is exclusive’
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Mathematical Reasoning, , 907, , (vii) Let p: For getting a driving licence you should have a ration card., q: For getting a driving licence you should have a passport., (p or q) is true. ‘or is inclusive’, (viii) Let p: 100 is a multiple of 6., q: 100 is a multiple of 8., (p or q) is false., (ix) Let p: Square of an integer is positive., q: Square of an integer is negative., (p or q) is false. [Hint: 02 is not positive.], (x) Let p: Sun rises., q: Moon sets., (p or q) is true. ‘or is exclusive’, 3. (i) {1, 2, 3, 4, 5, 6, 7}, 4. (i) True, , (ii) True, , (ii) φ, (iii) False, , (iii) N, (iv) True, , (v) True, , (vi) False, , CONDITIONAL AND BICONDITIONAL STATEMENTS, CONDITIONAL STATEMENTS Two statements connected by ‘if ... then’ is called a, conditional statement., For any statements p and q, the statement ‘If p, then q’ is denoted by ‘p ⇒ q’,, which means ‘p implies q’., , Here p is called the antecedent and q is called the consequent., We may write ‘if p, then q’ in following ways:, (i) p implies q, written as p ⇒ q., (ii) p only if q., (iii) q is a necessary condition for p., (iv) p is a sufficient condition for q., (v) ~ q ⇒ ~ p., EXAMPLE 1, , SOLUTION, , Rewrite the following statement, with ‘if ..., then’ in five different ways, conveying the same meaning:, If a natural number is even, then its square is even., We may rewrite it in following ways:, (i) A natural number is even implies that its square is even., (ii) A natural number is even only if its square is even., (iii) For a natural number to be even it is necessary that its square, is even., (iv) For the square of a natural number to be even, it is sufficient that, the number is even., (v) If the square of a natural number is not even, then the natural, number is not even.
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908, EXAMPLE 2, , SOLUTION, , Senior Secondary School Mathematics for Class 11, , Write each of the following statements in the form ‘if ... then’:, (i) A quadrilateral is a parallelogram if its diagonals bisect each other., (ii) The banana tree will bloom if it stays warm for a month., (iii) There is traffic jam whenever it rains., (iv) It is necessary to have a password to log on to the server., (v) You can access the website only if you pay a subscription fee., We can write them as under:, (i) If the diagonals of a quadrialateral bisect each other, then it is a, parallelogram., (ii) If the banana tree stays warm for a month, then it will bloom., (iii) If it rains, then there is traffic jam., (iv) If you have a password, then you can log on to the server., (v) If you can access the website, then you pay a subscription fee., , CONVERSE AND CONTRAPOSITIVE OF STATEMENTS, , Let p ⇒ q. Then,, (i) (converse of p ⇒ q) is ( q ⇒ p), (ii) (contrapositive of p ⇒ q) is (~ q ⇒~ p), EXAMPLE 3, , SOLUTION, , Write the converse and contrapositive of each of the following statements:, (i) If n is an even number, then n2 is even., (ii) If two integers a and b are such that a > b , then ( a − b) is always a, positive integer., (iii) If a CABC is right angled at B, then AB 2 + BC 2 = AC 2 ., (iv) If CABC and CDEF are congruent, then they are equiangular., (v) You cannot comprehend geometry if you do not know how to reason, deductively., (vi) Something is cold implies that it has low temperature., (i) Its converse is:, If a number n2 is even, then n is even., Its contrapositive is:, If a number n2 is not even, then n is not even., (ii) Its converse is:, If a and b are two integers such that ( a − b) is a positive integer,, then a > b., Its contrapositive is:, If two integers a and b are such that ( a − b) is not a positive integer,, then a is not greater than b., (iii) Its converse is:, In a CABC, if AB 2 + BC 2 = AC 2 , then it is right angled at B.
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Mathematical Reasoning, , 909, , Its contrapositive is:, In a CABC, if ( AB 2 + BC 2 ) is not equal to AC 2 , then it is not right, angled at B., (iv) Its converse is:, If CABC and CDEF are equiangular, then they are congruent., Its contrapositive is:, If CABC and CDEF are not equiangular, then they are not, congruent., (v) Its converse is:, If you do not know how to reason deductively, then you cannot, comprehend geometry., Its contrapositive is:, If you know how to reason deductively, then you can, comprehend geometry., (vi) Given statement is:, Something is cold ⇒ it has low temperature., Its converse is:, If something has low temperature, then it is cold., Its contrapositive is, If something does not have a low temperature, then it is not cold., BICONDITIONAL STATEMENT Two simple sentences connected by ‘if and only if ’, form a biconditional statement., We use the symbol ‘⇔’ for ‘if and only if’., , ( p ⇔ q) is the same as ( p ⇒ q) and ( q ⇒ p)., In CABC, ∠B = ∠C ⇔ AC = AB., This statement is the same as:, , EXAMPLE, , ‘In CABC:( ∠B = ∠C ⇒ AC = AB) and ( AC = AB ⇒ ∠B = ∠C)’., EXAMPLE 4, , SOLUTION, , EXAMPLE 5, , Rewrite the following statements in the form:, ‘p if and only if q’., p: If a quadrilateral is equiangular, then it is a rectangle., q: If a quadrilateral is a rectangle, then it is equiangular., The required statement is:, A quadrilateral is equiangular if and only if it is a rectangle., Given below are pairs of statements. In each case, combine them using ‘if, and only if ’., (i) p: In CABC, ∠B = ∠C., q: In CABC, AC = AB., (ii) p: A and B are two sets such that A ⊆ B and B ⊆ A., q: A = B.
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910, , Senior Secondary School Mathematics for Class 11, , SOLUTION, , (iii) p: CABC is equilateral., q: CABC is equiangular., (iv) p: {a ∈ R such that| a | < 2}., q: {a ∈ R such that (a > −2 and a < 2)}., We can combine the given statements as given below., (i) In CABC, ∠B = ∠C ⇔ AC = AB., (ii) For any sets A and B, A = B ⇔ ( A ⊆ B and B ⊆ A)., (iii) A CABC is equilateral ⇔ it is equiangular., (iv) For every real a,| a |< 2 ⇔ ( a > −2 and a < 2)., VARIOUS CASES SHOWING VARIOUS CONDITIONS, , SOME CONDITIONS ARE NECESSARY BUT NOT SUFFICIENT, , (i) Having four sides is necessary but not sufficient for being a square (since, a rectangle has four sides but it is not a square)., (ii) Being educated is necessary but not sufficient for getting the job of a, teacher (as certain minimum qualifications are required to be recruited as, a teacher)., (iii) Having a difference of 2 between two numbers is necessary but not, sufficient for them to be twin-primes (since they must be prime also)., , EXAMPLES, , SOME CONDITIONS ARE SUFFICIENT BUT NOT NECESSARY, , (i) Having a son is sufficient but not necessary for being a parent (since a, parent could have only a daughter)., (ii) It is obvious that being a multiple of 10 is sufficient but not necessary for, a number to be a composite number., (iii) Having 6 at the unit's place is sufficient but not necessary for a number, to be even (since numbers having 0, 2, 4, 8 in the unit's place are, also even)., (iv) Having the corresponding sides equal is sufficient but not necessary for, two triangles to be similar (since the triangles having the corresponding, sides proportional are also similar), , EXAMPLES, , SOME CONDITIONS ARE BOTH SUFFICIENT AND NECESSARY, , (i) Having four sides is both sufficient and necessary for being a, quadrilateral., (ii) Having all the three sides of equal measure is both sufficient and, necessary for a triangle to be equilateral., (iii) Having one number as zero out of two numbers to be multiplied is both, sufficient and necessary for the product to be zero., , EXAMPLES, , SOME CONDITIONS ARE NEITHER SUFFICIENT NOR NECESSARY, EXAMPLES, , (i) Being a tall person is neither sufficient nor necessary for being a, successful person.
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Mathematical Reasoning, , 911, , (ii) Being handsome is neither sufficient nor necessary for getting married., (iii) Having both the diagonals equal is neither sufficient nor necessary for a, quadrilateral to be a parallelogram., , EXERCISE 29C, 1. Rewrite the following statement in five different ways conveying the same, meaning., If a given number is a multiple of 6, then it is a multiple of 3., 2. Write each of the following statements in the form ‘if .... then’:, (i) A rhombus is a square only if each of its angles measures 90°., (ii) When a number is a multiple of 9, it is necessarily a multiple of 3., (iii) You get a job implies that your credentials are good., (iv) Atmospheric humidity increases only if it rains., (v) If a number is not a multiple of 3, then it is not a multiple of 6., 3. Write the converse and contrapositive of each of the following:, (i) If x is a prime number, then x is odd., (ii) If a positive integer n is divisible by 9, then the sum of its digits is, divisible by 9., (iii) If the two lines are parallel, then they do not intersect in the same, plane., (iv) If the diagonals of a quadrilateral bisect each other, then it is a, parallelogram., (v) If A and B are subsets of X such that A ⊆ B , then (X − B) ⊆ (X − A)., (vi) If f ( 2) = 0, then f ( x) is divisible by ( x − 2)., (vii) If you were born in India, then you are a citizen of India., (viii) If it rains, then I stay at home., 4. Given below are some pairs of statements. Combine each pair using if and, only if:, (i) p: If a quadrilateral is equiangular, then it is a rectangle., q: If a quadrilateral is a rectangle, then it is equiangular., (ii) p: If the sum of the digits of a number is divisible by 3, then the, number is divisible by 3., q: If a number is divisble by 3, then the sum of its digits is divisible, by 3., (iii) p: A quadrilateral is a parallelogram if its diagonals bisect each, other., q: If the diagonals of a quadrilateral bisect each other, then it is a, parallelogram., (iv) p: If f ( a) = 0, then ( x − a) is a factor of polynomial f ( x)., q: If ( x − a) is a factor of polynomial f ( x), then f ( a) = 0.
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912, , Senior Secondary School Mathematics for Class 11, , (v) p: If a square matrix A is invertible, then |A| is nonzero., q: If A is a square matrix such that |A| is nonzero, then A is, invertible., 5. Write each of the following using ‘if and only if’:, (i) In order to get A grade, it is necessary and sufficient that you do all, the homework regularly., (ii) If you watch television then your mind is free and if your mind is, free then you watch television., ANSWERS (EXERCISE 29C), , 1. (i) A given number is a multiple of 6 implies that it is a multiple of 3., , (ii) A given number is a multiple of 6 only if it is a multiple of 3., (iii) For a given number to be a multiple of 6, it is necessary that it is a, multiple of 3., (iv) For a given number to be a multiple of 3, it is sufficient that the number, is a multiple of 6., (v) If a given number is not a multiple of 3, then it is not a multiple of 6., 2. (i) If each angle of a rhombus measures 90°, then it is a square., , (ii), (iii), (iv), (v), , If a number is a multiple of 9, then it is a multiple of 3., If you get a job, then your credentials are good., If it rains, then atmospheric humidity increases., If a number is a multiple of 6, then it is a multiple of 3., , 3. (i) Converse If x is an odd number, then it is prime., Contrapositive If x is not an odd number, then it is not prime., (ii) Converse If the sum of the digits of a positive integer n is divisible by 9,, then it is divisible by 9., Contrapositive If the sum of the digits of a positive integer n is not, divisble by 9, then it is not divisible by 9., (iii) Converse If the two lines do not intersect in the same plane, then they, are parallel., Contrapositive If the two lines intersect in the same plane, then they are, not parallel., (iv) Converse If a quadrilateral is a parallelogram, then its diagonals bisect, each other., Contrapositive If a quadrilateral is not a parallelogram, then its diagonals, do not bisect each other., (v) Converse If A and B are subsets of X such that (X − B) ⊆ (X − A), then, A ⊆ B., Contrapositive If A and B are subsets of X such that (X − B) is not a subset, of (X − A), then A is not a subset of B., (vi) Converse If f ( x) is divisible by ( x − 2), then f ( 2) = 0., Contrapositive If f ( x) is not divisible by( x − 2), then f ( 2) is not equal to 0.
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Mathematical Reasoning, , 913, , (vii) Converse If you are a citizen of India, then you were born in India., Contrapositive If you are not a citizen of India, then you were not born in, , India., (viii) Converse If I stay at home, then it rains., Contrapositive If I don’t stay at home, then it does not rain., , 4. (i) A quadrilateral is a rectangle if and only if it is equiangular., , (ii) A number is divisible by 3 if and only if the sum of its digits is divisible, by 3., (iii) A quadrilateral is a parallelogram if and only if its diagonals bisect each, other., (iv) ( x − a) is a factor of polynomial f ( x) if and only if f ( a) = 0., (v) A square matrix A is invertible if and only if |A| is nonzero., 5. (i) You get an A grade if and only if you do your homework regularly., , (ii) You watch television if and only if your mind is free., , VALIDATING STATEMENTS, , We shall list some general rules for checking whether a given compound, statement is true or not., Rule 1., , Statements with ‘And’:, Let p and q be two statements. Then, in order to prove that ‘p and q’ is, true, the following steps are taken., , Rule 2., , Step 1., , Show that p is true., , Step 2., , Show that q is true., , Statements with ‘or’:, Let p and q be two statements. Then, in order to prove that ‘p or q’ is true,, we proceed as under., , Rule 3., , Case 1., , Assume that p is false and show that q is true., , Case 2., , Assume that q is false and show that p is true., , Statements with ‘If .... then’:, Let p and q be two statements. Then, in order to prove that ‘p ⇒ q’ is, true, we need to prove any one of the following., , DIRECT METHOD, , Assume that p is true and prove that q is true., , CONTRAPOSITIVE METHOD, , Assume that q is false and prove that p is false.
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914, Rule 4., , Senior Secondary School Mathematics for Class 11, , Statements with ‘If and only if ’:, In order to prove that ‘p ⇒ q’, we need to show that:, (i) If p is true, then q is true., (ii) If q is true, then p is true., , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Prove that the following statement is true:, If x, y ∈ Z such that x and y are odd, then xy is odd., Let p: x, y ∈ Z such that x and y are odd., And q: xy is odd., Then, we have to prove that xy is odd., DIRECT METHOD, , We assume that p is true and show that q is true., p is true means x and y are odd integers. Then,, x = ( 2m + 1) for some integer m, and y = ( 2n + 1) for some integer n., ∴ xy = ( 2m + 1)( 2n + 1), = ( 4mn + 2m + 2n + 1), = 2( 2mn + m + n) + 1, which is clearly odd., Thus, p ⇒ q., Hence, the given statement is true., CONTRAPOSITIVE METHOD, , We will show that ~ q ⇒ ~ p., Here ~ q: It is false that both x and y are odd., This means at least one of x and y is even., Let x be even. Then, x = 2n for some integer n., ∴ xy = 2ny for some integer n., This shows that xy is even., Thus, ~ p is true [Q ~ p : xy is even]., ∴ ~ q ⇒ ~ p., So, the given statement is true., EXAMPLE 2, , SOLUTION, , Prove that the necessary and sufficient condition for an integer n to be, odd is that n2 is odd., Let p: The integer n is odd., And q: n2 is odd., First we prove that p ⇒ q., Let n be odd. Then, n = ( 2k + 1) for some integer k., Then, n2 ( 2k + 1) 2 = ( 4k 2 + 4k + 1) = 2( 2k 2 + 2k) + 1
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Mathematical Reasoning, , 915, , Thus, n2 is 1 more than an even number and therefore, it is odd., Thus, p ⇒ q., Now, in order to prove that q ⇒ p, it is sufficient to show that, ~ p ⇒ ~ q [Contrapositive method]., Clearly, ~ p: The integer n is even., Let n = 2k for some integer k., Then, n2 = 4k 2 , which is even., ∴ n is even ⇒ n2 is even., Consequently, ~ p ⇒ ~ q and therefore, q ⇒ p., Hence, p ⇔ q , i.e., the integer n is odd if and only if n2 is odd., PROVING RESULTS BY CONTRADICTION, , In order to check whether a given statement p is true, we assume that p is false, i.e., ~ p, is true., From this we arrive at a result which contradicts our assumption. Hence, we, conclude that p is true., By using the method of contradiction verify that p: 5 is irrational., a, If possible, let 5 be rational and let its simplest form be ⋅, b, Then, a and b are integers having no common factor other than 1 and, b ≠ 0., a, a2, Now, 5 = ⇒ 5 = 2 [on squaring both sides], b, b, ... (i), ⇒ 5 b2 = a 2, , EXAMPLE 3, PROOF, , ⇒ 5 divides a 2 [Q 5 divides 5b 2 ], ⇒ 5 divides a [Q 5 is prime and divides a 2 ⇒ 5 divides a]., Let a = 5 c for some integer C., Putting a = 5 c in (i), we get:, 5 b 2 = 25 c2 ⇒ b 2 = 5 c2, ⇒ 5 divides b 2 [Q 5 divides 5c 2 ], ⇒ 5 divides b [Q 5 is prime and divides b 2 ⇒ 5 divides b]., Thus, 5 is a common factor of a and b., But, this contradicts the fact that a and b have no common factor other, than 1., Hence, 5 is irrational., EXAMPLE 4, , PROOF, , By using the method of contradiction, prove that the sum of an irrational, number and a rational number is irrational., , Let a be irrational and b be rational.
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916, , Senior Secondary School Mathematics for Class 11, , Then, we have to prove that ( a + b) is irrational., If possible, let ( a + b) be rational. Then,, ( a + b) is rational, b is rational., ⇒ {( a + b) − b} is rational [Q difference of rationals is rational], ⇒ a is rational., This contradicts the fact that a is irrational., Since the contradiction arises by assuming that ( a + b) is rational,, hence ( a + b) is irrational., Using contradiction method, check the validity of the following statement:, If n is a real number with n > 3, then n2 > 9., Let n be a real number such that n > 3 and if possible, let n2 be not, greater than 9., Then, n2 ≤ 9., Let n2 = ( 9 − a), where a is a real number such that a ≥ 0., , EXAMPLE 5, , PROOF, , Then, n = n2 = 9 − a ≤ 3 [Q ( 9 − a) ≤ 9]., But, this is a contradiction since n > 3., Since the contradiction arises by assuming that n2 is not greater than 9,, therefore, n2 > 9., Hence, the given statement is valid., EXAMPLE 6, , SOLUTION, , Consider the statement:, p: If x a real number such that x 3 + 4x = 0, then x = 0., Prove that p is a true statement, using:, (i) direct method, (ii) method of contradiction, (iii) method of contrapositive, (i) Direct Method:, Let x 3 + 4x = 0, where x ∈ R. Then,, x 3 + 4x = 0 ⇒ x( x 2 + 4) = 0, ⇒ x = 0 [Q x 2 + 4 ≠ 0 for x ∈ R ], Hence, p is a true statement., (ii) Method of contradiction:, If possible, let x 3 + 4x = 0 and x ≠ 0. Then,, x( x 2 + 4) = 0 and x ≠ 0 ⇒ x 2 + 4 = 0., But, this is a contradiction, since x 2 + 4 ≠ 0 for x ∈ R., Since, the contradiction arises by assuming that, x 2 + 4 = 0 and x ≠ 0, so x = 0., Hence, x 3 + 4x = 0 ⇒ 0 is a true statement., (iii) Method of contrapositive:, We have to prove that x 3 + 4x = 0 ⇒ x = 0., Let p: x 3 + 4x = 0 and q: x = 0.
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Mathematical Reasoning, , 917, , We shall prove that ~ q ⇒ ~ p., Let x ≠ 0. Then, x 3 + 4x = x( x 2 + 4) ≠ 0 [Q x ≠ 0 and x 2 + 4 ≠ 0]., Thus, ~ q ⇒ ~ p and therefore, p ⇒ q., , EXERCISE 29D, 2, , 1. Let p: If x is an integer and x is even, then x is even., , Using the method of contrapositive, prove that p is true., 2. Consider the statement:, , q: For any real numbers a and b, a 2 = b 2 ⇒ a = b., By giving a counter-example, prove that q is false., 3. By giving a counter-example, show that the statement is false:, , p: If n is an odd positive integer, then n is prime., 4. Use contradiction method to prove that:, , p:, , 3 is irrational, , is a true statement., 5. By giving a counter-example, show that the following statement is false:, , p: If all the sides of a triangle are equal, then the triangle is obtuse angled., r
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30, Statistics, INTRODUCTION, , In earlier classes we have studied the methods of representing data, graphically and in tabular form., Mean, median and mode are three measures of central tendency, giving, us a rough idea about the points where data are centred., In order to have a better idea as to how the data are scattered, we make a, study of (i) mean deviation and (ii) standard deviation., , MEAN DEVIATIONS (MD), MEAN DEVIATION ABOUT THE MEAN The AM of the numerical deviations of the, observations from the mean of the data is called the mean deviation about the mean., MEAN DEVIATION ABOUT THE MEDIAN The AM of the numerical deviations of the, observations from the median of the data is called the mean deviation about the, median., Statistics, , MEAN DEVIATIONS FOR UNGROUPED DATA, , Let x1 , x 2 , x 3 , … , x n be the given n observations. Let x be the AM and M be the, median. Then,, n, , ∑ |x i − x|, i =1, (i) MD ( x) =, , where x = mean., n, n, , ∑ |x i − M|, i =1, (ii) MD ( M) =, , where M = median., n, SUMMARY, n, , ∑ |x i − x|, i =1, , where x = mean., (i) MD ( x) =, n, n, , ∑ |x i − M|, i =1, , where M = median., (ii) MD ( M) =, n, 918
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Statistics, , Find the mean deviation about the mean for the following data:, 15 , 17 , 10 , 13 , 7 , 18 , 9 , 6 , 14 , 11, , EXAMPLE 1, , SOLUTION, , Let the mean of the given data be x. Then,, n, , å xi, , 120, =, = 12 [Q n = 10 ]., n, 10, The values of ( x i - x ) are:, 3 , 5 , - 2 , 1, - 5 , 6 , - 3 , - 6 , 2 , - 1., x=, , i =1, , So, the values of|x i - x|are:, 3, 5, 2, 1, 5, 6, 3, 6, 2, 1., n, , \, , MD ( x ) =, , å |x i - x|, , i =1, , n, Hence, MD ( x ) = 3.4., , 34, = 3.4., 10, , Find the mean deviation about the median for the data given below:, 11, 3 , 8 , 7 , 5 , 14 , 10 , 2 , 9, , EXAMPLE 2, , SOLUTION, , =, , Arranging the given data in an ascending order, we get:, 2, 3, 5, 7, 8, 9, 10, 11, 14., Here n = 9, which is odd., 1, \ median = ( n + 1)th observation, 2, 1, = ( 9 + 1)th observation = 5th observation = 8., 2, Thus, M = 8., The values of ( x i - M ) are:, -6 , - 5 , - 3 , - 1, 0 , 1, 2 , 3 , 6 ., \, , 9, , å |x i - M| = ( 6 + 5 + 3 + 1 + 0 + 1 + 2 + 3 + 6 ) = 27, , i =1, , 9, , Þ, , MD ( M ) =, , å |x i - M|, , i =1, , Hence, MD ( M ) = 3., EXAMPLE 3, , SOLUTION, , 9, , =, , 27, = 3., 9, , Find the mean deviation about the median for the data given below., 45, 36, 50, 60, 53, 46, 51, 48, 72, 42, Arranging the given data in an ascending order, we get:, 36, 42, 45, 46, 48, 50, 51, 53, 60, 72., Here n = 10, which is even., , 919
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920, , Senior Secondary School Mathematics for Class 11, , 1 ⎧n, ⎫, ⎛n, ⎞, ⋅ ⎨ th observation + ⎜ + 1⎟ th observation ⎬, ⎝2, ⎠, 2 ⎩2, ⎭, 1, = (5th observation + 6th observation), 2, 1, 98, = ( 48 + 50) =, = 49., 2, 2, Thus, M = 49., , ∴, , median =, , The values of ( x i − M) are, −13 , − 7 , − 4, − 3 , − 1, 1, 2, 4, 11, 23., 10, , ∴, , ∑ |x i − M | = (13 + 7 + 4 + 3 + 1 + 1 + 2 + 4 + 11 + 23) = 69, , i =1, , 10, , MD ( M ) =, , ⇒, , ∑ |x i − M |, , i =1, , 10, , =, , 69, = 6.9., 10, , MEAN DEVIATION FOR DISCRETE FREQUENCY DISTRIBUTION, , Let the given data consist of n distinct values x1 , x 2 , … , x n occurring with, frequencies f1 , f 2 , … , f n respectively., (i) Mean deviation about mean:, , We have:, n, , x=, , n, , ∑ fixi, , i =1, n, , ⇒ x=, , ∑ fi, , ∑ fixi, , i =1, , N, , n, , , where N = ∑ f i ., i =1, , i =1, , n, , ∴ MD ( x) =, , ∑ f i|x i − x|, , i =1, , n, , ∑ fi, , n, , =, , ∑ f i |x i − x|, , i =1, , N, , ⋅, , i =1, , (ii) Mean deviation about median:, n, , Let N = ∑ f i . First we find the cumulative frequencies and then, identify the, i =1, , observation whose cumulative frequency is equal to or just greater than, This value gives median. Then,, n, , MD ( M) =, , ∑ f i|x i − M|, , i =1, , N, , ⋅, , N, ⋅, 2
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Statistics, , 921, , SUMMARY, , For discrete frequency distribution, we have, n, , å fi|xi - x|, n, i =1, , where å fi = N., (i) MD ( x ) =, i =1, N, n, , å fi|x i - M|, n, i =1, , where å fi = N., (ii) MD ( M ) =, i =1, N, EXAMPLE 4, , SOLUTION, , Find the mean deviation about the mean for the following data:, xi, , 3, , 5, , 7, , 9, , 11, , 13, , fi, , 6, , 8, , 15, , 25, , 8, , 4, , We have, 6, , N = å fi = ( 6 + 8 + 15 + 25 + 8 + 4 ) = 66 ., i =1, 6, , å fi x i, , ( 6 ´ 3 ) + ( 8 ´ 5 ) + (15 ´ 7 ) + ( 25 ´ 9 ) + ( 8 ´ 11) + ( 4 ´ 13 ), 66, 528, =, = 8., 66, Now, we prepare the table given below:, x=, , i =1, , =, , N, , 6, , å fi|xi - x|, , xi, , fi, , |xi - x|, , 3, , 6, , 5, , 30, , 5, , 8, , 3, , 24, , 7, , 15, , 1, , 15, , 9, , 25, , 1, , 25, , 11, , 8, , 3, , 24, , 13, , 4, , 5, , 20, , N = 66, , i =1, , sum = 138, , 6, , å fi|x i - x |, , 138 23, =, =, = 2 .09., N, 66 11, Hence, mean deviation about the mean = 2.09., , \, , EXAMPLE 5, , MD ( x ) =, , i =1, , Find the mean deviation about the median for the following data:, xi, , 3, , 5, , 7, , 9, , 11, , 13, , fi, , 6, , 8, , 15, , 3, , 8, , 4
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922, , Senior Secondary School Mathematics for Class 11, , SOLUTION, , We have, 6, , N = å fi = ( 6 + 8 + 15 + 3 + 8 + 4 ) = 44., i =1, , 6, , å fi x i, , ( 6 ´ 3 ) + ( 8 ´ 5 ) + (15 ´ 7 ) + ( 3 ´ 9 ) + ( 8 ´ 11) + ( 4 ´ 13 ), N, 44, (18 + 40 + 105 + 27 + 88 + 52 ) 330 15, =, =, =, = 7.5., 44, 44, 2, , x=, , i =1, , =, , xi, , 3, , 5, , 7, , 9, , 11, , 13, , fi, , 6, , 8, , 15, , 3, , 8, , 4, , cf, , 6, , 14, , 29, , 32, , 40, , 44, , \, , N = 44, which is even., 1 ìN, N, ü, \ median = × í th observation + æç + 1 ö÷th observation ý, 2 î2, è 2, ø, þ, 1, = {22nd observation + 23rd observation}, 2, 1, = (7 + 7 ) = 7 ., 2, Thus, M = 7., Now, we have:, , \, , |x i - M |, , 4, , 2, , 0, , 2, , 4, , 6, , fi, , 6, , 8, , 15, , 3, , 8, , 4, , fi|x i - M |, , 24, , 16, , 0, , 6, , 32, , 24, , 6, , å fi = 44 and, , i =1, , 6, , å fi|x i - M| = 102 ., , i =1, , 6, , å fi |x i - M|, , 102, =, = 2.32 ., N, 44, Hence, the mean deviation about the median is 2.32., , \, , MD ( M ) =, , i =1, , MEAN DEVIATION ABOUT THE MEAN AND MEDIAN FOR CONTINUOUS, FREQUENCY DISTRIBUTION, I. MEAN DEVIATION ABOUT THE MEAN FOR GROUPED DATA, Short Cut Method (Step-deviation method):, Step 1, , Find the class mark x i of each class., , Step 2, , Assume that the mean is A., , Step 3, , Find deviations, di =, , ( x i - A), , where h = class size., h
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Statistics, , 923, , Step 4, , ⎧ f d, ⎫, Calculate, x = A + ⎨ i i × h⎬, where N = ∑ f i ., ⎭, ⎩ N, , Step 5, , Calculate, MD ( x) =, , EXAMPLE 6, , SOLUTION, , ∑ f i |x i − x |, ⋅, N, , Find the mean deviation about the mean for the following data:, Marks, obtained, , 10−20, , 20−30, , 30−40, , 40−50, , 50−60, , 60−70, , Number of, students, , 8, , 6, , 12, , 5, , 2, , 7, , Here h = 10. Let the assumed mean be A = 35., Then, we prepare the table given below., , Marks Number of Midpoint d i =, obtained students ( f i ), xi, , x i − 35, 10, , f i di, , |x i − x | f i × |x i − x |, , 10−20, , 8, , 15, , −2, , −16, , 22, , 176, , 20−30, , 6, , 25, , −1, , −6, , 12, , 72, , 30−40, , 12, , 35 = A, , 0, , 0, , 2, , 24, , 40−50, , 5, , 45, , 1, , 5, , 8, , 40, , 50−60, , 2, , 55, , 2, , 4, , 18, , 36, , 60−70, , 7, , 65, , 3, , 21, , 28, , 196, , N = ∑ fi, = 40, , ∴, , ∑ f i di = 8, , ∑ f i × |x i − x |, = 544, , ⎧∑ f d, ⎫, ⎧8, ⎫, N = ∑ f i = 40, x = A + ⎨ i i × h⎬ = 35 + ⎨ × 10⎬ = 37., N, 40, ⎭, ⎭, ⎩, ⎩, ∑ f i × |x i − x | 544 136, MD ( x ) =, =, =, = 13.6., N, 40, 10, , II. MEAN DEVIATION ABOUT THE MEDIAN FOR GROUPED DATA, Short Cut Method:, , Step 1, , Find the class mark x i of each class., , Step 2, , Find N = ∑ f i ., , Step 3, , Now, median class is the class interval whose cumulative frequency is, greater than or equal to ( N/ 2).
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924, Step 4, , Step 5, , Senior Secondary School Mathematics for Class 11, , Calculate median by using the formula, ⎧, ⎫, ⎛N, ⎞, ⎜ − c⎟, ⎪⎪, ⎪⎪, ⎝ 2, ⎠, M = median = ⎨L +, × h⎬, where, f, ⎪, ⎪, ⎪⎩, ⎪⎭, L = lower limit of the median class;, f = frequency of the median class;, h = width of the median class;, andc = cumulative frequency of the class just preceding the, median class., ∑ f i |x i − M |, Calculate MD ( M) =, ⋅, N, , EXAMPLE 7, , Calculate the mean deviation about the median for the following data:, , Height (in cm), , 95−105, , Number of boys, , 9, , SOLUTION, , 105−115 115−125 125−135 135−145 145−155, 13, , 25, , 30, , 13, , 10, , First we find the median., Class, , Frequency f i, , Cumulative, frequency (cf ), , Midpoint x i, , 95−105, , 9, , 9, , 100, , 105−115, , 13, , 22, , 110, , 115−125, , 25, , 47, , 120, , 125−135, , 30, , 77, , 130, , 135−145, , 13, , 90, , 140, , 145−155, , 10, , 100, , 150, , N = ∑ f i = 100, Thus N = 100 and therefore,, ⇒, ⇒, ∴, , ∴, , N, = 50, 2, , median class is 125−135, L = 125 , f = 30, h = 10 and c = 47., ⎛N, ⎞, ⎜ − c⎟, ⎝ 2, ⎠, median = L +, ×h, f, (50 − 47), ⎧, ⎫, = ⎨125 +, × 10⎬ = (125 + 1) = 126., 30, ⎭, ⎩, M = 126.
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Statistics, , 925, , Now, we prepare the table given below., fi, , xi, , |x i − M|, , f i × |x i − M|, , 9, , 100, , 26, , 234, , 13, , 110, , 16, , 208, , 25, , 120, , 6, , 150, , 30, , 130, , 4, , 120, , 13, , 140, , 14, , 182, , 10, , 150, , 24, , 240, , N = 100, , 1134, , ∑ f i × |x i − M| 1134, =, = 11.34., N, 100, Hence, the mean deviation about the median is 11.34., , ∴, , MD ( M) =, , Calculate the mean deviation about the median for the following data:, , EXAMPLE 8, , Class, , 16−20, , 21−25, , 26−30, , 31−35, , 36−40, , 41−45, , 46−50, , 51−55, , Frequency, , 5, , 6, , 12, , 14, , 26, , 12, , 16, , 9, , SOLUTION, , Converting the given series into an exclusive series, we prepare the, table, given below:, Class, , Frequency ( f i ), , cf, , Midpoint ( x i ), , 15.5−20.5, , 5, , 5, , 18, , 20.5−25.5, , 6, , 11, , 23, , 25.5−30.5, , 12, , 23, , 28, , 30.5−35.5, , 14, , 37, , 33, , 35.5−40.5, , 26, , 63, , 38, , 40.5−45.5, , 12, , 75, , 43, , 45.5−50.5, , 16, , 91, , 48, , 50.5−55.5, , 9, , 100, , 53, , N = 100, N, = 50, 2, median class is 35.5−40.5, L = 35.5, f = 26, h = 5 and c = 37., ⎛N, ⎞, ⎜ − c⎟, ⎝ 2, ⎠, median = L +, ×h, f, , Thus N = 100 and therefore,, ⇒, ⇒, ∴
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926, , Senior Secondary School Mathematics for Class 11, , (50 − 37), ⎧, ⎫, = ⎨ 35.5 +, × 5 ⎬ = ( 35.5 + 2.5) = 38., 26, ⎭, ⎩, Thus, M = 38., Now, we prepare the table given below., fi, , xi, , |x i − M|, , f i × |x i − M|, , 5, , 18, , 20, , 100, , 6, , 23, , 15, , 90, , 12, , 28, , 10, , 120, , 14, , 33, , 5, , 70, , 26, , 38, , 0, , 0, , 12, , 43, , 5, , 60, , 16, , 48, , 10, , 160, , 9, , 53, , 15, , 135, , N = 100, , 735, , Thus, ∑ f i × |x i − M| = 735 and N = 100., ∑ f i × |x i − M| 735, =, = 7.35., ∴ MD ( M) =, N, 100, Hence, the mean deviation about the median is 7.35., , EXERCISE 30A, Find the mean deviation about the mean for the following data:, 1. 7, 8, 4, 13, 9, 5, 16, 18, 2. 39, 72, 48, 41, 43, 55, 60, 45, 54, 43, 3. 17, 20, 12, 13, 15, 16, 12, 18, 15, 19, 12, 11, Find the mean deviation about the median for the following data:, 4. 12, 5, 14, 6, 11, 13, 17, 8, 10, 5. 4, 15, 9, 7, 19, 13, 6, 21, 8, 25, 11, 6. 34, 23, 46, 37, 40, 28, 32, 50, 35, 44, 7. 70, 34, 42, 78, 65, 45, 54, 48, 67, 50, 56, 63, Find the mean deviation about the mean for the following data:, 8., , xi, , 6, , 12, , 18, , 24, , 30, , 36, , fi, , 5, , 4, , 11, , 6, , 4, , 6
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Statistics, , 9., , 10., , 927, , xi, , 2, , 5, , 6, , 8, , 10, , 12, , fi, , 2, , 8, , 10, , 7, , 8, , 5, , xi, , 3, , 5, , 7, , 9, , 11, , 13, , fi, , 6, , 8, , 15, , 25, , 8, , 4, , Find the mean deviation about the median for the following data:, 11., , 12., , 13., , xi, , 15, , 21, , 27, , 30, , 35, , fi, , 3, , 5, , 6, , 7, , 8, , xi, , 5, , 7, , 9, , 11, , 13, , 15, , 17, , fi, , 2, , 4, , 6, , 8, , 10, , 12, , 8, , xi, , 10, , 15, , 20, , 25, , 30, , 35, , 40, , 45, , fi, , 7, , 3, , 8, , 5, , 6, , 8, , 4, , 9, , Find the mean deviation about the mean for the following data:, 14., , 15., , Mark, , 0−10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , Number of students, , 6, , 8, , 14, , 16, , 4, , 2, , Height (in cm), , 95−105 105–115 115–125 125–135 135–145 145–155, , Number of boys, 16., , 9, , 16, , 23, , 30, , Class, , 30−40, , 40–50, , 50–60, , 60–70, , 70–80, , Frequency, , 3, , 7, , 12, , 15, , 8, , 12, , 10, , 80–90 90−100, 3, , 2, , Find the mean deviation about the median for the following data:, 17., , 18., , Class, , 0−10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , Frequency, , 6, , 7, , 15, , 16, , 4, , 2, , Class, , 0−10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , Frequency, , 6, , 8, , 11, , 18, , 5, , 2
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928, , Senior Secondary School Mathematics for Class 11, ANSWERS (EXERCISE 30A), , 1. 4.25, , 2. 8.2, , 3. 2.5, , 4. 3, , 5. 5.36, , 6. 6.5, , 7. 10.5, , 8. 8, , 9. 2.3, , 10. 2.09, , 11. 5.1, , 12. 2.72, , 13. 10.1, , 14. 10.24, , 15. 11.6, , 16. 11.36, , 17. 10.16, , 18. 10.8, , VARIANCE AND STANDARD DEVIATION, Mean of the squares of the deviations from the mean is called the variance,, to be denoted by s 2 ., 1 n, The variance of n observations x 1 , x 2 , ¼, x n is given by s 2 = × å ( x i - x )2 ., n i =1, , VARIANCE, , STANDARD DEVIATION The positive square root of variance is called standard, deviation and it is denoted by s ., 1 n, s=, × å ( x i - x )2 ., n i =1, , Find the mean, variance and the standard deviation for the following data:, 5, 9, 8, 12, 6, 10, 6, 8, , EXAMPLE 1, , SOLUTION, , Here n = 8., 1, 64, Mean, x = ( 5 + 9 + 8 + 12 + 6 + 10 + 6 + 8 ) =, = 8., 8, 8, The values of ( x i - x ) are, -3 , 1, 0 , 4 , - 2 , 2 , - 2 , 0., ( x i - x )2 = ( 9 + 1 + 0 + 16 + 4 + 4 + 4 + 0 ) = 38 ., å( x i - x )2 38 19, \ variance, s 2 =, =, =, = 4.75., n, 8, 4, and, standard deviation, s = 4.75 = 2.17., \ mean = 8, variance = 4.75 and standard deviation = 2.17., , \, , Short Cut Method, , For simple ungrouped data, we have, 1 ì n, ü, × í å ( x i - x )2 ý, n îi = 1, þ, 1 ì n 2, ü, = × í å ( x i - 2 x i x + x 2 )ý, n îi = 1, þ, å x i2, å xi ö n x 2, æ, =, - 2x ç, ÷+, n, n, è n ø, , s2 =
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Statistics, , =, , 929, , ∑ x i2, ∑ x i2, − 2 x ⋅ x + x2 =, − x 2., n, n, , Thus, for short cut method, we have, ∑ x i2, σ2 =, − x 2., n, EXAMPLE 2, , SOLUTION, , Find the mean, standard deviation and variance of first n natural, numbers., First n natural numbers are 1, 2, 3, …, n., (1 + 2 + 3 + … + n) 1 1, 1, Mean, x =, = ⋅ n(n + 1) = (n + 1), n, n 2, 2, 1, ⎡, ⎤, ⎢⎣Q (1 + 2 + 3 + … + n) = 2 n(n + 1) ⎥⎦, , ∑ x i2, − x2, n, 2, ∑ n2 ⎧ 1, ⎫, =, − ⎨ (n + 1) ⎬, n, ⎭, ⎩2, 1 n(n + 1)( 2n + 1) 1, = ⋅, − (n + 1) 2, n, 6, 4, 1, ⎡, ⎤, 2, ⎢⎣Q ∑ n = 6 n(n + 1)( 2n + 1) ⎥⎦, ⎧(n + 1)( 2n + 1) (n + 1) 2 ⎫, =⎨, −, ⎬, 6, 4 ⎭, ⎩, ⎧( 2n + 1) (n + 1) ⎫, = (n + 1) ⋅ ⎨, −, ⎬, 4 ⎭, ⎩ 6, (n + 1)(n − 1) (n2 − 1), =, =, ⋅, 12, 12, (n2 − 1), ∴ variance, σ 2 =, ⋅, 12, n2 − 1 1, n2 − 1, Standard deviation, σ =, = ⋅, ⋅, 12, 2, 3, ∴, , EXAMPLE 3, SOLUTION, , variance, σ 2 =, , Find the mean, standard deviation and variance of first 10 multiples of 3., First 10 multiples of 3 are 3, 6, 9, 12, …, 30., This is an AP with a = 3 , n = 10 and l = 30., n, 10, ∴ sum = ( a + l) =, ⋅ ( 3 + 30) = 165., 2, 2, 165, = 16.5., ∴ mean, x =, 10
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930, , Senior Secondary School Mathematics for Class 11, , ∑ x i2, − ( x) 2, n, ⎧ 3 2 + 62 + 92 + … + 302, ⎫, =⎨, − (16.5) 2 ⎬, 10, ⎭, ⎩, 3 2 × {12 + 22 + 3 2 + … + 102}, =, − (16.5) 2, 10, 9 × 10 × (10 + 1)( 2 × 10 + 1), =, − (16.5) 2, 6 × 10, n(n + 1)( 2n + 1) ⎤, ⎡, 2, ⎢⎣Q ∑ n =, ⎥⎦, 6, , Variance, σ 2 =, , ⎛ 9 × 10 × 11 × 21, ⎞ ⎛ 693, ⎞, − 272.25⎟ = ⎜, =⎜, − 272.25⎟, ⎠, 6 × 10, ⎝, ⎠ ⎝ 2, = ( 346.5 − 272.25) = 74.25., ∴ variance, σ 2 = 74.25., Standard deviation, σ = 74.25 = 8.61., VARIANCE AND, DISTRIBUTION, , STANDARD, , DEVIATION, , OF, , A, , DISCRETE, , FREQUENCY, , Let the given frequency distribution be:, Variable, , x1, , x2, , x3, , …, , xn, , Frequency, , f1, , f2, , f3, , …, , fn, , Then, we define:, (i) Variance, σ 2 =, , n, 1 n, ⋅ ∑ f i ( x i − x) 2 , where N = ∑ f i, i =1, N i =1, , (ii) Standard deviation, σ =, , 1, ⋅ ∑ f i ( x i − x) 2, N, , SUMMARY, , Variance, σ 2 =, , 1 n, ⋅ ∑ f i ( x i − x) 2, N i =1, , Standard deviation, σ =, , EXAMPLE 4, , 1 n, ⋅ ∑ f i ( x i − x) 2, N i =1, , Find the variance and standard deviation for the following data:, xi, , 10, , 15, , 18, , 20, , 25, , fi, , 3, , 2, , 5, , 8, , 2
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Statistics, , 931, , We may prepare the table given below:, , SOLUTION, , ( x i − x) = ( x i − 18) ( x i − x) 2, , f i ( x i − x) 2, , xi, , fi, , fi xi, , 10, , 3, , 30, , −8, , 64, , 192, , 15, , 2, , 30, , −3, , 9, , 18, , 18, , 5, , 90, , 0, , 0, , 0, , 20, , 8, , 160, , 2, , 4, , 32, , 7, , 49, , 98, , 25, , 2, 50, ∑ f i x i 360, x = Mean =, =, = 18, ∑ fi, 20, ∴, , 5, , ∑ f i ( x i − x) 2, = 340, 5, , N = ∑ f i = 20, ∑ f i ( x i − x) 2 = 340., i =1, , i =1, , 1, ⎛ 1, ⎞, ∴ variance, σ =, ⋅ ∑ f i ( x i − x) 2 = ⎜, × 340⎟ = 17., ⎝ 20, ⎠, N, Standard deviation, σ = 17 = 412, . ., 2, , Short Cut Method:, , Let A be the assumed mean and let d i = ( x i − A)., ⎧ ∑ f d 2 ⎛ ∑ f i di ⎞ 2 ⎫, Then, σ 2 = ⎨ i i − ⎜, ⎟ ⎬⋅, ⎝ N ⎠ ⎭⎪, ⎩⎪ N, EXAMPLE 5, , SOLUTION, , Find the mean, variance and standard deviation for the following data, using short cut method:, xi, , 60, , 61, , 62, , 63, , 64, , 65, , 66, , 67, , 68, , fi, , 2, , 1, , 12, , 29, , 25, , 12, , 10, , 4, , 5, , Let the assumed mean be A = 64., Then, we may present the table given below:, xi, , fi, , 60, 61, 62, 63, 64 = A, 65, 66, 67, 68, , 2, 1, 12, 29, 25, 12, 10, 4, 5, 100, , d i = ( x i − A), = ( x i − 64), , d i2, , f i di, , f i d i2, , −4, −3, −2, −1, 0, 1, 2, 3, 4, , 16, 9, 4, 1, 0, 1, 4, 9, 16, , −8, −3, −24, −29, 0, 12, 20, 12, 20, 0, , 32, 9, 48, 29, 0, 12, 40, 36, 80, 286
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932, , Senior Secondary School Mathematics for Class 11, , ∴, , N = ∑ f i = 100, ∑ f i d i = 0 and, , ∴, , mean = A +, , Variance, σ 2 =, , ∑ f i d i2, , ∑ f i di ⎛, 0 ⎞, = ⎜ 64 +, ⎟ = 64., ⎝, ∑ fi, 100⎠, , ∑ f i d i2 ⎛ ∑ f i d i ⎞, −⎜, ⎟, ∑ fi, ⎝ ∑ fi ⎠, , 2, , 2, , =, ∴, , 286 ⎛ 0 ⎞, 286, −⎜, = 2.86., ⎟ =, 100 ⎝ 100⎠, 100, , standard deviation, σ = 2.86 = 1.69., , When Mean is a Decimal Fraction, , In this case, we use the formulae given below., 2, n, 1 ⎧, ⎛ n, ⎞ ⎫, (i) Variance, σ 2 = 2 ⋅ ⎨N ⋅ ∑ f i x i2 − ⎜ ∑ f i x i ⎟ ⎬, ⎝ i =1, ⎠ ⎭⎪, i =1, N ⎩⎪, (ii) Standard deviation, σ =, EXAMPLE 6, , SOLUTION, , n, 1, ⎛, ⋅ N ⋅ ∑ f i x i2 − ⎜, ⎝, i =1, N, , n, ⎞, ∑ fi xi⎟, ⎠, i =1, , 2, , Find the standard deviation for the following data:, xi, , 3, , 8, , 13, , 18, , 23, , fi, , 6, , 10, , 14, , 10, , 10, , We have, Mean, x =, , ∑ f i x i ( 18 + 80 + 182 + 180 + 230) 690, =, =, = 13.8., ∑ fi, 50, 50, , This value being in decimal form, the calculation will become, tedious., 1, So, we use the formula, σ =, ⋅ N ⋅ ∑ f i x i2 − ( ∑ f i ) 2 ., N, Now, we prepare the table given below:, xi, , fi, , fi xi, , x i2, , f i x i2, , 3, , 6, , 18, , 9, , 54, , 8, , 10, , 80, , 64, , 640, , 13, , 14, , 182, , 169, , 2366, , 18, , 10, , 180, , 324, , 3240, , 23, , 10, , 230, , 529, , 5290, , 50, , 690, , 11590
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Statistics, , 933, , ∴, , N = ∑ f i = 50, ∑ f i x i = 690 and ∑ f i x i2 = 11590, 1, ⋅ { N ⋅ ∑ f i x i2 − ( ∑ f i x i ) 2 }, ∴ standard deviation, σ =, N, 1, 1, ⋅ 50 × 11590 − ( 690) 2 =, × 579500 − ( 690) 2, ⇒ σ=, 50, 50, 1, 1, 321.5, =, 579500 − 476100 =, × 103400 =, = 6.43., 50, 50, 50, Hence, σ = 6.43., Shorter Method For Standard Deviation of Continuous Frequency Distribution, , Let A be the assumed mean and let y i =, , ( x i − A), ⋅, h, , Then, we have:, Variance, σ 2 =, , h2, N2, , 2, n, ⎧, ⎛ n, ⎞ ⎫, ⋅ ⎨N ⋅ ∑ f i y i2 − ⎜ ∑ f i y i ⎟ ⎬, ⎝i =1, ⎠ ⎭⎪, i =1, ⎪⎩, , Standard deviation, σ =, , n, h, ⎛ n, ⎞, ⋅ N ⋅ ∑ f i y i2 − ⎜ ∑ f i y i ⎟, ⎝i =1, ⎠, i =1, N, , Calculate mean, variance and standard deviation for the following, frequency distribution:, , EXAMPLE 7, , Class, , 0−30, , 30−60, , 60−90, , Frequency, , 2, , 3, , 5, , SOLUTION, , 2, , 90−120 120−150 150−180 180−210, 10, , 3, , 5, , 2, , y i2, , fi yi, , f i y i2, , Here h = 30. Let the assumed mean be A = 105., Now, we prepare the table given below., Midpoint y i =, xi, , ( x i − 105), 30, , Class, , Frequency, fi, , 0−30, , 2, , 15, , −3, , 9, , −6, , 18, , 30−60, , 3, , 45, , −2, , 4, , −6, , 12, , 60−90, , 5, , 75, , −1, , 1, , −5, , 5, , 90−120, , 10, , 105 = A, , 0, , 0, , 0, , 0, , 120−150, , 3, , 135, , 1, , 1, , 3, , 3, , 150−180, , 5, , 165, , 2, , 4, , 10, , 20, , 180−210, , 2, , 195, , 3, , 9, , 6, , 18, , Total, , N = 30, , 2, , 76
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934, , Senior Secondary School Mathematics for Class 11, , ∴, , A = 105 , h = 30, N = ∑ f i = 30, ∑ f i y i = 2 and ∑ f i y i2 = 76., ∑ fi yi, 2, ⎛, ⎞, ⎛, ⎞, ∴ x = ⎜A +, × h⎟ ⇒ x = ⎜105 +, × 30⎟ = (105 + 2) = 107., ⎝, ⎠, ⎝, ⎠, N, 30, Thus, mean = 107., h2, Variance, σ 2 = 2 ⋅ {N ⋅ ∑ f i y i2 − ( ∑ f i y i ) 2}, N, ( 30) 2, =, ⋅ { 30 × 76 − ( 2) 2} = ( 2280 − 4) = 2276., ( 30) 2, Standard deviation, σ = 2276 = 47.71., EXAMPLE 8, , Given below are the diameters of circles (in mm) drawn in a design., Diameter, , 33−36, , 37−40, , 41−44, , 45−48, , 49−52, , Number of circles, , 15, , 17, , 21, , 22, , 25, , Calculate the mean diameter of the circles, variance and standard, deviation., SOLUTION, , Converting the given series into an exclusive series, we prepare the, table given below., Midpoint y i = ( x i − 42.5), 4, xi, , Class, , Frequency, fi, , 32.5−36.5, , 15, , 34.5, , 36.5−40.5, , 17, , 40.5−44.5, , y i2, , fi yi, , f i y i2, , −2, , 4, , −30, , 60, , 38.5, , −1, , 1, , −17, , 17, , 21, , 42.5 = A, , 0, , 0, , 0, , 0, , 44.5−48.5, , 22, , 46.5, , 1, , 1, , 22, , 22, , 48.5−52.5, , 25, , 50.5, , 2, , 4, , 50, , 100, , 25, , 199, , N = 100, ∴, , A = 42.5, h = 4, N = ∑ f i = 100, ∑ f i y i = 25 and ∑ f i y i2 = 199., ∑ fi yi, 25, ⎛, ⎞, ⎛, ⎞, × h⎟ ⇒ x = ⎜ 42.5 +, ∴ x = ⎜A +, × 4⎟ = 43.5., ⎝, ⎠, ⎝, ⎠, N, 100, Thus, mean = 43.5 mm., h2, Variance, σ 2 = 2 ⋅ {N ⋅ ∑ f i y i2 − ( ∑ f i y i ) 2}, N, 16, 16, =, × {100 × 199 − ( 25) 2} =, × (19900 − 625), 10000, 10000, ⎛ 16, ⎞ 3084, =⎜, × 19275⎟ =, = 30.84., ⎝ 10000, ⎠, 100, Standard deviation, σ = 30.84 = 5.55., Hence, mean = 43.5 mm, variance = 30.84, and, standard deviation = 5.55.
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Statistics, , 935, , EXERCISE 30B, 1. Find the mean, variance and standard deviation for the numbers 4, 6, 10, 12,, 7, 8, 13, 12., 2. Find the mean, variance and standard deviation for first six odd natural, numbers., Using short cut method, find the mean, variance and standard deviation for the data:, 3., , 4., , 5., , 6., , 7., , 8., , 9., , xi, , 4, , 8, , 11, , 17, , 20, , 24, , 32, , fi, , 3, , 5, , 9, , 5, , 4, , 3, , 1, , xi, , 6, , 10, , 14, , 18, , 24, , 28, , 30, , fi, , 2, , 4, , 7, , 12, , 8, , 4, , 3, , xi, , 10, , 15, , 18, , 20, , 25, , fi, , 3, , 2, , 5, , 8, , 2, , xi, , 92, , 93, , 97, , 98, , 102, , 104, , 109, , fi, , 3, , 2, , 3, , 2, , 6, , 3, , 3, , Class, , 0−10, , 10−20, , 20−30, , 30−40, , 40−50, , Frequency, , 5, , 8, , 15, , 16, , 6, , Class, , 30−40, , 40−50, , 50−60, , 60−70, , 70−80, , Frequency, , 3, , 7, , 12, , 15, , 8, , 80−90 90−100, 3, , 2, , Class, , 25−35, , 35−45, , 45−55, , 55−65, , 65−75, , Frequency, , 64, , 132, , 153, , 140, , 51, , ANSWERS (EXERCISE 30B), , 1. Mean = 9, Variance = 9.25 and SD = 3.04, 2. Mean = 6, Variance = 11.67 and SD 3.41, 3. Mean = 14, Variance = 45.8 and SD = 6.77, 4. Mean = 19, Variance = 43.4 and SD = 6.59, 5. Mean = 18, Variance = 17 and SD = 4.12, 6. Mean = 100, Variance = 29.09 and SD = 5.39, 7. Mean = 27, Variance = 132 and SD = 11.49, 8. Mean = 62, Variance = 201 and SD 14.17, 9. Mean = 49.67, Variance = 135.44 and SD = 11.64
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936, , Senior Secondary School Mathematics for Class 11, MISCELLANEOUS WORD PROBLEMS, , EXAMPLE 1, , SOLUTION, , The mean and variance of seven observations are 8 and 16 respectively. If, five of these are 2, 4, 10, 12, 14, find the remaining two observations., Let the remaining two observations be x and y. Then,, 2 + 4 + 10 + 12 + 14 + x + y, mean = 8 ⇒, =8, 7, ⇒ 42 + x + y = 56, ... (i), ⇒ x + y = 14., Also, variance = 16, 1 2, ⇒, ( 2 + 42 + 102 + 122 + 142 + x 2 + y 2 ) − 82 = 16, 7, ⎡, ⎤, ∑ x i2, 2, − ( x) 2 ⎥, ⎢Q σ =, n, ⎣, ⎦, 1, 2, 2, ( 460 + x + y ) = 80, ⇒, 7, ⇒ 460 + x 2 + y 2 = 560, ⇒, , x 2 + y 2 = 100., , ... (ii), , Now, ( x + y) 2 + ( x − y) 2 = 2( x 2 + y 2 ), ⇒ ( x − y) 2 = 2( x 2 + y 2 ) − ( x + y) 2, ⇒ ( x − y) 2 = ( 2 × 100) − (14) 2 = ( 200 − 196) = 4, ⇒ x − y = ±2., Now, x + y = 14, x − y = 2 ⇒ x = 8, y = 6;, x + y = 14 , x − y = −2 ⇒ x = 6, y = 8., Hence, the remaining two observations are 6 and 8., EXAMPLE 2, , SOLUTION, , The mean and variance of six observations are 8 and 16 respectively. If, each observation is multiplied by 3, find the new mean and new variance, of the resulting observations., Let the given observations be x1 , x 2 , x 3 , x 4 , x5 , x 6., 1, Then, mean = 8 ⇒ ( x1 + x 2 + x 3 + x 4 + x5 + x 6) = 8, 6, ⇒ x1 + x 2 + x 3 + x 4 + x5 + x 6 = 48., , ... (i), , Also, variance = 16, ⇒, , 1 2, ( x1 + x 22 + x 32 + x 42 + x52 + x 62 ) − 8 2 = 16, 6, , ⇒, , x12 + x 22 + x 32 + x 42 + x52 + x 62 = 480., , ⎡, ⎤, ∑ x i2, 2, − ( x) 2 ⎥, ⎢Q σ =, n, ⎣, ⎦, ... (ii), , When each observation is multiplied by 3, then new observations are, 3 x1 , 3 x 2 , 3 x 3 , 3 x 4 , 3 x5 and 3 x 6.
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Statistics, , 937, , 1, ( 3 x1 + 3 x 2 + 3 x 3 + 3 x 4 + 3 x5 + 3 x 6), 6, 3, ⎛1, ⎞, = ( x1 + x 2 + x 3 + x 4 + x5 + x 6) = ⎜ × 48⎟ = 24 [using (i)], ⎝2, ⎠, 6, 2, 2, 2, 2, ( 3 x1 ) + ( 3 x 2 ) + ( 3 x 3 ) + ( 3 x 4 ) + ( 3 x5 ) 2 + ( 3 x 6) 2, ∴ new variance =, 6, − ( 24) 2, 9 2, = ( x1 + x 22 + x 32 + x 42 + x52 + x 62 ) − 576, 6, ⎛9, ⎞, [using (ii)]., = ⎜ × 480⎟ − 576 = (720 − 576) = 144, ⎝6, ⎠, Hence, new mean = 24 and new variance = 144., ∴, , EXAMPLE 3, , new mean =, , The mean and standard deviation of 20 observations are found to be 10, and 2 respectively. On rechecking it was found that an observation 8 was, incorrect. Calculate the correct mean and standard deviation when, (i) the wrong item is omitted,, (ii) the wrong item is replaced by 12., 20, , SOLUTION, , Mean = 10 ⇒, ⇒, , ∑ xi, , i =1, , 20, , = 10, , 20, , ∑ x i = 200., , i =1, , SD = 2 ⇒ σ = 4, 2, , 20, , ⇒, , ∑ x i2, , i =1, , 20, , − (10) 2 = 4, , 20, , ⇒ ∑ x i2 = 2080., i =1, , ⎛ 20 ⎞, ⎛ 20 ⎞, Thus, incorrect ⎜ ∑ x i2 ⎟ = 200 and incorrect ⎜ ∑ x i2 ⎟ = 2080., ⎝i =1 ⎠, ⎝i =1 ⎠, CASE (i), , When the wrong item is omitted, , On omitting 8, we are left with 19 observations., ∴, , ⎛ 20 ⎞, ⎛ 19 ⎞, correct ⎜ ∑ x i ⎟ = incorrect ⎜ ∑ x i ⎟ − 8, ⎝i =1 ⎠, ⎝i =1 ⎠, , = ( 200 − 8) = 192., ⎛ 19 ⎞, Thus, correct ⎜ ∑ x i ⎟ = 192, ⎝i =1 ⎠, ∴, , correct mean =, , 192, = 10.105., 19, , ... (i)
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938, , Senior Secondary School Mathematics for Class 11, , ⎛ 19 ⎞, ⎛ 20 ⎞, Also, correct ⎜ ∑ x i2 ⎟ = incorrect ⎜ ∑ x i2 ⎟ − 64, ⎝i =1 ⎠, ⎝i =1 ⎠, ∴, , = ( 2080 − 64) = 2016., 19, 1 ⎛, 2⎞, 2, correct variance =, ⎜ correct ∑ x i ⎟ − (correct mean), ⎠, i =1, 19 ⎝, 2, , ∴, , ∴, , ⎛1, ⎞ ⎛ 192⎞, ⎛ 2016 36864⎞, = ⎜ × 2016⎟ − ⎜, −, ⎟ =⎜, ⎟, ⎝ 19, ⎠ ⎝ 19 ⎠, ⎝ 19, 361 ⎠, ( 38304 − 36864) 1440, =, =, ⋅, 361, 361, 1440 12 10 (12 × 3.162), correct SD =, =, =, 361, 19, 19, 37.944, =, = 1.997., 19, mean = 10.105 and SD = 1.997., , CASE (ii), , When incorrect observation 8 is replaced by 12., , ⎛ 20 ⎞, Incorrect ⎜ ∑ x i ⎟ = 200, ⎝i =1 ⎠, ⎛ 20 ⎞, correct ⎜ ∑ x i ⎟ = ( 200 − 8 + 12) = 204, ⎝i =1 ⎠, 204, ⇒ correct mean =, = 10.2., 20, ⎛ 20 ⎞, Also, incorrect ⎜ ∑ x i2 ⎟ = 2080, ⎝i =1 ⎠, ⇒, , ⇒, , ⎛ 20 ⎞, correct ⎜ ∑ x i2 ⎟ = ( 2080 − 82 + 122 ) = 2160, ⎝i =1 ⎠, , ⎛ 20 ⎞, correct ⎜ ∑ x i2 ⎟, ⎝i =1 ⎠, ⇒ correct variance =, − (correct mean) 2, 20, ⎧ 2160, ⎫, =⎨, − (10.2) 2 ⎬ = (108 − 104.04) = 3.96., 20, ⎭, ⎩, ⇒ correct SD = 3.96 = 1.989, Thus, in this case, we have mean = 10.2 and SD = 1.989., EXAMPLE 4, , The mean and standard deviation of 100 observations were calculated as, 40 and 5.1 respectively by a student who took by mistake 50 instead of 40, for one observation. What are the correct mean and standard deviation?, 100, , SOLUTION, , Mean = 40 ⇒, , ∑ xi, , i =1, , 100, 100, , = 40, , ⇒ ∑ x i = 4000., i =1
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Statistics, , 939, , SD = 5.1 ⇒ σ 2 = (5.1) 2, 100, , ⇒, , ∑ x i2, , i =1, , 100, , − ( 40) 2 = 26.01, , 100, , ⇒ ∑ x i2 = 162601., i =1, , ⎛ 100 ⎞, ⎛ 100 ⎞, Thus, incorrect ⎜ ∑ x i ⎟ = 4000 and incorrect ⎜ ∑ x i2 ⎟ = 162601, ⎝i =1 ⎠, ⎝i =1 ⎠, ⎛ 100 ⎞, Now, incorrect ⎜ ∑ x i ⎟ = 4000, ⎝i =1 ⎠, ⇒, ⇒, , ⎛ 100 ⎞, correct ⎜ ∑ x i ⎟ = ( 4000 − 50 + 40) = 3990, ⎝i =1 ⎠, 3990, correct mean =, = 39.9., 100, , ... (i), , ⎛ 100 ⎞, And, incorrect ⎜ ∑ x i2 ⎟ = 162601, ⎝i =1 ⎠, ⇒, , ⇒, , ⎛ 100 ⎞, correct ⎜ ∑ x i2 ⎟ = {162601 − (50) 2 + ( 40) 2} = 161701, ⎝i =1 ⎠, ⎛ 100 ⎞, correct ⎜ ∑ x i2 ⎟, ⎝i =1 ⎠, correct variance =, − (correct mean) 2, 100, ⎧161701, ⎫, =⎨, − ( 39.9) 2 ⎬ = {1617.01 − ( 40 − 0.1) 2}, ⎭, ⎩ 100, = (1617.01) − {1600 + 0.01 − 8}, = (1617.01 − 1592.01) = 25, , ⇒, , correct SD = 25 = 5., , Hence, correct mean = 39.9 and correct SD = 5., EXAMPLE 5, , SOLUTION, , If each of the observations x1 , x 2 , x 3 , … , x n is increased by an amount a,, where a is a negative or positive number, then show that the variance, remains unchanged., Let x be the mean of x1 , x 2 , x 3 , … , x n . Then,, 1, x = ( x1 + x 2 + x 3 + … + x n )., n, Let y i = x i + a for each i = 1, 2, 3 , … , n., Let y be the mean of y1 , y 2 , y 3 , … , y n . Then,, 1, y = ( y1 + y 2 + y 3 + … + y n ), n
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940, , Senior Secondary School Mathematics for Class 11, , 1, ( x1 + a + x 2 + a + x 3 + a + … + x n + a), n, 1, 1, = ( x1 + x 2 + x 3 + … + x n ) + ( a + a + a + … n times), n, n, 1, = x + (na) = ( x + a)., n, y = ( x + a)., =, , ∴, , Now, the variance of the new observations is given by, variance ( y) = σ 2 =, , 1 n, ⋅ ∑ ( y i − y) 2, n i =1, , 1 n, ⋅ ∑ {( x i + a) − ( x + a)} 2, n i =1, [Q y i = ( x i + a), 1 n, = ⋅ ∑ ( x i + a − x − a) 2, n i =1, 1 n, = ⋅ ∑ ( x i − x) 2, n i =1, =, , and y = ( x + a)], , = variance ( x)., Hence, the variance of the new observations is the same as the, variance of the original observations., EXAMPLE 6, , SOLUTION, , If the mean and variance of the observations x1 , x 2 , x 3 , … , x n are x and, σ 2 respectively and a be a nonzero real number, then show that the mean, and variance of ax1 , ax 2 , ax 3 , … , ax n are ax and a 2 σ 2 respectively., Let x be the mean of x1 , x 2 , x 3 , … , x n and a be a nonzero real number., 1, Then, x = ( x1 + x 2 + x 3 + … + x n )., n, Let y i = ax i for each i = 1, 2, 3 , … , n. Then,, y=, , 1, ( y1 + y 2 + y 3 + … + y n ), n, , 1, 1, ( ax1 + ax 2 + ax 3 + … + ax n ) = a ⋅ ( x1 + x 2 + x 3 + … + x n ) = ax., n, n, Thus, y = ax., Now, the variance of new observations is given by, =, , variance ( y) = σ12, 1 n, ⋅ ∑ ( y i − y) 2, n i =1, 1 n, = ⋅ ∑ ( ax i − ax) 2 [Q y i = ax i for each i and y = ax ], n i =1, =
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Statistics, , 941, , 1 n, ⋅ ∑ ( xi − x ) 2, n i =1, = a 2 ⋅ {variance ( x)} = a 2 σ 2 ., new variance = a 2 σ 2 ., = a2 ⋅, , ∴, , σ1 = a 2 σ 2 = |a|⋅ σ., , REMARK, , EXERCISE 30C, 1. If the standard deviations of the numbers 2, 3, 2x, 11 is 3.5, calculate the, possible values of x., 2. The variance of 15 observations is 6. If each observation is increased by 8,, find the variance of the resulting observations., 3. The variance of 20 observations is 5. If each observation is multiplied by 2,, find the variance of the resulting observations., 4. The mean and variance of five observations are 6 and 4 respectively. If three, of these are 5, 7 and 9, find the other two observations., 5. The mean and variance of five observations are 4.4 and 8.24 respectively. If, three of these are 1, 2 and 6, find the other two observations., 6. The mean and standard deviation of 18 observations are found to be 7 and 4, respectively. On rechecking it was found that an observation 12 was, misread as 21. Calculate the correct mean and standard deviation., 7. For a group of 200 candidates, the mean and standard deviations of scores, were found to be 40 and 15 respectively. Later on it was discovered that the, score of 43 was misread as 34. Find the correct mean and standard, deviation., 8. The mean and standard deviations of a group of 100 observations were, found to be 20 and 3 respectively. Later on it was found that three, observations 21, 12 and 18 were incorrect. Find the mean and standard, deviation if the incorrect observations were omitted., ANSWERS (EXERCISE 30C), , 1. 3 and, , 7, 3, , 5. 4 and 9, , 2. 6, , 3. 20, , 4. 3 and 6, , 6. 6.5, 2.5, , 7. 40.045, 14.995, , 8. 20.06, 14.4, , HINTS TO SOME SELECTED QUESTIONS, 1. σ 2 =, , ∑ xi2, − ( x ) 2 ⇒ 3 x 2 − 16 x + 21 = 0., n
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942, , Senior Secondary School Mathematics for Class 11, , COEFFICIENT OF VARIATION, COEFFICIENT OF VARIATION, , Let the mean and standard deviation of the given data be x and σ respectively., Then, the coefficient of variation is defined as, ⎛σ, ⎞, CV = ⎜ × 100⎟ , where x ≠ 0., ⎝x, ⎠, For comparing the dispersion of two series, we calculate the coefficient of, variation for each series., The series whose CV is more, is more variable than the other., COMPARISON OF TWO FREQUENCY DISTRIBUTIONS WITH SAME MEAN, , Let there be two frequency distributions with the same mean x and having, standard deviations σ1 and σ 2 respectively., ⎛σ, ⎞, Then, CV for 1st distribution = ⎜ 1 × 100⎟ %., ⎝ x, ⎠, ⎛σ, ⎞, And, CV for 2nd distribution = ⎜ 2 × 100⎟ %,, ⎝ x, ⎠, σ1, , where CV1 and CV2 denote the CV for 1st and 2nd, σ2, distributions respectively., Thus,, , CV1, , CV2, , =, , Thus, in two series having the same mean, the one having more standard, deviation is more variable., EXAMPLE 1, , The following results show the number of workers and the wages paid to, them in two factories A and B., Factory, , Number of workers, Mean wages, Variance of distribution, of wages, , A, , B, , 4000, , 5000, , Rs 3500, , Rs 3500, , 64, , 81, , Which factory has more variation in wages?, SOLUTION, , We have:, Variance of distribution of wages in factory A , σ12 = 64., Variance of distribution of wages in factory B, σ 22 = 81., ∴ ( σ12 = 64 and σ 22 = 81) ⇒ ( σ1 = 8 and σ 2 = 9)., Thus, SD of distribution of wages in factory A = 8., And, SD of distribution of wages in factory B = 9.
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Statistics, , 943, , Since the monthly mean wages in two factories are the same and, σ1 < σ 2 , hence, the factory B has more variation in wages., EXAMPLE 2, , SOLUTION, , EXAMPLE 3, , Coefficient of variation of two distributions are 60% and 75%, and their, standard deviations are 18 and 15 respectively. Find their arithmetic, means., Given: CV1 = 60, CV2 = 75 , σ1 = 18 and σ 2 = 15., Let x1 and x 2 be the means of 1st and 2nd distribution respectively., Then,, σ, σ × 100, CV1 = 1 × 100 ⇒ x1 = 1, x1, CV1, 18 × 100, ⇒ x1 =, = 30., 60, σ, σ × 100, CV2 = 2 × 100 ⇒ x 2 = 2, x2, CV2, 15 × 100, ⇒ x2 =, = 20., 75, Hence, x1 = 30 and x 2 = 20., The mean and variance of the heights and weights of the students of a, class are given below., , Mean, Variance, , Height, , Weight, , 160 cm, , 50.4 kg, , 116.64 cm 2, , 17.64 kg 2, , Show that the weights are more variable than heights., SOLUTION, , For height, we have:, Variance, σ 2 = 116.64 cm 2 ⇒ SD, σ = 116.64 cm = 10.8 cm., Also, mean height = 160 cm., ∴, , ⎛ SD, ⎞, ⎛ 10.8, ⎞, CV1 = ⎜, × 100⎟ % = ⎜, × 100⎟ % = 4.75%, ⎝ mean, ⎠, ⎝ 160, ⎠, , For weight, we have:, Variance, σ 2 = 17.64 kg 2 ⇒ SD, σ = 17.64 kg = 4.2 kg., And, mean weight = 50.4 kg., ∴, , 25, 1, ⎛ SD, ⎞, ⎛ 4.2, ⎞, CV2 = ⎜, × 100⎟ % = ⎜, × 100⎟ % =, % = 8 %., ⎝ mean, ⎠, ⎝ 50.4, ⎠, 3, 3, , Clearly, (CV in weights) > (CV in heights)., Hence, weights are more variable than heights.
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944, , Senior Secondary School Mathematics for Class 11, , EXERCISE 30D, 1. The following results show the number of workers and the wages paid to, them in two factories F1 and F2 ., Factory, , A, , B, , 3600, , 3200, , Rs 5300, , Rs 5300, , 100, , 81, , Number of workers, Mean wages, Variance of distribution, of wages, , Which factory has more variation in wages?, 2. Coefficient of variation of two distributions are 60% and 80% respectively,, and their standard deviations are 21 and 16 respectively. Find their, arithmetic means., 3. The mean and variance of the heights and weights of the students of a class, are given below:, Heights, , Weights, , Mean, , 63.2 inches, , 63.2 kg, , SD, , 11.5 inches, , 5.6 kg, , Which shows more variability, heights or weights?, 4. The following results show the number of workers and the wages paid to, them in two factories A and B of the same industry., Firms, , A, , B, , 560, , 650, , Rs 5460, , Rs 5460, , 100, , 121, , Number of workers, Mean monthly wages, Variance of distribution, of wages, , (i) Which firm pays larger amount as monthly wages?, (ii) Which firm shows greater variability in individual wages?, 5. The sum and the sum of squares of length x (in cm) and weight y (in g) of 50, plant products are given below:, 50, , 50, , 50, , i =1, , i =1, , i =1, , ∑ x i = 212, ∑ x i2 = 902.8, ∑ y i = 261 and, , Which is more variable, the length or weight?, , 50, , ∑ y i2 = 1457.6., , i =1
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Statistics, , 945, ANSWERS (EXERCISE 30D), , 1. A, , 2. 35, 20, , 3. Heights, , 4. (i) A (ii) B, , 5. Weight, HINTS TO SOME SELECTED QUESTIONS, 2, , 2, , 2, , 1 50 2 ⎛ 1 50 ⎞, 902.8 ⎞ ⎛ 212 ⎞, 451.4 ⎛ 106 ⎞, −⎜, ⋅ ∑ xi − ⎜ ⋅ ∑ xi ⎟ = ⎛⎜, ⎟−⎜, ⎟ =, ⎟, ⎝ 25 ⎠, ⎝ 50 ⎠ ⎝ 50 ⎠, 50 i = 1, 25, ⎝ 50 i = 1 ⎠, 451.4 11236 ⎞ ⎛ 11285 − 11236 ⎞, = ⎛⎜, −, ⎟=⎜, ⎟, ⎝ 25, ⎠, 625 ⎠ ⎝, 625, 49, =, = 0.078., 625, 2, ⎧⎪ 1457.6 ⎛ 261 ⎞ 2 ⎫⎪, 1 50 2 ⎛ 1 50 ⎞, var ( y ) =, ⋅ ∑ yi − ⎜ ⋅ ∑ yi ⎟ = ⎨, −⎜, ⎟ ⎬, ⎝ 50 ⎠ ⎪, 50 i = 1, ⎝ 50 i = 1 ⎠, ⎪⎩ 50, ⎭, = {29.152 − (5.22 ) 2 } = 29.152 − 27.2484, , 5. var ( x ) =, , ∴ var (x) < var (y)., , = 1.9036., , q
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31, Probability, INTRODUCTION Historically, the theory of probability began to develop with, the study of games of chance such as roulette and cards. Apart from games,, uncertainty also prevails in other walks of life such as business, economy, and, even in day-to-day activities., , Probability is a concept which numerically measures the degree of uncertainty, and, therefore, of certainty of the occurrence of an event., Before defining some terms related to probability, we shall give certain, concepts used therein., EXPERIMENT An operation which can produce some well-defined outcomes is known, as an experiment., , In science or engineering if we perform an experiment and repeat it under, identical conditions, we get almost the same result every time. Such, experiments are called deterministic experiments., But, there are experiments, which when repeated under identical conditions,, do not produce the same outcome every time. For example, if we toss a fair, coin, we may get a head or a tail. Now, if we make further trials, i.e., toss the, coin again and again, the outcome of each trial depends on chance, and it is, not the same each time. Sometimes the head appears and sometimes, the tail., Such experiments are called random experiments., RANDOM EXPERIMENT If in each trial of an experiment, conducted under identical, conditions, the outcome is not unique, but may be any of the several possible outcomes, then such an experiment is known as a random experiment., , In a random experiment, the outcome of each trial depends on chance,, which is beyond our control, and as such it cannot be predicted with certainty., EXAMPLES, , NOTE 1, , Tossing a fair coin, rolling an unbiased die, drawing a card from a, well-shuffled pack of cards are all examples of random, experiments., , A die is a solid cube, the six faces of which are marked with 1, 2, 3, 4, 5, and 6 dots respectively. In throwing a die, the outcome is the number, of dots on the uppermost face., The plural of die is dice., 946
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Probability, NOTE 2, , 947, , A pack of cards consists of 52 cards in four suits, called spades, clubs,, hearts and diamonds. Out of these, spades and clubs are black-faced, cards, while hearts and diamonds are red-faced cards. The aces,, kings, queens and jacks are known as face cards., , SAMPLE SPACE The set of all possible outcomes in a random experiment is called a, sample space and it is generally denoted by S., , Each element of a sample space is called a sample point., EXAMPLES OF SAMPLE SPACES, EXAMPLE 1, SOLUTION, , EXAMPLE 2, SOLUTION, , EXAMPLE 3, SOLUTION, , EXAMPLE 4, , SOLUTION, , List the sample space in tossing a fair coin., In tossing a fair coin, there are two possible outcomes, namely, head, (H) and tail (T)., Hence, the sample space in this experiment is given by, S = {H , T }., List the sample space in throwing a die., When we throw a die, it can result in any of the six numbers, namely,, 1, 2, 3, 4, 5, 6., ∴ the sample space is S = {1, 2, 3, 4, 5, 6}., Two coins are tossed together. List the sample space., When two coins are tossed together, the sample space is, S = {HT , TH , HH , TT }., Here, HT shows a head on the first coin and a tail on the second., Similarly, TH means a tail on the first and a head on the second; HH, means a head on each, and TT means a tail on each., Three coins are tossed simultaneously. List the sample space for the, event., [CBSE 1999], When three coins are tossed simultaneously, the sample space is, given by, S = {HHH , HHT , HTH , THH , HTT , TTH , THT , TTT }., , EXAMPLE 5, SOLUTION, , List the sample space in a simultaneous toss of a die and a coin., Clearly, the sample space is given by, S = {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T),, (4, T), (5, T), (6, T)}., , EXAMPLE 6, , SOLUTION, , Two well-balanced dice are rolled and the numbers that turn up are, observed. Determine the sample space., [CBSE 1999], When two dice are rolled, there are ( 6 × 6) = 36 outcomes. The set of, all these outcomes is the sample space, given by
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948, , Senior Secondary School Mathematics for Class 11, , S = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1),, (1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2),, (1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3),, (1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4),, (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5),, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)}., From a group of 3 boys and 2 girls, we select two children. What would, be the sample space for this experiment?, , EXAMPLE 7, , SOLUTION, , Let us name the boys as B1 , B2 and B3 , and the girls as G1 and G2 . Then, the sample space is given by, S = {B1 B2 , B1 B3 , B1G1 , B1G2 , B2 B3 , B2G1 , B2G2 , B3G1 , B3G2 , G1G2}., A coin is tossed twice. If the second throw results in a tail then a die is, thrown. Describe the sample space., , EXAMPLE 8, , SOLUTION, , Clearly, the sample space is given by, S = {HH , TH , HT 1, HT 2, HT 3 , HT 4, HT5 , HT 6, TT 1, TT 2, TT 3 ,, TT 4, TT 5 , TT 6}., A coin is tossed. If it shows a tail, we draw a ball from a box which, contains 2 red and 3 black balls; if it shows a head, we throw a die. Find, the sample space for this experiment., , EXAMPLE 9, , SOLUTION, , Let the red balls be named as R1 and R 2 , and the black balls be, named as B1 , B2 and B3 ., Then, the sample space is given by, S = {TR 1 , TR 2 , TB1 , TB2 , TB3 , H1, H 2, H 3 , H 4, H 5 , H 6}., , EXAMPLE 10, , SOLUTION, , Clearly, the sample space is given by, S = {2H , 2T , 4H , 4T , 6H , 6T , 1HH , 1HT , 1TH , 1TT , 3 HH , 3 HT ,, 3 TH , 3 TT , 5 HH , 5 HT , 5 TH , 5 TT }., , EXAMPLE 11, , SOLUTION, , An experiment consists of rolling a die and then tossing a coin once if the, number on the die is even. If the number on the die is odd, the coin is, tossed twice. Write the sample space for this experiment., , A bag contains 4 identical red balls and 3 identical black balls. The, experiment consists of drawing one ball, then putting it into the bag and, again drawing a ball. What are the possible outcomes of this experiment?, , Let us denote a red ball by R and a black ball by B. Then, clearly the, sample space is given by, , S = {RR , RB , BR , BB }., Now, we define some terms related to probability., EVENT, , Every subset of a sample space is called an event., , EXAMPLE, , In a single throw of a die, we have, sample space S = {1, 2, 3, 4, 5, 6}.
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Probability, , 949, , The event of getting a prime number is given by, E = {2, 3, 5}., Clearly, E ⊆ S., IMPOSSIBLE EVENT, , SURE EVENT, , EXAMPLE, , Let S be a sample space., Since φ ⊂ S, φ is an event, called an impossible event., , Let S be a sample space., Since S ⊆ S, S is an event, called a sure event., In a throw of a die, we have, sample space S = {1, 2, 3, 4, 5, 6}., Let E1 = event of getting a number less than 1., And, E2 = event of getting a number less than 7., Clearly, no outcome can be less than 1., ∴ E1 is an impossible event., Also, each outcome is a number less than 7., ∴ E2 is a sure event., , SIMPLE EVENT An event containing only a single element of the sample space is, called a simple or an elementary event., COMPOUND EVENT, , An event which is not simple is called a compound or composite, , or mixed event., EXAMPLE, , In a simultaneous toss of two coins, we have, sample space S = {HH , HT , TH , TT }., Then E1 = event of getting a tail on both the coins = {TT },, is a simple event., And, E2 = event of getting at least 1 tail = {HT , TH , TT },, is a compound event., , MUTUALLY EXCLUSIVE EVENTS, , exclusive if E1 ∩ E2 = φ., , Two events E1 and E2 are said to be mutually, , However, if E1 ∩ E2 ≠ φ then E1 and E2 are called compatible events., EXAMPLES, , (i) In throwing a die, we have S = {1, 2, 3, 4, 5, 6}., Let E1 = event of getting a number less than 3., And, E2 = event of getting a number more than 4., Then E1 = {1, 2} and E2 = {5, 6}., Clearly, E1 ∩ E2 = φ., Hence, E1 and E2 are mutually exclusive., (ii) In a simultaneous toss of two coins, we have, S = {HH, HT, TH, TT }.
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950, , Senior Secondary School Mathematics for Class 11, , Let E1 = event of getting a head on the first coin = {HH , HT }., And, E2 = event of getting a tail on the second coin = {HT , TT }., Clearly, E1 ∩ E2 ≠ φ., Hence, E1 and E2 are compatible events., EXAMPLE 1, , SOLUTION, , Two dice are rolled. Let A, B, C be the events of getting a sum of 2, a sum, of 3 and a sum of 4 respectively. Then, show that, (i) A is a simple event, (ii) B and C are compound events, (iii) A and B are mutually exclusive, Clearly, we have, A = {(1, 1)}, B = {(1, 2), (2, 1)}, and C = {(1, 3), (3, 1), (2, 2)}., (i) Since A consists of a single sample point, it is a simple event., (ii) Since both B and C contain more than one sample point,, each one of them is a compound event., (iii) Since A ∩ B = φ, A and B are mutually exclusive., , EXAMPLE 2, , SOLUTION, , From a group of 2 boys and 3 girls, two children are selected at random., Describe the events:, (i) A = event that both the selected children are girls, (ii) B = event that the selected group consists of one boy and one girl, (iii) C = event that at least one boy is selected, Which pairs of events are mutually exclusive?, Let us name the boys as B1 and B2 , and the girls as G1 , G2 and G3 ., Then, S = {B1 B2 , B1G1 , B1G2 , B1G3 , B2G1 , B2G2 , B2G3 , G1G2 , G1G3 , G2G3}, We have, (i) A = {G1G2 , G1G3 , G2G3}, (ii) B = {B1G1 , B1G2 , B1G3 , B2G1 , B2G2 , B2G3}, (iii) C = {B1G1 , B1G2 , B1G3 , B2G1 , B2G2 , B2G3 , B1 B2}, Clearly, A ∩ B = φ and A ∩ C = φ., Hence, ( A , B) and ( A , C) are mutually exclusive events., , MUTUALLY EXCLUSIVE AND EXHAUSTIVE SYSTEM OF EVENTS, , Let E1 , E2 , … , En be, , subsets of a sample space S. Then, we say that, the events E1 , E2 , … , En form a mutually exclusive and exhaustive system if, (i) Ei ∩ Ej ≠ φ for i ≠ j , and (ii) E1 ∪ E2 ∪ … ∪ En = S., EXAMPLE, , Suppose we draw a card from a well-shuffled pack of 52 cards., Let E1 , E2 , E3 and E4 be the events of drawing a spade, drawing a club,, drawing a heart and drawing a diamond respectively.
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Probability, , 951, , As the card drawn is necessarily one of the four types, one of these, events is sure to occur., When one of these events occurs then none of the others occur., Thus, E1 , E2 , E3 , E4 form a mutually exclusive and exhaustive system, of events., EXAMPLE, , SOLUTION, , Two dice are rolled. A is the event that the sum of the numbers shown on, the two dice is 5, and B is the event that at least one of the dice shows up a, 3. Are the two events (i) mutually exclusive, (ii) exhaustive? Give, arguments in support of your answer., [CBSE 2001C ], When two dice are rolled, we have n( S) = ( 6 × 6) = 36., Now, A = {(1, 4), (2, 3), (4, 1), (3, 2)}, and, B = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (1, 3), (2, 3),, (4, 3), (5, 3), (6, 3)}., (i) A ∩ B = {( 2, 3), ( 3 , 2)} ≠ φ., Hence, A and B are not mutually exclusive., (ii) Also, A ∪ B ≠ S., ∴ A and B are not exhaustive events., , INDEPENDENT EVENTS Two events E1 and E2 are said to be independent if the, occurrence of one does not depend upon the occurrence of the other., , If the two events are not independent, they are known as dependent events., EXAMPLE, , Suppose we toss two unbiased coins., Let E1 = event of getting a head on the first coin., And, E2 = event of getting a head on the second coin., Clearly, the occurrence of a head on the second coin does not, depend upon the occurrence of a head on the first coin., ∴ E1 and E2 are independent events., , OCCURRENCE OF AN EVENT In a random experiment, let S be the sample space, and let E ⊆ S. Then, E is an event., , Let w be an outcome of a trial., If w ∈ E, we say that the event E has occurred., If w ∉ E, we say that the event E has not occurred., EXAMPLE, , In a throw of a die, we have, S = {1, 2, 3, 4, 5, 6}., Let E be the event of getting an even number., Then E = {2, 4, 6}., In a trial, let the outcome be 4., Since 4 ∈E, we say that the event E has occurred., In another trial, let the outcome be 5., Since 5 ∉E, in this case, the event E has not occurred.
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952, , Senior Secondary School Mathematics for Class 11, , EQUALLY LIKELY EVENTS A given number of events are said to be equally likely if, none of them is expected to occur in preference to the others., EXAMPLE, , If we roll an unbiased die, each outcome is equally likely to, happen., If, however, a die is so formed that a particular face occurs most, often then the die is biased. In this case, the outcomes are not, equally likely to happen., , COMPLEMENTARY EVENTS In a random experiment, let S be the sample space, and let E be an event. Then E ⊆ S., , Now, E ⊆ S ⇒ Ec ⊆ S., Thus, Ec is also an event, called the complement of E., We denote it by Ec or E′ or E and call it not E., Clearly, E occurs when E does not occur., And, E occurs when E does not occur., Thus, in a trial, one of two events, E or E, is sure to occur., ALGEBRA OF EVENTS, , In a random experiment, let S be the sample space., , Let E ⊆ S and F ⊆ S., Then, E as well as F is an event., We say that, (i) (E ∩ F ) is an event that occurs only when each one of E and F, occurs, (ii) (E ∪ F ) is an event that occurs only when at least one of E, and F occurs, (iii) E is an event that occurs only when E does not occur, , PROBABILITY OF AN EVENT, INTRODUCTION Suppose a bag contains 90 red and 10 white balls, which are, similar in shape and size., , If the balls are mixed thoroughly and then one ball is drawn at random, it, will be either red or white., Clearly, the ball drawn is more likely to be red than white., We express it by saying that the event of drawing a red ball is more, probable than the event of drawing a white ball., To every event associated with a random experiment, we try to attach a, numerical value, called its probability, in such a manner that for any two, events, the event which is more likely to happen has a higher probability.
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Probability, , 953, , PROBABILITY OF AN EVENT In a random experiment, let S be the sample space and let, E ⊆ S. Then, E is an event., The probability of occurrence of E is defined as, number of outcomes favourable to occurrence of E, P(E ) =, number of all possible outcomes, number of distinct elements in E, =, number of distinct elements in S, n(E ), =, ⋅, n( S), n(E ), ⋅, ∴ P(E ) =, n( S), , SOLVED EXAMPLES, EXAMPLE 1, SOLUTION, , A coin is tossed once. Find the probability of getting a head., When a coin is tossed once, the sample space is given by S = {H , T }., Let E be the event of getting a head., Then, E = {H }., ∴ n(E ) = 1 and n( S) = 2, ⇒ P (getting a head) = P(E ) =, , EXAMPLE 2, , SOLUTION, , n(E ) 1, = ⋅, n( S) 2, , Two coins are tossed once. Find the probability of, (i) getting 2 heads, (ii) getting at least 1 head, (iii) getting no head, (iv) getting 1 head and 1 tail, When two coins are tossed once, the sample space is given by, S = {HH , HT , TH , TT } and, therefore, n( S) = 4., (i) Let E1 = event of getting 2 heads. Then,, E1 = {HH } and, therefore, n(E1 ) = 1., ∴ P(getting 2 heads) = P(E1 ) =, , n(E1 ), n( S), , 1, ⋅, 4, , =, , (ii) Let E2 = event of getting at least 1 head. Then,, E2 = {HT , TH , HH } and, therefore, n(E2 ) = 3., ∴ P(getting at least 1 head) = P(E2 ) =, , n(E2 ), n( S), , (iii) Let E3 = event of getting no head. Then,, E3 = {TT } and, therefore, n(E3 ) = 1., ∴ P(getting no head) = P(E3 ) =, , n(E3 ), n( S), , =, , 1, ⋅, 4, , =, , 3, ⋅, 4
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954, , Senior Secondary School Mathematics for Class 11, , (iv) Let E4 = event of getting 1 head and 1 tail. Then,, E4 = {HT , TH } and, therefore, n(E4 ) = 2., n(E4 ) 2 1, = = ⋅, ∴ P(getting 1 head and 1 tail) = P(E4 ) =, n( S) 4 2, EXAMPLE 3, , SOLUTION, , Three unbiased coins are tossed once. What is the probability of getting, (i) all heads?, (ii) two heads?, (iii) one head?, (iv) at least 1 head?, (v) at least 2 heads?, In tossing three coins, the sample space is given by, S = {HHH , HHT , HTH , THH , HTT , THT , TTH , TTT }., And, therefore, n( S) = 8., (i) Let E1 = event of getting all heads. Then,, E1 = {HHH} and, therefore, n(E1 ) = 1., n(E1 ), , 1, = ⋅, n( S) 8, (ii) Let E2 = event of getting 2 heads. Then,, ∴ P(getting all heads) = P(E1 ) =, , E2 = {HHT , HTH , THH} and, therefore, n(E2 ) = 3., n(E2 ) 3, = ⋅, ∴ P(getting 2 heads) = P(E2 ) =, n( S), 8, (iii) Let E3 = event of getting 1 head. Then,, E3 = {HTT , THT , TTH } and, therefore, n(E3 ) = 3., n(E3 ) 3, ∴ P(getting 1 head) = P(E3 ) =, = ⋅, n( S), 8, (iv) Let E4 = event of getting at least 1 head. Then,, E4 = {HTT , THT , TTH , HHT , HTH , THH , HHH }., And, therefore, n(E4 ) = 7., n(E4 ) 7, = ⋅, n( S) 8, (v) Let E5 = event of getting at least 2 heads. Then,, ∴ P(getting at least 1 head) = P(E4 ) =, , E5 = {HHT , HTH , THH , HHH } and, therefore, n(E5 ) = 4., ∴ P(getting at least 2 heads) = P(E5 ) =, EXAMPLE 4, , SOLUTION, , n(E5 ), n( S), , =, , 4 1, = ⋅, 8 2, , A die is tossed once. What is the probability of getting, (i) the number 4?, (ii) an even number?, (iii) a number less than 5?, (iv) a number greater than 4?, (v) the number 8?, (vi) a number less than 8?, In tossing a die once, the sample space is given by, S = {1, 2, 3 , 4, 5 , 6} and, therefore, n( S) = 6.
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Probability, , 955, , (i) Let E1 = event of getting the number 4., Then, E1 = {4} and, therefore, n(E1 ) = 1., n(E1 ) 1, = ⋅, ∴ P(getting the number 4) = P(E1 ) =, n( S) 6, (ii) Let E2 = event of getting an even number. Then,, E2 = {2, 4, 6} and, therefore, n(E2 ) = 3., n(E2 ) 3 1, = = ⋅, ∴ P(getting an even number) = P(E2 ) =, n( S), 6 2, (iii) Let E3 = event of getting a number less than 5. Then,, E3 = {1, 2, 3 , 4} and, therefore, n(E3 ) = 4., n(E3 ) 4 2, = = ⋅, ∴ P(getting a number less than 5) = P(E3 ) =, n( S), 6 3, (iv) Let E4 = event of getting a number greater than 4. Then,, E4 = {5 , 6} and, therefore, n(E4 ) = 2., ∴ P(getting a number greater than 4), n(E4 ) 2 1, = P(E4 ) =, = = ⋅, n( S) 6 3, (v) Let E5 = event of getting the number 8., Since no face of the die is marked with the number 8,, we have E5 = φ and, therefore, n(E5 ) = 0., n(E5 ) 0, = = 0., ∴ P(getting the number 8) = P(E5 ) =, n( S) 8, (vi) Let E6 = event of getting a number less than 8. Then,, E6 = {1, 2, 3 , 4, 5 , 6} and, therefore, n(E6) = 6., n(E6) 6, = = 1., ∴ P(getting a number less than 8) = P(E6) =, n( S) 6, EXAMPLE 5, SOLUTION, , EXAMPLE 6, , In a single throw of two dice, find the probability of obtaining ‘a total of 8’., We know that in a single throw of two dice, the total number of, possible outcomes is ( 6 × 6) = 36., Let S be the sample space. Then, n( S) = 36., Let E = event of getting a total of 8. Then,, E = {( 2, 6), ( 3 , 5), ( 4, 4), (5 , 3), ( 6, 2)} and, therefore, n(E ) = 5., n(E ), 5, =, ⋅, ∴ P(getting a total of 8) = P(E ) =, n( S) 36, Two dice are thrown simultaneously. Find the probability of getting, (i) a doublet, (ii) an even number as the sum, (iii) a prime number as the sum, (iv) a multiple of 3 as the sum
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956, , Senior Secondary School Mathematics for Class 11, , (v) a total of at least 10, (vi) a doublet of even numbers, (vii) a multiple of 2 on one die and a multiple of 3 on the other die, SOLUTION, , We know that in a single throw of two dice, the total number of, possible outcomes is ( 6 × 6) = 36., Let S be the sample space. Then, n( S) = 36., (i) Let E1 = event of getting a doublet. Then,, E1 = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}., ∴ n(E1 ) = 6., n(E1 ), 6 1, =, = ⋅, ∴ P(E1 ) =, n( S), 36 6, (ii) Let E2 = event of getting an even number as the sum. Then,, E2 = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 3), (3, 1),, (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2),, (6, 4), (6, 6)}., ∴ n(E2 ) = 18., n(E2 ) 18 1, =, = ⋅, ∴ P(E2 ) =, n( S), 36 2, (iii) Let E 3 = event of getting a prime number as the sum. Then,, E 3 = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2),, (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}., ∴ n(E3 ) = 15., n(E3 ) 15 5, =, =, ⋅, ∴ P(E3 ) =, n( S), 36 12, (iv) Let E4 = event of getting a multiple of 3 as the sum. Then,, E4 = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5),, (5, 1), (5, 4), (6, 3), (6, 6)}., ∴ n(E4 ) = 12., n(E4 ) 12 1, =, = ⋅, ∴ P(E4 ) =, n( S), 36 3, (v) Let E5 = event of getting a total of at least 10. Then,, E5 = event of getting a total of 10 or 11 or 12, = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}., ∴ n(E5 ) = 6., n(E5 ), 6 1, =, = ⋅, ∴ P(E5 ) =, n( S), 36 6, (vi) Let E6 = event of getting a doublet of even numbers. Then,, E6 = {(2, 2), (4, 4), (6, 6)}.
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Probability, , 957, , ∴ n(E6) = 3., n(E6), 3, 1, =, =, ⋅, ∴ P(E6) =, n( S), 36 12, (vii) Let E7 = event of getting a multiple of 2 on one die and a, multiple of 3 on the other die. Then,, E7 = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), (3, 4),, (3, 6), (6, 2), (6, 4)}., ∴ n(E7 ) = 11., n(E7 ) 11, =, ⋅, ∴ P(E7 ) =, n( S), 36, EXAMPLE 7, , SOLUTION, , 20 cards are numbered from 1 to 20. One card is then drawn at random., What is the probability that the number on the card drawn is, (i) a prime number?, (ii) an odd number?, (iii) a multiple of 5?, (iv) not divisible by 3?, Clearly, the sample space is given by, S = {1, 2, 3 , 4, 5 , … , 19, 20} and, therefore, n( S) = 20., (i) Let E1 = event of getting a prime number. Then,, E1 = {2, 3 , 5 , 7 , 11, 13 , 17 , 19} and, therefore, n(E1 ) = 8., n(E1 ), 8 2, =, = ⋅, ∴ P(getting a prime number) = P(E1 ) =, n( S) 20 5, (ii) Let E2 = event of getting an odd number. Then,, E2 = {1, 3 , 5 , 7 , 9, 11, 13 , 15 , 17 , 19} and, therefore, n(E2 ) = 10., n(E2 ) 10 1, =, = ⋅, ∴ P(getting an odd number) = P(E2 ) =, 20 2, n( S), (iii) Let E 3 = event of getting a multiple of 5. Then,, E 3 = {5 , 10, 15 , 20} and, therefore, n(E 3 ) = 4., n(E 3 ), 4 1, =, = ⋅, ∴ P(getting a multiple of 5) = P(E3 ) =, 20 5, n( S), (iv) Let E4 = event of getting a number which is not divisible by 3., Then, E4 = {1, 2, 4, 5 , 7 , 8, 10, 11, 13 , 14, 16, 17 , 19, 20} and, so, n(E4 ) = 14., ∴ P(getting a number which is not divisible by 3), n(E4 ) 14 7, = P(E4 ) =, =, =, ⋅, n( S) 20 10, , EXAMPLE 8, , From a well-shuffled deck of 52 cards, a card is drawn at random. Find the, probability of getting, (i) an ace, (ii) a heart, (iii) an eight of hearts, (iv) a club, (v) a red card, (vi) a face card, (vii) a diamond, (viii) a jack, (ix) a black card
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958, SOLUTION, , Senior Secondary School Mathematics for Class 11, , Let S denote the sample space. Then, n( S) = 52., (i) Let E1 = event of drawing an ace., Since, the number of all aces is 4, so n(E1 ) = 4., n(E1 ), 4, 1, =, =, ⋅, ∴ P(getting an ace) = P(E1 ) =, n( S) 52 13, (ii) Let E2 = event of getting a heart. Then, n(E2 ) = 13., n(E2 ) 13 1, =, = ⋅, ∴ P(getting a heart) = P(E2 ) =, n( S) 52 4, (iii) Let E3 = event of getting an eight of hearts. Then, n(E3 ) = 1., n(E3 ) 1, =, ⋅, ∴ P(getting an eight of hearts) = P(E3 ) =, n( S) 52, (iv) Let E4 = event of getting a club. Then, n(E4 ) = 13., n(E4 ) 13 1, =, = ⋅, ∴ P(getting a club) = P(E4 ) =, n( S) 52 4, (v) Let E5 = event of getting a red card. Then, n(E5 ) = 26., n(E5 ) 26 1, =, = ⋅, ∴ P(getting a red card) = P(E5 ) =, n( S) 52 2, (vi) Let E6 = event of getting a face card. Then, n(E6) = 16., n(E6) 16 4, =, =, ⋅, ∴ P(getting a face card) = P(E6) =, n( S) 52 13, (vii) Let E 7 = event of getting a diamond. Then, n(E 7 ) = 13., n(E 7 ) 13 1, =, = ⋅, ∴ P(getting a diamond) = P(E 7 ) =, n( S) 52 4, (viii) Let E 8 = event of getting a jack. Then, n(E 8 ) = 4., n(E 8 ), 4, 1, =, =, ⋅, ∴ P(getting a jack) = P(E 8 ) =, n( S) 52 13, (ix) Let E 9 = event of getting a black card. Then, n(E 9 ) = 26., n(E 9 ) 26 1, =, = ⋅, ∴ P(getting a black card) = P(E 9 ) =, n( S), 52 2, , EXAMPLE 9, , SOLUTION, , From a well-shuffled deck of 52 cards, a card is drawn at random. Find, the probability that the card drawn is, (i) red and a king, (ii) either red or a king, Let S denote the sample space. Then, n( S) = 52., (i) Let E1 = event of drawing a red card which is a king., We know that the number of red kings is 2. So, n(E1 ) = 2.
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Probability, , ∴ P(getting a red king) = P(E1 ) =, , 959, , n(E1 ), n( S), , =, , 2, 1, =, ⋅, 52 26, , (ii) Let E2 = event of drawing a card which is either red or a king., There are 26 red cards (including 2 red kings) and there are 2, more kings., ∴ n(E2 ) = ( 26 + 2) = 28, n(E2 ) 28 7, =, =, ⋅, ∴ P(getting a red card or a king) = P(E2 ) =, n( S) 52 13, EXAMPLE 10, , SOLUTION, , A bag contains 9 red and 12 white balls. One ball is drawn at random., Find the probability that the ball drawn is red., , Total number of balls = ( 9 + 12) = 21., Let S be the sample space. Then,, n( S) = number of ways of selecting 1 ball out of 21 = 21., Let E be the event of drawing a red ball. Then,, n(E ) = number of ways of selecting 1 red ball out of 9 = 9., n(E ), 9, 3, ∴ P(getting a red ball) = P(E ) =, =, = ⋅, n( S) 21 7, , ODDS OF AN EVENT Let there be m outcomes favourable to an event E and n outcomes, unfavourable to E. Then,, m, (i) odds in favour of E = or (m : n), n, n, (ii) odds against E = or (n : m), m, m, (iii) P(E ) =, (m + n), EXAMPLE 11 The odds in favour of occurrence of an event are 5 : 12 . Find the, probability of the occurrence of this event., SOLUTION, , Number of favourable outcomes = 5., Number of unfavourable outcomes = 12., Total number of outcomes = (5 + 12) = 17., Let E be the event. Then,, number of favourable outcomes 5, =, ⋅, P(E ) =, 17, total number of outcomes, , EXAMPLE 12, , SOLUTION, , If the probability of the occurrence of a certain event E is 3 11 , find (i) the, odds in favour of its occurrence, and (ii) the odds against its occurrence., , Let there bem outcomes favourable toEandn outcomes unfavourable, to E. Then,, m, m, 3, 3, (given)], P(E ) =, ⇒, =, [Q P(E ) =, m+n, m + n 11, 11, ⇒ m = 3 and n = 8.
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960, , Senior Secondary School Mathematics for Class 11, , m 3, = ⋅, n 8, n 8, (ii) odds against E = = ⋅, m 3, (i) odds in favour of E =, , PROBABILITY OF OCCURRENCE OF A COMPLEMENTARY EVENT, Let E be an event and E be the complement of E ⋅ Then, P( E ) = 1 − P(E )., , THEOREM 1, , PROOF, , Let there be m outcomes favourable to E and n outcomes unfavourable, to E. Then,, number of possible outcomes = (m + n)., m, n, and P ( E ) = P(not E) =, ∴ P(E ) =, (m + n), (m + n), ⎛ m, n ⎞ (m + n), ⇒ P(E ) + P( E ) = ⎜, +, =1, ⎟ =, ⎝ m + n m + n⎠ (m + n), ⇒ P( E ) = 1 − P(E )., Hence, P( E ) = 1 − P(E )., , NOTE, , There is an alternative proof of this theorem, which we will soon discuss., , EXAMPLE 13, , SOLUTION, , Let E be the event. Then,, 3⎞ 7, 3, ⎛, P(E ) =, ⇒ P ( E ) = {1 − P(E )} = ⎜1 − ⎟ =, ⋅, ⎝, 10⎠ 10, 10, 7, Hence, P(not E) =, ⋅, 10, , EXAMPLE 14, , SOLUTION, , If 3 10 is the probability that an event will happen, what is the probability, that it will not happen?, , Three dice are thrown together. Find the probability of getting a total of, at least 6., , In throwing 3 dice together, the number of all possible outcomes is, ( 6 × 6 × 6) = 216., Let E = event of getting a total of at least 6., Then, E = event of getting a total of less than 6, = event of getting a total of 3 or 4 or 5., = {(1, 1, 1), (1, 1, 2), (1, 2, 1), ( 2, 1, 1), (1, 1, 3), (1, 3 , 1),, ( 3 , 1, 1), (1, 2, 2), ( 2, 1, 2), ( 2, 2, 1)}, n (E ), 10, 5, Now, n( E ) = 10 ⇒ P(not E) = P( E ) =, =, =, ⋅, n( S), 216 108, 5 ⎞ 103, ⎛, ⇒ P(E ) = 1 − P(not E) = ⎜1 −, ⋅, ⎟ =, ⎝, 108⎠ 108, 103, Hence, the required probability is, ⋅, 108
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Probability, , 961, , For an event E,, , THEOREM 2, , P(E ), 1 − P(E ), 1 − P(E ), (ii) odds against E =, P(E ), (i) odds in favour of E =, , PROOF, , Let there be m outcomes favourable to E and n outcomes unfavourable, to E., Then, there are m outcomes favourable to E and n outcomes favourable, to E., n(E ), m, n, and P ( E ) =, =, ⋅, ∴ P(E ) =, n( S) (m + n), (m + n), m, m (m + n), P(E ), P(E ), (i) Odds in favour of E = =, =, =, ⋅, n, 1 − P(E ), n, P (E ), (m + n), (ii) Odds against E =, , EXAMPLE 15, , SOLUTION, , If the odds in favour of an event be, occurrence of the event., , 3, , 5, , , find the probability of the, , Let the given event be E and let P(E ) = x. Then,, P(E ), odds in favour of E =, 1 − P(E ), P(E ), 3, x, 3, ⇔, =, ⇔, =, 1 − P(E ) 5, (1 − x) 5, 3, ⇔ 5 x = 3 − 3 x ⇔ 8x = 3 ⇔ x = ⋅, 8, 3, ∴ required probability = ⋅, 8, , EXAMPLE 16, , SOLUTION, , n 1 − P(E ), =, ⋅, m, P(E ), , Two dice are thrown. Find (i) the odds in favour of getting the sum 5, and, (ii) the odds against getting the sum 6., , Let S be the sample space. Then, n( S) = 36., (i) Let E1 be the event of getting the sum 5. Then,, E1 = {(1, 4), ( 2, 3), ( 3 , 2), ( 4, 1)} ⇒ n(E1 ) = 4., n(E1 ), 4 1, =, =, ∴ P(E1 ) =, n( S), 36 9, P(E1 ), ( 1 9), 1, ⇒ odds in favour of E1 =, =, = ⋅, 1 − P(E1 ) (1 − 1 9) 8
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962, , Senior Secondary School Mathematics for Class 11, , (ii) Let E2 be the event of getting the sum 6. Then,, E2 = {(1, 5), ( 2, 4), ( 3 , 3), ( 4, 2), (5 , 1)} ⇒ n(E2 ) = 5., n(E2 ), 5, =, ∴ P(E2 ) =, n( S), 36, 5⎞, ⎛, 1− ⎟, 1 − P(E2 ) ⎜⎝, 31, 36⎠, =, ⋅, =, ⇒ odds against E2 =, (5/ 36), P(E2 ), 5, EXAMPLE 17, , SOLUTION, , A card is drawn from a well-shuffled deck of 52 cards. Find (i) the odds in, favour of getting a face card, and (ii) the odds against getting a spade., , When a card is drawn from a well-shuffled deck of 52 cards, the, number of possible outcomes is 52., Now, let S be the sample space. Then, n( S) = 52., (i) Let E1 be the event of getting a face card., Since there are 16 face cards, we have n(E1 ) = 16., n(E1 ) 16 4, =, =, ∴ P(E1 ) =, n( S) 52 13, ⇒ odds in favour of getting a face card, P(E1 ), 4, ( 4/13), =, =, = ⋅, 4⎞ 9, {1 − P(E1 )} ⎛, ⎜1 − ⎟, ⎝, 13 ⎠, (ii) Let E2 be the event of getting a spade. Then, n(E2 ) = 13., n(E2 ) 13 1, =, =, ∴ P(E2 ) =, n( S) 52 4, {1 − P(E2 )}, ⇒ odds against getting a spade =, P(E2 ), 1⎞, ⎛, ⎜1 − ⎟, ⎝, 3, 4⎠, =, = ⋅, (1/4), 1, , EXAMPLE 18, , SOLUTION, , Two cards are drawn at random from a pack of 52 cards. What is the, probability that both the drawn cards are aces?, , Let S be the sample space. Then,, n( S) = number of ways of selecting 2 cards out of 52, (52 × 51), = 52C 2 =, = 1326., ( 2 × 1), Let E = event that both the cards drawn are aces. Then,, n(E ) = number of ways of drawing 2 aces out of 4, ( 4 × 3), = 4C 2 =, = 6., ( 2 × 1)
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Probability, , P(getting 2 aces) = P(E) =, , 963, , n(E ), 6, 1, =, =, ⋅, n( S) 1326 221, , EXERCISE 31A, 1. A coin is tossed once. Find the probability of getting a tail., 2. A die is thrown. Find the probability of, (i) getting a 5, (ii) getting a 2 or a 3, (iii) getting an odd number, (v) getting a multiple of 3, , (iv) getting a prime number, (vi) getting a number between 3 and 6, , 3. In a single throw of two dice, find the probability of, (i) getting a sum less than 6, (ii) getting a doublet of odd numbers, (iii) getting the sum as a prime number, 4. In a single throw of two dice, find, (i) P (an odd number on the first die and a 6 on the second), (ii) P (a number greater than 3 on each die), (iii) P (a total of 10), (iv) P (a total greater than 8), (v) P (a total of 9 or 11), 5. A bag contains 4 white and 5 black balls. A ball is drawn at random from the, bag. Find the probability that the ball drawn is white., 6. An urn contains 9 red, 7 white and 4 black balls. A ball is drawn at random., Find the probability that the ball drawn is, (i) red, (ii) white, (iii) red or white, (iv) white or black, (v) not white, 7. In a lottery, there are 10 prizes and 25 blanks. Find the probability of getting, a prize., 8. If there are two children in a family, find the probability that there is at least, one boy in the family., 9. Three unbiased coins are tossed once. Find the probability of getting, (i) exactly 2 tails, (ii) exactly one tail, (iii) at most 2 tails, (iv) at least 2 tails, (v) at most 2 tails or at least 2 heads, [CBSE 2005], 10. In a single throw of two dice, determine the probability of not getting the, same number on the two dice., 11. If a letter is chosen at random from the English alphabet, find the, probability that the letter chosen is (i) a vowel, and (ii) a consonant.
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964, , Senior Secondary School Mathematics for Class 11, , 12. A card is drawn at random from a well-shuffled pack of 52 cards. What is, the probability that the card bears a number greater than 3 and less than 10?, 13. Tickets numbered from 1 to 12 are mixed up together and then a ticket is, withdrawn at random. Find the probability that the ticket has a number, which is a multiple of 2 or 3., 14. What is the probability that an ordinary year has 53 Tuesdays?, 15. What is the probability that a leap year has 53 Sundays?, 16. What is the probability that in a group of two people, both will have the, same birthday, assuming that there are 365 days in a year and no one has, his/her birthday on 29th February?, 17. Which of the following cannot be the probability of occurrence of an event?, −3, 3, 4, (i) 0, (ii), (iii), (iv), 4, 4, 3, 18. If 7 10 is the probability of occurrence of an event, what is the probability that, it does not occur?, 19. The odds in favour of the occurrence of an event are 8 : 13. Find the, probability that the event will occur., 20. If the odds against the occurrence of an event be 4 : 7, find the probability of, the occurrence of the event., 21. If 5 14 is the probability of occurrence of an event, find, (i) the odds in favour of its occurrence, (ii) the odds against its occurrence, 22. Two dice are thrown. Find, (i) the odds in favour of getting the sum 6, (ii) the odds against getting the sum 7, 23. A combination lock on a suitcase has 3 wheels, each labelled with nine, digits from 1 to 9. If an opening combination is a particular sequence of, three digits with no repeats, what is the probability of a person guessing the, right combination?, 24. In a lottery, a person chooses six different numbers at random from 1 to 20., If these six numbers match with the six numbers already fixed by the, lottery committee, he wins the prize. What is the probability of winning the, prize in the game?, 25. In a single throw of three dice, find the probability of getting (i) a total of 5, (ii) a total of at most 5., [CBSE 2005], ANSWERS (EXERCISE 31A), , 1, 1., 2, , 1, 1, 2. (i), (ii), 6, 3, , 1, 1, 1, (iii), (iv), (v), 2, 2, 3, , (vi), , 1, 3
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Probability, , 965, , 5, 1, 5, 1, 1, 1, 5, 1, (ii), (iii), 4. (i), (ii), (iii), (iv), (v), 18, 12, 12, 12, 4, 12, 18, 6, 4, 9, 7, 4, 11, 13, 2, 6. (i), (ii), (iii) (iv), (v), 7., 8., 9, 20, 20, 5, 20, 20, 7, 3, 3, 7, 1, 7, 5, 5, 21, (iii) (iv), (v), 10., 11. (i), (ii), 12., (i) (ii), 8, 8, 8, 2, 8, 6, 26, 26, 2, 1, 2, 1, 3, 14., 15., 16., 17. (ii) and (iv), 18., 19., 3, 7, 7, 365, 10, 7, 21. (i) 5 : 9 (ii) 9 : 5, 22. (i) 5 : 31 (ii) 5 : 1, 23., 11, 1, 1, 5, 25. (i), (ii), 38760, 36, 108, , 3. (i), 5., 9., 13., 20., 24., , 3, 4, 6, 13, 8, 21, 1, 504, , HINTS TO SOME SELECTED QUESTIONS, 7. There are 10 prizes and the total number of outcomes is 35., 8. S = {B1 B2 , B1G 2 , G 1 B2 , G 1G 2 } and E = {B1G 2 , G 1 B2 , B1 B2 }., n( E ), ∴ required probability =, n( S ), 10. n( S ) = ( 6 × 6 ) = 36 ., Let E = event of getting the same number on two dice, or E = {( 1, 1), ( 2 , 2 ), ( 3 , 3 ), ( 4 , 4 ), (5 , 5 ), ( 6 , 6 )}., n( E ), 6, 1, =, =, ∴ P( E ) =, n( S ) 36 6, ⇒ P(not getting the same number on the two dice) = ⎛⎜ 1 −, ⎝, , 1⎞ 5, ⎟= ⋅, 6⎠ 6, , 11. There are 26 letters in the English alphabet, out of which there are 5 vowels and 21, consonants., 12. There are four suits in all, and the number of cards bearing a number greater than 3, and less than 10 in each suit is 6., ∴ total number of favourable cases = ( 6 × 4 ) = 24., Total number of cards = 52., 13. Clearly, n( S ) = 12 and E = {2 , 4 , 6 , 8 , 10 , 12 , 3 , 9} ⇒ n( E ) = 8 ., 14. An ordinary year has 365 days, i.e., 52 weeks and 1 day., Now, 52 weeks have 52 Tuesdays and the remaining one day can be any of the 7 days., 1, ∴ required probability = probability of this day being a Tuesday = ⋅, 7, 15. A leap year has 366 days, and, therefore, 52 weeks and 2 days. These 2 days may be, (i) Sunday and Monday, or (ii) Monday and Tuesday, or (iii) Tuesday and Wednesday or (iv) Wednesday and Thursday, or (v) Thursday and Friday, or (vi) Friday and Saturday, or (vii) Saturday and Sunday., Thus, out of 7 possibilities, 2 favour the event that one of the two days is a Sunday., 2, ∴ required probability = ⋅, 7
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966, , Senior Secondary School Mathematics for Class 11, , 22. n( S ) = ( 6 × 6 ) = 36 ., (i) Let E1 = event of getting the sum 6, = {( 1, 5 ), ( 2 , 4 ), ( 3 , 3 ), ( 4 , 2 ), (5 , 1)}, ⇒ n( E1 ) = 5 ., Thus, E1 occurs in 5 ways and does not occur in ( 36 − 5 ) = 31 ways., ∴ odds in favour of getting the sum 6 = 5 : 31., (ii) Let E2 = event of getting the sum 7, = { ( 1, 6 ), ( 2 , 5 ), ( 3 , 4 ), ( 4 , 3 ), (5 , 2 ), ( 6 , 1)}, ⇒ n( E2 ) = 6 ., Thus, E2 occurs in 6 ways and does not occur in ( 36 − 6 ) = 30 ways., ∴ odds against getting the sum 7 = 30 : 6 = 5 : 1., 23. Let us fill 3 places with different digits from 1 to 9., Total number of ways = ( 9 × 8 × 7 ) = 504., Thus, we have in all 504 combinations., 1, ⋅, P(choosing one opening combination) =, 504, 24. n( S ) = number of ways of choosing 6 numbers out of 20, = 20C 6 ., There is a fixed set of 6 numbers, which can be chosen in 1 way only. Let E be the, event of getting the fixed set of 6 numbers., Then, n( E ) = 1., n( E ), 1, ∴ P( E ) =, =, ⋅, n( S ) 20C 6, 25. n( S ) = ( 6 × 6 × 6 ) = 216, (i) E1 = {( 1, 1, 3 ), ( 1, 3 , 1), ( 3 , 1, 1), ( 1, 2 , 2 ), ( 2 , 1, 2 ), ( 2 , 2 , 1)}, 6, 1, ∴ P( E1 ) =, =, ⋅, 216 26, (ii) E2 = {( 1, 1, 1), ( 1, 1, 2 ), ( 1, 2 , 1), ( 2 , 1, 1), ( 1, 1, 3 ), ( 1, 3 , 1), ( 3 , 1, 1), ( 1, 2 , 2 ), ( 2 , 1, 2 ),, ( 2 , 2 , 1)}, 10, 5, ∴ P( E2 ) =, =, ⋅, 216 108, , RESULTS ON PROBABILITY, THEOREM 1, , PROOF, , Let S be the sample space and let E be an event. Then,, (i) P(E ) ≥ 0, (ii) P(φ) = 0, (iii) P( S) = 1, , (i) Since E is an event, we have E ⊆ S., n(E ), ∴ P(E ) =, ≥ 0 [Q n(E ) ≥ 0 ]., n( S), (ii) P( φ) =, , n( φ), 0, =, = 0., n( S) n( S)
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970, , Senior Secondary School Mathematics for Class 11, , Then, E1 = {(1, 1), ( 2, 2), ( 3 , 3), ( 4, 4), (5 , 5), ( 6, 6)}, and E2 = {( 2, 4), ( 3 , 3), ( 4, 2)}., ∴ (E1 ∩ E2 ) = {( 3 , 3)}., Thus, n(E1 ) = 6, n(E2 ) = 3 and n(E1 ∩ E2 ) = 1., ∴ P(E1 ) =, , n(E1 ), n( S), , and P(E1 ∩ E2 ) =, , =, , n(E2 ), 6 1, 3, 1, = , P(E2 ) =, =, =, n( S), 36 6, 36 12, , n(E1 ∩ E2 ), n( S), , =, , 1, ⋅, 36, , ∴ P(getting a doublet or a total of 6), = P(E1 or E2 ) = P(E1 ∪ E2 ), = P(E1 ) + P(E2 ) − P(E1 ∩ E2 ), , [by the addition theorem], , 1, 1⎞, 8, 2, ⎛1, =⎜ +, − ⎟ =, = ⋅, ⎝ 6 12 36⎠ 36 9, Hence, the required probability is, EXAMPLE 3, , SOLUTION, , 2, ⋅, 9, , In a single throw of two dice, find the probability that neither a doublet, nor a total of 10 will appear., [CBSE 1998, 2003], Let S be the sample space. Then, n( S) = 36., Let E1 = event that a doublet appears,, and E2 = event of getting a total of 10., Then, E1 = {(1, 1), ( 2, 2), ( 3 , 3), ( 4, 4), (5 , 5), ( 6, 6)} ,, and E2 = {( 4, 6), (5 , 5), ( 6, 4)}., ∴ (E1 ∩ E2 ) = {(5 , 5)}., Thus, n(E1 ) = 6, n(E2 ) = 3 and n(E1 ∩ E2 ) = 1., ∴ P(E1 ) =, , n(E1 ), n( S), , and P(E1 ∩ E2 ) =, , =, , n(E2 ), 6 1, 3, 1, = , P(E2 ) =, =, =, n( S), 36 6, 36 12, , n(E1 ∩ E2 ), n( S), , =, , 1, ⋅, 36, , ∴ P(getting a doublet or a total of 10), = P(E1 or E2 ) = P(E1 ∪ E2 ), = P(E1 ) + P(E2 ) − P(E1 ∩ E2 ), 1, 1⎞, 8, 2, ⎛1, =⎜ +, − ⎟ =, = ⋅, ⎝ 6 12 36⎠ 36 9
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Probability, , 971, , ∴ P(getting neither a doublet nor a total of 10), = P( E1 and E2 ) = P( E1 ∩ E2 ), 2⎞ 7, ⎛, = P(E1 ∪ E2 ) = 1 − P(E1 ∪ E2 ) = ⎜1 − ⎟ = ⋅, ⎝, 9⎠ 9, 7, Hence, the required probability is ⋅, 9, EXAMPLE 4, , SOLUTION, , A natural number is chosen at random from among the first 500. What is, the probability that the number so chosen is divisible by 3 or 5?, Let S be the sample space. Then, clearly n( S) = 500., Let E1 = event of getting a number divisible by 3,, and E2 = event of getting a number divisible by 5. Then,, (E1 ∩ E2 ) = event of getting a number divisible by both 3 and 5, = event of getting a number divisible by 15., ∴ E1 = { 3 , 6, 9, … , 495 , 498} , E2 = {5 , 10, 15 , … , 495 , 500}, and (E1 ∩ E2 ) = {15 , 30, 45 , … , 495}., ⎛ 498⎞, ⎛ 500⎞, ∴ n(E1 ) = ⎜, ⎟ = 166, n(E2 ) = ⎜, ⎟ = 100, ⎝ 3 ⎠, ⎝ 5 ⎠, ⎛ 495 ⎞, and n(E1 ∩ E2 ) = ⎜, ⎟ = 33., ⎝ 15 ⎠, ∴ P(E1 ) =, , n(E1 ), n( S), , and P(E1 ∩ E2 ) =, , =, , n(E2 ) 100 1, 166 83, , P(E2 ) =, =, =, =, n( S) 500 5, 500 250, , n(E1 ∩ E2 ), n( S), , =, , 33, ⋅, 500, , ∴ P(the chosen number is divisible by 3 or 5), = P(E1 or E2 ) = P(E1 ∪ E2 ), = P(E1 ) + P(E2 ) − P(E1 ∩ E2 ), 1 33 ⎞ 233, ⎛ 83, =⎜, + −, ⋅, ⎟ =, ⎝ 250 5 500⎠ 500, 233, Hence, the required probability is, ⋅, 500, EXAMPLE 5, , SOLUTION, , A card is drawn at random from a well-shuffled deck of 52 cards. Find the, probability of its being a spade or a king., Let S be the sample space. Then, n( S) = 52., Let E1 = event of getting a spade,, and E2 = event of getting a king., Then, (E1 ∩ E2 ) = event of getting a king of spades., Clearly, n(E1 ) = 13 , n(E2 ) = 4 and n(E1 ∩ E2 ) = 1.
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972, , Senior Secondary School Mathematics for Class 11, , ∴ P(E1 ) =, , n(E1 ), , =, , n( S), , n(E2 ), 13 1, 4, 1, = , P(E2 ) =, =, =, n( S) 52 13, 52 4, , n(E1 ∩ E2 ), , and P(E1 ∩ E2 ) =, , n( S), , 1, ⋅, 52, , =, , ∴ P(getting a spade or a king), = P(E1 or E2 ) = P(E1 ∪ E2 ), = P(E1 ) + P(E2 ) − P(E1 ∩ E2 ) [by the addition theorem for two, events], 1, 1 ⎞ 16 4, ⎛1, =⎜ +, − ⎟ =, =, ⋅, ⎝ 4 13 52⎠ 52 13, Hence, the required probability is, EXAMPLE 6, , SOLUTION, , 4, ⋅, 13, , Two cards are drawn at random from a well-shuffled pack of 52 cards., What is the probability that either both are red or both are kings?, Let S be the sample space. Then,, n( S) = number of ways of drawing 2 cards out of 52, = 52C 2 ., Let E1 = event that both are red cards,, and E2 = event that both are kings., Then, (E1 ∩ E2 ) = event of getting 2 red kings., ∴ n(E1 ) = number of ways of drawing 2 red cards out of 26 red, cards = 26C 2 ., n(E2 ) = number of ways of drawing 2 kings out of 4 kings, = 4C 2 ., ∴ n(E1 ∩ E2 ) = number of ways of drawing 2 red kings out of 2, red kings = 2C 2 = 1., 4, n(E2 ), n(E1 ) 26C 2, C, = 52 ; P(E2 ) =, = 52 2, ∴ P(E1 ) =, n( S), n( S), C2, C2, and P(E1 ∩ E2 ) =, , n(E1 ∩ E2 ), n( S), , =, , 52, , 1, ⋅, C2, , ∴ P(drawing both red cards or both kings), = P(E1 or E2 ) = P(E1 ∪ E2 ), = P(E1 ) + P(E2 ) − P(E1 ∩ E2 ), ⎛ 26C, = ⎜ 52 2 +, ⎝ C2, , 4, , C2, −, C2, , 52, , 52, , 1 ⎞ ( 26C 2 + 4C 2 − 1), ⎟ =, 52, C2 ⎠, C2
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Probability, , =, , ( 325 + 6 − 1), 330, 55, =, =, ⋅, 1326, 1326 221, , Hence, the required probability is, EXAMPLE 7, , SOLUTION, , 973, , 55, ⋅, 221, , A box contains 100 bolts and 50 nuts. It is given that 50% bolts and 50%, nuts are rusted. Two objects are selected from the box at random. Find, the probability that either both are bolts or both are rusted. [CBSE 2003C], Total number of objects = (100 + 50) = 150., Let S be the sample space. Then,, n( S) = number of ways of selecting 2 objects out of 150, = 150C 2 ., Number of rusted objects, = (50% of 100) + (50% of 50) = (50 + 25) = 75., Let E1 = event of selecting 2 bolts out of 100 bolts,, and E2 = event of selecting 2 rusted objects out of 75 rusted objects., ∴ (E1 ∩ E2 ) = event of selecting 2 rusted bolts out of 50 rusted, bolts, ∴ n(E1 ) = number of ways of selecting 2 bolts out of 100, = 100C 2 ., ∴ n(E2 ) = number of ways of selecting 2 rusted objects out of 75, =, , 75, , C2 ., , ∴ n(E1 ∩ E2 ) = number of ways of selecting 2 rusted bolts out of 50, = 50C 2 ., ∴ P(E1 ) =, , n(E1 ), n( S), , and P(E1 ∩ E2 ) =, , =, , 100, , 75, , 150, , 150, , n(E2 ), C2, , P(E2 ) =, =, n( S), C2, , n(E1 ∩ E2 ), n( S), , =, , C2, C2, , 50, , C2, ⋅, C2, , 150, , P(selecting both bolts or both rusted objects), = P(E1 or E2 ) = P(E1 ∪ E2 ), = P(E1 ) + P(E2 ) − P(E1 ∩ E2 ), =, =, , 100, , C2, +, 150, C2, , 75, , C2, −, 150, C2, , C 2 (100C 2 + 75C 2 − 50C 2 ), =, 150, 150, C2, C2, 50, , ( 4950 + 2775 − 1225), 6500, 260, =, =, = 0.58., 11175, 11175 447, , Hence, the required probability is 0.58.
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976, , Senior Secondary School Mathematics for Class 11, , 1 1 1 1 1, 1 ⎞ 28 7, ⎛ 1, =⎜, + + −, − −, + ⎟ =, =, ⋅, ⎝ 13 4 2 52 4 26 52⎠ 52 13, 7, Hence, the required probability is, ⋅, 13, , EXERCISE 31B, 1. If A and B are two events associated with a random experiment for which, P( A) = 0.60, P( A or B) = 0.85 and P( A and B) = 0.42, find P( B). [CBSE 1998], 2. Let A and B be two events associated with a random experiment for which, P( A) = 0.4, P( B) = 0.5 and P( A or B) = 0.6. Find P( A and B)., 3. In a random experiment, let A and B be events such that P( A or B) = 0.7,, P( A and B) = 0.3 and P( A ) = 0.4. Find P( B)., 4. If A and B are two events associated with a random experiment such that, P( A) = 0.25, P( B) = 0.4 and P( A or B) = 0.5, find the values of, (i) P( A and B), (ii) P( A and B ), 5. If A and B be two events associated with a random experiment such that, P( A) = 0.3 , P( B) = 0.2 and P( A ∩ B) = 0.1, find, (i) P( A ∩ B), (ii) P( A ∩ B ), 6. If A and B are two mutually exclusive events such that P( A) = (1/2) and, P( B) = (1/ 3), find P( A or B)., 7. Let A and B be two mutually exclusive events of a random experiment such, that P(not A) = 0.65 and P(A or B) = 0.65, find P(B)., 8. A , B , C are three mutually exclusive and exhaustive events associated with, a random experiment., If P( B) = ( 3/2) P( A) and P(C) = (1/2) P( B), find P( A)., 9. The probability that a company executive will travel by plane is ( 2/5) and, that he will travel by train is (1/ 3). Find the probability of his travelling by, plane or train., 10. From a well-shuffled pack of 52 cards, a card is drawn at random. Find the, probability of its being a king or a queen., 11. From a well-shuffled pack of cards, a card is drawn at random. Find the, probability of its being either a queen or a heart., 12. A card is drawn at random from a well-shuffled deck of 52 cards. Find the, probability of its being a spade or a king., 13. A number is chosen from the numbers 1 to 100. Find the probability of its, being divisible by 4 or 6., 14. A die is thrown twice. What is the probability that at least one of the two, throws comes up with the number 4?
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Probability, , 977, , 15. Two dice are tossed once. Find the probability of getting an even number on, the first die or a total of 8., [CBSE 2004], 16. Two dice are thrown together. What is the probability that the sum of the, numbers on the two faces is neither divisible by 3 nor by 4., 17. In a class, 30% of the students offered mathematics, 20% offered chemistry, and 10% offered both. If a student is selected at random, find the, probability that he has offered mathematics or chemistry., 18. The probability that Hemant passes in English is (2/3) and the probability, that he passes in Hindi is (5/9). If the probability of his passing both the, subjects is (2/5), find the probability that he will pass in at least one of these, subjects., 19. The probability that a person will get an electrification contract is (2/5) and, the probability that he will not get a plumbing contract is (4/7). If the, probability of getting at least one contract is (2/3), what is the probability, that he will get both?, [CBSE 2005C], 20. The probability that a patient visiting a dentist will have a tooth extracted is, 0.06, the probability that he will have a cavity filled is 0.2, and the, probability that he will have a tooth extracted or a cavity filled is 0.23. What, is the probability that he will have a tooth extracted as well as a cavity, filled?, 21. In a town of 6000 people, 1200 are over 50 years old and 2000 are females. It, is known that 30% of the females are over 50 years. What is the probability, that a randomly chosen individual from the town is either female or over, 50 years?, ANSWERS (EXERCISE 31B), , 1. 0.67, 7. 0.3, 15., , 5, 9, , 2. 0.3, , 3. 0.4, , 4, 13, 4, 16., 9, , 11, 15, 2, 17., 5, , 8., , 9., , 4. (i) 0.15 (ii) 0.1, 2, 13, 37, 18., 45, , 10., , 4, 13, 17, 19., 105, , 11., , 5. (i) 0.1 (ii) 0.2, 33, 100, 13, 20. 0.03 21., 30, 12., , 4, 13, , 13., , 5, 6, 11, 14., 36, 6., , HINTS TO SOME SELECTED QUESTIONS, 4. (ii) P( A and B ) = P( A ∩ B ) = P( A ) − P( A ∩ B)., 5. (i) P( A ∩ B) = P( B) − P( A ∩ B)., 6. A ∩ B = φ ⇒ P( A ∪ B) = P( A ) + P( B)., 8. A , B, C being mutually exclusive exhaustive events, we have, 3, 1 3, P( A ) + P( B) + P(C ) = 1 ⇒ P( A ) + P( A ) + ⋅ ⎧⎨ P( A )⎫⎬ = 1, 2, 2 ⎩2, ⎭, ⇒, , 13, 4, ⋅ P( A ) = 1 ⇒ P( A ) =, ⋅, 4, 13
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978, , Senior Secondary School Mathematics for Class 11, , 9. The events of travelling by plane and that by train are mutually exclusive., ∴ P( E1 ∪ E 2 ) = P( E1 ) + P( E 2 )., 10. The probability of getting a king and that of getting a queen are mutually exclusive., 4, 4, ∴ P(a king or a queen) = P(a king) + P(a queen) = ⎛⎜ + ⎞⎟ ⋅, ⎝ 52 52 ⎠, 11. Clearly, n( S) = 52., Let E1 = event of getting a queen,, and E 2 = event of getting a heart., Then, E1 ∩ E 2 = event of getting a queen of hearts., ∴ n( E1 ) = 4 , n( E 2 ) = 13 and n( E1 ∩ E 2 ) = 1., Now, use P( E1 ∪ E 2 ) = P( E1 ) + P( E 2 ) − P( E1 ∩ E 2 )., 13. Here, S = {1, 2 , 3 , 4 , … , 99 , 100 } ⇒ n( S ) = 100 ., Let E1 = event of getting a number divisible by 4., And, E 2 = event of getting a number divisible by 6., Then, E1 ∩ E 2 = event of getting a number divisible by both 4 and 6, i.e., divisible, by 12 (LCM of 4 and 6)., ∴ E1 = {4 , 8 , 12 , … , 100 } ⇒ n( E1 ) = 25 ., E 2 = {6 , 12 , 18 , … , 96 } ⇒ n( E2 ) = 16 ., ( E1 ∩ E 2 ) = {12 , 24 , 36 , … , 96 } ⇒ n( E1 ∩ E 2 ) = 8 ., Now, use P( E1 ∪ E 2 ) = P( E1 ) + P( E 2 ) − P( E1 ∩ E 2 )., 14. Clearly, n( S ) = ( 6 × 6 ) = 36 ., Let E1 = event that the first throw shows a 4,, and E 2 = event that the second throw shows a 4., Then, ( E1 ∩ E 2 ) = event that each throw shows a 4., ∴ E1 = {( 4 , 1), ( 4 , 2 ), ( 4 , 3 ), ( 4 , 4 ), ( 4 , 5 ), ( 4 , 6 )},, and E 2 = {( 1, 4 ), ( 2 , 4 ), ( 3 , 4 ), ( 4 , 4 ), (5 , 4 ),( 6 , 4 )}., ∴ ( E1 ∩ E 2 ) = {( 4 , 4 )}., Now, use P( E1 ∪ E 2 ) = P( E1 ) + P( E 2 ) − P( E1 ∩ E 2 )., 15. Clearly, n( S ) = ( 6 × 6 ) = 36 ., Let E1 = event of getting an even number on the first die,, and E 2 = event of getting a total of 8 on both dice. Then,, E1 = {( 2 , 1), ( 2 , 2 ), … , ( 2 , 6 ), ( 4 , 1), ( 4 , 2 ), … , ( 4 , 6 ), ( 6 , 1), ( 6 , 2 ), … , ( 6 , 6 )},, E 2 = {( 2 , 6 ), ( 3 , 5 ), ( 4 , 4 ), (5 , 3 ), ( 6 , 2 )}., ∴ ( E1 ∩ E 2 ) = {( 2 , 6 ), ( 4 , 4 ), ( 6 , 2 )}., ∴ n( E1 ) = 18 , n( E 2 ) = 5 and n( E1 ∩ E 2 ) = 3., Now, use P( E1 ∪ E 2 ) = P( E1 ) + P( E 2 ) − P( E1 ∩ E 2 )., 16. Clearly, n( S ) = ( 6 × 6 ) = 36 ., Let E1 = event that the sum of numbers on two faces is divisible by 3,, and E 2 = event that the sum of numbers on two faces is divisible by 4., Then, E1 ∩ E 2 = event that the sum of the numbers on two faces is divisible by both, 3 and 4, i.e., by 12.
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Probability, , 979, , ∴ E1 = {( 1, 2 ), ( 1, 5 ), ( 2 , 1), ( 2 , 4 ), ( 3 , 3 ), ( 3 , 6 ), ( 4 , 2 ), ( 4 , 5 ), (5 , 1), (5 , 4 ), ( 6 , 3 ), ( 6 , 6 )},, E 2 = {1, 3 ), ( 2 , 2 ), ( 2 , 6 ), ( 3 , 1), ( 3 , 5 ), ( 4 , 4 ), (5 , 3 ), ( 6 , 2 ), ( 6 , 6 )}., ∴ E1 ∩ E 2 = {( 6 , 6 )}., Thus, n( E1 ) = 12 , n( E 2 ) = 9 and n( E1 ∩ E 2 ) = 1., Required probability = P[(not E1 ) and (not E 2 )] = P( E1 ∩ E 2 ), = P( E1 ∪ E 2 ) = 1 − P( E1 ∪ E 2 ), = 1 − {P( E1 ) + P( E 2 ) − P( E1 ∩ E 2 )}., 17. Here, n( S) = 100 , n( M ) = 30 , n(C ) = 20 and n( M ∩ C ) = 10 ., 30, 3, 20 1, 10, 1, ∴ P( M ) =, =, ; P(C ) =, = and P( M ∩ C ) =, =, ⋅, 100 10, 100 5, 100 10, Use P( M ∪ C ) = P( M ) + P(C ) + P( M ∩ C )., 18. P( E) =, , 2, 5, 2, , P( H ) = and P( E ∩ H ) = ⋅ Find P( E ∪ H )., 3, 9, 5, , 19. Let E1 = event of getting an electrification contract,, and E 2 = event of getting a plumbing contract., 2, 4, 4, 3, Then, P( E1 ) = , P( not E 2 ) = ⇒ P( E 2 ) = ⎛⎜ 1 − ⎞⎟ = ⋅, ⎝, 5, 7, 7⎠ 7, And, P( E1 ∪ E 2 ) = P( E1 or E2 ) =, , 2, ⋅, 3, , ∴ required probability = P( E1 ∩ E 2 ), = P( E1 ) + P( E 2 ) − P( E1 ∪ E 2 )., 20. P( E1 ) = 0.06 , P( E 2 ) = 0.2 and P( E1 ∪ E 2 ) = 0.23., P( E1 ∩ E 2 ) = P( E1 ) + P( E 2 ) − P( E1 ∪ E 2 )., 21. Let E1 = event of person being a female,, and E 2 = event of person being 50 years old., Then, n( E1 ) = 2000 , n( E 2 ) = 1200,, and n( E1 ∩ E 2 ) = ( 30% of 2000) = 600., Now, n( E1 ∪ E 2 ) = n( E1 ) + n( E 2 ) − n( E1 ∩ E 2 ) = 2600., 2600 13, =, ⋅, ∴ P( E1 ∪ E 2 ) =, 6000 30, , q
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Logarithm, LOGARITHM, , If a is a positive real number and a ≠ 1 such that a x = m , then x is called the logarithm, of m to the base a, written as log a m., ∴ a x = m ⇔ log a m = x., LAWS OF LOGARITHM, , (i) log a ( mn) = ( log a m) + ( log a n), ⎛ m⎞, (ii) log a ⎜ ⎟ = ( log a m) − ( log a n), ⎝ n⎠, (iii) log a ( mk ) = k ⋅ ( log a m), (iv) log a 1 = 0, (v) log a a = 1, ⎛ log m⎞ ⎛ log b m⎞, (vi) log a m = ⎜, ⎟=⎜, ⎟, ⎝ log a ⎠ ⎝ log b a ⎠, (vii) log a m =, , 1, log ba, , COMMON LOGARITHMS, , log10 a is called common logarithm, written as log a., EXAMPLES, , (i) log 10 = 1, (ii) log 100 = log (10) 2 = 2, 1, ⎛ ⎞, (iii) log ⎜ ⎟ = log (10) −1 = −1, etc., ⎝ 10⎠, , Standard Form of a Decimal, , Any number in decimal form expressed as the product of a number between 1 and 10,, and an integral power of 10, is in standard form., , (i), (ii), (iii), (iv), (v), , Number, 27.3, 106.34, 0.49, 0.08, 0.00037, , Standard form, 2.73 × 10 1, 1.0634 × 10 2, 4.9 × 10 −1, 8 × 10 − 2, 3.7 × 10 − 4, 980
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Logarithm, , 981, , CHARACTERISTIC AND MANTISSA OF A LOGARITHM, , log m consists of two parts:, (i) the integral part is called characteristic,, (ii) the decimal part is called mantissa., Mantissa is always taken as positive, while the characteristic may be, positive or negative., Instead of −2, we write, 2., REMARK 2 .7384 means ( −2 + 0.7384)., CHARACTERISTIC OF log m, , (i) If m ≥ 1, then its characteristic is one less than the number of digits to the, left of the decimal point., (ii) If m < 1, then its characteristic is a negative number whose numerical, value is one more than the number of zeros between the decimal point, and the first nonzero figure., p, ANOTHER METHOD Write the number in standard form: m × 10 . Then, its, characteristic = p., EXAMPLES, , Number, , Standard form, , Characteristic, , 5376.4, 53.764, 5.3764, 0.53764, 0.00537, , 5. 3764 × 10, 5. 3764 × 10 1, 5. 3764 × 10 0, 5. 3764 × 10 −1, 5. 37 × 10 − 3, , 3, 1, 0, −1, −3, , 3, , MANTISSA OF log m, , We find mantissa from log table., Position of the decimal point in a number is immaterial for mantissa., Mantissa of log (56), log (.56), log (.0056) is the same., For finding mantissa of 4385 from log table, we proceed in the row headed, by 43 and in this row, we find the number under the column headed by 8., Now, to this number, we add the mean difference headed by 5 and in the same, row headed by 43., Thus, mantissa for 4385 is ( 6415 + 5) = .6420., REMARK 1 For finding the mantissa of 43 from log table, we find the number in, the row headed by 43 under the column 0. It is 0.6335., REMARK 2 For finding the mantissa of 4 from log table, we find the number in, the row headed by 40 and under the column 0. It is 0.6021., Thus, we have:, (i) log 4385 = 3.6420, (ii) log 43.85 = 1.6420, (iii) log 4.385 = 0.6420, (iv) log 0.4385 = 1.6420, (v) log 0.04385 = 2.6420, etc.
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982, , Senior Secondary School Mathematics for Class 11, , ANTILOGARITHM, EXAMPLE, , If log m = n, then m = antilog n., , log 1000 = 3 ⇒ antilog 3 = 1000., , Finding Antilogarithm of a Number, , For finding the antilog of a number, we use its decimal part and read the, antilog table in the same manner as log table., After finding the number from antilog table, we insert the decimal point, as under., (i) When the characteristic is n, the decimal point is inserted after (n + 1)th, digit., (ii) When the characteristic is n , the decimal point is inserted in such a way, that the first significant digit is at the nth place., EXAMPLES, , Number, , Antilog, , Number, , Antilog, , 0.2364, , 1.724, , 1.4356, , 0.2727, , 1.4216, , 26.40, , 2 .0318, , 0.01076, , 2.6019, , 399.08, , 3.1459, , 0.001399, , 5.2612, , 182500, , 4 .7912, , 0.0006183, , SOME EXAMPLES, EXAMPLE 1, , SOLUTION, , EXAMPLE 2, , SOLUTION, , EXAMPLE 3, , Find the value of:, (i) log 3, (ii) log 23.45, (iv) log 0. 325, (v) log 0.0954, Using log table, we have, (i) log 3 = 0.4771, (ii) log 23.45 = 1.3701, (iii) log 4.17 = 0.6201, (iv) log 0.325 = 1.5119, (v) log 0.0954 = 2.9795, (vi) log 0.00056 = 4.7482, Find the value of:, (i) antilog 0.654, (ii) antilog 1.204, (iii) antilog 1. 3612, (iv) antilog 3.4568, We have, (i) antilog (0.654) = 4.508, (ii) antilog (1.204) = 16.00, (iii) antilog (1.3612) = 0.2297, (iv) antilog (3.4568) = 0.002863, Evaluate, , 8492 × 3.72, ⋅, 47.8 × 52.24, , (iii) log 4.17, (vi) log 0.00056
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Logarithm, , SOLUTION, , 8492 × 3.72, ⋅ Then,, 47.8 × 52.24, log x = log 8492 + log 3.72 − log 47.8 − log 52.24, , Let x =, , = 3.9290 + 0.5705 − 1.6794 − 1.7180, = 4.4995 − 3.3974 = 1.1021, ⇒ x = antilog (1.1021) = 12.65., Hence, the value of the given expression is 12.65., EXAMPLE 4, , SOLUTION, , 563.4 × 3 0.4573, ⋅, (6.15) 3, , Evaluate, Let x =, , 563.4 × 3 0.4573, ⋅ Then,, ( 615, . )3, , log x = log 563.4 +, , 1, log ( 0.4573) − 3 log ( 6.15), 3, , 1, ( 1.6602) − 3 × 0.7889, 3, 1, = 2.7508 + ( −0.3398) − 2.3667, 3, = 2.7508 − 0.1133 − 2.3667, = 0.2708, ⇒ x = antilog ( 0.2708) = 1.865., Hence, the value of the given expression is 1.865., = 2.7508 +, , EXAMPLE 5, SOLUTION, , Evaluate, , 3, , 0.08034., , Let x = 0.08034. Then,, 1, log x = log (0.08034)., 3, 1, 1, 1, = ( 2.9049) = ( −2 + 0.9049) = ( −1.0951), 3, 3, 3, = − 0.3650 = −1 + (1 − 0.3650) = 1.6350, 3, , ⇒ x = antilog ( 1.6350) = 0.4315., EXAMPLE 6, SOLUTION, , Evaluate, , 7, , 0.00003587 ., , Let x = 7 0.00003587 . Then,, 1, log x = log (0.00003587), 7, 1, 1, = ( 5 .5548) = ( −5 + 0.5548), 7, 7, 1, = ( −4.4452) = −0.6350, 7, = −1 + (1 − 0.6350) = 1. 3650, , 983
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984, , Senior Secondary School Mathematics for Class 11, , ⇒ x = antilog ( 1. 3650) = 0.2317., Hence, the required value is 0.2317., , SOLUTION, , (76.24)5 × 3 65, ⋅, ( 3.2) 7 × 17, , Evaluate, , EXAMPLE 7, , (76.24)5 × 3 65, ⋅ Then,, ( 3.2) 7 × 17, 1, ⎧, ⎫, 5, 1, ⎪(76.24) × ( 65) 3 ⎪, log x = ⋅ log ⎨, 1 ⎬, 2, ⎪ (3.2) 7 × (17) 2 ⎪, ⎭, ⎩, 1, 1, 1 ⎧, ⎫, = ⋅ ⎨5 log (76.24) + log 65 − 7 log (3.2) − log 17 ⎬, 2, 3, 2 ⎩, ⎭, 1, 1, 1 ⎧, ⎫, = ⋅ ⎨5 × 1.8822 + × 1.8129 − 7 × 0.5051 − × 1.2304⎬, 2, 3, 2 ⎩, ⎭, 1, = ⋅ {9.4110 + 0.6043 − 3.5357 − 0.6152}, 2, 1, 1, = × (10.0153 − 4.1509) = × 5.8644 = 2.9322, 2, 2, ⇒ x = antilog (2.9322) = 855.5., Hence, the required value of the given expression is 855.5., , Let x =, , EXERCISE 1, Using log table evaluate the following:, 1. 69.13 × 0.34 × 0.014, , 2. 0.7625 × 0.000357, 8.67 × 99, 4., 1.78, 6. 5 8.0125, , 3. 358.6 × 0.078 × 0.5943, 5. (0.09634) 3, 7., , 7, , 142.71, , 8., , 0.9876 × (16.42) 2, 1, , (4.567) 3, 1, , 9., , (6.45) 3 × (0.00034) 3 × (981.4), 1, , (9.37) 2 × (8.93) 4 × (0.0617), ANSWERS (EXERCISE 1), , 1. 0.3291, , 2. 0.0002723, , 3. 16.63, , 4. 19.19, , 5. 0.0008941, , 6. 7.517, , 7. 2.032, , 8. 160.4, , 9. 1963, q
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Appendix, , Mathematical Modelling, , WHAT IS MATHEMATICAL MODELLING?, , Mathematical modelling consists of translating a real-world problem into a, mathematical problem, solving it and interpreting the solution in the language of, real world., There are four steps involved in mathematical modelling., Step 1, , Convert the given problem into a mathematical problem., , Step 2, , Solve the mathematical problem., , Step 3, , Interpret the result for the real situation., , Step 4, , If need arises, modify the model., , The technique of modelling can be expressed through the diagram given, below., , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , The angle of elevation of the top of a tower from a point P on the ground,, which is 600 m away from the foot of the tower, measured by a sextant, is, 30°. Find the height of the tower., B, Step 1. First we try to understand the problem., AB is a tower. Let its height be h metres., Then, AB = h metres., , h, , Let P be the position of the observer., Then, PA = 600 m., , P, , 985, , 30º, d, , A
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986, , Senior Secondary School Mathematics for Class 11, , Let B be the point on the top of the tower. Then, ∠APB = 30°., The real problem is to find the value of h in metres., The three quantities involved in the problem are, height AB = h metres, distance PA = 600 m and ∠APB = 30°, h, We use the formula: tan θ =, ⇔ h = d tan θ., d, , Step 2., , Step 3., , In equation (i), we put d = 600 m and θ = 30°., , ⎛, 1 ⎞, 1, ⎛, ×, ∴ h = 600 m × tan 30° = ⎜ 600 ×, ⎟ m = ⎜ 600 ×, ⎝, 3⎠, 3, ⎝, = ( 200 ×, Step 4., , 3⎞, ⎟ m, 3⎠, , 3 ) m = ( 200 × 1.732) m = 346.4 m., , Hence, the height of the tower is 346.4 m., , OUR PROCEEDINGS, , Studied the real-life problem and identified three, parameters: height = h m, distance AP = d m and ∠APB = θ °., , Step 1., Step 2., , Using trigonometry, we applied the formula, h = d tan θ., , Step 3., , Putting the known values of d and θ, we find h., , This gives us the solution of the problem., EXAMPLE 2, , Suppose the present population of a city is 100000 and we want to find its, population, say after 10 years., , FORMULATION, , Let P(t) be the population in a certain year t., , Let B(t) be the number of births and D(t) be the number of deaths, between the years t and t + 1. Then,, P(t + 1) = P(t) + B(t) − D(t)., B(t), D(t), Let, = b and, = d. Then,, P(t), P(t), , … (i), , P(t + 1) = P(t) + b P(t) − d P(t), ⇒ P(t + 1) = (1 + b − d) P(t)., , … (ii), , Putting t = 0 in (ii), we get P(1) = (1 + b − d) P( 0)., , … (iii), , Putting t = 1 in (ii), we get P( 2) = (1 + b − d) P(1), = (1 + b − d) 2 P( 0), , [using (iii)]., , Thus, P( 2) = (1 + b − d) P( 0)., 2, , Continuing in this way, we get:, P(t) = (1 + b − d) t P( 0) for t = 0, 1, 2, …, ⇒ P(t) = P( 0) × r t, where (1 + b − d) = r.
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Appendix, , Mathematical Modelling, , 987, , SOLUTION OF GIVEN PROBLEM, , Suppose it is given that P( 0) = 100000, b = 0.02 and d = 0.01., Then, P(10) = (1.01)10 × 100000, [let (1.01)10 = 1.104622125 be given], = (1.104622125 × 100000) = 1104622.125., VALIDATION AND INTERPRETATION, , Since we cannot have the number of persons in decimal fraction, the above, result is not valid., So, we take the population as 1104622 approximately., r
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OBJECTIVE QUESTIONS, , Complex Numbers, EXERCISE 1, Mark (3) against the correct answer in each of the following., , 1. i 91 = ?, (a) –1, 2. i, , 326, , (b) 1, , (c) i, , (d) –i, , (b) –1, , (c) i, , (d) –i, , (b) –i, , (c) 1, , (d) –1, , (b) 1, , (c) i, , (d) –i, , (b) –1, , (c) i, , (d) –i, , (c) 1, , (d) –1, , =?, , (a) 1, 3. i 273 = ?, (a) i, 4. i, , 124, , =?, , (a) –1, 5. i, , −75, , =?, , (a) 1, 6. i, , −38, , =?, , (a) i, , (b) –i, , 7. For any positive integer n, ( − −1), (a) 1, , 4n+ 3, , =?, , (c) i, , (b) –1, , (d) –i, , 8. If i = −1 and n is a positive integer then (i + i, n, , (b) i n, , (a) i, 9. (i, , 109, , +i, , 114, , +i, , 119, , +i, , 10. 3i, , + 5i, , 27, , − 2i, , 38, , (a) 1, , + 5i, , +i, , n+ 2, , + i n+ 3) = ?, , (c) 1, , (d) 0, , (c) −2i, , (d) 2, , (c) − 4i, , (d) 10i, , ) =?, , (b) i, , (a) 0, 34, , 124, , n +1, , 41, , =?, , (b) –1, , 11. If a and b are integers then a × b = ab is true only when, (a) a and b are both positive, (b) a and b are both negative, (c) a and b are both zero, (d) at least one of a and b is non-negative, 12., , −9 × −25 = ?, (a) 15, , (c) 15i, , (b) –15, 988, , (d) −15i
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Objective Questions, , 989, , 13. Which of the following statements is correct?, (a) (5 + 7i ) > ( 3 + 4i ), (b) (5 + 7i ) < ( 3 + 4i ), (c) ( 3 + 5i ) > ( 4 + 3i ), (d) none of these, 14. Which of the following statements is correct?, (a) 3i > 2i, (b) 1 + 3i > 2i, (c) i > 0, 15. Which of the following statements is correct?, (a) ( 2 + 3i ) > ( 2 − 3i ), (c) (5 + 4i ) > ( −5 − 4i ), , (d) none of these, , (b) ( 3 + 2i ) > ( −3 + 2i ), (d) none of these, , 2, , ⎛ 1 − i⎞, 16. ⎜, ⎟ =?, ⎝ 1 + i⎠, (a) 1, 17. If ( a + i b) =, , (b) –1, , (c), , −1, 2, , (d), , 1, 2, , 1+i, then the value of ( a 2 + b 2 ) is, 1 −i, , (a) 1, , (b) –1, , (c) 2, , (d) –2, n, , ⎛ 1 + i⎞, 18. The smallest positive integer n for which ⎜, ⎟ = 1 is, ⎝ 1 − i⎠, (a) 2, 19. If ( a 2 + b 2 ) = 1 then, (a) ( a + ib), , (b) 3, , (c) 4, , (d) 6, , (b) ( b + ia), , (c) i( a + b), , (d) none of these, , (b) quadrant II, , (c) quadrant III, , (d) quadrant IV, , ( a 2 + b2 ), ( c2 − d 2 ), , (d) none of these, , ⎛ −1 1 ⎞, (b) ⎜, + i⎟, ⎝ 2 2⎠, , ⎛1 1 ⎞, (c) ⎜ − i⎟, ⎝2 2 ⎠, , (d) none of these, , 4 ⎞, ⎛ −3, (b) ⎜, +, i⎟, ⎝ 25 25 ⎠, , ⎛ −3 4 ⎞, (c) ⎜, − i⎟, ⎝ 25 25 ⎠, , (d) none of these, , (1 + b + ia), =?, (1 + b − ia), , ⎛ 1 + 2i ⎞, 20. ⎜, ⎟ lies in, ⎝ 1 −i ⎠, (a) quadrant I, , ⎛ a + ib ⎞, 2, 2, 21. If ( x + iy) = ⎜, ⎟ then ( x + y ) = ?, ⎝ c + id ⎠, (a), , ( a 2 + b2 ), ( c2 + d 2 ), , (b), , ( a 2 − b2 ), ( c2 + d 2 ), , (c), , 22. (1 + i ) −1 = ?, (a) ( 2 − i ), 23. (1 − 2i ) −2 = ?, 4 ⎞, ⎛ 3, (a) ⎜, − i⎟, ⎝ 25 25 ⎠
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990, , Senior Secondary School Mathematics for Class 11, , 24. (1 − i ) −3 = ?, ⎛1 1 ⎞, (a) ⎜ − i⎟, ⎝4 4 ⎠, , ⎛ −1 1 ⎞, (b) ⎜, + i⎟, ⎝ 4 4⎠, , ⎛ −1 1 ⎞, (c) ⎜, − i⎟, ⎝ 4 4⎠, , (d) none of these, , (c) ( − 6 + 17i ), , (d) none of these, , ⎛ 21 1 ⎞, (c) ⎜ + i⎟, ⎝ 13 13 ⎠, , (d) none of these, , 25. ( 2 − 3i )( −3 + 4i ) = ?, (a) ( 6 + 17i ), , (b) ( 6 − 17i ), , 26. ( 3 − 5i ) ÷ ( −2 + 3i ) = ?, ⎛ 21 1 ⎞, ⎛ −21 1 ⎞, (a) ⎜ − i⎟, (b) ⎜, + i⎟, ⎝ 13 13 ⎠, ⎝ 13 13 ⎠, ⎛ 2 − −9 ⎞, 27. If ⎜, ⎟ = ( x + iy) then, ⎝ 1 − −4 ⎠, 2, 3, (a) x = , y =, 5, 5, 8, 1, (c) x = , y =, 5, 5, , (b) x =, , 3, 2, ,y=, 5, 5, , (d) none of these, , 28. (1 − −1)(1 + −1)(5 − −7 )(5 + −7 ) = ?, (a) ( 25 + 7i ), , (b) ( 32 + 5 i ), , (c) ( 29 − 3 i ), , 29. The multiplicative inverse of ( −2 + 5i ) is, ⎛ 2 5 ⎞, ⎛ −2 5 ⎞, ⎛ −2 5 ⎞, (a) ⎜, (b) ⎜ − i⎟, (c) ⎜, + i⎟, − i⎟, ⎝ 29 29 ⎠, ⎝ 29 29 ⎠, ⎝ 29 29 ⎠, 30. The multiplicative inverse of ( 3 + 2i ) 2 is, 12 ⎞, 12 ⎞, 12 ⎞, ⎛ −5, ⎛ 5, ⎛ 5, (a) ⎜, +, −, +, i⎟ (b) ⎜, i⎟ (c) ⎜, i⎟, ⎝ 169 169 ⎠, ⎝ 169 169 ⎠, ⎝ 169 169 ⎠, 31. The complex number z such that, (a) the x-axis, , (a) the x-axis, , (b) the y-axis, , 33. The complex number z such that, (a) the x-axis, , (b) the y-axis, , 5 ⎞, ⎛ 2, (d) ⎜ + i⎟, ⎝ 29 29 ⎠, , (d) none of these, , z −i, = 1 lies on, z+i, , (b) the line y = 1 (c) a circle, , 32. The complex number z such that, , (d) none of these, , (d) none of these, , z − 5i, = 1 lies on, z + 5i, (c) a circle, , (d) none of these, , z−2, = 2 lies on, z+2, (c) a circle, , (d) none of these, , 34. –2, when expressed in polar form, is, π, π⎞, ⎛, (b) 2(cos π + i sin π ), (a) −2 ⎜ cos + i sin ⎟, ⎝, 2, 2⎠, (c) 2(cos 2π + i sin 2π ), , (d) none of these
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Objective Questions, , 35. −3i, when expressed in polar form, is, ⎡, ⎛ π⎞, ⎛ π⎞ ⎤, (a) 3 ⎢cos ⎜ − ⎟ + i sin ⎜ − ⎟ ⎥, ⎝, ⎠, ⎝ 2⎠ ⎦, 2, ⎣, π, π⎤, ⎡, (c) −3 ⎢cos + i sin ⎥, 2, 2⎦, ⎣, , π, π⎤, ⎡, (b) 3 ⎢cos + i sin ⎥, 2, 2⎦, ⎣, (d) none of these, , 36. (1 + i ), when expressed in polar form, is, π, π⎤, π, π⎤, ⎡, ⎡, (a) 2 ⎢cos + i sin ⎥, (b) 2 ⎢cos + i sin ⎥, 2, 2⎦, 4, 4⎦, ⎣, ⎣, π, π, ⎡, ⎤, (d) none of these, (c) 2 ⎢cos + i sin ⎥, 4, 4⎦, ⎣, 37. ( −1 + i 3 ), when expressed in polar form, is, (a) 2(cos π + i sin π ), , π, π⎞, ⎛, (b) 2 ⎜ cos + i sin ⎟, ⎝, 3, 3⎠, , 2π, 2π ⎞, ⎛, (c) 2 ⎜ cos, + i sin, ⎟, ⎝, 3, 3⎠, , ⎡, ⎛ −2π ⎞, ⎛ −2π ⎞ ⎤, (d) 2 ⎢cos ⎜, ⎟ + i sin ⎜, ⎟, ⎝, ⎠, ⎝ 3 ⎠ ⎥⎦, 3, ⎣, , 38. ( 3 + i ), when expressed in polar form, is, π, π⎞, π, π⎞, ⎛, ⎛, (a) 2 ⎜ cos + i sin ⎟, (b) 2 ⎜ cos + i sin ⎟, ⎝, ⎝, 3, 3⎠, 6, 6⎠, 2π, 2π ⎞, ⎛, (c) 2 ⎜ cos, + i sin, ⎟, ⎝, 3, 3⎠, , (d) none of these, , 39. (1 + i ), when expressed in polar form, is, π, π⎞, π, π⎞, ⎛, ⎛, (a) 2 ⎜ cos + i sin ⎟, (b) 2 ⎜ cos + i sin ⎟, ⎝, ⎝, 4, 4⎠, 6, 6⎠, π, π⎞, ⎛, (c) 2 ⎜ cos + i sin ⎟, ⎝, 3, 3⎠, , (d) none of these, , 40. ( −1 − i ), when expressed in polar form, is, ⎧, ⎛ −π ⎞, ⎛ −π ⎞ ⎫, (a) 2 ⎨cos ⎜ ⎟ + i sin ⎜ ⎟ ⎬, ⎝ 4⎠, ⎝ 4 ⎠⎭, ⎩, , ⎧, ⎛ −3 π ⎞, ⎛ −3 π ⎞ ⎫, (b) 2 ⎨cos ⎜, ⎟ + i sin ⎜, ⎟, ⎝ 4 ⎠, ⎝ 4 ⎠ ⎬⎭, ⎩, , ⎧, ⎛ −2π ⎞, ⎛ −2π ⎞ ⎫, (c) 2 ⎨cos ⎜, ⎟ + i sin ⎜, ⎟, ⎝, ⎠, ⎝ 3 ⎠ ⎬⎭, 3, ⎩, , ⎧, ⎛ −π ⎞, ⎛ −π ⎞ ⎫, (d) 2 ⎨cos ⎜ ⎟ + i sin ⎜ ⎟ ⎬, ⎝, ⎠, ⎝ 2 ⎠⎭, 2, ⎩, , 41. ( −1 − 3i ), when expressed in polar form, is, ⎡, ⎛ −π ⎞, ⎛ −π ⎞ ⎤, (a) 2 ⎢cos ⎜ ⎟ + i sin ⎜ ⎟ ⎥, ⎝, ⎠, ⎝ 3 ⎠⎦, 3, ⎣, , ⎡, ⎛ −2π ⎞, ⎛ −2π ⎞ ⎤, (b) 2 ⎢cos ⎜, ⎟ + i sin ⎜, ⎟, ⎝, ⎠, ⎝ 3 ⎠ ⎥⎦, 3, ⎣, , ⎡, ⎛ −3 π ⎞, ⎛ −3 π ⎞ ⎤, (c) 2 ⎢cos ⎜, ⎟ + i sin ⎜, ⎟, ⎝, ⎠, ⎝ 4 ⎠ ⎥⎦, 4, ⎣, , (d) none of these, , 991
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992, , Senior Secondary School Mathematics for Class 11, , 42. (1 − i ), when expressed in polar form, is, π, π⎞, ⎛, (a) 2 ⎜ cos − i sin ⎟, ⎝, 4, 4⎠, , ⎡, ⎛ −π ⎞, ⎛ −π ⎞ ⎤, (b) 2 ⎢cos ⎜ ⎟ + i sin ⎜ ⎟ ⎥, ⎝, ⎠, ⎝ 4 ⎠⎦, 4, ⎣, , 3π, 3π ⎞, ⎛, (c) 2 ⎜ cos, (d) none of these, + i sin, ⎟, ⎝, 4, 4 ⎠, (5 − i ), 43., , when expressed in polar form, is, ( 2 − 3i ), π, π⎤, 2π, 2π ⎤, ⎡, ⎡, (b) 3 ⎢cos, (a) 2 ⎢cos + i sin ⎥, + i sin, 3, 3⎦, 3, 3 ⎥⎦, ⎣, ⎣, π, π⎤, 3π, 3π ⎤, ⎡, ⎡, (c) 2 ⎢cos + i sin ⎥, (d) 2 ⎢cos, + i sin, 4, 4⎦, 4, 4 ⎥⎦, ⎣, ⎣, 44., , (1 − 3i ), , when expressed in polar form, is, (1 + 2i ), π, π⎞, 5π, 5π ⎞, ⎛, ⎛, (a) 2 ⎜ cos + i sin ⎟, (b) 2 ⎜ cos, + i sin, ⎟, ⎝, ⎝, 4, 4⎠, 4, 4⎠, 3π, 3π ⎞, ⎛, (c) 2 ⎜ cos, + i sin, ⎟, ⎝, 4, 4 ⎠, , (d) none of these, , 45. ( − 4 + 4 3 i ), when expressed in polar form, is, π, π⎞, π, π⎞, ⎛, ⎛, (a) 4 ⎜ cos + i sin ⎟, (b) 8 ⎜ cos + i sin ⎟, ⎝, ⎝, 3, 3⎠, 6, 6⎠, 2π, 2π ⎞, ⎛, (c) 8 ⎜ cos, + i sin, ⎟, ⎝, 3, 3⎠, , (d) none of these, , 46. (sin 30° + i cos 30°), when expressed in polar form, is, (a) (cos 60° + i sin 60°), (c) (cos 150° + i sin 150°), , (b) (cos 60° − i sin 60°), (d) none of these, , 47. (sin 120° − i cos 120°), when expressed in polar form, is, (a) (cos 60° + i sin 60°), (b) (cos 30° + i sin 30°), (c) (cos 150° + i sin 150°), (d) none of these, 48. 3(cos 300° − i sin 30°), when expressed in polar form, is, 3 ⎡, ⎛ −π ⎞, cos ⎜ ⎟ + i sin, ⎢, ⎝ 4⎠, 2⎣, , (a), , 3 ⎛, π, π⎞, ⎜ cos + i sin ⎟, ⎝, 4, 4⎠, 2, , (b), , (c), , 3 ⎛, 3π, 3π ⎞, + i sin, ⎜ cos, ⎟, ⎝, 4, 4 ⎠, 2, , (d) none of these, , ⎛ −π ⎞ ⎤, ⎜ ⎟⎥, ⎝ 4 ⎠⎦, , 49. If z = ( 2 + −5 ) then|z| = ?, (a) 9, (b) 7, 50. If z = ( 3i − 1) 2 then|z| = ?, (a) 8, , (b) 10, , (c) 3, , (d) none of these, , (c) 4, , (d) 10
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Objective Questions, , (1 − i 3 ), then|z| = ?, 2(1 − i ), 1, 1, (a), (b), 2, 2 2, , 993, , 51. If z =, , (c), , 1, 3, , (d) none of these, , 52. If z = ( 3 + 2 i ) then zz = ?, (a) 5, , (b) 7, , (c) 11, , (d) 11, , 53. arg (1 + i ) = ?, (a) π, , (b), , π, 2, , (c), , π, 3, , (b), , 2π, 3, , (c) π, , (b), , π, 4, , (c), , (d), , π, 4, , 54. arg ( −1 + i 3 ) = ?, (a), , π, 3, , (d) none of these, , ⎛ 2 + 6 3 i⎞, 55. arg ⎜, ⎟ =?, ⎝ 5 + 3i ⎠, (a), , π, 3, , 2π, 3, , 56. For any complex number z, arg (z) + arg (z ) = ?, π, (a) 0, (b), (c) π, 2, , (d) π, , (d) 2π, , 57. If z1 and z2 be two complex numbers such that |z1 + z2| = |z1 − z2| then, amp (z1 ) − amp (z2 ) = ?, π, π, π, (a), (b), (c), (d) π, 4, 3, 2, ⎧ 1, 3 ⎫ ⎛ 3 + 4i ⎞, 58. ⎨, +, ⎟ =?, ⎬⎜, (, 1, −, 2, ), (, 1, +, i, i ) ⎭ ⎝ 2 − 4i ⎠, ⎩, ⎛3 9⎞, (a) ⎜ + i⎟, ⎝4 4⎠, , ⎛3 5 ⎞, (b) ⎜ − i⎟, ⎝4 4⎠, , ⎛1 3 ⎞, (c) ⎜ + i⎟, ⎝2 2 ⎠, , ⎛1 9 ⎞, (d) ⎜ + i⎟, ⎝4 4 ⎠, , 59. Compare List I with List II and choose the correct answer using codes, given below:, List I (Complex number), , List II (Conjugate), , (i) 3 + −4, , (p) ( −2 − 3i), , (ii) −3 − 2i, , (q) ( 2 + 3i), , (iii) −3i − 2, , (r) ( 3 − 2i), , (iv) −3i + 2, , (s) ( −3 + 2i), , (a) (i)–(s), (ii)–(r), (iii)–(p), (iv)–(q), (c) (i)–(r), (ii)–(s), (iii)–(p), (iv)–(q), , (b) (i)–(r), (ii)–(s), (iii)–(q), (iv)–(p), (d) (i)–(p), (ii)–(s), (iii)–(r), (iv)–(q)
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994, , Senior Secondary School Mathematics for Class 11, , 60. Compare List I with List II and choose the correct answer using codes, given below:, List I (Complex number), , (i) ( 4 − 3i ), , List II (Its modulus), , (p) 10, 1, (q), 5, , (ii) ( 8 + 6i ), (iii), , 1, ( 3 + 4i ), , (r) 1, , (iv), , ( 3 − 4i ), ( 3 + 4i ), , (s) 5, , (a) (i)–(p), (ii)–(s), (iii)–(r), (iv)–(q), (c) (i)–(s), (ii)–(p), (iii)–(r), (iv)–(q), , (b) (i)–(s), (ii)–(p), (iii)–(q), (iv)–(r), (d) (i)–(r), (ii)–(p), (iii)–(s), (iv)–(q), , 61. For any complex numbers z1 and z2 , compare List I with List II and choose, the correct answer, using codes given below:, List I, , List II, , π, 2, , (i) arg (z1 z2 ), , (p), , ⎛z ⎞, (ii) arg ⎜ 1 ⎟, ⎝ z2 ⎠, , (q) arg (z1 ) − arg (z2 ), , (iii) arg (z) + arg (z ), , (r) arg (z1 ) + arg (z2 ), , (iv) arg (i), , (s) 2π, , (a) (i)–(q), (ii)–(r), (iii)–(s), (iv)–(p), (c) (i)–(r), (ii)–(q), (iii)–(s), (iv)–(p), , (b) (i)–(r), (ii)–(q), (iii)–(p), (iv)–(s), (d) none of these, , 62. The solution set of x 2 + 2 = 0 is, (a) { 2 , − 2}, (b) { 2 i , − 2}, 63. The solution set of x 2 + 13 = 4x is, , (c) { 2 , − 2 i }, , (a) { 3 + 2i , 3 − 2i } (b) {2 + 3i , 2 − 3i } (c) { 3i , − 3i }, 64. The solution set of x 2 + 2x + 2 = 0 is, (a) {i , − i }, , (b) {1 + i , − i }, , (d) { 2 i , − 2 i }, (d) none of these, , (c) {1 + i , 1 − i }, , (d) {−1 + i , − 1 − i }, , ANSWERS (EXERCISE 1), , 1. (d), 9. (a), 17. (a), 25. (a), 33. (c), , 2. (b), 10. (b), 18. (c), 26. (b), 34. (b), , 3. (a), 11. (d), 19. (b), 27. (c), 35. (a), , 4. (b), 12. (b), 20. (b), 28. (d), 36. (c), , 5. (c), 13. (d), 21. (a), 29. (c), 37. (c), , 6. (d), 14. (d), 22. (c), 30. (b), 38. (b), , 7. (c), 15. (d), 23. (b), 31. (a), 39. (a), , 8. (d), 16. (b), 24. (b), 32. (a), 40. (b)
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Objective Questions, , 995, , 41. (b), , 42. (b), , 43. (c), , 44. (b), , 45. (c), , 46. (a), , 47. (b), , 48. (b), , 49. (c), 57. (c), , 50. (b), 58. (d), , 51. (a), 59. (c), , 52. (c), 60. (b), , 53. (d), 61. (c), , 54. (b), 62. (d), , 55. (a), 63. (b), , 56. (d), 64. (d), , HINTS TO SOME SELECTED QUESTIONS, 1. i 91 = (i 4 ) 22 × i 3 = 1 × i 2 × i = 1 × ( −1) × i = −i., 2. i 326 = (i 4 ) 81 × i 2 = ( 1) 81 × ( −1) = 1 × ( −1) = −1., 3. i 273 = (i 4 ) 68 × i = ( 1) 68 × i = ( 1 × i ) = i., 4. i 124 = (i 4 ) 31 = ( 1) 31 = 1., 1, 1, 1, 1, 5. i − 75 =, =, ×i =, ×i =, × i = 1× i = i ., (i 4 )19, i 75 i 75 × i, i 76, 6. i − 38 =, , 1, i, , 38, , ×, , i2, i, , 2, , =, , −1, i, , 40, , =, , −1, 4 10, , (i ), , =, , −1, 10, , ( 1), , =, , −1, = −1., 1, , 7. ( − −1 ) 4 n + 3 = ( −i ) 4 n + 3 = {( −i ) 4 } n ⋅ ( −i ) 3 = 1 × ( −i ) × ( −i ) × ( −i ) = i 2 × ( −i ) = −1 × ( −i ) = i ., 8. (i n + i n + 1 + i n + 2 + i n + 3 ) = i n ( 1 + i + i 2 + i 3 ) = i n ( 1 + i − 1 − i ) = i n × 0 = 0., 9. i109 + i114 + i119 + i124 = i109 ( 1 + i5 + i10 + i15 ), = i109 × [1 + i 4 × i + (i 4 ) 2 × i 2 + (i 4 ) 3 × i 3 ] = i109 [1 + i + i 2 + i 3 ], = i109 × [1 + i − 1 − i] = i109 × 0 = 0., 10. 3i 34 + 5i 27 − 2i 38 + 5i 41 = 3 × (i 4 ) 8 × i 2 + 5 × (i 4 ) 6 × i 3 − 2 × (i 4 ) 9 × i 2 + 5 × (i 4 )10 × i, = 3 × 1 × ( −1) + 5 × 1 × ( −i ) − 2 × 1 × ( −1) + 5 × 1 × i, = −3 − 5i + 2 + 5i = −1., 11., , a × b = ab is true only when at least one of a and b is non-negative., , 12., , −9 × −25 = ( 3i ) × (5i ) = ( 15 × i 2 ) = [15 × ( −1)] = −15., , 13. Two complex numbers cannot be compared., 14. Two complex numbers cannot be compared., 15. Two complex numbers cannot be compared., 16., , ( 1 − i ) ( 1 − i ) ( 1 − i ) ( 1 − i ) 2 1 + i 2 − 2 i 1 − 1 − 2 i −2 i, =, =, ×, =, =, =, = −i, ( 1 + i) ( 1 + i) ( 1 − i) ( 1 − i2 ), ( 1 + 1), 2, 2, 2, , ⇒, , ⎛ 1 − i ⎞ = ( −i ) 2 = i 2 = −1., ⎜, ⎟, ⎝ 1+ i⎠, , 17. ( a + ib ) =, ⇒, 18., , 1+ i, ⇒ ( a − ib ) =, 1−i, , ( a + ib )( a − ib ) =, , 1+ i, ×, 1−i, , 1−i, 1+ i, 1−i, = 1 ⇒ ( a2 + b 2 ) = 1., 1+ i, , ( 1 + i ) ( 1 + i ) ( 1 + i ) ( 1 + i )2 1 + i 2 + 2 i 1 − 1 + 2 i 2 i, ×, =, =, =, =, =, = i., ( 1 − i) ( 1 − i) ( 1 + i) ( 1 − i2 ), ( 1 + 1), 2, 2, Clearly, 4 is the smallest positive integer for which i 4 = 1., ∴, , n = 4.
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996, 19., , Senior Secondary School Mathematics for Class 11, ( 1 + b + ia) 2, ( 1 + b + ia) ( 1 + b + ia) ( 1 + b + ia), =, =, ×, ( 1 + b − ia) ( 1 + b − ia) ( 1 + b + ia) ( 1 + b ) 2 − (ia) 2, =, =, , 20., , 1 + b 2 − a2 + 2 b + 2iab + 2ia, 2, , 1 + b + 2b + a, , 2, , =, , 2 b 2 + 2 b + 2iab + 2ia, 2b + 2, , [Q a2 + b 2 = 1], , 2 b( b + 1) + 2ia( b + 1) 2( b + 1)( b + ia), =, = ( b + ia)., 2( b + 1), 2( b + 1), , ( 1 + 2 i ) ( 1 + 2 i ) ( 1 + i ) ( 1 + 2 i )( 1 + i ) 1 + 3i + 2i 2, =, ×, =, =, ( 1 − i), ( 1 − i) ( 1 + i), ( 1 + 1), ( 12 − i 2 ), 1 − 2 + 3i − 1 + 3 i ⎛ − 1 3 ⎞, =, =⎜, + i⎟ ⋅, ⎝ 2, 2, 2, 2 ⎠, , =, , −1 3, Clearly, ⎛⎜ , ⎞⎟ lies in quadrant II., ⎝ 2 2⎠, 21. ( x + iy ) =, , ( a + ib ), a + ib |a + ib|, ⇒ |x + iy| =, =, c + id |c + id|, ( a + id ), ⇒ |x + iy|2 =, , 22. ( 1 + i ) −1 =, , =, , =, , 2, , |c + id|, , ⇒ ( x2 + y2 ) =, , ( a2 + b 2 ), ( c2 + d2 ), , ⋅, , ( 1 − i), ( 1 − i), ( 1 − i) ⎛ 1 1 ⎞, 1, 1, =, =, = ⎜ − i⎟ ⋅, =, ×, ⎝2 2 ⎠, 2, ( 1 + i ) ( 1 + i ) ( 1 − i ) ( 12 − i 2 ), , 23. ( 1 − 2 i ) −2 =, , 24. ( 1 − i ) −3 =, , |a + ib|2, , 1, ( 1 − 2i ), , =, , 2, , ( −3 + 4 i ), ( 9 − 16 i 2 ), , 1, ( 1 − i )3, , =, , =, , 1, 2, , ( 1 + 4i − 4i ), ( −3 + 4 i ), 25, , =, , ( −3 + 4 i ), 1, 1, =, ×, ( 1 − 4 − 4 i ) ( −3 − 4 i ) ( −3 + 4 i ), , −3, 4 ⎞, i⎟ ⋅, = ⎛⎜, +, ⎝ 25 25 ⎠, , 1, 1 − i 3 − 3i ( 1 − i ), , =, , 1, 1 + i − 3i − 3, , ( −2 + 2 i ) ( −2 + 2 i ) ( −2 + 2 i ) ⎛ −1 1 ⎞, 1, =, =⎜, =, ×, + i⎟ ⋅, ⎝ 4, 8, 4 ⎠, ( −2 − 2 i ) ( −2 + 2 i ), 4 − 4i 2, , 25. ( 2 − 3 i )( −3 + 4 i ) = ( − 6 + 8 i + 9i − 12 i 2 ) = ( − 6 + 17 i + 12 ) = ( 6 + 17 i )., ( 3 − 5i ), ( 3 − 5 i ) ( −2 − 3 i ) ( 3 − 5 i )( −2 − 3 i ), 26., =, ×, =, ( −2 + 3 i ) ( −2 + 3 i ) ( −2 − 3 i ), ( −2 ) 2 − ( 3 i ) 2, =, , − 6 − 9i + 10 i + 15 i 2, ( 4 − 9i 2 ), , =, , − 6 + i − 15 ( −21 + i ) ⎛ −21, 1 ⎞, i⎟ ⋅, =, =⎜, +, ⎝ 13, 13, ( 4 + 9), 13 ⎠, , ⎛ 2 − −9 ⎞ ⎛ 2 − 3 i ⎞ ( 2 − 3 i ) ( 1 + 2 i ) ( 2 − 3 i )( 1 + 2 i ) ( 2 + 4 i − 3 i − 6 i 2 ), ⎟=⎜, 27. ⎜⎜, =, =, ×, ⎟=, ⎟, ( 1 − 4i 2 ), ⎝ 1 − − 4 ⎠ ⎝ 1 − 2 i ⎠ ( 1 − 2 i ) ( 1 + 2 i ) ( 1 − 2 i )( 1 + 2 i ), =, , ( 2 + i + 6) ( 8 + i ) ⎛ 8 1 ⎞, =, = ⎜ + i⎟ ⋅, ⎝5 5 ⎠, ( 1 + 4), 5, , 28. Given expression = ( 1 − i )( 1 + i )(5 − 7 i )(5 + 7 i ), = ( 1 − i 2 )( 25 − 7 i 2 ) = ( 1 + 1)( 25 + 7 ) = ( 2 × 32 ) = 64 .
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Objective Questions, 29. z = ( −2 + 5 i ) ⇒ z−1 =, , ( −2 − 5 i ) ( −2 − 5 i ) ( −2 − 5 i ) ( −2 − 5 i ), 1, 1, =, =, =, =, ×, 29, z ( −2 + 5 i ) ( −2 − 5 i ) ( 4 − 25 i 2 ) ( 4 + 25 ), , −2 5 ⎞, ⇒ z−1 = ⎛⎜, −, i⎟ ⋅, ⎝ 29 29 ⎠, 30. z = ( 3 + 2 i ) 2 = ( 9 + 4 i 2 + 12 i ) = ( 9 − 4 + 12 i ) = (5 + 12 i ), (5 − 12 i ), (5 − 12 i ), (5 − 12 i ), (5 − 12 i ), 1, =, ⇒ z −1 =, ×, =, =, (5 + 12 i ) (5 − 12 i ) ( 25 − 144 i 2 ) ( 25 + 144 ), ( 169 ), ⇒, 31., , z−i, z−i, =1⇒, z+ i, z+ i, ⇒, ⇒, , 32., , 33., , 5, 12 ⎞, z−1 = ⎛⎜, −, i⎟ ⋅, ⎝ 169 169 ⎠, , x + iy − i, x + iy + i, , 2, , =1, , 2, , 2, , =1⇒, , x 2 + ( y − 1) 2, x 2 + ( y + 1) 2, , |x + i ( y − 1)|2, x + i ( y − 1), =1⇒, =1, x + i ( y + 1), |x + i ( y + 1)|2, , = 1 ⇒ x 2 + ( y − 1) 2 = x 2 + ( y + 1) 2, , ⇒, , ( y + 1) 2 − ( y − 1) 2 = 0 ⇒ 4 y = 0 ⇒ y = 0, , ⇒, , z lies on the x-axis., , z − 5i, z − 5i, =1⇒, z + 5i, z + 5i, , 2, , =1⇒, , |z − 5 i|2, |z + 5 i|2, , =1, , ⇒, , |z − 5 i|2 =|z + 5 i|2 ⇒ |x + iy − 5 i|2 =|x + iy + 5 i|2, , ⇒, , |x + i ( y − 5 )|2 =|x + i ( y + 5 )|2 ⇒ x 2 + ( y − 5 ) 2 = x 2 + ( y + 5 ) 2, , ⇒, , ( y + 5 ) 2 − ( y − 5 ) 2 = 0 ⇒ 20 y = 0 ⇒ y = 0, , ⇒, , z lies on the x-axis., , z− 2, z− 2, =2 ⇒, z+ 2, z+ 2, , 2, , =4 ⇒, , |z − 2|2, |z + 2|2, , =4, , ⇒, , |x + iy − 2|2 = 4|x + iy + 2|2 ⇒ ( x − 2 ) 2 + y 2 = 4{( x + 2 ) 2 + y 2 }, , ⇒, , x 2 − 4 x + 4 + y 2 = 4{x 2 + 4 x + 4 + y 2 }, , ⇒, , 3( x 2 + y 2 ) + 20 x + 12 = 0 , which is a circle., , 34. Let z = −2 + 0 ⋅ i and let z = r (cos θ + i sin θ)., Then, r cos θ = −2 and r sin θ = 0., Squaring and adding, we get r 2 = 4 ⇒ r = 2., ∴ [cos θ = −1and sin θ = 0] ⇒ θ = π ., ∴, −2 = 2[cos π + i sin π]., 35. Let z = 0 − 3 i and let z = r (cos θ + i sin θ)., Then, r cos θ = 0 and r sin θ = −3., Squaring and adding, we get r 2 = 9 ⇒ r = 3., , 997
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998, , Senior Secondary School Mathematics for Class 11, , ∴, , [cos θ = 0 and sin θ = −1] ⇒ θ = −, , ∴, , ⎡, −π ⎞, ⎛ −π ⎞ ⎤ ⋅, −3 i = 3 ⎢ cos ⎛⎜, ⎟ + i sin ⎜, ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠ ⎥⎦, ⎣, , π, ⋅, 2, , 36. Let z = ( 1 + i ) and let z = r (cos θ + i sin θ)., Then, r cos θ = 1and r sin θ = 1., Squaring and adding, we get r 2 = 2 ⇒ r = 2 ., ∴, ∴, , π, 1, 1 ⎫, ⎧, and sin θ =, ⎨ cos θ =, ⎬ ⇒ θ= ⋅, 4, 2, 2⎭, ⎩, π, π⎤, ⎡, ( 1 + i ) = 2 ⎢ cos + i sin ⎥ ⋅, 4, 4⎦, ⎣, , 37. Let z = ( −1 + i 3 ) and let z = r (cos θ + i sin θ)., Then, r cos θ = −1and r sin θ =, , 3., 2, , Squaring and adding, we get r = 4 ⇒ r = 2., ∴, ∴, , ⎡, 3⎤, 2π, −1, and sin θ =, ⋅, ⎥ ⇒ θ=, ⎢ cos θ =, 2, 2 ⎦, 3, ⎣, 2π, 2π ⎤, ⎡, ( −1 + i 3 ) = 2 ⎢ cos, + i sin, ⋅, 3, 3 ⎥⎦, ⎣, , 38. Let z = ( 3 + i ) and let z = r (cos θ + i sin θ)., Then, r cos θ =, , 3 and r sin θ = 1., , Squaring and adding, we get r 2 = 4 ⇒ r = 2., ∴, ∴, , ⎡, 3, 1⎤, π, and sin θ = ⎥ ⇒ θ = ⋅, ⎢ cos θ =, 2, 2⎦, 6, ⎣, π, π⎞, ⎛, ( 3 + i ) = 2 ⎜ cos + i sin ⎟ ⋅, ⎝, 6, 6⎠, , 39. Let z = ( 1 + i ) and let z = r (cos θ + i sin θ)., Then, r cos θ = 1and r sin θ = 1., Squaring and adding, we get r 2 = 2 ⇒ r = 2 ., ∴, ∴, , 1, ⎡, ⎢ cos θ = 2 and sin θ =, ⎣, π, ( 1 + i ) = 2 ⎛⎜ cos + i sin, ⎝, 4, , 1 ⎤, π, ⇒ θ= ⋅, 4, 2 ⎥⎦, π⎞, ⎟⋅, 4⎠, , 40. Let z = ( −1 − i ) and let z = r (cos θ + i sin θ)., Then, r cos θ = −1and r sin θ = −1., Squaring and adding, we get r 2 = 2 ⇒ r = 2 ., ∴, ∴, , −1, −1 ⎤, −3 π, ⎡, ⎢ cos θ = 2 and sin θ = 2 ⎥ ⇒ θ = 4 ⋅, ⎣, ⎦, −3 π ⎞, ⎧, ⎛ −3 π ⎞ ⎫ ⋅, ( −1 − i ) = 2 ⎨ cos ⎛⎜, ⎟⎬, ⎟ + i sin ⎜, ⎝ 4 ⎠⎭, ⎝ 4 ⎠, ⎩, , 46. (sin 30 ° + i cos 30 ° ) = sin ( 90 ° − 60 ° ) + i cos ( 90 ° − 60 ° ) = (cos 60 ° + i sin 60 ° ).
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1000, 58., , Senior Secondary School Mathematics for Class 11, , ( 1 + 2i ) ( 1 + 2i ) ⎛ 1 3 ⎞, 1, 1, = ⎜ + i⎟, =, =, ×, ( 1 − 2i ) ( 1 − 2i ) ( 1 + 2i ) ( 1 − 4i 2 ) ⎝ 5 5 ⎠, ( 1 − i ) ( 3 − 3i ) ( 3 − 3i ) ⎛ 3 3 ⎞, 3, 3, = ⎜ − i⎟, =, =, =, ×, ( 1 + i) ( 1 + i) ( 1 − i) ( 1 − i2 ), ( 1 + 1) ⎝ 2 2 ⎠, ( 3 + 4 i ) ( 3 + 4 i ) ( 2 + 4 i ) −10 + 20 i −10 + 20 i ⎛ −10 20 ⎞ ⎛ −1 ⎞, =, ×, + i⎟ ⋅, =, =⎜, +, =, i⎟ = ⎜, ⎠, ⎝ 20, ( 2 − 4 i ) ( 2 − 4 i ) ( 2 + 4 i ) ( 4 − 16 i 2 ) ( 4 + 16 ), 20 ⎠ ⎝ 2, ∴, , 3 3 ⎫ −1 ⎞, ⎧ 1 2, given expression = ⎨ ⎛⎜ + i ⎞⎟ + ⎛⎜ − i ⎞⎟ ⎬ ⎛⎜, + i⎟, ⎠, ⎩⎝ 5 5 ⎠ ⎝ 2 2 ⎠ ⎭⎝ 2, 2 3 ⎫ −1 ⎞ ⎛ 17 11i ⎞ ⎛ −1 ⎞, ⎧ 1 3, −, + i⎟ = ⎜, = ⎨ ⎛⎜ + ⎞⎟ + ⎛⎜ − i ⎞⎟ ⎬ ⎛⎜, + i⎟, ⎟⎜, ⎠ ⎝ 10 10 ⎠ ⎝ 2, ⎠, ⎩⎝ 5 2 ⎠ ⎝ 5 2 ⎠ ⎭⎝ 2, −17 11 ⎞ ⎛ 17 11 ⎞, ⎛ 5 + 45 i ⎞ = ⎛ 1 + 9 i ⎞ ⋅, +, = ⎛⎜, +, ⎟ ⎜, ⎟i = ⎜, ⎟+⎜, ⎟, ⎝ 20 20 ⎠ ⎝ 4 4 ⎠, ⎝ 20, 10 ⎠ ⎝ 10 20 ⎠, , 59. (i) → ( 3 + 2 i ) = ( 3 − 2 i ), (ii) → ( −3 − 2 i ) = ( −3 + 2 i ), (iii) → ( −2 − 3 i ) = ( −2 − 3 i ), and (iv) → ( 2 − 3 i ) = ( 2 + 3 i )., ∴, , the correct answer is (i)–(r), (ii)–(s), (iii)–(p), (iv)–(q)., , 60. (i) → |4 − 3 i| = 16 + 9 = 25 = 5 , (ii) → |8 + 6 i|= 64 + 36 = 100 = 10 ,, (iii) →, ∴, , 1, =, |3 + 4 i|, , 1, =, 9 + 16, , 3 − 4 i |3 − 4 i|, 1, 1, = , (iv) →, =, =, 3 + 4 i |3 + 4 i|, 25 5, , 25 5, = = 1., 25 5, , the correct answer is (i)–(s), (ii)–(p), (iii)–(q), (iv)–(r)., , ⎧, ⎫, ⎛z ⎞, 61. (i) → {arg (z1z2 ) = arg (z1 ) + arg (z2 )} → 3 , (ii) → ⎨ arg ⎜ 1 ⎟ = arg (z1 ) − arg (z2 )⎬ → 2,, ⎝ z2 ⎠, ⎩, ⎭, π⎫, ⎧, (iii) → {arg (z) + arg ( z ) = 2 π} → 4, (iv) → ⎨ arg (i ) = ⎬ → 1., 2⎭, ⎩, ∴, , the correct answer is (i)–(r), (ii)–(q), (iii)–(s), (iv)–(p)., , 2, , 62. x + 2 = 0 ⇒ x 2 = −2 ⇒ x = ± 2 i., Solution set is { 2 i , − 2 i }., 63. x 2 + 13 = 4 x ⇒ x 2 − 4 x + 13 = 0, 4±, , 16 − 52 4 ± −36 4 ± 6 i, =, =, = ( 2 ± 3i ), 2, 2, 2, , ∴, , x=, , ∴, , solution set = {2 + 3 i , 2 − 3 i }., , 64. x 2 + 2 x + 2 = 0 ⇒ x =, ∴, , −2 ±, , 4−8, 2, , =, , −2 ±, , −4, 2, , solution set = {−1 + i , − 1 − i }., , q, , =, , −2 ± 2 i, ⇒ x = −1 ± i, 2
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Permutations and Combinations, REVIEW OF FACTS AND FORMULAE, 1. The Factorial, We define, n ! = n(n − 1)(n − 2) … 3 ⋅ 2 ⋅ 1., Examples (i) 5 ! = (5 × 4 × 3 × 2 × 1) = 120., (5 !) × 6 × 7 × 8 × 9 9 !, (ii) 6 × 7 × 8 × 9 =, = ⋅, 5!, 5!, 2. Fundamental Principles of Counting, (i) Multiplication Principle, If an event can occur in m different ways and if following it, a second, event can occur in n different ways, then the two events in, succession can occur in (m × n) different ways., (ii) Addition Principle, If two events can occur independently in exactly m ways and n ways, respectively, then either of the two events can occur in(m + n) ways., 3. (i) Number of all permutations of n dissimilar things taken r at a time is, n!, given by n Pr = n(n − 1)(n − 2) … (n − r + 1) =, ⋅, (n − r) !, (ii) n Pn = n !., (iii) 0 ! = 1., (iv) n Pn = n Pn −1 ., (vi) n Pr =, 4., , n −1, , Pr + r ⋅, , (v) n Pr = n ⋅, n −1, , n −1, , Pr −1 ., , Pr −1 ., , (i) Number of all permutations of n different things taken r at a time, when a particular thing is always included in each arrangement, = r ⋅ n −1 Pr −1 ., (ii) Number of all permutations of n different things taken r at a time, when a particular thing is never taken = n −1 Pr ., (iii) If there are n objects out of which m are alike of one kind and (n − m), objects are alike of another kind, then total number of permutations, n!, =, ⋅, (m !) ⋅ (n − m) !, , 5., , (i) Number of ways in which n persons can be seated round a table, = (n − 1) !., (ii) Number of ways in which n different beads can be arranged, 1, = (n − 1) !., 2, 1001
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1002, , Senior Secondary School Mathematics for Class 11, , COMBINATIONS, , Each of the different groups or selections which can be formed by taking some, or all of a number of objects, irrespective of their arrangements, is called a, combination., Examples, , (i) The different combinations formed of three letters a, b, c taken, two at a time are ab, bc, ac., (ii) The only combination formed of three letters a, b, c taken all at a, time is abc., (iii) Groups of 2 out of 4 persons A, B, C, D are AB, AC, AD, BC,, BD, CD., , ab and ba are two different permutations but each gives the same, combination., 1. (i) Number of all combinations of n things taken r at a time = nC r ., , NOTE, , (ii) n C r =, , n!, ⋅, (r !) × (n − r) !, , (iii), , (iv) n C r = nC n − r ., n, , Cr =, , n, , Pr, ⋅, r!, , (v) n C r + nC r −1 =, , Cr, n −r + 1, =, ⋅, r, C r −1, n, , (vi), , n, , (vii), , n, , C r + nC r, , +1, , n+1, , Cr ., , =, , n+1, , Cr, , + 1., , (viii) n C p = nC q ⇒ p + q = n., , EXERCISE 1, Mark (3) against the correct answer in each of the following., , 1, 1, x, +, = , then x = ?, 6! 7 ! 8!, (a) 32, (b) 48, , 1. If, , n −1, , P3 : P4 = 1 : 9, then n = ?, (a) 12, (b) 11, , 2. If, , (c) 56, , (d) 64, , (c) 9, , (d) 10, , n, , n, , 3., , Pn, =?, Pn − 2, 1, (a), 2, , n, , (b) 2, , (c), , 1, n−2, , (d) n(n − 1), , 4. If 15 Pr = 2730, then r = ?, (a) 3, (b) 4, , (c) 5, , (d) 6, , 5. P3 = ?, (a) 105, , (c) 210, , (d) 175, , (c) 10, , (d) 11, , 7, , (b) 140, , 6. If P5 = 20 ⋅ P3 , then n = ?, (a) 8, (b) 9, n, , n
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Objective Questions, , 1003, , 7. If 15 Pr −1 : 16Pr − 2 = 3 : 4, then r = ?, (a) 8, (b) 10, , (c) 12, , (d) 14, , 8. If C10 = C14 , then n = ?, (a) 4, (b) 24, , (c) 14, , (d) 10, , 9. If C 3 = 220, then n = ?, (a) 9, (b) 10, , (c) 11, , (d) 12, , C x , then x = ?, (b) r, , (c) r + 1, , (d) n, , (b) 612, , (c) 630, , (d) none of these, , n, , n, , n, , 10. If C r + C r + 1 =, (a) r − 1, n, , 11., , n, , C 34 = ?, (a) 1224, , n +1, , 36, , n, , 12., , Cr, =?, C r −1, n −r, (a), r, , n, , (b), , n −r −1, r, , 13. If n C18 = nC12 , then 32 C n = ?, (a) 248, (b) 496, 14., , (c), , n −r + 1, r, , (c) 992, , (d) none of these, , (d) none of these, , C 60 = ?, , 60, , (a) 60 !, , (b) 1, , (c), , 1, 60, , (d) none of these, , 15. In how many ways can 5 persons occupy 3 seats?, (a) 15, (b) 20, (c) 30, , (d) 60, , 16. In how many ways can 5 children stand in a queue?, (a) 5, (b) 25, (c) 60, , (d) 120, , 17. In how many ways can 4 different books be arranged on a shelf?, (a) 4, (b) 8, (c) 24, (d) 16, 18. Ten students are participating in a race. In how many different ways can, the first prize be won?, (a) 30, (b) 60, (c) 120, (d) 720, 19. Three different rings are to be worn in 4 fingers. In how many ways can, this be done?, (a) 12, (b) 24, (c) 64, (d) 81, 20. There are 6 periods on each working day of a school. In how many ways can, one arrange 5 subjects such that each subject is allowed at least one period?, (a) 360, (b) 720, (c) 3600, (d) 1800, 21. How many words with or without meaning can be formed by using all the, letters of the word, ‘DELHI’, using each letter exactly once?, (a) 20, , (b) 60, , (c) 120, , (d) 5
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1004, , Senior Secondary School Mathematics for Class 11, , 22. It is required to seat 5 men and 4 women in a row so that the women, occupy the even places. How many such arrangements are possible?, (a) 24, (b) 120, (c) 720, (d) 2880, 23. How many 3-digit numbers are there?, (a) 648, , (b) 729, , (c) 900, , (d) 1000, , 24. How many 3-digit numbers are there with no digit repeated?, (a) 648, , (b) 720, , (c) 729, , (d) none of these, , 25. How many 4-digit numbers can be formed with no digit repeated by using, the digits 3, 4, 5, 6, 7, 8 and 0?, (a) 280, (b) 560, (c) 720, (d) 840, 26. How many 3-digit even numbers can be formed with no digit repeated by, using the digits 0, 1, 2, 3, 4 and 5?, (a) 50, (b) 52, (c) 54, (d) 56, 27. The number of positive integers greater than 6000 and less than 7000 which, are divisible by 5 with no digit repeated, is, (a) 28, (b) 56, (c) 84, (d) 112, 28. How many 10-digit numbers can be formed by using the digits 1 and 2?, (b) 10C 2, (c) 210, (d) 10 !, (a) 10P2, 29. How many words beginning with T and ending with E can be made with, no letter repeated out of the letters of the word ‘TRIANGLE’?, (a) 8 P6, (b) 720, (c) 722, (d) 1440, 30. How many words can be formed from the letters of the word ‘DAUGHTER’, so that the vowels always come together?, (a) 720, (b) 2160, (c) 4320, (d) none of these, 31. How many words can be formed from the letters of the word ‘LAUGHTER’, so that the vowels are never together?, (a) 3600, (b) 4320, (c) 36000, (d) 40320, 32. In how many ways can the letters of the word ‘MACHINE’ be arranged so, that the vowels may occupy only odd positions?, (a) 288, (b) 576, (c) 5040, (d) none of these, 33. In how many ways can the letters of the word ‘PENCIL’ be arranged so that, N is always next to E?, (a) 120, (b) 240, (c) 720, (d) 1440, 34. In how many ways can the letters of the word ‘APPLE’ be arranged?, (a) 6, (b) 60, (c) 90, (d) 120, 35. How many words can be formed by using all the letters of the word, ‘ALLAHABAD’?, (a) 9 !, (b) 1890, (c) 3780, (d) 7560
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Objective Questions, , 1005, , 36. How many words can be formed using the letter A thrice, the letter B twice, and the letter C once?, (a) 6, (b) 60, (c) 90, (d) 120, 37. In how many ways can 10 books be arranged on a shelf so that a particular, pair of books shall be always together?, (a) 8 !, , (b) 9 !, , (c) 2 × 8 !, , (d) 2 × 9 !, , 38. In how many ways can 10 books be arranged on a shelf so that a particular, pair of books shall be never together?, (a) 8 !, , (b) 9 !, , (c) 2 × 9 !, , (d) 8 × 9 !, , 39. How many 4-digit numbers are there when a digit may be repeated any, number of times in each number?, (a) 5040, , (b) 9000, , (c) 10000, , (d) 4500, , 40. In how many ways can 6 boys be arranged in a row?, (a) 5 !, (b) 6 !, (c) 6, , (d) 2 × 6 !, , 41. In how many ways can 6 girls be seated in a circle?, (a) 6, (b) 6 !, (c) 5 !, , (d) 2 × 5 !, , 42. How many diagonals are there in a polygon of n sides?, 1, 1, 1, 1, (b) n(n − 2), (c) n(n − 3), (d) n(n + 1), (a) n(n − 1), 2, 2, 2, 2, 43. How many diagonals are there in an octagon?, (a) 28, , (b) 24, , (c) 20, , (d) 36, , 44. A polygon has 54 diagonals. Number of sides in this polygon is, (a) 9, (b) 12, (c) 15, (d) 16, 45. There are 10 points in a plane, out of which 4 points are collinear. The, number of line segments obtained from the pairs of these points is, (a) 39, (b) 40, (c) 41, (d) 45, 46. There are 10 points in a plane, out of which 4 points are collinear. The, number of triangles formed with vertices as these points is, (a) 20, (b) 120, (c) 116, (d) none of these, 47. Out of 7 consonants and 4 vowels, how many words of 3 consonants and, 2 vowels can be formed?, (a) 330, (b) 1050, (c) 6300, (d) 25200, 48. In how many ways can a committee of 5 members be selected from 6 men, and 5 ladies, consisting of 3 men and 2 ladies?, (a) 25, (b) 50, (c) 100, (d) 200, 49. Out of 5 men and 2 women, a committee of 3 is to be formed. In how many, ways can it be formed if at least one woman is included in each committee?, (a) 21, (b) 25, (c) 32, (d) 50
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1006, , Senior Secondary School Mathematics for Class 11, , 50. A committee of 5 is to be formed out of 6 gents and 4 ladies. In how many, ways can this be done when each committee may have at the most 2 ladies?, (a) 120, (b) 160, (c) 180, (d) 186, 51. How many different teams of 7 players can be chosen out of 10 players?, (a) 720, , (b) 70, , (c) 120, , (d) none of these, , 52. 12 persons meet in a room and each shakes hands with all the others. How, many handshakes are there?, (a) 144, , (b) 132, , (c) 72, , (d) 66, , 53. In how many ways can we select 9 balls out of 6 red balls, 5 white balls and, 5 blue balls if 3 balls of each colour are selected?, (a) 40, (b) 200, (c) 2000, (d) 400, 54. In how many ways can a cricket team be chosen out of a batch of 15 players,, if a particular player is always chosen?, (a) 1364, (b) 364, (c) 1001, (d) none of these, 55. In how many ways can a cricket team be chosen out of a batch of 15 players,, if a particular player is never chosen?, (a) 364, , (b) 1001, , (c) 1364, , (d) none of these, , 56. For the post of 5 teachers, there are 23 applicants. 2 posts are reserved for, SC candidates and there are 7 SC candidates among the applicants. In how, many ways can the selection be made?, (a) 5880, (b) 11760, (c) 3920, (d) none of these, 57. In how many ways can 5 white balls and 3 black balls be arranged in a row, so that no two black balls are together?, (a) 192, (b) 40, (c) 20, (d) 120, 58. In an examination, a candidate has to pass in each of the five subjects. In, how many ways can he fail?, (a) 5, (b) 10, (c) 21, (d) 31, 59. An examination paper contains 12 questions consisting of two parts, A, and B. Part A contains 7 questions and part B contains 5 questions. A, candidate is required to attempt 8 questions, selecting at least 3 from each, part. In how many ways can the candidate select the questions?, (a) 210, (b) 175, (c) 420, (d) none of these, ANSWERS (EXERCISE 1), , 1. (d), , 2. (c), , 3. (b), , 4. (a), , 5. (c), , 6. (a), , 7. (d), , 8. (b), , 9. (d) 10. (c), , 11. (c) 12. (c) 13. (b) 14. (b) 15. (d) 16. (d) 17. (c) 18. (d) 19. (b) 20. (c), 21. (c) 22. (d) 23. (c) 24. (a) 25. (c) 26. (b) 27. (d) 28. (c) 29. (b) 30. (c), 31. (c) 32. (b) 33. (a) 34. (b) 35. (d) 36. (b) 37. (d) 38. (d) 39. (b) 40. (b)
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Objective Questions, , 1007, , 41. (c) 42. (c) 43. (c) 44. (b) 45. (b) 46. (c) 47. (d) 48. (d) 49. (b) 50. (c), 51. (c) 52. (d) 53. (c) 54. (c) 55. (a) 56. (b) 57. (c) 58. (d) 59. (c), HINTS TO SOME SELECTED QUESTIONS, 1., , 1, 1, x, 8×7, 8, x, +, =, ⇒, +, =, 6! 7 ! 8!, 8 × 7 × ( 6 !) 8 × (7 !) 8 !, ⇒, n −1, , 2., 3., , n, , P3, , n, , P4, , n, , Pn, , (n − 1) !, (n − 4 ) ! 1, 1, 1 1, ×, = ⇒ = ⇒ n = 9., ⇒, n!, n 9, 9, (n − 1 − 3 ) !, 9, , =, , = (n !) ×, , Pn − 2, , 56, 8, x, +, =, ⇒ x = 56 + 8 = 64., 8! 8! 8!, , {n − (n − 2 )} !, = 2 ! = 2., n!, , 4. 2730 = ( 15 × 14 × 13 ) = 15 P3 ⇒ r = 3., 7!, 7 × 6 × 5 × ( 4 !), 5. 7 P3 =, =, = 210., (7 − 3 ) !, 4!, 6., , n, , P5, , n, , P3, , n!, (n − 3 ) !, ×, = 20, n!, (n − 5 ) !, , = 20 ⇒, ⇒, , (n − 3 )(n − 4 ){(n − 5 ) !}, = 20 ⇒ (n − 3 )(n − 4 ) = 20, (n − 5 ) !, , ⇒ n2 − 7 n − 8 = 0 ⇒ (n − 8 )(n + 1) = 0 ⇒ n = 8., 15, , 7., , 16, , Pr −1, , =, , Pr − 2, , 15 !, {16 − (r − 2 )} ! 3, 3, ⇒, ×, =, 4, 16 !, 4, {15 − (r − 1)} !, ⇒, , 1 ( 18 − r ) ! 3, = ⇒ ( 18 − r )( 17 − r ) = 12, ×, 16 ( 16 − r ) ! 4, , ⇒ r 2 − 35r + 294 = 0 ⇒ (r − 21)(r − 14 ) = 0 ⇒ r = 14 [Q r ≤ 16]., 8., , n, , C p = nC q ⇒ p + q = n., n, , ∴, 9., , n, , C 10 = nC 14 ⇒ n = ( 10 + 14 ) = 24., , C 3 = 220 ⇒, , ⇒, , n(n − 1)(n − 2 ), = 220, 6, , n(n − 1)(n − 2 ) = 1320 ⇒ n = 12 [Q 12 × 11 × 10 = 1320]., , 10. We know that n C r + nC r + 1 =, , n +1, , C r +1 ., , So, x = (r + 1)., 11., , 36, , n, , 12., , 36 × 35, = 630., 2, (r − 1) ! × (n − r + 1) !, n!, =, ×, (r !) × (n − r ) !, n!, , C 34 =, , n, , Cr, , C r −1, , 36, , =, , C (36− 34 ) =, , 36, , C2 =, , (r − 1) ! × (n − r + 1) × (n − r ) ! (n − r + 1), =, ⋅, r ⋅ (r − 1) ! × (n − r ) !, r
Page 1019 :
1008, 13., , n, , Senior Secondary School Mathematics for Class 11, C 18 = nC 12 ⇒ n = ( 18 + 12 ) = 30., 32, , ∴, 14., , Cn =, , 32, , C 30 =, , 32, , C2 =, , 32 × 31, = 496., 2, , 60, , C 60 = 1 [Q nC n = 1]., , 15. Required number of ways = 5 P3 =, , 5!, 5× 4× 3× 2× 1, = 60., =, (5 − 3 ) !, 2× 1, , 16. Required number of ways = 5 P5 = 5 ! = (5 × 4 × 3 × 2 × 1) = 120., 17. Required number of ways = 4 ! = ( 4 × 3 × 2 × 1) = 24., 18. Required number of ways = 10P3 = ( 10 × 9 × 8 ) = 720., 19. Required number of ways = ( 4 × 3 × 2 ) = 24., 20. Out of 6 periods, 5 may be arranged for 5 subjects in 6P5 ways., Remaining 1 period may be arranged for any one of the five subjects in 5 P1 ways., ∴, , required number of ways = ( 6 P5 × 5 P1 ) = ( 6 × 5 × 4 × 3 × 2 × 5 ) = 3600., , 21. Required number of words = number of arrangements of 5 letters taken all at a time, = 5 P5 = 5 ! = (5 × 4 × 3 × 2 × 1) = 120., 22. In a row of 9 seats, the 2nd, 4th, 6th and 8th are the even places., These 4 places can be occupied by 4 women in 4 P4 ways = 24 ways., Remaining 5 places can be occupied by 5 men in 5 P5 ways = 120 ways., total number of seating arrangements = ( 24 × 120 ) = 2880., ∴, 23. The hundreds place can be filled by any of the 9 nonzero digits., So, there are 9 ways of filling this place. The tens place can be filled by any of the 10, digits. So, there are 10 ways of filling it., The units place can be filled by any of the 10 digits. So, there are 10 ways of filling it., total number of 3-digit numbers = ( 9 × 10 × 10 ) = 900., ∴, 24. The hundreds place can be filled by any of the 9 nonzero digits., So, there are 9 ways of filling the hundreds place. The tens digit can be filled by any, of the remaining 9 digits. So, there are 9 ways of filling the tens place., The units place can now be filled by any of the remaining 8 digits., So, there are 8 ways of filling the units digit., Required number of numbers = ( 9 × 9 × 8 ) = 648., 25. Thousands place can be filled by any of the 6 nonzero digits., So, there are 6 ways to fill this place., Hundreds place can be filled by any of the remaining 6 digits., So, there are 6 ways to fill this place., Tens place can be filled by any of the remaining 5 digits., So, there are 5 ways to fill this place., Units place can be filled by any of the remaining 4 digits., So, there are 4 ways to fill this place., Required number of numbers = ( 6 × 6 × 5 × 4 ) = 720.
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Objective Questions, , 1009, , 26. Numbers with 0 at units place = (5 × 4 × 1) = 20., Numbers with 2 at units place = ( 4 × 4 × 1) = 16., Numbers with 4 at units place = ( 4 × 4 × 1) = 16., Total numbers = (20 + 16 + 16) = 52., 27. Clearly, thousands digit is 6., Number of numbers with units digit 0 = ( 1 × 8 × 7 × 1) = 56., Number of numbers with units digit 5 = ( 1 × 8 × 7 × 1) = 56., Required number of numbers = (56 + 56) = 112., 28. Each place of the number can be filled in 2 ways., required number of numbers = 210., ∴, 29. Fixing T at the beginning and E at the end, the remaining 6 letters can be arranged in, 6 places in 6 ! = 720 ways., ∴, , required number of words = 720., , 30. Take all the vowels A U E together and take them as one letter., Then, the letters to be arranged are D, G, H, T, R, (A U E)., These 6 letters can be arranged in 6 places in 6 ! ways., Now, 3 letters A, U, E among themselves can be arranged in 3 ! = 6 ways., ∴, , required number of words = ( 6 !) × 6 = ( 6 × 5 × 4 × 3 × 2 × 1 × 6 ) = 4320., , 31. Total number of words formed by using all the 8 letters at a time = 8 P8 = 8 ! = 40320., Number of words in which vowels are never together, = (total number of words) – (no. of words in which vowels are always together), = (40320 – 4320) = 36000., 32. The given word has 7 letters out of which there are 3 vowels and 4 consonants. Let us, mark the positions of these letters as (1) (2) (3) (4) (5) (6) (7)., Now, the 3 vowels can be placed at any of the 3 places out of four marked 1, 3, 5, 7., ∴, , number of ways of arranging vowels = 4 P3 = ( 4 × 3 × 2 ) = 24., , Now, 4 consonants may be arranged at the remaining four positions in 4 P4 = 4 ! = 24, ways., Required number of ways = ( 24 × 24 ) = 576., 33. Keeping EN together and considering it as one letter, we have to arrange 5 letters at, 5 places., This can be done in 5 P5 = 5 ! = 120 ways., 34. There are in all 5 letters out of which there are 2P, 1A, 1L and 1E., 5!, required number of ways =, ∴, = 60., ( 2 !)( 1 !)( 1 !)( 1 !), 35. There are 9 letters in all. Out of these A is repeated 4 times, L is repeated 2 times and, the rest are different., 9!, Required number of words =, = 7560., ( 4 !) ⋅ ( 2 !)
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1010, , Senior Secondary School Mathematics for Class 11, , 36. There are 6 letters in all. Out of these A is repeated thrice, 8 is repeated twice and C is, taken only once., 6!, 6×5× 4× 3× 2× 1, required number of words =, = 60., ∴, =, ( 3 !)( 2 !)( 1 !), 3× 2× 2, 37. Let us keep the two particular books together and treat them as one., Now, 9 books can be arranged among themselves in 9 ! ways., Also, 2 books can be arranged among themselves in 2 ! = 2 ways., required number of ways = 2 × 9 !., ∴, 38. Number of ways in which 10 books may be arranged = 10 !., Number of ways in which 10 books may be arranged with two particular books, together = ( 2 × 9 !)., Required number of ways in which 2 particular books are never together, = ( 10 !) − ( 2 × 9 !) = ( 10 × 9 !) − ( 2 × 9 !) = ( 8 × 9 !)., 39. Clearly, 0 cannot be placed at the thousands place., So, this place can be filled in 9 ways., Each of the hundreds, tens and units digits can be filled in 10 ways., required number of numbers = ( 9 × 10 × 10 × 10 ) = 9000., ∴, 40. 6 boys can be arranged in a row in 6 ! ways., 41. 6 girls can be arranged in a circle in 5 ! ways., 42. In a polygon of n sides, number of diagonals =, , 1, n(n − 3 )., 2, , 1, 43. Putting n = 8 in n(n − 3 ), we get 20., 2, ∴, , number of diagonals in an octagon = 20., , 1, 44. n(n − 3 ) = 54 ⇒ n(n − 3 ) = 108 ⇒ n2 − 3n − 108 = 0, 2, ⇒ n2 − 12n + 9n − 108 = 0 ⇒ n(n − 12 ) + 9(n − 12 ) = 0, ⇒ (n − 12 )(n + 9 ) = 0 ⇒ n = 12 ., 45. Number of line segments formed by joining pairs of points out of 10, 10 × 9, = 10C 2 =, = 45., 2, 4× 3, Number of line segments formed by joining pairs of 4 points = 4 C 2 =, = 6., 2, But, these points being collinear give only one line., ∴, , required number of line segments = ( 45 − 6 + 1) = 40., , 46. Number of triangles obtained from 10 points = 10C 3 =, , 10 × 9 × 8, = 120., 3× 2× 1, , Number of triangles obtained from 4 points = 4 C 3 = 4 C 1 = 4., But, these 4 points being collinear will give no triangle., ∴, , required number of triangles = (120 – 4) = 116.
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Objective Questions, , 1011, , 47. Number of ways of selecting 3 consonants out of 7 and 2 vowels out of 4, 7 × 6× 5 4× 3⎞, = ( 7 C 3 × 4 C 2 ) = ⎛⎜, ×, ⎟ = 210., ⎝ 3× 2× 1 2× 1⎠, Now, 5 letters can be arranged among themselves in 5 ! ways = 120 ways., Required number of words = ( 210 × 120 ) = 25200., 48. Number of ways of selecting 3 men out of 6 and 2 ladies out of 5 = ( 6C 3 × 5C 2 ), 6× 5× 4 5× 4⎞, = ⎛⎜, ×, ⎟ = 200., ⎝ 3 × 2 × 1 2 × 1⎠, 49. We may have:, (i) 1 woman and 2 men or, ∴, , (ii) 2 women and 1 man., , 5× 4⎞, required number of ways = ( 2 C 1 × 5C 2 ) + ( 2 C 2 × 5C 1 ) = ⎛⎜ 2 ×, ⎟ + ( 1 × 5), ⎝, 2 × 1⎠, = ( 20 + 5 ) = 25., , 50. We may have:, (i) (1 lady out of 4) and (4 gents out of 6), or (ii) (2 ladies out of 4) and (3 gents out of 6)., ∴, , required number of ways, 4× 3 6×5× 4⎞, = ( 4 C 1 × 6C 4 ) + ( 4 C 2 × 6C 3 ) = ( 4 × 6C 2 ) + ⎛⎜, ×, ⎟, ⎝ 2 × 1 3 × 2 × 1⎠, 6× 5⎞, = ⎛⎜ 4 ×, ⎟ + ( 6 × 20 ) = ( 60 + 120 ) = 180., ⎝, 2 × 1⎠, , 51. Required number of teams = 10C 7 = 10C 3 =, , 10 × 9 × 8, = 120., 3× 2× 1, , 12 × 11, = 66., 2, 53. Required number of ways = ( 6C 3 × 5C 3 × 5C 3 ), , 52. Number of handshakes = 12C 2 =, , 6× 5× 4 5× 4 5× 4⎞, = ( 6C 3 × 5C 2 × 5C 2 ) = ⎛⎜, ×, ×, ⎟ = 2000., ⎝ 3 × 2 × 1 2 × 1 2 × 1⎠, 54. When a particular player is always chosen, then we have to select 10 players out of 14., 14 × 13 × 12 × 11, required number of ways = 14 C 10 = 14 C 4 =, ∴, = 1001., 4× 3× 2× 1, 55. When a particular player is never chosen, then we have to select 11 players out of 14., 14 × 13 × 12, required number of ways = 14 C 11 = 14 C 3 =, ∴, = 364., 3× 2× 1, 56. We have to select 2 posts out of 7 SC and 3 posts out of 16., 7 × 6 16 × 15 × 14 ⎞, Required number of ways = ( 7 C 2 × 16C 3 ) = ⎛⎜, ×, ⎟ = 11760., ⎝ 2, 3× 2× 1 ⎠, 57. Let us arrange the white balls (shown by W) and leave a space in between every pair, as shown below., XWXWXWXWXWX, 6×5× 4, Now, 3 black balls may be arranged in 6 places in 6C 3 ways =, = 20., 3× 2× 1
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1012, , Senior Secondary School Mathematics for Class 11, , 58. The candidate can fail by failing in 1 or 2 or 3 or 4 or 5 subjects out of 5 in each case., ∴, , required number of ways = 5C 1 + 5C 2 + 5C 3 + 5C 4 + 5C 5, = 5C 1 + 5C 2 + 5C (5 − 3 ) + 5C (5 − 4 ) + 1, = 5C 1 + 5C 2 + 5C 2 + 5C 1 + 1, , 59. He may select:, , 5× 4⎞, = 2 ( 5 C 1 + 5C 2 ) + 1 = 2 ⎛⎜ 5 +, ⎟ + 1 = ( 30 + 1) = 31., ⎝, 2 × 1⎠, , (i) (3 out of 7 from A) and (5 out of 5 from B), or (ii) (4 out of 7 from A) and (4 out of 5 from B), or (iii) (5 out of 7 from A) and (3 out of 5 from B)., The number of ways of these selections are:, (i) 7C 3 × 5C 5 = ( 35 × 1) = 35, (ii) 7C 4 × 5C 4 = ( 35 × 5 ) = 175, (iii) 7C 5 × 5C 3 = ( 21 × 10 ) = 210., ∴, , required number of ways = (35 + 175 + 210) = 420., , q
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Binomial Theorem, REVIEW OF FACTS AND FORMULAE, BINOMIAL EXPANSIONS, , 1., , (i) ( x + y )n =, , n, , ∑, , n, , C r xn − r y r ., , r =0, , (ii) Tr + 1 = nC r xn − r y r ., (iii) Number of terms in the expansion of ( x + y) n = (n + 1)., (iv) When n is even, middle term in the expansion of ( x + y) n is ⎛⎜, , n, + 1⎞⎟ th, ⎝2, ⎠, , term., , (v) When n is odd, middle terms in the expansion of ( x + y) n are, ⎛⎜ n +, ⎝ 2, , 1⎞, n + 3⎞, ⎟ th term and ⎛⎜, ⎟ th term., ⎠, ⎝ 2 ⎠, , (vi) pth term from the end = ( n − p + 2)th term., 2., , (i) ( 1 + x)n =, (ii) Tr + 1 =, , 3., , n, , ∑, , n, , C r xr ., , r =0, n, C r xr ., n, , (i) ( x − y )n =, , ∑ ( −1 ), , C r xn − r y r ., , r n, , r =0, , (ii) Tr, 4., , +1, , = ( −1)r nC r xn − r y r ., , (i) ( 1 − x)n =, , n, , ∑ ( −1 ), , r, , ⋅ nC r xr ., , r =0, r n, , (ii) Tr + 1 = ( −1) ⋅ C r xr ., 5. (i) ( x + y )n + ( x − y )n = 2[ n C 0 xn y 0 + nC 2 xn − 2 y 2 + nC 4 xn − 4 y 4 + …]., (ii) ( x + y )n − ( x − y )n = 2[ n C1 xn −1 y + nC 3 xn − 3 y 3 + …]., 6., , (i) When n is odd, then each one of [( x + y) n + ( x − y) n ] and, [( x + y) n − ( x − y) n ] has, , n+ 1, terms., 2, , n, (ii) When n is even, then [( x + y) n + ( x − y) n ] has ⎛⎜ + 1⎞⎟ terms and, , [( x + y) n − ( x − y) n ] has, 7., , n, terms., 2, , ⎝2, , ⎠, , RESULTS ON BINOMIAL COEFFICIENTS, , We denote n C 0 by C 0 ; nC1 by C1 ; nC 2 by C 2 , … and n C r by C r ., (i) C 0 + C1 + C 2 + … + C n = 2n ., 1013
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1014, , Senior Secondary School Mathematics for Class 11, , (ii) C 0 + C 2 + C 4 + … = C1 + C 3 + C5 + … = 2n −1 ., (iii) C 02 + C12 + C 22 + … + C n2 =, , ( 2n) !, ⋅, (n !) 2, , (iv) C 0 C1 + C1 C 2 + C 2 C 3 + … + C n −1 C n =, , n ⋅ 2n ⋅ [1 ⋅ 3 ⋅ 5 ⋅… ⋅ ( 2n − 1)], ⋅, (n + 1) !, , EXERCISE 1, Mark (3) against the correct answer in each of the following., , 1. The number of terms in the expansion of ( 2x + 3 y)17 is, (a) 16, (b) 17, (c) 18, (d) 34, ⎛, 5⎞, 2. The number of terms in the expansion of ⎜ 3 x − ⎟, y⎠, ⎝, (a) 9, (b) 11, (c) 10, , 10, , is, (d) 8, , 3. The number of terms in the expansion of ( x + y ) + ( x − y ) 8 is, (a) 8, (b) 7, (c) 5, (d) 9, 8, , 4. The number of terms in the expansion of {( 2x + 3 y) 9 + ( 2x − 3 y) 9} is, (a) 10, (b) 8, (c) 4, (d) 5, 5. The number of terms in the expansion of {(5 x + 2y) 7 − (5 x − 2y) 7} is, (a) 4, (b) 8, (c) 6, (d) 3, 6. The number of terms in the expansion of {( 2x +, (a) 11, (b) 9, (c) 6, , 3 y)10 − ( 2x − 3 y)10} is, (d) 5, , 7. The number of terms in the expansion of {( x + a)16 + ( x − a)16} is, (a) 7, (b) 8, (c) 9, (d) 17, 8. The 7th term in the expansion of ( x + y )10 is, (b) 210x 2 y 3, (c) 210x 3 y 4, (a) 120xy 2, , (d) none of these, , 12, , 1⎞, ⎛, 9. The 10th term in the expansion of ⎜ 2x 2 + 2 ⎟ is, ⎝, x ⎠, −16, −12, (a) 264x, (b) 1760x, (c) 792x −14, , (d) none of these, , 10. The 4th term in the expansion of ( x − 2y) is, (a) 1760x 8 y 6, (b) − 440x 7 y5, (c) −1760x 9 y 3, , (d) none of these, , 12, , 18, , 1 ⎞, ⎛, 11. The 13th term in the expansion of ⎜ 9x −, ⎟ , x ≠ 0 is, ⎝, 3 x⎠, (a) 16854, (b) 18564, (c) 17954, (d) 18832, 9, , ⎛ 4x 5 ⎞, 12. The 6th term in the expansion of ⎜, − ⎟ is, ⎝ 5 2x ⎠, (a) −2520x, , (b) −5040x, , (c), , −5040, x, , (d), , −560, x2
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Objective Questions, , 1015, 9, , ⎛ x3, 2⎞, 13. The 5th term from the end in the expansion of ⎜, − ⎟ is, ⎝ 2 x2 ⎠, (a) 252x 2, (b) −252x 2, (c) −2520x 3, (d) 2520x 3, 6, , 1⎞, ⎛, 14. The 3rd term from the end in the expansion of ⎜ x + ⎟ is, ⎝, x⎠, 15, 30, 12, 24, (b) 3, (c) 2, (d) 3, (a) 2, x, x, x, x, 1⎞, ⎛, 15. The 5th term from the end in the expansion of ⎜ x − ⎟, ⎝, x⎠, (a) 99x 4, , (b) 495 x 4, , 12, , is, , (c) −99x 4, , (d), , 495, x4, , 16. The 14th term from the end in the expansion of ( x − y )17 is, (a) 17 C 6( x )11 ⋅ y 3 (b) − 17C5 x 6( y )5 (c) 17 C 4 x13 / 2 ⋅ y 2 (d) none of these, ⎛, 1⎞, 17. The middle term in the expansion of ⎜ x − ⎟, 2y ⎠, ⎝, (a), , −63 5 −5, x y, 8, , (b), , −21 6 − 6, x y, 4, , (c), , 10, , is, , 63 4 −4, x y, 8, , (d), , −63 4 −4, x y, 8, , 10, , 2⎞, ⎛, 18. The middle term in the expansion of ⎜ x 2 − ⎟ is, ⎝, x⎠, (a) 8064x5, (b) −8064x5, (c) 6720x 4, 1⎞, ⎛, 19. The coefficient of x 7 in the expansion of ⎜ x 2 + ⎟, ⎝, x⎠, (a) 154, (b) 231, (c) 462, , (d) −6720x 4, , 11, , is, (d) 924, 6, , 1⎞, ⎛, 20. The coefficient of x 2 in the expansion of ⎜ 3 x − ⎟ is, ⎝, x⎠, (a) 405, , (b) 1215, , (c) 2430, , (d) 3645, 9, , 1 ⎞, ⎛, 21. The coefficient of x 6 in the expansion of ⎜ 3 x 2 −, ⎟ is, ⎝, 3x⎠, (a) 378, , (b) 756, , (c) 189, , (d) 567, 10, , 1⎞, ⎛, 22. The term independent of x in the expansion of ⎜ x + ⎟ is, ⎝, x⎠, (a) 210, (b) 252, (c) 504, (d) 756, 10, , 1 ⎞, ⎛, 23. The term independent of x in the expansion of ⎜ x +, ⎟ is, ⎝, 3x2 ⎠, (a) 35, (b) 32, (c) 15, (d) 5
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1016, , Senior Secondary School Mathematics for Class 11, , 1⎞, ⎛, 24. The term independent of x in the expansion of ⎜ x − ⎟, ⎝, x⎠, (a) 924, , (b) 462, , (b) 546, , is, , (c) 231, , (d) 693, , 1⎞, ⎛, 25. The coefficient of x 32 in the expansion of ⎜ x 4 − 3 ⎟, ⎝, x ⎠, (a) 273, , 12, , 15, , is, , (c) 1365, , (d) 1092, 5, , a⎞, ⎛, 26. The coefficient of x in the expansion of ⎜ x 2 + ⎟ is, ⎝, x⎠, (a) 12a 2, , (b) 10a 3, , (c) 9a 4, , (d) none of these, 9, , 1 ⎞, ⎛3, 27. The term independent of x in the expansion of ⎜ x 2 −, ⎟ is, ⎝2, 3x⎠, 1, 7, 8, 5, (a), (b), (c), (d), 24, 18, 27, 36, 11, , 3⎞, ⎛, 28. Which term contains x 7 in the expansion of ⎜ 2x 2 − ⎟ ?, ⎝, x⎠, (a) 6th, , (b) 7th, , (c) 8th, , (d) none of these, 8, , 1⎞, ⎛, 29. Which term contains x 3 in the expansion of ⎜ 3 x − ⎟ ?, ⎝, 2x ⎠, (a) 2nd, , (b) 3rd, , (c) 4th, , (d) none of these, 8, , ⎛ 4x 5 ⎞, 30. The coefficient of x −4 in the expansion of ⎜, + ⎟ is, ⎝5, 2x ⎠, (a) 625, , (b) 1875, , (c) 4375, , (d) none of these, 10, , 1 ⎞, ⎛, 31. The coefficient of x −15 in the expansion of ⎜ 3 x 2 −, ⎟ is, ⎝, 3x3 ⎠, 40, −40, 80, −80, (a), (b), (c), (d), 27, 27, 9, 9, 32. The total number of terms in the expansion of ( x + k)100 + ( x − k)100 after, simplification is, (a) 50, , (b) 51, , (c) 101, , (d) 202, , 33. The number of terms in the expansion of (1 + 3 2x) + (1 − 3 2x) 9 is, 9, , (a) 5, , (b) 7, , (c) 9, , (d) 10, , 34. If the coefficients of the second, third and fourth terms in the expansion of, (1 + x) n are in AP, then n = ?, (a) 5, , (b) 6, , (c) 7, , (d) 9
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Objective Questions, , 1017, , 35. If p and q are positive integers, then the coefficients of x p and x q in the, expansion of (1 + x) p + q are, (a) equal, (b) equal with opposite signs, (c) reciprocal to each other, (d) none of these, 36. Let the coefficient of x n in the expansion of (1 + x) 2 n be p and the coefficient, of x n in the expansion of (1 + x) 2 n − 1 be q. Then,, (a) p = 2q, (b) 2p = q, (c) 2p = 3 q, (d) 3 p = 2q, 37. In the expansion of (1 + x) n , the binomial coefficients of three consecutive, terms are respectively 220, 495 and 792. The value of n is, (a) 10, (b) 11, (c) 12, (d) 13, n, , x⎞, ⎛, 38. If the coefficients of x 7 and x 8 are equal in the expansion of ⎜ 2 + ⎟ , then, ⎝, 3⎠, n =?, (a) 15, (b) 45, (c) 55, (d) 56, 39. Total number of terms in the expansion of ( a + b + c) n is, 1, 1, (d) (n + 1)(n + 2), (a) (n + 1), (b) ( 2n + 1), (c) n(n + 1), 2, 2, 40. Total number of terms in the expansion of ( a + b + c)10 is, (a) 11, , (b) 21, , (c) 55, , (d) 66, , 41. If the total number of terms in the expansion of( a − b + 2c) n is 45, thenn = ?, (a) 7, , (b) 8, , (c) 9, , (d) 22, , 42. The coefficient of x 3 in the expansion of (1 + 2x) 6(1 − x) 7 is, (a) 43, (b) − 43, (c) 63, (d) – 63, 43. In the expansion of ( x + a) n , if the sum of odd terms be P and the sum of, even terms be Q, then 4PQ = ?, (b) ( x + a) n − ( x − a) n, (a) ( x + a) n + ( x − a) n, 2n, 2n, (d) none of these, (c) ( x + a) − ( x − a), 44. If the coefficients of the second, third and fourth terms in the expansion of, (1 + x) 2 n are in AP, then, (b) 2n2 − 7n + 5 = 0, (a) n2 − 3n + 8 = 0, (d) none of these, (c) 2n2 − 9n + 7 = 0, 45. IfC r denotes n C r in the expansion of(1 + x) n , thenC 0 + C1 + C 2 + … + C n = ?, 1, 1, (c) n( 2n + 1), (d) 2n, (a) 2n, (b) n(n + 1), 2, 3, 46. (C 0 + C 2 + C 4 + C 6 + …) = ?, (b) 2n, (a) 2n−1, , (c) 4n, , (d) 4n−1
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1018, , Senior Secondary School Mathematics for Class 11, , ⎧C, C, C, C ⎫, 47. ⎨ 1 + 2 2 + 3 3 + … + n ⋅ n ⎬ = ?, C, C, C, C, ⎩ 0, 1, 2, n −1 ⎭, (c) 2n, , 1, (d) n(n + 1), 2, , 48. {C1 + 2C 2 + 3C 3 + … + nC n } = ?, (b) n ⋅ 2n −1, (a) n ⋅ 2n, , (c) (n − 1) ⋅ 2n, , (d) (n + 1) ⋅ 2n, , 49. {C 0 + 2C1 + 3C 2 + … + (n + 1)C n } = ?, (b) n ⋅ 2n −1, (a) (n + 1) ⋅ 2n, , (c) (n + 2) ⋅ 2n −1, , (d) (n + 2) ⋅ 2n, , (a) 2n, , (b) 2n−1, , 50. {C 0 + 3C1 + 5C 2 + … + ( 2n + 1)C n } = ?, (b) (n + 2) ⋅ 2n −1, (a) (n + 1) 2n, (c) (n − 1)(n + 2), , +1, , (d) none of these, , ANSWERS (EXERCISE 1), , 1. (c), , 2. (b), , 3. (c), , 4. (d), , 5. (a), , 6. (d), , 7. (c), , 8. (b), , 9. (b) 10. (c), , 11. (b) 12. (c) 13. (b) 14. (a) 15. (d) 16. (c) 17. (a) 18. (b) 19. (c) 20. (b), 21. (a) 22. (b) 23. (d) 24. (a) 25. (c) 26. (b) 27. (b) 28. (a) 29. (d) 30. (c), 31. (b) 32. (b) 33. (a) 34. (c) 35. (a) 36. (a) 37. (c) 38. (c) 39. (d) 40. (d), 41. (b) 42. (b) 43. (c) 44. (c) 45. (d) 46. (a) 47. (d) 48. (b) 49. (c) 50. (a), HINTS TO SOME SELECTED QUESTIONS, 1. The expansion of ( 2 x + 3 y )17 has 18 terms., ⎛, 2. The expansion of ⎜ 3 x −, ⎝, , 5⎞, ⎟, y⎠, , 3. The expansion of {( x +, , 10, , has 11 terms., , 8, y ) 8 + ( x − y ) 8 } has ⎛⎜ + 1⎞⎟ = 5 terms., ⎝2, ⎠, , 9 + 1⎞, 4. The expansion of {( 2 x + 3 y ) 9 + ( 2 x − 3 y ) 9 } has ⎛⎜, ⎟ = 5 terms., ⎝ 2 ⎠, 7 + 1⎞, ⎟ = 4 terms., 2 ⎠, 10, 6. The expansion of {( 2 x + 3 y )10 − ( 2 x − 3 y )10} has, = 5 terms., 2, 16, 7. The expansion of {( x + a)16 + ( x − a)16} has ⎛⎜, + 1⎞⎟ = 9 terms., ⎝ 2, ⎠, 5. The expansion of {(5 x + 2 y ) 7 − (5 x − 2 y ) 7 } has ⎛⎜, ⎝, , 8. In the expansion of ( x +, , y )10 , we have, , Tr + 1 = 10C r ⋅ ( x )10− r ⋅ ( y ) r, ⇒, , T7 = T6+ 1 = 10C 6( x ) 4 ⋅ ( y ) 6 = 10C 4 x 2 y 3 =, , 10 × 9 × 8 × 7 2 3, x y = 210 x 2 y 3 ., 4× 3× 2× 1
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Objective Questions, 1⎞, 9. In the expansion of ⎛⎜ 2 x 2 +, ⎟, ⎝, x2 ⎠, , 1019, , 12, , , we have, r, , 1, Tr + 1 = 12C r ⋅ ( 2 x 2 )12 − r ⋅ ⎛⎜ ⎞⎟, ⎝ x2 ⎠, , 9, , ⇒, , 1, 1, T10 = T9 + 1 = 12C 9 ⋅ ( 2 x 2 ) (12 − 9 ) ⋅ ⎛⎜ ⎞⎟ = 12C 3 ⋅ ( 2 x 2 ) 3 ⋅ ⎛⎜ ⎞⎟, ⎝ x2 ⎠, ⎝ x2 ⎠, , 9, , 12 × 11 × 10, 1, 1760, × 8x 6 ⋅, =, = 1760 x −12 ., 3× 2× 1, x18, x12, 10. In the expansion of ( x − 2 y )12 , we have, =, , Tr + 1 = 12C r ⋅ x (12 − r ) ⋅ ( −2 y ) r, ⇒, , T4 = T3 + 1 = 12C 3 ⋅ x (12 − 3 ) ⋅ ( −2 y ) 3 =, , 1 ⎞, 11. In the expansion of ⎛⎜ 9 x −, ⎟, ⎝, 3 x⎠, Tr, , +1, , 12 × 11 × 10, × ( −8 )x 9 y 3 = −1760 x 9 y 3 ., 3× 2× 1, , 18, , , we have, , ⎛ −1 ⎞, = 18C r ( 9 x ) (18 − r ) ⎜, ⎟, ⎝ 3 x⎠, , r, , 12, , ⇒, , 12, , ⎛ −1 ⎞, 18, 6 ⎛ −1 ⎞, T13 = T12 + 1 = 18C 12 ( 9 x ) (18 −12 ) ⋅ ⎜, ⎟, ⎟ = C 6( 9 x ) ⋅ ⎜, ⎝ 3 x⎠, ⎝ 3 x⎠, 18 × 17 × 16 × 15 × 14 × 13, 1, =, × 96x6 ⋅, = 18564 ., 6×5× 4× 3× 2× 1, 312 x 6, 9, , 4x 5 ⎞, 12. In the expansion of ⎛⎜, −, ⎟ , we have, ⎝ 5, 2x ⎠, 4x, Tr + 1 = 9C r ⋅ ⎛⎜ ⎞⎟, ⎝ 5 ⎠, ⇒, , (9 − r ), , 4x, T6 = T5 + 1 = 9C 5 ⋅ ⎛⎜ ⎞⎟, ⎝ 5 ⎠, =, , −5, ⋅ ⎛⎜ ⎞⎟, ⎝ 2x ⎠, (9 −5 ), , r, , 5, , 4, , −55, −5, 4x, ⋅ ⎛⎜ ⎞⎟ = 9C 4 ⋅ ⎛⎜ ⎟⎞ ×, ⎝ 2x ⎠, ⎝ 5 ⎠, 25 x5, , 9 × 8 × 7 × 6 256 x 4 ( −3125 ) −5040, ×, ×, =, ⋅, 4× 3× 2× 1, 625, x, 32 x5, 9, , ⎛ x3, 2 ⎞, 13. Given expansion is ⎜, −, ⎟ ⋅, x2 ⎠, ⎝ 2, pth term from the end = (n − p + 2 )th term from the beginning., ∴, , 5th term from the end = ( 9 − 5 + 2 )th term = 6th term., ⎛ x3 ⎞, Tr + 1 = ( −1) r ⋅ 9C r ⎜ ⎟, ⎝ 2 ⎠, , ⇒, , (9 − r ), , 2, ⋅ ⎛⎜ ⎞⎟, ⎝ x2 ⎠, , ⎛ x3 ⎞, T6 = T( 5 + 1 ) = ( −1)5 ⋅ 9C 5 ⋅ ⎜ ⎟, ⎝ 2 ⎠, =−, , r, , (9 −5 ), , 4, , 5, 5, ⎛ x3 ⎞, 2, 2, ⋅ ⎛⎜ ⎞⎟ = − 9 C 4 ⋅ ⎜ ⎟ ⋅ ⎛⎜ ⎞⎟, 2, 2, ⎝x ⎠, ⎝ 2 ⎠ ⎝x ⎠, , 9 × 8 × 7 × 6 x12 32, ⋅, ×, = −252 x 2 ., 4 × 3 × 2 × 1 16 x10
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1020, , Senior Secondary School Mathematics for Class 11, 6, , 1, 14. Given expansion is ⎛⎜ x + ⎞⎟ ⋅, ⎝, x⎠, pth term from the end = (n − p + 2 )th term from the beginning., 3rd term from the end = ( 6 − 3 + 2 )th term = 5th term., 1, Tr + 1 = 6C r x (6 − r ) ⋅ ⎛⎜ ⎞⎟, ⎝ x⎠, , r, , 4, , ⇒, , 1, 1 6×5 1, 15, =, ×, T5 = T4 + 1 = 6C 4 ⋅ x (6 − 4 ) ⋅ ⎛⎜ ⎞⎟ = 6C 2 ⋅ x 2 ⋅, =, ⋅, ⎝ x⎠, x4 2 × 1 x2 x2, , 1, 15. Given expansion is ⎛⎜ x − ⎞⎟, ⎝, x⎠, , 12, , ⋅, , pth term from the end = (n − p + 2 )th term from the beginning., 5th term from the end = ( 12 − 5 + 2 )th term = 9th term., 1, Tr + 1 = ( −1) r ⋅ 12C r x ( 12 − r ) ⋅ ⎛⎜ ⎞⎟, ⎝ x⎠, ⇒, , r, , T9 = T8 + 1 = ( −1) 8 ⋅ 12C 8 ⋅ x (12 − 8 ) ⋅, , 1, x8, , = 12C 4 ⋅ x 4 ⋅, , 1, x8, , =, , 12 × 11 × 10 × 9 1 495, ⋅, =, ⋅, 4 × 3 × 2 × 1 x4, x4, , 16. Given expansion is ( x − y )17 ., pth term from the end = (n − p + 2 )th term., 14th term from the end = ( 17 − 14 + 2 )th term = 5th term., Tr + 1 = ( −1) r ⋅ 17C r ⋅ ( x ) (17 − r ) ⋅ ( y ) r, ⇒, , T5 = T4 + 1 = ( −1) 4 ⋅ 17C 4 ⋅ ( x ) (17 − 4 ) ⋅ ( y ) 4 = 17C 4 x13 / 2 ⋅ y 2 ., , ⎛, 1⎞, 17. Given expansion is ⎜ x −, ⎟, 2y ⎠, ⎝, , 10, , ⋅, , Total number of terms in this expansion is 11., ⎛ 1⎞, Tr + 1 = ( −1) r ⋅ 10C r x ( 10 − r ) ⋅ ⎜ ⎟, ⎝ 2y ⎠, , r, , 5, , ∴, , ⎛ 1⎞, −63 5 −5, middle term = 6th term = T5 + 1 = ( −1)5 ⋅ 10C 5 ⋅ x (10 − 5 ) ⋅ ⎜ ⎟ =, x y ., 2, y, 8, ⎝ ⎠, , 2, 18. Given expansion is ⎛⎜ x 2 − ⎞⎟, ⎝, x⎠, , 10, , ⋅, , Total number of terms = 11., Middle term = 6th term., 2, Now, Tr + 1 = ( −1) r ⋅ 10C r ⋅ ( x 2 ) ( 10 − r ) ⋅ ⎛⎜ ⎞⎟, ⎝ x⎠, , r, , 5, , ∴, , 2, middle term = T6 = T5 + 1 = ( −1)5 ⋅ 10C 5 ⋅ ( x 2 ) ( 10 − 5 ) ⋅ ⎛⎜ ⎞⎟, ⎝ x⎠, 25, 10 × 9 × 8 × 7 × 6 10 32, = −10C 5 ⋅ x10 ⋅, =−, ⋅x ⋅, = −8064 x5 ., 5× 4× 3× 2× 1, x5, x5
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Objective Questions, 1, 19. General term in the expansion of ⎛⎜ x 2 + ⎞⎟, ⎝, x⎠, , 1021, , 11, , is, , r, , 1, Tr + 1 = 11C r ( x 2 ) ( 11 − r ) ⋅ ⎛⎜ ⎞⎟ = 11C r x (22 − 2 r − r ) = 11C r x (22 − 3 r ) ., ⎝ x⎠, Putting 22 − 3r = 7 , we get 3r = 15 ⇒ r = 5., T5 = 11C 5 x (22 −15 ) = 11C 5 x 7 ., ∴, , 11 × 10 × 9 × 8 × 7, = 462., 5× 4× 3× 2× 1, , coefficient of x 7 = 11C 5 =, , 6, , 1, 20. General term in the expansion of ⎛⎜ 3 x − ⎞⎟ is, ⎝, x⎠, r, , 1, Tr + 1 = ( −1) r ⋅ 6C r ( 3 x ) (6 − r ) ⋅ ⎛⎜ ⎞⎟ = ( −1) r ⋅ 6C r ⋅ 3 (6 − r ) x (6 − 2 r )., ⎝ x⎠, Putting 6 − 2r = 2 , we get 2r = 4 ⇒ r = 2., ∴, , T3 = ( −1) 2 ⋅ 6C 2 ⋅ 3 (6− 2 ) ⋅ x 2 = 6C 2 × 3 4 =, , 6×5, × 81 = 1215., 2× 1, , 9, , 1⎞, 21. In the expansion of ⎛⎜ 3 x 2 −, ⎟ , we have, ⎝, 3x ⎠, r, , 1⎞, r 9, (9 − 2 r ), Tr + 1 = ( −1) r ⋅ 9C r ⋅ ( 3 x 2 ) (9 − r ) ⋅ ⎛⎜, ⋅ x ( 18 − 2 r − r ) ., ⎟ = ( −1) ⋅ C r ⋅ 3, ⎝ 3x ⎠, Putting 18 − 3r = 6 , we get 3r = 12 ⇒ r = 4., ∴, ⇒, , T5 = ( −1) 4 ⋅ 9C 4 ⋅ 31 ⋅ x 6, 9× 8×7 × 6, coefficient of x 6 = 3 ×, = 378., 4× 3× 2× 1, 10, , 1, 22. In the expansion of ⎛⎜ x + ⎞⎟ , we have, ⎝, x⎠, r, , 1, Tr + 1 = 10C r ⋅ x ( 10 − r ) ⋅ ⎛⎜ ⎞⎟ = 10C r x ( 10 − 2 r ) ., ⎝ x⎠, Putting 10 − 2r = 0 , we get 2r = 10 ⇒ r = 5., 10 × 9 × 8 × 7 × 6, = 252., T6 = 10C 5 =, 5× 4× 3× 2× 1, 10, , 1 ⎞, 23. In the expansion of ⎛⎜ x +, ⎟ , we have, ⎝, 3x 2 ⎠, ⎛, , 1, , ⎞, , ⎜5 − r −2 r⎟, 1 ⎞, 1, ⎠, ⎝, 10, 2, Tr + 1 = 10C r ( x ) (10 − r ) ⋅ ⎛⎜, ⋅, ⋅, ⎟ = Cr x, 2, ⎝ 3x ⎠, 3r, 1, 5r, 5r, Now, 5 − r − 2r = 0 ⇒ 5 −, =0 ⇒, = 5 ⇒ r = 2., 2, 2, 2, 1, 10 × 9 1, =, × = 5., T3 = 10C 2 x 0 ⋅, ∴, 2× 1 9, 32, r, , 1, 24. In the expansion of ⎛⎜ x − ⎞⎟, ⎝, x⎠, , 12, , , we have, r, , 1, Tr + 1 = ( −1) r ⋅ 12C r x ( 12 − r ) ⋅ ⎛⎜ ⎞⎟ = ( −1) r ⋅ 12C r − x ( 12 − 2 r) ., ⎝ x⎠
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1022, , Senior Secondary School Mathematics for Class 11, , Now, 12 − 2r = 0 ⇒ 2r = 12 ⇒ r = 6., 12 × 11 × 10 × 9 × 8 × 7, T7 = ( −1) 6 ⋅ 12C 6 = 12C 6 =, = 924 ., 6×5× 4× 3× 2× 1, 1⎞, 25. In the expansion of ⎛⎜ x 4 −, ⎟, ⎝, x3 ⎠, , 15, , , we have, r, , 1, Tr + 1 = ( −1) r ⋅ 15C r ( x 4 ) ( 15 − r ) ⋅ ⎛⎜ ⎞⎟ = ( −1) r ⋅ 15C r x (60 − 4 r − 3 r) ., ⎝ x3 ⎠, Putting 60 − 7 r = 32 , we get 7 r = 28 ⇒ r = 4., ∴, , T5 = ( −1) 4 ⋅ 15C 4 x 32 ., , ∴, , coefficient of x 32 = 15C 4 =, , 15 × 14 × 13 × 12, = 1365., 4× 3× 2× 1, , 5, , a, 26. In the expansion of ⎛⎜ x 2 + ⎞⎟ , we have, ⎝, x⎠, r, , a, Tr + 1 = 5C r ⋅ ( x 2 ) (5 − r ) ⋅ ⎛⎜ ⎞⎟ = 5C r ⋅ x ( 10 − 3 r ) ⋅ ar ., ⎝ x⎠, Putting 10 − 3r = 1, we get 3r = 9 ⇒ r = 3., 5× 4× 3 3, coefficient of x = 5C 3 ⋅ a3 =, a = 10 a3 ., ∴, 3× 2× 1, 9, , 3, 1⎞, 27. In the expansion of ⎛⎜ x 2 −, ⎟ , we have, ⎝2, 3x ⎠, 3, Tr + 1 = 9C r ⋅ ⎛⎜ x 2 ⎞⎟, ⎝2 ⎠, ⇒, , ( 9 −r ), , r, , 1⎞, r, ⋅ ⎛⎜, ⎟ ⋅ ( −1), ⎝ 3x ⎠, , Tr + 1 = ( −1) r ⋅ 9C r ⋅ 3 (9 − 2 r ) ⋅, , 1, 2 (9 − r ), , ⋅ x ( 18 − 3 r )., , Putting 18 − 3r = 0 , we get r = 6., ∴, , the term independent of x = T7 = ( −1) 6 ⋅ 9C 6 ⋅ 3 −3 ⋅, , 1, , 23, 1, 9× 8×7, 1, 7, = 9C 3 ⋅, =, ×, =, ⋅, 8 × 27 3 × 2 × 1 8 × 27 18, , 3, 28. In the expansion of ⎛⎜ 2 x 2 − ⎞⎟, ⎝, x⎠, , 11, , , we have, r, , 3, Tr + 1 = ( −1) r ⋅ 11C r ⋅ ( 2 x 2 ) ( 11 − r ) ⋅ ⎛⎜ ⎞⎟ = ( −1) r ⋅ 11C r ⋅ 2 ( 11 − r ) ⋅ 3 r ⋅ x 22 − 3 r ., ⎝ x⎠, Putting 22 − 3r = 7 , we get 3r = 15 ⇒ r = 5., Hence, the 6th term contains x 7 ., 8, , 1⎞, 29. In the expansion of ⎛⎜ 3 x −, ⎟ , we have, ⎝, 2x ⎠, r, , 1 (8 − 2 r ), 1, Tr + 1 = ( −1) r ⋅ 8C r ( 3 x ) (8 − r ) ⋅ ⎛⎜ ⎞⎟ = ( −1) r ⋅ 8C r ⋅ 3 (8 − r ) ⋅, x, ., ⎝ 2x ⎠, 2r
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Objective Questions, Putting 8 − 2r = 3 , we get 2r = 5 ⇒ r =, ∴, , 1023, , 5, ⋅, 2, , no term contains x 3 ., 8, , 4x, 5 ⎞, 30. In the expansion of ⎛⎜, +, ⎟ , we have, ⎝ 5, 2x ⎠, 4x, Tr + 1 = 8C r ⋅ ⎛⎜ ⎞⎟, ⎝ 5 ⎠, , (8 − r ), , 5, 2 ( 16 − 2 r − r ), ⋅, ⋅ ⎛⎜ ⎞⎟ = 8C r x (8 − 2 r ) ⋅, ⎝ 2x ⎠, 5 (8 − 2 r ), r, , Putting 8 − 2r = − 4 , we get 2r = 12 ⇒ r = 6., T7 = 8C 6x −4 ⋅, ∴, , 2 ( 16 − 3 × 6), 5, , (8 − 2 × 6), , = 8C 2 x −4 ⋅, , 54, 22, , =, , 8 × 7 625 −4, ×, x = 4375 x −4 ., 2× 1, 4, , coefficient of x −4 is 4375., 10, , 1 ⎞, 31. In the expansion of ⎛⎜ 3 x 2 −, ⎟ , we have, ⎝, 3x 3 ⎠, r, , 1 ⎞, ( 10 − 2 r ), r 10, Tr + 1 = ( −1) r ⋅ 10C r ⋅ ( 3 x 2 ) ( 10 − r ) ⋅ ⎛⎜, ⋅ x (20 − 2 r − 3 r ) ., ⎟ = ( −1) ⋅ C r ⋅ 3, ⎝ 3x 3 ⎠, Putting 20 − 5r = −15 , we get 5r = 35 ⇒ r = 7 ., T8 = ( −1) 7 ⋅ 10C 7 ⋅ 3 (10 −14 ) ⋅ x −15 = −10C 3 ⋅, ∴, , coefficient of x −15 is, , 1, 34, , x −15 =, , − ( 10 × 9 × 8 ) 1 −15 −40 −15, ×, =, x, x ., ( 3 × 2 × 1), 81, 27, , −40, ⋅, 27, , n, 32. When n is even, the number of terms in {( x + y ) n + ( x − y ) n } is ⎛⎜ + 1⎞⎟ ⋅, ⎝2, ⎠, ∴, , 100, ( x + k )100 + ( x − k )100 has ⎛⎜, + 1⎞⎟ = 51 terms., ⎝ 2, ⎠, , n+, 33. When n is odd, the number of terms in {( x + y ) n + ( x − y ) n } is ⎛⎜, ⎝ 2, ( 9 + 1), ∴ ( 1 + 3 2 x ) 9 + ( 1 − 3 2 x ) 9 contains, = 5 terms., 2, 34. In the expansion of ( 1 + x ) n , we have Tr + 1 = nC r x r ., ∴, , 1⎞, ⎟⋅, ⎠, , T2 = nC 1 x1 = nx , T3 = nC 2 x 2 and T4 = nC 3 x 3 ., , n(n − 1)(n − 2 ), 6, ⇒ (n − 2 )(n − 7 ) = 0 ⇒ n = 7 [Q number of terms is 4 or more]., 35. Tp + 1 = (p + q)C p ⋅ x p and Tq+ 1 = (p + q)C q ⋅ x q., ∴, , But,, , 2 × nC 2 = nC 1 + nC 3 ⇒ n(n − 1) = n +, , ( p + q), , ( p + q), , Cp =, , C ( p + q) − p =, p, , (p + q), , C q., , q, , So, the coefficients of x and x are equal., 36. In the expansion of ( 1 + x ) 2 n , we have, t r +1 =, , C r x 2 n − r ⇒ Tn + 1 =, , 2n, , 2n, , C nxn ⇒, , 2n, , In the expansion of ( 1 + x ) 2 n −1 , we have Tr + 1 =, , C n = p., , 2 n −1, , C r x 2 n −1 − r .
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1024, , Senior Secondary School Mathematics for Class 11, , Putting 2n − 1 − r = n, we get r = n − 1., ∴, , Tn = T(n −1 ) + 1 =, , ∴, , p, =, q, , 2n, 2 n −1, , Cn, C n −1, , =, , 2 n −1, , C n −1 x n ⇒, , 2 n −1, , C n −1 = q., , ( 2n) !, {(n − 1) !}(n !) 2n, = 2 ⇒ p = 2 q., ×, =, n, (n !)(n !), ( 2n − 1) !, , 37. Let the three successive terms be Tr , Tr +1 and Tr + 2 . Then,, n, , ⇒, ⇒, ⇒, ⇒, , C r −1 = 220 , nC r = 495 and n C r + 1 = 792, n, , n, 495 9, C r + 1 792 8, = and, =, =, n, 220, 4, 495 5, C r −1, Cr, n − (r + 1) + 1 8, n−r + 1 9, =, = and, 5, r+1, r, 4, 4n − 13r + 4 = 0 and 5n − 13r − 8 = 0, n = 12 and r = 4., , Cr, , n, , =, , n, , x, 38. In the expansion of ⎛⎜ 2 + ⎞⎟ , we have:, ⎝, 3⎠, r, , x, Tr + 1 = nC r ⋅ 2 (n − r ) ⋅ ⎛⎜ ⎞⎟ = nC r ⋅ 2 (n − r ) ⋅ 3 − r ⋅ x r ., ⎝ 3⎠, Putting r = 7 and 8, we get, T8 = nC 7 ⋅ 2 (n − 7 ) ⋅ 3 −7 ⋅ x 7 and T9 = nC 8 ⋅ 2 (n − 8 ) ⋅ 3 −8 ⋅ x 8 ., ∴, , n, , C 7 ⋅ 2 (n − 7 ) ⋅ 3 −7 = nC 8 ⋅ 2 (n − 8 ) ⋅ 3 −8 ⇒, , n, n, , C 8 2 ( n − 8 ) 3 −8, ⋅, ⋅, =1, C 7 2 ( n − 7 ) 3 −7, , (n − 8 + 1) 1 1, × × = 1 ⇒ n − 7 = 48 ⇒ n = 55., 8, 2 3, n, 39. ( a + b + c ) = {a + ( b + c )} n, ⇒, , = an + nC 1 an −1 ( b + c ) + nC 2 ⋅ an − 2 ( b + c ) 2 + … + nC n ( b + c ) n ., 1, Total number of terms in this expansion = 1 + 2 + 3 + … + (n + 1) = (n + 1)(n + 2 )., 2, 1, 40. Putting n = 10, required number of terms = × 11 × 12 = 66., 2, 1, 41. (n + 1)(n + 2 ) = 45 ⇒ (n + 1)(n + 2 ) = 90, 2, ⇒ n2 + 3n − 88 = 0 ⇒ (n − 8 )(n + 11) = 0 ⇒ n = 8., 42. ( 1 + 2 x ) 6( 1 − x ) 7, = {1 + 6C 1 ⋅ 2 x + 6C 2 ⋅ ( 2 x ) 2 + 6C 3 ⋅ ( 2 x ) 3 + … } × {1 − 7C 1 ⋅ x + 7C 2 ⋅ x 2 − 7C 3 ⋅ x 3 + …}, = ( 1 + 12 x + 60 x 2 + 160 x 3 + … )( 1 − 7 x + 21x 2 − 35 x 3 + … )., ∴, , coefficient of x 3 = 1 × ( −35 ) + ( 12 × 21) + 60 × ( −7 ) + ( 160 × 1), , = ( −35 + 252 − 420 + 160 ) = −43., 43. P + Q = ( x + a) n and P − Q = ( x − a) n, ⇒, , 4 PQ = ( P + Q ) 2 − ( P − Q ) 2 = ( x + a) 2 n − ( x − a) 2 n ., , 44. Tr + 1 =, Now,, , 2n, , 2n, , C r x r ⇒ T2 =, , C1 ,, , 2n, , C 2 and, , 2n, , 2n, , C 1 x , T3 =, , 2n, , C 3 are in AP, , C 2 x 2 and T4 =, , 2n, , C 3 x3.
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Objective Questions, , 1025, , 2n( 2n − 1), 2n( 2n − 1)( 2n − 2 ), = 2n +, 2, 6, , ⇒, , 2⋅, , ⇒, , 2n2 − 9n + 7 = 0 ., , 45. We know that C 0 + C 1 + C 2 + … + C n = 2 n ., 46. We know that C 0 + C 2 + C 4 + … = 2 n −1 ., Cr, n−r + 1, 47. We know that, =, ⋅, C r −1, r, Putting r = 1, 2 , 3 , … , n, we get, C1, C, C, Cn, 1, + 2 ⋅ 2 + 3 ⋅ 3 + … + n⋅, = n + (n − 1) + (n − 2 ) + … + 1 = n(n + 1)., C0, C1, C2, C n −1, 2, 48. C 1 + 2C 2 + 3C 3 + … + nC n = n + 2 ⋅, , n(n − 1), n(n − 1)(n − 2 ), + 3⋅, + …+ n, 2, 3!, , ⎡, ⎤, (n − 1)(n − 2 ), = n ⋅ ⎢ 1 + (n − 1) +, + … + 1⎥, 2, !, ⎣, ⎦, = n ⋅ [ (n −1 )C 0 +, = n ⋅ ( 1 + 1), , n −1, , (n −1 ), , C1 +, , = n⋅ 2, , n −1, , (n −1 ), , C2 + … +, , (n −1 ), , C n −1 ], , ., , 49. C 0 + 2C 1 + 3C 2 + … + (n + 1)C n, = (C 0 + C 1 + C 2 + … + C n ) + (C 1 + 2C 2 + 3C 3 + … + nC n ], = 2 n + n ⋅ 2 n −1 = (n + 2 ) ⋅ 2 n −1 ., 50. C 0 + 3C 1 + 5C 2 + … + ( 2n + 1)C n, = (C 0 + C 1 + C 2 + … + C n ) + 2(C 1 + 2C 2 + … + nC n ), = 2 n + 2(n ⋅ 2 n −1 ) = (n + 1) ⋅ 2 n ., , q
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Sequences and Series (AP, GP, HP), REVIEW OF FACTS AND FORMULAE, 1. ARITHMETIC PROGRESSION (AP), , I. a , a + d , a + 2d , … , a + (n − 1) d , … is an AP in which, first term = a and common difference = d., In this AP, we have:, (i) nth term, Tn = a + (n − 1) d., (ii) nth term from the end = l − (n − 1) d., n, (iii) sum to n terms, Sn = ⋅ {2a + (n − 1) d}., 2, n, (iv) sum to n terms, Sn = ( a + l ), where l is the last term., 2, II. (i) Three numbers in AP are taken as ( a − d), a , ( a + d)., (ii) Four numbers in AP are taken as ( a − 3 d), ( a − d), ( a + d), ( a + 3 d)., (iii) Five numbers in AP are taken as ( a − 2d), ( a − d), a , ( a + d), ( a + 2d)., III., , 1, (i) AM between a and b = ( a + b)., 2, (ii) If a , A1 , A 2 , A 3 , b are in AP, we say that A1 , A 2 , A 3 are the three, arithmetic means between a and b., , 2. GEOMETRICAL PROGRESSION (GP), , I. a , ar , ar 2 , … , ar n −1 , … is a GP in which, first term = a and common ratio = r., In this GP, we have, (i) nth term, Tn = ar n −1 ., l, (ii) nth term from the end = n −1 ⋅, r, (iii) sum to n terms of the GP is given by, ⎧ a(1 − r n ), , when r < 1, ⎪, ⎪ (1 − r), Sn = ⎨ n, ⎪ a(r − 1) , when r > 1., ⎪⎩ (r − 1), a, II. (i) Three numbers in GP are taken as , a , ar., r, a a, (ii) Four numbers in GP are taken as 3 , , ar , ar 3 ., r r, 1026
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Objective Questions, , 1027, , (i) If a, G, b are in GP then G is the GM between a and b and G = ab ., (ii) AM ≥ GM., , III., , 3. HARMONIC PROGRESSION (HP), , a , b , c are in HP ⇒, 4., , 1 1 1, , , are in AP., a b c, , 1, (i) (1 + 2 + 3 + … + n) = n(n + 1), 2, 1, (ii) (12 + 22 + 3 2 + … + n2 ) = n(n + 1)( 2n + 1), 6, ⎧1, ⎫, (iii) (13 + 23 + 3 3 + … + n3 ) = ⎨ n(n + 1) ⎬, ⎩2, ⎭, , 2, , EXERCISE 1, Mark (3) against the correct answer in each of the following., , 3, , a , 2 are in AP then a = ?, 4, 3, 5, (b), (a), 8, 8, , 1. If, , (c), , 7, 8, , 2. What is the next term of the AP 8 , 18 ,, (a) 40, , (b) 48, , 3. What is the next term of the AP 2 ,, (a) 24, , (b) 28, , (d), , 11, 8, , 32 , …?, , (c) 50, , (d) 54, , 8 , 18 , …?, (c) 30, , (d) 32, , 4. Which term of the AP 5, 8, 11, 14, … is 323?, (a) 108th, , (b) 107th, , (c) 106th, , (d) 54th, , 5. Which term of the AP 92, 88, 84, … is 0?, (a) 24th, , (b) 27th, , (c) 23rd, , (d) 28th, , 6. Which term of the AP 27, 24, 21, 18, … is –81?, (a) 39th, , (b) 38th, , (c) 37th, , (d) 29th, , 7. Which term of the AP 40, 35, 30, … is the first negative term?, (a) 9th, , (b) 10th, , (c) 12th, , (d) 14th, , 2, 5, 1, 8. Which term of the AP , 1, 1 , … is 4 ?, 3, 6, 6, (a) 26th, , (b) 25th, , (c) 24th, , (d) 19th, , 9. The 5th and 13th terms of an AP are 5 and –3 respectively. The 24th term of, this AP is, (a) – 6, (b) –8, (c) –11, (d) –14
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1028, , Senior Secondary School Mathematics for Class 11, , 10. The 2nd, 31st and the last terms of an AP are 7, How many terms are there in this AP?, (a) 53, , (b) 56, , 3 1, 1, , and − 6 respectively., 4 2, 2, , (c) 59, , (d) 62, , 11. The 10th term from the end of the AP 7, 10, 13, …, 184 is, (a) 151, , (b) 154, , (c) 160, , (d) 157, , 12. The 12th term from the end of the AP 17, 14, 11, …, – 61 is, (a) –31, , (b) –28, , (c) –34, , (d) –37, , 13. If 4, x1 , x 2 , x 3 , 28 are in AP then x 3 = ?, (a) 24, , (b) 22, , (c) 20, (d) cannot be determined, 14. The sides of a right triangle are in AP. The ratio of their lengths is, (a) 1 : 2 : 3, (b) 2 : 3 : 4, (c) 3 : 4 : 5, (d) 5 : 8 : 3, 15. How many 2-digit numbers are there which are divisible by 6?, (a) 14, (b) 15, (c) 16, (d) 17, 16. How many numbers are there between 102 and 750 which are divisible, by 8?, (a) 75, (b) 78, (c) 81, (d) 84, 17. If the nth term of a progression be a linear expression in n then the given, progression is, (a) an AP, (b) a GP, (c) an HP, (d) none of these, 18. In a given AP if pth term is q and the qth term is p then its nth term is, (a) ( p + q + n), , (b) ( p + q − n), , (c) ( p − q + n), , (d) ( p − q − n), , 1, 1, and nth term is then its (mn)th term is, n, m, 1, (d) 1, (b) mn, (c), mn, , 19. In a given AP if mth term is, ⎛ 1 1⎞, (a) ⎜ + ⎟, ⎝ m n⎠, , 20. In an AP if m times the mth term is equal to n times the nth term then its, (m + n)th term is, (a) − (m + n), (b) –1, (c) 1, (d) 0, 21. (5 + 9 + 13 + 17 + … up to 23 terms) = ?, (a) 1123, (b) 1127, (c) 1131, 22. (0.7 + 0.71 + 0.72 + … up to 100 terms) = ?, , (d) 1135, , (a) 117.5, (b) 118.5, 23. (25 + 28 + 31 + … + 100) = ?, , (c) 119.5, , (d) 121.5, , (a) 1545, (b) 1585, 24. (1 + 3 + 5 + 7 + … + 999) = ?, , (c) 1625, , (d) 1525, , (c) 249496, , (d) 250000, , (a) 251001, , (b) 249500
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Objective Questions, , 1029, , 25. (101 + 99 + 97 + … + 47) = ?, (a) 2076, , (b) 2072, , (c) 2177, , (d) 2173, , (c) 36, , (d) 54, , (c) –24, , (d) –30, , 26. If (1 + 6 + 11 + … + x) = 148 then x = ?, (a) 8, , (b) 48, , 27. If ( 26 + 21 + 16 + … + x) = 11 then x = ?, (a) –12, , (b) –18, , 28. How many terms of the AP −5 ,, (a) 21, , −9, , − 4, … will give the sum 0?, 2, , (b) 18, , (c) 23, , (d) 16, , 29. The 3rd term of an AP is 1 and its 6th term is –11. The sum of 32 terms of, this AP is, (a) 1696, (b) –1696, (c) 848, (d) –848, 30. The sum of first 7 terms of an AP is 10 and the sum of next 7 terms is 17., What is the 3rd term of the AP?, 2, 3, 5, (b) 1, (c) 1, (d) 2, (a) 1, 7, 7, 7, 31. The sum of first 80 natural numbers is, (a) 3236, , (b) 3240, , (c) 3248, , (d) 3250, , 32. The sum of all even natural numbers between 300 and 400 is, (a) 17350, , (b) 17250, , (c) 17150, , (d) 17400, , 33. The sum of all odd numbers between 100 and 200 is, (a) 7500, , (b) 7450, , (c) 7560, , (d) 7600, , 34. The sum of all positive integral multiples of 3 less than 100 is, (a) 1686, , (b) 1683, , (c) 1680, , (d) 1677, , 35. How many terms of the AP 6, 12, 18, 24, … must be taken to make the, sum 816?, (a) 16, (b) 18, (c) 14, (d) 22, 36. The sum of all 3-digit numbers divisible by 5 is, (a) 97650, , (b) 98550, , (c) 95850, , (d) 96950, , 37. The sum of n terms of an AP is ( 3n + 2n). Its common difference is, 2, , (a) 5, , (b) –5, , (c) 6, , (d) – 6, , 38. The sum of n terms of an AP is ( 3n + 5n). Which of its terms is 164?, 2, , (a) 28th, , (b) 27th, , (c) 26th, , (d) 29th, , 39. If the sum of first m terms of an AP is the same as the sum of its first n terms, then the sum of its first (m + n) terms is, (a) 0, (b) 1, (c) 2(m + n), (d) none of these
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1030, , Senior Secondary School Mathematics for Class 11, , ⎛ 1⎞, ⎛ 1⎞, 40. If the mth term of an AP is ⎜ ⎟ and the nth term is ⎜ ⎟ then the sum of its, ⎝ n⎠, ⎝ m⎠, mn terms is, 1, (a) (mn + 1), 2, , (b) 2(m + n), , (c) − (m + n), , (d) none of these, , 41. If the sum of first m terms is n and the sum of first n terms is m, then the sum, of first (m + n) terms is, (a) 0, , (b) (m + n), , (c) − (m + n), , (d) −2(m + n), , 42. If the sum of n terms of a progression be a quadratic expression in n then, it is, (a) an AP, (b) a GP, (c) an HP, (d) none of these, 43. The sum of n terms of an AP is given by Sn = ( 3n2 + 4n). Its rth term is, (a) ( 3r + 4), , (b) ( 4r + 3), , (c) (5r + 2), , (d) ( 6r + 1), , 44. If S1 , S2 , S3 be the sums of n, 2n and 3n terms of an AP respectively and, ( S2 − S1 ) = kS3 then k = ?, 1, 1, (a), (b), (c) 2, (d) 3, 2, 3, 45. The sum of all 2-digit numbers is, (a) 4750, , (b) 4895, , (c) 4905, , (d) 4850, , 46. The second and 7th terms of an AP are 2 and 22 respectively. The sum of its, first 35 terms is, (a) 2160, (b) 2240, (c) 2310, (d) 2470, 47. The first and 5th terms of an AP are –14 and 2 respectively and the sum of, its n terms is 40. Then, n = ?, (a) 8, (b) 10, (c) 12, (d) 13, ( a n + 1 + bn + 1 ), is the arithmetic mean between unequal numbers a and b, ( a n + bn), then n = ?, , 48. If, , (a) 0, , (b) 1, , (c) 2, , (d) 4, , 49. The ratio between the sums of n terms of two arithmetic progressions is, (7n + 1) : ( 4n + 27). The ratio of their 11th terms is, (a) 136 : 117, (b) 124 : 105, (c) 148 : 111, (d) 78 : 71, 50. The ratio between the sums of n terms of two arithmetic progressions is, ( 3n + 8) : (7n + 15). The ratio of their 12th terms is, (a) 9 : 14, (b) 7 : 16, (c) 8 : 15, (d) 6 : 11, 51. If there are ( 2n + 1) terms in an AP then the ratio of the sum of all odd terms, and that of all even terms is, (a) (n + 1) : n, (b) (n + 2) : n, (c) (n − 1) : n, (d) 1 : 1
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Objective Questions, , 1031, , 52. If the ratio of the sums of m and n terms of an AP be (m 2 : n2 ) then the ratio, of mth and nth terms is, (a) m : n, (b) ( 2m + 1) : ( 2n + 1), (c) ( 2m − 1) : ( 2n − 1), , (d) m 2 : n2, 53. Three numbers are in AP whose sum is 24 and whose product is 440. The, largest of these numbers is, (a) 13, (b) 11, (c) 9, (d) 12, 54. The angles of a quadrilateral are in AP whose common difference is 10°., The largest of these angles is, (a) 110°, (b) 120°, (c) 125°, (d) 130°, 55. If, , ( 3 + 5 + 7 + … to n terms), = 7 then n = ?, (5 + 8+ 11+ … to 10 terms), , (a) 35, , (b) 34, , (c) 32, , (d) 30, , 56. How many numbers lie between 10 and 300, which when divided by 4, leave a remainder 3?, (a) 71, , (b) 72, , (c) 73, , (d) 74, , 57. Two APs have the same common difference and their first terms are –1 and, –8 respectively. The difference between their 4th terms is, (a) 4, (b) 5, (c) 6, (d) 7, 58. If ( k − 1), ( 2k + 1), ( 6k + 3) are in GP then k = ?, (a) 7, , (b) 4, , (c) –2, , 59. For what values of x are the numbers, (a) –1 and 1, , (b) –1 and 2, , 60. The 17th term of the GP 2,, (a) 256, , (b) 512, , 8 , 4,, , (d) 0, , −2, −7, in GP?, , x,, 7, 2, (c) –2 and 1, , (d) –2 and 2, , 32 , … is, (c) 128 2, , 4 4, 61. The nth term of the GP 12, 4, , , … is, 3 9, 4, 4, 4, (b) n− 2, (c) n− 3, (a) n−1, 3, 3, 3, , (d) 256 2, , (d) none of these, , 62. Which term of the GP 5, 10, 20, 40, … is 5120?, (a) 9th, , (b) 10th, , (c) 11th, , (d) 12th, , 63. Which term of the GP 3 , 3 , 3 3 , … is 729?, (a) 11th, , (b) 12th, , (c) 10th, , (d) 13th, , 64. If the 4th and 9th terms of a GP are 54 and 13122 respectively then its 6th, term is, (a) 243, (b) 1458, (c) 486, (d) 729
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1032, , Senior Secondary School Mathematics for Class 11, , 1, −1, and, respectively. Its first term is, 18, 486, 2, −2, 3, −3, (a), (b), (c), (d), 3, 3, 2, 2, 66. The 8th term from the end of the GP 3, 6, 12, 24, …, 12288 is, 65. The 4th and 7th terms of a GP are, , (a) 96, , (b) 192, , (c) 48, , (d) 288, , 1, is, 67. The 6th term from the end of the GP 8, 4, 2, … ,, 1024, 1, 1, 1, 1, (b), (c), (d), (a), 64, 32, 128, 16, 68. The 4th, 7th and 10th terms of a GP are in, (b) GP, (a) AP, 69. If a, x, b are in GP then, , (c) HP, , (d) none of these, , 1, 1, (d) d = ( a + b), ab, 2, 2, 70. The arithmetic mean of two numbers is 34 and their geometric mean is 16., The numbers are, (a) x = ab, , (b) x 2 = ab, , (c) x =, , (a) 64 and 4, , (b) 52 and 16, , (c) 56 and 12, , 71. If, , (d) 60 and 8, , 1, , x1 , x 2 , 9 are in GP then x 2 = ?, 3, , (a) 1, (c) 6, , (b) 3, (d) cannot be determined, 72. If the nth term of a GP is 2n , the sum of its first 6 terms is, (a) 124, (b) 126, (c) 190, 73. (1 + 3 + 3 + 3 3 + … up to 10 terms) = ?, , (d) 254, , (a) 81( 3 + 1), (b) 100( 3 + 1), (c) 121( 3 + 1), 74. (0.15 + 0.015 + 0.0015 + … to 8 terms) = ?, 1 ⎞, 1 ⎞, 1 ⎞, 1⎛, 1⎛, 1⎛, (a) ⎜1 − 7 ⎟, (b) ⎜1 + 8 ⎟, (c) ⎜1 − 8 ⎟, 6 ⎝ 10 ⎠, 6⎝, 6 ⎝ 10 ⎠, 10 ⎠, , (d) none of these, , ⎛ 1 1 1, ⎞, 75. ⎜1 − + − + … to 9 terms⎟ = ?, ⎝, ⎠, 2 4 8, 151, 161, (b), (a), 196, 225, , (c), , 171, 256, , 1 ⎞, 1⎛, (d) ⎜1 + 7 ⎟, 6⎝, 10 ⎠, , (d), , 181, 256, , 76. (3 + 6 + 12 + … + 1536) = ?, (a) 1023, (b) 2046, 77. (2 + 6 + 18 + 54 + … + 4374) = ?, , (c) 3069, , (d) 4092, , (a) 6450, (b) 6560, (c) 6670, (d) 6380, 78. In a GP it is given that a = 3 , Tn = 96 and Sn = 189. The value of n is, (a) 7, , (b) 8, , (c) 6, , (d) 5
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Objective Questions, , 1033, , 79. How many terms of the GP 2, 6, 18, … will make the sum 728?, (a) 6, , (b) 9, , (c) 8, (d) 7, 2 1 1, 80. How many terms of the GP − + − … must be taken to make the sum, 9 3 2, 55, ?, 72, (a) 6, (b) 5, (c) 7, (d) 8, 81. If the sum of n terms of a GP is ( 2n − 1) then its common ratio is, 1, −1, (d), (a) 2, (b) 3, (c), 2, 2, 82. In a GP, the ratio between the sum of first 3 terms and the sum of first 6, terms is 125 : 152. The common ratio is, 1, 2, 3, 5, (b), (c), (d), (a), 2, 3, 5, 6, 83. If the nth term of the GP 3 ,, (a) 12, , 3 , 1, … is, , (b) 13, , 1, then n = ?, 243, , (c) 14, , (d) 15, , 84. For any two positive numbers, we have, (a) AM ≤ GM, , (b) AM ≥ GM, , (c) AM =, , 3, GM, 4, , (d) none of these, , 85. The AM between two positive numbers a and b ( a > b) is twice their GM., Then, a : b = ?, (b) ( 2 + 3 ) : ( 2 − 3 ), (a) ( 3 + 2) : ( 3 − 2), (c) 2 : 3, (d) none of these, 86. GM between 27 and 243 is, (a) 135, (b) 3 30, 87. GM between 0.15 and 0.0015 is, (a) 1.5, , (b) 0.015, , (c) 81, , (d) 40.5, , (c) 0.15, , (d) none of these, , 1 1 1, ⎛, ⎞, 88. Sum of the infinite GP ⎜1 + + +, + … ∞⎟ = ?, ⎝, ⎠, 3 9 27, 2, 5, 3, 4, (a), (b), (c), (d), 3, 9, 2, 9, 1, 1, 1, ⎛, ⎞, 89. The sum of the infinite geometric series ⎜1 − + 2 − 3 + … ∞⎟ = ?, ⎝, ⎠, 3 3, 3, 3, 2, 2, 4, (b), (c), (d), (a), 4, 3, 9, 9, 5, ⎛ −5 5, ⎞, 90. The sum of the infinite geometric series ⎜, +, −, + … ∞⎟ = ?, ⎝ 4 16 64, ⎠, 1, −1, 5, (a), (b), (c), (d) –1, 4, 4, 8
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1034, , Senior Secondary School Mathematics for Class 11, , 91. The sum of the infinite geometric series {( 2 + 1) + 1 + ( 2 − 1) + … ∞} = ?, (a), , ( 2 + 3 2), 2, , (b), , ( 4 + 3 2), 2, , (c), , ( 3 + 2 2), 2, , (d), , ( 3 + 2), 2, , 2, 3, ⎛2 3, ⎞, 92. The sum of the infinite series ⎜ + 2 + 3 + 4 + … ∞⎟ = ?, ⎝5 5, ⎠, 5, 5, 11, 13, 11, 13, (b), (c), (d), (a), 15, 15, 24, 24, 4, 80, 93. The sum of an infinite GP is and its common ratio is − ⋅ The first term of, 5, 9, the GP is, (a) 8, , (b) 12, , (c) 16, , (d) 20, , 94. The sum of an infinite series is 8 and its second term is 2. Its common, ratio is, 1, 1, 2, 3, (b), (c), (d), (a), 2, 4, 3, 4, 95. The sum of an infinite geometric series is 15 and the sum of the squares of, these terms is 45. The first term of the series is, (a) 6, (b) 7, (c) 5, (d) 9, 96. If a, b, c are in GP and a1/x = b1/y = c1/z then x, y, z are in, (a) AP, , (b) GP, , (c) HP, , (d) none of these, , 97. If a, b, c are in GP then log a , log b , log c are in, (b) GP, , (a) AP, , 98. If y = x + x + x + … ∞ then x = ?, y, y, (a), (b), (1 − y), (1 + y), 2, , (c) HP, , (d) none of these, , 3, , (c), , 1, y, , (d), , 1+ y, y, , 99. If x = 1 + a + a 2 + … ∞ and y = 1 + b + b 2 + … ∞ , where|a| < 1 and|b| < 1 then, (1 + ab + a 2 b 2 + … ∞) = ?, xy, x+y, xy, xy, (c), (a), (b), (d), x+y, xy, x+ y+1, x + y −1, a a, c, c, ⎛, ⎛, ⎛ b b, ⎞, ⎞, ⎞, 100. If x = ⎜ a + + 2 + … ∞⎟ , y = ⎜ b − + 2 − … ∞⎟ and z = ⎜ c + 2 + 4 + … ∞⎟, ⎝ r r, ⎠, ⎝, ⎠, ⎝ r r, ⎠, r, r, xy, then, =?, z, (a), , ab, c, , (b), , c, ab, , (c) c ab, , (d), , c( a + b), ab, , 101. If a, b, c are in AP; a, x, b and b, y, c are in GP then x 2 , b 2 , y 2 are in, (b) GP, (c) HP, (d) none of these, (a) AP, 102. If a, b, c are in AP and a, b, d are in GP then a , ( a − b), ( d − c) are in, (a) AP, , (b) GP, , (c) HP, , (d) none of these
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Objective Questions, , 1035, , 103. If pth, qth and rth terms of a GP be a, b, c respectively then a (q− r ) , b (r − p ) ,, c( p − q ) = ?, (a) 0, , (b) 1, , (c) –1, , (d) none of these, , 104. If a, b, c are in AP then bc, ca, ab are in, (b) GP, (c) HP, (a) AP, 2, 2, 2, 105. If a , b , c are in AP then ( b + c), ( c + a), ( a + b) are in, , (d) none of these, , (b) GP, (c) HP, (d) none of these, a( b + c) b( c + a) c( a + b), are in, 106. If a, b, c are in AP then, ,, ,, bc, ca, ab, (a) AP, , (b) GP, , (a) AP, 107. If, , (c) HP, , (d) none of these, , ( b + c − a) ( c + a − b) ( a + b − c), are in AP then a, b, c are in, ,, ,, a, b, c, , (b) GP, (a) AP, 108. (1 + 2 + 3 + … + 80) = ?, (a) 6480, , (b) 3240, , (c) HP, , (d) none of these, , (c) 1620, , (d) 2430, , (c) 63920, , (d) 71930, , (c) 672400, , (d) none of these, , 109. (1 + 2 + 3 + … + 60 ) = ?, 2, , 2, , 2, , 2, , (a) 73810, (b) 36905, 110. (13 + 23 + 3 3 + … + 403 ) = ?, (a) 635400, , (b) 628400, , ANSWERS (EXERCISE 1), , 1. (d) 2. (c) 3. (d) 4. (b) 5. (a) 6. (c) 7. (b) 8. (c) 9. (d) 10. (c), 11. (d) 12. (b) 13. (b) 14. (c) 15. (b) 16. (c) 17. (a) 18. (b) 19. (d) 20. (d), 21. (b) 22. (c) 23. (c) 24. (d) 25. (b) 26. (c) 27. (c) 28. (a) 29. (b) 30. (a), 31. (b) 32. (c) 33. (a) 34. (b) 35. (a) 36. (b) 37. (c) 38. (b) 39. (a) 40. (a), 41. (c) 42. (a) 43. (d) 44. (b) 45. (b) 46. (c) 47. (b) 48. (a) 49. (c) 50. (b), 51. (a) 52. (c) 53. (b) 54. (b) 55. (a) 56. (c) 57. (d) 58. (b) 59. (a) 60. (b), 61. (b) 62. (c) 63. (b) 64. (c) 65. (d) 66. (a) 67. (b) 68. (b) 69. (b) 70. (a), 71. (b) 72. (b) 73. (c) 74. (c) 75. (c) 76. (c) 77. (b) 78. (c) 79. (a) 80. (b), 81. (a) 82. (c) 83. (b) 84. (b) 85. (b) 86. (c) 87. (b) 88. (c) 89. (a) 90. (d), 91. (b) 92. (d) 93. (c) 94. (a) 95. (c) 96. (a) 97. (a) 98. (b) 99. (d) 100. (a), 101. (a) 102. (b) 103. (b) 104. (c) 105. (c) 106. (a) 107. (c) 108. (b) 109. (a) 110. (c), HINTS TO SOME SELECTED QUESTIONS, 3, 1. Since , a, 2 are in AP, we have, 4, a−, , 3, 3, 11, 11, ⇒ a=, ⋅, = 2 − a ⇒ 2 a = ⎛⎜ 2 + ⎞⎟ =, ⎝, 4, 4⎠ 4, 8
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1036, , Senior Secondary School Mathematics for Class 11, , 2. Given AP is 2 2 , 3 2 , 4 2 . Its next term is 5 2 = 25 × 2 = 50 ., 3. Given AP is 2 , 2 2 , 3 2 . Its next term is 4 2 = 16 × 2 =, , 32 ., , 4. Let Tn = 323. Then, a + (n − 1)d = 323 , where a = 5 and d = 3., 318, = 106 ⇒ n = 107 ., ∴ 5 + (n − 1) × 3 = 323 ⇒ (n − 1) =, 3, ∴, , 107th term is 323., , 5. Let Tn = 0. Then, a + (n − 1)d = 0 , where a = 92 and d = − 4., ∴, , 92 + (n − 1) × ( − 4 ) = 0 ⇒ 4n = 96 ⇒ n = 24., , ∴, , 24th term is 0., , 6. Let Tn = −81. Then, a + (n − 1)d = −81, where a = 27 and d = −3., ∴, , 27 + (n − 1) × ( −3 ) = −81 ⇒ 3n = ( 30 + 81) = 111 ⇒ n = 37 ., , ∴, , 37th term is –81., , 7. Let Tn < 0. Then, {a + (n − 1)d } < 0 , where a = 40 and d = −5., ∴, , {40 + (n − 1) × ( −5 )} < 0 ⇒ 45 < 5n ⇒ 5n > 45 ⇒ n > 9., , ∴, , 10th term is the first negative term., 5, 1, 14, 8. This is an AP in which a = , d = and Tn =, ⋅, 6, 6, 3, 14, 5, 1 14, a + (n − 1)d =, ⇒, + (n − 1) × =, ∴, 6, 3, 3, 6, ∴, , 1, 14 5 1 ⎞ ( 28 − 5 + 1) 24, =, = 4 ⇒ n = 24 ., n = ⎛⎜, − + ⎟=, ⎝ 3 6 6⎠, 6, 6, 6, , Hence, 24th term is 4, 9. a + 4 d = 5, , 2, ⋅, 3, , ... (i), , and a + 12 d = −3, , ... (ii), , On solving (i) and (ii), we get a = 9 and d = −1., ∴, , T24 = a + 23 d = ( 9 − 23 ) = −14 ., 31, 1, 10. a + d =, ... (i), and a + 30 d =, 4, 2, −1, On solving (i) and (ii), we get a = 8 and d =, ⋅, 4, −13, −13, Tn =, ⇒ a + (n − 1)d =, 2, 2, ( −1) ⎛ −13, −29, −13, −1, ⇒ (n − 1) ×, =⎜, − 8 ⎞⎟ =, ⇒ 8 + (n − 1) × ⎛⎜ ⎞⎟ =, ⎝ 2, ⎠, ⎝ 4⎠, 4, 2, 2, ⇒ (n − 1) =, ∴, , −29, 4, ×, = 58 ⇒ n = 59., 2, ( −1), , the given AP has 59 terms., , 11. 10th term from the end = l − (n − 1)d , where l = 184 , n = 10 , d = 3, = ( 184 − 9 × 3 ) = ( 184 − 27 ) = 157 ., 12. 12th term from the end = l − (n − 1)d , where l = − 61, n = 12 , d = −3, = − 61 − 11 × ( −3 ) = ( − 61 + 33 ) = −28., , ... (ii)
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Objective Questions, , 1037, , 13. a = 4 and T5 = 28 ⇒ a + 4 d = 28 ⇒ 4 + 4 d = 28 ⇒ d = 6 ., ∴, , x3 = a + 3 d = 4 + 3 × 6 = 22., , 14. Let the sides of the triangle be a − d , a and a + d., Then, ( a + d ) 2 = ( a − d ) 2 + a2, ⇒, , a2 = ( a + d ) 2 − ( a − d ) 2 ⇒ a2 = 4 ad ⇒ a = 4 d, , ⇒, , the sides are ( 4 d − d ), ( 4 d ) and ( 4d + d ), i.e., 3 d , 4 d , 5 d, , ⇒, , ratio of the sides = 3 : 4 : 5., , 15. The required numbers are 12, 18, 24, …, 96., This is an AP in which a = 12 , d = 6 and t n = 96., a + (n − 1)d = 96 ⇒ 12 + (n − 1) × 6 = 96 ⇒ (n − 1) =, , 84, = 14 ⇒ n = 15., 6, , So, there are 15 such numbers., 16. The required numbers are 104, 112, 120, …, 744., This is an AP in which a = 104 , d = 8 and Tn = 744., a + (n − 1)d = 744 ⇒ 104 + (n − 1) × 8 = 744 ⇒ (n − 1) =, , 640, = 80 ⇒ n = 81., 8, , So, there are 81 such numbers., 17. Let Tn = an + b , where a and b are constants., ∴, , ( Tn − Tn −1 ) = ( an + b ) − [a(n − 1) + b] = a, which is a constant., , ∴, , the given progression is an AP., , 18. Given a + ( p − 1)d = q, , ... (i), , and a + ( q − 1)d = p, , ... (ii), , On subtracting, we get ( p − q)d = ( q − p ) ⇒ d = −1., Putting d = −1in (i), we get a = ( p + q − 1)., ∴, , Tn = a + (n − 1)d ⇒ Tn = ( p + q − 1) + (n − 1) × ( −1) = ( p + q − n)., 1, 1, 19. a + (m − 1)d =, ... (i), and a + (n − 1)d =, n, m, 1, 1, and a =, On solving (i) and (ii), we get d =, ⋅, mn, mn, 1, 1, Tmn = a + (mn − 1)d =, + (mn − 1) ×, = 1., ∴, mn, mn, Hence, (mn)th term is 1., 20. m ⋅ Tm = n ⋅ Tn, ⇒, , m ⋅ [a + (m − 1)d] = n ⋅ [a + (n − 1)d], , ⇒, , [(m 2 − n2 ) − (m − n)] ⋅ d = (n − m )a, , ⇒, , (m + n − 1)d = − a ⇒ a + (m + n − 1)d = 0 ⇒ Tm + n = 0., , Hence, the (m + n)th term is 0., 21. In this AP, we have a = 5 and d = 4., n, 23, Sn = [2 a + (n − 1)d] ⇒ S23 =, ⋅ [2 × 5 + 22 × 4], 2, 2, 23, ⇒ S23 = ⎛⎜ × 98 ⎞⎟ = ( 23 × 49 ) = 1127 ., ⎝ 2, ⎠, , ... (ii)
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1038, , Senior Secondary School Mathematics for Class 11, , 22. In this AP, we have a = 0.7 and d = (0.71 − 0.70) = 0.01., Sn =, , n, 100, [2 a + (n − 1)d] ⇒ S100 =, ⋅ [2 × 0.7 + 99 × 0.1], 2, 2, ⇒ S100 = 50 × (1.4 + 0.99) = (50 × 2.39) = 119.5., , 23. In this AP, we have a = 25 , d = 3 and l = 100., Let Tn = 100 ⇒ a + (n − 1)d = 100 ⇒ 25 + (n − 1) × 3 = 100, ⇒ (n − 1) =, S26 =, , 75, = 25 ⇒ n = 26 ., 3, , 26, n, ( a + l) =, ⋅ ( 25 + 100 ) = ( 13 × 125 ) = 1625., 2, 2, , 24. This is an AP in which a = 1, d = 2 and l = 999., Let Tn = 999 ⇒ a + (n − 1)d = 999, ⇒ 1 + (n − 1) × d = 999 ⇒ (n − 1) =, ∴, , Sn =, , 998, = 499 ⇒ n = 500 ., 2, , n, 500, ( a + l) =, ⋅ ( 1 + 999 ) = ( 250 × 1000 ) = 250000., 2, 2, , 25. This is an AP in which a = 101, d = −2 and l = 47., Let Tn = 47. Then, a + (n − 1)d = 47 ., ∴, , 101 + (n − 1) × ( −2 ) = 47 ⇒ 2n = ( 103 − 47 ) = 56 ⇒ n = 28., , ∴, , S28 =, , 28, n, ( a + l) =, ⋅ ( 101 + 47 ) = ( 14 × 148 ) = 2072., 2, 2, , 26. Let the number of terms be n. Then,, , ⇒, , n, ⋅ [2 a + (n − 1)d] = 148, 2, n, ⋅ [2 × 1 + (n − 1) × 5] = 148 ⇒ n(5n − 3 ) = 296, 2, , ⇒, , 5n2 − 3n − 296 = 0 ⇒ n =, , ⇒, , n=, , ⇒, , n= 8, , ( 3 + 77 ), 10, , or n =, , 3±, , 9 + 5920 3 ± 77, =, 10, 10, , ( 3 − 77 ), −37, ⇒ n = 8 or n =, 10, 5, [Q number of terms cannot be a fraction], , T8 = ( a + 7 d ) = ( 1 + 7 × 5 ) = 36 ⇒ x = 36 ., 27. Let the number of terms be n. Then,, , ⇒, , n, ⋅ [2 a + (n − 1)d] = 11, 2, n, ⋅ [2 × 26 + (n − 1) × ( −5 )] = 11 ⇒ n(57 − 5n) = 22, 2, , ⇒, , 5n2 − 57 n + 22 = 0 ⇒ 5n2 − 55n − 2n + 22 = 0, , ⇒, , 5n(n − 11) − 2(n − 11) = 0 ⇒ (n − 11)(5n − 2 ) = 0 ⇒ n = 11, , T11 = ( a + 10 d ) = [26 + 10 × ( −5 )] = ( 26 − 50 ) = −24 ⇒ x = −24 ., , 2⎤, ⎡, ⎢⎣Q n ≠ 5 ⎥⎦
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Objective Questions, , 1039, , ∴, , n, ⋅ [2 a + (n − 1)d] = 0., 2, n ⎡, 1⎤, ⎡ −21 1 ⎤, ⋅ 2 × ( −5 ) + (n − 1) × ⎥ = 0 ⇒ n ⋅ ⎢, + n⎥ = 0, 2 ⎣⎢, 2⎦, 2 ⎦, ⎣ 2, 1, 21, 1, 21, ⎧, = 0 ⇒ n=, ⇒ n = 21⎫⎬, ⇒ n = 0 or ⎨ n −, 2, 2, 2, ⎩2, ⎭, , ∴, , sum of 21 terms is 0., , 28. Let Sn = 0. Then,, , 29. T3 = 1 and T6 = −11, ⇒, , a + 2d = 1, , ⇒, , d = −4 and a = 9, 32, ⋅ [2 a + ( 32 − 1)d] = 16 × [2 × 9 + 31 × ( − 4 )] = 16 × ( −106 ) = −1696., S32 =, 2, , ⇒, , ... (i), , a + 5 d = −11, , 30. Given S7 = 10 and S14 = 27 ., 7, 14, ⋅ ( 2 a + 6 d ) = 10 and, ⋅ ( 2 a + 13 d ) = 27, ∴, 2, 2, 10, 27, and 2 a + 13 d =, ⇒ a + 3d =, 7, 7, 1, ⇒ d = and a = 1, 7, 2, 2, T3 = ( a + 2 d ) = ⎛⎜ 1 + ⎞⎟ = 1 ⋅, ∴, ⎝, 7, 7⎠, 31. Given sum = 1 + 2 + 3 + … + 80., n, 80, S = ( a + l) =, × ( 1 + 80 ) = ( 40 × 81) = 3240 ., 2, 2, 32. Let Sn = 302 + 304 + … + 398., Let the number of terms be n. Then,, a + (n − 1)d = 398 ⇒ 302 + (n − 1) × 2 = 398 ⇒ n − 1 = 48 ⇒ n = 49., ∴, , Sn =, , 49, 49, n, ( a + l) =, ( 302 + 398 ) = ⎛⎜ × 700 ⎞⎟ = ( 49 × 350 ) = 17150., ⎝ 2, ⎠, 2, 2, , 33. Let Sn = 101 + 103 + 105 + … + 199., Let the number of terms be n. Then,, ∴, , a + (n − 1)d = 199 ⇒ 101 + (n − 1) × 2 = 199 ⇒ n − 1 = 49 ⇒ n = 50., n, 50, Sn = ( a + l ) =, ⋅ ( 101 + 199 ) = ( 25 × 300 ) = 7500., 2, 2, , 34. Let Sn = 3 + 6 + 9 + 12 + … + 99., Let the number of terms be n. Then,, a + (n − 1)d = 99 ⇒ 3 + (n − 1) × 3 = 99 ⇒ (n − 1) = 32 ⇒ n = 33., ∴, , Sn =, , 33, 33, n, ( a + l) =, ⋅ ( 3 + 99 ) = ⎛⎜, × 102 ⎞⎟ = ( 33 × 51) = 1683., ⎝ 2, ⎠, 2, 2, , 35. Here a = 6 and d = 6. Let Sn = 816. Then,, n, n, ⋅ [2 a + (n − 1)d] = 816 ⇒ ⋅ [2 × 6 + (n − 1) × 6] = 816, 2, 2, , ... (ii)
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1040, , Senior Secondary School Mathematics for Class 11, , ⇒, , n(n + 1) = 272 ⇒ n2 + n − 272 = 0 ⇒ n2 + 17 n − 16n − 272 = 0, , ⇒, , n(n + 17 ) − 16(n + 17 ) = 0 ⇒ (n + 17 )(n − 16 ) = 0 ⇒ n = 16., , Required number of terms = 16., 36. Sn = 100 + 105 + 110 + … + 995, a + (n − 1)d = 995 ⇒ 100 + (n − 1) × 5 = 995 ⇒ (n − 1) =, Sn =, , 895, = 179 ⇒ n = 180., 5, , n, 180, ( a + l) =, × ( 100 + 995 ) = ( 90 × 1095 ) = 98550., 2, 2, , 37. Sn = ( 3n2 + 2n) ⇒ Sn −1 = 3(n − 1) 2 + 2(n − 1) = ( 3n2 − 4n + 1), ∴, , Tn = ( Sn − Sn −1 ) = ( 3n2 + 2n) − ( 3n2 − 4n + 1) = ( 6n − 1), , ⇒, , T1 = ( 6 × 1 − 1) = 5 and T2 = ( 6 × 2 − 1) = 11, , ⇒, , d = ( T2 − T1 ) = ( 11 − 5 ) = 6., , 38. Sn = ( 3n2 + 5n) ⇒ Sn −1 = 3(n − 1) 2 + 5(n − 1) = ( 3n2 − n − 2 ), ∴, , Tn = ( Sn − Sn −1 ) = ( 3n2 + 5n) − ( 3n2 − n − 2 ) = ( 6n + 2 ), , ∴, , T1 = 8 , T2 = 14 and d = ( 14 − 8 ) = 6., , Let Tn = 164. Then a + (n − 1)d = 164 ., ∴, , 8 + (n − 1) × 6 = 164 ⇒ 6n = 162 ⇒ n = 27 ., , ∴, , 27th term is 164., m, n, ⋅ [2 a + (m − 1)d] = ⋅ [2 a + (n − 1)d], 2, 2, 1, 1, ⇒ ma + m(m − 1)d = na + n(n − 1)d, 2, 2, 1, ⇒ {(m 2 − n2 ) − (m − n)} ⋅ d = (n − m )a, 2, , 39. Sm = Sn ⇒, , ⇒ (m + n − 1)d = −2 a., (m + n), (m + n), Sm + n =, ⋅ [2 a + (m + n − 1)d] =, ⋅ [2 a − 2 a] = 0., 2, 2, 1, 1, 40. a + (m − 1)d =, … (i), and a + (n − 1)d =, n, m, 1, 1, and a =, On solving (i) and (ii), we get d =, ⋅, mn, mn, ∴, 41., , ... (ii), , mn ⎞, ⎛ mn ⎞ ⎡ 2 + (mn − 1) ⎤ = 1 (mn + 1)., Smn = ⎛⎜, ⎟ [2 a + (mn − 1)d] = ⎜, ⎟, ⎝ 2 ⎠ ⎢⎣ mn, ⎝ 2 ⎠, mn ⎥⎦ 2, , m, n, ⋅ [2 a + (m − 1)d] = n and ⋅ [2 a + (n − 1)d] = m, 2, 2, ⇒, , m ⋅ [2 a + (m − 1)d] = 2n, , ... (i), , and n ⋅ [2 a + (n − 1)d] = 2 m, , ... (ii), , On subtracting, we get, 2 a(m − n) + [(m 2 − n2 ) − (m − n)] ⋅ d = 2 (n − m ), ⇒, , 2 a(m − n) + (m − n)(m + n − 1)d = 2 (n − m ), , ⇒, , 2 a + (m + n − 1)d = −2, , ... (iii)
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Objective Questions, ∴, , m + n⎞, Sm + n = ⎛⎜, ⎟ ⋅ [2 a + (m + n − 1)d], ⎝ 2 ⎠, =, , (m + n), ⋅ ( −2 ) = − (m + n), 2, , [using (iii)]., , 42. Let Sn = an2 + bn + c . Then,, Sn−1 = a(n − 1) 2 + b(n − 1) + c, ∴, , Tn = ( Sn − Sn −1 ) = a[n2 − (n − 1) 2 ] + b[n − (n − 1)] = a( 2n − 1) + b, = 2an + ( b − a), which is a linear expression in n., , Hence, the given progression is an AP., 43. Sn = ( 3n2 + 4n) and Sn−1 = 3(n − 1) 2 + 4(n − 1), ∴, , Tn = ( Sn − Sn −1 ) = 3{n2 − (n − 1) 2 } + 4{n − (n − 1)} = 3( 2n − 1) + 4 = ( 6n + 1)., , ∴, , rth term = Tr = ( 6r + 1)., 2n, n, 44. ( S2 − S1 ) =, ⋅ [2 a + ( 2n − 1)d] − ⋅ [2 a + (n − 1)d], 2, 2, n, n, = ⋅ [4 a + ( 4n − 2 )d] − ⋅ [2 a + (n − 1)d], 2, 2, n, = ⋅ [( 4 a − 2 a) + {( 4n − 2 ) − (n − 1)}d], 2, n, 1 3n, 1, = ⋅ [2 a + ( 3n − 1)d] = ⋅ ⎧⎨ ⋅ [2 a + ( 3n − 1)d]⎫⎬ = S3 ., 2, 3 ⎩ 2, ⎭ 3, ∴, , k=, , 1, ⋅, 3, , 45. Sn = 11 + 12 + 13 + … + 99., Here, n = ( 99 − 10 ) = 89 , d = 1., ∴, , Sn =, , 89, 89 × 110 ⎞, n, ( a + l) =, ⋅ ( 11 + 99 ) = ⎛⎜, ⎟ = ( 89 × 55 ) = 4895., ⎝, ⎠, 2, 2, 2, , 46. a + d = 2 and a + 6 d = 22., On solving, we get d = 4 and a = −2., 35, 35, ⋅ [2 a + 34 d] =, ⋅ [2 × ( −2 ) + 34 × 4], ∴, S35 =, 2, 2, 35, = ⎛⎜, × 132 ⎞⎟ = ( 35 × 66 ) = 2310., ⎝ 2, ⎠, 47. a = −14 and ( a + 4 d = 2 ⇒ − 14 + 4 d = 2 ⇒ 4 d = 16 ⇒ d = 4 ), ∴, a = −14 and d = 4., n, Sn = 40 ⇒ ⋅ [2 a + (n − 1)d] = 40, 2, n, ⋅ [2 × ( −14 ) + (n − 1) × 4] = 40, ⇒, 2, ⇒, , n( 4n − 32 ) = 80 ⇒ n(n − 8 ) = 20, , ⇒, , n2 − 8n − 20 = 0 ⇒ n2 − 10 n + 2 n − 20 = 0, , ⇒, , n(n − 10 ) + 2(n − 10 ) = 0 ⇒ (n − 10 )(n + 2 ) = 0 ⇒ n = 10., , Hence, n = 10., , 1041
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1042, , Senior Secondary School Mathematics for Class 11, , 48. Given, , ( an + 1 + b n + 1 ), ( an + b n ), , =, , a+ b, 2, , ⇒, , ( an + b n )( a + b ) = 2( an + 1 + b n + 1 ), , ⇒, , an + 1 + b n + 1 + an b + b n a = 2 an + 1 + 2 b n + 1, , ⇒, , an + 1 − an b + b n + 1 − b n a = 0 ⇒ an ( a − b ) − b n ( a − b ) = 0, , ⇒, , ( a − b )( an − b n ) = 0 ⇒ an − b n = 0, , ⇒, , n, , [Q a − b ≠ 0], , n, , a = b and a ≠ b ⇒ n = 0., , n, [2 a + (n − 1)d]., 2, n, And, Sn′ = [2 A + (n − 1)D]., 2, 2 a + (n − 1)d, 7n + 1, 7n + 1, Sn, ⇒, =, =, ∴, 4, 27, 2, (, 1, ), 4, +, +, −, n, A, n, D, n + 27, S′, , 49. Let Sn =, , ... (i), , n, , 2 a + ( 21 − 1)d, T11, a + 10 d, 2 a + 20 d, =, =, =, t11, A + 10 D 2 A + 20 D 2 A + ( 21 − 1)D, 7 × 21 + 1 148, [putting n = 21in (i)]., =, =, 4 × 21 + 27 111, ∴, , required ratio = 148 : 111., n, 50. Let Sn = ⋅ [2 a + (n − 1)d]., 2, n, And, Sn′ = ⋅ [2 A + (n − 1)D]., 2, 2 a + (n − 1)d, 3n + 8, 3n + 8, Sn, Now,, ⇒, =, =, 7, 15, 2, (, 1, ), 7, +, +, −, n, A, n, D, n + 15, ′, S, , ... (i), , n, , ∴, , 2 a + ( 23 − 1)d, T12, a + 11d, 2 a + 22 d, =, =, =, t12, A + 11D 2 A + 22 D 2 A + ( 23 − 1)D, =, , ∴, , 3 × 23 + 8, 77, 7, =, =, 7 × 23 + 15 176 16, , [putting n = 23 in (i)]., , required ratio = 7 : 16., , 51. Let a be the first term and d be the common difference of the given AP., Then, ak = a + ( k − 1)d, , ... (i), , Let S1 = sum of all odd terms and S2 = sum of all even terms., Then, S1 = a1 + a3 + a5 + … + a2 n + 1, (n + 1), (n + 1), =, ⋅ {a1 + a2 n + 1 } =, ⋅ {a + a + ( 2n + 1 − 1)d }, 2, 2, (n + 1), =, ⋅ 2( a + nd ) = (n + 1)( a + nd ), 2, And, S2 = a2 + a4 + a6 + … + a2 n, n, n, = ⋅ {a2 + a2 n } = ⋅ {( a + d ) + a + ( 2n − 1)d }, 2, 2, n, = ⋅ 2{a + nd } = n( a + nd ), 2, , ... (ii), , ... (iii)
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Objective Questions, ∴, , 1043, , S1 (n + 1)( a + nd ) n + 1, =, ⇒ S1 : S2 = (n + 1) : n., =, S2, n( a + nd ), n, , 52. Let a be the first term and d be the common difference., m, n, Then, Sm = ⋅ [2 a + (m − 1)d] and Sn = ⋅ [2 a + (n − 1)d]., 2, 2, m, [2 a + (m − 1)d], S, m2, 2, = m =, ∴, n, S, n2, n, [2 a + (n − 1)d], 2, 2 a + (m − 1)d m, =, ⇒ 2 an + n(m − 1)d = 2 am + m(n − 1)d, ⇒, 2 a + (n − 1)d n, ⇒, ∴, , (m − n)d = 2 a(m − n) ⇒ d = 2 a., Tm a + (m − 1)d a + (m − 1) × 2 a, [Q d = 2 a], =, =, Tn, a + (n − 1)d, a + (n − 1) × 2 a, =, , ⇒, , 1 + 2(m − 1) 2m − 1, ⋅, =, 1 + 2(n − 1), 2n − 1, , Tm : Tn = ( 2m − 1) : ( 2n − 1)., , 53. Let the numbers be ( a − d ), a and ( a + d ). Then,, ( a − d ) + a + ( a + d ) = 24 ⇒ 3 a = 24 ⇒ a = 8., And, ( 8 − d ) × 8 × ( 8 + d ) = 440 ⇒ ( 64 − d 2 ) = 55 ⇒ d 2 = 9 ⇒ d = ±3., So, the numbers are {( 8 − 3 ), 8 and ( 8 + 3 )} or {( 8 + 3 ), 8 and ( 8 − 3 )}., The largest of these numbers is 11., 54. Let the given angles be ( a − 3 d ), ( a − d ), ( a + d ), ( a + 3 d )., Then, a − 3 d + a − d + a + d + a + 3 d = 360 ⇒ 4 a = 360 ⇒ a = 90., Also, d = 10 ° (given), ∴, , the angles are ( 90 ° − 3 × 10 ° ), ( 90 ° − 10 ° ), ( 90 ° + 10 ° ), ( 90 ° + 3 × 10 ° ), , i.e., 60 ° , 80 ° , 100 ° and 120°., The largest of these angles is 120°., ( 3 + 5 + 7 + … to n terms ), 55., =7, (5 + 8 + 11 + … to 10 terms), ⇒, , n, ⋅ [6 + (n − 1) × 2], 2, = 7 ⇒ n(n + 2 ) = 1295, 10, ⋅ [10 + 9 × 3], 2, , ⇒, , n2 + 2n − 1295 = 0 ⇒ n2 + 37 n − 35n − 1295 = 0, , ⇒, , n(n + 37 ) − 35(n + 37 ) = 0 ⇒ (n + 37 )(n − 35 ) = 0 ⇒ n = 35., , 56. The required numbers are 11, 15, 19, …, 299., This is an AP in which a = 11, d = 4 and Tn = 299., Tn = 299 ⇒ a + (n − 1)d = 299, ⇒ 11 + (n − 1) × 4 = 299 ⇒ (n − 1) × 4 = 288 ⇒ n − 1 = 72 ⇒ n = 73., So, there are 73 such numbers.
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1044, , Senior Secondary School Mathematics for Class 11, , 57. Let the common difference of each AP be d. Then,, T4 = −1 + 3 d and t 4 = −8 + 3 d, ⇒, , ( T4 − t 4 ) = ( −1 + 3 d ) − ( −8 + 3 d ) = 7 ., , 58. ( k − 1), ( 2 k + 1), ( 6 k + 3 ) are in GP, ( 2 k + 1) ( 6 k + 3 ) 3( 2 k + 1), =3, =, =, ⇒, ( k − 1), ( 2 k + 1), ( 2 k + 1), ⇒, , 2k + 1 = 3k − 3 ⇒ k = 4 ., , ⎛ −7 ⎞, ⎜, ⎟, ⎝ 2 ⎠, x, −2, −7, 59., are in GP ⇒, , x,, =, x, 7, 2, ⎛ −2 ⎞, ⎜ ⎟, ⎝ 7 ⎠, −7 ⎞, −2, ⇒ x 2 = ⎛⎜ ⎞⎟ × ⎛⎜, ⎟ = 1 ⇒ x = −1 or, ⎝ 7 ⎠ ⎝ 2 ⎠, ∴, , required values of x are –1 and 1., , 60. Given GP is 2 , 2 2 , 4 , 4 2 , … ., 2 2, = 2., 2, , Here, a = 2 and r =, ∴, , T17 = ar 16 = 2 × ( 2 )16 = 2 × 2 8 = 2 9 = 512 ., , Hence, 17th term is 512., 4 4, 61. Given GP is 12 , 4 , , , … ., 3 9, 4, 1, Here, a = 12 and r =, = ⋅, 12 3, ∴, , Tn = ar n −1 = 12 × ⎛⎜, ⎝, , 1⎞, ⎟, 3⎠, , n −1, , = 4× 3×, , 1, 3 n −1, , =, , 4, 3 (n − 2 ), , ⋅, , 62. Given GP is 5, 10, 20, 40, … ., 10, Here, a = 5 and r =, = 2., 5, Let Tn = 5120. Then, ar n −1 = 5120 ⇒ 5 × 2 n −1 = 5120, ∴, , 2 n −1 =, , 5120, = 1024 = 210 ⇒ n − 1 = 10 ⇒ n = 11., 5, , Hence, the 11th term is 5120., 63. Given GP is 3 , 3 , 3 3 , … ., 3 and r =, , Here, a =, , 3, =, 3, , 3., , Let Tn = 729. Then, ar n −1 = 729 ⇒, ∴, , n, , 12, , ( 3) = ( 3), , ⇒ n = 12 ., , Hence, the 12th term is 729., 64. Given T4 = 54 and T9 = 13122 ., ∴, , ar 3 = 54 and ar 8 = 13122, , 3 × ( 3 ) n −1 = 729 = 3 6, , x = 1.
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Objective Questions, ar 8, , 1045, , 13122, = 243 ⇒ r 5 = 35 ⇒ r = 3 ., 54, 54, ∴, a × 3 3 = 54 ⇒ a =, = 2., 27, Thus, a = 2 and r = 3., ⇒, , ∴, , ar 3, , =, , T6 = ar 5 = ( 2 × 35 ) = ( 2 × 243 ) = 486., , Hence, the 6th term is 486., 1, −1, and T7 =, ⋅, 18, 486, 1, −1, and ar 6 =, ∴, ar 3 =, 18, 486, , 65. Given T4 =, , ⇒, , ar 6, ar, , 3, , =, , −1, −1, −1 ⎛ −1 ⎞, × 18 =, ⇒ r3 =, =⎜ ⎟, 486, 27, 27 ⎝ 3 ⎠, , 3, , ⇒ r=, , −1, ⋅, 3, , 3, , 1, 1, 1, −3, −1, −1, a × ⎛⎜ ⎞⎟ =, ⇒ a × ⎛⎜ ⎞⎟ =, ⇒ a=, × ( −27 ) =, ⋅, ⎝ 3⎠, ⎝ 27 ⎠ 18, 18, 18, 2, −3, Hence, the first term is, ⋅, 2, 6, 66. Here, a = 3 , r = = 2 and l = 12288., 3, l, l, 12288 12288, 8th term from the end =, =, =, =, = 96 ., ∴, 128, r ( 8 −1 ) r 7, 27, Hence, the 8th term from the end is 96., 1, 67. Given GP is 8 , 4 , 2 , … ,, ⋅, 1024, 4 1, 1, Here, r = = and l =, ⋅, 8 2, 1024, ∴, , ∴, , 6th term from the end =, , l, r, , ( 6 −1 ), , =, , l, r, , 5, , =, , 1, 1, 25, 32, 1, ⋅, =, =, =, ⋅, 5, 1024 ⎛ 1 ⎞, 1024 1024 32, ⎜ ⎟, ⎝ 2⎠, , 68. Let a be the first term and r be the common ratio of the given GP. Then,, T4 = ar 3 , T7 = ar 6 and T10 = ar 9 ., T, T, T, T, Clearly, 7 = r 3 and 10 = r 3 ⇒ 7 = 10 ⋅, T4, T7, T4 T7, ∴, , T4 , T7 and T10 are in GP., x b, 69. a, x , b are in GP ⇒, = ⇒ x 2 = ab., a x, 70. Let the required numbers be a and b. Then,, ⎛ a + b = 34 ⇒ a + b = 68 ⎞ and ab = 16 ⇒ ab = ( 16 ) 2 = 256., ⎜, ⎟, ⎝ 2, ⎠, ( a − b ) 2 = ( a + b ) 2 − 4 ab = ( 68 ) 2 − 4 × 256 = ( 4624 − 1024 ) = 3600, ⇒, , a− b =, , 3600 = 60., , On solving a + b = 68 and a − b = 60, we get a = 64 , b = 4 ., ∴, , required numbers are 64 and 4.
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1046, , Senior Secondary School Mathematics for Class 11, , 1, and let r be the common ratio., 3, 1, 1, 1, Then, x1 = ar = r , x2 = ar 2 = r 2 and 9 = ar 3 = r 3 ., 3, 3, 3, , 71. Here, a =, , ∴, , r 3 = 27 = 3 3 ⇒ r = 3., , Hence, x2 =, , 1, 1, × 3 2 = × 9 = 3., 3, 3, , 72. S6 = 2 + 2 2 + 2 3 + 2 4 + 25 + 2 6., Here, a = 2 and r = 2., Sn =, , a(r n − 1), 2( 2 6 − 1), ⇒ S6 =, = 2 × ( 64 − 1) = 2 × 63 = 126., (r − 1), ( 2 − 1), , 73. Here, a = 1and r =, ∴, , Sn =, , 3, =, 1, , 3 > 1., , a(r n − 1), a(r 10 − 1) 1 × [( 3 )10 − 1] ( 35 − 1) ( 3 + 1), ⇒ S10 =, =, =, ×, (r − 1), (r − 1), ( 3 − 1), ( 3 − 1) ( 3 + 1), , ( 243 − 1), 242, × ( 3 + 1) =, ⋅ ( 3 + 1) = 121( 3 + 1)., 2, 2, 0.015, 0.015, 1, 74. Here, a = 0.15 and r =, =, =, < 1., 0.15 0.015 × 10 10, 1 ⎤, ⎡, 0.15 ⎢ 1 −, ⎥, a( 1 − r n ), ⎣, 10 n ⎦ = ⎛ 15 × 10 ⎞ ⎛ 1 − 1 ⎞ = 1 ⎛ 1 − 1 ⎞ ⋅, =, Sn =, ∴, ⎜, ⎟⎜, ⎜, ⎟, ⎟, ⎝ 100 9 ⎠ ⎝, (1 − r ), ⎛1− 1 ⎞, 10 n ⎠, 10 n ⎠ 6 ⎝, ⎜, ⎟, ⎝, 10 ⎠, ⇒ S10 =, , ∴, , S8 =, , 1⎛, 1 ⎞, ⎜1− 8 ⎟ ⋅, 6⎝, 10 ⎠, , 75. This is a GP in which a = 1and r =, , −1, ⋅, 2, , 9, ⎡, −1 ⎤, 1 × ⎢ 1 − ⎛⎜ ⎞⎟ ⎥, ⎝ 2⎠ ⎥ 2, ⎢⎣, a( 1 − r ), a( 1 − r ), ⎦ = × ⎛1+ 1 ⎞, =, ⇒ S9 =, Sn =, ⎜, ⎟, (1 − r ), (1 − r ), 3 ⎝, ⎛ 1 + 1⎞, 29 ⎠, ⎜, ⎟, ⎝, 2⎠, n, , 9, , ⇒ S9 =, , 2 ⎛, 1 ⎞ 2 513 171, =, ⋅, × ⎜1+, ⎟= ×, 3 ⎝, 512 ⎠ 3 512 256, , 6, = 2 and l = 1536., 3, ( lr − a) ( 1536 × 2 − 3 ), required sum =, = ( 3072 − 3 ) = 3069., =, (r − 1), ( 2 − 1), , 76. This is a GP in which a = 3 , r =, ∴, , 6, = 3 and l = 4374., 2, ( lr − a) ( 4374 × 3 − 2 ) ( 13122 − 2 ) 13120, required sum =, =, =, = 6560., =, (r − 1), ( 3 − 1), 2, 2, , 77. This is a GP in which a = 2 , r =, ∴
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Objective Questions, , 1047, , 78. Let the common ratio be r. Then,, Sn =, ∴, , ( lr − a), ( 96r − 3 ), ⇒, = 189, (r − 1), (r − 1), , ( 96r − 3 ) = 189r − 189 ⇒ ( 189r − 96r ) = 186 ⇒ 93r = 186 ⇒ r = 2 ., , Now, l = ar n −1 ⇒ 3 × 2 (n −1 ) = 96 ⇒ 2 (n −1 ) = 32 = 25 ⇒ n − 1 = 5 ⇒ n = 6., Hence, n = 6., 79. Let the required number of terms be n., 6, Here, a = 2 , r = = 3 > 1 and Sn = 728., 2, 2( 3 n − 1), a(r n − 1), = 728, ⇒, Sn =, (r − 1), ( 3 − 1), ∴, , 3 n − 1 = 728 ⇒ 3 n = 729 = 3 6 ⇒ n = 6., , ∴, , required number of terms = 6., , 80. Let the required number of terms be n., Here, a =, , 2, −1 9 −3, 55, ,r =, × =, < 1 and Sn =, ⋅, 9, 3 2, 2, 72, , a( 1 − r n ), Sn =, ⇒, (1 − r ), , 2 ⎧⎪, ⎛ −3 ⎞, ⎨1 − ⎜ ⎟, ⎝ 2 ⎠, 9 ⎩⎪, , n⎫, , ⎪, ⎬, ⎭⎪, , ⎛ 1+ 3⎞, ⎜, ⎟, ⎝, 2⎠, , =, , 55, 72, , n, n, 2 ⎞ ⎧⎪, ⎛ −3 ⎞ ⎫⎪ 55 ⇒ 1 − ⎛ −3 ⎞ = 55 × 45 = 275, ⎜ ⎟, ⎟ ⎨1 − ⎜ ⎟ ⎬ =, ⎝ 2 ⎠, ⎝ 2 ⎠ ⎪ 72, 5 ⎠ ⎩⎪, 72 4, 32, ⎭, , ⇒, , ⎛2×, ⎜, ⎝9, , ⇒, , ⎛ −3 ⎞ = ⎛ 1 − 275 ⎞ = ( 32 − 275 ) = −243 = ⎛ −3 ⎞, ⎜ ⎟, ⎜, ⎟, ⎜ ⎟, ⎝ 2 ⎠, ⎝, ⎝ 2 ⎠, 32 ⎠, 32, 32, , ∴, , required number of terms = 5., , n, , 5, , ⇒ n = 5., , 81. Sn = ( 2 n − 1) ⇒ S1 = ( 21 − 1) = 1, S2 = ( 2 2 − 1) = ( 4 − 1) = 3., ⇒ T1 = 1, T2 = ( S2 − S1 ) = ( 3 − 1) = 2, 2, T, ⇒ r = 2 = = 2., T1 1, Hence, the common ratio is 2., 82. S3 =, ∴, , a(r 3 − 1), a(r 6 − 1), and S6 =, ⋅, (r − 1), (r − 1), (r − 1), a(r 3 − 1), 125, S3 125, 125, 1, ×, =, ⇒, =, ⇒, =, (r − 1), S6 152, (r 3 + 1) 152, a(r 6 − 1) 152, ⇒ 125r 3 + 125 = 152 ⇒ 125r 3 = ( 152 − 125 ) = 27, ⇒ r3 =, , ∴, , 27 ⎛ 3 ⎞, =⎜ ⎟, 125 ⎝ 5 ⎠, , the common ratio is, , 3, ⋅, 5, , 3, , ⇒ r=, , 3, ⋅, 5
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1048, , Senior Secondary School Mathematics for Class 11, 3, 1, =, ⋅, 3, 3, , 83. In this GP, we have a = 3 and r =, ∴, , Tn =, , n −1, , 1, 1, ⎛ 1 ⎞, ⇒ ar n −1 =, ⇒ 3× ⎜, ⎟, ⎝ 3⎠, 243, 243, ⎛ 1 ⎞, ⇒ ⎜, ⎟, ⎝ 3⎠, ⇒, , n −1, , =, , ⎛ n −1 ⎞, ⎟, 2 ⎠, , 1, 1 ⎜⎝, ⇒ ⎛⎜ ⎞⎟, ⎝ 3⎠, 729, , =, , 1, 243, , 1, = ⎛⎜ ⎞⎟, ⎝ 3⎠, , 6, , n− 1, = 6 ⇒ n = 13., 2, , 84. AM ≥ GM is always true., 1, a+ b 2, 85. AM = 2 × GM ⇒ ( a + b ) = 2 ab ⇒, =, 2, 2 ab 1, ⇒, ⇒, ⇒, , a + b + 2 ab 2 + 1 3, =, =, a + b − 2 ab 2 − 1 1, ( a+, , b )2, , ( a − b), , 2, , =, , ( 3 )2, ( 1) 2, , ⇒, , ( a+, , b) + ( a − b), , ( a+, , b) −( a − b), , ⇒, , a, =, b, , a+ b, 3, =, 1, a− b, 3+1, 2 a, ⇒, =, 3−1, 2 b, , =, , 3+1, a ( 3 + 1) 2 2 +, ⇒, =, =, b ( 3 − 1) 2 2 −, 3−1, , 3+1, 3−1, , 3, ⋅, 3, , 86. GM = 27 × 243 = 81 × 81 = 81., 15, 15, 15 × 15, 15, 15, ×, =, =, =, = 0.015., 100 10000, 10 6, 10 3 1000, 1, 88. In the given GP, we have a = 1and r = ⋅, 3, 3, 1, a, = ⋅, S∞ =, =, ∴, 1⎞ 2, (1 − r ) ⎛, ⎜1− ⎟, ⎝, 3⎠, , 87. GM = 0.15 × 0.0015 =, , 89. In the given geometric series a = 1and r =, ∴, , S∞ =, , 3, 1, a, = ⋅, =, 1⎞ 4, (1 − r ) ⎛, ⎜1+ ⎟, ⎝, 3⎠, , 90. In the given geometric series a =, Clearly,|r| =, , ∴, , −1, 1, ⋅ Clearly,|r| = < 1., 3, 3, , 5 ( −4 ) −1, −5, and r =, ×, =, ⋅, 16, 5, 4, 4, , 1, < 1., 4, , ⎛ −5 ⎞, ⎜ ⎟, ⎝ 4 ⎠, a, −5, S∞ =, =, = ⎛⎜, ×, 1⎞ ⎝ 4, (1 − r ) ⎛, ⎜1+ ⎟, ⎝, 4⎠, , 4⎞, ⎟ = −1., 5⎠
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Objective Questions, 91. Here a = ( 2 + 1) and r =, ∴, , S∞ =, , 1049, , ( 2 − 1) ( 2 − 1), 1, =, = ( 2 − 1) and|r| < 1., ×, 1, ( 2 + 1) ( 2 − 1), , ( 2 + 1), ( 2 + 1) ( 2 +, a, ×, =, =, ( 1 − r ) {1 − ( 2 − 1)} ( 2 − 2 ) ( 2 +, , 2), 2), , =, , 4+ 3 2 4+ 3 2, =, ⋅, ( 4 − 2), 2, , 2, 2, 2, 3, 3, 92. Given series = ⎧⎨ +, +, + … ∞ ⎫⎬ + ⎧⎨ +, + … ∞ ⎫⎬, ⎩ 5 5 3 55, ⎭ ⎩5 2 5 4, ⎭, ⎛ 3⎞, ⎛ 2⎞, ⎛ 3⎞, ⎛ 2⎞, ⎜ ⎟, ⎜ ⎟, ⎜ ⎟, ⎜ 2⎟, ⎝5⎠, ⎝ 25 ⎠, ⎝5⎠, ⎝5 ⎠, =, +, =, +, ⎧ 1 − 2 × 5 ⎫ ⎪⎧, 3 5 2 ⎫⎪ ⎛⎜ 1 − 1 ⎞⎟ ⎛⎜ 1 − 1 ⎞⎟, ⎨, ⎬, 1−, × ⎬ ⎝, 125 2 ⎭ ⎨⎪, 25 ⎠ ⎝, 25 ⎠, ⎩, 3 ⎭⎪, 54, ⎩, ( 10 + 3 ) 13, 2 25, 3 25, 5, 1, + ⎞⎟ =, =, ⋅, = ⎛⎜ × ⎞⎟ + ⎛⎜ × ⎞⎟ = ⎛⎜, ⎝ 5 24 ⎠ ⎝ 25 24 ⎠ ⎝ 12 8 ⎠, 24, 24, −4, 4, and so|r| = < 1., 5, 5, 80, 80, 80, a, a, =, S∞ =, ⇒, =, ⇒, (1 − r ) 9, 9, 9, ⎛ 1+ 4⎞, ⎜, ⎟, ⎝, 5⎠, , 93. Here, r =, , ⇒, ∴, , 5 a 80, 80, =, ⇒ a = ⎛⎜ ×, ⎝ 9, 9, 9, , 9⎞, ⎟ = 16., 5⎠, , the first term = 16., , 94. Let the given geometric series be ( a + ar + ar 2 + … ∞ )., Then, ar = 2 and, Putting, r =, , 2, , we get, a, , a, ⎛1−, ⎜, ⎝, , a, = 8., (1 − r ), , 2⎞, ⎟, a⎠, , =8 ⇒, , a2, = 8 ⇒ a2 = 8 a − 16, ( a − 2), , ⇒ a2 − 8 a + 16 = 0 ⇒ ( a − 4 ) 2 = 0 ⇒ a − 4 = 0 ⇒ a = 4 ., ∴, , r=, , 2 1, = ⋅, 4 2, , Hence, the common ratio is, , 1, ⋅, 2, , 95. Let ( a + ar + ar 2 + … ∞ ) = 15 and ( a2 + a2r 2 + a2r 4 + … ∞ ) = 45., Then,, , a, a2, = 45., = 15 and, (1 − r ), (1− r 2 ), , On dividing, we get, , a2, (1 − r 2 ), , ×, , ( 1 − r ) 45, a, =, ⇒, =3, 15, (1 + r ), a, ⇒, , 15( 1 − r ), =3, (1 + r ), , [using, , a, = 15], (1 − r )
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1050, , Senior Secondary School Mathematics for Class 11, , ⇒, , 3 + 3r = 15 − 15r ⇒ 18r = 12 ⇒ r =, , ∴, , a, = 15 ⇒ 3 a = 15 ⇒ a = 5., ⎛1− 2⎞, ⎜, ⎟, ⎝, 3⎠, , 12 2, = ⋅, 18 3, , Hence, the first term is 5., 96. Let a1 / x = b1 / y = c1 / z = k. Then, a = k x , b = k y and c = k z ., Now, a, b, c are in GP., ⇒, , b 2 = ac ⇒ ( k y ) 2 = ( k x × k z ), , ⇒, , k 2 y = k ( x+ z ) ⇒ 2 y = x + z, , ⇒ x, y, z are in AP., 97. a, b, c are in GP, b 2 = ac ⇒ log ( b 2 ) − log ( ac ), , ⇒, ⇒, , 2 log b = log a + log c ⇒ log a, log b , log c are in AP., 1, 98. y = x( 1 + x + x 2 + … ∞ ) = x ⋅, [sum of infinite GP], ( 1 − x), ⇒, , ( 1 − x )y = x ⇒ y − xy = x ⇒ y = x + xy = x( 1 + y ), y, x=, ⋅, ( 1 + y), , ⇒, 99. x =, , 1, 1, and y =, ⋅, ( 1 − a), ( 1 − b), , ∴, , ( 1 + ab + a2 b 2 + … ∞ ) =, , 1, ⋅, ( 1 − ab ), , ⎧ 1, 1 ⎫, ⋅, ⎨, ⎬, ( 1 − a) ( 1 − b ) ⎭, xy, ⎩, Now,, =, ( x + y − 1) ⎧ 1, ⎫, 1, +, − 1⎬, ⎨, ⎩ ( 1 − a) ( 1 − b ), ⎭, ⎧, ⎫, ( 1 − a)( 1 − b ), 1, 1, =⎨, ×, ⋅, ⎬=, (, 1, a, )(, 1, b, ), (, 1, b, ), (, 1, a, ), (, 1, a, )(, 1, b, ), −, −, −, +, −, −, −, −, ⎩, ⎭ ( 1 − ab ), ∴, 100. x =, , ∴, , ( 1 + ab + a2 b 2 + … ∞ ) =, a, ⎛ 1 − 1⎞, ⎜, ⎟, ⎝, r⎠, , =, , xy, ⋅, ( x + y − 1), , ar, b, br, c, cr 2, and z =, ,y=, =, =, ⋅, 2, (r − 1), ⎛ 1 + 1 ⎞ (r + 1), ⎛ 1 − 1 ⎞ (r − 1), ⎜, ⎟, ⎜, ⎟, ⎝, ⎝, r⎠, r2 ⎠, , xy, (r 2 − 1) ab, ar, br, =, ×, ×, =, ⋅, z (r − 1) (r + 1), c, cr 2, , 101. a, b, c are in AP ⇒ 2b = ( a + c )., a, x, b are in GP ⇒ x 2 = ab., b, y, c are in GP ⇒ y 2 = bc.
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Objective Questions, ∴, , 1051, , x 2 + y 2 = ab + bc = b( a + c ) = 2 b 2, , ⇒ x 2 , b 2 , y 2 are in AP., 102. a, b, c are in AP ⇒ 2b = ( a + c )., a, b, d are in GP ⇒ b 2 = ad., ∴, , a( d − c ) = ad − ac = b 2 − ac = b 2 − a( 2 b − a) = a2 + b 2 − 2 ab = ( a − b ) 2 ., , ∴, , a, ( a − b ), ( d − c ) are in GP., , 103. Let A be the first term and R be the common ratio of the GP., Then, Tp = AR (p −1 ) , Tq = AR (q−1 ) and Tr = AR (r −1 )., ∴, , a = AR (p −1 ) , b = AR (q−1 ) and c = AR (r −1 )., , ∴, , a(q− r ) ⋅ b (r − p ) ⋅ c (p − q) = A (q− r ) ⋅ R (p −1 )(q− r ) ⋅ A (r − p ) R (r − p )(q−1 ) ⋅ A (p − q) R (r −1 )(p − q), = A (p − q) ⋅ R (p − q)(r −1 ) = A 0R 0 = 1., a, b, c, are in AP, ,, ,, abc abc abc, , 104. a, b, c are in AP ⇒, , 1 1 1, are in AP, , ,, bc ca ab, , ⇒, , ⇒ bc , ca, ab are in HP., 105. a2 , b 2 , c 2 are in AP ⇒ ( a2 + ab + bc + ca), ( b 2 + ab + bc + ca), ( c 2 + ab + bc + ca) are in AP, ⇒ ( a + b )( c + a), ( a + b )( b + c ), ( c + a)( b + c ) are in AP, ⇒, , 1, 1, 1, are in AP, ,, ,, b + c c+ a a+ b, , ⇒ ( b + c ), ( c + a), ( a + b ) are in HP., 106. a, b, c are in AP ⇒, , a, b, c, are in AP, ,, ,, abc abc abc, , ⇒, , 1 1 1, are in AP, , ,, bc ca ab, , ⇒, , ( ab + bc + ca) ( ab + bc + ca) ( ab + bc + ca), are in AP, ,, ,, bc, ca, ab, , ( ab + bc + ca), ( ab + bc + ca), ⎧ ( ab + bc + ca) − 1⎫, ⇒ ⎧⎨, − 1⎫⎬ , ⎧⎨, − 1⎫⎬ , ⎨, ⎬, ab, bc, ca, ⎭, ⎩, ⎭ ⎩, ⎭ ⎩, are in AP, ⇒, 107., , a( b + c ) b( c + a) c ( a + b ), are in AP., ,, ,, bc, ca, ab, , b + c− a c+ a− b a+ b − c, are in AP, ,, ,, a, b, c, b+ c−a, c+ a− b, a+ b − c, + 2 ⎫⎬ , ⎧⎨, + 2 ⎫⎬ , ⎧⎨, + 2 ⎫⎬ are in AP, ⇒ ⎧⎨, a, b, c, ⎩, ⎭ ⎩, ⎭ ⎩, ⎭, ⇒, , ( a + b + c) ( a + b + c) ( a + b + c), 1 1 1, are in AP ⇒ , , are in AP, ,, ,, a, b, c, a b c, , ⇒ a, b, c are in HP.
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1052, , Senior Secondary School Mathematics for Class 11, , 108. ( 1 + 2 + 3 + … + n) =, ∴, , 1, n(n + 1)., 2, , 1, the given sum = ( 1 + 2 + 3 + … + 80 ) = ⎛⎜ × 80 × 81⎞⎟ = 3240., ⎝2, ⎠, , 109. ( 12 + 2 2 + 3 2 + … + n2 ) =, ∴, , 1, n(n + 1)( 2n + 1)., 6, , 1, the given sum = ( 12 + 2 2 + 3 2 + … + 60 2 ) = ⎛⎜ × 60 × 61 × 121⎞⎟ = 73810., ⎝6, ⎠, , 1, 110. ( 13 + 2 3 + 3 3 + … + n3 ) = ⎧⎨ n(n + 1)⎫⎬, 2, ⎩, ⎭, , 2, , 2, , ∴, , 1, the given sum = ( 13 + 2 3 + 3 3 + … + 40 3 ) = ⎛⎜ × 40 × 41⎞⎟ = ( 820 ) 2 = 672400 ., ⎝2, ⎠, , q
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Trigonometry, REVIEW OF FACTS AND FORMULAE, 1. Measurement of angles, I. Sexagesimal system, 1 right angle = 90°., 1° = 60 minutes = 60′ ., 1′ = 60 seconds = 60′′ ., II. Circular measures, π radian, written as π c = 180°., 2. θ =, , s, arc, =, ⋅, r radius, , 3. In a clock, we have:, (i) Angle between two consecutive digits, π, = 30° = radian., 6, (ii) Angle traced by the hour hand in 1 hr = 30°., (iii) Angle traced by the minute hand in 1 min = 6°., 4. T-ratios, , C, , perpendicular BC, (i) sin θ =, =, ⋅, hypotenuse, AC, (ii) cos θ =, , base, AB, =, ⋅, hypotenuse AC, , (iii) tan θ =, , perpendicular BC, =, ⋅, base, AB, , (iv) cosec θ =, 5., , 1, ⋅, sin θ, , (i) sin 2θ + cos2θ = 1., , θ, A, , (v) sec θ =, , 1, ⋅, cos θ, , (ii) 1 + tan 2θ = sec2θ., , (vi) cot θ =, , B, , 1, ⋅, tan θ, , (iii) 1 + cot 2θ = cosec2θ., , 6. Extremum values of T-ratios, (i) −1 ≤ sin θ ≤ 1 ⇒ |sin θ| ≤ 1., , (ii) −1 ≤ cos θ ≤ 1 ⇒ |cos θ| ≤ 1., , (iii) Either sec θ ≤ −1 or sec θ ≥ 1., ∴ |sec θ| ≥ 1., (iv) Either cosec θ ≤ −1 or cosec θ ≥ 1., ∴ |cosec θ| ≥ 1., 1053
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Objective Questions, , 12., , (i) sin 15 ° = cos 75 ° =, , ( 3 − 1), ⋅, 2 2, , (iii) tan 15 ° = cot 75 ° = ( 2 − 3 )., 13., , (i) sin 18° = cos 72° =, , ( 5 − 1), ⋅, 4, , (iii) sin 36° = cos 54° =, 14., , 15., , 16., , 19., , (iv) cot 15 ° = tan 75 ° = ( 2 +, (ii) cos 18° = sin 72° =, , 3 )., , 10 + 2 5, ⋅, 4, , 10 − 2 5, ( 5 + 1), ⋅ (iv) cos 36° = sin 54° =, ⋅, 4, 4, , (iii) tan ( A + B) =, , tan A + tan B, ⋅, 1 − tan A tan B, , (iv) cot ( A + B) =, , cot A cot B − 1, ⋅, cot A + cot B, , (i) sin ( A − B) = sin A cos B − cos A sin B., (ii) cos ( A − B) = cos A cos B + sin A sin B., (iii) tan ( A − B) =, , tan A − tan B, ⋅, 1 + tan A tan B, , (iv) cot ( A − B) =, , cot A cot B + 1, ⋅, cot B − cot A, , sin ( A + B) ⋅ sin ( A − B) = sin 2 A − sin 2 B = cos2 B − sin 2 A., cos ( A + B) ⋅ cos ( A − B) = cos2 A − sin 2 B = cos2 B − sin 2 A., sin ( A + B) + sin ( A − B) = 2 sin A cos B., sin ( A + B) − sin ( A − B) = 2 cos A sin B., cos ( A + B) + cos ( A − B) = 2 cos A cos B., cos ( A − B) − cos ( A + B) = 2 sin A sin B., , (i) sin C + sin D = 2 sin, , (C + D), (C − D), cos, ⋅, 2, 2, , (ii) cos C + cos D = 2 cos, , (C + D), (C − D), cos, ⋅, 2, 2, , (i) sin C − sin D = 2 cos, , (C + D), (C − D), sin, ⋅, 2, 2, , (ii) cos C − cos D = 2 sin, , (C + D), ( D − C), sin, ⋅, 2, 2, , T-RATIOS OF MULTIPLE ANGLES, , 20., , ( 3 + 1), ⋅, 2 2, , (i) sin ( A + B) = sin A cos B + cos A sin B., (ii) cos ( A + B) = cos A cos B − sin A sin B., , (i), (ii), 17. (i), (ii), (iii), (iv), 18., , (ii) cos 15 ° = sin 75 ° =, , 1055, , (i) sin 2A = 2 sin A cos A =, , 2 tan A, ⋅, 1 + tan 2 A
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1056, , Senior Secondary School Mathematics for Class 11, , (ii) cos 2A = (cos2 A − sin 2 A) = (1 − 2 sin 2 A) = ( 2 cos2 A − 1) =, (iii) tan 2A =, 21., , 2 tan A, ⋅, 1 − tan 2 A, , (i) sin 3 A = 3 sin A − 4 sin 3 A., (iii) tan 3 A =, , 1 − tan 2 A, ⋅, 1 + tan 2 A, , (ii) cos 3 A = 4 cos3 A − 3 cos A., , 3 tan A − tan 3 A, ⋅, 1 − 3 tan 2 A, , TRIGONOMETRIC EQUATIONS, , 22., , 23., , (i) sin θ = 0 ⇔ θ = nπ , n ∈ I., π, (ii) cos θ = 0 ⇔ θ = ( 2n + 1) , n ∈ I., 2, (iii) tan θ = 0 ⇔ θ = nπ , n ∈ I., (i) sin θ = sin α ⇒ θ = nπ + ( −1) n α , n ∈ I., (ii) cos θ = cos α ⇒ θ = 2nπ ± α , n ∈ I., (iii) tan θ = tan α ⇒ θ = nπ + α , n ∈ I., , 24., , (i) sin 2θ = sin 2α ⇒ θ = nπ ± α , n ∈ I., (ii) cos2θ = cos2α ⇒ θ = nπ ± α , n ∈ I., (iii) tan 2θ = tan 2α ⇒ θ = nπ ± α , n ∈ I., , SOLUTION OF TRIANGLES, , 25. Sine rule, a, b, c, =, =, ⋅, sin A sin B sin C, 26. Cosine formulae, (i) a 2 = b 2 + c2 − 2bc cos A., , (ii) b 2 = c2 + a 2 − 2ca cos B., , (iii) c2 = a 2 + b 2 − 2ab cos C., 27. Projection formulae, (i) a = a cos B + b cos C., , (ii) b = a cos C + c cos A., , (iii) c = a cos B + b cos A., 28. Half-angle formulae, (i) sin, , A, (s − b)(s − c), =, ⋅, 2, bc, , (ii) sin, , (iii) sin, , C, (s − a)(s − b), =, ⋅, 2, ab, , (iv) cos, , A, s(s − a), =, ⋅, 2, bc, , B, s(s − b), =, ⋅, 2, ca, , (vi) cos, , C, s(s − c), =, ⋅, 2, ab, , (v) cos, , B, (s − c)(s − a), =, ⋅, 2, ca
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Objective Questions, , (i) tan, , A, (s − b)(s − c), =, ⋅, 2, s(s − a), , (iii) tan, , C, (s − a)(s − b), ⋅, =, 2, s(s − c), , 29., , 1057, , (ii) tan, , B, (s − c)(s − a), =, ⋅, 2, s(s − b), , 30. Area of a triangle is:, 1, 1, 1, a = ab sin C = bc sin A = ac sin B., 2, 2, 2, 31. Napier’s analogies, (i) tan, , ( B − C) ( b − c), A, =, cot ⋅, 2, ( b + c), 2, , (iii) tan, , ( A − B) ( a − b), C, =, cot ⋅, 2, ( a + b), 2, , (ii) tan, , (C − A) ( c − a), B, =, cot ⋅, 2, ( c + a), 2, , EXERCISE 1, Mark (3) against the correct answer in each of the following., , 1. 25° when measured in radians is, ⎛ 5π ⎞, (a) ⎜ ⎟, ⎝ 18 ⎠, , c, , ⎛ 5π ⎞, (b) ⎜ ⎟, ⎝ 24 ⎠, , c, , ⎛ 5π ⎞, (c) ⎜ ⎟, ⎝ 36 ⎠, , c, , ⎛ 4π ⎞, (c) ⎜ ⎟, ⎝ 3⎠, , c, , (d) none of these, , 2. 162° when measured in radians is, ⎛ 7π ⎞, (a) ⎜ ⎟, ⎝ 10 ⎠, , c, , ⎛ 9π ⎞, (b) ⎜ ⎟, ⎝ 10 ⎠, , c, , ⎛ 5π ⎞, (d) ⎜ ⎟, ⎝ 4⎠, , c, , c, , ⎛ 8π ⎞, 3. ⎜ ⎟ = ?, ⎝ 5 ⎠, (a) 272°, , (b) 302°, , (c) 288°, , (d) 316°, , (b) 372°, , (c) 418°, , (d) 630°, , (b) 57 °16′ 22′′, , (c) 55 °18′ 32′′, , (d) 57 °26′ 32′′, , 4. 11 = ?, c, , (a) 315°, 5. 1 = ?, c, , (a) 56°27′ 22′′, , 6. 3 °45′ expressed in radians is, ⎛ π⎞, (a) ⎜ ⎟, ⎝ 36⎠, , c, , ⎛ π⎞, (b) ⎜ ⎟, ⎝ 54⎠, , c, , ⎛π⎞, (c) ⎜ ⎟, ⎝ 48⎠, , c, , ⎛ 5π ⎞, (d) ⎜ ⎟, ⎝ 96 ⎠, , c, , ⎛ 9π ⎞, (c) ⎜ ⎟, ⎝ 32 ⎠, , c, , ⎛ 11π ⎞, (d) ⎜, ⎟, ⎝ 36 ⎠, , 7. 50° 37′ 30′′ = ?, ⎛ 5π ⎞, (a) ⎜ ⎟, ⎝ 16 ⎠, , c, , ⎛ 7π ⎞, (b) ⎜ ⎟, ⎝ 18 ⎠, , c, , c
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1058, , Senior Secondary School Mathematics for Class 11, c, , ⎛π⎞, 8. In a right triangle, the difference between two acute angles is ⎜ ⎟ ⋅ The, ⎝ 15 ⎠, measure of the smallest angle is, (a) 40°, , (b) 45°, , (c) 36°, , (d) 39°, , 9. The angles of a triangle are in AP and the greatest angle is double the least., The largest angle measures, (a) 60°, , (b) 80°, , (c) 75°, , (d) 90°, , 10. The angles of a triangle are in AP and the ratio of the number of degrees, in the least to the number of radians in the greatest is 60 : π . The smallest, angle is, (a) 15°, , (b) 30°, , (c) 45°, , (d) 60°, , 11. In a circle, the central angle of 45° intercepts an arc of length 33 cm. The, radius of the circle is, (a) 21 cm, , (b) 35 cm, , (c) 42 cm, , (d) 14 cm, , 12. In a circle of radius 14 cm an arc subtends an angle of 36° at the centre. The, length of the arc is, (a) 6.6 cm, , (b) 7.7 cm, , (c) 8.8 cm, , (d) 9.1 cm, , 13. The minute hand of a watch is 1.4 cm long. How far does its tip move in, 45 minutes?, (a) 6 cm, , (b) 6.3 cm, , (c) 6.6 cm, , (d) 7 cm, , 14. If the arcs of the same length in two circles subtend angles of 60° and 75° at, their respective centres, the ratio of their radii is, (a) 4 : 5, , (b) 5 : 4, , (c) 3 : 5, , (d) 5 : 3, , 15. A wire of length 121 cm is bent to form an arc of a circle of radius 180 cm., The angle subtended at the centre by the arc is, (a) 36°20′, , (b) 34°40′, , (c) 38° 30′, , (d) 39°10′, , 16. A horse is tied to a post by a rope. If the horse moves along a circular path,, always keeping the rope tight and describes 88 m when it traces 72° at the, centre, then the length of the rope is, (a) 35 m, , (b) 70 m, , (c) 17.5 m, , (d) 22 m, , 17. A pendulum swings through an angle of 42° in describing an arc of length, 55 cm. The length of the pendulum is, (a) 56 cm, , (b) 60 cm, , (c) 75 cm, , (d) 88 cm, , 18. The radius of a circle is 30 cm. The length of the arc of this circle whose, chord is 30 cm long, is, (a) 9π cm, , (b) 10π cm, , (c) 12π cm, , (d) 13.6π cm, , 19. A wheel makes 180 revolutions in 1 minute. How many radians does it, turn in 1 second?, (a) ( 3π ) c, , (b) ( 4π ) c, , (c) ( 6π ) c, , (d) (12π ) c
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Objective Questions, , 1059, , 20. A railway train is moving on a circular curve of radius 1500 m at a speed of, 90 km/hr. Through what angle has it turned in 11 seconds?, (a) 10° 30′, , (b) 11°40′, , (d) 16° 30′, , (c) 12°, , 21. When a clock shows the time 7 : 20, what is the angle between its minute, hand and the hour hand?, (a) 60°, , (b) 80°, , (c) 100°, , (d) 120°, , 22. The angle between the hour hand and the minute hand of a clock at half, past three is, (a) 54°, , (b) 63°, , (c) 72°, , (d) 75°, , 23. The angle between the minute hand and the hour hand of a clock when the, time is 8 : 25 am, is, (a) 107 °15′, , (b) 105°, , (c) 102° 30′, , (d) 92°45′, , 24. The length of a pendulum is 60 cm. The angle through which it swings, when its tip describes an arc of length 16.5 cm is, (a) 15 ° 30′, , (b) 15 °45′, , (c) 16°15′, , (d) 16° 30′, , 25. The angles of a quadrilateral in degrees are in AP and the greatest angle is, 120°. The smallest angle is, ⎛ π⎞, (a) ⎜ ⎟, ⎝ 4⎠, , c, , ⎛ π⎞, (b) ⎜ ⎟, ⎝ 3⎠, , c, , ⎛ π⎞, (c) ⎜ ⎟, ⎝5⎠, , c, , ⎛ π⎞, (d) ⎜ ⎟, ⎝ 6⎠, , c, , 26. The perimeter of a sector of a circle is equal to half the circumference of the, circle. The angle of the sector is, ⎛ π⎞, (a) ⎜ ⎟, ⎝ 4⎠, 27. sin, (a), 28. cos, (a), , c, , ⎛ π⎞, (b) ⎜ ⎟, ⎝ 2⎠, , c, , (c) ( π − 2) c, , (d) ( π + 2) c, , 25 π, =?, 3, 1, 2, , (b), , 1, 2, , (c), , 3, 2, , (d), , − 3, 2, , (b), , −1, 2, , (c), , 3, 2, , (d), , − 3, 2, , (b) − 3, , (c), , 1, 3, , (d), , −1, 3, , (b) 1, , (c) 3, , (d), , 1, 3, , 41π, =?, 4, 1, 2, , ⎛ −16π ⎞, 29. tan ⎜, ⎟ =?, ⎝ 3 ⎠, (a) 3, ⎛ 29π ⎞, 30. cot ⎜, ⎟ =?, ⎝ 4 ⎠, (a) –1
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1060, , Senior Secondary School Mathematics for Class 11, , ⎛ −19π ⎞, 31. sec ⎜, ⎟ =?, ⎝ 3 ⎠, (a) –2, , (b), , −1, 2, , (c) 2, , (d), , 1, 2, , (d), , −1, 2, , ⎛ −33 π ⎞, 32. cosec ⎜, ⎟ =?, ⎝ 4 ⎠, (a) − 2, 33. cos 15π = ?, (a) 1, , 1, 2, , (b) 2, , (c), , (b) –1, , (c) 0, , (d) none of these, , (b) –1, , (c) 2, , (d) − 2, , (c) 1, , (d) –1, , 34. sec 6π = ?, (a) 1, 5π, 35. tan, =?, 4, (a) 3, , (b), , 1, 3, , (b), , − 3, 2, , (c), , 1, 2, , (d), , (b) − 3, , (c), , −1, 3, , (d) none of these, , 36. sin (765°) = ?, (a), , 3, 2, , −1, 2, , 37. cot ( −600°) = ?, (a) –1, , −2 6, and x lies in quadrant III, then cot x = ?, 5, 1, −1, 3, −3, (a), (b), (c), (d), 2 6, 2 6, 2 6, 2 6, , 38. If sin x =, , 39. If cos x =, (a), , 3, 4, , π, − 15, and < x < π , then sin x = ?, 2, 4, −3, 1, (b), (c), 4, 4, , 40. If sec x = −2 and π < x <, (a), , 3, 2, , 41. If cosec x =, (a), , 1, 3, , (b), , (d), , −1, 4, , (d), , −1, 2, , 3π, , then sin x = ?, 2, , − 3, 2, , (c), , 1, 2, , −2, and x lies in quadrant IV, then tan x = ?, 3, −1, (d) − 3, (b), (c) 3, 3
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1062, , Senior Secondary School Mathematics for Class 11, , 51. cos 135° = ?, 1, (a), 2, , (b), , −1, 2, , (c), , 1, 2, , (d), , −1, 2, , 52. sec 120° = ?, (a) 2, , (b) − 2, , (c) 2, , (d) –2, , (b) 2, , (c) 2, , (d) − 2, , 53. cosec 150° = ?, (a) –2, 54. sin 315° = ?, 1, (a), 2, , (b), , −1, 2, , (c) 2, , (d) − 2, , (b), , 1, 2, , (c) − 2, , (d), , (b), , 1, 3, , (c) 3, , (d) − 3, , (c) − 3, , (d) 3, , 55. cos 405° = ?, (a) 2, 11π, =?, 6, −1, (a), 3, , −1, 2, , 56. tan, , 57. cot 675° = ?, (a) –1, , (b) 1, , ⎛ 31π ⎞, 58. sin ⎜, ⎟ =?, ⎝ 3 ⎠, (a), , 1, 2, , (b), , −1, 2, , (c), , 3, 2, , (d), , − 3, 2, , (b), , −1, 3, , (c) 3, , (d), , 1, 3, , (c) 2, , (d) –2, , 59. cot ( −600°) = ?, (a) − 3, , 60. cosec ( −1110°) = ?, 2, −2, (a), (b), 3, 3, ⎛ −33 π ⎞, 61. sec ⎜, ⎟ =?, ⎝ 4 ⎠, (a) − 2, , (b) 2, , (c), , − 3, 2, , (d), , 3, 2, , (b) 3, , (c), , −1, 3, , (d), , 1, 3, , ⎛ −25 π ⎞, 62. tan ⎜, ⎟ =?, ⎝ 3 ⎠, (a) − 3
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Objective Questions, , 63. The values of cot, , 1063, , π, π, π, , cot , cot are in, 3, 4, 6, , (a) AP, , (b) GP, , (c) HP, , (d) none of these, , 64. Which is smaller, sin 64° or cos 64°?, (a) sin 64°, , (b) cos 64°, , (c) both are equal, , (d) cannot be compared, , 65. Which is larger, sin 24° or cos 24°?, (a) sin 24°, , (b) cos 24°, , (c) both are equal, , (d) cannot be compared, , 66. The extremum values of sin θ are, (a) 0 and 1, , (b) –1 and 0, , (c) –1 and 1, , (d), , 1, − 3, and, 2, 2, , (c) –1 and 1, , (d), , − 3, and 2, 2, , 67. The extremum values of cos θ are, (a) 0 and 1, , (b) –1 and 0, , 68. The value of sec θ can, (a) never be greater than 1, , (b) never be less than 1, , (c) never be equal to 1, , (d) never lie between –1 and 1, , 69. tan 150° = ?, 1, (a), 3, 70. sec 150° = ?, 2, (a), 3, 71. cot 120° = ?, −1, (a), 3, , (b), , −1, 3, , (b) 2, , (b), , (c) 3, , (c), , −2, 3, , 1, 3, , (c) − 3, , 1, 2, , (c), , (d) − 3, , (d) –2, , (d) 3, , 72. (sin 105 ° + cos 105 °) = ?, (a) 2, , (b), , 1, 3, , (d), , 2, 3, , 73. sin 15° = ?, (a), , 3, 2 2, , (b), , ( 3 + 1), 2 2, , (c), , ( 3 − 1), 2 2, , (d), , ( 2 − 1), 2, , (b), , ( 3 − 1), 2 2, , (c), , ( 3 + 1), 2, , (d), , ( 3 − 1), 2, , 74. cos 15° = ?, (a), , ( 3 + 1), 2 2
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1064, , Senior Secondary School Mathematics for Class 11, , 75. tan 15° = ?, (a), , ( 3 + 1), ( 3 − 1), , (b), , ( 3 − 1), ( 3 + 1), , (c), , ( 2 + 1), ( 2 − 1), , (d), , ( 2 − 1), ( 2 + 1), , (b), , ( 3 − 1), 2 2, , (c), , ( 2 + 1), 2 2, , (d), , ( 2 − 1), 2 2, , (b), , ( 3 − 1), 2 2, , (c), , ( 2 − 1), 2 2, , (d), , ( 2 + 1), 2 2, , 76. sin 75° = ?, (a), , ( 3 + 1), 2 2, , 77. cos 75° = ?, ( 3 + 1), 2 2, 13 π, 78. tan, =?, 12, (a), , (a) ( 2 +, , 3), , (b) (1 + 2), , (c) ( 2 − 3 ), , (d) ( 2 − 1), , 79. (sin 70° cos 10° − cos 70° sin 10°) = ?, (a), , 1, 2, , (b), , 1, 2, , − 3, 2, , (c), , 3, 2, , (d), , (c), , 3, 2, , (d) 1, , (c), , 1, 2, , (d) 1, , 80. (sin 36° cos 9° + cos 36° sin 9°) = ?, (a), , 1, 2, , (b), , 1, 2, , 81. cos 80° cos 20° + sin 80° sin 20° = ?, (a), , 1, 2, , (b), , 3, 2, , 82. cos 50° cos 10° − sin 50° sin 10° = ?, (a), , 1, 2, , (b), , 3, 2, , (c), , 1, 2, , (d) 1, , 83. sin ( 40° + θ) cos (10° + θ) − cos ( 40° + θ) sin (10° + θ) = ?, (a), 84. sin, , 3, 2, , (a), , 1, 2, , (c) 2, , (d) none of these, , 3, 2, , (d) none of these, , 7π, π, 7π, π, cos − cos, sin = ?, 12, 4, 12, 4, , (a) 2, 85. sin, , (b), , (b), , 1, 2, , (c), , π, π, π, π, cos + cos sin, =?, 4, 12, 4, 12, 3, 2, , (b) 2, , (c), , 1, 2, , (d) none of these
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Objective Questions, , 2π, π, 2π, π, cos − sin, sin = ?, 3, 4, 3, 4, ( 3 + 1), − ( 3 + 1), (b), (a), 2, 2 2, , 1065, , 86. cos, , π, =?, 12, ( 3 − 1), (a), 2 2, , (c), , ( 3 + 1), 2 2, , (d), , − ( 3 + 1), 2 2, , 87. sin, , 5π, =?, 12, ( 3 − 1), (a), 2 2, , (b), , − ( 3 − 1), 2 2, , (c), , ( 3 + 1), 2 2, , (d) none of these, , (b), , ( 3 + 1), 2 2, , (c), , − ( 3 − 1), 2 2, , (d), , 88. sin, , − ( 3 + 1), 2 2, , 15, 12, and cos φ =, , where θ and φ both lie in quadrant I, then, 17, 13, sin (θ + φ) = ?, 171, 180, 220, 181, (b), (c), (d), (a), 221, 221, 221, 221, , 89. If sin θ =, , 3, −12, and cos φ =, , where θ and φ both lie in quadrant II, then, 5, 13, sin (θ − φ) = ?, 16, −16, 33, −33, (b), (c), (d), (a), 65, 65, 65, 65, , 90. If sin θ =, , 4, 12, and cos φ =, , where θ and φ both lie in quadrant IV, then, 5, 13, cos (θ + φ) = ?, 33, −33, 16, −16, (b), (c), (d), (a), 65, 65, 65, 65, , 91. If cos θ =, , 1, −5, and sec φ =, , where θ lies in quadrant III and φ lies in, 2, 3, quadrant II, then tan (θ + φ) = ?, 5, 2, −6, 10, (b), (c), (d), (a), 11, 11, 11, 11, , 92. If cot θ =, , 93. cos 15 ° − sin 15 ° = ?, 1, 1, (b), (a), 2, 2, , (c), , ( 2 − 1), 2, , (d), , ( 2 + 1), 2, , (d), , ( 3 + 1), ( 3 − 1), , 94. cot 105 ° − tan 105 ° = ?, (a) 3, , (b) 2 3, , (c), , 3, 2
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Objective Questions, , 1067, , ⎛ 3π, ⎞, ⎛ 3π, ⎞, 105. cos ⎜, + x⎟ − cos ⎜, − x⎟ = ?, ⎝ 4, ⎠, ⎝ 4, ⎠, , 106., , 107., , (d), , sin 3 x − sin x, =?, cos x − cos 3 x, (a) tan 2x, (b) cot 2x, , (c) − tan 2x, , (d) − cot 2x, , cos 6x + cos 4x, =?, sin 6x − sin 4x, (a) cot x, (b) tan x, , (c) − cot x, , (d) − tan x, , (c) cot 2x, , (d) cot 3x, , (c) tan 2x, , (d) cot 2x, , (b) 2sin x, , cos 4x + cos 3 x + cos 2x, 108., =?, sin 4x + sin 3 x + sin 2x, (a) tan 2x, (b) tan 3x, 109., , 1, sin x, 2, , (c) − 2 sin x, , (a) 2 sin x, , sin 7 x − sin 5 x, =?, cos 7 x + cos 5 x, (a) tan x, (b) cot x, , 110. (sin 6x − sin 4x) = ?, (a) sin 10x, (b) sin 2x, 2, , 2, , (c) sin 10x sin 2x (d) none of these, , 111. cos 20° cos 40° cos 80° = ?, (a), , 1, 16, , (b), , 1, 8, , (c), , 3, 8, , (d), , 3, 16, , 112. sin 10° sin 50° sin 70° = ?, (a), , 3, 4, , (b), , (c), , 1, 8, , (d), , 1, 16, , (c), , ( 3 + 1), 2, , (d), , 3, 2, , 3, 2, , (c), , ( 3 + 1), 2, , (d) none of these, , 1, 2, , (c), , 1, 2 2, , (d), , 3, 2, , (d), , 1, 2 3, , 3, 8, , 113. 2 cos 45 ° cos 15 ° = ?, (a), , 3, 2, , (b), , ( 3 − 1), 2, , 114. 2 sin 75 ° sin 15 ° = ?, (a), , 1, 2, , (b), , 115. cos 15 ° − sin 15 ° = ?, (a), , 1, 2, , 116. If sin x =, (a), , 1, 2, , (b), , −1, and x lies in quadrant III, then sin 2x = ?, 2, 3, (c), (b) 3, 2
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1068, , Senior Secondary School Mathematics for Class 11, , −13, π, and < x < π , then cos 2x = ?, 12, 2, −120, 119, −120, (a), (b), (c), 169, 169, 119, , 117. If sec x =, , −3, 3π, and, < x < 2π , then tan 2x = ?, 4, 2, −24, 7, 24, (a), (b), (c), 25, 25, 7, , (d) none of these, , 118. If tan x =, , 119. If sin x =, (a) 1, 120. If cos x =, (a), , 3, 2, , 1, , then sin 3x = ?, 3, 1, (b), 9, 1, , then cos 3x = ?, 2, 1, (b), 6, , (c), , 7, 9, , (c) –1, , (d), , −24, 7, , (d), , 23, 27, , (d), , 2, 3, , (d), , −1, 2, , π, π⎞, ⎛, 121. ⎜ cos2 − sin 2 ⎟ = ?, ⎝, 12, 12⎠, (a), 122., , 1, 2, , (b), , 3, 2, , 1 + sin 2x − cos 2x, =?, 1 + sin 2x + cos 2x, (a) tan 2x, (b) tan x, , 123. If cos x =, (a), , −2, 5, , 124. If cos x =, (a), , 1, 5, , (c), , 3, 2, , (c) cot 2x, , −3, x, π, and < x < π , then sin = ?, 5, 2, 2, 2, −1, (b), (c), 5, 5, −3, x, π, and < x < π , then cos = ?, 5, 2, 2, −1, 2, (b), (c), 5, 5, , −4, 3π, x, and π < x <, , then cos = ?, 5, 2, 2, 1, −1, 3, (a), (b), (c), 10, 10, 10, , (d) cot x, , (d), , 1, 5, , (d), , −2, 5, , (d), , −3, 10, , (d), , −1, 3, , 125. If cos x =, , 126. If tan x =, (a) 3, , 3, 3π, x, and π < x <, , then tan = ?, 4, 2, 2, 1, (b), (c) –3, 3
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Objective Questions, , 127. If cos x =, , −1, x, and x lies in quadrant III, then tan = ?, 3, 2, (b) − 2, , (a) 2, 128. If sin x =, , 2+ 3, 2, 1 + cos x, 129., =?, 1 − cos x, x, (a) tan 2, 2, , 131., , (d) − 3, , (c) 3, , −1, x, and x lies in quadrant IV, then sin = ?, 2, 2, , (a), , 130., , 1069, , 1 + sin x, =?, 1 − sin x, x, (a) tan, 2, sin x, =?, 1 + cos x, x, (a) tan, 2, x, x, − tan = ?, 2, 2, (a) 2tan x, , (b), , 3+ 2, 2, , (b) cot 2, , x, 2, , (c), , 2− 3, 2, , (c) sec2, , x, 2, , (d) none of these, , (d) cosec2, , x, 2, , (b) cot, , x, 2, , ⎛ π x⎞, (c) tan ⎜ + ⎟, ⎝ 4 2⎠, , ⎛ π x⎞, (d) cot ⎜ + ⎟, ⎝ 4 2⎠, , (b) cot, , x, 2, , ⎛ π x⎞, (c) tan ⎜ + ⎟, ⎝ 4 2⎠, , (d) none of these, , (c) 2sin x, , (d) 2cos x, , 132. cot, , (b) 2cot x, , 133. sin 18° = ?, (a), , ( 5 − 1), 4, , (b), , ( 5 + 1), 4, , (c), , ( 3 + 1), 2, , (d), , ( 3 − 1), 2, , (c), , ( 5 + 1), 4, , (d) none of these, , (c), , ( 3 + 1), 2, , (d), , 134. cos 18° = ?, (a), , 10 − 2 5, 4, , (b), , 10 + 2 5, 4, , 135. cos 36° = ?, (a), , ( 5 − 1), 4, , (b), , ( 5 + 1), 4, , ( 3 − 1), 2, , 136. sin 36° = ?, (a), , ( 5 − 1), 4, , (b), , 10 + 2 5, 4, , (c), , 10 − 2 5, 4, , (d) none of these, , (c), , 10 + 2 5, 4, , (d), , 137. sin 54° = ?, (a), , ( 5 + 1), 4, , (b), , ( 5 − 1), 4, , 10 − 2 5, 4
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1070, , Senior Secondary School Mathematics for Class 11, , 138. cos 72° = ?, (a), , ( 5 + 1), 4, , (b), , ( 5 − 1), 4, , (c), , (2 + 5), 3, , (d), , (2 − 5), 3, , 139. cos 54° = ?, 10 + 2 5, 4, , (a), , (b), , 10 − 2 5, 4, , (c), , ( 5 + 1), 4, , (d), , ( 5 − 1), 4, , (b), , 10 − 2 5, 4, , (c), , ( 5 − 1), 4, , (d), , ( 5 + 1), 4, , 140. sin 72° = ?, 10 + 2 5, 4, , (a), , 1, 1, 141. 2 sin 22 ° cos 22 ° = ?, 2, 2, (a) 1, , (b), , 1, 2, , (c), , 1, 2, , (d) 2, , 142. ( 2 cos2 15 ° − 1) = ?, (a), , 3, 2, , (b), , 3, 2, , 3, 2, , (c) 2 3, , (d), , (c) 3 3, , (d) none of these, , (c) 1, , (d), , 143. ( 3 sin 40° − 4 sin 3 40°) = ?, 3, 2, , (a), , (b), , 3, 2, , 144. ( 8 cos3 20° − 6 cos 20°) = ?, (a), , 5, 2, , 145. If tan θ =, (a) a, , (b), , 5, 3, , a, , then a sin 2θ + b cos 2θ = ?, b, (b) b, (c) a + b, , 3, 2, , (d) a − b, , 146. cot x − 2 cot 2x = ?, (a) tan x, , (b) cos x, , (c) sin x, , (d) cos 2x, , (c) 1, , (d), , 147. cos 2x + 2 sin 2 x = ?, (a) 2, 148., , 149., , (b), , 1, 2, , 3, 2, , sin 2x, =?, 1 − cos 2x, (a) tan x, , (b) cot x, , (c) sec x, , (d) cosec x, , sin 2x, =?, 1 + cos 2x, (a) tan x, , (b) cosec x, , (c) sec x, , (d) cot x
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Objective Questions, , 1073, , 172. ( 3 sin 20° − 4 sin 3 20°) = ?, (a) –1, , (b), , 1, 2, , (c) 3, , 173. ( 4 cos3 15 ° − 3 cos 15 °) = ?, (a) 1, , (b) –1, , (c), , 174. ( 2 cos 75 ° cos 15 °) = ?, 1, (b) 2, (a), 2, , 1, 2, , (c) 2, , 3, 2, , (d), , (d) 0, , (d) 4, , 175. (sin 105 ° sin 75 °) = ?, (a), , (2 − 3), 4, , (b), , (2 + 3), 4, , 176. (sin 105 ° + cos 105 °) = ?, 1, (a), (b) 2, 2, , (c), , ( 5 + 1), 4, , (d), , ( 5 − 1), 4, , (c), , 1, 2, , (d), , 3, 2, , 177. If 0° < θ < 90° and (sin θ + cos θ) = 2 , then θ = ?, π, π, π, (a), (b), (c), 3, 4, 6, 178. If 0° < θ < 90° and (sin θ + cosec θ) =, (a), , π, 3, , (b), , π, 4, , 179. If 0° < θ < 90° and sin 2θ =, (a), , 2, 5, , (b), , 5, , then θ = ?, 2, π, (c), 6, , (d) π, , (d) π, , 1, , then (sin θ + cos θ) = ?, 5, , 3, 5, , (c), , 6, 5, , (d), , 7, 5, , 180. If θ is acute and sin θ = cos 2θ , then (sin θ + cos θ) = ?, (a), , ( 3 − 1), 2, , (b), , ( 3 + 1), 2, , (c), , ( 3 + 2), 2, , (d), , ( 3 − 2), 2, , 181. Ifθ is acute and(cos θ − sin θ) > 0, then(cos θ + sin θ) cannot be greater than, 1, 1, 1, (a), (b), (c), (d) 2, 2, 3, 2, 182., , 2 + 2 + 2 cos 4θ = ?, (a) 2sin θ, , (b) 2cos θ, , (c) sin 2θ, , (d) cos 2θ, , 183. If sin 2θ = cos 3θ and θ is acute, then sin θ = ?, (a), , ( 5 − 1), 4, , (b), , ( 5 + 1), 4, , (c), , ( 3 + 2), 4, , (d), , ( 3 − 1), 4
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1074, , Senior Secondary School Mathematics for Class 11, , 184. The least value of ( 2 sin 2θ + 3 cos2θ) is, (a) 1, (b) 2, (c) 3, 185. The greatest value of (sin 4θ + cos4θ) is, 1, (b) 1, (c) 2, (a), 2, 186. If cos2 θ + sec2 θ = p , then, (a) p < 1, (b) p = 1, (c) 1 < p < 2, 187. If sin 2θ + cos4θ = p , then, 3, 3, 13, (c) ≤ p ≤, (a) 1 ≤ p ≤ 2, (b) ≤ p ≤ 1, 4, 4, 16, 188. If 2 sin 2θ − cos θ = p , then, (a) 1 ≤ p ≤ 2, , (b) −1 ≤ p ≤ 3, , ⎧, ⎛ π A⎞, ⎛ π A⎞ ⎫, 189. ⎨sin 2 ⎜ + ⎟ − sin 2 ⎜ − ⎟ ⎬ = ?, ⎝ 8 2⎠, ⎝ 8 2⎠⎭, ⎩, 1, 1, (a), (b) sin A, sin A, 2, 2, , (c) −2 ≤ p ≤ 3, , (c) 2 sin, , π, 8, , (d) 5, (d) 3, (d) p ≥ 2, (d), , 13, ≤ p ≤1, 16, , (d) −2 ≤ p ≤ 4, , (d) sin A, , 190. (cos 1° cos 2° cos 3 °…… cos 178° cos 179°) = ?, (a) –1, , (b) 1, , (c) 0, , (d) 2, , 191. (tan 1° tan 2° tan 3 °…… tan 89°) = ?, 1, 2, 192. (tan 15 ° tan 25 ° tan 45 ° tan 65 ° tan 75 °) = ?, (a) 0, , (b) 1, , (c), , (a) 1, , (b) 2, (c) 3, π, 3π, 5π, 7π, 9π ⎞, ⎛, 193. ⎜ tan, tan, tan, tan, tan, ⎟ =?, ⎝, 20, 20, 20, 20, 20 ⎠, 1, (a) –1, (b) 1, (c), 2, 194. The general solution of the equation sin θ = 0 is, (a) θ = 2nπ , n ∈ I, (b) θ = nπ , n ∈ I, π, nπ, (d) θ =, (c) θ = ( 2n + 1) , n ∈ I, , n ∈I, 2, 2, 195. The general solution of the equation cos θ = 0 is, (a) θ = nπ , n ∈ I, (b) θ = 2nπ , n ∈ I, π, (c) θ = ( 2n + 1) , n ∈ I, (d) none of these, 2, 196. The general solution of the equation tan θ = 0 is, (a) θ = nπ , n ∈ I, π, (c) θ = ( 2n + 1) , n ∈ I, 2, , (b) θ = 2nπ , n ∈ I, (d) none of these, , (d) none of these, , (d) 3, , (d) none of these
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Objective Questions, , 1077, , 214. The general solution of the equation 4 sin 2θ = 1 is, π, π, (a) θ = nπ ± , n ∈ I, (b) θ = 2nπ ± , n ∈ I, 6, 6, nπ π, (d) none of these, (c) θ =, +, , n ∈I, 4 24, 215. The general solution of the equation 2 cos2θ = 1 is, π, nπ π, (a) θ = 2nπ + , n ∈ I, (b) θ =, ± , n ∈I, 4, 2, 8, π, (d) none of these, (c) θ = nπ ± , n ∈ I, 4, 216. The general solution of the equation cot 2θ = 3 is, π, π, (a) θ = nπ + , n ∈ I, (b) θ = nπ ± , n ∈ I, 6, 6, π, (d) none of these, (c) θ = 2nπ + , n ∈ I, 6, 217. In a a ABC, if a = 2, b = 3 , c = 4, then cos A = ?, 2, 5, 6, (a), (b), (c), 9, 7, 7, , (d), , 7, 8, , 218. In a a ABC, if a = 2, b = 6 and c = ( 3 + 1), then ∠A = ?, (a) 18°, , (b) 30°, , (c) 45°, , (d) 60°, , 219. In a a ABC, if a = 2 , b = 6 , c = 8, then ∠A = ?, (a) 60°, , (b) 45°, , (c) 30°, , (d) 90°, , 220. The angles of a a ABC are in the ratio 1 : 2 : 3. The ratio of their, corresponding sides is, (a) 1 : 2 : 3, , (b) 3 : 2 : 1, , (c) 1 : 2 :, , 3, , (d) 1 :, , 3 :2, , 221. In a a ABC, if a = ( 3 + 1), b = 2 and ∠C = 60° , then c = ?, (a) 2, , (b) 3, , (c) 5, , (d) 6, , ⎛ 2⎞, 222. In a a ABC, if a = 2, b = 3 and sin A = ⎜ ⎟ , then ∠B = ?, ⎝ 3⎠, (a) 30°, , (b) 60°, , (c) 90°, , 223. The angles of a a ABC are in AP and b : c =, (a) 30°, , (b) 45°, , 224. In a a ABC, if ∠A = 30° and b : c = 2 :, (a) 30°, , (b) 45°, , (d) 120°, 3 : 2 , then ∠C = ?, , (c) 60°, , (d) 90°, , 3 , then ∠B = ?, (c) 60°, , (d) 90°, , 225. In a right a ABC, the sides are in AP, then their ratio is, (a) 2 : 3 : 4, , (b) 3 : 4 : 5, , (c) 4 : 5 : 6, , (d) 2 : 3 : 5
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1078, , Senior Secondary School Mathematics for Class 11, , 226. In a a ABC, if ∠A = 45 ° , ∠B = 60° , ∠C = 75 ° , then a : b : c = ?, (a) 2 :, , 3: 5, , (b) 3 : 4 : 5, , (c) 2 : 6 : ( 3 + 1) (d) none of these, , 227. In a a ABC, if ∠A = 30° , ∠C = 105 ° and b = 3 2, then a = ?, (a) 2, , (b) 3, , (c) 2, , (d) 3 2, , 228. In a a ABC, if a = 5 , c = 2 2 and ∠B = 45 °, then b = ?, (a) 6, , (b) 3, , (c) 13, , (d) 2 3, , 229. In a a ABC, if a = ( 3 + 1), ∠B = 30° , ∠C = 45 °, then c = ?, (a) 5, , (b) 4, , (c) 3, , (d) 2, , 230. In a a ABC , b = 5 , c = 5 3 and ∠A = 30°, then a ABC is, (a) isosceles, , (b) equilateral, , (c) right angled, , (d) scalene, , 231. In a a ABC, it is given that ∠A : ∠B : ∠C = 3 : 4 : 5 , then a : c = ?, (a) 3 : 5, , (b) 3 : 5, , (c) 2 : ( 3 + 1), , (d) 3 : ( 2 + 1), , 232. In a a ABC, if b = 20, c = 15 and ∠A = 150°, then ar(a ABC) = ?, (a) 60 sq units, , (b) 75 sq units, , (c) 90 sq units, , (d) 120 sq units, , 233. In a a ABC, if a = 18, b = 24 and c = 30, then ar(a ABC) = ?, (a) 180 sq units, , (b) 210 sq units, , (c) 216 sq units, , (d) 225 sq units, , 234. In a a ABC, if a = 16, c = 9 and ∠B = 30°, then ar(a ABC) = ?, (a) 36 sq units, , (b) 48 sq units, , (c) 18 3 sq units (d) 72 sq units, , 235. In a a ABC, if a = 15 , b = 14, c = 13 , then its circumradius is, 21, 17, 55, 65, (a), (b), (c), (d), 2, 2, 4, 8, 236. In a a ABC, if a = 18, b = 24, c = 30, then the area of its circumcircle is, (a) 121π sq units, , (b) 144π sq units (c) 225π sq units (d) 256π sq units, , 237. In a a ABC, if a = 4, b = 13 , c = 15, then the radius of the incircle is, 3, 5, (a) 3, (b) 2, (c), (d), 2, 2, 238. In a a ABC, if a = 13 , b = 14 and c = 15, then the area of the incircle is, (a) (16π ) sq units (b) ( 25π ) sq units (c) ( 36π ) sq units (d) none of these, 239. In an equilateral a ABC, we have:, (a) R = 2r, , (b) R = 3r, , (c) R = 4r, , (d) 2R = 3r, , 240. In an equilateral triangle of side 2 3 , the circumradius is, (a) 1, , (b) 2, , (c) 3, , (d) 3, , 241. The sides of a a ABC are in the ratio 3 : 7 : 8. Then, ( R : r) = ?, (a) 7 : 5, , (b) 8 : 5, , (c) 7 : 2, , (d) 7 : 3
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Objective Questions, , 1079, , 242. The perimeter of a a ABC is 27 cm and its area is 81 cm 2 . Its inradius is, (a) 4.5 cm, , (b) 6 cm, , (c) 7.5 cm, , (d) 7 cm, , 243. In a right a ABC, we have: sin A + sin B + sin C = ?, 2, , (a) 1, , (b) 2, , 244. In a a ABC, we have:, (a), , ( b 2 − c2 ), a2, , 2, , 2, , (c) 3, , (d) 0, , sin ( B − C), =?, sin ( B + C), , (b), , ( b 2 + c2 ), a2, , (c), , b 2 − c2, b 2 + c2, , (d) none of these, , 245. In a a ABC, we have: a( b cos C − c cos B) = ?, (b) a( b 2 − c2 ), , (a) a 2, , 246. In a a ABC, we have:, (a), , a 2 + b2, a 2 + c2, , (d) 0, , 1 + cos ( A − B) cos C, =?, 1 + cos ( A − C) cos B, , (b), , b 2 + c2, c2 + a 2, , 247. In a a ABC , ( a − b) 2 cos2, (a) a 2, , (c) ( b 2 − c2 ), , (c), , c2 − a 2, a 2 + b2, , (d) none of these, , C, C, + ( a + b) 2 sin 2 = ?, 2, 2, , (b) b 2, , (c) c2, , (d) 2( a 2 + b 2 ), , 248. In a a ABC , ( b + c) cos A + ( c + a) cos B + ( a + b) cos C = ?, (a) a + b + c, , (b) abc, , (c) ab + bc + ca, , (d) none of these, , 249. In a a ABC , 2( bc cos A + ca cos B + ab cos C) = ?, (a) ( a 2 + b 2 − c2 ), , (b) ( a 2 + c2 − b 2 ), , (c) ( b 2 + c2 − a 2 ), , (d) ( a 2 + b 2 + c2 ), , (c) ( c2 − a 2 ), , (d) none of these, , ( A − B), 2, (c), ( A + B), tan, 2, , (d) none of these, , (c) 3s, , (d) 4s, , 250. In a a ABC , ( ac cos B − bc cos A) = ?, (a) ( a 2 − b 2 ), 251. In a a ABC ,, , (b) ( b 2 − c2 ), a−b, =?, a+b, , sin ( A − B), (a), sin ( A + B), , 252. In a a ABC , c cos2, (a) s, , cos ( A − B), (b), cos ( A + B), A, C, + a cos2 = ?, 2, 2, (b) 2s, , tan, , ( b − c), A ( c − a), B ( a − b), C, cos2 +, cos2 +, ⋅ cos2 = ?, 2, 2, 2, a, b, c, (a) a + b + c, (b) abc, (c) 0, (d) none of these, , 253. In a a ABC ,
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1080, , Senior Secondary School Mathematics for Class 11, , A, B, C⎞, ⎛, 254. In a a ABC , a ⎜ cot, + cot + cot ⎟ = ?, ⎝, 2, 2, 2⎠, (b) s2, , (a) s, 255. In a a ABC, bc cos2, , A, B, C, + ca cos2 + ab cos2 = ?, 2, 2, 2, (b) (s − a) 2, , (a) s2, , (d) 2s2, , (c) 2s, , (c) (s − b) 2, , (d) (s − c) 2, , A, B, C, cos cos = ?, 2, 2, 2, (b) ( b + c − a) a, (c) ( c + a − b) a, , 256. In a a ABC , 2abc cos, (a) ( a + b − c) a, 257. In a a ABC, if cos, , A, =, 2, , (a) c2 + a 2 = b 2, 258. In a a ABC, if cot, (a) 6 : 5 : 4, , b+ c, , then, 2c, , (b) a 2 + b 2 = c2, , (c) b 2 + c2 = a 2, , (d) none of these, , A, B, C, : cot : cot = 3 : 5 : 7 , then a : b : c = ?, 2, 2, 2, (b) 4 : 5 : 6, (c) 5 : 6 : 4, (d) 5 : 4 : 6, , 259. If a = 16, b = 24 and c = 20, then cos, (a), , (d) ( a + b + c) a, , 1, 2, , (b), , B, =?, 2, , 1, 3, , (c), , 1, 4, , (d), , 3, 4, , 260. In a a ABC, if a cos A = b cos B , then a ABC is, (a) either equilateral or right angled, (b) either isosceles or right angled, (c) isosceles as well as right angled, (d) none of these, ANSWERS (EXERCISE 1), , 1. (c), , 2. (b), , 3. (c), , 4. (d), , 5. (b), , 6. (c), , 7. (c), , 8. (d), , 9. (b) 10. (b), , 11. (c) 12. (c) 13. (c) 14. (b) 15. (c) 16. (b) 17. (c) 18. (b) 19. (c) 20. (a), 21. (c) 22. (d) 23. (c) 24. (b) 25. (b) 26. (c) 27. (c) 28. (a) 29. (b) 30. (b), 31. (c) 32. (a) 33. (b) 34. (a) 35. (c) 36. (c) 37. (c) 38. (a) 39. (c) 40. (b), 41. (d) 42. (c) 43. (a) 44. (a) 45. (a) 46. (b) 47. (b) 48. (d) 49. (c) 50. (a), 51. (b) 52. (d) 53. (b) 54. (b) 55. (b) 56. (a) 57. (a) 58. (c) 59. (b) 60. (d), 61. (b) 62. (a) 63. (b) 64. (b) 65. (b) 66. (c) 67. (c) 68. (d) 69. (b) 70. (c), 71. (a) 72. (b) 73. (c) 74. (a) 75. (b) 76. (a) 77. (b) 78. (c) 79. (c) 80. (a)
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Objective Questions, , 1081, , 81. (a) 82. (c) 83. (b) 84. (c) 85. (a) 86. (d) 87. (a) 88. (b) 89. (c) 90. (b), 91. (a) 92. (b) 93. (b) 94. (b) 95. (c) 96. (a) 97. (c) 98. (d) 99. (b) 100. (a), 101. (c) 102. (b) 103. (a) 104. (b) 105. (c) 106. (b) 107. (a) 108. (d) 109. (a) 110. (c), 111. (b) 112. (c) 113. (c) 114. (a) 115. (b) 116. (c) 117. (b) 118. (d) 119. (d) 120. (c), 121. (b) 122. (b) 123. (b) 124. (a) 125. (b) 126. (c) 127. (b) 128. (c) 129. (b) 130. (c), 131. (a) 132. (b) 133. (a) 134. (b) 135. (b) 136. (c) 137. (a) 138. (b) 139. (b) 140. (a), 141. (c) 142. (b) 143. (a) 144. (c) 145. (b) 146. (a) 147. (c) 148. (b) 149. (a) 150. (c), 151. (a) 152. (b) 153. (a) 154. (b) 155. (a) 156. (b) 157. (d) 158. (b) 159. (a) 160. (b), 161. (c) 162. (a) 163. (b) 164. (b) 165. (d) 166. (b) 167. (b) 168. (b) 169. (b) 170. (d), 171. (b) 172. (d) 173. (c) 174. (a) 175. (b) 176. (a) 177. (b) 178. (c) 179. (c) 180. (b), 181. (d) 182. (b) 183. (a) 184. (b) 185. (b) 186. (d) 187. (b) 188. (b) 189. (a) 190. (c), 191. (b) 192. (a) 193. (b) 194. (b) 195. (c) 196. (a) 197. (b) 198. (b) 199. (b) 200. (c), 201. (a) 202. (b) 203. (c) 204. (c) 205. (c) 206. (c) 207. (a) 208. (c) 209. (c) 210. (a), 211. (c) 212. (a) 213. (b) 214. (a) 215. (c) 216. (b) 217. (d) 218. (c) 219. (c) 220. (d), 221. (d) 222. (c) 223. (b) 224. (d) 225. (b) 226. (c) 227. (b) 228. (c) 229. (d) 230. (a), 231. (c) 232. (b) 233. (c) 234. (a) 235. (d) 236. (c) 237. (c) 238. (a) 239. (a) 240. (b), 241. (c) 242. (b) 243. (b) 244. (a) 245. (c) 246. (a) 247. (c) 248. (a) 249. (d) 250. (a), 251. (c) 252. (a) 253. (c) 254. (b) 255. (a) 256. (d) 257. (b) 258. (a) 259. (d) 260. (b), , HINTS TO SOME SELECTED QUESTIONS, c, , π ⎞, 1. 180 ° = π c ⇒ 1° = ⎛⎜, ⎟, ⎝ 180 ⎠, , c, , π, 5π, ⇒ 25 ° = ⎛⎜, × 25 ⎞⎟ = ⎛⎜ ⎞⎟ ⋅, ⎝ 180, ⎠, ⎝ 36 ⎠, , c, , π ⎞, 2. 180 ° = π c ⇒ 1° = ⎛⎜, ⎟, ⎝ 180 ⎠, , c, , π, 9π ⎞, ⇒ 162 ° = ⎛⎜, × 162 ⎞⎟ = ⎛⎜, ⎟ ⋅, ⎝ 180, ⎠, ⎝ 10 ⎠, , c, , c, , c, 180 ⎞ °, ⎛ 8 π ⎞ = ⎛ 180 × 8 π ⎞ ° = 288°., 3. π c = 180 ° ⇒ 1c = ⎛⎜, ⎟ ⇒ ⎜, ⎟, ⎜, ⎟, ⎝ π ⎠, ⎝ 5 ⎠, ⎝ π, 5 ⎠, 180 ⎞ °, c, ⎛ 180 × 11⎞ ° = ⎛ 180 × 7 × 11⎞ ° = 630 °., 4. π c = 180 ° ⇒ 1c = ⎛⎜, ⎟ ⇒ 11 = ⎜, ⎟, ⎜, ⎟, ⎝ π ⎠, ⎝ π, ⎠, ⎝, ⎠, 22, °, °, °, 180, 7, 630, ⎞ = ⎛ 180 × ⎞ = ⎛, ⎞ = 57 ° 16 ′ 22 ′ ′ (app.)., 5. π c = 180 ° ⇒ 1c = ⎛⎜, ⎟, ⎜, ⎟, ⎜, ⎟, ⎝ π ⎠, ⎝, ⎝ 11 ⎠, 22 ⎠, 45 °, 3 °, 15 °, 6. 3 ° 45 ′ = ⎛⎜ 3 ⎞⎟ = ⎛⎜ 3 ⎞⎟ = ⎛⎜ ⎞⎟ ⋅, ⎝ 60 ⎠, ⎝ 4⎠, ⎝ 4⎠, , π ⎞, 180 ° = π c ⇒ 1° = ⎛⎜, ⎟, ⎝ 180 ⎠, , c, , π, 15 °, π, 15, ⇒ ⎛⎜ ⎞⎟ = ⎛⎜, × ⎞⎟ = ⎛⎜ ⎞⎟ ⋅, ⎝ 48 ⎠, ⎝ 4⎠, ⎝ 180 4 ⎠, c, , c
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1082, , Senior Secondary School Mathematics for Class 11, , 30 ⎞ ′, ⎛ 75 ⎞ ° = ⎛ 50 5 ⎞ ° = ⎛ 405 ⎞ °⋅, ⎛ 75 ⎞ ′, 7. 50 ° 37 ′ 30 ′ ′ = 50 ° + ⎛⎜ 37, ⎟, ⎜, ⎟, ⎜, ⎟, ⎟ = 50 ° + ⎜ ⎟ = 50 ° + ⎜, ⎝ 2 × 60 ⎠, ⎝ 8⎠, ⎝ 8 ⎠, ⎝, ⎠, ⎝, ⎠, 60, 2, π ⎞, 180 ° = π c ⇒ 1° = ⎛⎜, ⎟, ⎝ 180 ⎠, , c, , 405 ⎞ ° ⎛ π, 405 ⎞, ⎛ 9π ⎞ ⋅, ⇒ ⎛⎜, ×, ⎟, ⎟ =⎜, ⎟ =⎜, ⎝ 32 ⎠, ⎝ 8 ⎠, ⎝ 180, 8 ⎠, c, , c, , 8. Let ∠C = 90 °. Then, A + B = 90 °., π 180 ⎞ °, A − B = ⎛⎜ ×, ⎟ = 12 ° ⇒ A − B = 12 ° ., ⎝ 15, π ⎠, On solving A + B = 90 and A − B = 12, we get: A = 51° and B = 39 °., ∴, , the measure of the smallest angle is 39°., , 9. Let the angles be ( a − d )° , a° and ( a + d )°. Then,, a − d + a + a + d = 180 ⇒ 3 a = 180 ⇒ a = 60., So, the angles are ( 60 − d )° , 60 ° and ( 60 + d )°., ∴, , 60 + d = 2( 60 − d ) ⇒ 3 d = 60 ⇒ d = 20., , ∴, , the largest angle = ( 60 + 20 )° = 80 °., , 10. Let the angles be ( a − d )° , a° and ( a + d )°. Then,, a − d + a + a + d = 180 ⇒ 3 a = 180 ⇒ a = 60., So, the angles are ( 60 − d )° , 60 ° and ( 60 + d )° ., c, , ( 60 + d )π ⎫, π, 180 ° = π c ⇒ ( 60 + d )° = ⎧⎨, × ( 60 + d )⎫⎬ = ⎧⎨, ⎬, ⎩ 180, ⎭, ⎩ 180 ⎭, ∴, , ( 60 − d ), π, ( 60 + d ) ⋅, 180, , =, , c, , 3( 60 − d ), 60, ⇒, =1, ( 60 + d ), π, ⇒ 180 − 3 d = 60 + d ⇒ 4 d = 120 ⇒ d = 30., , ∴, , the smallest angle = ( 60 − 30 )° = 30 ° ., c, , c, , π, π, 11. Here, l = 33 cm and θ = ⎛⎜, × 45 ⎞⎟ = ⎛⎜ ⎞⎟ ⋅, ⎝ 180, ⎠, ⎝ 4⎠, θ=, , l, l, 4, 7, ⇒ r = = ⎛⎜ 33 × ⎞⎟ = ⎛⎜ 33 × 4 × ⎞⎟ = 42 cm., 22 ⎠, r, θ ⎝, π⎠ ⎝, c, , c, , π ⎞, ⎛ π⎞, 12. θ = ⎛⎜ 36 ×, ⎟ = ⎜ ⎟ and r = 14 cm., ⎝, ⎝5⎠, 180 ⎠, ∴, , π, 22 1 ⎞, 44, cm = 8.8 cm., l = r θ = ⎛⎜ 14 × ⎞⎟ cm = ⎛⎜ 14 ×, × ⎟ cm =, ⎝, ⎝, 5⎠, 7 5⎠, 5, , 13. Angle traced by the minute hand in 60 min = ( 2π ) c ., c, , c, , 2π, 3π ⎞, Angle traced by the minute hand in 45 min = ⎛⎜, × 45 ⎞⎟ = ⎛⎜, ⎟ ⋅, ⎝ 60, ⎠, ⎝ 2 ⎠, c, , ∴, , 3π ⎞, r = 1.4 cm and θ = ⎛⎜, ⎟, ⎝ 2 ⎠, , ⇒, , 3π ⎞, 3 22 ⎞, ⎛, l = r θ = ⎛⎜ 1.4 ×, ⎟ cm = ⎜ 1.4 × × ⎟ cm = 6.6 cm., ⎝, ⎝, 2 ⎠, 2 7 ⎠
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Objective Questions, c, , c, , 1083, c, , c, , π ⎞, π ⎞, ⎛ π⎞, ⎛, ⎛ 5π ⎞, 14. θ1 = 60 ° = ⎛⎜ 60 ×, ⎟ = ⎜ ⎟ and θ 2 = 75 ° = ⎜ 75 ×, ⎟ =⎜ ⎟ ⋅, ⎝, ⎝ 3⎠, ⎝, ⎝ 12 ⎠, 180 ⎠, 180 ⎠, ∴, , l = r1θ1 = r2θ 2, , ⇒, , r1 ×, , π, 5π, 5, 5, r, = r2 ×, ⇒ 1 = ⎛⎜ × 3 ⎞⎟ = ⇒ r1 : r2 = 5 : 4 ., ⎠ 4, 3, 12, r2 ⎝ 12, , 15. l = 121cm and r = 180 cm., l ⎛ 121 ⎞, ⎛ 121 × 180 ⎞ ° = ⎛ 121 × 7 ⎞ ° = ⎛ 77 ⎞ ° = 38 ° 30 ′ ., =⎜, ⎟ =⎜, ⎟, ⎜, ⎟, ⎜ ⎟, ⎝ 180, ⎝, ⎝ 2⎠, 22 ⎠, r ⎝ 180 ⎠, π ⎠, c, , ∴, , θ=, , c, , c, , c, , c, , π ⎞, ⎛ 2 π ⎞ and l = 88 m., 16. θ = 72 ° = ⎛⎜ 72 ×, ⎟ =⎜, ⎟, ⎝, ⎝ 5 ⎠, 180 ⎠, l, 5 ⎞, 5 7 ⎞, ⎛, ∴ r = = ⎛⎜ 88 ×, ⎟ m = ⎜ 88 × × ⎟ m = 70 m., ⎝, 2 22 ⎠, θ ⎝, 2π ⎠, π ⎞, ⎛ 7 π ⎞ and l = 55 cm., 17. θ = 42 ° = ⎛⎜ 42 ×, ⎟ =⎜, ⎟, ⎝, ⎝ 30 ⎠, 180 ⎠, l, 30 ⎞, 30 7 ⎞, ⎛, × ⎟ cm = 75 cm., ∴ r = = ⎛⎜ 55 ×, ⎟ cm = ⎜ 55 ×, ⎝, 7, 22 ⎠, θ ⎝, 7π⎠, 18. Let O be the centre of the circle and AB be the chord., Then, OA = OB = AB = 30 cm., ∴, , aOAB is an equilateral triangle and therefore each of, its angles is 60°., c, , π ⎞, ⎛ π⎞, θ = ∠AOB = 60 ° = ⎛⎜ 60 ×, ⎟ = ⎜ ⎟ and r = 30 cm., ⎝, ⎝ 3⎠, 180 ⎠, , ∴, , π, l = r θ = ⎛⎜ 30 × ⎞⎟ cm = ( 10 π ) cm., ⎝, 3⎠, , 19. Number of revolutions made in 1 second =, , 180, = 3., 60, , Angle turned in 1 revolution = ( 2π ) c ., Angle turned in 3 revolutions = ( 3 × 2 π ) c = ( 6 π ) c ., 5, 20. Speed = 90 km/hr = ⎛⎜ 90 × ⎞⎟ m/sec = 25 m/sec., ⎝, 18 ⎠, Distance moved in 11 sec = ( 25 × 11) m = 275 m., l = 275 m and r = 1500 m., l ⎛ 275 ⎞, ⎛ 275 × 180 ⎞ ° = ⎛ 33 × 7 ⎞ ° = ⎛ 21 ⎞ ° = 10 ° 30 ′ ., =⎜, ⎟ =⎜, ⎟, ⎜, ⎟, ⎜ ⎟, ⎝ 1500, ⎝, ⎝ 2⎠, 22 ⎠, r ⎝ 1500 ⎠, π ⎠, c, , θ=, , 60º, , c, , ∴, , ∴, , O, , 21. Angle traced by the hour hand in 12 hours = 360°., 22, 360 22 ⎞ °, Angle traced by the hour hand in, hours = ⎛⎜, × ⎟ = 220 ° ., ⎝ 12, 3, 3⎠, Angle traced by the minute hand in 60 min = 360°., °, 360, Angle traced by the minute hand in 20 min = ⎛⎜, × 20 ⎞⎟ = 120 ° ., ⎝ 60, ⎠, Angle between the two hands = ( 220 ° − 120 ° ) = 100 ° ., , A, , B
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1084, , Senior Secondary School Mathematics for Class 11, , 22. Angle traced by the hour hand in 12 hours = 360°., Angle traced by the hour hand in, , 7, 360 7 ⎞ °, hours = ⎛⎜, × ⎟ = 105 ° ., ⎝ 12 2 ⎠, 2, , Angle traced by the minute hand in 60 min = 360°., °, 360, Angle traced by the minute hand in 30 min = ⎛⎜, × 30 ⎞⎟ = 180 ° ., ⎝ 60, ⎠, Angle between the two hands = ( 180 ° − 105 ° ) = 75 ° ., 23. Angle traced by the hour hand in 12 hours = 360 ° ., Angle traced by the hour hand in 8, , 25, 360 101 ⎞ ° ⎛ 505 ⎞ °, hours = ⎛⎜, ×, ⎟ =⎜, ⎟ = 252.5 ° ., ⎝ 12, ⎝ 2 ⎠, 60, 12 ⎠, , Angle traced by the minute hand in 60 min = 360°., °, 360, Angle traced by the minute hand in 25 min = ⎛⎜, × 25 ⎞⎟ = 150 ° ., ⎝ 60, ⎠, Angle between the two hands = (252.5 − 150)° = 102 ° 30 ′ ., 24. Here, r = 60 cm and l = 16.5 cm., l ⎛ 16.5 ⎞, ⎛ 16.5 × 180 ⎞ ° = ⎛ 16.5 × 180 × 7 ⎞ ° = ⎛ 63 ⎞ ° = 15 ° 45 ′., =⎜, ⎜, ⎟, ⎜ ⎟, ⎟ =⎜, ⎟, ⎝ 60, ⎝ 4⎠, ⎝, ⎠, ⎝ 60, 22 ⎠, 60, r, π ⎠, c, , ∴, , θ=, , 25. Let the angles be ( a − 3 d )° , ( a − d )° , ( a + d )° and ( a + 3 d )°. Then,, a − 3 d + a − d + a + d + a + 3 d = 360 ⇒ 4 a = 360 ⇒ a = 90., ∴, , the angles are ( 90 − 3 d )° , ( 90 − d )° , ( 90 + d )° and ( 90 + 3 d )° ., , ∴, , 90 + 3 d = 120 ⇒ 3 d = 30 ⇒ d = 10., , ∴, , π, the smallest angle = ( a − 3 d )° = ( 90 − 3 × 10 )° = 60 ° = ⎛⎜ ⎞⎟ ⋅, ⎝ 3⎠, , c, , 26. Let the radius be r cm. Then,, 1, r + r + arc length = × 2πr = πr., 2, l, l = πr − 2r = ( π − 2 ) r ⇒ θ = = ( π − 2 ) c ., ∴, r, 25 π, π, π, 3, 27. sin, = sin ⎛⎜ 8 π + ⎞⎟ = sin =, [Q sin ( 2nπ + θ) = sin θ]., ⎝, 3, 3⎠, 3, 2, 41π, π, π, 1, 28. cos, = cos ⎛⎜ 10 π + ⎞⎟ = cos =, [Q cos ( 2nπ + θ) = cos θ]., ⎝, 3, 4⎠, 4, 2, −16 π ⎞, 16 π, 29. tan ⎛⎜, [Q tan ( −θ) = − tan θ], ⎟ = − tan, ⎝ 3 ⎠, 3, π, π, = − tan ⎛⎜ 5 π + ⎞⎟ = − tan = − 3, [Q tan (nπ + θ) = tan θ]., ⎝, 3⎠, 3, 29 π ⎞, 30. cot ⎛⎜, ⎟ = cot, ⎝ 4 ⎠, , ⎛ 7 π + π ⎞ = cot π = 1, [Q cot (nπ + θ) = cot θ]., ⎜, ⎟, ⎝, 4⎠, 4, −19 π ⎞, 19 π, 31. sec ⎛⎜, [Q sec ( −θ) = sec θ], ⎟ = sec, ⎝ 3 ⎠, 3, π, π, = sec ⎛⎜ 6 π + ⎞⎟ = sec = 2, [Q sec ( 2nπ + θ) = sec θ]., ⎝, 3⎠, 3
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Objective Questions, −33 π ⎞, 33 π, π, = − cosec ⎛⎜ 8 π + ⎞⎟, ⎟ = − cosec, ⎝, 4 ⎠, 4, 4⎠, , 32. cosec ⎛⎜, ⎝, , = − cosec, , π, =− 2, 4, , [Q cosec ( 2nπ + θ) = cosec θ]., , 33. cos 15 π = cos ( 14 π + π ) = cos π = −1 [Q cos ( 2nπ + θ) = cos θ]., 34. sec 6 π = sec ( 6 π + 0 ) = sec 0 = 1, [Q sec ( 2nπ + θ) = sec θ]., 5π, π⎞, π, ⎛, 35. tan, = tan ⎜ π + ⎟ = tan = 1, [Q tan ( π + θ) = tan θ]., ⎝, 4, 4⎠, 4, 1, 36. sin (765 ° ) = sin ( 360 × 2 + 45 )° = sin 45 ° =, ⋅, 2, 37. cot ( −600 ° ) = − cot 600 ° = − cot ( 360 + 240 )° = − cot 240 °, −1, = − cot ( 180 ° + 60 ° ) = − cot 60 ° =, ⋅, 3, 38. In quadrant III, we know that cos x is negative., 24, 1, −1, cos 2 x = ( 1 − sin 2 x ) = ⎛⎜ 1 − ⎞⎟ =, ⇒ cos x =, ⋅, ⎝, 25 ⎠ 25, 5, ∴, , cot x =, , cos x −1, 5, 1, =, ×, =, ⋅, sin x 5 −2 6 2 6, , 15, 1, 39. sin 2 x = ( 1 − cos 2 x ) = ⎛⎜ 1 − ⎞⎟ =, ⋅, ⎝, 16 ⎠ 16, When, , π, < x < π , then x lies in quadrant II., 2, , And sin x > 0 in quadrant II., ∴, , 1, 1, = ⋅, 16 4, , sin x =, , 3π, ⇒ x lies in quadrant III and sin x < 0 in quadrant III., 2, −1, sec x = −2 ⇒ cos x =, ⋅, 2, , 40. π < x <, , sin 2 x = ( 1 − cos 2 x ) = ⎛⎜ 1 −, ⎝, ∴, , sin x = −, , 1⎞ 3, ⎟= ⋅, 4⎠ 4, , 3 − 3, =, ⋅, 4, 2, , 41. In quadrant IV, cos x > 0., Now, cosec x =, , −2, − 3, ⇒ sin x =, ⋅, 2, 3, , 3, 1, 1 1, cos 2 x = ( 1 − sin 2 x ) = ⎛⎜ 1 − ⎞⎟ = ⇒ cos x = +, = ⋅, ⎝, 4⎠ 4, 4 2, ∴, , tan x =, , sin x ⎛ − 3 2 ⎞, =⎜, × ⎟ = − 3., 1⎠, cos x ⎝ 2, , 1085
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Objective Questions, 63. cot, , π, 1, π, π, =, , cot = 1, cot =, 3, 4, 6, 3, , Now,, , 3., , 1, , 1, 3 are in GP with common ratio, 3, , 3., , 64. In quadrant I, sin θ is increasing., Now, cos 64 ° = cos ( 90 ° − 26 ° ) = sin 26 ° ., Clearly, sin 26 ° < sin 64 ° ⇒ cos 64 ° < sin 64 ° ., 65. cos 24 ° = cos ( 90 ° − 66 ° ) = sin 66 ° ., In quadrant I, sin θ is increasing., ∴, , sin 66 ° > sin 24 ° ⇒ cos 24 ° > sin 24 ° ., , 66. The extremum values of sin θ are –1 and 1., 67. The extremum values of cos θ are –1 and 1., 68. |sec θ| ≥ 1 ⇒ (sec θ ≤ −1) or (sec θ ≥ 1)., ∴, , value of sec θ can never lie between –1 and 1., −1, ⋅, 3, −2, 70. sec 150 ° = sec ( 180 ° − 30 ° ) = − sec 30 ° =, ⋅, 3, −1, 71. cot 120 ° = cot ( 180 ° − 60 ° ) = − cot 60 ° =, ⋅, 3, 69. tan 150 ° = tan ( 180 ° − 30 ° ) = − tan 30 ° =, , 72. sin 105 ° + cos 105 ° = sin ( 60 ° + 45 ° ) + cos ( 60 ° + 45 ° ), = (sin 60 ° cos 45 ° + cos 60 ° sin 45 ° ) + (cos 60 ° cos 45 ° − sin 60 ° sin 45° ), ⎛ 3, 1 ⎞ ⎛1, 1 ⎞ ⎛1, 1 ⎞ ⎛ 3, 1 ⎞, =⎜, ×, ×, ⎟+⎜ ×, ⎟, ⎟+⎜ ×, ⎟−⎜, ⎝, ⎠, ⎝, ⎠, 2, 2, 2, 2, 2⎠, 2, 2, 2⎠, ⎝, ⎝, =, , 3, 1, 1, 3, 2, 1, +, +, −, =, =, ⋅, 2 2 2 2 2 2 2 2 2 2, 2, , 73. sin 15 ° = sin ( 45 ° − 30 ° ) = (sin 45 ° cos 30 ° − cos 45 ° sin 30 ° ), ⎛ 1, 3⎞ ⎛ 1, 1⎞ ⎛ 3, 1 ⎞ ( 3 − 1), =⎜, ×, × ⎟=⎜, −, ⋅, ⎟−⎜, ⎟=, 2 ⎠ ⎝ 2 2⎠ ⎝ 2 2 2 2 ⎠, 2 2, ⎝ 2, 74. cos 15 ° = cos ( 45 ° − 30 ° ) = cos 45 ° cos 30 ° + sin 45 ° sin 30 °, ⎛ 1, 3⎞ ⎛ 1, 1 ⎞ ( 3 + 1), =⎜, ×, × ⎟=, ⋅, ⎟+⎜, 2 ⎠ ⎝ 2 2⎠, 2 2, ⎝ 2, 1 ⎞, ⎛, ⎜1−, ⎟, ⎝, tan 45 ° − tan 30 °, 3 ⎠ = ( 3 − 1) ⋅, 75. tan 15 ° = tan ( 45 ° − 30 ° ) =, =, 1 ⎞ ( 3 + 1), 1 + tan 45 ° tan 30 ° ⎛, ⎜1+, ⎟, ⎝, 3⎠, 76. sin 75 ° = sin ( 90 ° − 15 ° ) = cos 15 ° =, 77. cos 75 ° = cos ( 90 ° − 15 ° ) = sin 15 ° =, , ( 3 + 1), 2 2, ( 3 − 1), 2 2, , ⋅, ⋅, , 1089
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1090, 78. tan, , Senior Secondary School Mathematics for Class 11, 13 π, π⎞, π, = tan ⎛⎜ π +, ⎟ = tan, ⎝, 12, 12 ⎠, 12, = tan 15 ° =, , ( 3 − 1), ( 3 + 1), , =, , [Q tan ( π + θ) = tan θ], , ( 3 − 1), ( 3 + 1), , ×, , ( 3 − 1), ( 3 − 1), , ( 3 + 1− 2 3) (4 − 2 3), =, =, = (2 −, 2, 2, , =, , ( 3 − 1) 2, ( 3 − 1), , 3 )., , 79. (sin 70 ° cos 10 ° − cos 70 ° sin 10 ° ) = sin (70 ° − 10 ° ), [Q sin A cos B − cos A sin B = sin ( A − B)], 3, = sin 60 ° =, ⋅, 2, 80. (sin 36 ° cos 9 ° + cos 36 ° sin 9 ° ) = sin ( 36 ° + 9 ° ), [Q sin A cos B + cos A sin B = sin ( A + B)], 1, = sin 45 ° =, ⋅, 2, 81. cos 80 ° cos 20 ° + sin 80 ° sin 20 ° = cos ( 80 ° − 20 ° ), [Q cos A cos B + sin A sin B = cos ( A − B)], 1, = cos 60 ° = ⋅, 2, 82. cos 50 ° cos 10 ° − sin 50 ° sin 10 ° = cos (50 ° + 10 ° ), [Q cos A cos B − sin A sin B = cos ( A + B)], 1, = cos 60 ° = ⋅, 2, 83. sin ( 40 ° + θ) cos ( 10 ° + θ) − cos ( 40 ° + θ) sin ( 10 ° + θ), = sin {( 40 °+ θ) − ( 10 ° + θ)}, [Q sin A cos B − cos A sin B = sin ( A − B)], 1, = sin 30 ° = ⋅, 2, 7π, π, 7π, π, 7π π⎞, 84. sin, cos − cos, sin = sin ⎛⎜, − ⎟ [Q sin A cos B − cos A sin B = sin ( A − B)], ⎝ 12 4 ⎠, 12, 4, 12, 4, = sin, , 4π, π, 3, = sin =, ⋅, 12, 3, 2, , 85. sin, , π, π, π, π, π, π⎞, 4π, π, 3, = sin =, cos, + cos sin, = sin ⎛⎜ +, ⋅, ⎟ = sin, ⎝ 4 12 ⎠, 4, 12, 4, 12, 12, 3, 2, , 86. cos, , π, 2π, π, 2π, π, π, π, π, cos − sin, sin = cos ⎛⎜ π − ⎞⎟ cos − sin ⎛⎜ π − ⎞⎟ sin, ⎝, ⎝, 4, 3, 4, 3, 4, 3⎠, 4, 3⎠, = − cos, , π, π, π, π, cos − sin sin, 3, 4, 3, 4, , 1 ⎞ ⎛ 3, 1 ⎞ − ( 3 + 1), ⎛ 1, = ⎜− ×, ×, ⋅, ⎟=, ⎟−⎜, ⎝ 2, ⎠, 2, 2, 2⎠, 2 2, ⎝, 87. sin, , π, π π, π, π, π, π, = sin ⎛⎜ − ⎞⎟ = sin cos − cos sin, ⎝ 4 6⎠, 12, 4, 6, 4, 6, ⎛ 1, 3⎞ ⎛ 1, 1 ⎞ ( 3 − 1), =⎜, ×, × ⎟=, ⋅, ⎟−⎜, 2 ⎠ ⎝ 2 2⎠, 2 2, ⎝ 2
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1092, , Senior Secondary School Mathematics for Class 11, , 92. In quadrant III, sin θ < 0 , cos θ < 0 and tan θ > 0., In quadrant II, sin φ > 0 , cos φ < 0 and tan φ < 0., 1, Now, cot θ = ⇒ tan θ = 2, 2, −5, −3, sec φ =, ⇒ cos φ =, ⋅, 3, 5, 16 4, = ⋅, 25 5, , ∴, , 9, 16, ⇒ sin φ =, sin 2 φ = ( 1 − cos 2 φ) = ⎛⎜ 1 − ⎞⎟ =, ⎝, 25 ⎠ 25, , ∴, , 4 5 ⎞ −4, tan φ = ⎛⎜ ×, ⋅, ⎟=, ⎝ 5 −3 ⎠, 3, , ∴, , ⎛ 2⎞, ⎛2 − 4⎞, ⎜ ⎟, ⎜, ⎟, ⎝ 3⎠, tan θ + tan φ, ⎝, 2 3, 2, 3⎠, =, = ⎛⎜ × ⎞⎟ =, tan (θ + φ) =, =, ⋅, ⎝, ⎠, 8, 4, 1 − tan θ tan φ ⎧, −, 3, 11, 11, ⎛, ⎞, ⎫, ⎛, ⎞, ⎟⎬ ⎜1+ ⎟, ⎨1 − ⎜ 2 ×, ⎝, 3⎠, 3 ⎠⎭ ⎝, ⎩, , 93. cos 15 ° − sin 15 ° = cos ( 45 ° − 30 ° ) − sin ( 45 ° − 30 ° ), = (cos 45 ° cos 30 ° + sin 45 ° sin 30 ° ) − (sin 45 ° cos 30 ° − cos 45 ° sin 30° ), ⎧⎪ ⎛ 1, 1 ⎞ ⎫⎪, 3⎞ ⎛ 1, 1 ⎞ ⎫⎪ ⎧⎪ ⎛ 1, 3⎞ ⎛ 1, = ⎨⎜, ×, × ⎟ ⎬ − ⎨⎜, ×, × ⎟⎬, ⎟+⎜, ⎟−⎜, ⎝, ⎝, ⎠, 2 ⎠, 2 ⎠, 2 2 ⎪⎭ ⎩⎪ ⎝ 2, 2 2 ⎠ ⎪⎭, ⎩⎪ ⎝ 2, =, , ( 3 + 1), 2 2, , −, , ( 3 − 1), 2 2, , =, , 2, 1, =, ⋅, 2 2, 2, , ⎛ 1, ⎞, × 1 − 1⎟, ⎜, ⎠ (1−, cot 60 ° cot 45 ° − 1 ⎝ 3, 94. cot 105 ° = cot ( 60 ° + 45 ° ) =, =, =, cot 60 ° + cot 45 °, ⎛ 1, ⎞, (1+, + 1⎟, ⎜, ⎝ 3, ⎠, tan 105 ° = tan ( 60 ° + 45 ° ) =, ∴, , cot 105 ° − tan 105 ° =, =, , 95. 2 sin, , 3), 3), , ⋅, , (tan 60 ° + tan 45 ° ) ( 3 + 1), =, ⋅, 1 − tan 60 ° tan 45 ° ( 1 − 3 ), , (1 −, , 3), , (1+, , 3), , (1−, , 3 )2 − ( 1 +, ( 1 − 3), , −, , (1+, , 3), , (1−, , 3), 3 )2, , =, , −4 3, = 2 3., −2, , 5π, π, sin, = cos ( A − B) − cos ( A + B), 12, 12, 5π π ⎞, 5π, π ⎞⎫, ⎧, +, = ⎨ cos ⎛⎜, − ⎟ − cos ⎛⎜, ⎟⎬, ⎝, ⎠⎭, ⎝, ⎠, 12, 12, 12, 12, ⎩, π, π, 1, 1, = ⎛⎜ cos − cos ⎞⎟ = ⎛⎜ − 0 ⎞⎟ = ⋅, ⎠ 2, ⎝, 3, 2⎠ ⎝ 2, , 96. 2 cos, , 5π, π⎞, 5π, π, ⎛ 5π − π ⎞, cos, = cos ( A + B) + cos ( A − B) = cos ⎛⎜, +, ⎟, ⎟ + cos ⎜, ⎝ 12 12 ⎠, ⎝ 12 12 ⎠, 12, 12, π, π, 1, 1, = ⎛⎜ cos + cos ⎞⎟ = ⎛⎜ 0 + ⎞⎟ = ⋅, ⎝, 2⎠ 2, 2, 3⎠ ⎝
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1094, , Senior Secondary School Mathematics for Class 11, , sin 3 x − sin x, =, cos x − cos 3 x, , 4x, 2x, 2 cos ⎛⎜ ⎞⎟ sin ⎛⎜ ⎞⎟, ⎝ 2⎠, ⎝ 2 ⎠ cos 2 x sin x, =, = cot 2 x., 4x, 2x, sin 2 x sin x, 2 sin ⎛⎜ ⎞⎟ sin ⎛⎜ ⎞⎟, ⎝ 2⎠, ⎝ 2⎠, , (C + D ), (C − D ), cos, 2, 2, (C + D ), (C − D ), and sin C − sin D = 2 cos, sin, , we get:, 2, 2, 10 x ⎞, ⎛ 2x ⎞, 2 cos ⎛⎜, ⎟ cos ⎜ ⎟, ⎝ 2 ⎠, ⎝ 2 ⎠ cos 5 x cos x cos x, cos 6 x + cos 4 x, =, = cot x., =, =, 10 x ⎞, 2 x ⎞ cos 5 x sin x sin x, sin 6 x − sin 4 x, ⎛, ⎛, 2 cos ⎜, ⎟ sin ⎜ ⎟, ⎝ 2 ⎠, ⎝ 2⎠, ( 4x − 2x), ( 4 x + 2 x), cos, + cos 3 x, 2 cos, (cos 4 x + cos 2 x ) + cos 3 x, 2, 2, 108. Given exp. =, =, (sin 4 x + sin 2 x ) + sin 3 x 2 sin ( 4 x + 2 x ) cos ( 4 x − 2 x ) + sin 3 x, 2, 2, 2 cos 3 x cos x + cos 3 x cos 3 x( 2 cos x + 1) cos 3 x, =, = cot 3 x., =, =, 2 sin 3 x cos x + sin 3 x sin 3 x( 2 cos x + 1) sin 3 x, , 107. Using cos C + cos D = 2 cos, , (7 x + 5 x ), (7 x − 5 x ), 2 cos, sin, sin x, sin 7 x − sin 5 x, 2, 2, 109. Given exp. =, =, = tan x., =, cos 7 x + cos 5 x 2 cos (7 x + 5 x ) cos (7 x − 5 x ) cos x, 2, 2, 110. sin 2 6 x − sin 2 4 x = sin ( 6 x + 4 x ) sin ( 6 x − 4 x ) [Q sin 2 x − sin 2 y = sin ( x + y ) sin ( x − y )], = sin 10 x sin 2 x., 1, 111. Given exp. = ( 2 cos 20 ° cos 80 ° ) cos 40 °, 2, 1, = [cos ( 80 ° + 20 ° ) + cos ( 80 ° − 20 ° )] cos 40 °, 2, ⎤, 1, 1⎡, 1, = [(cos 100 ° + cos 60 ° ) cos 40 ° ] = ⎢ ⎛⎜ cos 100 ° + ⎞⎟ cos 40° ⎥, 2⎠, 2 ⎣⎝, 2, ⎦, 1, 1, ( 2 cos 100 ° cos 40 ° ) + cos 40 °, 4, 4, 1, 1, = [cos ( 100 ° + 40 ° ) + cos ( 100 ° − 40 ° )] + cos 40 °, 4, 4, 1, 1, 1, 1 1, = (cos 140 ° + cos 60 ° ) + cos 40 ° = (cos 140 ° + cos 40 ° ) + ⎛⎜ × ⎞⎟, ⎝ 4 2⎠, 4, 4, 4, =, , 1, 1 1, 1 1, [cos ( 180 ° − 40 ° ) + cos 40 ° ] + = ( − cos 40 ° + cos 40 ° ) + = ⋅, 4, 8 4, 8 8, 1, 1, 112. Given exp. = ( 2 sin 70 ° sin 50 ° ) ⋅ sin 10 ° = [cos (70 ° − 50 ° ) − cos (70 ° + 50 ° )] sin 10 °, 2, 2, =, , =, , 1, 1, 1, (cos 20 ° − cos 120 ° ) sin 10 ° = ⎛⎜ cos 20 ° + ⎞⎟ sin 10 °, 2, 2⎝, 2⎠, , =, , 1, 1, ( 2 cos 20 ° sin 10 ° ) + sin 10 °, 4, 4
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Objective Questions, =, , 1, 1, [sin ( 20 ° + 10 ° ) − sin ( 20 ° − 10 ° )] + sin 10 °, 4, 4, , =, , 1, 1 1, 1, sin 30 ° = ⎛⎜ × ⎞⎟ = ⋅, ⎝ 4 2⎠ 8, 4, , 1095, , 113. Using 2 cos A cos B = cos ( A + B) + cos ( A − B), we get:, 2 cos 45 ° cos 15 ° = cos ( 45 ° + 15 ° ) + cos ( 45 ° − 15 ° ), ⎛1, 3 ⎞ ( 3 + 1), = (cos 60 ° + cos 30 ° ) = ⎜ +, ⋅, ⎟=, 2 ⎠, 2, ⎝2, 114. Using 2 sin A sin B = cos ( A − B) − cos ( A + B), we get:, 2 sin 75 ° sin 15 ° = cos (75 ° − 15 ° ) − cos (75 ° + 15 ° ), 1, 1, = (cos 60 ° − cos 90 ° ) = ⎛⎜ − 0 ⎞⎟ = ⋅, ⎝2, ⎠ 2, 115. cos 15 ° − sin 15 ° = cos 15 ° − cos ( 90 ° − 15 ° ) = cos 15 ° − cos 75 °, ( 15 ° + 75 ° ), (75 ° − 15 ° ), = 2 sin, sin, 2, 2, (C + D ), (D − C )⎤, ⎡, sin, ⎢Q cos C − cos D = 2 sin, ⎥, 2, 2, ⎣, ⎦, 1, 1⎞, 1, ⎛, = 2 sin 45 ° sin 30 ° = ⎜ 2 ×, × ⎟=, ⋅, ⎝, 2 2⎠, 2, 116. In quadrant III, we have: sin x < 0 and cos x < 0., sin x =, , 1, 3, −1, ⇒ cos x = ( 1 − sin 2 x ) = ⎛⎜ 1 − ⎞⎟ =, ⎝, 2, 4⎠ 4, ⇒ cos x = −, , ∴, , 3, 3, =−, ⋅, 4, 2, , sin 2 x = 2 sin x cos x = 2 ×, , ( −1) ( − 3 ), 3, ×, =, ⋅, 2, 2, 2, , 117. Here, x lies in quadrant II, where sin x > 0 and cos x < 0., −13, −12, sec x =, ⇒ cos x =, ⋅, 12, 13, 2, ⎧⎪, ⎫⎪ ⎛, ( 288 − 169 ) 119, −12 ⎞, 144, Now, cos 2 x = ( 2 cos 2 x − 1) = ⎨ 2 × ⎛⎜, =, ⋅, − 1⎞⎟ =, ⎟ − 1⎬ = ⎜ 2 ×, ⎝, ⎠, ⎝, ⎠, 169, 169, 69, 13, 1, ⎩⎪, ⎭⎪, 118. Here, x lies in quadrant IV, where cos x > 0 , sin x < 0 and tan x < 0., tan x =, , 9 ⎞ 25, −3, ⇒ sec2 x = ( 1 + tan 2 x ) = ⎛⎜ 1 +, ⎟=, ⎝, 4, 16 ⎠ 16, ⇒ sec x = +, , 25 5, 4, = ⇒ cos x = ⋅, 16 4, 5, , −3, 16, 9, 9, Now, sin 2 x = ( 1 − cos 2 x ) = ⎛⎜ 1 − ⎞⎟ =, ⇒ sin x = −, =, ⋅, ⎝, 25 ⎠ 25, 25, 5, ∴, , tan x =, , sin x ⎛ −3 5 ⎞ −3, ⋅, =⎜, × ⎟=, 4, cos x ⎝ 5 4 ⎠
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1096, , Senior Secondary School Mathematics for Class 11, , ∴, , ( −3 ) ⎛ −3 ⎞, ⎜ ⎟, 8, 4 = ⎝ 2 ⎠ = −3 × 16 = −24 ⋅, =, tan 2 x =, 2, 7, 7, 1 − tan x ⎛ 1 − 9 ⎞ ⎛ 7 ⎞ 21, ⎜, ⎟ ⎜ ⎟, ⎝, 16 ⎠ ⎝ 16 ⎠, 2 tan x, , 2×, , 1, 1⎞ ⎛, 4 ⎞ 23, 119. sin 3 x = ( 3 sin x − 4 sin 3 x ) = ⎛⎜ 3 × − 4 ×, ⋅, ⎟=, ⎟ = ⎜1−, ⎝, 3, 27 ⎠ ⎝, 27 ⎠ 27, 1, 1, 1 3, 120. cos 3 x = ( 4 cos 3 x − 3 cos x ) = ⎛⎜ 4 × − 3 × ⎞⎟ = ⎛⎜ − ⎞⎟ = −1., ⎝, 8, 2⎠ ⎝ 2 2 ⎠, π, π, π, 121. cos 2, [Q cos 2 x − sin 2 x = cos 2 x], − sin 2, = cos ⎛⎜ 2 × ⎞⎟, ⎝, 12, 12, 12 ⎠, = cos, 122. Given exp. =, =, 123., , π, 3, =, ⋅, 6, 2, , ( 1 − cos 2 x ) + sin 2 x 2 sin 2 x + 2 sin x cos x, =, ( 1 + cos 2 x ) + sin 2 x 2 cos 2 x + 2 sin x cos x, 2 sin x (sin x + cos x ) sin x, =, = tan x., 2 cos x (cos x + sin x ) cos x, , π, π x π, π, x, lies in quadrant I ⇒ sin > 0., <x<π ⇒, < < ⇒, 2, 4 2 2, 2, 2, Now, 2 sin 2, , x, 3, 8, x, 8 4, = ⋅, = ( 1 − cos x ) = ⎛⎜ 1 + ⎞⎟ = ⇒ sin 2 =, ⎝, 2, 5⎠ 5, 2 10 5, , 4, =, 5, π, π, 124., <x<π ⇒, <, 2, 4, ∴, , sin, , x, =, 2, , 2, ⋅, 5, x π, x, x, < ⇒ lies in quadrant I ⇒ cos > 0., 2 2, 2, 2, , x, 3, 2, x 1, x, 1, ⋅, = ( 1 + cos x ) = ⎛⎜ 1 − ⎞⎟ = ⇒ cos 2 = ⇒ cos =, ⎝, 2, 5⎠ 5, 2 5, 2, 5, 3π, π x 3π, x, x, 125. π < x <, ⇒, < <, ⇒ lies in quadrant II ⇒ cos < 0., 2, 2 2, 4, 2, 2, 2 cos 2, , 2 cos 2, , x, 4, 1, x, 1, = ( 1 + cos x ) = ⎛⎜ 1 − ⎞⎟ = ⇒ cos 2 =, ⎝, 2, 5⎠ 5, 2 10, ⇒ cos, , x, −1, =, 2, 10, , x, ⎡, ⎤, ⎢⎣Q cos 2 < 0 ⎥⎦ ⋅, , 126. Since x lies in quadrant III, we have: cos x < 0., Now, tan x =, , 9 ⎞ 25, 3, ⇒ sec2 x = ( 1 + tan 2 x ) = ⎛⎜ 1 +, ⎟=, ⎝, 4, 16 ⎠ 16, ⇒ cos 2 x =, , Also, π < x <, , 16, 16 −4, ⇒ cos x = −, =, ⋅, 25, 25, 5, , 3π, π x 3π, x, ⇒, < <, ⇒ lies in quadrant II, 2, 2 2, 4, 2, x, x, ⇒ sin > 0 and cos < 0., 2, 2
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Objective Questions, 2 sin 2, ∴, , sin, , x, 4, 9, x 9, ⋅, = ( 1 − cos x ) = ⎛⎜ 1 + ⎞⎟ = ⇒ sin 2 =, ⎝, 2, 5⎠ 5, 2 10, , x, 9, =+, =, 2, 10, , 2 cos 2, , x, ⎡, ⎤, ⎢⎣Q sin 2 > 0 ⎥⎦ ⋅, , 3, 10, , x, 4, 1, x, 1, = ( 1 + cos x ) = ⎛⎜ 1 − ⎞⎟ = ⇒ cos 2 =, ⎝, 2, 5⎠ 5, 2 10, , x, 1, −1, =−, =, ⋅, 2, 10, 10, , ∴, , cos, , ∴, , x, tan =, 2, , x, sin ⎛⎜ ⎞⎟, ⎝ 2⎠, =, x, cos ⎛⎜ ⎞⎟, ⎝ 2⎠, , 3, 10, ×, = −3., −1, 10, , 127. Since x lies in quadrant III, we have: π < x <, ∴, , π x 3π, x, < <, ⇒ lies in quadrant II, 2 2, 4, 2, x, x, x, ⇒ sin > 0 , cos < 0 and tan < 0., 2, 2, 2, 2 sin 2, , ∴, , 3π, ⋅, 2, , sin, , x, 1, 4, x 2, = ( 1 − cos x ) = ⎛⎜ 1 + ⎞⎟ = ⇒ sin 2 =, ⎝, 2, 3⎠ 3, 2 3, , x, =, 2, , 2 cos 2, , 2, ⋅, 3, , x, 1, 2, x 1, = ( 1 + cos x ) = ⎛⎜ 1 − ⎞⎟ = ⇒ cos 2 =, ⎝, 2, 3⎠ 3, 2 3, , x, 1, =−, ⋅, 2, 3, , ∴, , cos, , ∴, , x, tan =, 2, , x, sin ⎛⎜ ⎞⎟, ⎝ 2⎠, =, x, cos ⎛⎜ ⎞⎟, ⎝ 2⎠, , 2, × ( − 3 ) = − 2., 3, , 128. Since x lies in quadrant IV, we have: cos x > 0., Now, cos 2 x = ( 1 − sin 2 x ) = ⎛⎜ 1 −, ⎝, , 1⎞ 3, 3, 3, =, ⋅, ⎟ = ⇒ cos x = +, 4⎠ 4, 4, 2, , 3π, 3π x, x, < x < 2π ⇒, < < π ⇒ lies in quadrant II., 2, 4, 2, 2, x, sin > 0., 2, , Now,, ∴, , 2 sin 2, ⇒ sin 2, , ⎛, x, 3 ⎞ (2 − 3), = ( 1 − cos x ) = ⎜ 1 −, ⎟=, 2 ⎠, 2, 2, ⎝, , x (2 − 3), x, =, ⇒ sin =, 2, 4, 2, , 2− 3, ⋅, 2, , 1097
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1098, , Senior Secondary School Mathematics for Class 11, , 1 + cos x, 129., =, 1 − cos x, , x, 2 cos 2 ⎛⎜ ⎞⎟, ⎝ 2⎠, x, = cot 2 ⋅, x, 2, 2 sin 2 ⎛⎜ ⎞⎟, ⎝ 2⎠, 1, , π, π x ⎫2, ⎧, 1, 1 − cos ⎛⎜ + x ⎞⎟ ⎪ 2 sin 2 ⎛⎜ + ⎞⎟ ⎪, 1 + sin x, ⎝ 4 2⎠ ⎪ ⎧, ⎝2, ⎠ ⎪, x⎞ ⎫2, ⎛ π x⎞, 2⎛ π, 130., =, =⎨, ⎬ = ⎨ tan ⎜ + ⎟ ⎬ = tan ⎜ + ⎟ ⋅, ⎝ 4 2⎠ ⎭, ⎝ 4 2⎠, π x, π, 1 − sin x, 1 + cos ⎛⎜ + x ⎞⎟ ⎪ 2 cos 2 ⎛⎜ + ⎞⎟ ⎪ ⎩, ⎝ 4 2 ⎠ ⎪⎭, ⎝2, ⎠ ⎪⎩, x, x, 2 sin cos, sin x, 2, 2 = tan x ⋅, 131., =, x, 1 + cos x, 2, 2 cos 2, 2, ⎧ 2 ⎛ x⎞, x, x, 2 ⎛ x⎞ ⎫, cos ⎛⎜ ⎞⎟ sin ⎛⎜ ⎞⎟ 2 ⎨ cos ⎜ ⎟ − sin ⎜ ⎟ ⎬, ⎝ 2 ⎠ ⎭ 2 cos x, ⎝ 2⎠, ⎝ 2⎠, ⎝ 2⎠, x, x, ⎩, 132. cot − tan =, =, =, = 2 cot x., −, x, x, x⎞, x⎞, sin x, 2, 2, ⎛, ⎛, 2 sin cos, sin ⎜ ⎟ cos ⎜ ⎟, ⎝ 2⎠, ⎝ 2⎠, 2, 2, ( 5 − 1), ⋅, 4, ⎧⎪, ( 5 − 1) 2 ⎪⎫ ⎧, ( 6 − 2 5 ) ⎫ ( 10 + 2 5 ), 134. cos 2 18 ° = ( 1 − sin 2 18 ° ) = ⎨ 1 −, ⎬ = ⎨1 −, ⎬=, 16, 16, 16, ⎪⎩, ⎪⎭ ⎩, ⎭, 133. Remember sin 18 ° =, , ⇒ cos 18 ° =, , 10 + 2 5, ⋅, 4, , ⎧⎪, ( 5 − 1) 2 ⎫⎪ ⎧, ( 6 − 2 5 ) ⎫ ( 5 + 1), 135. cos 36 ° = ( 1 − 2 sin 2 18 ° ) = ⎨ 1 − 2 ⋅, ⋅, ⎬ = ⎨1 −, ⎬=, 16, 8, 4, ⎭, ⎩⎪, ⎭⎪ ⎩, ( 5 + 1) 2 ⎪⎫ ( 10 − 2 5 ), ⎪⎧, 136. sin 2 36 ° = ( 1 − cos 2 36 ° ) = ⎨ 1 −, ⇒ sin 36 ° =, ⎬=, 16, 16, ⎪⎩, ⎪⎭, ( 5 + 1), ⋅, 4, ( 5 − 1), 138. cos 72 ° = cos ( 90 ° − 18 ° ) = sin 18 ° =, ⋅, 4, , 137. sin 54 ° = sin ( 90 ° − 36 ° ) = cos 36 ° =, , 139. cos 54 ° = cos ( 90 ° − 36 ° ) = sin 36 ° =, , 10 − 2 5, ⋅, 4, , 140. sin 72 ° = sin ( 90 ° − 18 ° ) = cos 18 ° =, , 10 + 2 5, ⋅, 4, , 141. Using 2 sin A cos A = sin 2 A , we get, 1, 1, 45 °, 1, 2 sin 22 ° cos 22 ° = sin ⎛⎜ 2 × ⎞⎟ = sin 45 ° =, ⋅, ⎝, 2, 2, 2⎠, 2, 142. Using ( 2 cos 2θ − 1) = cos 2θ, we get, ( 2 cos 2 15 ° − 1) = cos ( 2 × 15 ° ) = cos 30 ° =, , 3, ⋅, 2, , 10 − 2 5, ⋅, 4
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Objective Questions, , 1099, , 143. Using ( 3 sin θ − 4 sin 3θ) = sin 3θ, we get, ( 3 sin 40 ° − 4 sin 3 40 ° ) = sin ( 3 × 40 ° ) = sin 120 °, 3, ⋅, 2, , = sin ( 180 ° − 60 ° ) = sin 60 ° =, 144. Using ( 4 cos 3θ − 3 cos θ) = cos 3θ, we get, ( 8 cos 3 20 ° − 6 cos 20 ° ) = 2( 4 cos 3 20 ° − 3 cos 20 ° ), , 1, = 2 × cos ( 3 × 20 ° ) = 2 cos 60 ° = ⎛⎜ 2 × ⎞⎟ = 1., ⎝, 2⎠, 145. Using sin 2θ =, , 2 tan θ, 2, , ( 1 + tan θ), , a sin 2θ + b cos 2θ =, , =, , =, , and cos 2θ =, , 2 a tan θ, 2, , ( 1 + tan θ), , +, , ( 1 − tan 2θ), ( 1 + tan 2θ), , b( 1 − tan 2θ), ( 1 + tan 2θ), 2, , 2 a tan θ + b( 1 − tan θ), ( 1 + tan 2 θ), ⎛ 2 a2 a2 ⎞, ⎜, − ⎟+b, b⎠, ⎝ b, 2, , 2, , (a + b ), , , we get, , =, , =, , ⎛ 2a ×, ⎜, ⎝, , ⎛ a2, ⎞, ⎜, + b⎟ b2, ⎝ b, ⎠, 2, , 2, , (a + b ), , ⎛, a⎞, a2 ⎞, ⎟ + b⎜1− 2 ⎟, ⎠, b, b ⎠, ⎝, ⎛, a2 ⎞, ⎜1+, ⎟, b2 ⎠, ⎝, , =, , ( a2 + b 2 )b, ( a2 + b 2 ), , = b., , b2, ⎧⎪ 1, 2 ( 1 − tan 2 x ) ⎫⎪ 2 tan 2 x, 1, 2, 146. cot x − 2 cot 2 x =, = tan x., −, =⎨, −, ⎬=, 2 tan x ⎪⎭ 2 tan x, tan x tan 2 x ⎪⎩ tan x, 147. cos 2 x + 2 sin 2 x = cos 2 x − sin 2 x + 2 sin 2 x = cos 2 x + sin 2 x = 1., 2 sin x cos x cos x, sin 2 x, 148., =, =, = cot x., 1 − cos 2 x, sin x, 2 sin 2 x, sin 2 x, 2 sin x cos x sin x, 149., =, =, = tan x., 1 + cos 2 x, cos x, 2 cos 2 x, ⎛ sin 2 x ⎞, ⎜, ⎟, 2 sin x cos x sin x, tan 2 x, ⎝ cos 2 x ⎠, sin 2 x, 150., =, =, =, = tan x., =, cos x, 1 + sec 2 x ⎛, 2 cos 2 x, 1 ⎞ ( 1 + cos 2 x ), 1, +, ⎜, ⎟, cos 2 x ⎠, ⎝, sin x ( 2 sin x + 1), 1 − cos 2 x + sin x, 2 sin 2 x + sin x, = tan x., =, =, sin 2 x + cos x, 2 sin x cos x + cos x cos x ( 2 sin x + 1), π, π x, π x, sin ⎛⎜ + x ⎞⎟, 2 sin ⎛⎜ + ⎞⎟ cos ⎛⎜ + ⎞⎟, ⎝ 4 2⎠, cos x, ⎝2, ⎠, ⎝ 4 2⎠, π x, 152., = tan ⎛⎜ + ⎞⎟ ⋅, =, =, ⎝ 4 2⎠, π, x⎞, ( 1 − sin x ), ⎛, ⎞, 2⎛ π, 2 cos ⎜ + ⎟, 1 + cos ⎜ + x ⎟, ⎝ 4 2⎠, ⎝2, ⎠, , 151., , 153. (sin 2α cos 2β + cos 2α sin 2β + sin 2α sin 2β + cos 2α cos 2β ), = (sin 2α cos 2β + sin 2α sin 2β ) + (cos 2α sin 2β + cos 2α cos 2β ), = sin 2α(cos 2β + sin 2β ) + cos 2α(sin 2β + cos 2β ) = (sin 2α + cos 2α ) = 1.
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1102, , Senior Secondary School Mathematics for Class 11, , 169. cosec2θ − cot 2θ = 1 ⇒ ( cosec θ − cot θ)( cosec θ + cot θ) = 1, 1, ∴ ( cosec θ − cot θ) × 6 = 1 ⇒ ( cosec θ − cot θ) = ⋅, 6, 1, On solving ( cosec θ + cot θ) = 6 and ( cosec θ − cot θ) = , we get, 6, 37, 35, and cot θ =, cosec θ =, ⋅, 12, 12, 170. ( 3 sin θ + 5 cos θ) 2 + (5 sin θ − 3 cos θ) 2 = 9 (sin 2θ + cos 2θ) + 25(cos 2θ + sin 2θ), = ( 9 × 1) + ( 25 × 1) = 34., ⇒ (5 ) 2 + (5 sin θ − 3 cos θ) 2 = 34 ⇒ (5 sin θ − 3 cos θ) 2 = ( 34 − 25 ) = 9, ⇒ (5 sin θ − 3 cos θ) = 3., ( 1 + tan 15 ° ) (tan 45 ° + tan 15 ° ), 171., =, = tan ( 45 ° + 15 ° ) = tan 60 ° =, ( 1 − tan 15 ° ) ( 1 − tan 45 ° tan 15 ° ), , 3., , 172. Using ( 3 sin θ − 4 sin 3θ) = sin 3θ, we get, ( 3 sin 20 ° − 4 sin 3 20 ° ) = sin ( 3 × 20 ° ) = sin 60 ° =, , 3, ⋅, 2, , 173. We know that ( 4 cos 3θ − 3 cos θ) = cos 3θ., ∴ ( 4 cos 3 15 ° − 3 cos 15 ° ) = cos ( 3 × 15 ° ) = cos 45 ° =, , 1, ⋅, 2, , 174. Using 2 cos A cos B = cos ( A + B) + cos ( A − B), we get, 2 cos 75 ° cos 15 ° = cos (75 ° + 15 ° ) + cos (75 ° − 15 ° ), 1 1, = cos 90 ° + cos 60 ° = 0 + = ⋅, 2 2, 175. Using 2 sin A sin B = cos ( A − B) − cos ( A + B), we get, 2 sin 105 ° sin 75 ° = cos ( 105 ° − 75 ° ) − cos ( 105 ° + 75 ° ), = (cos 30 ° − cos 180 ° ) =, ⇒ sin 105 ° sin 75 ° =, , ⎛ 3, ⎞ ( 3 + 2), 3, + cos 0 = ⎜, + 1⎟ =, 2, 2, ⎝ 2, ⎠, , ( 3 + 2), ⋅, 4, , 176. sin 105 ° + cos 105 ° = sin 105 ° + cos ( 90 ° + 15 ° ), ( 105 ° + 15 ° ), ( 105 ° − 15 ° ), = sin 105 ° − sin 15 ° = 2 cos, sin, 2, 2, 1, 1 ⎞, 1, ⎛, = 2 cos 60 ° sin 45 ° = ⎜ 2 × ×, ⋅, ⎟=, ⎝, 2, 2⎠, 2, 177. sin θ + cos θ = 2 ⇒ sin θ + sin ( 90 ° − θ) = 2, ⇒ 2 sin, , {( 90 ° − θ) + θ}, {θ + ( 90 ° − θ)}, cos, = 2, sin θ, 2, , ⇒ 2 sin 45 ° cos ( 45 ° − θ) = 2 ⇒ 2 ×, , 1, cos ( 45 ° − θ) = 2, 2, , ⇒ cos ( 45 ° − θ) = 1 ⇒ 45 ° − θ = 0 ⇒ θ = 45 ° =, , π, ⋅, 4
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1106, , Senior Secondary School Mathematics for Class 11, , 218. cos A =, , 2 3 ( 3 + 1), ( b 2 + c 2 − a2 ) 6 + ( 3 + 1) 2 − 4, 6+ 2 3, 1, =, =, =, =, 2 bc, 2 6 ( 3 + 1), 2 6 ( 3 + 1) 2 6 ( 3 + 1), 2, , ⇒ A = 45 ° ., 219. cos A =, , ( b 2 + c 2 − a2 ) ( 6 + 8 − 2 ), 12, 3, 3, =, =, =, =, ⇒ θ = 30 ° ., 2, 2 bc, 2 6 8, 4 12 2 3, , 220. Let the angles of a ABC be x ° , 2 x ° and 3x°., Then, x + 2 x + 3 x = 180 ⇒ 6 x = 180 ⇒ x = 30., So, the angles are 30 ° , 60 ° and 90°., ∴, , ratio of sides = sin 30 ° : sin 60 ° : sin 90 ° =, , 221. cos C =, ⇒, , 1, 3, :, : 1 = 1: 3 : 2., 2 2, , ( a2 + b 2 − c 2 ), ( 4 + 2 3 + 4 − c2 ), ⇒ cos 60 ° =, 2 ab, 4( 3 + 1), , 8 + 2 3 − c2 1, = ⇒ 16 + 4 3 − 2 c 2 = 4 3 + 4, 2, 4( 3 + 1), ⇒ 2 c 2 = 12 ⇒ c 2 = 6 ⇒ c = 6 ., , 222., , 2, 3, a, b, =, ⇒ sin B = 1 ⇒ B = 90 ° ., =, ⇒, sin A sin B, ⎛ 2 ⎞ sin B, ⎜ ⎟, ⎝ 3⎠, , 223. A, B, C are in AP ⇒ 2 B = A + C ⇒ 3 B = A + B + C = 180 ° ⇒ B = 60 ° ., ∴, , b, c, 3, 2, =, ⇒, =, ⇒, sin B sin C, sin 60 ° sin C, ⇒ sin C =, , 224. Let b = 2 k and c =, cos A =, , 2, , 2, ×, 3, , 3=, , 2, sin C, , 2, 1, =, ⇒ C = 45 ° ., 2, 2, , 3 k. Then,, 2, , ( b + c − a2 ), 4 k 2 + 3 k 2 − a2, ⇒ cos 30 ° =, 2 bc, 4 3k 2, , ∴ 7 k 2 − a2 = 4 3 k 2 ×, , 3, 2, , ⎡, 3⎤, ⎢Q cos 30 ° =, ⎥, 2, ⎣, ⎦, , ⇒ k 2 = a 2 ⇒ a = k ⇒ b = 2 a, c =, , 3 a., , a, b, a, 2a, Now,, =, ⇒, =, ⇒ sin B = 1 ⇒ B = 90 ° ., sin A sin B, sin 30 ° sin B, 225. Let the sides be ( a − d ), a and ( a + d ). Then,, ( a + d ) 2 = ( a − d ) 2 + a2 ⇒ a2 = 4 ad ⇒ a = 4 d., ∴, , the sides are 3 d , 4 d , 5 d. Their ratio is 3 : 4 : 5., , 226. a : b : c = sin A : sin B : sin C = sin 45 ° : sin 60 ° : sin 75 °, =, , 1, 3 ( 3 + 1), :, :, = 2 : 6 : ( 3 + 1)., 2 2, 2 2
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Objective Questions, , 1107, , 227. ∠A = 30 ° , ∠C = 105 ° , ∠B = 180 ° − ( 30 ° + 105 ° ) = 45 ° ., ∴, , a, b, a, 3 2, =, ⇒, =, ⇒ 2 a = 6 ⇒ a = 3., sin A sin B, sin 30 ° sin 45 °, , 228. cos B =, ∴, , a2 + c 2 − b 2, 25 + 8 − b 2, ⇒ cos 45 ° =, ⇒, 2 ac, 2×5× 2 2, , 1, 33 − b 2, =, 2, 20 2, , 33 − b 2 = 20 ⇒ b 2 = 13 ⇒ b = 13 ., , 229. ∠B = 30 ° , ∠C = 45 ° , ∠A = 180 ° − ( 30 ° + 45 ° ) = 105 ° ., ∴, , ∴, , ( 3 + 1), ( 3 + 1), a, c, c, c, =, ⇒, =, ⇒, =, sin A sin C, sin 105 ° sin 45 °, cos 15 °, ⎛ 1 ⎞, ⎜, ⎟, ⎝ 2⎠, c = ( 3 + 1) ⋅, , 1, 2 2, ⋅, = 2., 2 ( 3 + 1), , 230. b = 5 , c = 5 3 and ∠A = 30 ° ., cos A =, , 25 + 75 − a2, b 2 + c 2 − a2, ⇒ cos 30 ° =, 2 bc, 50 3, ⇒, , 100 − a2, 3, =, ⇒ 200 − 2 a2 = 150 ⇒ a2 = 25 ⇒ a = 5 ., 2, 50 3, , Thus, a = b and hence a ABC is isosceles., 231. Let the angles be 3 x ° , 4 x ° and 5x°. Then,, 3 x + 4 x + 5 x = 180 ⇒ 12 x = 180 ⇒ x = 15., ∴, , ∠A = 45 ° , ∠B = 60 ° and ∠C = 75 °., , ∴, , a : c = sin 45 ° : sin 75 ° =, , 1, 3+1, :, = 2 : ( 3 + 1)., 2 2 2, , 232. Here, b = 20 , c = 15 , ∠A = 150 ° ., ∴ a=, 233. s =, , 1, 1, bc sin A = × 20 × 15 × sin 150 ° = 150 × sin 30 ° = ⎛⎜ 150 ×, ⎝, 2, 2, , 1⎞, ⎟ = 75 sq units., 2⎠, , 1, ( 18 + 24 + 30 ) = 36 , (s − a) = ( 36 − 18 ) = 18 , (s − b ) = ( 36 − 24 ) = 12, 2, , and (s − c ) = ( 36 − 30 ) = 6., ∴ a = s(s − a)(s − b )(s − c ) =, =, , 36 × 18 × 12 × 6, 36 × 36 × 36 = ( 36 × 6 ) = 216 sq units., , 234. Here, a = 16 , c = 9 and ∠B = 30 ° ., ∴ a=, 235. s =, , 1, 1, 1, ac sin B = ⎛⎜ × 16 × 9 × sin 30 ° ⎞⎟ = ⎛⎜ 72 × ⎞⎟ = 36 sq units., ⎝2, ⎠ ⎝, 2⎠, 2, , 1, ( 15 + 14 + 13 ) = 21, (s − a) = 6 , (s − b ) = 7 , (s − c ) = 8., 2, , ∴ a = s(s − a)(s − b )(s − c ) = 21 × 6 × 7 × 8 = ( 21 × 4 ) = 84 sq units., ∴, , R=, , abc 15 × 14 × 13 65, =, =, ⋅, 4a, 4 × 84, 8
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1108, , Senior Secondary School Mathematics for Class 11, , 236. s =, , 1, ( 18 + 24 + 30 ) = 36 , (s − a) = ( 36 − 18 ) = 18 , (s − b ) = ( 36 − 24 ) = 12, 2, , and (s − c ) = ( 36 − 30 ) = 6., ∴ a = s(s − a)(s − b )(s − c ) =, ∴, , 36 × 18 × 12 × 6 = 216 sq units., , abc ( 18 × 24 × 30 ), = 15., R=, =, 4a, 4 × 216, , Area of the circumcircle = πR 2 = ( π × 15 × 15 ) = ( 225 π ) sq units., 237. s =, , 1, ( 4 + 13 + 15 ) = 16 , (s − a) = 12 , (s − b ) = 3 , (s − c ) = 1., 2, , ∴ a = s(s − a)(s − b )(s − c ) = 16 × 12 × 3 × 1 = 24 sq units., ∴ r=, 238. s =, , a 24 3, =, = ⋅, s 16 2, , 1, ( 13 + 14 + 15 ) = 21, (s − a) = 8 , (s − b ) = 7 , (s − c ) = 6 ., 2, , ∴ a = s(s − a)(s − b )(s − c ) = 21 × 8 × 7 × 6 = 21 × 21 × 16 = ( 21 × 4 ) = 84 ., ∴ r=, , a 84, =, = 4⋅, s 21, , Area of the incircle = πr 2 = ( π × 4 × 4 ) = ( 16 π ) sq units., 239. r = 4 R sin, , A, B, C, sin sin, 2, 2, 2, , ⇒ r = 4 R sin 30 ° ⋅ sin 30 ° ⋅ sin 30 °, 1 1 1, ⇒ r = 4R × × × ⇒ R = 2 r ., 2 2 2, 240. a =, ∴, , [Q A = B = C = 60 ° ], , ⎛1, 3⎞, 1, × a × a × sin 60 ° = ⎜ × 2 3 × 2 3 ×, ⎟ = 3 3., 2, 2, 2, ⎝, ⎠, R=, , ( 2 3 )3, abc a3, 24 3, =, =, =, = 2., 4a 4a 4 × 3 3 12 3, , 241. Let a = 3 k , b = 7 k , c = 8 k. Then, s = 9 k , (s − a) = 6 k , (s − b ) = 2 k , (s − c ) = k., R abc s, abc, 3k × 7 k × 8k ⎞ 7, =, ⋅ =, = ⎜⎛, ⎟= ⋅, r, 4a a 4(s − a)(s − b )(s − c ) ⎝ 4 × 6 k × 2 k × k ⎠ 2, Hence, R : r = 7 : 2 ., 242. s =, , 27, cm and a = 81 cm 2 ., 2, , ∴ r=, , a ⎛, 2 ⎞, = ⎜ 81 ×, ⎟ = 6 cm., 27 ⎠, s ⎝, , 243. Let C = 90 °. Then, B = ( 90 ° − A )., sin 2 A + sin 2 B + sin 2C = sin 2 A + sin 2 ( 90 ° − A ) + sin 2 90 °, = (sin 2 A + cos 2 A + 1) = 2 .
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Objective Questions, , 244., , 1109, , sin ( B − C ) sin ( B − C ) ⋅ sin ( B + C ) sin 2 B − sin 2 C, =, =, sin ( B + C ), sin 2 ( B + C ), sin 2 ( π − A ), =, , sin 2 B − sin 2C, 2, , sin A, , =, , ⎡, a, b, c ⎤, =, =, ⎥⋅, ⎢Q, ⎣ sin A sin B sin C ⎦, , ( b 2 − c2 ), a2, , 245. a( b cos C − c cos B) = ( b cos C + c cos B)( b cos C − c cos B), = ( b 2 cos 2C − c 2 cos 2 B) = b 2 ( 1 − sin 2C ) − c 2 ( 1 − sin 2 B), ⎡, b, c ⎤, = ( b 2 − c 2 ) + ( c 2 sin 2 B − b 2 sin 2C ) = ( b 2 − c 2 ) ⎢Q, =, ⎥⋅, sin, B, sin, C⎦, ⎣, 246., , 1 + cos ( A − B) cos C 1 + cos ( A − B) cos [π − ( A + B)], =, 1 + cos ( A − C ) cos B 1 + cos ( A − C ) cos [π − ( A + C )], =, =, , 1 − cos ( A − B) cos ( A + B) 1 − (cos 2 A − sin 2 B), =, 1 − cos ( A − C ) cos ( A + C ) 1 − (cos 2 A − sin 2C ), sin 2 A + sin 2 B, 2, , 2, , sin A + sin C, , 247. Given exp. = ( a2 + b 2 − 2 ab ) cos 2, , =, , a2 + b 2, a2 + c 2, , ⋅, , C, C, + ( a2 + b 2 + 2 ab ) sin 2, 2, 2, , C, C, C, C, C, C, = a2 ⎛⎜ cos 2 + sin 2 ⎞⎟ + b 2 ⎛⎜ cos 2 + sin 2 ⎞⎟ − 2 ab ⎛⎜ cos 2 − sin 2 ⎞⎟, ⎝, ⎝, ⎝, 2, 2⎠, 2, 2⎠, 2, 2⎠, = a2 + b 2 − 2 ab cos C, = a2 + b 2 − 2 ab ⋅, , ⎡, ⎤, 2C, 2C, ⎢⎣Q cos 2 − sin 2 = cos C ⎥⎦, , ( a2 + b 2 − c 2 ), = c2., 2 ab, , 248. Given exp. = ( b cos C + c cos B) + ( c cos A + a cos C ) + ( a cos B + b cos A ), = ( a + b + c), , [by projection formulae]., , 249. Given exp. = 2 bc cos A + 2 ca cos B + 2 ab cos C, = 2 bc ⋅, , ( b 2 + c 2 − a2 ), ( c 2 + a2 − b 2 ), ( a2 + b 2 − c 2 ), + 2 ca ⋅, + 2 ab ⋅, 2ab, 2 bc, 2 ca, , = ( b 2 + c 2 − a2 ) + ( c 2 + a2 − b 2 ) + ( a2 + b 2 − c 2 ) = ( a2 + b 2 + c 2 )., 250. Given exp. = ac cos B − bc cos A = ac ⋅, =, 251., , ( c 2 + a2 − b 2 ), ( b 2 + c 2 − a2 ), − bc ⋅, 2 ac, 2 bc, , ( c 2 + a2 − b 2 ) − ( b 2 + c 2 − a2 ), = ( a2 − b 2 )., 2, , a − b k sin A − k sin B k(sin A − sin B ), =, =, a + b k sin A + k sin B k(sin A + sin B), =, , ( A + B), ( A − B), ( A − B), 2 cos, sin, tan, (sin A − sin B ), 2, 2, 2, =, =, ⋅, (sin A + sin B ) 2 sin ( A + B) cos ( A − B) tan ( A + B), 2, 2, 2
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1110, , Senior Secondary School Mathematics for Class 11, , 252. c cos 2, , s(s − c ), s(s − a), A, C, + a cos 2 = c ⋅, + a⋅, ab, bc, 2, 2, ⎧ s(s − a) s(s − c ) ⎫ s, =⎨, +, ⎬ = ⋅ {(s − a) + (s − c )}, b ⎭ b, ⎩ b, , s, s, s, ⋅ {2s − ( a + c )} = ⋅ {( a + b + c ) − ( a + c )} = × b = s., b, b, b, ( b − c), ( b − c ) s(s − a) s(s − a)( b − c ), 2 A, 253., ⋅ cos, =, ⋅, =, ⋅, a, a, bc, abc, 2, ( c − a), B ( c − a) s(s − b ) s(s − b )( c − a), Similary,, ⋅ cos 2 =, ⋅, =, ⋅, b, b, ac, abc, 2, ( a − b), C ( a − b ) s(s − c ) s(s − c )( a − b ), And,, ⋅ cos 2 =, ⋅, =, ⋅, c, c, ab, abc, 2, s(s − a)( b − c ) s(s − b )( c − a) s(s − c )( a − b ), ∴ given exp. =, +, +, abc, abc, abc, s, =, ⋅ [(s − a)( b − c ) + (s − b )( c − a) + (s − c )( a − b )], abc, s, =, ⋅ [s( b − c + c − a + a − b ) − {a( b − c ) − b( c − a) − c( a − b )}] = 0., abc, ⎡ s(s − a) s(s − b ) s(s − c ) ⎤, 254. Given exp. = a ⋅ ⎢, +, +, a, a ⎥⎦, ⎣ a, =, , = s(s − a) + s(s − b ) + s(s − c ) = s ⋅ [(s − a) + (s − b ) + (s − c )], = s[ 3s − ( a + b + c )] = s[ 3s − 2s] = (s × s ) = s 2 ., 255. bc cos 2, , s(s − a), s(s − b ), s(s − c ), A, B, C, + ca cos 2 + ab cos 2 = bc ⋅, + ca ⋅, + ab ⋅, 2, 2, 2, bc, ca, ab, = s(s − a) + s(s − b ) + s(s − c ), = s ⋅ [(s − a) + (s − b ) + (s − c )] = s ⋅ [ 3s − ( a + b + c )], = s ⋅ [ 3s − 2s] = (s × s ) = s 2 ., , 256. 2 abc cos, , s(s − a) s(s − b ) s(s − c ), A, B, C, cos cos = 2 abc ⋅, ⋅, ⋅, 2, 2, 2, bc, ac, ab, , 1, ⋅ s(s − a)(s − b )(s − c ) ⋅ s = 2sa = ( a + b + c )a ., abc, s(s − a) b + c, A b+ c, 257. cos 2 =, ⇒, =, ⇒ 2s( 2s − 2 a) = 2 b( b + c ), bc, 2, 2c, 2c, = 2 abc ⋅, , ⇒ ( a + b + c )( a + b + c − 2 a) = 2 b( b + c ), ⇒ ( b + c + a)( b + c − a) = 2 b 2 + 2 bc, ⇒ ( b + c ) 2 − a2 − 2 b 2 − 2 bc = 0, ⇒ c 2 − a2 = b 2 ⇒ a2 + b 2 = c 2 ., A, B, C, 258. cot : cot : cot = 3 : 5 : 7, 2, 2, 2, s(s − a) s(s − b ) s(s − c ), ⇒, :, :, = 3:5:7, a, a, a
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Objective Questions, , 1111, , ⇒ ( s − a) : ( s − b ) : ( s − c ) = 3 : 5 : 7, b + c− a c+ a− b a+ b − c, :, :, = 3:5:7, ⇒, 2, 2, 2, ⇒ ( b + c − a) : ( c + a − b ) : ( a + b − c ) = 3 : 5 : 7., Let b + c − a = 3 k , c + a − b = 5 k and a + b − c = 7 k., Adding, we get: a + b + c = 15 k., ∴, , 2 a = 12 k , 2 b = 10 k and 2 c = 8 k, , ⇒ a = 6 k , b = 5 k and c = 4 k, ⇒ a : b : c = 6 : 5 : 4., 1, ( 16 + 24 + 20 ) = 30 , (s − a) = 14 , (s − b ) = 6 and (s − c ) = 10., 2, , 259. s =, ∴, , cos, , s(s − b ), B, =, =, 2, ca, , 30 × 6, =, 20 × 16, , 9, 3, = ⋅, 16 4, , 260. a cos A = b cos B ⇒ k sin A cos A = k sin B cos B, ⇒ sin A cos A = sin B cos B ⇒ 2 sin A cos A = 2 sin B cos B, ⇒ sin 2 A − sin 2 B = 0 ⇒ 2 cos ( A + B) sin ( A − B) = 0, ⇒ cos ( A + B) = 0 or, ⇒ A + B = 90 ° or, , sin ( A − B) = 0, , A − B = 0 ⇒ ∠C = 90 ° or, , ⇒ a ABC is either right angled or isosceles., , q, , A=B
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Coordinate Geometry, SUMMARY OF THE RESULTS, 1. If A( x1 , y1 ) and B( x 2 , y 2 ) be two given points, then, AB 2 = {( x2 − x1 )2 + (y 2 − y1 )2 }., , 2., , (i) Let A( x1 , y1 ), B( x 2 , y 2 ) and C( x 3 , y 3 ) be the vertices of a ABC, then, a=, , 1, ⋅|x1 (y 2 − y 3 ) + x2 (y 3 − y1 ) + x3 (y1 − y 2 )| sq units., 2, , (ii) Points, A( x1 , y1 ), B( x 2 , y 2 ), C( x 3 , y 3 ), ⇔ ar(a ABC) = 0., 3., , are, , collinear, , (i) If P( x , y) divides the join of A( x1 , y1 ) and B( x 2 , y 2 ) in the ratio m : n,, (mx 2 + nx1 ), (my 2 + ny1 ), then x =, ,y=, ⋅, (m + n), (m + n), (ii) If P( x , y) is the midpoint of the join of A( x1 , y1 ) and B( x 2 , y 2 ), then, ( x + x2 ), ( y + y2 ), and y = 1, x= 1, ⋅, 2, 2, , 4. A quadrilateral is a, (i) rectangle, if its opposite sides are equal and diagonals are equal., (ii) square, if its all sides are equal and diagonals are equal., (iii) ||gm but not a rectangle, if its opposite sides are equal and diagonals, are not equal., (iv) rhombus but not a square, if its all sides are equal and the diagonals, are not equal., 5., , (i) Inclination of a line:, The angleθ which the given line makes with the positive direction of, the x-axis. Clearly, 0° ≤ θ ≤ 180°., (ii) Slope of a line: m = tan θ., (iii) Slope of a horizontal line is 0., (iv) Slope of a vertical line is not defined., (v) Slope of a line passing through the points A( x1 , y1 ) and B( x 2 , y 2 ) is, ( y − y1 ), m= 2, ⋅, ( x 2 − x1 ), (vi) L 1|| L 2 ⇔ m1 = m2 ., (vii) L 1 ⊥ L 2 ⇔ m1 m2 − 1., (viii) tan α =, , m1 − m2, , where 1 + m1 m2 ≠ 0., 1 + m1 m2, 1112
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Objective Questions, , 6., , 1113, , (i) Equation of the x-axis is y = 0., (ii) Equation of the y-axis is x = 0., (iii) Equation of a line parallel to the y-axis, x = a ., (iv) Equation of a line parallel to the x-axis, y = b., , 7. Equation of a line, (i) Point-slope form: m =, (ii) Two-point form:, , ( y − y1 ), ⋅, ( x − x1 ), , ( y − y1 ) ( y 2 − y1 ), =, ⋅, ( x − x1 ) ( x 2 − x1 ), , (iii) Slope-intercept form: y = mx + c., (iv) Slope = m , x-intercept = d, then y = m( x − d)., (v) Intercept form:, , x y, + = 1., a b, , (vi) Normal form: x cos α + y sin α = p., 8. Reduction of general form Ax + By + C = 0 to, ( −A), ( − C), ⋅x +, ⋅, B, B, x, y, (ii) intercept form:, +, = 1., −, C, −, ⎛, ⎞ ⎛ C⎞, ⎜, ⎟ ⎜, ⎟, ⎝ A⎠ ⎝ B⎠, (i) slope intercept form: y =, , (iii) normal form:, 9., , − Ax, A +B, 2, , 2, , − By, , +, , A +B, 2, , 2, , C, , =, , A + B2, 2, , ⋅, , (i) Length of perpendicular from a point P( x1 , y1 ) on a line, Ax + By + C = 0 is given by, d=, , |Ax1 + By1 + C|, A 2 + B2, , (ii) Distance between two parallel, Ax + By + C 2 = 0 is given by, d=, , lines, , |C 2 − C1|, A 2 + B2, , ⋅, , Ax + By + C1 = 0 and, , ⋅, , (iii) Distance between two parallel lines y = mx + C1 and y = mx + C 2 is, given by, d=, , |C 2 − C1|, ⋅, 1 + m2
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1114, , 10., , Senior Secondary School Mathematics for Class 11, , (i) Equation of a circle with centre C( h, k) and radius = r is given by, ( x − h) 2 + ( y − k) 2 = r 2 ., (ii) Equation of a circle described on the line joining A( x1 , y1 ) and, B( x 2 , y 2 ) as diameter is ( x − x1 )( x − x 2 ) + ( y − y1 )( y − y 2 ) = 0., , 11. Parabolas, I. Right-handed parabola (y 2 = 4ax, a > 0), (i) Vertex is O( 0, 0)., (ii) Focus is F( a , 0)., (iii) Directrix is x + a = 0., (iv) Axis is y = 0., (v) Length of latus rectum = 4a., (vi) Equation of latus rectum is, x = a., , II. Left-handed parabola (y 2 = −4ax, a > 0), (i) Vertex is O( 0, 0)., (ii) Focus is F( − a , 0)., (iii) Directrix is x − a = 0., (iv) Axis is y = 0., (v) Length of latus rectum = 4a., (vi) Equation of latus rectum is, x = − a., , III. Upward parabola ( x2 = 4ay , a > 0), (i) Vertex is O( 0, 0)., (ii) Focus is F( 0, a)., (iii) Directrix is y + a = 0., (iv) Axis is x = 0., (v) Length of latus rectum = 4a., (vi) Equation of latus rectum is y = a.
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Objective Questions, , 1115, , IV. Downward parabola ( x2 = −4ay , a > 0), (i) Vertex is O( 0, 0)., (ii) Focus is F( 0, − a)., (iii) Directrix is y = a., (iv) Axis is x = 0., (v) Length of latus rectum = 4a., (vi) Equation of latus rectum is y = − a., , MAIN FACTS ABOUT ALL TYPES OF PARABOLAS, Parabola, (Equation), , Focus Vertex Equation Equation Length Equation, of, of axis, of latus of latus, directrix, rectum, rectum, , (i) y 2 = 4ax , a > 0, (Right-handed), , ( a , 0), , (0, 0), , x+a=0, , y=0, , 4a, , x−a = 0, , (ii) y 2 = − 4ax , a > 0, (Left-handed), , ( −a , 0) (0, 0), , x−a = 0, , y=0, , 4a, , x+a=0, , (iii) x 2 = 4ay , a > 0, (Upward), , ( 0, a), , (0, 0), , y+a=0, , x=0, , 4a, , y−a = 0, , ( 0, −a) (0, 0), , y−a = 0, , x=0, , 4a, , y+a=0, , (iv) x 2 = − 4ay , a > 0, (Downward), 12. Ellipse, , HORIZONTAL ELLIPSE, 2, , 2, , x, y, +, = 1 , b < a., a2 b 2, , VERTICAL ELLIPSE, , x2 y2, +, = 1 , b < a., b 2 a2
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1116, , Senior Secondary School Mathematics for Class 11, , MAIN FACTS ABOUT HORIZONTAL AND VERTICAL ELLIPSES, Properties, , Horizontal Ellipse, x2 y 2, +, =1, a2 b 2, 0 < b < a and, c2 = ( a2 − b 2 ), , Vertical Ellipse, x2 y 2, +, =1, b 2 a2, 0 < b < a and, c2 = ( a2 − b 2 ), , (i) Centre, , (0, 0), , (0, 0), , (ii) Vertices, , A( − a , 0), B( a , 0), , A( 0, − a), B( 0, a), , F1 ( − c, 0) and F2 ( c, 0) or, ( −ae, 0) and ( ae, 0), , F1 ( 0, − c) and F2 ( 0, c) or, ( 0, −ae) and ( 0, ae), , (iv) Length of the major, axis, , 2a, , 2a, , (v) Length of the minor, axis, , 2b, , 2b, , (iii) Foci, , (vi) Equation of, major axis, , the, , y=0, , x=0, , (vii) Equation of, minor axis, , the, , x=0, , y=0, , 2b 2, a, , 2b 2, a, , (viii) Length of the latus, rectum, e=, , (ix) Eccentricity, , c, =, a, , a 2 − b2, a, , 13. Hyperbola, I. Horizontal hyperbola, (i) Its equation is, , x2 y 2, −, = 1., a2 b 2, , (ii) Centre is O( 0, 0)., (iii) X′ OX, is, the, transverse axis and, YOY′ is the conjugate, axis., (iv) Length of the transverse axis = 2a., (v) Length of the conjugate axis = 2b., (vi) Foci are F1 ( − c, 0) and F2 ( c, 0), where c = ae., (vii) Vertices are A( − a , 0) and B( a , 0)., , e=, , c, =, a, , a 2 − b2, a
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Objective Questions, , (viii) Eccentricity, e =, , c, =, a, , 1117, , a 2 + b2, ⋅, a, , (ix) Length of latus rectum =, , 2b 2, ⋅, a, , II. Vertical hyperbola, (i) Its equation is, , y2 x2, −, = 1., a 2 b2, , (ii) Its centre is O( 0, 0)., (iii) YOY′ is the transverse axis and X′ OX is the, conjugate axis., (iv) Length of transverse axis = 2a., (v) Length of conjugate axis = 2b., (vi) Its vertices are A( 0, − a) and B( 0, a)., (vii) Its foci are F1 ( 0, − c) and F2 ( 0, c), where c = ae., (viii) Eccentricity, e =, , c, =, a, , a 2 + b2, ⋅, a, , (ix) Length of latus rectum =, , 2b 2, ⋅, a, , EXERCISE 1, Mark (3) against the correct answer in each of the following., , 1. The distance between the points A( 2, − 3) and B( −6, 3) is, (a) 4 units, (b) 6 units, (c) 8 units, (d) 10 units, 2. The distance between the points P( 5 + 1,, (a) 3 2 units, , (b) 4 3 units, , 3 − 1) and Q( 5 − 2,, , (c) 3 5 units, , 3 + 2) is, , (d) 2 6 units, , 3. If the distance between the points A( −3 , 4) and B( x , 7) is 5 units, then x = ?, (a) –1 or 7, (b) 1 or –7, (c) 5 or –3, (d) –5 or 3, 4. If the distance between the points A( x , 3) and B( 3 , 5) is 4 units, then x = ?, (a) 2 ± 3 3, , (b) 3 ± 2 3, , (c) 3 ± 3 2, , (d) 2 ± 2 2, , 5. The distance of the point A( 6, − 6) from the origin is, (a) 2 3 units, , (b) 6 units, , (c) 12 units, , (d) 6 2 units, , 6. If P( x , y) is equidistant from A( 6, −1) and B( 2, 3), then, (a) 3 x − y = 3, (b) x − 3 y = 3, (c) x − y = 3, (d) x + y = 3, 7. A point P on the x-axis which is equidistant from the points A(7 , 6) and, B( −3 , 4) is, (a) P( 3 , 0), (b) P( 0, 3), (c) P( −3 , 0), (d) P( 0, − 3)
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1118, , Senior Secondary School Mathematics for Class 11, , 8. A point P on the y-axis which is equidistant from the points A( −4, 3) and, B(5 , 2) is, (a) P( −2, 0), , (b) P( 0, − 2), , (c) P( 2, 0), , (d) P( 0, 2), , 9. A is a point on the x-axis with abscissa –8 and B is a point on the y-axis with, ordinate 15. Distance AB = ?, (a) 13 units, , (b) 15 units, , (c) 17 units, , (d) 23 units, , 10. If the points A( −2, 3), B(1, 2) and C( k , 0) are collinear, then k = ?, (a) 5, , (b) 6, , (c) 7, , (d) 8, , 11. The vertices of a a ABC are A( 3 , 8), B( −4, 2) and C(5 , −1). The area of, a ABC is, 1, 1, (a) 57 sq units, (b) 75 sq units, (c) 28 sq units (d) 37 sq units, 2, 2, 12. The area of quadrilateral ABCD with A(1, 1), B(7 , − 3), C(12, 2) and D(7 , 21), as its vertices is, (a) 35 sq units, , (b) 65 sq units, , (c) 85 sq units, , (d) 115 sq units, , 13. In a ABC having vertices A( −1, 3), B(1, −1) and C(5 , 1), the length of the, median AD is, (a) 3 units, , (b) 4 units, , (c) 5 units, , (d) 6 units, , 14. In a ABC, its vertices are A(7 , − 3), B( 3 , −1) and C(5 , 3). If BE is one of its, medians, then BE = ?, (a) 2 2 units, , (b) 3 units, , (c) 10 units, , (d) none of these, , 15. In a ABC with vertices A( −1, 0), B(5 , − 2) and C( 8, 2), the centroid is, (a) (4, 0), , (b) (0, 4), , (c) (6, 0), , (d) (0, 6), , 16. In a ABC having vertices A(5 , 2) and B( −7 , − 4), the centroid is G( −2, 3)., Then, the third vertex C is, (a) C( −6, 9), , (b) C( 8, 11), , (c) C( −4, 11), , 17. The points A(1, 1), B( −1, −1) and C( − 3 ,, , (d) C( −6, 11), , 3 ) are the vertices of, , (a) an equilateral triangle, , (b) an isosceles triangle, , (c) a right triangle, , (d) none of these, , 18. The points A( 2a , 4a), B( 2a , 6a) and C{( 2 +, , 3 ) a , 5 a} are the vertices of, , (a) an isosceles triangle, , (b) an equilateral triangle, , (c) a right triangle, , (d) none of these, , 19. The points A(7 , 10), B( −2, 5) and C( 3 , − 4) are the vertices of, (a) an isosceles right triangle, , (b) a scalene right triangle, , (c) an equilateral triangle, , (d) none of these, , 20. The points A( 2, 6), B(5 , 1), C( 0, − 2) and D( −3 , 3) are the vertices of, (a) a rhombus, , (b) a square, , (c) a trapezium, , (d) none of these
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Objective Questions, , 1119, , 21. The points A( −1, 0), B( 0, 3), C(1, 3) and D( 0, 0) are the vertices of, (a) a parallelogram, , (b) a rectangle, , (c) a rhombus, , (d) none of these, , 22. Three consecutive vertices of a parallelogram ABCD are A( 3 , − 2), B( 4, 0), and C( 6, − 3). The 4th vertex D is, (a) D(5 , −5), , (b) D( −5 , 5), , (c) D( −3 , 8), , (d) D( 3 , − 8), , 23. The vertices of a a ABC are A( 8, 6), B( 8, − 2) and C( 2, − 2). The coordinates, of its circumcentre are, (a) (5, –2), , (b) (5, 2), , (c) (–5, 2), , (d) (–5, –2), , 24. A point P divides the join of A(5 , − 2) and B( 9, 6) in the ratio 3 : 1. The, coordinates of P are, ⎛7 ⎞, (d) (8, 4), (a) (4, –7), (b) (–4, 7), (c) ⎜ , 4⎟, ⎝2 ⎠, 25. The ratio in which the point P(1, 2) divides the join of A( −2, 1) and B(7 , 4) is, (a) 1 : 2, , (b) 2 : 1, , (c) 3 : 2, , (d) 2 : 3, , 26. The ratio in which the line segment joining the points A( −12, 2) and B( 8, 3), is divided by the y-axis is, (a) 2 : 1, , (b) 1 : 2, , (c) 2 : 3, , (d) 3 : 2, , 27. The ratio in which the line segment joining the points P( 4, 5) and, Q( −10, − 2) is divided by the x-axis is, (a) 3 : 2, , (b) 2 : 3, , (c) 5 : 2, , (d) 2 : 5, , 28. The vertices ofa ABC are A( 0, 6), B( 8, 12) andC( 8, 0). The coordinates of its, incentre are, ⎛ 16 ⎞, (a) (– 4, 3), (b) (5, 6), (c) (8, 11), (d) ⎜ , 6⎟, ⎝ 3 ⎠, 29. A circle passes through the points A( 2, − 9), B(5 , − 8) and C( 2, 1). The centre, of the circle is, ⎛ −16⎞, (a) (2, – 4), (b) (–3, 4), (c) ⎜ 3 ,, (d) none of these, ⎟, ⎝, 3 ⎠, 30. One end of the diameter of a circle is A( 4, 1) and its centre is C( 3 , 3). The, coordinates of the other end of the diameter is, (a) B(5 , 2), , (b) B( −4, 1), , (c) B( 4, − 1), , (d) B( 2, 5), , 31. The slope of the line AB passing through the points A( −2, 3) and B( 8, −5) is, 4, −4, 5, −5, (a), (b), (c), (d), 5, 5, 4, 4, 32. The inclination of the line joining the points A( x , − 3) and B( 2, 5) is 135°., Then, the value of x is, (a) 8, , (b) –8, , (c) 10, , (d) –10
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1120, , Senior Secondary School Mathematics for Class 11, , 33. The angle between the x-axis and the line joining the points A( 3 , −1) and, B( 4, − 2) is, (a) 45°, , (b) 90°, , (c) 135°, , (d) 150°, , 34. If the line AB with A( −2, 6) and B( 4, 8) is perpendicular to the line CD with, C( 8, 12) and D( x , 24), then x = ?, (a) –4, , (b) 4, , (c) –6, , (d) 6, , 35. If the points A( x , −1), B( 2, 1) and C( 4, 5) are collinear, then x = ?, (a) 1, , (b) –2, , (c) –4, , (d) –5, , 36. If A( −2, 1), B( 2, 3) and C( −2, − 4) be the vertices of a a ABC, then tan B = ?, 1, 3, 2, 4, (a), (b), (c), (d), 3, 4, 3, 5, 37. If the angle between two lines is, slope of the other line is, −1, −1, (b) 3 or, (a) 2 or, 2, 3, , π, 1, and the slope of one line is , then the, 4, 2, (c) 4 or, , −1, 4, , (d), , 1, −1, or, 2, 3, , 38. If θ is the angle between AB and CD with A( 0, 0), B( 2, 3), C( 2, − 2) and, D( 3 , 5), then tan θ = ?, 9, 11, 8, 10, (b), (c), (d), (a), 13, 23, 17, 19, 39. The slopes of two lines AB and CD are ( 2 − 3 ) and ( 2 +, The angle between these lines is, (a) 30°, , (b) 45°, , (c) 60°, , 3 ) respectively., , (d) 120°, , 40. If the slope of the line joining the points A( x , 2) and B( 6, − 8) is, x =?, (a) –2, , (b) 2, , (c) –3, , −5, , then, 4, , (d) 3, , 41. If A( 3 , x), B( 2, 7), C( −1, 4) and D( 0, 6) are the points such that AB ||CD ,, then x = ?, (a) 6, , (b) 8, , (c) 9, , (d) 12, , 42. If A( −2, 6), B( 4, x), C( 3 , − 3) and D(5 , − 9) are the points such that AB is, perpendicular to CD, then x = ?, (a) 9, , (b) 8, , (c) 7, , (d) 6, , 43. The vertices of a parallelogram ABCD are A( 0, 2), B( −2, −1), C( 4, 0) and, D( 2, 3) andθ is the angle between the diagonals AC and BD. Then, tan θ = ?, 3, 2, (b), (c) 2, (d) 3, (a), 2, 3, 44. If A( −1, 8), B( 4, − 2) and C( −5 , − 3) are the vertices of a a ABC, then, equation of AB is, (a) 2x − y + 6 = 0 (b) 2x + y − 6 = 0 (c) 3 x − y + 9 = 0 (d) 3 x + y − 9 = 0
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Objective Questions, , 1121, , 45. The vertices of a a ABC are A( 2, 5), B( −4, 9) and C( −2, −1). The equation of, the median BE is, (a) x − 5 y + 23 = 0 (b) 8x − y + 15 = 0 (c) 7 x + 4y − 8 = 0 (d) none of these, 46. A line passes through the point P( 0, 5) and its inclination with the x-axis is, 30°. The equation of the line is, (a) x + 3 y − 5 3 = 0, (c) x − 3 y + 5 3 = 0, , (b) x − 3 y − 5 3 = 0, (d) none of these, , 47. The equation of the perpendicular bisector of the line joining the points, A( 2, 3) and B( 6, −5) is, (a) x + 2y − 6 = 0 (b) x − 2y − 6 = 0 (c) x + 2y + 6, (d) x − 2y + 6 = 0, 48. A line passes through the points A(1, 3 ) and B( 2 ,, AB makes with the x-axis is, (a) 30°, (b) 45°, (c) 60°, , 6). The angle which, (d) 90°, , 49. If A( 2, 1), B( −2, 3) and C( 4, 5) are the vertices of a a ABC, then the equation, of altitude through B is, (a) x − 2y + 4 = 0 (b) x + 2y − 4 = 0 (c) x − 2y − 4 = 0 (d) x + 2y + 4 = 0, 50. If A(1, 4), B( 2, − 3) and C( −1, − 2) are the vertices of a a ABC, then the, equation of the right bisector of BC is, (a) 3 x − y − 4 = 0 (b) 3 x + y − 4 = 0 (c) 3 x − y + 4 = 0 (d) none of these, 51. The equation of a line which is equidistant from the lines x = −2and x = 6is, (a) x = 4, (b) x = 2, (c) x = 3, (d) none of these, 52. The equation of a line which is equidistant from the lines y = 8and y = −2is, (a) y = 6, (b) y = 4, (c) y = 3, (d) none of these, 53. The equation of a line passing through the point (3, – 4) and parallel to the, x-axis is, (a) x − 4 = 0, (b) x + 4 = 0, (c) y − 4 = 0, (d) y + 4 = 0, 1, −5, and y-intercept, is, 2, 4, (a) 2x − 4y − 5 = 0 (b) 2x − 4y + 5 = 0 (c) 2x + 4y − 5 = 0 (d) 2x + 4y + 5 = 0, , 54. The equation of a line with slope, , 55. A line cuts the y-axis at a distance of 3 units from the origin and makes, an angle of 30° with the positive direction of the x-axis. The equation of the, line is, (a) y − 3 x + 3 3 = 0, (c) x + 3 y − 3 3 = 0, , (b) x − 3 y + 3 3 = 0, (d) none of these, , 56. A line cuts off an intercept of 4 units on negative direction of the y-axis and, makes an angle of 120° with the positive direction of the x-axis. The, equation of the line is, (a) 3 x − y − 4 = 0, (c) 3 x + y + 4 = 0, , (b) 3 x + y − 4 = 0, (d) none of these
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1122, , Senior Secondary School Mathematics for Class 11, , 1, and x-intercept = 5 is, 3, (a) x − 3 y − 5 = 0 (b) x + 3 y − 5 = 0 (c) x + 3 y + 5 = 0 (d) none of these, , 57. The equation of a line for which tan θ =, , 58. A line cuts the x-axis at a distance of 3 units to the left of the origin and has, slope –2. The equation of the line is, (a) 2x − y + 6 = 0 (b) 2x + y − 6 = 0 (c) 2x + y + 6 = 0 (d) 2x − y − 6 = 0, 59. The lines 2x + 3 y + 7 = 0 and 27 x − 18y + 25 = 0 are, (a) parallel to each other, (b) coincident, (c) perpendicular to each other, (d) none of these, 60. The lines x + 2y − 9 = 0 and 2x + 4y + 5 = 0 are, (a) parallel to each other, (b) coincident, (c) perpendicular to each other, (d) none of these, 61. The angle made by the line x +, the x-axis is, (a) 45°, (b) 60°, , 3 y − 6 = 0 with the positive direction of, (c) 120°, , (d) 150°, , 62. A line passes through the point (2, –5) and is parallel to the line 2x − 3 y = 7., The equation of the line is, (a) 2x + 3 y − 14 = 0, (b) 2x − 3 y − 21 = 0, (c) 2x − 3 y − 19 = 0, (d) none of these, 63. A line passes through the point (–2, –4) and is perpendicular to the line, 3 x − y + 5 = 0. The equation of the line is, (a) x + 3 y + 15 = 0, (b) 3 x + y + 15 = 0, (c) x + 3 y − 14 = 0, (d) x + 3 y + 14 = 0, 64. The y-intercept of a line is –3 and it is perpendicular to the line, 3 x − 2y + 5 = 0. The equation of the line is, (a) 2x + 3 y + 9 = 0 (b) 2x − 3 y + 9 = 0 (c) 2x − 3 y − 9 = 0 (d) none of these, 65. The equation of the line perpendicular to the line x − 7 y + 5 = 0 and having, x-intercept 3 is, (a) 7 x + y − 21 = 0 (b) x + 7 y − 21 = 0 (c) 7 x + y − 15 = 0 (d) none of these, 66. The equation of the line making intercepts 2 and –3 on the x-axis and y-axis, respectively is, (a) 3 x + 2y − 6 = 0 (b) 3 x − 2y + 6 = 0 (c) 3 x − 2y − 6 = 0 (d) none of these, 67. The equation of the line so that the line segment intercepted between the, axes is bisected at the point (2, 3), is, (a) 3 x + 2y − 12 = 0 (b) 2x + 3 y − 12 = 0 (c) 3 x + 2y + 12 = 0 (d) none of these, 68. If the straight line, B(12, −15), then, (a) a = 3 , b = 2, , x y, + = 1 passes through the points A( 8, − 9) and, a b, , (b) a = 2, b = 3, , (c) a = 4, b = 9, , (d) a = 9, b = 4
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Objective Questions, , 1123, , 69. The equation of the line passing through the point P( 2, 2) and cutting off, intercepts on the axis having sum 9 is, (a) x + 2y − 6 = 0 or 2x + y − 6 = 0, , (b) x − 2y + 6 = 0 or 2x + y + 6 = 0, , (c) x − 2y − 6 = 0 or 2x − y + 6 = 0, , (d) none of these, , 70. The equation of a line for which p = 5 and α = 135 ° is, (a) x + y + 3 2 = 0, , (b) x − y + 5 2 = 0, , (c) x + y + 5 2 = 0, , (d) none of these, , 71. The equation of a line for which p = 8 and α = 150° is, (a) 3 x − y + 8 = 0, , (b) 3 x + y − 16 = 0, , (c) 3 x − y + 16 = 0, , (d) none of these, , 72. The slope of the line 3 x + y + 2 = 0 is, 1, −1, (a), (b), (c) 3, 3, 3, , (d) − 3, , 73. The perpendicular distance of a line from the origin is 5 units and the angle, between the positive direction of the x-axis and the perpendicular is 30°., The equation of the line is, (a) 3 x + y − 10 = 0, , (b) 3 x − y + 10 = 0, , (c) 3 x − y − 10 = 0, , (d) none of these, , 74. The equation 3 x − 2y + 4 = 0 when reduced to intercept form takes the, x y, form + = 1, where, a b, 4, −4, −4, (a) a = , b = 2 (b) a =, , b = −2 (c) a =, , b = 2 (d) none of these, 3, 3, 3, 75. The distance of the point P( 4, 1) from the line 3 x − 4y + 12 = 0 is, (a) 4 units, (b) 5 units, (c) 3 units, (d) 6 units, 76. The distance of the point P( −1, 1) from the line 12x − 5 y + 82 = 0 is, (a) 8 units, , (b) 6 units, , (c) 5 units, , (d) 7 units, , 77. The length of perpendicular from the origin to the line 4x + 3 y − 2 = 0 is, 2, 2, 4, 4, (b) unit, (c) unit, (d) unit, (a) unit, 3, 5, 3, 5, 78. The distance between the parallel lines 3 x − 4y + 9 = 0 and 6x − 8y − 17 = 0 is, 5, 7, (a) units, (b) 3 units, (c) units, (d) 4 units, 2, 2, 79. What are the points on the x-axis whose perpendicular distance from the, x y, line + = 1 is 4 units?, 3 4, (a) (8, 0) and (2, 0), , (b) (–8, 0) and (2, 0), , (c) (8, 0) and (–2, 0), , (d) (–8, 0) and (–2, 0)
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1124, , Senior Secondary School Mathematics for Class 11, , 80. The perpendicular distance of a line from the origin is 5 units and its slope, is –1. The equation of the line is, (a) x − y + 5 2 = 0 or x − y − 5 2 = 0 (b) x + y + 5 2 = 0 or x + y − 5 2 = 0, (c) x − y + 2 5 = 0 or x − y − 2 5 = 0 (d) none of these, 81. The distance between the parallel lines p( x + y) + q = 0and p( x + y) − r = 0is, |q + r|, |q + r|, |q − r|, |q − r|, (a), (b), (c), (d), 2p, 2p, 2p, 2p, 82. The point of intersection of the lines 5 x + 7 y = 3 and 2x − 3 y = 7 is, (a) (2, –1), (b) (–2, 1), (c) (–2, –1), (d) (2, 1), 83. The equation of the line parallel to the y-axis and drawn through the point, of intersection of the lines x − 7 y + 15 = 0 and 2x + y = 0 is, (a) x − 1 = 0, , (b) x + 1 = 0, , (c) x − 2 = 0, , (d) x + 2 = 0, , 84. The equation of the line passing through the point of intersection of the, lines x + 2y + 3 = 0 and 3 x + 4y + 7 = 0 and parallel to the line y − x = 6 is, (a) x − y = 0, , (b) x + y = 1, , (c) x + 2y − 3 = 0 (d) 2x + y − 3 = 0, , 85. The image of the point P( 3 , 8) in the line x + 3 y − 7 = 0 is, (a) (1, 4), (b) (–1, 4), (c) (1, – 4), (d) (–1, – 4), 86. The point of intersection of the lines x − y = 6, 4x − 3 y = 20 and, 6x + 5 y + 8 = 0 is, (a) (2, 4), (b) (2, – 4), (c) (–2, 4), (d) (–2, – 4), 87. The area of the triangle formed by the lines y − x = 0, x + y = 0 and x − c = 0is, 1, (c) c2 sq units, (d) 3 c2 sq units, (a) c2 sq units (b) 2c2 sq units, 2, 88. If the lines 3 x + y = 2, kx + 2y = 3 and 2x − y = 3 are concurrent, then k = ?, (a) –5, (b) 5, (c) –3, (d) 3, 89. The centre of the circle x 2 + y 2 − 6x + 4y − 12 = 0 is, (a) (–3, 2), (b) (3, 2), (c) (3, –2), , (d) (–3, –2), , 90. Radius of the circle x 2 + y 2 − 4x + 2y − 45 = 0 is, (a) 5 2 units, , (b) 4 2 units, , (c) 3 5 units, , (d) 4 5 units, , 91. The end points of a diameter of a circle are A( 2, − 3) and B( −3 , 5). The, equation of the circle is, (a) x 2 + y 2 + 2x − y − 21 = 0, , (b) x 2 + y 2 + x − 2y − 21 = 0, , (c) x + y + x − 2y + 21 = 0, , (d) none of these, , 2, , 2, , 92. The centre of a circle is C( 2, −5) and the circle passes through the point, A( 3 , 2). The equation of the circle is, (a) x 2 + y 2 − 4x + 10y − 21 = 0, , (b) x 2 + y 2 + 4x + 6y − 21 = 0, , (c) x 2 + y 2 + 4x − 10y + 21 = 0, , (d) none of these
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Objective Questions, , 1125, , 93. For the circle ( x + 5) 2 + ( y − 3) 2 = 20, the centre and radius are respectively, (a) (5 , − 3), 2 5, , (b) ( −5 , 3), 5 2, , (c) ( −5 , 3), 2 5, , (d) none of these, , 94. If A( −1, 3) and B(α , β ) be the extremities of the diameter of the circle, x 2 + y 2 − 6x + 5 y − 7 = 0, then, (a) α = −7 , β = 8 (b) α = 7 , β = − 8 (c) α = − 6, β = 7, , (d) α = 6, β = −7, , 95. For the parabola y = 8x , the focus and vertex are respectively, 2, , (a) F( 2, 0) and O( 0, 0), , (b) F( −2, 0) and O( 0, 0), , (c) F( 4, 0) and O( 0, 0), , (d) F( − 4, 0) and O( 0, 0), , 96. In the parabola y = −12x , the focus and the equation of directrix are, respectively, 2, , (a) F( 3 , 0), x = −3, , (b) F( −3 , 0), x = 3, , (c) F( −3 , 0), x = −3, , (d) none of these, , 97. For the parabola y 2 = − 8x , the focus and the equation of directrix are, respectively, (a) F( −2, 0), x = 2, , (b) F( 2, 0), x = −2, , (c) F( 2, 0), x = 2, , (d) F( −2, 0), x = −2, , 98. For the parabola x 2 = −16y , the focus and the equation of directrix are, respectively, (a) F( 0, 4), y = 4, , (b) F( 0, − 4), y = 4, , (c) F( 0, 4), y = −4, , (d) none of these, , 99. For the parabola x 2 = 6y , the focus and the equation of directrix are, respectively, 3, 3, ⎛ −3 ⎞, ⎛ 3⎞, (a) F⎜ 0,, (b) F⎜ 0, ⎟ , y =, ⎟, y =, ⎠, ⎠, ⎝, ⎝, 2, 2, 2, 2, −3, ⎛ 3⎞, (c) F⎜ 0, ⎟ , y =, ⎝ 2⎠, 2, , (d) none of these, , 100. If the parabola y 2 = 4ax passes through the point P( 3 , 2), then the length of, its latus rectum is, 1, 2, 4, (a), (b), (c), (d) 4, 3, 3, 3, 101. If A( 2, 0) is the vertex and the y-axis is the directrix of a parabola, then its, focus is, (a) F( 2, 0), , (b) F( −2, 0), , (d) F( − 4, 0), , (c) F( 4, 0), , 102. An equilateral triangle is inscribed in a parabola y = 4ax , whose vertex is, at the vertex of the parabola. The length of each side of the triangle is, 2, , (a) 2a 3, , (b) 4a 3, , (c) 6a 3, , (d) 8a 3
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1126, , Senior Secondary School Mathematics for Class 11, , 103. In the parabola y 2 = 4ax , the length of the chord passing through the vertex, ⎛ π⎞, and inclined to the axis at an angle ⎜ ⎟ is, ⎝ 4⎠, (a) 2 a, , (b) 2a 2, , (d) 4a 2, , (c) 2a, , 104. The focal distance of a point P on the parabola y = 12x is 4. The abscissa of, P is, 2, , (a) –1, , (b) 1, , (c) –2, , (d) 2, , 105. The equation of a parabola with vertex at O( 0, 0) and focus at F( 0, 2) is, (a) y 2 = 8x, , (b) x 2 = 8y, , (c) y 2 = 4x, , (d) x 2 = 4y, , 106. A parabola has its vertex at the origin; its axis lies along the x-axis and it, passes through the point P( 2, 3). The equation of the parabola is, 4, 9, 3, (b) y 2 = x, (c) y 2 = x, (d) y 2 = x, (a) 4y 2 = 9x, 9, 2, 2, 107. A parabola has its vertex at the origin and passes through the point, P( 3 , − 4) and it is symmetric about the y-axis. Its equation is, (a) 4x 2 + 9y = 0, , (b) 4x + 9y 2 = 0, 2, , (c) 9x 2 + 4y = 0, , (d) none of these, , 2, , x, y, +, = 1., 16, 9, Fill in the blanks given below:, , 108. The equation of an ellipse is, , (i) Length of the major axis = …… units., (ii) Length of the minor axis = …… units., (iii) Coordinates of its vertices are A(……, ……) and B(……, ……)., (iv) Coordinates of its foci are F1 (……, ……) and F2 (……, ……)., (v) Eccentricity, e = …… ., (vi) Length of the latus rectum = …… units., 109. The equation of an ellipse is 4x 2 + 9y 2 = 144., Fill in the blanks given below:, (i) Length of the major axis = …… units., (ii) Length of the minor axis = …… units., (iii) Coordinates of its vertices are A(……, ……) and B(……, ……)., (iv) Coordinates of its foci are F1 (……, ……) and F2 (……, ……)., (v) Eccentricity, e = …… ., (vi) Length of the latus rectum = …… units., x2 y2, +, = 1., 4, 36, Fill in the blanks given below:, , 110. The equation of an ellipse is, , (i) Length of the major axis = …… units., (ii) Length of the minor axis = …… units.
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Objective Questions, , 1127, , (iii) Coordinates of its vertices are A(……, ……) and B(……, ……)., (iv) Coordinates of its foci are F1 (……, ……) and F2 (……, ……)., (v) Eccentricity, e = …… ., (vi) Length of the latus rectum = …… units., 111. The equation of an ellipse is 4x 2 + y 2 = 100., Fill in the blanks given below:, (i) Length of the major axis = …… units., (ii) Length of the minor axis = …… units., (iii) Coordinates of its vertices are A(……, ……) and B(……, ……)., (iv) Coordinates of its foci are F1 (……, ……) and F2 (……, ……)., (v) Eccentricity, e = …… ., (vi) Length of the latus rectum = …… units., Mark (3) against the correct answer in each of the following., , 112. The vertices of an ellipse are ( ±5 , 0) and its foci are ( ±4, 0). The equation of, the ellipse is, (a), , x2 y2, +, =1, 25, 9, , (b), , x2 y2, +, =1, 9 25, , (c), , x2 y2, +, =1, 25 16, , 113. The foci of an ellipse are ( ±4, 0) and its eccentricity is, ellipse is, (a), , x2 y2, +, =1, 16, 9, , (b), , (d), , x2 y2, +, =1, 16 25, , 1, ⋅ The equation of the, 3, , x2, y2, x2 y2, +, = 1 (c), +, =1, 144 128, 144 81, , (d) none of these, , 114. The foci of an ellipse are ( 0, ± 5) and its vertices are ( 0, ± 13). The equation, of the ellipse is, (a), , x2, y2, x2 y2, +, = 1 (b), +, =1, 169 144, 169 25, , (c), , x2, y2, +, =1, 144 169, , (d) none of these, , 115. The foci of an ellipse are ( 0, ± 6) and the length of its minor axis is 16. The, equation of the ellipse is, (a), , x2 y2, +, =1, 16 36, , (b), , x2 y2, +, =1, 36 64, , (c), , x2 y2, +, =1, 100 64, , (d), , x2 y2, +, =1, 64 100, , 116. The foci of an ellipse are ( 0, ± 5) and the length of its major axis is 20. The, equation of the ellipse is, (a), , x2 y2, +, =1, 25 40, , (b), , x2 y2, +, =1, 25 100, , (c), , x2 y2, +, = 1 (d) none of these, 75 100, 4, 117. The vertices of an ellipse are ( 0, ± 10) and e = ⋅ The equation of the, 5, ellipse is, (a), , x2 y2, +, =1, 100 36, , (b), , x2 y2, +, =1, 36 100, , (c), , x2 y2, +, =1, 36 64, , (d) none of these
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1128, , Senior Secondary School Mathematics for Class 11, , 118. The equation of a hyperbola is, , x2 y2, −, = 1., 36 64, , Fill in the blanks given below:, (i) Length of its transverse axis = …… units., (ii) Length of its conjugate axis = …… units., (iii) Coordinates of its vertices are A(……, ……) and B(……, ……)., (iv) Coordinates of its foci are F1 (……, ……) and F2 (……, ……)., (v) Eccentricity, e = …… ., (vi) Length of the latus rectum = …… units., 119. The equation of a hyperbola is 9x 2 − 16y 2 = 144., Fill in the blanks given below:, (i) Length of its transverse axis = …… units., (ii) Length of its conjugate axis = …… units., (iii) Coordinates of its vertices are A(……, ……) and B(……, ……)., (iv) Coordinates of its foci are F1 (……, ……) and F2 (……, ……)., (v) Eccentricity, e = …… ., (vi) Length of the latus rectum = …… units., Mark (3) against the correct answer in each of the following., , 120. The vertices of a hyperbola are ( ±2, 0) and its foci are ( ±3 , 0). The equation, of the hyperbola is, (a), , x2 y2, −, =1, 2, 3, , (b), , x2 y2, −, =1, 3, 4, , (c), , x2 y2, −, =1, 4, 5, , (d) none of these, , 121. The foci of a hyperbola are ( ±5 , 0) and its transverse axis is of length 8. The, equation of the hyperbola is, (a), , x2 y2, −, =1, 16 9, , (b), , x2 y2, −, =1, 9 16, , (c), , x2 y2, −, =1, 25 16, , (d), , x2 y2, −, =1, 16 25, , 122. The foci of a hyperbola are ( ±4, 0) and the length of its latus rectum is, 12 units. The equation of the hyperbola is, (a), , x2 y2, −, =1, 4, 6, , (b), , x2 y2, −, =1, 4 12, , 123. The equation of a hyperbola is, , (c), , x2 y2, −, =1, 4, 8, , (d) none of these, , y2 x2, −, = 1., 4, 9, , Fill in the blanks given below:, (i) Length of its transverse axis = …… units., (ii) Length of its conjugate axis = …… units., (iii) The coordinates of its vertices are A(……, ……) and B(……, ……)., (iv) The coordinates of its foci are F1 (……, ……) and F2 (……, ……).
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Objective Questions, , 1129, , (v) Its eccentricity, e = …… ., (vi) Length of its latus rectum = …… units., Mark (3) against the correct answer in each of the following., , 124. The vertices of a hyperbola are ( 0, ± 3) and its foci are ( 0, ± 5). The equation, of the hyperbola is, (a), , y2 x2, −, =1, 16 9, , (b), , y2 x2, −, =1, 9 16, , (c), , y2 x2, −, =1, 9 25, , (d) none of these, , 125. The vertices of a hyperbola are ( 0, ± 6) and its eccentricity is, equation of the hyperbola is, (a), , y2 x2, −, =1, 36 64, , (b), , y2 x2, −, =1, 36 25, , (c), , x2 y2, −, =1, 36 64, , 5, ⋅ The, 3, , (d) none of these, , 126. One focus of a hyperbola is at (0, 4) and the length of its transverse axis is 6., The equation of the hyperbola is, (a), , x2 y2, −, =1, 7, 9, , (b), , y2 x2, −, =1, 9, 7, , (c), , y2 x2, −, =1, 4, 9, , (d) none of these, , 127. Foci of a hyperbola are ( 0, ± 6) and the length of its conjugate axis is 2 11., The equation of the hyperbola is, (a), , y2 x2, −, =1, 36 11, , (b), , x2 y2, −, =1, 36 11, , (c), , y2 x2, −, =1, 25 11, , (d) none of these, , ANSWERS (EXERCISE 1), , 1. (d), , 2. (a), , 3. (b), , 4. (b), , 5. (d), , 6. (c), , 7. (a), , 8. (b), , 9. (c) 10. (c), , 11. (d) 12. (c) 13. (c) 14. (c) 15. (a) 16. (c) 17. (a) 18. (b) 19. (a) 20. (b), 21. (a) 22. (a) 23. (b) 24. (d) 25. (b) 26. (d) 27. (c) 28. (b) 29. (a) 30. (d), 31. (b) 32. (c) 33. (c) 34. (b) 35. (a) 36. (c) 37. (b) 38. (b) 39. (c) 40. (a), 41. (c) 42. (b) 43. (d) 44. (b) 45. (c) 46. (c) 47. (b) 48. (c) 49. (b) 50. (a), 51. (b) 52. (c) 53. (d) 54. (a) 55. (b) 56. (c) 57. (a) 58. (c) 59. (c) 60. (a), 61. (d) 62. (c) 63. (d) 64. (a) 65. (a) 66. (c) 67. (a) 68. (b) 69. (a) 70. (b), 71. (c) 72. (d) 73. (a) 74. (c) 75. (a) 76. (c) 77. (b) 78. (c) 79. (c) 80. (b), 81. (b) 82. (a) 83. (b) 84. (a) 85. (d) 86. (b) 87. (c) 88. (b) 89. (c) 90. (a), 91. (b) 92. (a) 93. (c) 94. (b) 95. (a) 96. (b) 97. (a) 98. (b) 99. (c) 100. (c), 101. (c) 102. (d) 103. (d) 104. (b) 105. (b) 106. (c) 107. (a), 108. (i) 8 (ii) 6 (iii) A( − 4, 0) and B( 4, 0) (iv) F1 ( − 7 , 0), F2 ( 7 , 0) (v) e =, (vi) 6, , 7, 4
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1130, , Senior Secondary School Mathematics for Class 11, , 109. (i) 6 (ii) 4 (iii) A( −3 , 0) and B( 3 , 0) (iv) F1 ( 5 , 0), F2 ( − 5 , 0) (v) e =, (vi), , 5, 3, , 8, 3, (iii) A( 0, − 6) and B( 0, 6), 4, (vi), 3, , 110. (i) 12, , (ii) 4, 2 2, (v) e =, 3, , (iv) F1 ( 0, − 4 2), F2 ( 0, 4 2), , (ii) 10 (iii) A( 0, −10) and B( 0, 10), 3, (vi) 5, (v) e =, 2, , 111. (i) 20, , 112. (a), , 113. (b), , 114. (c), , (iv) F1 ( 0, −5 3 ), F2 ( 0, 5 3 ), , 115. (d), , 116. (c), , 117. (b), 5, (v) e =, 3, , 118. (i) 12 (ii) 16 (iii) A( − 6, 0) and B( 6, 0) (iv) F1 ( −10, 0), F2 (10, 0), 64, (vi), 3, 5, 119. (i) 8 (ii) 6 (iii) A( − 4, 0) and B( 4, 0) (iv) F1 ( −5 , 0), F2 (5 , 0) (v) e =, 4, 9, (vi), 2, 120. (c), , 121. (a), , 123. (i) 4, (v) e =, 124. (b), , (iii) A( 0, − 2) and B( 0, 2), , (ii) 6, 13, 2, , 122. (b), (iv) F1 ( 0, − 13 ), F2 ( 0, 13 ), , (vi) 9, , 125. (a), , 126. (b), , 127. (c), HINTS TO SOME SELECTED QUESTIONS, , 2, , 2, , 2, , 2, , 2, , 1. AB = ( −6 − 2 ) + ( 3 + 3 ) = ( −8 ) + 6 = ( 64 + 36 ) = 100, ⇒, , AB = 100 units = 10 units., 2, , 2. PQ = {( 5 − 2 ) − ( 5 + 1)} 2 + {( 3 + 2 ) − ( 3 − 1)} 2 = ( −3 ) 2 + 3 2 = ( 9 + 9 ) = 18, ⇒, , PQ = 18 units = 3 2 units., , 3. AB 2 = 5 2 ⇒ ( x + 3 ) 2 + (7 − 4 ) 2 = 25, ⇒ x 2 + 6 x + 9 + 9 = 25 ⇒ x 2 + 6 x − 7 = 0, ⇒ ( x + 7 )( x − 1) = 0 ⇒ x = 1 or –7., 4. AB 2 = 4 2 ⇒ ( 3 − x ) 2 + (5 − 3 ) 2 = 4 2, ⇒ x 2 − 6 x + 9 + 4 = 16 ⇒ x 2 − 6 x − 3 = 0, ⇒ x=, , 6±, , 36 + 12 6 ± 48 6 ± 4 3, =, =, = 3 ± 2 3., 2, 2, 2
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Objective Questions, , 1131, , 5. OA 2 = ( 6 − 0 ) 2 + ( −6 − 0 ) 2 = 6 2 + ( −6 ) 2 = ( 36 + 36 ) = 72, OA = 72 = 6 2 units., , ⇒, 2, , 6. AP = BP 2 ⇒ ( x − 6 ) 2 + ( y + 1) 2 = ( x − 2 ) 2 + ( y − 3 ) 2, ⇒ x 2 − 12 x + 36 + y 2 + 2 y + 1 = x 2 − 4 x + 4 + y 2 − 6 y + 9, ⇒ 37 − 12 x + 2 y = 13 − 4 x − 6 y ⇒ 8 x − 8 y = 24 ⇒ x − y = 3., 7. Let the required point be P( x , 0 ). Then,, AP 2 = BP 2 ⇒ ( x − 7 ) 2 + ( 6 − 0 ) 2 = ( x + 3 ) 2 + ( 0 − 4 ) 2, ⇒ x 2 − 14 x + 49 + 36 = x 2 + 6 x + 9 + 16, ⇒ 20 x = ( 85 − 25 ) = 60 ⇒ x = 3., ∴, , the point is P( 3 , 0 )., , 8. Let the required point be P( 0 , y ). Then,, AP 2 = BP 2 ⇒ ( − 4 − 0 ) 2 + ( 3 − y ) 2 = (5 − 0 ) 2 + ( 2 − y ) 2, ⇒ ( − 4)2 + ( 3 − y)2 = 5 2 + ( 2 − y)2, ⇒ 16 + 9 − 6 y + y 2 = 25 + 4 + y 2 − 4 y ⇒ 2 y = − 4 ⇒ y = −2., Hence, the required point is P( 0 , − 2 )., 9. The given points are A( −8 , 0 ) and B( 0 , 15 )., ∴, , AB 2 = ( 0 + 8 ) 2 + ( 15 − 0 ) 2 = 8 2 + ( 15 ) 2 = ( 64 + 225 ) = 289, , ⇒, , AB = 289 units = 17 units., , 10. Here, ( x1 = −2 , y1 = 3 ); ( x2 = 1, y2 = 2 ) and ( x3 = k , y3 = 0 )., ar(a ABC ) = 0 ⇒ x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) = 0, ⇒ − 2( 2 − 0 ) + 1 ⋅ ( 0 − 3 ) + k ⋅ ( 3 − 2 ) = 0 ⇒ − 4 − 3 + k = 0 ⇒ k = 7 ., 11. Here, ( x1 = 3 , y1 = 8 ), ( x2 = − 4 , y2 = 2 ) and ( x3 = 5 , y3 = −1)., 1, ar(a ABC ) = {x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 )}, ∴, 2, 1, 1, = { 3( 2 + 1) − 4( −1 − 8 ) + 5( 8 − 2 )} = ( 9 + 36 + 30 ), 2, 2, 75, 1, sq units = 37 sq units., =, 2, 2, 1, 12. Area of the a ABC = ⋅ [1 ⋅ ( −3 − 2 ) + 7 ⋅ ( 2 − 1) + 12 ⋅ ( 1 + 3 )] = 25 sq units,, 2, 1, area of the a BCD = ⋅ [7 ⋅ ( 2 − 21) + 12 ⋅ ( 21 + 3 ) + 7 ⋅ ( −3 − 2 )] = 60 sq units., 2, ∴, , area of the quad. ABCD = ar(a ABC ) + ar(a BCD ) = 85 sq units., , 1 + 5 −1 + 1 ⎞, 13. Midpoint of BC is D ⎛⎜, ,, ⎟ , i.e., D( 3 , 0 )., ⎝ 2, 2 ⎠, ∴, , median AD = ( −1 − 3 ) 2 + ( 3 − 0 ) 2 = ( −4 ) 2 + 3 2 = 16 + 9 = 25 units = 5 units., , 7 + 5 −3 + 3 ⎞, 14. Midpoint of AC is E ⎛⎜, ,, ⎟ , i.e., E( 6 , 0 )., ⎝ 2, 2 ⎠, ∴, , median BE = ( 6 − 3 ) 2 + ( 0 + 1) 2 =, , 3 2 + 12 = 9 + 1 = 10 units.
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1132, , Senior Secondary School Mathematics for Class 11, , −1 + 5 + 8 0 − 2 + 2 ⎞, 15. Coordinates of centroid are G ⎛⎜, ,, ⎟ , i.e., G( 4 , 0 )., ⎝, ⎠, 3, 3, 16. Let the 3rd vertex be C ( x , y ). Then,, 5 − 7 + x 2 − 4 + y⎞, ⎛ x − 2 y − 2⎞, centroid is G ⎛⎜, ,, ,, ⎟⋅, ⎟ , i.e., ⎜, ⎝ 3, ⎝, ⎠, 3 ⎠, 3, 3, y−2, x−2, = −2 and, = 3 ⇒ x = ( − 6 + 2 ) = − 4 , y = ( 9 + 2 ) = 11., 3, 3, , ∴, , Hence, the 3rd vertex of a ABC is C( −4 , 11)., 17. AB 2 = ( −1 − 1) 2 + ( −1 − 1) 2 = ( −2 ) 2 + ( −2 ) 2 = ( 4 + 4 ) = 8 ;, BC 2 = ( −1 +, 2, , 3 ) 2 + ( −1 −, , 3 )2 = ( 1 + 3 − 2 3 ) + ( 1 + 3 + 2 3 ) = 8;, , 2, , AC = ( − 3 − 1) + ( 3 − 1) 2 = ( 3 + 1 + 2 3 ) + ( 3 + 1 − 2 3 ) = 8., AB = BC = AC = 8., , ∴, , Hence, a ABC is equilateral., 18. AB 2 = ( 2 a − 2 a) 2 + ( 6 a − 4 a) 2 = 0 2 + ( 2 a) 2 = 4 a2 ;, BC 2 = ( 2 a +, 2, , AC = ( 2 a +, ∴, , 3 a − 2 a) 2 + ( 5 a − 6 a) 2 = ( 3 a) 2 + ( − a) 2 = ( 3 a 2 + a 2 ) = 4 a 2 ;, 3 a − 2 a) 2 + ( 5 a − 4 a) 2 = ( 3 a) 2 + a 2 = ( 3 a 2 + a 2 ) = 4 a 2 ., , AB = BC = AC = 2 a., , Hence, a ABC is an equilateral triangle., 19. AB 2 = (7 + 2 ) 2 + ( 10 − 5 ) 2 = ( 9 2 + 5 2 ) = ( 81 + 25 ) = 106 ;, BC 2 = ( −2 − 3 ) 2 + (5 + 4 ) 2 = ( −5 ) 2 + 9 2 = ( 25 + 81) = 106 ;, AC 2 = (7 − 3 ) 2 + ( 10 + 4 ) 2 = 4 2 + ( 14 ) 2 = ( 16 + 196 ) = 212., ∴, , AB 2 + BC 2 = AC 2 and AB = BC ≠ AC ., , Hence, a ABC is an isosceles right triangle., 20. AB 2 = (5 − 2 ) 2 + ( 1 − 6 ) 2 = 3 2 + ( −5 ) 2 = ( 9 + 25 ) = 34 ;, BC 2 = ( 0 − 5 ) 2 + ( −2 − 1) 2 = ( −5 ) 2 + ( −3 ) 2 = ( 25 + 9 ) = 34 ;, CD 2 = ( −3 − 0 ) 2 + ( 3 + 2 ) 2 = ( −3 ) 2 + 5 2 = ( 9 + 25 ) = 34 ;, AD 2 = ( −3 − 2 ) 2 + ( 3 − 6 ) 2 = ( −5 ) 2 + ( −3 ) 2 = ( 25 + 9 ) = 34., ∴, , AB = BC = CD = AD =, , 34., , Diag. AC 2 = ( 0 − 2 ) 2 + ( −2 − 6 ) 2, = ( −2 ) 2 + ( −8 ) 2 = ( 4 + 64 ) = 68., 2, , Diag. BD = ( −3 − 5 ) 2 + ( 3 − 1) 2 = ( −8 ) 2 + 2 2 = ( 64 + 4 ) = 68., diagonal AC = diagonal BD., ∴, Hence, ABCD is a square., 21. AB 2 = ( 0 + 1) 2 + ( 3 − 0 ) 2 = ( 12 + 3 2 ) = ( 1 + 9 ) = 10 ;, BC 2 = ( 1 − 0 ) 2 + ( 3 − 3 ) 2 = ( 12 + 0 2 ) = ( 1 + 0 ) = 1;, CD 2 = ( 0 − 1) 2 + ( 0 − 3 ) 2 = ( −1) 2 + ( −3 ) 2 = ( 1 + 9 ) = 10 ;, AD 2 = ( −1 − 0 ) 2 + ( 0 − 0 ) 2 = ( −1) 2 + 0 2 = ( 1 + 0 ) = 1., ∴, , AB = CD = 10 and BC = AD = 1 = 1., , Diag. AC 2 = ( 1 + 1) 2 + ( 3 − 0 ) 2 = 2 2 + 3 2 = ( 4 + 9 ) = 13;
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Objective Questions, Diag. BD 2 = ( −3 − 5 ) 2 + ( 3 − 1) 2 = ( −8 ) 2 + 2 2 = ( 64 + 4 ) = 68., ∴, , diag. AC = 13 and diag. BD = 68., , Hence, ABCD is a||gm., 22. Let the 4th vertex be D( x , y ). Then,, ∴, , midpoint of AC = midpoint of BD., 0 + y −2 − 3, 4+ x 3+ 6, and, =, =, 2, 2, 2, 2, , ⇒, , x = ( 9 − 4 ) = 5 and y = −5, , ⇒, , point D is D(5 , − 5 )., , 23. Let the circumcentre be P( x , y )., PA 2 = PB 2 ⇒ ( x − 8 ) 2 + ( y − 6 ) 2 = ( x − 8 ) 2 + ( y + 2 ) 2, ⇒ ( y − 6)2 − ( y + 2)2 = 0, ⇒ ( y − 6 + y + 2 )( y − 6 − y − 2 ) = 0, ⇒ − 8( 2 y − 4 ) = 0 ⇒ 2 y − 4 = 0 ⇒ y = 2., PB 2 = PC 2 ⇒ ( x − 8 ) 2 + ( y + 2 ) 2 = ( x − 2 ) 2 + ( y + 2 ) 2, ⇒ ( x − 8 ) 2 − ( x − 2 ) 2 = 0 ⇒ ( x − 8 + x − 2 )( x − 8 − x + 2 ) = 0, ⇒ ( 2 x − 10 )( −6 ) = 0 ⇒ 2 x − 10 = 0 ⇒ x = 5., ∴, , the required point is P(5 , 2 )., , 24. Coordinates of P are P( x , y ), where, 3 × 6 + 1 × ( −2 ) 16, 3 × 9 + 1 × 5 32, =, = 4., =, = 8, y =, x=, 3+ 1, 4, 3+ 1, 4, ∴, , the required point is P( 8 , 4 )., , 25. Let the required ratio be λ : 1. Then,, −2 λ + 7, = 1 ⇒ − 2λ + 7 = λ + 1 ⇒ 3λ = 6 ⇒ λ = 2 ., λ+ 1, the required ratio is 2 : 1., ∴, 26. Let the required ratio be k : 1., Then, abscissa of the point lying on the y-axis is 0., 8 k − 12, 3, ∴, = 0 ⇒ 8 k − 12 = 0 ⇒ 8 k = 12 ⇒ k = ⋅, k+1, 2, ∴, , the required ratio =, , 3, : 1 = 3 : 2., 2, , 27. Let the required ratio be k : 1., Then, the ordinate of the point lying on the x-axis is 0., −2 k + 5, 5, = 0 ⇒ − 2k + 5 = 0 ⇒ 2k = 5 ⇒ k = ⋅, ∴, k+1, 2, ∴, , the required ratio is, , 5, : 1, i.e., 5 : 2., 2, , 28. a = BC = ( 8 − 8 ) 2 + ( 12 − 0 ) 2 = 0 + 144 = 144 = 12 ., b = AC = ( 8 − 0 ) 2 + ( 0 − 6 ) 2 = 8 2 + ( − 6 ) 2 = 64 + 36 = 100 = 10 ., , 1133
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1134, , Senior Secondary School Mathematics for Class 11, , c = AB = ( 8 − 0 ) 2 + ( 12 − 6 ) 2 = 8 2 + 6 2 = 64 + 36 = 100 = 10., ax + bx2 + cx3 ay1 + by2 + cy3 ⎞, Incentre is ⎛⎜ 1, ,, ⎟, ⎝, ⎠, a+ b + c, a+ b + c, ⎛ 12 × 0 + 10 × 8 + 10 × 8 , 12 × 6 + 10 × 12 + 10 × 0 ⎞ = (5 , 6 )., ⎜, ⎟, ⎝, ⎠, 12 + 10 + 10, 12 + 10 + 10, 29. Let O ( x , y ) be the centre of the circle. Then,, OA 2 = OB 2 = OC 2 ⇒ ( x − 2 ) 2 + ( y + 9 ) 2 = ( x − 5 ) 2 + ( y + 8 ) 2, ⇒ − 4 x + 4 + 18 y + 81 = −10 x + 25 + 16 y + 64, ⇒ 6x + 2y = 4 ⇒ 3x + y = 2, And, ( x − 2 ) 2 + ( y + 9 ) 2 = ( x − 2 ) 2 + ( y − 1) 2, ⇒, , − 4 x + 4 + 18 y + 81 = −4 x + 4 + 1 − 2 y, , ⇒, , 20 y = −80 ⇒ y = − 4., , Putting y = − 4 in (i), we get x = 2., Hence, the centre is O( 2 , − 4 )., 30. Let the other end of the dimeter be B( x , y )., 1+ y, 4+ x, Then,, = 3 ⇒ x = 2 , y = 5., = 3 and, 2, 2, ∴, , the other end of the diameter is B( 2 , 5 )., ( y − y1 ) ( −5 − 3 ) −8 −4, 31. Slope of AB = 2, =, =, =, ⋅, ( x2 − x1 ) ( 8 + 2 ), 10, 5, 32. Slope of AB =, ∴, , 5 − ( −3 ), 8, =, ⋅, 2−x, ( 2 − x), , 8, = tan 135 ° = tan ( 180 ° − 45 ° ) = − tan 45 ° = −1, 2−x, , ⇒, , x − 2 = 8 ⇒ x = 10., −2 − ( −1) −2 + 1, 33. Slope, m =, =, = −1., 4− 3, 1, tan θ = m = −1 = − tan 45 ° = tan ( 180 ° − 45 ° ) = tan 135 ° ⇒ θ = 135 ° ., ( 8 − 6), 2 1, 34. m1 = slope of AB =, = = ⋅, 4 − ( −2 ) 6 3, m 2 = slope of CD =, , ( 24 − 12 ), 12, =, ⋅, ( x − 8), ( x − 8), , AB ⊥ CD ⇒ m1 m 2 = −1 ⇒, , 1, 12, ×, = −1 ⇒ − x + 8 = 4 ⇒ x = 4., 3 ( x − 8), , 35. Since the given points are collinear, we have, 1 − ( −1) (5 − 1) 4, slope of AB = slope of BC ⇒, = =2, =, ( 4 − 2) 2, 2−x, ∴, , 2, = 2 ⇒ 2 = 4 − 2 x ⇒ 2 x = 2 ⇒ x = 1., 2−x, , ... (i)
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Objective Questions, , 1135, , 36. Clearly, B is the angle between BA and BC., m1 = slope of BA =, , 3−1 2 1, = = ⋅, 2+ 2 4 2, , m 2 = slope of BC =, , − 4 − 3 −7 7, = ⋅, =, −2 − 2 − 4 4, , ∴, , tan B =, , 37. tan θ =, , 7 1, −, 2, m 2 − m1, = 4 2 = ⋅, 7 1 3, 1 + m1 m 2, 1+ ×, 4 2, , 1, m 2 − m1, π, , where m1 = , m 2 = m and θ = ⋅, 2, 1 + m1 m 2, 4, , ∴, , 1, 2 = tan π = 1 ⇒ 2m − 1 = 1, 1, 2+m, 4, 1+ m, 2, , ∴, , 2m − 1, = +1 or, 2+m, , ⇒, , 2m − 1 = 2 + m or, , 38. m1 =, , m−, , 2m − 1, = −1, 2+m, 2m − 1 = −2 − m ⇒ m = 3 or m =, , −1, ⋅, 3, , ( 3 − 0) 3, 5+ 2 7, = and m 2 =, =, ( 2 − 0) 2, 3−2 1, , 3, 7−, m 2 − m1, 11 2, 11, 2, =, =, ×, =, ⋅, tan θ =, 3, 2 23 23, 1 + m1 m 2, 1+ ×7, 2, 39. Here, m1 = ( 2 −, ∴, , tan θ =, , ⇒, , θ = 60 ° ., , 3 ) and m 2 = ( 2 +, , 3 )., , (2 − 3) − (2 + 3), 2 3, m1 − m 2, −2 3, =, =, =, =, 1 + m1 m 2, 2, 1 + ( 2 − 3 )( 2 + 3 ) 1 + ( 4 − 3), , −8 − 2, −10, −5, 40. m =, ⇒, =, ⇒ 5( 6 − x ) = 40, 6−x, 6−x, 4, ⇒ 30 − 5 x = 40 ⇒ 5 x = −10 ⇒ x = −2., 41. Since AB||CD , we have: slope of AB = slope of CD., ∴, 42. m1 =, , 7 −x 6−4, 7−x 2, =, ⇒, = ⇒ 7 − x = −2 ⇒ x = 9., 2− 3 0+ 1, −1, 1, −9 + 3 − 6, x−6 x−6, and m 2 =, =, = −3., =, 4+ 2, 6, 5− 3, 2, , Since AB ⊥ CD , we have: m1 m 2 = −1., ∴, , x−6, × ( −3 ) = −1 ⇒ x − 6 = 2 ⇒ x = 8., 6, , 3
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1136, , Senior Secondary School Mathematics for Class 11, , 43. Let m1 = slope of the diagonal AC =, , 0 − 2 −2 −1, =, =, 4−0, 4, 2, , and m 2 = slope of the diagonal BD =, , ( 3 + 1) 4, = = 1., ( 2 + 2) 4, , ∴, , m − m2, tan θ = 1, =, 1 + m1 m 2, , 44. Equation of AB is, , ⎛ −3 ⎞, 1, ⎜ ⎟, −1, ⎝ 2 ⎠, 2, = 3 ⇒ tan θ = 3., =, −1, ⎛ 1⎞, 1 + ⎛⎜ ⎞⎟ × 1, ⎜ ⎟, ⎝ 2⎠, ⎝ 2⎠, −, , y − y1 y2 − y1, =, x − x1 x2 − x1, , i.e.,, , y − 8 −2 − 8, y − 8 −10, ⇒, =, = −2, =, 5, x+ 1 4+ 1, x+ 1, , ⇒, , y − 8 = −2( x + 1) ⇒ y − 8 = −2 x − 2 ⇒ 2 x + y − 6 = 0 ., , 2 − 2 5 − 1⎞, 45. Clearly, E is the midpoint of AC given by E ⎛⎜, ,, ⎟ , i.e., E( 0 , 2 )., ⎝ 2, 2 ⎠, Equation of the median BE is, , y − y1 y2 − y1, =, x − x1 x2 − x1, , i.e.,, , y − 9 2 − 9 −7, =, ⇒ 4( y − 9 ) = −7 ( x + 4 ), =, 4, x+ 4 0+ 4, , ⇒, , 4 y − 36 = −7 x − 28 ⇒ 7 x + 4 y − 8 = 0., , 46. The equation of the line is, ⇒, , y−5, 1, = tan 30 ° =, x−0, 3, , 3y − 5 3 = x ⇒ x −, , 3 y + 5 3 = 0., , 47. End points of AB are A( 2 , 3 ) and B( 6 , − 5 )., 2 + 6 3 − 5⎞, Midpoint of AB is M ⎛⎜, ,, ⎟ , i.e., M( 4 , − 1)., ⎝ 2, 2 ⎠, Slope of AB =, , −5 − 3 −8, =, = −2., 6−2, 4, , Let the slope of LM be m. Then,, LM ⊥ AB ⇒ − 2 × m = −1 ⇒ m =, ∴, , slope of LM =, , 1, ⋅, 2, , Required equation is, 48. Slope of AB =, ∴, , tan θ =, , 1, ⋅, 2, , y − ( −1) 1, = ⇒ x − 4 = 2( y + 1) ⇒ x − 2 y − 6 = 0., x−4, 2, , 3 ( 2 − 1), 6− 3, =, =, 2−1, ( 2 − 1), , 3, , 3 ⇒ θ = 60 ° ., , Hence, AB makes an angle of 60° with the x-axis.
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Objective Questions, 49. Let A( 2 , 1), B( −2 , 3 ) and C( 4 , 5 ) be the vertices of a ABC., Let BL ⊥ AC ., Slope of AC =, , (5 − 1) 4, = = 2., ( 4 − 2) 2, , Let the slope of BL be m. Then,, −1, 2 × m = −1 ⇒ m =, ⋅, 2, y − 3 −1, equation of BL is, ∴, =, x+ 2 2, ⇒ 2 y − 6 = − x − 2 ⇒ x + 2 y − 4 = 0., 50. Let A( 1, 4 ), B( 2 , − 3 ) and C( −1, − 2 ) be the vertices of a a ABC., Let D be the midpoint of BC and let DE ⊥ BC ., 2 − 1 −3 − 2 ⎞, ⎛ 1 −5 ⎞, Then, the coordinates of D are ⎛⎜, ,, ⎟ , i.e., ⎜ , ⎟ ⋅, ⎝ 2, ⎝2 2 ⎠, 2 ⎠, −2 + 3 −1, Slope of BC =, =, ⋅ Let the slope of DE be m., 3, −1 − 2, −1, Then, m ×, = −1 ⇒ m = 3. Thus, slope of DE is 3., 3, 5, y+, 2 = 3 ⇒ 2y + 5 = 3, equation of DE is, ∴, 1, 2x − 1, x−, 2, ⇒ 2 y + 5 = 6 x − 3 ⇒ 2 y − 6 x = −8 ⇒ y − 3 x + 4 = 0 ⇒ 3 x − y − 4 = 0., −2 + 6 ⎞, 51. The midpoint of the join of A( −2 , 0 ) and B( 6 , 0 ) is M ⎛⎜, , 0 ⎟ , i.e., M( 2 , 0 )., ⎝ 3, ⎠, So, the required line is x = 2., 8 − 2⎞, 52. The midpoint of the join of A( 0 , 8 ) and B( 0 , − 2 ) is M ⎛⎜ 0 ,, ⎟ , i.e., M( 0 , 3 )., ⎝, 2 ⎠, So, the required line is y = 3., 53. Since the given line is parallel to the x-axis, we have m = 0., y+ 4, So, the required equation is, = 0 ⇒ y + 4 = 0., y− 3, 54. Here, m =, ∴, , 1, −5, and c =, ⋅, 2, 4, , the required equation is y =, , 55. Here, m = tan 30 ° =, ∴, or, , 1, 5, x − ⇒ 4 y = 2 x − 5 ⇒ 2 x − 4 y − 5 = 0., 2, 4, , 1, and c = 3., 3, , the equation of the line is y =, 3y = x + 3 3 ⇒ x −, , 1, x+ 3, 3, , 3 y + 3 3 = 0., , 56. Here, m = tan 120 ° = tan ( 180 ° − 60 ° ) = − tan 60 ° = − 3 and c = − 4., ∴, , the required equation is y = − 3 x − 4 ⇒, , 3 x + y + 4 = 0., , 1137
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1138, , Senior Secondary School Mathematics for Class 11, , 1, and x-intercept, d = 5., 3, So, the equation of the line is y = m( x − d )., 1, the required equation is y = ( x − 5 ) ⇒ x − 5 = 3 y ⇒ x − 3 y − 5 = 0., ∴, 3, 58. Here, m = −2 and d = −3., , 57. Here, m = tan θ =, , ∴, , the required equation is y = m( x − d ), , or, , y = −2( x + 3 ) ⇒ 2 x + y + 6 = 0., , −2, 7, x− ⋅, 3, 3, 3, 25, 27 x − 18 y + 25 = 0 ⇒ 18 y = 27 x + 25 ⇒ y = x +, ⋅, 2, 18, −2, 3, and m 2 = ⋅, ∴ m1 =, 3, 2, −2 3, × = −1, ⇒ m1 m 2 =, 3 2, ⇒ the given lines are perpendicular to each other., 1, 9, 60. x + 2 y − 9 = 0 ⇒ 2 y = 9 − x ⇒ y = − x + ⋅, 2, 2, −1, 5, 2 x + 4 y + 5 = 0 ⇒ 4 y = −2 x − 5 ⇒ y =, x− ⋅, 2, 4, 1, ∴ m1 = m 2 = − ⋅, 2, 59. 2 x + 3 y + 7 = 0 ⇒ 3 y = −2 x − 7 ⇒ y =, , Hence, the given lines are parallel to each other., −1, 6, 3y − 6 = 0 ⇒ 3y = − x + 6 ⇒ y =, x+, ⋅, 3, 3, −1, −1, ⇒ tan θ =, = − tan 30 ° = tan ( 180 ° − 30 ° ) = tan 150 ° ., ∴ m=, 3, 3, ∴ θ = 150 ° ., 2, 7, 62. 2 x − 3 y = 7 ⇒ 3 y = 2 x − 7 ⇒ y = x − ⋅, 3, 3, 2, slope of the required line = slope of given line = ⋅, ∴, 3, y − ( −5 ) 2, the required equation is, =, ∴, 3, x−2, y+5 2, = ⇒ 2 x − 4 = 3 y + 15 ⇒ 2 x − 3 y − 19 = 0., ⇒, x−2 3, 63. 3 x − y = 5 = 0 ⇒ y = 3 x + 5., , 61. x +, , Slope of the given line = 3., Let the slope of the required line be m., −1, Then, 3m = −1 ⇒ m =, ⋅, 3, y − ( −4 ) −1, y + 4 −1, the required equation is, =, ⇒ 3 y + 12 = − x − 2, =, ⇒, ∴, 3, 3, x − ( −2 ), x+ 2, ⇒ x + 3 y + 14 = 0.
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Objective Questions, 64. 3 x − 2 y + 5 = 0 ⇒ 2 y = 3 x + 5 ⇒ y =, Slope of the given line =, , 1139, , 3, 5, x+ ⋅, 2, 2, , 3, ⋅, 2, , Let the slope of the required line be m., 3, −2, Then, m × = −1 ⇒ m =, ⋅, 2, 3, Also, y-intercept = –3., −2, the required equation is y =, x − 3 ⇒ 2 x + 3 y + 9 = 0., ∴, 3, 1, 5, 65. x − 7 y + 5 = 0 ⇒ 7 y = x + 5 ⇒ y = x + ⋅, 7, 7, 1, Slope of this line = ⋅, 7, Let the slope of the required line be m., 1, Then, m × = −1 ⇒ m = −7., 7, Also, x-intercept, d = 3., ∴, , the required equation is y = m( x − d ), i.e., y = −7 ( x − 3 ), i.e., y + 7 x − 21 = 0., x y, 66. The required equation is + = 1, where a = 2 and b = −3., a b, x y, ∴, − = 1 ⇒ 3 x − 2 y − 6 = 0., 2 3, x y, 67. Let the equation be + = 1., a b, Then, the end points of the line segment are A( a, 0 ) and B( 0 , b ) and the midpoint of AB, is M( 2 , 3 )., a+ 0, 0+ b, a, b, ∴, = 2 and, =3 ⇒, = 2 and = 3 ⇒ a = 4 and b = 6., 2, 2, 2, 2, x y, the required equation is + = 1 ⇒ 6 x + 4 y − 24 = 0 ⇒ 3 x + 2 y − 12 = 0., ∴, 4 6, x y, 68. Since + = 1passes through A( 8 , − 9 ) and B( 12 , − 15 ), we have:, a b, 8 9, 12 15, ... (i), and, ... (ii), − =1, −, =1, a b, a, b, On solving, we get: a = 2 , b = 3., y, x, 69. Let the required equation be +, = 1., a ( 9 − a), Since it passes through ( 2 , 2 ), we have:, 2, 2, +, = 1 ⇒ 2( 9 − a) + 2 a = a( 9 − a), a 9−a, ⇒ 9 a − a2 = 18 ⇒ a2 − 9 a + 18 = 0, ⇒ ( a − 6 )( a − 3 ) = 0 ⇒ a = 6 or, , a = 3.
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1140, , Senior Secondary School Mathematics for Class 11, , ∴, , the required equation is, , ⇒, , x + 2 y − 6 = 0 or, , x y, + = 1 or, 6 3, , x y, + =1, 3 6, , 2 x + y − 6 = 0., , 70. The required equation is x cos α + y sin α = p , where α = 135 ° and p = 5, i.e., x cos 135 ° + y sin 135 ° = 5, ⇒, , ⎛ −1 ⎞, ⎛ 1 ⎞, x× ⎜, ⎟ + y× ⎜, ⎟ =5, ⎝ 2⎠, ⎝ 2⎠, , ⇒, , − x + y = 5 2 ⇒ x − y + 5 2 = 0., , 71. The required equation is x cos α + y sin α = p , where α = 150 ° and p = 8., ∴, , x cos 150 ° + y sin 150 ° = 8 ⇒ x cos ( 180 ° − 30 ° ) + y sin ( 180 ° − 30 ° ) = 8, , ⇒, , − x cos 30 ° + y sin 30 ° = 8 ⇒ − x ×, , ⇒, , − 3 x + y = 16 ⇒, , 3, 1, + y× = 8, 2, 2, , 3 x − y + 16 = 0., , 3x + y + 2 = 0 ⇒ y = − 3x − 2, , 72., ∴, , y = mx + c , where m = − 3., , 73. Here, p = 5 and α = 30 ° ., So, the equation of the line is, x cos 30 ° + y sin 30 ° = 5 ⇒ x ⋅, ⇒, , 3, 1, + y⋅ =5, 2, 2, , 3 x + y − 10 = 0., , ⎛ −2 ⎞, y, 3, x, x+ ⎜, + = 1., ⎟y = 1 ⇒, ( − 4) 2, ( − 4), ⎝ − 4⎠, 3, −4, x-intercept =, , y-intercept = 2., ∴, 3, −4, Hence, a =, and b = 2., 3, , 74. 3 x − 2 y = − 4 ⇒, , 75. Distance of the point P( 4 , 1) from the line 3 x − 4 y + 12 = 0 is, |3 × 4 − 4 × 1 + 12| 20, units = 4 units., =, d=, 5, 3 2 + ( −4 ) 2, 76. Distance of the point P( −1, 1) from the line 12 x − 5 y + 82 = 0 is, |12 × ( −1) − 5 × 1 + 82|, 65, 65, =, =, = 5 units., 169 13, ( 12 ) 2 + ( −5 ) 2, 77. Length of the perpendicular from O( 0 , 0 ) to the line 4 x + 3 y − 2 = 0 is, |4 × 0 + 3 × 0 − 2|, 2, 2, =, = unit., d=, 25 5, 42 + 32, 78. Converting each of the given lines to the form y = mx + c , we get, 3, 9, 3, 17, y = x + and y = x −, ⋅, 4, 4, 4, 8
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Objective Questions, Here, m =, , 1141, , −17, 3, 9, ⋅, , c1 = and c2 =, 8, 4, 4, , Distance between given lines =, , −17 9, −35, −, 35 4 ⎞ 7, 8, 4, 8, =, =, = ⎛⎜, × ⎟ = units., 2, ⎝, 8 5⎠ 2, 9, 25, 1+ m, 1+, 16, 16, , |c2 − c1|, , 79. Let the required point be P( x , 0 )., |4 x + 3 × 0 − 12|, Then,, = 4 ⇒ |x − 3| = 5, 42 + 32, ⇒, , x − 3 = 5 or, , x − 3 = −5 ⇒ x = 8 or, , x = −2., , Hence, the required points are (8, 0) and (–2, 0)., 80. Let the required equation be y = − x + c. Then,, |0 + 0 − c|, = 5 ⇒ |c| = 5 2 ⇒ c = 5 2 or, 12 + 12, ∴, , required equation is x + y = −5 2, , i.e., x + y + 5 2 = 0 or, , or, , c = −5 2., , x+ y=5 2, , x + y − 5 2 = 0., , ⎛ − q⎞, 81. Putting x = 0 in px + py + q = 0, we get y = ⎜ ⎟ ⋅, ⎝ p⎠, ∴, , ⎛ − q⎞, required distance = length of perp. from P ⎜ 0, ⎟ on px + py − r = 0, p⎠, ⎝, ( − q), p× 0 + p×, −r, p, |− q − r| |q + r|, =, =, =, ⋅, 2p, 2p, p2 + p2, , 82. On solving 5 x + 7 y − 3 = 0, , ... (i), , 2x − 3y − 7 = 0, y, y, x, 1, x, 1, we get, =, =, ⇒, =, =, ⋅, − 49 − 9 − 6 + 35 −15 − 14, −58 29 −29, ∴, , ... (ii), , 1 ⎞, 1 ⎞, ⎛, x = ⎛⎜ −58 ×, ⎟ = 2 and y = ⎜ 29 ×, ⎟ = −1., ⎝, ⎝, −29 ⎠, −29 ⎠, , Hence, the required point is (2, –1)., 83. On solving x − 7 y + 15 = 0 and 2 x + y = 0 , we get x = −1and y = 2., So, the given lines intersect at the point P( −1, 2 )., The line parallel to the y-axis and passing through P( −1, 2 ) is x = −1, i.e., x + 1 = 0., 84. On solving x + 2 y + 3 = 0 and 3 x + 4 y + 7 = 0 , we get x = −1and y = −1., So, the given lines intersect at P( −1, − 1)., Now, y − x = 8 ⇒ y = x + 8., Slope of this line = 1 = slope of the required line., ∴, , the required equation is, , y+ 1, = 1 ⇒ y + 1 = x + 1 ⇒ x − y = 0., x+ 1
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1142, , Senior Secondary School Mathematics for Class 11, , 85. Equation of the line AB is y =, Slope of AB =, , −1, ⋅, 3, , −1, 7, x+ ⋅, 3, 3, , Let PN ⊥ AB and let PQ = 2 × PN ., Let the slope of PQ be m., −1, Then, m ×, = −1 ⇒ m = 3., 3, y−8, Equation of PQ is, = 3 ⇒ y − 8 = 3 x − 9 ⇒ 3 x − y − 1 = 0., x− 3, On solving x + 3 y = 7 and 3 x − y = 1, we get x = 1, y = 2., Thus, AB and PQ intersect at N( 1, 2 )., 8 + y1, 3 + x1, Since N bisects PQ, we have, = 1 and, = 2., 2, 2, ∴, x1 = −1, y1 = − 4 ., Required image is Q( −1, − 4 )., 86. On solving x − y − 6 = 0 and 4 x − 3 y − 20 = 0 , we get x = 2 and y = −4., Also, x = 2 , y − 4 satisfies 6 x + 5 y + 8 = 0 as ( 6 × 2 ) + [5 × ( − 4 )] + 8 = 0., point of intersection of the given lines is (2, – 4)., ∴, 87. Let the sides AC, AB and BC of a ABC be given by, y − x = 0 , x + y = 0 and x − c = 0 respectively., On solving these equations pairwise, we get A( 0 , 0 ), B( c , − c ), and C ( c , c )., 1, area of a ABC = ⋅|x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 )|, 2, 1, = ⋅|0 ⋅ ( − c − c ) + c( c − 0 ) + c( 0 + c )| = c 2 sq units., 2, 88. On solving 3 x + y = 2 and 2 x − y = 3 , we get x = 1and y = −1., Putting x = 1and y = −1in kx + 2 y = 3, we get k = 5., 89. Given circle is x 2 + y 2 − 6 x + 4 y − 12 = 0., This is of the form x 2 + y 2 + 2 gx + 2 fy + c = 0 , where, ( 2 g = −6 and 2 f = 4 ) ⇒ ( g = −3 and f = 2)., ∴, , centre is ( − g , − f ) = ( 3 , − 2 )., , 90. Given circle is x 2 + y 2 − 4 x + 2 y − 45 = 0., This is of the form x 2 + y 2 + 2 gx + 2 fy + c = 0 , where, ( 2 g = − 4 and 2 f = 2 ) ⇒ ( g = −2 and f = 1). Also, c = − 45., ∴, , radius = g 2 + f 2 − c = ( −2 ) 2 + 12 + 45 = 50 = 5 2 ., , 91. Given end points are ( x1 , y1 ) = ( 2 , − 3 ) and ( x2 , y2 ) = ( −3 , 5 )., ∴, , equation of the circle is ( x − x1 )( x − x2 ) + ( y − y1 )( y − y2 ) = 0, , ⇒, ⇒, , ( x − 2 )( x + 3 ) + ( y + 3 )( y − 5 ) = 0, x 2 + y 2 + x − 2 y − 21 = 0.
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Objective Questions, 92. Radius of the circle is AC = ( 3 − 2 ) 2 + ( 2 + 5 ) 2 = 1 + 49 = 50 ., ∴, , equation of the circle is ( x − 2 ) 2 + ( y + 5 ) 2 = 50, , ⇒, , x 2 + y 2 − 4 x + 10 y − 21 = 0., , 93. Given circle is ( x + 5 ) 2 + ( y − 3 ) 2 = 20., Its centre is (–5, 3) and radius = 20 = 2 5., 94. Given equation is x 2 + y 2 − 6 x + 5 y − 7 = 0., This is of the form x 2 + y 2 + 2 gx + 2 fy + c = 0 , where, , ∴, , 5, ( 2 g = − 6 , 2 f = 5 and c = −7 ) ⇒ ⎛⎜ g = −3 , f = , c = −7 ⎞⎟, ⎝, ⎠, 2, −5, centre is ( − g , − f ) = ⎛⎜ 3 , ⎞⎟ ⋅, ⎝, 2 ⎠, , But, C is the midpoint of AB, where A( −1, 3 ) and B(α , β ) are given., ⎛ 3 = −1 + α and −5 = 3 + β ⎞ ⇒ (α = 7 and β = −8 )., ∴, ⎜, ⎟, ⎝, 2, 2, 2 ⎠, 95. General equation is y 2 = 4 ax. Here, 4 a = 8 ⇒ a = 2., ∴, , focus of the parabola is F( a, 0 ) = F( 2 , 0 ) and its vertex is O( 0 , 0 )., , 96. Given equation is y 2 = − 4 ax , where a = 3., focus is F( − a, 0 ) = F( −3 , 0 )., ∴, And, the directrix is x = a ⇒ x = 3., 97. Given equation is y 2 = − 4 ax , where a = 2., ∴, , its focus is F( − a, 0 ) = F( −2 , 0 )., , Its directrix is x = a ⇒ x = 2., 98. Given equation is x 2 = − 4 ay , where a = 4., Its focus is F( 0 , − a) = F( 0 , − 4 )., Its directrix is y = a ⇒ y = 4., 99. Given equation is x 2 = 4 ay , where 4 a = 6 ⇒ a =, , 3, ⋅, 2, , 3, Its focus is F( 0 , a) = F ⎛⎜ 0 , ⎞⎟ ⋅, ⎝ 2⎠, −3, Its directrix is y = − a ⇒ y =, ⋅, 2, 100. Since the point P( 3 , 2 ) lies on y 2 = 4 ax , we have, 1, 4a × 3 = 2 2 ⇒ a = ⋅, 3, 1, 4, ∴ latus rectum = 4 a = ⎛⎜ 4 × ⎞⎟ = ⋅, ⎝, 3⎠ 3, 101. Let ZM be the directrix, where coordinates, of Z are (0, 0)., Let the focus be F( a, 0 )., Then, vertex A is the midpoint of ZF., a+ 0, ∴, = 2 ⇒ a = 4., 2, ∴ focus is F( 4 , 0 )., , 1143
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1144, , Senior Secondary School Mathematics for Class 11, , 102. Let aOAB be the equilateral triangle inscribed, in a parabola y 2 = 4 ax., Then, ∠COB = 30 ° ., Let OA = OB = AB = b., ⎛ 3b b ⎞, Then, we have B( b cos 30 ° , b sin 30 ° ) = B ⎜, , ⎟⋅, ⎝ 2 2⎠, This point B lies on y 2 = 4 ax., ∴, , b2, = 2 3 ab ⇒ b = 8 3 a., 4, , π, 103. Let AP be the chord of length p, making an angle of ⎛⎜ ⎞⎟ with the axis of the parabola., ⎝ 4⎠, Then, coordinates of P are, p ⎞, π, π, ⎛ p, P ⎛⎜ p cos , p sin ⎞⎟ = P ⎜, ,, ⎟⋅, ⎝, ⎝ 2, 4, 4⎠, 2⎠, p, p ⎞, 2, Since P ⎛⎜, ,, ⎟ lies on y = 4 ax , we have, ⎝ 2, 2⎠, p2, p, = 4a ×, ⇒ p = 4 a 2., 2, 2, 104. Given parabola is y 2 = 4 ax , where a = 3., Its directrix is x = − a ⇒ x = −3 ⇒ x + 3 = 0., Focal distance of P( x1 , y1 ) = distance of P( x1 , y1 ) from x + 3 = 0, x + 3 x1 + 3, = 1, =, = x1 + 3., 1, 12, ∴, , x1 + 3 = 4 ⇒ x1 = 1., , 105. Let O( 0 , 0 ) be the vertex and F( 0 , 2 ) be the focus., Clearly, it is a case of upward parabola., Let its equation be x 2 = 4 ay., Since focus is F( 0 , a), so a = 2., ∴, , its equation is x 2 = 4 × 2 × y ⇒ x 2 = 8 y., , 106. The vertex of the parabola is O( 0 , 0 ) and its axis lies along the x-axis. So, its equation, is y 2 = 4 ax or y 2 = − 4 ax., But, it passes through the point P( 2 , 3 ), so its equation is y 2 = 4 ax.., Since it passes through P( 2 , 3 ), we have 4 a × 2 = 9 ⇒ a =, ∴, , its equation is y 2 = 4 ×, , 9, 9, x ⇒ y 2 = x., 8, 2, , 9, ⋅, 8
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Objective Questions, , 1145, , 107. It is given that the vertex of the parabola is O( 0 , 0 ) and it is symmetric about the, y-axis., So, its equation is x 2 = 4 ay or, , x 2 = − 4 ay., , Since it passes through the point P( 3 , − 4 ), so it lies in quadrant IV., ∴, , it is a downward parabola., , Let its equation be x 2 = − 4 ay., Since it passes through P( 3 , − 4 ), we have 16 a = 9 ⇒ a =, ∴, , its equation is x 2 = − 4 ×, , 108. Given equation is, This is of the form, , 9, ⋅, 16, , 9, y ⇒ 4 x 2 + 9 y = 0., 16, , x2 y2, +, = 1., 16, 9, x2, a, , 2, , +, , y2, b2, , = 1, where a2 > b 2 ., , Here, ( a2 = 16 ⇒ a = 4 ) and ( b 2 = 9 ⇒ b = 3 )., ∴, , c = a2 − b 2 = 16 − 9 = 7 ., , Thus, a = 4 , b = 3 , c = 7 ., (i) Length of the major axis = 2 a = 8 units., (ii) Length of the minor axis = 2 b = 6 units., (iii) Coordinates of its vertices are A( − a, 0 ) and B( a, 0 ), i.e., A( −4 , 0 ) and B( 4 , 0 )., (iv) Coordinates of its foci are F1 ( − c , 0 ) and F2 ( c , 0 ), i.e., F1 ( − 7 , 0 ) and F2 ( 7 , 0 )., (v) Eccentricity, e =, , c, 7, =, ⋅, 4, a, , (vi) Length of the latus rectum =, 109. Given equation is, This is of the form, , x2 y2, +, = 1., 9, 4, x2, a, , 2, , +, , y2, b2, , 2b 2 ( 2 × 9), =, = 6 units., a, 3, , = 1, where a2 > b 2 ., , Here, ( a2 = 9 ⇒ a = 3 ) and ( b 2 = 4 ⇒ b = 2 )., ∴, , c = a2 − b 2 = 9 − 4 = 5., , Thus, a = 3 , b = 2 and c = 5., (i) Length of the major axis = 2 a = 6 units., (ii) Length of the minor axis = 2 b = 4 units., (iii) Coordinates of its vertices are A( − a, 0 ) and B( a, 0 ), i.e., A( −3 , 0 ) and B( 3 , 0 )., (iv) Coordinates of its foci are F1 ( − c , 0 ) and F2 ( c , 0 ), i.e., F1 ( 5 , 0 ) and F2 ( − 5 , 0 )., (v) Eccentricity, e =, , c, 5, =, ⋅, a, 3, , (vi) Length of the latus rectum =, , 2b 2 2 × 4 8, =, = units., a, 3, 3
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1146, , Senior Secondary School Mathematics for Class 11, x2 y2, +, = 1., 4, 36, x2 y2, This is of the form, +, = 1, where a2 > b 2 ., b2, a2, So, it is an equation of a vertical ellipse., , 110. Given equation is, , Now, ( b 2 = 4 and a2 = 36 ) ⇒ ( b = 2 and a = 6)., ∴ c = a2 − b 2 = 36 − 4 =, Thus, a = 6 , b = 2 , c = 4 2 ., , 32 = 4 2., , (i) Length of the major axis = 2 a = 12 units., (ii) Length of the minor axis = 2 b = 4 units., (iii) Coordinates of its vertices are A( 0 , − a) and B( 0 , a), i.e., A( 0 , − 6 ) and B( 0 , 6 )., (iv) Coordinates of its foci are F1 ( 0 , − c ) and F2 ( 0 , c ), i.e., F1 ( 0 , − 4 2 ) and F2 ( 0 , 4 2 )., c 4 2 2 2, =, =, ⋅, 6, 3, a, 2b 2 ( 2 × 4) 4, (vi) Length of the latus rectum =, =, = units., a, 6, 3, 2, 2, y, x, 111. Given equation is, +, = 1., 25 100, x2 y2, This is of the form, +, = 1, where a2 > b 2 ., b2, a2, ∴ it is an equation of a vertical ellipse., (v) Eccentricity, e =, , Now, ( b 2 = 25 , a2 = 100 ) ⇒ ( b = 5 , a = 10 )., ∴, , c = a2 − b 2 = 100 − 25 = 75 = 5 3., , Thus, a = 10 , b = 5 and c = 5 3., (i) Length of the major axis = 2 a = 20 units., (ii) Length of the minor axis = 2 b = 10 units., (iii) Coordinates of its vertices are A( 0 , − a) and B( 0 , a), i.e., A( 0 , − 10 ) and B( 0 , 10 )., (iv) Coordinates of its foci are F1 ( 0 , − c ) and F2 ( 0 , c ), i.e., F1 ( 0 , − 5 3 ) and F2 ( 0 , 5 3 )., c 5 3, 3, =, =, ⋅, 10, 2, a, 2 b 2 ( 2 × 25 ), (vi) Length of the latus rectum =, =, = 5 units., a, 10, 112. Since the vertices of the given ellipse are on the x-axis, so it is a horizontal ellipse., (v) Eccentricity, e =, , Let the equation of the ellipse be, , x2, 2, , +, , y2, , a, b2, Its vertices are ( ± a, 0 ) = ( ±5 , 0 ) ⇒ a = 5., , = 1, where a2 > b 2 ., , Its foci are ( ± c , 0 ) = ( ±4 , 0 ) ⇒ c = 4 ., c 2 = ( a2 − b 2 ) ⇒ b 2 = ( a2 − c 2 ) = ( 25 − 16 ) = 9 ⇒ b = 3., ∴, , required equation of the ellipse is, , x2 y2, +, = 1., 25, 9
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Objective Questions, 113. Since the foci of the given ellipse are on the x-axis, so it is a horizontal ellipse., x2, , y2, , = 1, where a2 > b 2 ., a, b2, Let its foci be ( ± c , 0 ). Then, c = 4., 4, c, c, Also, e = ⇒ a = =, = 12., a, e ⎛ 1⎞, ⎜ ⎟, ⎝ 3⎠, Let its equation be, , 2, , +, , c 2 = ( a2 − b 2 ) ⇒ b 2 = ( a2 − c 2 ) = ( 144 − 16 ) = 128., ∴, , a2 = ( 12 ) 2 = 144 and b 2 = 128., , y2, x2, +, = 1., 144 128, 114. Since the vertices of the given ellipse lie on the y-axis, it is a vertical ellipse., ∴, , the required equation of the ellipse is, , x2, , y2, , = 1, where a2 > b 2 ., b, a2, Its vertices are ( 0 , ± a) = ( 0 , ± 13 ) ⇒ a = 13., Let the required equation be, , 2, , +, , Its foci are ( 0 , ± c ) = ( 0 , ±5 ) ⇒ c = 5., ∴, , b 2 = ( a2 − c 2 ) = ( 169 − 25 ) = 144 ., , Thus, b 2 = 144 and a2 = 169., ∴, , the required equation is, , y2, x2, +, = 1., 144 169, , 115. Since the foci of the given ellipse lie on the y-axis, it is a vertical ellipse., Let its equation be, , x2, b, , 2, , +, , y2, a2, , = 1, where a2 > b 2 ., , Let c 2 = ( a2 − b 2 )., Then, its foci are ( 0 , ± c ) = ( 0 , ± 6 ) ⇒ c = 6., 1, b = length of semi-minor axis = ⎛⎜ × 16 ⎞⎟ = 8., ⎝2, ⎠, ∴, , a2 = ( c 2 + b 2 ) = ( 36 + 64 ) = 100., , Thus, b 2 = 64 and a2 = 100., Hence, the required equaton is, , y2, x2, +, = 1., 64 100, , 116. Since the foci of the given ellipse lie on the y-axis, so it is a vertical ellipse., Let its equation be, , x2, b2, , +, , y2, a2, , = 1, where a2 > b 2 ., , Let c 2 = ( a2 − b 2 )., Its foci are ( 0 , ± c ) = ( 0 , ±5 ) ⇒ c = 5., 1, a = length of semi-major axis = ⎛⎜ × 20 ⎞⎟ = 10., ⎝2, ⎠, c 2 = ( a2 − b 2 ) ⇒ b 2 = ( a2 − c 2 ) = ( 100 − 25 ) = 75., , 1147
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1148, , Senior Secondary School Mathematics for Class 11, , ∴, , a2 = ( 10 ) 2 = 100 and b 2 = 75., , y2, x2, +, = 1., 75 100, 117. Since the vertices of the ellipse lie on the y-axis, it is a vertical ellipse., Hence, the required equation is, , Let the required equation be, , x2, b, , 2, , +, , y2, a2, , = 1, where a2 > b 2 ., , Its vertices are ( 0 , ± a) = ( 0 , ± 10 ) ⇒ a = 10., Let c 2 = ( a2 − b 2 )., Now, e =, , 4, c, ⇒ c = ae = ⎛⎜ 10 × ⎞⎟ = 8., ⎝, 5⎠, a, , ∴, , c 2 = ( a2 − b 2 ) ⇒ b 2 = ( a2 − c 2 ) = ( 100 − 64 ) = 36., , ∴, , a2 = ( 10 ) 2 = 100 and b 2 = 36., , Hence, the required equation is, 118. Given equation is, , x2 y2, −, = 1., 36 64, , Comparing it with, ∴, , x2, a, , c = a2 + b 2 =, , 2, , −, , y2, b2, , y2, x2, +, = 1., 36 100, , = 1, we get a = 6 , b = 8., , 36 + 44 = 100 = 10., , Thus, we have, (i) Length of the transverse axis = 2 a = 12 units., (ii) Length of the conjugate axis = 2 b = 16 units., (iii) Coordinates of its vertices are A( − a, 0 ) and B( a, 0 ), i.e., A( − 6 , 0 ) and B( 6 , 0 )., (iv) Coordinates of its foci are F1 ( − c , 0 ) and F2 ( c , 0 ), i.e., F1 ( −10 , 0 ) and F2 ( 10 , 0 )., c 10 5, (v) Eccentricity, e = =, = ⋅, 3, a 6, 2 b 2 2 × 64 64, units., (vi) Length of the latus rectum =, =, =, a, 6, 3, x2 y2, 119. 9 x 2 − 16 y 2 = 144 ⇒, −, = 1., 16, 9, Comparing it with, ∴, , x2, a, , 2, , −, , y2, b2, , = 1, we get a = 4 , b = 3., , c = a2 + b 2 = 16 + 9 = 25 = 5., , Thus, we have, (i) Length of the transverse axis = 2 a = 8 units., (ii) Length of the conjugate axis = 2 b = 6 units., (iii) Its vertices are A( − a, 0 ) and B( a, 0 ), i.e., A( − 4 , 0 ) and B( 4 , 0 )., (iv) Its foci are F1 ( − c , 0 ) and F2 ( c , 0 ), i.e., F1 ( −5 , 0 ) and F2 (5 , 0 )., c 5, (v) Eccentricity, e = = ⋅, a 4
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Objective Questions, , 1149, , 2b 2 2 × 9 9, =, = units., a, 4, 2, , (vi) Length of the latus rectum =, , 120. Since the vertices of the given hyperbola are of the form ( ± a, 0 ), so it is a horizontal, hyperbola., x2 y2, Let the required equation be, −, = 1., a2 b 2, Then, its vertices are ( ± a, 0 ) = ( ±2 , 0 ) ⇒ a = 2., Let its foci be ( ± c , 0 ) = ( ±3 , 0 ) ⇒ c = 3., ∴, , b 2 = ( c 2 − a2 ) = ( 3 2 − 2 2 ) = ( 9 − 4 ) = 5., , ∴, , a2 = 2 2 = 4 , b 2 = 5., , Hence, the required equation is, , x2 y2, −, = 1., 4, 5, , 121. Since the foci of the given hyperbola are ( ±5 , 0 ), so it is a horizontal hyperbola., Let the required equation be, , x2, 2, , −, , y2, , a, b2, 2, Now, 2 a = 8 ⇒ a = 4 ⇒ a = 16., , = 1., , Let its foci be ( ± c , 0 ) = ( ±5 , 0 ) ⇒ c = 5., ∴, , b 2 = ( c 2 − a2 ) = (5 2 − 4 2 ) = ( 25 − 16 ) = 9., , Thus, a2 = 16 and b 2 = 9., Hence, the required equation is, , x2 y2, −, = 1., 16, 9, , 122. Since the foci of the given hyperbola are ( ±4 , 0 ), so it is a horizontal hyperbola., x2, , ∴, , 2, , −, , y2, , = 1., a, b2, Its foci are ( ± c , 0 ) = ( ±4 , 0 ) ⇒ c = 4 ., Let its equation be, , c 2 = a2 + b 2 ⇒ a2 + b 2 = 4 2 = 16., 2b 2, = 12 ⇒ b 2 = 6 a., a, a2 + 6 a = 16 ⇒ a2 + 6 a − 16 = 0 ⇒ ( a + 8 )( a − 2 ) = 0 ⇒ a = 2., , Length of the latus rectum = 12 ⇒, ∴, , b 2 = ( 6 × 2 ) = 12 and a2 = 2 2 = 4 ., x2 y2, Hence, the required equation is, −, = 1., 4, 12, , ∴, , y2 x2, −, = 1, which is a vertical hyperbola., 4, 9, Here, ( a2 = 4 , b 2 = 9 ) ⇒ ( a = 2 , b = 3 )., , 123. Given equation is, , ∴, , c = a2 + b 2 = 4 + 9 = 13., , (i) Length of the transverse axis = 2 a = 4 units., (ii) Length of the conjugate axis = 2 b = 6 units., (iii) Its vertices are A( 0 , − a) and B( 0 , a), i.e., A( 0 , − 2 ) and B( 0 , 2 )., (iv) Its foci are F1 ( 0 , − c ) and F2 ( 0 , c ), i.e., F1 ( 0 , − 13 ) and F2 ( 0 , 13 ).
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1150, , Senior Secondary School Mathematics for Class 11, c, 13, =, ⋅, a, 2, , (v) Eccentricity, e =, , (vi) Length of the latus rectum =, , 2b 2 2 × 9, =, = 9 units., a, 2, , 124. Since the vertices of the given hyperbola are of the form ( 0 , ± a), it is a vertical, y2 x2, hyperbola. Let its equation be, −, = 1., a2 b 2, Its vertices are ( 0 , ± a) = ( 0 , ± 3 ) ⇒ a = 3., Its foci are ( 0 , ± c ) = ( 0 , ±5 ) ⇒ c = 5., b 2 = ( c 2 − a2 ) = ( 25 − 9 ) = 16 , a2 = 9., ∴, , its equation is, , y2 x2, −, = 1., 9, 16, , 125. Since the vertices of the given hyperbola are ( 0 , ± a), so it is a vertical hyperbola. Let, y2 x2, its equation be, −, = 1., a2 b 2, Its vertices are ( 0 , ± a) = ( 0 , ± 6 ) ⇒ a = 6 ⇒ a2 = 36., Now, e =, , 5, 5, and c = ae = ⎛⎜ 6 × ⎞⎟ = 10 ⇒ c 2 = 100., ⎝, 3⎠, 3, , b 2 = ( c 2 − a2 ) = ( 100 − 36 ) = 64 ., ∴, , the required equation is, , y2 x2, −, = 1., 36 64, , 126. Since the foci are ( 0 , ± c ), so it is a case of vertical hyperbola., y2, 2, , −, , x2, , = 1., a, b2, Length of the transverse axis = 6 ⇒ 2 a = 6 ⇒ a = 3., Let its equation be, , Also, c = 4 (given)., ∴, , c 2 = ( a2 + b 2 ) ⇒ b 2 = ( c 2 − a2 ) = ( 4 2 − 3 2 ) = ( 16 − 9 ) = 7 ., , Thus, a2 = 3 2 = 9 and b 2 = 7 ., Hence, the required equation is, , y2 x2, −, = 1., 9, 7, , 127. Since the foci of the hyperbola are ( 0 , ± c ), so it is a vertical hyperbola., y2, 2, , −, , x2, , = 1., a, b2, Its foci are ( 0 , ± c ) = ( 0 , ± 6 ) ⇒ c = 6., , Let its equation be, , 2 b = 2 11 ⇒ b = 11 ⇒ b 2 = 11., c 2 = ( a2 + b 2 ) ⇒ a2 = ( c 2 − b 2 ) = ( 36 − 11) = 25., ∴, , a2 = 25 and b 2 = 11., , ∴, , the required equation is, , y2 x2, −, = 1., 25 11, , q
