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CBSE, New Pattern, Mathematics, Class 11, , (Term I)

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ARIHANT PRAKASHAN (School Division Series), , © Publisher, No part of this publication may be re-produced, stored in a retrieval system or by any, means, electronic, mechanical, photocopying, recording, scanning, web or otherwise, without the written permission of the publisher. Arihant has obtained all the information, in this book from the sources believed to be reliable and true. However, Arihant or its, editors or authors or illustrators don’t take any responsibility for the absolute accuracy of, any information published and the damage or loss suffered thereupon., , All disputes subject to Meerut (UP) jurisdiction only., Administrative & Production Offices, Regd. Office, ‘Ramchhaya’ 4577/15, Agarwal Road, Darya Ganj, New Delhi -110002, Tele: 011- 47630600, 43518550, , Head Office, Kalindi, TP Nagar, Meerut (UP) - 250002, Tel: 0121-7156203, 7156204, , Sales & Support Offices, Agra, Ahmedabad, Bengaluru, Bareilly, Chennai, Delhi, Guwahati,, Hyderabad, Jaipur, Jhansi, Kolkata, Lucknow, Nagpur & Pune., , ISBN : 978-93-25793-67-5, PO No : TXT-XX-XXXXXXX-X-XX, Published by Arihant Publications (India) Ltd., For further information about the books published by Arihant, log on to, www.arihantbooks.com or e-mail at info@arihantbooks.com, Follow us on, , CBSE, New Pattern

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Contents, Chapter, Sets, , -, , Chapter, Relations and Functions, , -, , Chapter, Complex Numbers, , -, , Chapter, Sequence and Series, , -, , Chapter, Straight Lines, , -, , Chapter, Limits, , -, , Chapter, Statistics, , -, , Practice Papers, , CBSE, New Pattern, , -, , -

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Syllabus, THOERY, No., , UNITS, , Marks, , I, , Sets and Functions, , II, , Algebra, , III, , Coordinate Geometry, , IV, , Calculus, , V, , Statistics, , Probability, , Total, Internal Assessement, Total, No chapter-wise weightage. Care to be taken to cover all the chapters., , UNIT I SETS AND FUNCTIONS, . Sets, Sets and their representations. Empty set. Finite and Infinite sets., Equal sets. Subsets. Subsets of a set of real numbers especially, intervals with notations . Power set. Universal set. Venn diagrams., Union and Intersection of sets., . Relations Functions, 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 R x R only .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., , CBSE, New Pattern

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UNIT II ALGEBRA, . Complex Numbers and Quadratic Equations, Need for complex numbers, especially√− , to be motivated by, inability to solve some of the quardratic equations. Algebraic, properties of complex numbers. Argand plane. Statement of, Fundamental Theorem of Algebra, solution of quadratic equations, with real coefficients in the complex number system., , . Sequence and Series, Sequence and Series. Arithmetic Progression A.P. . Arithmetic Mean, A.M. Geometric Progression G.P. , general term of a G.P., sum of n, terms of a G.P., infinite G.P. and its sum, geometric mean G.M. ,, relation between A.M. and G.M., , UNIT III COORDINATE GEOMETRY, . Straight Lines, Brief recall of two dimensional geometry from earlier classes. 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, twopoint form, intercept form and normal form. General equation of a, line. Distance of a point from a line., , UNIT IV CALCULUS, . Limits, Intuitive idea of limit. Limits of polynomials and rational functions,, trigonometric, exponential and logarithmic functions, , UNIT V STATISTICS AND PROBABILITY, . Statistics, Measures of Dispersion: Range, mean deviation, variance and, standard deviation of ungrouped grouped data., , Internal Assessment, , Marks, , Periodic Test, , Marks, , Mathematical Activities : Activity file record + Term end, Assessment of one activity Viva, , Marks, , CBSE, New Pattern

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CBSE New Pattern ~ Mathematics XI (Term I), , 1, , 01, Sets, Quick Revision, A well-defined collection of objects, is called a set., Sets are denoted by capital letters A, B , C , X , Y , Z, etc. and elements of a set are denoted by, a , b , c , x , y , z etc., If a is an element of set A, then we say that a, belongs to A. The phrase ‘belongs to’ denoted by, the Greek symbol ∈ (epsilon). Thus, we write and, written as a ∈ A and b does not belongs to set A is, written as b ∉ A., , Representation of Sets, There are two ways of representing a set, (i) Roster form or Tabular form or Listing, method In the roster form, we list all the, elements of the set within curly braces {} and, separate them by commas., (ii) Set-builder form or Rule method In the, set-builder form, we list the property or, properties satisfied by all the elements of the, sets., , Types of Sets, (i) Empty set A set which does not contain any, element is called an empty set or the void set, or the null set and it is denoted by {} or φ ., (ii) Singleton set A set consisting of a single, element, is called a singleton set., (iii) Finite and infinite sets A set which is empty, or consists of a finite number of elements is, called a finite set, otherwise, the set is called, an infinite set., , (iv) Equivalent sets Two finite sets A and B are, said to be equal, if they have equal number of, elements, i.e. n ( A ) = n ( B )., (v) Equal sets Two 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, said to be unequal and we write A ≠ B ., , Subset, A set A is said to be a subset of a set B, if every, element of A is also an element of B. In symbols,, we can write, A ⊂ B , if x ∈ A ⇒ x ∈ B, Also, if A ⊂ B and A ≠ B , then A is called a proper, subset of B and B is called superset of A., Note, (i) Every set is a subset of itself., (ii) The empty set is a subset of every sets., (iii) The total number of subsets of a finite set, containing n elements is 2n ., Subsets of the Set of Real Numbers, We know that, every real number is either a, rational or an irrational number and the set of real, numbers is denoted by R. There are many, important subsets of set of real numbers which are, given below, (i) Natural numbers The numbers being used in, counting as 1, 2, 3, 4,..., called natural numbers., The set of natural numbers is denoted by N., Thus,, N = {1, 2, 3, 4, ...}

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2, , CBSE New Pattern ~ Mathematics XI (Term I), , (ii) Whole numbers The natural numbers along, with number 0 (zero) form the set of whole, numbers i.e. 0, 1, 2, 3, ... , are whole numbers., The set of whole numbers is denoted by W., Thus,, W = {0, 1, 2, 3, ...}, (iii) Integers The natural numbers, their negatives, and zero make the set of integers and it is, denoted by Z., Z = {..., − 5, − 4 , − 3, − 2, − 1, 0, 1, 2, 3, 4 , ...}, p, (iv) Rational numbers A number of the form ,, q, where p and q both are integers and q ≠ 0, (division by 0 is not permissible), is called, a rational number., The set of rational numbers is generally, denoted by Q., , p, Thus, Q =  : p , q ∈ Z and q ≠ 0, , q, (v) Irrational numbers A number which cannot, be written in the form p/q, where p and q both, are integers and q ≠ 0, is called an irrational, number., The set of irrational numbers is denoted by T ., Thus, T = { x : x ∈ R and x ∉Q }, Diagrammatical Representation All the subsets, can be represented diagrammatically as given, below, Integers, , Real, , number, , Whole, number, , Rational, number, Irrational, number, , Natural, number, , Intervals as Subsets of R, Let a and b be two given real numbers such that, a < b , then, (i) the set of real numbers { x : a < x < b } is called, an open interval and is denoted by (a , b )., (ii) the set of real numbers { x : a ≤ x ≤ b } is called, a closed interval and is denoted by [a, b]., (iii) intervals closed at one end and open at the, other are known as semi-open or semi-closed, intervals. [ a , b ) = { x : a ≤ x < b } is an open, , interval from a to b which includes a but, excludes b. (a , b ] = { x : a < x ≤ b } is an open, interval from a to b which excludes a but, includes b., , Universal Set, 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 and is denoted by U., , Power Set, The collection of all subsets of a set A is called the, power set of A. It is denoted by P ( A ). If the, number of elements in A, i.e. n ( A ) = m , then the, number of elements in P ( A ) i. e. n [P ( A )] = 2m ., Properties of Power Sets, (i) If A ⊆ B , then P ( A ) ⊆ P ( B )., (ii) P ( A ) ∩ P ( B ) = P ( A ∩ B ), (iii) P ( A ∪ B ) ≠ P ( A ) ∪ P ( B ), , Venn Diagrams, Venn diagrams are the diagrams, which represent, the relationship between sets. In Venn diagrams,, the universal set is represented usually by a, rectangular region and its subset are represented, usually by circle or a closed geometrical figure, inside the universal set. Also, an element of a set is, represented by a point within the circle of set., e.g. If U = {1, 2, 3, 4 , ..., 10} and A = {1, 2, 3}, then, its Venn diagram is as shown in the figure, U, , 10 A, , 4, , 9, 1, , 5, 6, , 2, , 8, , 3, 7, , Operations on Sets, (i) Union of sets The union of two sets A and B, is the set of all those elements which belong to, either in A or in B or in both A and B. It is, denoted by A ∪ B ., Thus, A ∪ B = { x : x ∈ A or x ∈ B }, (ii) Intersection of sets The intersection of two, sets A and B is the set of all those elements,, which are common to both A and B. It is, denoted by A ∩ B ., Thus, A ∩ B = { x : x ∈ A and x ∈ B }

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CBSE New Pattern ~ Mathematics XI (Term I), , (iii) Disjoint sets Two sets A and B are said to be, disjoint sets, if they have no common elements, i.e. if A ∩ B = φ., , 3, Results on Number of Elements in Sets, (i) When number of elements of two sets are, given, , Laws of Algebra of Sets, (i) Idempotent laws For any set A, we have, (a) A ∪ A = A, (b) A ∩ A = A, (ii) Identity laws For any set A, we have, (a) A ∪ φ = A, (b) A ∩ U = A, (iii) Commutative laws For any two sets A, and B, we have, (a) A ∪ B = B ∪ A, (b) A ∩ B = B ∩ A, (iv) Associative laws For any three sets A, B, and C , we have, (a) A ∪ ( B ∪ C ) = ( A ∪ B ) ∪ C, (b) A ∩ ( B ∩ C ) = ( A ∩ B ) ∩ C, (v) Distributive laws If A, B and C are three, sets, then, (a) A ∪ ( B ∩ C ) = ( A ∪ B ) ∩ ( A ∪ C ), (b) A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ), , B, , A, a, , b, , U, , c, , Then, n ( A ∪ B ) = a + b + c, (ii) When number of elements of three sets are, given, A, , B, , a, , b c, e f g, d, , U, , C, , Then,, n (A ∪ B ∪ C ) = a + b + c + d + e + f + g, , Objective Questions, Multiple Choice Questions, 1. If A ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then, insert appropriate symbol ∈or ∉ in each, of the following blank spaces., (i) 4 ... A (ii) − 4 ... A (iii) 12 ... A are, (a) ∈, ∈, ∈, (c) ∈, ∉, ∉, , (b) ∈, ∉, ∈, (d) ∉, ∉, ∉, , 2. The following set in Roster form is, {x : x is positive integer and a divisor, of 9}, (a) {1, 3, 9}, (c) {9, 8, 27}, , (b) {1, 3, 8}, (d) None of these, , 3. The set of all natural numbers x such, that 4x + 9 < 50 in roster form is, , (a), (b), (c), (d), , {1, 2, 4, 6, 8, 10}, {1, 3, 5, 7, 9}, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, None of the above, , 4. List of the elements of the following set, {x : x =, , n, , 1 + n2, n ∈ N }, is, , and 1 ≤ n ≤ 3, where, ,  3 1 2, (a)  , , , 10 5 3 ,  1 2 3, (c)  , , , 2 5 5 , , 1 2 3 , (b)  , , , 2 3 10 , 1 2 3 , (d)  , , , 2 5 10 , , 5. The set A = {14, 21, 28, 35, 42, ..., 98} in, set-builder form is, (a), (b), (c), (d), , A = { x : x = 7n, n ∈N and 1 ≤ n ≤ 15}, A = { x : x = 7n, n ∈N and 2 ≤ n ≤ 14}, A = { x : x = 7n, n ∈N and 3 ≤ n ≤ 13}, A = { x : x = 7n, n ∈N and 4 ≤ n ≤ 12}, , 6. If a set is denoted as A = φ, then, number of elements in A is, (a) 0, (c) 2, , (b) 1, (d) 3

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4, , CBSE New Pattern ~ Mathematics XI (Term I), , 7. A set B = {5} is called, (a) singleton set, (c) infinite set, , 16. If A = {2, 4, 6}, B = {1, 3, 5} and, (b) empty set, (d) None of these, , 8. The set of months of a year is …X…, set. Here, X refers to, (a) empty, (c) infinite, , (b) finite, (d) singleton, , 9. Let A = {x : x is a square of a natural, number and x is less than 100} and B is, a set of even natural numbers. The, cardinality of A ∩ B is, (a) 4, (c) 9, , (b) 5, (d) None of these, , 10. The set {1, 2, 3, ...} is …Y… set. Here,, Y refers to, (a) null, (c) infinite, , (b) finite, (d) singleton, , 11. From the following sets given below,, pair the equivalent sets., A = {1, 2, 3}, B = {t , p, q , r , s }, C = {α, β, γ }, and D = {a, e, i, o, u}, (a) Sets A, C and A, D, (c) Sets A, C and B, D, , (b) Sets A, B and B, D, (d) Sets A, C and B, C, , 12. If A = the set of letters in ‘ALLOY’ and, , B = the set of letters in ‘LOYAL’, then A, and B are …X… . Here, X refers to, (a) equal, (c) disjoint, , (b) unequal, (d) None of these, , 13. If X = {8 n – 7n – 1 | n ∈ N } and, [NCERT Exemplar], , (b) Y ⊂ X, (d) X ∩ Y = φ, , 14. Two finite sets have m and n elements., The number of subsets of the first set is, 112 more than that of the second set., The values of m and n are, respectively, [NCERT Exemplar], , (a) 4, 7, , (b) 7, 4, , (c) 4, 4, , (d) 7, 7, , 15. Write the {x : x ∈ R, − 5 < x ≤ 6} as, interval, then the length of the interval is, (a) 9, , (b) 10, , (c) 11, , (a), (b), (c), (d), , {0, 7}, {1, 2, 3, 4, 5, 6}, {0,1, 3, 5, 7}, {0, 1, 2, 3, 4, 5, 6, 7}, , 17. Total number of elements in the power, set of A containing 15 elements is, (a) 2 15, (c) 2 15 − 1, , (b) 152, (d) 2 15 − 1, , 18. If A = P ({1, 2}), where P denotes the, power set, then which one of the, following is correct?, (a) {1, 2 } ⊂ A, (c) φ ∉A, , (b) 1 ∈A, (d) {1, 2 } ∈A, , 19. If A = {1, 3, 5, 7}, then what is the, cardinality of the power set P ( A )?, (a) 8, (c) 16, , (b) 15, (d) 17, , 20. If A and B are two sets, then, A ∩ ( A ∪ B ) equals to, (a) A, (c) φ, , [NCERT Exemplar], , (b) B, (d) A ∩ B, , 21. If A = {2, 4, 6, 8} and B = {6, 8, 10, 12},, then A ∪ B is, (a) {2, 4, 6, 8}, (c) {6, 8}, , (b) {6, 8, 10, 12}, (d) {2, 4, 6, 8, 10, 12}, , 22. If A ={1, 3, 9} and B ={2, 4, 5, 6}, then, A ∪ B is, , y = {49n – 49 | n ∈N }. Then,, (a) X ⊂ Y, (c) X = Y, , C = {0, 7},then the universal set will be, , (d) 12, , (a) {1, 3, 2, 4, 9}, (c) {1, 2, 3, 4, 5, 9}, , (b) {1, 2, 3, 4, 5, 6}, (d) {1, 2, 3, 4, 5, 6, 9}, , 23. Let A = {a, e , i , o, u } and B = {a, i , u }., Then, A ∪ B is …X… . Here, X refers, to, (a) A, (c) A and B, , (b) B, (d) None of these, , 24. If A = {(x , y ): x 2 + y 2 = 25} and, B = {(x , y ): x 2 + 9 y 2 = 144}, then A ∩ B, contains, (a) one point, (c) two points, , (b) three points, (d) four points

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CBSE New Pattern ~ Mathematics XI (Term I), , 25. Let X = {Ram, Geeta, Akbar} be the set, of students of class XI, who are in, school hockey team and Y = {Geeta,, David, Ashok} be the set of students, from class XI, who are in the school, football team. Then, X ∩ Y is, (a) {Ram, Geeta}, (c) {Geeta}, , (b) {Ram}, (d) None of these, , 26. Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and, B = {2, 3, 5, 7}. Then, which of, following is true?, (a) A ∩ B = A, (c) A ∩ B ⊄ B, , (b) A ∩ B = B, (d) None of these, , 27. If S = {x | x is a positive multiple of 3, , less than 100} and P = {x | x is a prime, number less than 20}. Then, n( S ) + n(P ), [NCERT Exemplar], is equal to, (a) 34, (c) 33, , (b) 31, (d) 41, , 5, , 32. In a town with a population of 5000,, 3200 people are egg-eaters, 2500, meat-eaters and 1500 eat both egg and, meat. Then, the number of pure, vegetarians is, (a), (b), (c), (d), , 800, 1000, 1600, 2000, , 33. In a group of students, 100 students, know Hindi, 50 know English and 25, know both. Each of the students knows, either Hindi or English. The number of, students in the group is, (a) 50, (c) 75, , (b) 125, (d) 175, , 34. In a school sports event, 50 students, , A = {x : x is the boys of your school}, B = {x : x is the girls of your school}, , participate for football, 30 students, participate for cricket and 15 students, participate for both. Then, the number, of students who participated for either, football or for cricket, is, , (a) Yes, (c) Can’t say, , (a) 60, (c) 65, , 28. Are the given sets disjoint?, , (b) No, (d) Insufficient data, , 29. Let A and B be two sets such that, , n ( A ) = 0.16, n (B ) = 0.14 and, n ( A ∪ B ) = 0.25. Then, n ( A ∩ B ) is, equal to, (a) 0.3, (c) 0.05, , (b) 0.5, (d) None of these, , 30. If X and Y are two sets such that X has, 40 elements, X ∪ Y has 60 elements, and X ∩ Y has 10 elements, then the, number of elements does Y have, (a) 10, (c) 30, , (b) 20, (d) 40, , 31. In a committee, 50 people speak, French, 20 speak Spanish and 10 speak, both Spanish and French. The number, of people who speak atleast one of, these two languages, is, (a) 40, (c) 20, , (b) 60, (d) 80, , (b) 55, (d) 75, , Assertion-Reasoning MCQs, Directions (Q. Nos. 35-49) Each of these, questions contains two statements, Assertion (A) and Reason (R). Each of the, questions has four alternative choices, any, one of the which is the correct answer. You, have to select one of the codes (a), (b), (c) and, (d) given below., (a) A is true, R is true; R is a correct, explanation of A., (b) A is true, R is true; R is not a correct, explanation of A., (c) A is true; R is false., (d) A is false; R is true., , 35. Assertion (A) ‘The collection of all, natural numbers less than 100’ is a set., Reason (R) A set is a well-defined, collection of the distinct objects.

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6, , CBSE New Pattern ~ Mathematics XI (Term I), , 36. Assertion (A) The set D = {x : x is a, prime number which is a divisor of 60}, in roster form is {1, 2, 3, 4, 5}., Reason (R) The set E = the set of all, letters in the word, ‘TRIGONOMETRY’, in the roster, form is {T, R, I, G, O, N, M, E, Y}., , 37. Assertion (A) The set {1, 4, 9, ... 100}, , in the set-builder form is {x : x = n 2 ,, where n ∈ N and 1 ≤ n ≤ 10}., Reason (R) In roster form, the order in, which the elements are listed is,, immaterial., , 38. Assertion (A) The set {x : x is a month, of a year not having 31 days} in roster, form is {February, April, June,, September, November}., Reason (R) The set F = {x : x is a, consonant in the English alphabet which, precedes k } in roster form is, F = {b , c , d , f , g , h, j }., , 39. Assertion (A) The set A = {x : x is an, even prime number greater than 2} is, the empty set., Reason (R) The set B = {x : x 2 = 4, x is, odd} is not an empty set., , 43. Assertion (A) Let A = {1, 2, 3} and, B = {1, 2, 3, 4}. Then, A ⊂ B ., Reason (R) If every element of X is also, an element of Y, then X is a subset of Y., , 44. Assertion (A) The interval {x : x ∈ R,, −4 < x ≤ 6} is represented by ( −4, 6]., Reason (R) The interval {x : x ∈ R, −12, < x < −10} is represented by [−12 , − 10]., , 45. Assertion (A) Set of English alphabets, is the universal set for the set of vowels, in English alphabets., Reason (R) The set of vowels is the, subset of set of consonants in the, English alphabets., , 46. Assertion (A) The power set of the set, {1, 2} is the set {φ, {1}, {2}, {1, 2}}., Reason (R) The power set is set of all, subsets of the set., , 47. Assertion (A) If A ⊂ B for any two sets, A and B., ∪, A, , B, , 40. Assertion (A) The set, , A = {a, b , c , d , e , g } is finite set., Reason (R) The set B = { men living, presently in different parts of the world}, is finite set., , 41. Assertion (A) The set of positive, integers greater than 100 is infinite., Reason (R) The set of prime numbers, less than 99 is finite., , 42. Assertion (A) If A = set of prime, numbers less than 6 and B = set of, prime factors of 30, then A = B ., Reason (R) If P = {1, 2, 3} and Q = {2, 2,, 1, 3, 3}, then P and Q are not equal., , Then, above Venn diagram represents, correct relationship between A and B., Reason (R) If A ⊂ B , then all elements, of A is also in B., , 48. Assertion (A) Let A = {a, b } and, B = {a, b , c }. Then, A ⊄ B ., Reason (R) If A ⊂ B , then A ∪ B = B ., , 49. Assertion (A) If n( A ) = 3, n(B ) = 6 and, A ⊂ B , then the number of elements in, A ∪ B is 9., Reason (R) If A and B are disjoint, then, n( A ∪ B ) is n( A ) + n(B ).

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CBSE New Pattern ~ Mathematics XI (Term I), , Case Based MCQs, 50. A class teacher Mamta Sharma of class, XI write three sets A, B and C are such, that A = {1, 3, 5, 7, 9}, B = {2, 4, 6, 8}, and C = {2, 3, 5, 7, 11}., Answer the following questions which, are based on above sets., (i) Find A ∩ B ., (a) {3, 5, 7}, (c) {1, 5, 7}, , (b) φ, (d) {2, 5, 7}, , (ii) Find A ∩ C ., (a) {3, 5, 7}, (c) {1, 5, 7}, , (b) φ, (d) {3, 4, 7}, , (iii) Which of the following is correct for, two sets A and B to be disjoint?, (a) A ∩ B = φ, (c) A ∪ B = φ, , (b) A ∩ B ≠ φ, (d) A ∪ B ≠ φ, , (iv) Which of the following is correct for, two sets A and C to be intersecting?, (a) A ∩ C = φ, (c) A ∪ C = φ, , (b) A ∩ C ≠ φ, (d) A ∪ C ≠ φ, , (v) Write the n [P (B )]., (a) 8, (c) 16, , (b) 4, (d) 12, , 51. The school organised a farewell party, for 100 students and school, management decided three types of, drinks will be distributed in farewell, party i.e. Milk (M), Coffee (C) and, Tea (T)., , Organiser reported that 10 students had, all the three drinks M,C,T. 20 students, had M and C; 30 students had C and T;, 25 students had M and T. 12 students, , 7, had M only; 5 students had C only; 8, students had T only., Based on the above information,, answer the following questions., (i) The number of students who did not, take any drink, is, (a) 20, (c) 10, , (b) 30, (d) 25, , (ii) The number of students who prefer, Milk is, (a) 47, (c) 53, , (b) 45, (d) 50, , (iii) The number of students who prefer, Coffee is, (a) 47, (c) 45, , (b) 53, (d) 50, , (iv) The number of students who prefer, Tea is, (a) 51, (c) 50, , (b) 53, (d) 47, , (v) The number of students who prefer, Milk and Coffee but not tea is, (a) 12, , (b) 10, , (c) 15, , (d) 20, , 52. In a library, 25 students read physics,, chemistry and mathematics books. It, was found that 15 students read, mathematics, 12 students read physics, while 11 students read chemistry., 5 students read both mathematics and, chemistry, 9 students read physics and, mathematics. 4 students read physics, and chemistry and 3 students read all, three subject books.

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8, , CBSE New Pattern ~ Mathematics XI (Term I), , Based on the above information,, answer the following questions., (i) The number of students who reading, only chemistry is, (a) 5, (c) 2, , (b) 4, (d) 1, , (ii) The number of students who reading, only mathematics is, (a) 4, (c) 5, , (b) 3, (d) 11, , (iii) The number of students who reading, only one of the subjects is, (a) 5, (c) 11, , (b) 8, (d) 6, , (iv) The number of students who reading, atleast one of the subject is, (a) 20, (c) 23, , (b) 22, (d) 21, , (v) The number of students who reading, none of the subject is, (a) 2, (c) 3, , (b) 4, (d) 5, , 53. In an University, out of 100 students 15, , (iii) The number of students who offered, statistics is, (a) 31, (c) 39, , (b) 35, (d) 34, , (iv) The number of students who offered, mathematics and statistics but not, physics is, (a) 7, (c) 5, , (b) 6, (d) 4, , (v) The number of students who did not, offer any of the above three subjects, is, (a) 4, (c) 5, , (b) 1, (d) 3, , 54. The school organised a cultural event, for 100 students. In the event, 15, students participated in dance, drama, and singing. 25 students participated in, dance and drama; 20 students, participated in drama and singing; 30, students participated in dance and, singing. 8 students participated in dance, only; 5 students in drama only and 12, students in singing only., , students offered Mathematics only, 12, students offered Statistics only,, 8 students offered only Physics, 40, students offered Physics and, Mathematics, 20 students offered, Physics and Statistics, 10 students, offered Mathematics and Statistics, 65, students offered Physics., Based on the above information answer, the following questions, (i) The number of students who offered, all the three subjects is, (a) 4, (c) 2, , (b) 3, (d) 5, , (ii) The number of students who offered, Mathematics is, (a) 62, (c) 55, , (b) 65, (d) 60, , Based on the above information,, answer the following questions., (i) The number of students who, participated in dance, is, (a) 18, (c) 40, , (b) 30, (d) 48

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CBSE New Pattern ~ Mathematics XI (Term I), , 9, , (ii) The number of students who, participated in drama, is, (a) 35, , (b) 30, , (c) 25, , (d) 20, , (iii) The number of students who, participated in singing, is, (a) 42, (c) 47, , (b) 45, (d) 37, , (iv) The number of students who, participated in dance and drama but, not in singing, is, (a) 20, (c) 10, , Based on the above information answer, the following questions, (i) The number of employees who, offered all three floors., , (b) 5, (d) 15, , (a) 5, , (v) The number of students who did not, participate in any of the events, is, (a) 20, (c) 25, , (b) 3, , (c) 4, , (d) 6, , (ii) The number of employees who, offered ground floor., , (b) 30, (d) 35, , (a) 50, , (b) 60, , (c) 65, , (d) 70, , (iii) The number of employees who, offered first floor., , 55. In a company, 100 employees offered, to do a work. In out of them,, 10 employees offered ground floor, only, 15 employees offered first floor, only, 10 employees offered second floor, only, 30 employees offered second, floor and ground floor to work,, 25 employees offered first and second, floor, 15 employees offered ground and, first floor, 60 employees offered second, floor., , (a) 40, , (b) 45, , (c) 50, , (d) 55, , (iv) The number of employees who, offered ground and first floor but not, second floor., (a) 10, , (b) 15, , (c) 20, , (d) 25, , (v) The number of employees who did, not offer any of the above three floors., (a) 15, , (b) 10, , (c) 5, , (d) 0, , ANSWERS, Multiple Choice Questions, 1. (c), 11. (c), , 2. (a), 12. (a), , 3. (c), 13. (a), , 4. (d), 14. (b), , 5. (b), 15. (c), , 6. (a), 16. (d), , 7. (a), 17. (a), , 8. (b), 18. (d), , 9. (a), 19. (c), , 10. (c), 20. (a), , 21. (d), 31. (b), , 22. (d), 32. (a), , 23. (a), 33. (b), , 24. (d), 34. (c), , 25. (c), , 26. (b), , 27. (d), , 28. (a), , 29. (c), , 30. (c), , 38. (b), 48. (d), , 39. (c), 49. (d), , 40. (b), , 41. (b), , 42. (c), , 43. (a), , 44. (c), , Assertion-Reasoning MCQs, 35. (a), 45. (c), , 36. (d), 46. (a), , 37. (b), 47. (d), , Case Based MCQs, 50. (i) - (b); (ii) - (a); (iii) - (a); (iv) - (b); (v) - (c), 52. (i) - (a); (ii) - (a); (iii) - (c); (iv) - (c); (v) - (a), 54. (i) - (d); (ii) - (a); (iii) - (c); (iv) - (c); (v) - (b), , 51. (i) - (a); (ii) - (a); (iii) - (c); (iv) - (b); (v) - (b), 53. (i) - (b); (ii) - (a); (iii) - (c); (iv) - (a); (v) - (b), 55. (i) - (a); (ii) - (a); (iii) - (c); (iv) - (a); (v) - (c)

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10, , CBSE New Pattern ~ Mathematics XI (Term I), , SOLUTIONS, 1. Given, A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, (i) Since, 4 is an element of A, therefore 4 ∈ A., (ii) Since, − 4 is not an element of A, therefore, − 4 ∉ A., (iii) Since, 12 is not an element of A, therefore, 12 ∉ A., , 2. Here, x is a positive integer and a divisor of 9., So, x can take values 1, 3, 9., ∴ {x : x is a positive integer and a divisor of 9}, = {1, 3, 9}, , 3. We have, 4 x + 9 < 50, ⇒, , 4 x + 9 − 9 < 50 − 9, [subtracting 9 from both sides], 41, ⇒, 4 x < 41 ⇒ x <, 4, ∴, x < 10.25, Since, x is a natural number, so x can take, values 1, 2, 3, 4, 5, 6, 7, 8, 9, 10., ∴ Required set = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, n, 4. Given, {x : x =, and 1 ≤ n ≤ 3,, 1 + n2, where n ∈ N }, n, Here, x =, , 1 ≤ n ≤ 3, n ∈ N, 1 + n2, n, , n = 1, 2, 3, ⇒, x=, 1 + n2, 1 2 3, 1, 2, 3, ,, ,, x=, = , ,, ⇒, 2 5 10, 1 + 12 1 + 22 1 + 32, 1 2 3 , ∴ Required set is  , , .,  2 5 10 , , 5. Let x represents the elements of given set., Given numbers are natural numbers greater than, 13, less than 99 and multiples of 7., Thus, A = {x : x is a natural number greater than, 13, less than 99 and a multiple of 7}, which is, the required set-builder form of given set., This can also be written as, A = { x : x is a natural number, a multiple of 7, and 13 < x < 99}, or A = { x : x = 7n, n ∈ N and 2 ≤ n ≤ 14}, , 6. The empty set contains no element., Hence, the number of elements in A will, be 0., , 7. If a set A has only one element, we call it a, singleton set. Thus, {a } is a singleton set., , 8. A year contains 12 months. So, the set of, months of a year is finite., , 9. Given, A = {1, 4, 9, 16, 25, 36, 49, 64, 81}, and B = { 2, 4, 6, K }, Now, A ∩ B = {4, 16, 36, 64}, ∴Cardinality of ( A ∩ B ), = Number of elements in ( A ∩ B ) = 4, , 10. The set {1, 2, 3, ...} contains infinite number, of elements. So, it is infinite., , 11. Given, A = {1, 2, 3} ⇒ n( A ) = 3, B = {t , p, q , r , s } ⇒ n( B ) = 5, C = {α, β, γ} ⇒ n(C ) = 3, D = {a, e, i, o, u} ⇒ n( D ) = 5, Here, n( A ) = n(C ) = 3 and n( B ) = n( D ) = 5, ∴The sets A , C and B , D are equivalent sets., , 12. A = {A, L, O, Y}, B = {L, O, Y, A}, Thus, A and B are equal., , 13. X = {8n – 7n – 1 | n ∈ N } = { 0, 49, 490, ...}, Y = { 49n – 49 | n ∈ N } = { 0, 49, 98, 147, ...}, Clearly, every elements of X is in Y but every, element of Y is not in X., ∴, X ⊂Y, , 14. Since, number of subsets of a set containing, m elements is 112 more than the subsets of, the set containing n elements., Q, 2m – 2n = 112, ⇒, , 2n ⋅ ( 2m – n – 1) = 2 4 ⋅ 7, , ⇒ 2n = 24 and 2m – n – 1 = 7, ⇒, n = 4 and 2m – n = 8, ⇒, ⇒, ⇒, ∴, , 2m – n = 23, m –n = 3, m –4=3 ⇒m =4+3, m =7

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CBSE New Pattern ~ Mathematics XI (Term I), , 15. { x : x ∈ R, − 5 < x ≤ 6} is the set that does not, , contain −5 but contains 6. So, it can be, written as a semi-closed interval whose first, end point is open and last end point is closed, i.e. ( −5, 6 ]., Length of the interval is 6 − ( −5) = 11., , 11, 24. Clearly, A is the set of all points on the circle, x 2 + y 2 = 25 and B is the set of all points on, the ellipse x 2 + 9 y 2 = 144., These two intersect at four points P, Q, R, and S., Y, , 16. If there are some sets under consideration,, then a set can be chosen arbitrarily which is, a superset of each one of the given sets. Such, a set is known as the universal set and it is, denoted by U., Q A = {2, 4, 6}, B = {1, 3, 5} and C = {0, 7}, ∴ U = {0, 1, 2, 3, 4, 5, 6, 7}, , X′, , 18. Let B = {1, 2}, Then, A = P ( B ) = { φ, {1}, { 2}, {1, 2}}, Clearly, {1, 2} ∈ A, , 19. Given that, A = {1, 3, 5, 7}, Here, n ( A ) = 4, ∴ Number of elements on power set of A, = 2n ( A ) = 24 = 16, ∴ Cardinality of the power set, P ( A ) = 16, A ∩ (A ∪ B) = A, , 20. Q, , U, A, , B, , 21. A = { 2, 4, 6, 8 } and B = {6, 8, 10, 12}, The common elements of A and B are 6 and 8, which have been taken only once., ∴, A ∪ B = { 2, 4, 6, 8, 10, 12}, , 22. A = {1, 3, 9}, B = { 2, 4, 5, 6},, then A ∪ B = {1, 2, 3, 4, 5, 6, 9}, , 23. A = {a , e , i , o, u}, B = {a , i , u}, A ∪ B = {a , e , i , o, u} = A, We observe that B ⊂ A ., Hence, if B ⊂ A , then A ∪ B = A ., , X, S, , R, , x 2 + 9y 2 = 144, , Y′, , Hence, A ∩ B contains four points., , 25. Here, X = {Ram, Geeta, Akbar}, and Y = {Geeta, David, Ashok}, Then, X ∩ Y = {Geeta }, , 26. A = {1, 2, 3, 4, K , 10}, and B = { 2, 3, 5, 7}, ∴ A ∩ B = { 2, 3, 5, 7} = B, 27. Q S = { x | x is a positive multiple of 3 less, than 100}, ∴ n ( S ) = 33, and P = { x | x is a prime number less than 20}, ∴, n (P ) = 8, n ( S ) + n ( P ) = 33 + 8 = 41, , 28. Here, A = {b1 , b 2 , ... bn } and B = { g 1 , g 2 , ... g m }, where, b1 , b 2 , ... , bn are the boys and g 1 , g 2 ,…,, g m are the girls of school., Clearly,, , A ∩ (A ∪ B), , P, O, , 17. If a set A has n elements, then its power set, will contain 2n elements., ∴Total number of elements in power set, of A = 215 ., , x 2 + y 2 = 25, , Q, , A ∩B = φ, , Hence, this pair of set is disjoint set., , 29. Given, n( A ) = 016, . , n( B ) = 014, ., and n( A ∪ B ) = 0.25, Let n ( A ∩ B ) = a, ∴ n ( A ∪ B ) = 0.16 − a + a + 0.14 − a, ⇒, ⇒, , 0.25 = 0.30 − a, a = 0.5, A, , B, , U, , 0.16 – a a 0.14 – a

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12, , CBSE New Pattern ~ Mathematics XI (Term I), , 30. Given, n ( X ) = 40, n ( X ∪ Y ) = 60, and, , n ( X ∩ Y ) = 10, Y, , X, , 10, , 30, , 75, , a, , 31. Given, n ( F ) = 50, n ( S ) = 20, and n ( F ∩ S ) = 10, S, , 40, , 10, , U, , ∴ n ( F ∪ S ) = 40 + 10 + 10 = 60, Hence, there are 60 people who speak atleast, one of these two languages., , 32. Let E be the set of people who are egg-eaters, M be set of people who are meat-eaters and, U be the set of people in the town., ∴ n ( E ) = 3200, n ( M ) = 2500, and n ( E ∩ M ) = 1500, M, , E, , U, , 1000, , 1700, , 1500, , ∴ n ( E ∪ M ) = 1700 + 1500 + 1000 = 4200, Q, n (U ) = 5000, ∴ Number of pure vegetarians = 5000 − 4200, = 800, , 33. Let H be the set of those students who know, Hindi and E be the set of those students who, know English., , 25, , U, , 25, , ∴, , n( H ∪ E ) = 75 + 25 + 25, = 125, Hence, the number of students in the group, is 125., , 34. Given, n ( F ) = 50, n (C ) = 30 and, n ( F ∩ C ) = 15, ∴ The number of students who participated, for either football or for cricket is 65., C, , F, , 35, , 10, , E, , H, , U, , Clearly, n ( X ∪ Y ) = 30 + 10 + a, ⇒, 60 = 30 + 10 + a, ⇒, 60 = 40 + a, ⇒, a = 60 − 40, ∴, a = 20, Hence, Y have 10 + a = 10 + 20 = 30 elements., , F, , Then, n( H ) = 100, n( E ) = 50, and n( H ∩ E ) = 25, , 15, , U, , 15, , 35. Assertion ‘The collection of all natural, numbers less than 100’, is a well-defined, collection. So, it is a set., Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., , 36. Assertion We can write 60 = 2 × 2 × 3 × 5, ∴Prime factors of 60 are 2, 3 and 5., Hence, the set D in roster form is {2, 3, 5}., Reason There are 12 letters in the word, ‘TRIGONOMETRY’ out of which three, letters T, R and O are repeated., Hence, set E in the roster form is, {T, R, I, G, O, N, M, E, Y}., Hence, Assertion is false and Reason is true., , 37. Assertion We see that each member in the, given set is the square of a natural number., Hence, the given set in set-builder form is, { x : x = n 2 , where n ∈ N and 1 ≤ n ≤ 10}., Therefore, Assertion is true., Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion.

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CBSE New Pattern ~ Mathematics XI (Term I), , 38. Assertion The months not containing 31, days are February, April, June, September,, November. So, the given set in roster form is, {February, April, June, September,, November}., Reason The given set can be represented in, roster form as F = {b , c , d , f , g , h , j }., Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion., , 39. Assertion 2 is the only even prime number., So, set A is the empty set., Reason The equation x 2 = 4 is not satisfied by, any odd value of x. So, set B is the empty set., Hence, Assertion is true and Reason is false., , 40. Assertion We know that, a set which is, empty or consists of a definite number of, elements, is called finite, otherwise the set is, called infinite. Since, set A contains finite, number of elements. So, it is a finite set., Reason We do not know the number of, elements in B, but it is some natural number., So, B is also finite., Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion., , 13, 43. Assertion Since, every element of A is in B,, , so A ⊂ B ., Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., , 44. Assertion The interval { x : x ∈ R, −4 < x ≤ 6}, is represented by ( −4, 6 ] ., Reason The interval { x : x ∈ R, − 12 < x < − 10}, is represented by ( −12, − 10 )., Hence, Assertion is true and Reason is false., , 45. Assertion Since, the set of vowels is the, subset of the set of English alphabets., So, the set of English alphabets is the, universal set for set of vowels in English, alphabets., Reason We know set of vowels is not the, subset of the set of consonants., Hence, Assertion is true and Reason is false., , 46. Assertion The subsets of the sets {1, 2} are, , φ, {1}, { 2} and {1, 2}., So, P ( A ) = { φ, {1}, { 2}, {1, 2}}, Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., , 47. Assertion If A ⊂ B , then every element of A, is in B. Hence, this can be represented by, Venn diagram as, U, , 41. Assertion There are infinite positive integer, greater than 100. So, the set of positive, integers greater than 100 is infinite., Reason There are 25 prime numbers less than, 99. So, the set of prime numbers less than 99, is finite., Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion., , 42. Assertion A = { 2, 3, 5}, Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30., So,, B = { 2, 3, 5}, ∴, A =B, Reason We know that, a set does not change,, if one or more elements of the set are, repeated., Hence, P and Q are equal., Hence, Assertion is true and Reason is false., , B, A, , Hence, Assertion is false and Reason is true., , 48. Assertion A = {a , b }, B = {a , b , c }, Since, all the elements of A are in B. So,, A ⊂B, Reason Q, A ⊂B, ⇒, , A ∪B =B, , Hence, Assertion is false and Reason is true., , 49. Assertion A ⊂ B, ⇒, , n( A ∪ B ) = n( B ) = 6, , Reason If A and B are disjoint, then, n( A ∪ B ) = n( A ) + n( B ), Hence, Assertion is false and Reason is true.

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14, , CBSE New Pattern ~ Mathematics XI (Term I), , 50. We have, A = {1, 3, 5, 7, 9},, B = {2, 4, 6, 8}, and, C = {2, 3, 5, 7, 11}, (i) A ∩ B = {1, 3, 5, 7, 9} ∩ {2, 4, 6, 8}, =φ, (ii) A ∩ C = {1, 3, 5, 7, 9} ∩ {2, 3, 5, 7, 11}, = {3, 5, 7}, (iii) Here, A ∩ B = φ, (iv) The correct option for intersecting of two, sets A and C is, A ∩C ≠ φ, (v) The number of elements in set B are 4., Therefore, the number of elements in, n [ P ( B )] are 24 i.e. 16., , 51. Consider the following Venn diagram, M, , C, a, e, , b, f, , (ii) Number of students who prefer Milk, = a + b + f + e = 12 + 10 + 10 + 15 = 47, (iii) Number of students who prefer Coffee, = b + c + f + g = 10 + 5 + 10 + 20 = 45, (iv) Number of students who prefer Tea, = d + e + f + g = 8 + 15 + 10 + 20 = 53, (v) Number of students who prefer Milk and, Coffee but not Tea is b, i.e. 10., , 52. Let M denotes set of student who reading, mathematics books, P denotes who reading, Physics books and C denotes who reading, chemistry books., We have,, n(U ) = 25, n( M ) = 15, n( P ) = 12, n(C ) = 11, n( M ∩ C ) = 5, n( M ∩ P ) = 9, n( P ∩ C ) = 4,, n( M ∩ P ∩ C ) = 3, , U, , C, , M, , c, , 4, , g, , 6, , 2, 3, , U, , 5, 1, , 2, d, , P, T, , (i) The number of students who reading only, Chemistry is 5., (ii) The number of students who reading only, Mathematics is 4., (iii) The number of students who reading only, one of the subject is 4 + 5 + 2 i.e. 11., (iv) The number of students who reading atleast, one of the subject is 4 + 6 + 3 + 2 + 5 + 1 + 2, i.e. 23., (v) The number of students who reading none, of the subject is 25 − 23 i.e. 2., , 53. Let M , S and P be the sets of students wo, offered Mathematics, Statistics and Physics, respectively. Let x be the number of students, who offered all the three subjects, then the, number of members in different regions are, shown in the following diagram., u, , M, , 15, , 10–x, , where,, a = Number of students who had M only, b = Number of students who had M and C, only, c = Number of students who had C only, d = Number of students who had T only, e = Number of students who had M and T, only, f = Number of students who had three, drinks M , C , T, and g = Number of students who had C and, T only, Then, we have, a = 12, b + f = 20, c = 5, d = 8, e + f = 25, f = 10 and g + f = 30, ⇒ a = 12, b = 10, c = 5, d = 8, e = 15, f =10, and g = 20, (i) Number of students who did not take any, drink, = 100 − ( a + b + c + d + e + f + g ), = 100 − (12 + 10 + 5 + 8 + 15 + 10 + 20 ), = 100 − 80 = 20, , S, 12, , x2, 40–x 0–x, 8, , P

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CBSE New Pattern ~ Mathematics XI (Term I), , 15, , From the Venn diagram, we get, the number, of students who offered Physics., [given], = ( 40 − x ) + x + ( 20 − x ) + 8 = 65, ⇒, 68 − x = 65, ⇒, x =3, (i) The number of students who offered all the, three subjects are 3., (ii) The number of students who offered, Mathematics, = 15 + (10 − x ) + x + ( 40 − x ), = 65 − x, = 65 − 3 = 62, [Q x = 3], (iii) The number of students who offered, Statistics, = 12 + (10 − x ) + x + ( 20 − x ), = 42 − x, = 42 − 3 = 39, [Q x = 3], (iv) 10 − x = 10 − 3 = 7, , e = Number of students who participated, in dance and singing only, f = Number of students who participated, in all three events dance,, drama and singing, and g = Number of students who participated, in drama and singing only, Then, we have, a = 8, b + f = 25, c = 5, d =12, e + f = 30,, f = 15 and g + f = 20, ⇒ a = 8, b = 10, c = 5, d = 12, e = 15, f = 15, and g = 5, (i) The number of students who participated, in dance = a + b + e + f, = 8 + 10 + 15 + 15, = 48, (ii) The number of students who participated, in drama = b + c + f + g, = 10 + 5 + 15 + 5, = 35, (iii) The number of students who participated, in singing = d + e + f + g, = 12 + 15 + 15 + 5, = 47, (iv) The number of students who participated in, dance and drama but not in singing = b = 10, (v) The number of students who did not, participats in any of the events, = 100 − ( a + b + c + d + e + f + g ), = 100 − ( 8 + 10 + 5 + 12 + 15 + 15 + 5), = 100 − ( 70 ) = 30, , (v) The number of students who offered, anyone of the three subjects, = 15 + 12 + 8 + (10 − x ) + ( 40 − x ), + ( 20 − x ) + x, = 105 − 2x, = 105 − 2 × 3 = 99, [Q x = 3], ∴The number of students who did not offer, anyone of the three subjects = 100 − 99 = 1, , 54. Consider the following Venn diagram, Drama, , Dance, a, e, , b, f, , U, , c, , 55. Let G, F and S be the sets of employees who, , g, , d, Singing, , U, , G, F, 10, , 15–x, , Where,, a = Number of students who participated, in dance only, b = Number of students who participated, in dance and drama only, c = Number of students who participated, in drama only, d = Number of students who participated, in singing only, , offered ground floor, first floor and second, floor respectively. Let x be the number of, employees who offered all three floors, then, the number of members in different regions, are shown in the following diagram., , 15, , x2, 30–x 5–x, 10, S

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16, (i) From the Venn diagram, we get the number, of employees who offered second floor, = ( 30 − x ) + x + ( 25 − x ) + 10 = 60 [given], ⇒ 65 − x = 60, ⇒, x =5, (ii) The number of employees who offered, ground floor, = 10 + (15 − x ) + x + ( 30 − x ), = 55 − x, = 55 − 5, = 50, (iii) The number of employees who offered first, floor, = 15 + (15 − x ) + x + ( 25 − x ), = 55 − x, = 55 − 5 = 50, , CBSE New Pattern ~ Mathematics XI (Term I), , (iv) The number of employees who offered, ground and first floor but not second floor, = 15 − x, = 15 − 5, = 10, (v) The number of employees who offer anyone, of the three floors, = 10 + 15 + 10 + (15 − x ) + ( 25 − x ), + ( 30 − x ) + x, = 105 − 2x, = 105 − 2 × 5, = 95, ∴The number of employees who did not, offer any of the three floors, = 100 − 95, =5

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17, , CBSE New Pattern ~ Mathematics XI (Term I), , 02, Relations and Functions, Quick Revision, Ordered Pair, If a pair of elements written in a small brackets and, grouped together in a particular order, then such a, pair is called ordered pair. The ordered pair of two, elements a and b is denoted by (a , b ), where a is, first element and b is second element., Two ordered pairs (a , b ) and (c , d ) are equal, if, their corresponding elements are equal i.e. a = c, and b = d ., , Cartesian Products of Sets, For any two non-empty sets A and B, the set of all, ordered pairs (a , b ) of elements a ∈ A and b ∈ B is, called the cartesian product of sets A and B and is, denoted by A × B ., Thus, A × B = {(a , b ) : a ∈ A and b ∈ B }, If A = φ or B = φ, then A × B = φ., Note A × B ≠ B × A, , Number of Elements in Cartesian, Product of Two Sets, (i) If there are p elements in set A and q elements, in set B, then there will be pq elements in, A × B., i.e. if n ( A ) = p and n ( B ) = q , then, n ( A × B ) = pq ., (ii) If A and B are non-empty sets and either A or, B is an infinite set, then A × B will also be an, infinite set., , (iii) If A or B is the null set or an empty set, then, A × B will also be an empty set., i.e. A × B = φ, , Relations, A relation R from a non-empty set A to a, non-empty set B is a subset of the cartesian, product A × B , i.e. R ⊆ A × B ., In (a , b ) ∈ A × B , the second element is called the, image of first element. The set of all first elements, in a relation R, is called the domain of the relation, R and the set of all second elements is called the, range of R. The set B is called the codomain of, relation R., Thus, if R = {(a , b ) :a ∈ A, b ∈ B }, then, domain (R ) = {a : (a , b ) ∈R }, and range (R ) = {b : (a , b ) ∈ R }, Note If n ( A ) = m , n ( B ) = n , then n ( A × B ) = mn and, the total number of possible relations from set A to, set B = 2mn, , Representation of a Relation, A relation can be represented algebraically by, roster form or by set-builder form and visually, it, can be represented by an arrow diagram., (i) Roster form In this form, we represent the, relation by the set of all ordered pairs belongs, to R.

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18, , CBSE New Pattern ~ Mathematics XI (Term I), , (ii) Set-builder form In this form, we represent, the relation R from set A to set B as, R = {(a , b ) : a ∈ A, b ∈ B and the rule which, relate the elements of A and B}, , Functions, A relation f from a non-empty set A to non-empty, set B is said to be function, if every element of set, A has one and only one image in set B., If f is a function from a set A to a set B, then we, write f : A → B and it is read as f is function from, A to B or f map A to B and (a , b ) ∈ f , then, f (a ) = b , where b is called image of a under f and a, is called the pre-image of b under f ., Domain, Codomain and Range, of a Function, , If f : A → B , then the set A is called the domain of, function f and the set B is called the codomain of f ., The subset of B containing the images of elements, of A is called the range of the function., Note Every function is a relation but converse is, not true., , Real Functions, A function f : A → B is called a real valued, function, if B is a subset of R (set of all real, numbers). If A and B both are subsets of R, then f, is called a real function., , Types of Functions, (i) Identity function The function f : R → R, defined by f ( x ) = x for each x ∈ R is called, identity function., Domain of f = R, and Range of f = R, (ii) Constant function The function f : R → R, defined by f ( x ) = c , ∀ x ∈ R , where c is a, constant ∈R, is called a constant function., Domain of f = R, and Range of f = {c }, , (iii) Polynomial function A real function, f : R → R defined by, f ( x ) = a 0 + a1 x + a 2 x 2 +…+ an x n , where n is a, non-negative integer and a 0 , a 1, a 2 , K , an ∈ R, for each x ∈ R , is called a polynomial function., If an ≠ 0, then n is called the degree of the, polynomial. The domain of a polynomial, function is R and range depends on the, polynomial representing the function., (iv) Rational function A function of the form, f (x ), , where f ( x ) and g ( x ) are polynomial, g (x ), functions of x defined in a domain and, g ( x ) ≠ 0, is called a rational function., (v) Modulus or Absolute value function, The real function f : R → R defined by,  − x , if x < 0, f (x ) = | x| = ,  x , if x ≥ 0, is called the modulus function., Domain of f = R ,, Range of f = R + ∪ {0} i.e. [0, ∞ ), (vi) Signum function The real function f : R → R, defined by, | x |, , if x ≠ 0, , f (x ) =  x, if x = 0,  0,,  − 1, if x < 0, , =  0, if x = 0,  1, if x > 0, , is called the signum function., Domain of f = R ; Range of f = { − 1, 0, 1}, (vii) Greatest integer or Step function The real, function f : R → R defined by f ( x ) = [x ], is, called the greatest integer function, where [x ] =, integral part of x or greatest integer less than, or equal to x., Domain of f = R ;, Range of f = The set of all integers

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19, , CBSE New Pattern ~ Mathematics XI (Term I), , Objective Questions, Multiple Choice Questions, 1. Which of the following is an ordered, pair?, (a), (b), (c), (d), , (p, q), p ∈ P and q ∈ Q, [p, q] , p ∈ P and q ∈ Q, { p, q }, p ∈ P and q ∈ Q, All of the above, , 8. If A = {1, 3 , 6} and B = {x , y }, then, representation of cartesian products by, an arrow diagrams of A × B is, , (a), , A, , B, , 1, , x, , 3, 6, , y, , B, , A, , x, , 1, , 2. The values of a and b, if ordered pair is, ( 2a − 5, 4 ) = (5, b + 6 ), (a) − 2, 5, (c) 5, 2, , (b) 2, 5, (d) 5, − 2, , 3. If A = {a1 , a 2 } and B = {b1 , b 2 , b 3 , b4 },, , y, , then A × B is equal to, (a), (b), (c), (d), , {(a 1, b 1), (a 2, b 2 )}, {(a 1, b 1), (a2 , b 2 ), (a 3, b 3), (a 4 , b 4 )}, {(a 1, b 1), (a 1, b 2 ), (a 1, b 3), (a 1, b 4 )}, {(a 1,b 1), (a 1, b 2 ), (a 1, b 3), (a 1, b 4 ), (a2 , b 1), (a2 , b 2 ),, (a2 , b 3), (a2 , b 4 )}, , (c), , 4. If A = {1, 2, 5, 6} and B = {1, 2, 3}, then, what is ( A × B ) ∩ (B × A ) equal to?, (a) {(1, 1), (2, 1), (6, 1), (3, 2)}, (c) {(1, 1), (2, 2)}, , (b) {(1, 1), (1, 2), (2, 1), (2, 2)}, (d) {(1, 1), (1, 2), (2, 5), (2, 6)}, , 5. If A × B = {(a, 1), (b , 3), (a, 3),, (b , 1), (a, 2), (b , 2)}. Then, A and B is, (a), (b), (c), (d), , A = {1, 3, 2 } and B = {a, b }, A = {a, 1, 2 } and B = {b , 3}, A = {a, b } and B = {1, 2, 3}, A = {a, b , 1} and B = {a, b , 2, 3}, , 6. Let n( A ) = m and n(B ) = n. Then, the, total number of non-empty relations, that can be defined from A to B is, (a) m n, (c) mn − 1, , (b) n m − 1, (d) 2 mn − 1, , 7. Let A = {a, b , c , d } and B = {x , y , z }. What, , 3, , (b), , 6, , A, , A, , 1, , 1, , 3, , 3, , 6, , 6, , B, , B, , x, , x, , y, , y, , (d), , 9. If A = {1, 2, 3, 4} and B = {5, 6, 7, 8},, then which of the following are, relations from A to B?, (a) R1 = {(1, 5), (2, 7), (3, 8)}, (b) R2 = {(5, 2), (3, 7), (4, 7)}, (c) R3 = {(6, 2), (3, 7), (4, 7)}, (d) All are correct, , 10. The figure shows a relation R between, the sets P and Q ., Q, P, 9, , is the number of elements in A × B ?, , 4, , (a) 6, (c) 12, , 25, , (b) 7, (d) 64, , 5, 3, 2, 1, –2, –3, –5

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20, , CBSE New Pattern ~ Mathematics XI (Term I), , The relation R in Roster form is, (a), (b), (c), (d), , {(9, 3), (4, 2), (25, 5)}, {(9, −3 ), (4, −2), (25, −5)}, {(9, −3 ), (9, 3), (4, −2), (4, 2), (25, −5), (25, 5)}, None of the above, , 11. The figure shows a relation R between, the sets P and Q ., Q, P, , 5, 3, 2, 1, –2, –3, –5, , 9, 4, 25, , The relation R in Set-builder form is, (a), (b), (c), (d), , {(x, y) : x ∈ P, y ∈ Q }, {(x, y) : x ∈ Q, y ∈ P }, {(x, y) : x is the square of y, x ∈ P, y ∈ Q}, {(x, y) : y is the square of x, x ∈ P, y ∈ Q}, , 12. If a relation R is defined on the set Z of, integers as follows, (a, b ) ∈ R ⇔ a 2 + b 2 = 25,, then domain ( R ) is equal to, (a) { 3, 4, 5 }, (c) {0, ± 3, ± 4, ± 5 }, , (b) {0, 3, 4, 5 }, (d) None of these, , 13. If A = {1, 2, 6} and R be the relation, , defined on A by R = {(a, b ): a ∈A, b ∈A, and a divides b}, then range of R is, equal to, (a) {1, 2}, (c) {1, 2, 6}, , (b) {2, 6}, (d) None of these, , 14. Let A = {1, 2, 3, 4} and B = {1, 4, 9, 16, 25}, and R be a relation defined from A to B,, as R = {(x , y ): x ∈A, y ∈B and y = x 2 }, then, domain of R and codomain of R is, (a), (b), (c), (d), , {1, 2, 3, 4} and {1, 4, 9, 16, 25}, {1, 4, 9, 16, 25} and {1, 2, 3, 4}, {1, 2, 3, 4} and {1, 2, 3, 4, 9, 16, 25}, None of the above, , 15. The inverse of the function, f ( x ) = log a ( x + x, , 2, , (where, a < 0, a ≠ 1) is, , + 1), , 1 x, (a − a − x ), 2, (c) defined for x > 0, , (b) not defined for all x, , (a), , (d) None of the above, , 16. If f ( x ) = 3x + 10 and g ( x ) = x 2 − 1, then, ( fog ) −1 is equal to, x − 7, (a) , ,  3 , , 1/ 2, , x + 7, (b) , ,  3 , , 1/ 2, , x − 3, (c) , ,  7 , , 1/ 2, , x + 3, (d) , ,  7 , , 1/ 2, , 17. Is the given relation a function?, {(3, 3), (4, 2), (5, 1), (6, 0), (7, 7)}, (a) Yes, (c) cannot say, , (b) No, (d) Insufficient data, , 18. There are three relations R1 , R2 and R3, , such that R1 = {( 2, 1), ( 3, 1)( 4, 2)},, R 2 = {( 2, 2), ( 2, 4 ), ( 3, 3), ( 4, 4 )} and, R 3 = {(1, 2), ( 2, 3), ( 3, 4 ), ( 4, 5), (5, 6 ),, (6, 7 )}., Then,, , (a), (b), (c), (d), , R1 and R2 are functions, R2 and R3 are functions, R1 and R3 are functions, Only R1 is a function, , 19. Domain of a 2 − x 2 (a > 0 ) is, (a) (−a, a), (c) [0, a], , 20. Range of f ( x ) =, , (b) [−a, a], (d) (−a, 0], , 1, is, 1 − 2 cos x, , 1, (a)  , 1,  3 , 1, (c) (−∞, − 1] ∪  , ∞, 3 , , 1, (b) −1, ,  3 , 1, (d) − , 1,  3 , , 21. f : R − {3} → R be defined by, x2 −9, , and g : R → R be defined, x−3, by g (x ) = x + 3. Then, f (x ) and g (x ) are, f (x ) =, , (a), (b), (c), (d), , Equal functions, not equal (domains are same), not equal (domains are not same), None of the above

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21, , CBSE New Pattern ~ Mathematics XI (Term I), , 22. The domain and range of the real, function f defined by f ( x ) =, , 4−x, x−4, , 27. If G represents the name of the function, in above graph, then G is a/an, , is, , Y, , given by, (a), (b), (c), (d), , Domain = R, Range = { −1, 1 }, Domain = R − { 1 }, Range = R, Domain = R − {4 }, Range = R − { −1 }, Domain = R − { −4 }, Range = { −1, 1 }, , G, X¢, , –6 –4 –2 O, , 2, , 4, , X, 6, , 23. The domain and range of the function, f given by f ( x ) = 2 − | x − 5 | is, , (a), (b), (c), (d), , Y¢, , +, , Domain = R , Range = (−∞, 1], Domain = R, Range = (−∞, 2], Domain = R, Range = (−∞, 2), Domain = R + , Range = (−∞, 2], , (a), (b), (c), (d), , 24. The domain of the function f , defined, by f (x ) =, , 1, x − |x|, , identity function, constant function, modulus function, None of the above, , 28. The graph of the functions,, f (x ) = | x − 2| is, , is, (b) R +, (d) None of these, , (a) R, (c) R −, , 25. The range of the function f (x ) =, (a) (−∞, ∞), 1 1, (c) − , ,  2 2 , , x, 1+ x 2, , is, , (a), , (2, 0), , (b), , (–2, 0), , (b) [−1, 1], (d) [ −2 , 2], , 26. The graph of an identity function on R is, Y, , (a), , (b), , 6, 4, 2, , X¢, , –6 –4 –2 O 2, –2, –4, , (c), , 4, , 6, , X X¢, , –6 –4 –2 O, –2, –4, , –6, Y¢, , –6, Y¢, , Y, , Y, , (d), , 6, 4, 2, , X¢, , Y, 6, 4, 2, , –6 –4 –2 O 2, –2, –4, –6, Y¢, , 4, , 6, , X X¢, , (c), , 2, , 4, , 6, , X, , 29. For each non-zero real number x,, let f (x ) =, , 6, 4, 2, , –6 –4 –2 O, –2, –4, –6, Y¢, , (d) None of these, , 2, , 4, , 6, , X, , x, ., |x |, , The range of f is, (a) a null set, (b) a set consisting of only one element, (c) a set consisting of two elements, (d) a set consisting of infinitely many elements

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22, , CBSE New Pattern ~ Mathematics XI (Term I), , 30., , Assertion-Reasoning MCQs, , Y, , Directions (Q. Nos. 36-50) Each of these, questions contains two statements, Assertion (A) and Reason (R). Each of the, questions has four alternative choices, any, one of the which is the correct answer. You, have to select one of the codes (a), (b), (c) and, (d) given below., , 3, 2, X′, , 1, , –3 –2 –1, , 1, O, –1, –2, , 2, , 3 4 5, , X, , –3, Y′, , f (x ) = [x ], Which of the following options identify, the above graph?, (a), (b), (c), (d), , 36. Assertion (A) If (x + 1, y − 2) = ( 3, 1) ,, , Modulus function, Greatest integer function, Signum function, None of the above, , 31. If [x] 2 − 5 [x] + 6 = 0, where [ ⋅ ] denote, the greatest integer function, then, (a) x ∈[3, 4], (c) x ∈[2, 3], , (b), , 1, 4, , 33. If f (x ) = x 3 −, , (c), , of two non-empty sets P and Q is, denoted as P × Q and, P × Q = {( p, q ) : p ∈ P , q ∈Q }., Reason (R) If A = {red, blue} and, B = {b , c , s }, then A × B = {(red, b), (red,, c), (red, s), (blue, b), (blue, c), (blue, s)}., , 38. Assertion (A) If ( 4x + 3, y ) = ( 3x + 5, − 2),, (d) 4, ,  1, , then f (x ) + f   is, x , x, , (a) 0, (c) 2, , 34. If y = e, , 1, 2, , 1, , 3, , (b) 1, (d) not defined, x3 − 2, , , then log y at x = 5 is, , (a) 321, (c) 125, , (b) 234, (d) 123, ,  2x , 1 + x , ,  , then f , 2 , 1 − x , 1 + x , , 35. If f (x ) = log e , is equal to, (a) [f (x)] 2, (c) 2f (x), , (b) [f (x)] 3, (d) 3 f (x), , then x = 2 and y = 3., Reason (R) Two ordered pairs are, equal, if their corresponding elements, are equal., , 37. Assertion (A) The cartesian product, , (b) x ∈(2, 3], (d) x ∈[2, 4), ,  2, x <0, x ,, 32. If f (x ) =  x , 0 ≤ x < 1,, 1, x ≥1,  x ,,  1, then the value of f   is,  2, (a) 2, , (a) A is true, R is true; R is a correct, explanation of A., (b) A is true, R is true; R is not a correct, explanation of A., (c) A is true; R is false., (d) A is false; R is true., , then x = 2 and y = − 2., Reason (R) If A = {−1, 3, 4}, then A × A, is {(−1, −1), (−1, 3), (−1, 4), (3, −1),, (4, −1), (3, 4)}., , 39. Assertion (A) If (x , 1), ( y , 2) and (z , 1) are, in A × B and n( A ) = 3, n(B ) = 2, then, A = {x , y , z } and B = {1, 2}., Reason (R) If n( A ) = 3 and n(B ) = 2,, then n( A × B ) = 6 ., , 40. Assertion (A) Let A = {1, 2} and, , B = {3, 4}. Then, number of relations, from A to B is 16., Reason (R) If n( A ) = p and n(B ) = q ,, then number of relations is 2 pq .

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23, , CBSE New Pattern ~ Mathematics XI (Term I), , 41. Let A = {1, 2, 3, 4, 6}. If R is the relation, on A defined by {(a, b) : a, b ∈ A, b is, exactly divisible by a}., , Assertion (A) The relation R in Roster, form is {(6, 3), (6, 2), (4, 2)}., Reason (R) The domain and range of R, is {1, 2, 3, 4, 6}., , 42. Consider the following statements, A, , B, , –2, –1, 0, 1, 2, , 0, 1, 4, , Assertion (A) The figure shows a, relationship between the sets A and B., Then, the relation in Set-builder form is, {(x , y ) : y = x 2 , x , y ∈ N and − 2 ≤ x ≤ 2}., Reason (R) The above Relation in, Roster form is {(−1, 1), (−2, 4), (0, 0),, (1, 1), (2, 4)}., , 43. Let R be a relation defined by, R = {(x , x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}, Then, consider the following, Assertion (A) The domain of R is, {0, 1, 2, 3, 4, 5}., Reason (R) The range of R is, {0, 1, 2, 3, 4, 5}., , 44. Assertion (A) The domain of the, , relation R = {(x + 2, x + 4) : x ∈ N , x < 8}, is {3, 4, 5, 6, 7, 8, 9}., Reason (R) The range of the relation, R = {(x + 2, x + 4) : x ∈ N , x < 8} is, {1, 2, 3, 4, 5, 6, 7}., , 45. Assertion (A) The following arrow, diagram represents a function., , P, , Q, , p, , 1, , q, , 2, , r, , 3, , s, , 4, , Reason (R) Let f : R − {2} → R be, x2 − 4, and g : R → R, defined by f (x ) =, x−2, be defined by g (x ) = x + 3. Then, f = g ., , 46. Assertion (A) The range of the, , function f (x ) = 2 − 3x , x ∈ R, x > 0 is R., Reason (R) The range of the function, f (x ) = x 2 + 2 is [2, ∞ )., , 47. Assertion (A) Let A = {1, 2, 3, 5},, , B = {4, 6, 9} and R = {(x, y) : | x − y | is, odd, x ∈ A, y ∈ B }. Then, domain of R is, {1, 2, 3, 5}., Reason (R) | x | is always positive ∀x ∈ R., , 48. Assertion (A) The domain of the real, , function f defined by f (x ) = x − 1 is, R − {1}., Reason (R) The range of the function, defined by f (x ) = x − 1 is [0, ∞ )., 1, 49. Assertion (A) If f (x ) = x + , then, x,  1, [ f (x )] 3 = f (x 3 ) + 3 f   ., x , Reason (R) If f (x ) = (x − a ) 2 (x − b ) 2 ,, then f (a + b ) is 0., , 50. Assertion (A) If f : R → R and, , g : R → R are defined by f (x ) = 2x + 3, and g (x ) = x 2 + 7, then the values of x, such that g { f (x )} = 8 are −1 and 2., Reason (R) If f : R → R be given by, 4x, for all x ∈ R, then, f (x ) = x, 4 +2, f (x ) + f (1 − x ) = 1.

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24, , CBSE New Pattern ~ Mathematics XI (Term I), , Case Based MCQs, 51. Method to Find the Sets When, Cartesian Product is Given, For finding these two sets, we write first, element of each ordered pair in first set, say A and corresponding second, element in second set B (say)., Number of Elements in Cartesian, Product of Two Sets, If there are p elements in set A and q, elements in set B, then there will be pq, elements in A × B i.e. if n ( A ) = p and, n (B ) = q , then n ( A × B ) = pq ., Based on the above two topic, answer, the following questions., (i) If A × B = {(a, 1), (b , 3), (a, 3),, (b , 1), (a, 2), (b , 2)}. Then, A and B are, (a) {1, 3, 2}, {a,b }, (c) {a,b },{1, 3,2 }, , (b) {a,b },{1, 3}, (d) None of these, , (ii) If the set A has 3 elements and set B, has 4 elements, then the number of, elements in A × B is, (a) 3, (c) 7, , (b) 4, (d) 12, , (iii) A and B are two sets given in such a, way that A × B contains 6 elements., If three elements of A × B are (1, 3),, (2, 5) and (3, 3), then A, B are, (a) {1, 2, 3}, {3, 5}, (c) {1, 2}, {3, 5}, , (b) {3, 5,}, {1, 2, 3}, (d) {1, 2, 3}, {5}, , (iv) The remaining elements of A × B, in (iii) is, (a), (b), (c), (d), , (5, 1), (3, 2), (3, 5), (1, 5), (2, 3), (3, 5), (1, 5), (3, 2), (5, 3), None of the above, , (v) The cartesian product P × P has 16, elements among which are found, (a, 1) and (b , 2). Then, the set P is, (a) {a,b }, (c) {a,b ,1,2 }, , (b) {1, 2}, (d) {a,b ,1,2,4 }, , 52. Ordered Pairs The ordered pair of, two elements a and b is denoted by, (a, b ) : a is first element (or first, component) and b is second element (or, second component)., Two ordered pairs are equal if their, corresponding elements are equal., i.e. (a, b ) = (c , d ) ⇒ a = c and b = d, Cartesian Product of Two Sets For, two non-empty sets A and B, the, cartesian product A × B is the set of all, ordered pairs of elements from sets, A and B., In symbolic form, it can be written as, A × B = {(a, b ) : a ∈ A, b ∈ B }, Based on the above topics, answer the, following questions., (i) If (a − 3, b + 7 ) = ( 3, 7 ), then the, value of a and b are, (a) 6, 0, (c) 7, 0, , (b) 3, 7, (d) 3, − 7, , (ii) If (x + 6, y − 2) = (0, 6 ), then the, value of x and y are, (a) 6, 8, (c) − 6,8, , (b) − 6, − 8, (d) 6,− 8, , (iii) If (x + 2, 4 ) = (5, 2x + y ), then the, value of x and y are, (a) −3,2, (c) −3, − 2, , (b) 3, 2, (d) 3,− 2, , (iv) Let A and B be two sets such that, A × B consists of 6 elements. If three, elements of A × B are (1, 4), (2, 6), and (3, 6), then, (a), (b), (c), (d), , (A × B) = (B × A), (A × B) ≠ (B × A), A × B = {(1, 4), (1, 6), (2, 4)}, None of the above, , (v) If n( A × B ) = 45, then n( A ) cannot be, (a) 15, (c) 5, , (b) 17, (d) 9

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25, , CBSE New Pattern ~ Mathematics XI (Term I), , 53. Representation of a Relation, A relation can be represented, algebraically by roster form or by, set-builder form and visually it can be, represented by an arrow diagram which, are given below, (i) Roster form In this form, we, represent the relation by the set of, all ordered pairs belongs to R., (ii) Set-builder form In this form, we, represent the relation R from set A, to set B as R = {(a, b ) : a ∈ A, b ∈ B, and the rule which relate the, elements of A and B}., (iii) Arrow diagram To represent a, relation by an arrow diagram, we, draw arrows from first element to, second element of all ordered pairs, belonging to relation R., Based on the above topics, answer the, following questions., (i) Expression of, R = {(a, b ): 2a + b = 5; a, b ∈W } as the, set of ordered pairs (in roster form), is, (a), (b), (c), (d), , R = {(5, 0), (3, 1), (1, 2)}, R = {(0, 5), (1, 3), (1, 2)}, R = {(0, 5), (1, 3), (2, 1)}, None of the above, , 9, 4, 25, , (a) R = {(x, y) : x ∈ P, y ∈ Q and x is the, square of y}, (b) R = {(x, y) : x ∈ P, y ∈ Q and y is the, square of x}, (c) R = {(x, y) : x ∈ P, y ∈ Q and x = ± y}, (d) None of the above, , (iv) If A = {a, b } and B = {2, 3}, then the, number of relations from A to B is, (a) 4, , (b) 8, , (c) 6, , (d) 16, , (v) If n ( A ) = 3 and B = {2, 3, 4, 6, 7, 8},, then the number of relations from A, to B is, (a) 2 3, , (b) 2 6, , (c) 2 18, , (d) 2 9, , 54. Function as a Relation A relation f, from a non-empty set A to a non-empty, set B is said to be a function, if every, element of set A has one and only one, image in set B., In other words, we can say that a, function f is a relation from a, non-empty set A to a non-empty set B, such that the domain of f is A and no, two distinct ordered pairs in f have the, same first element or component., If f is a function from a set A to a set B,, then we write, f, , (ii) The relation between sets P and Q, given by an arrow diagram in roster, form will be, P, , (iii) The relation given in (ii) can be, written in set-builder form as, , Q, 5, 4, 3, 2, 1, –2, –3, –5, , (a) R = {(9, 3),(9, − 3),(4,2),(4, − 2), (25,5),(25, − 5)}, (b) R = {(9, 3), (4, 2), (25, 5)}, (c) R = {(9, − 3),(4, − 2),(25, − 5)}, (d) None of the above, , f : A → B or A → B, and it is read as f is a function from A to, B or f maps A to B., Based on the above topic, answer the, following questions., (i) The given curve is a, Y, , X′, , O, , Y′, , X

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26, , CBSE New Pattern ~ Mathematics XI (Term I), , (a), (b), (c), (d), , (iii) If f (x ) = x 2 + 2x + 3, then among, f (1), f ( 2) and f ( 3), which one gives, the maximum value., , Function, Relation, Can’t say anything, Data not sufficient, , (a) f (1), (c) f (3), , (ii) The given curve is a, Y, , (b) f (2), (d) f (1) = f (2) = f (3), , (iv) If f (1 + x ) = x 2 + 1, then f ( 2 − h ) is, (a) h2 − 2h + 2, (c) h2 − 2h − 2, , X′, , O, , (v) If f (x ) =, , X, , 1, , then range ( f ) is, 2 − sin 3x, , equal to, , Y′, , (a), (b), (c), (d), , (b) h2 − 2h + 1, (d) h2 + 2h + 2, , 1 1, (b) − , ,  3 3 , −1, (d) − 1, , , 3 , , (a) [− 1, 1], , Function, Relation, Can’t say anything, Data not sufficient, , 1, (c)  , 1,  3 , , ANSWERS, Multiple Choice Questions, 1. (a), 11. (c), 21. (c), , 2. (d), 12. (c), 22. (c), , 3. (d), 13. (c), 23. (b), , 4. (b), 14. (a), 24. (d), , 5. (c), 15. (a), 25. (c), , 31. (c), , 32. (c), , 33. (a), , 34. (d), , 35. (c), , 39. (b), 49. (c), , 40. (a), 50. (d), , 6. (d), 16. (a), 26. (a), , 7. (c), 17. (a), 27. (b), , 8. (a), 18. (c), 28. (a), , 9. (a), 19. (b), 29. (c), , 10. (c), 20. (b), 30. (b), , 41. (d), , 42. (d), , 43. (c), , 44. (c), , 45. (c), , Assertion-Reasoning MCQs, 36. (a), 46. (d), , 37. (a), 47. (b), , 38. (c), 48. (d), , Case Based MCQs, 51. (i) - (c); (ii) - (d); (iii) - (a); (iv) - (b); (v) - (c), 53. (i) - (c); (ii) - (a); (iii) - (a); (iv) - (d); (v) - (c), , 52. (i) - (a); (ii) - (c); (iii) - (d); (iv) - (b); (v) - (b), 54. (i) - (b); (ii) - (a); (iii) - (c); (iv) - (a); (v) - (c)

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27, , CBSE New Pattern ~ Mathematics XI (Term I), , SOLUTIONS, 1. We know that, an ordered pair of elements, , Required arrow diagram is, , taken from any two sets P and Q is a pair, of elements written in small brackets and, grouped together in a particular order,, i.e. ( p, q ), p ∈ P and q ∈Q ., , A, , B, , 1, , x, , 2. We know that, two ordered pairs are equal, if, , 6, , their corresponding elements are equal., ( 2a − 5, 4 ) = ( 5, b + 6 ), ⇒, 2a − 5 = 5 and 4 = b + 6, [equating corresponding elements], ⇒, 2a = 5 + 5 and 4 − 6 = b, ⇒, 2a = 10 and − 2 = b, ⇒, a = 5 and b = − 2, , 3. If A = {a 1 , a 2 }, B = {b1 , b 2 , b 3 , b 4 }, then., A × B = {( a 1 , b1 ), ( a 1 , b 2 ), ( a 1 , b 3 ), ( a 1 , b 4 ),, ( a 2 , b1 ), ( a 2 , b 2 ), ( a 2 , b 3 ), ( a 2 , b 4 )}, , 4. Given, A = {1, 2, 5, 6}, and, B = {1, 2, 3}, A × B = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2),, (2, 3), (5, 1), (5, 2), (5, 3), (6, 1),, (6, 2), ( 6, 3)}, B × A = {(1, 1), (1, 2), (1, 5), (1, 6), (2, 1),, (2, 2), (2, 5), (2, 6), (3, 1), (3, 2),, (3, 5), ( 3, 6 )}, ∴( A × B ) ∩ ( B × A ) = {(1, 1), (1, 2), (2, 1), ( 2, 2)}, , 5. Here, first element of each ordered pair of, A × B gives the elements of set A and, corresponding second element gives the, elements of set B., ∴, A = {a , b } and B = {1, 3, 2}, , 6. We have, n ( A ) = m and n ( B ) = n, n( A × B ) = n ( A ) ⋅ n ( B ) = mn, Total number of relation from A to B, = 2mn − 1 = 2n ( A × B ) − 1, , 7. Here, n ( A ) = 4 and n ( B ) = 3, ∴n ( A × B ) = n ( A ) × n ( B ) = 4 × 3 = 12, , 8. We have, A = {1, 3, 6}, B = { x , y}, A × B = {1, 3, 6} × { x , y}, = {(1, x ), (1, y ), ( 3, x ), ( 3, y ),, ( 6, x ), ( 6, y )}, , 3, y, , 9. We have, A = {1, 2, 3, 4} and B = {5, 6, 7, 8}, A × B = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5),, (2, 6), (2, 7), (2, 8), (3, 5), (3, 6),, (3, 7), (3, 8), (4, 5), (4, 6),, (4, 7),(4, 8)}, (a) Since, R1 ⊆ A × B , therefore R1 is a, relation from A to B., (b) Since, ( 5, 2) ∈ R2 but (5, 2) ∉ A × B,, therefore R2 ⊆, / A × B . Thus, R2 is not a, relation from A to B., (c) Since, (6, 2) ∈ R3 but (6, 2) ∉A × B ,, therefore R3 ⊆, / A × B . Thus, R3 is not a, relation from A to B., , 10. In Roster form, R = {( 9, 3), ( 9, −3), ( 4, 2),, ( 4, − 2), ( 25, 5), ( 25, − 5)}., , 11. It is obvious that the relation R is ‘x is the, square of y’., In Set-builder form, R = {( x , y ): x is the square, of y, x ∈ P , y ∈Q }, , 12. We have, ( a , b ) ∈ R ⇔ a 2 + b 2 = 25, ⇒, Clearly,, , b = ± 25 − a 2, a = 0 ⇒b = ± 5, a = ± 3⇒b = ± 4, a = ± 4 ⇒b = ± 3, and, a = ± 5⇒b = 0, Hence, domain ( R ) = { 0, ± 3, ± 4, ± 5}., , 13. Let R = {( a , b ) : a ∈ A , b ∈ A , and a divides b}, ∴ R = {(1, 1), (1, 2), (1, 6), (2, 2), (2, 6), (6, 6)}, ∴ Range = {1, 2, 6} = A, , 14. Relation R = {(1, 1), (2, 4), (3, 9), (4, 16)}, Domain ( R ) = {1, 2, 3, 4}, Codomain ( R ) = {1, 4, 9, 16, 25}

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28, , CBSE New Pattern ~ Mathematics XI (Term I), , 15. Let f ( x ) = y, then, a y = x + x2 +1, x − x2 +1, ⇒, a −y =, =, −1, x + x2 +1, 1, ∴ a y − a − y = 2x ⇒ x = ( a y − a − y ), 2, 1 x, −1, −x, ∴ f ( x ) = (a − a ), 2, 1, , 16., , f ( x ) = 3x + 10 and g ( x ) = x 2 − 1, ∴ fog = f [ g ( x )] = 3 [ g ( x )] + 10, = 3( x 2 − 1) + 10 = 3x 2 + 7, y −7, Let 3x 2 + 7 = y ⇒ x 2 =, 3,  y − 7, x =, ,  3 , , 1/2, , x − 7, So, ( fog ) −1 = , ,  3 , , 1/2, , ⇒, , It is a function, because first element of each, ordered pair is different., , 18. Since 2, 3, 4 are the elements of domain of, R1 having their unique images, this relation, R1 is a function., Since, the same first element 2 corresponds to, two different images 2 and 4, this relation R2, is not a function., Since, every element has one and only one, image, this relation R3 is a function., , 19. Let, f (x) = a 2 − x2, f ( x ) is defined, if a 2 − x 2 ≥ 0, ⇒, ⇒, ∴, , x2 − a 2 ≤ 0, (x − a ) (x + a ) ≤ 0, −a ≤ x ≤ a, Domain of f = [ −a , a ], , 20. We know that,, ⇒, ⇒, ⇒, ⇒, , −1 ≤ − cos x ≤ 1, −2 ≤ − 2 cos x ≤ 2, 1 − 2 ≤ 1 − 2 cos x ≤ 1 + 2, −1 ≤ 1 − 2 cos x ≤ 3, 1, 1, −1 ≤, ≤, 1 − 2 cos x 3, , −1 ≤ f ( x ) ≤, , ∴, , 1, Range of f = −1, , 3 , , , 21. Here, f ( x ) =, , 1, 3, , x 2 − 9 ( x − 3) ( x + 3), =, = ( x + 3), x −3, x −3, , and g ( x ) = x + 3 ⇒ f ( x ) = g ( x ) = x + 3, But f ( x ) ≠ g ( x ) as domain off ( x ) is R − { 3}, and domain of g ( x ) is R., ⇒ Domain of f ( x ) ≠ Domain of g ( x ), ⇒, f ≠g, 4−x, 22. We have, f ( x ) =, x −4, f ( x ) is defined, if x − 4 ≠ 0 i.e. x ≠ 4, Domain of f = R − { 4}, ∴, Let f ( x ) = y, 4−x, y=, ⇒ xy − 4 y = 4 − x, ∴, x −4, , 17. {(3, 3), (4, 2), (5, 1), (6, 0), (7, 7)}, , ⇒, , ⇒, , [Q a > 0], , ⇒ xy + x = 4 + 4 y ⇒ x ( y + 1) = 4(1 + y ), 4 (1 + y ), ∴, x=, y +1, x assumes real values, if y + 1 ≠ 0 i.e. y ≠ −1., ∴ Range of f = R − { −1}, , 23. We have, f ( x ) = 2 − | x − 5 |, f ( x ) is defined for all x ∈ R, ∴ Domain of f = R, We know that,, | x − 5| ≥ 0 ⇒ −| x − 5| ≤ 0 ⇒ 2 − | x − 5| ≤ 2, ∴ f (x) ≤ 2, ∴ Range of f = ( −∞, 2], 1, 24. We have, f ( x ) =, x− x, where, x − x = x − x = 0, if x ≥ 0, x − ( − x ) = 2x , if x < 0, 1, is not defined for x ∈ R., Thus,, x− x, Hence, f is not defined for any x ∈ R., x, 25. We have, f ( x ) =, 1 + x2, Let,, ∴, ⇒, , f (x) = y, y=, , x, 1 + x2, , x 2 y + y = x ⇒ yx 2 − x + y = 0

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29, , CBSE New Pattern ~ Mathematics XI (Term I), , x assumes real values, if ∆ ≥ 0, ( −1) 2 − 4 ( y ) ( y ) ≥ 0, ⇒, , 4y 2 − 1 ≤ 0, , − x + 2 x ≤ 2, f (x) = ,  x −2 x >2, y=, , ( 2y + 1) ( 2y − 1) ≤ 0,  1 1, ⇒, y∈ − ,,  2 2 , 1 1, ∴ Range of f = − ,,  2 2 , , –, , x+, , 2, , y=, , ⇒, , 2, , 1 − 4y 2 ≥ 0, , x–, , ⇒, , 28. f ( x ) = | x − 2 |, , 26. Let R be the set of real numbers. Define the, , real valued function f : R → R by, y = f ( x ) = x for each x ∈ R., Such a function is called the identity function., Here, the domain and range of f are R. The, graph is a straight line as shown in figure, given below. It passes through the origin., Y, 8, 6, , y=x, , 4, 2, X¢, , O2, –2, –4, , –8 –6 –4 –2, , Y¢, , 4, , X, 6, , 8, , –6, –8, , 27. Constant function Define the function, , f : R → R by y = f ( x ) = c , x ∈ R, where c is a, constant., Here, domain of f is R and its range is {c }., Y, 8, 6, y=c, , 4, 2, X¢, , –8 –6 –4 –2 O, , Y¢, , 2, –2, –4, , 4, , 6, , 8, , X, , –6, –8, , e.g. The graph of f ( x ) = 3, is a line passing, through ( 0, 3) and parallel to X -axis., , 29. We have, f ( x ) =, , x, , for x =/ 0, |x|, , x, , if x > 0, , 1, if x > 0, i.e. f ( x ) =  x, =, x, −1, if x < 0,  , if x < 0,  −x, Thus, range of f = {1, − 1}., , 30. Greatest integer function The function, , f : R → R defined by f ( x ) = [ x ], x ∈ R, assumes the value of the greatest integer, less, than or equal to x. Such a function is called, the greatest integer function., From the definition of [ x ], we can see that, [ x ] = − 1 for − 1 ≤ x < 0, [ x ] = 0 for 0 ≤ x < 1, [ x ] = 1 for 1 ≤ x < 2, [ x ] = 2 for 2 ≤ x < 3 and so on., The graph of the function is given in the, question., , 31. We have,, ⇒, , [x ] 2 − 5 [x ] + 6 = 0, , [ x ]2 − 3[ x ] − 2 [ x ] + 6 = 0, , ⇒ [ x ] ([ x ] − 3) − 2([ x ] − 3) = 0, ⇒, ([ x ] − 3) ([ x ] − 2) = 0, ⇒, [ x ] = 2, 3, ∴, x ∈[ 2, 3],  2, x <0, x ,, 32. Given, f ( x ) =  x , 0 ≤ x < 1, 1 ,, x ≥1,  x, 1, At x = ⇒ f ( x ) = x, 2, 1, 1, , f   =,  2 2

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30, , CBSE New Pattern ~ Mathematics XI (Term I), , 33. Given, f ( x ) = x 3 −, , 1, x3, 1, , 1, 1,  1, = 3 − x3, f   = 3 −, 3,  x x, x,  1,  ,  x, 1, 1,  1, f (x) + f   = x3 − 3 + 3 − x3,  x, x, x, =0, , 34. Given, y = e x, , 3, , −2, , log y = log e x, , 3, , −2, , log y = ( x − 2) log e, log y = x 3 − 2, 3, , log y |x = 5 = 53 − 2 = 125 − 2 = 123, 1 + x , , 1 − x , , 35. We have, f ( x ) = loge , , 1 + 2x , , ,  2x , 1 + x2 , , =, f , log, , e, 2,  1 − 2x , 1 + x , , , 1 + x2 , , 1 + x 2 + 2x , = loge , , 2,  1 + x − 2x , 1 + x , = loge , , 1 − x , , 2, , 1 + x , = 2 loge , , 1 − x , [Q log a b = blog a ], = 2f ( x ), , 36. Assertion Two ordered pairs are equal, if, and only if the corresponding first elements, are equal and the second elements are also, equal., Given, ( x + 1, y − 2) = ( 3, 1)., Then, by the definition, x + 1 = 3 and y − 2 = 1, ⇒, x = 2 and y = 3, Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., , 37. Assertion P and Q are two non-empty sets., , The cartesian product P × Q is the set of all, ordered pairs of elements from P and Q , i.e., P × Q = {( p, q ) : p ∈ P and q ∈Q }, , Reason Now, A = {red, blue}, B = {b , c , s }, A × B = Set of all ordered pairs, = {(red, b), (red, c), (red, s ), (blue, b),, (blue, c), (blue, s )}, Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., , 38. Assertion Given, ( 4 x + 3, y ) = ( 3x + 5, − 2), Two ordered pairs are equal when their, corresponding elements are equal., 4 x + 3 = 3x + 5 and y = − 2, 4 x − 3x = 5 − 3, x =2, Reason Now, A = { −1, 3, 4}, ∴ A × A = {( −1, − 1), ( −1, 3), ( −1, 4 ), ( 3, − 1),, ( 3, 3), ( 3, 4 ), ( 4, − 1), ( 4, 3), ( 4, 4 )}, ∴ Assertion is true and Reason is false., , 39. Assertion A = Set of first elements = { x , y, z }, B = Set of second elements = {1, 2}, ∴ A is correct., Reason n( A ) = 3, n( B ) = 2, n( A × B ) = n( A ) × n( B ) = 3 × 2 = 6, Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion., , 40. Assertion The total number of relation that, can be defined from a set A to a set B is the, number of possible subset of A × B . If, n( A ) = p and n( B ) = q , then n( A × B ) = pq and, the total number of relation is 2 p q ., Given,, A = {1, 2} and B = { 3, 4}, ∴, A × B = {(1, 3), (1, 4 ), ( 2, 3), ( 2, 4 )}, Since, n( A × B ) = 4, the number of subsets of, A × B is 24 . Therefore, the number of relation, from A to B will be 24 = 16., Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., , 41. Assertion In Roster form R = {(1, 1), (1, 2),, (1, 3), (1, 4), (1, 6), (2, 4), (2, 6), (2, 2), (4, 4),, (6, 6), (3, 3), (3, 6)}, Reason Domain of R = Set of first element, of ordered pairs in R = {1, 2, 3, 4, 6}, Range of R = {1, 2, 3, 4, 6}, Hence, Assertion is false, Reason is true.

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31, , CBSE New Pattern ~ Mathematics XI (Term I), , 42. Assertion In Set-builder form,, R = {( x , y ) : y = x , x , y ∈ Z and −2 ≤ x ≤ 2}, 2, , [Q 0 ∉N ], Reason The relation shown in figure is, represented in Roster form as, R = {( −2, 4 ), ( −1, 1), ( 0, 0 ), (1, 1), ( 2, 4 )}, We observe that, second element of each, ordered pair is the square of first element., Hence, Assertion is false, Reason is true., , 43. Assertion The given relation in Roster form, is, R = {( 0, 5), (1, 6 ), ( 2, 7 ), ( 3, 8 ), ( 4, 9 ), ( 5, 10 )}, Domain of R = { 0, 1, 2, 3, 4, 5}, Reason Range of R = { 5, 6, 7, 8, 9, 10}, Hence, Assertion is true, Reason is false., , 44. Assertion The given relation in Roster form, , is, R = {( 3, 5), ( 4, 6 ), ( 5, 7 ), ( 6, 8 ), ( 7, 9 ),, ( 8, 10 ), ( 9, 11)}, ∴ Domain of R = { 3, 4, 5, 6, 7, 8, 9}, So, A is true., Reason Range of R = { 5, 6, 7, 8, 9, 10, 11}, So, R is false., Hence, Assertion is true, Reason is false., , 45. Assertion In arrow diagram, every element, of P has its unique image in Q. So, it, represent a function., Reason Domain of f = R − { 2}., Domain of g = R, Q, D f ≠ Dg, We know that, two functions are equal when, their domain and range are equal and same, element in their domain produce same image., ∴, f ≠g, Hence, Assertion is true and Reason is false., , 46. Assertion We have,, Let, ⇒, ⇒, Q, , f ( x ) = 2 − 3x , x ∈ R, x > 0, f ( x ) = y, then y = 2 − 3x, 3x = 2 − y, 2−y, x=, 3, , x> 0, , 2−y, > 0⇒2− y > 0⇒ 2> y, 3, ∴, y<2, Hence, range of f = ( − ∞, 2), ⇒, , Reason Now, f ( x ) = x 2 + 2, Let y = f ( x ), then, y = x2 + 2 ⇒ x = y − 2, x assumes real values, if y − 2 ≥ 0, ⇒, y ≥ 2 ⇒ y ∈ [ 2, ∞ ), ∴ Range of f = [ 2, ∞ ), Hence, Assertion is false, Reason is true., , 47. Assertion Given,, R = {( x , y ) : | x − y | is odd, x ∈ A , y ∈ B }, The relation R in Roster form is, R = {(1, 4 ), (1, 6 ), ( 2, 9 ), ( 3, 4 ), ( 3, 6 ),, ( 5, 4 ), ( 5, 6 )}, ∴ Domain of R = {1, 2, 3, 5}, So, A is true., Reason It is also true | x | is always positive., Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion., , 48. Assertion We have, f ( x ) = x − 1, f ( x ) is defined, if x − 1 ≥ 0 i.e. x ≥ 1, ∴ Domain of f = [1, ∞ ), Hence, A is incorrect., Reason Let f ( x ) = y, Then,, , y = x −1, , Q, x ≥1, ∴Range of f = [ 0 , ∞ )., Hence, Assertion is false, Reason is true., , 49. Assertion Given,, f (x) = x +, , 1, x, , f (x3) = x3 +, , 1, x3, , 1, , [ f ( x )]3 =  x + , , x, = x3 +, , 3, , 1, 1, , + 3 x + , , x, x3, , = f (x3) + 3 f (x),  1, = f ( x 3 ) + 3f  ,  x, , ,  1 1, Q f  x  = x + x = f ( x ) , , 

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32, , CBSE New Pattern ~ Mathematics XI (Term I), , Reason Now, we have,, f ( x ) = ( x − a )2 ( x − b )2, f ( a + b ) = ( a + b − a ) 2 ( a + b − b ) 2 = b 2a 2, Hence, Assertion is true, Reason is false., , 50. Assertion We have,, f ( x ) = 2x + 3, g ( x ) = x 2 + 7, ⇒, ⇒, , g [ f ( x )] = 8, g ( 2x + 3) = 8, ( 2x + 3) 2 + 7 = 8, , ⇒, , ( 2x + 3) 2 = 1, , ⇒, , 2x + 3 = ± 1,, 2x + 3 = − 1, or, 2x + 3 = 1, then, ⇒, x = − 1, x = − 2, 4x, Reason Now, f ( x ) = x, 4 +2, f (1 − x ) =, ∴ f ( x ) + f (1 − x ) =, , 41 − x, 41 − x + 2, 4x, 41 − x, +, 4 x + 2 41 − x + 2, x, , x, , 4, 4/4, +, x, 4, 4 +2, + 2, 4x, 4x, 4, = x, +, 4 + 2 4 + 2 ⋅ 4x, =, , =, , 4x, 2, +, 4 + 2 4x + 2, , =, , 4x + 2, =1, 4x + 2, , x, , Hence, Assertion is false, Reason is true., , 51. (i) Here, first element of each ordered pair of, , A × B gives the elements of set A and, corresponding second element gives the, elements of set B., ∴, A = {a , b } and B = {1, 3, 2}, Note We write each element only one time, in set, if it occurs more than one time., (ii) Given, n ( A ) = 3 and n ( B ) = 4., ∴The number of elements in A × B is, n ( A × B ) = n ( A ) × n ( B ) = 3 × 4 = 12, , (iii) It is given that (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 = {1, 2, 3} and B = { 3, 5}, (iv) Q, A = {1, 2, 3} and B = { 3, 5}, ∴, 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), (3, 5)., (v) Given, n( P × P ) = 16, ⇒, n( P ) ⋅n( P ) = 16, ... (i), ⇒, n( P ) = 4, Now, as, ( a , 1) ∈ P × P, ∴, a ∈ P and 1 ∈ P, Again,, ( b , 2) ∈ P × P, ∴, b ∈ P and 2 ∈ P, ⇒, a , b , 1, 2 ∈ P, From Eq. (i), it is clear that P has exactly, four elements., , 52. (i) We know that, two ordered pairs are equal,, if their corresponding elements are equal., ( a − 3, b + 7 ) = ( 3, 7 ) ⇒ a − 3 = 3, and, b + 7 =7, [equating corresponding elements], ⇒ a = 3 + 3 and b = 7 − 7, ⇒ a = 6 and b = 0, (ii) ( x + 6, y − 2) = ( 0, 6 ), x + 6 = 0⇒x = −6, y − 2=6⇒y =6 + 2=8, (iii) ( x + 2, 4 ) = ( 5, 2x + y ), x + 2= 5⇒x = 5− 2= 3, 4 = 2x + y ⇒ 4 = 2 × 3 + y, ⇒, y = 4 −6 = − 2, (iv) Since, (1, 4 ), ( 2, 6 ) and (3, 6) are elements of, A × B , it follows that 1, 2, 3 are elements of, A and 4, 6 are elements of B. It is given that, A × B has 6 elements., So, A = {1, 2, 3} and B = { 4, 6}, Hence, A × B = {1, 2, 3} × { 4, 6}, = {(1, 4 ), (1, 6 ), ( 2, 4 ), ( 2, 6 ),, ( 3, 4 ), ( 3, 6 )}

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33, , CBSE New Pattern ~ Mathematics XI (Term I), , and B × A = { 4, 6} × {1, 2, 3}, = {( 4, 1), ( 4, 2), ( 4, 3), ( 6, 1),, ( 6, 2), ( 6, 3)}, (v) We have,, n( A × B ) = 45, ⇒, n( A ) × n( B ) = 45, ⇒ n( A ) and n( B ) are factors of 45 such that, their product is 45., Here, n( A ) cannot be 17., , 54. (i) If we draw a vertical line, then it will, intersect the curve at two points. It shows, that given curve is a relation., Y, , X′, , 53. (i) Given, R = {( a , b ) : 2a + b = 5; a , b ∈W }, Here, W represent set of whole numbers., When, a = 0, b = 5, When, a = 1, b = 3, When, a = 2, b = 1, For a ≥ 3, the value of b given by the above, relation are not whole numbers., ∴, R = {( 0, 5), (1, 3), ( 2, 1)}, (ii) From arrow diagram, we have, P = {9, 4, 25}, and Q = { 5, 4, 3, 2, 1, − 2, − 3, − 5}, Here, the relation R is ‘x is the square of y’,, where x ∈ P and y ∈Q ., In roster form, it can be written as, R = {( 9, 3) ( 9, − 3) ( 4, 2) ( 4, − 2), ( 25, 5) ( 25, − 5)}, (iii) In set-builder form, R can be written as, R = {( x , y ) : x ∈ P , y ∈Q, and x is the square of y}, (iv) We have, A = {a , b } and B = { 2, 3},, ∴, n( A × B ) = n( A ) × n( B ), = 2× 2 = 4, Now, number of subsets of A × B, = 2n ( A × B ), = 24, = 16, Thus, the number of relations from A to B, is 16., (v) Given,, n( A ) = 3, and, B = {2, 3, 4, 6, 7, 8}, ⇒, n( B ) = 6, ∴ Number of relations from A toB, = 2n ( A ) × n ( B ), = 23 × 6 = 218, , X, , O, , Y′, , (ii) If we draw a vertical line, then it will, intersect the curve at only one point. It, shows that given curve is a function., Y, , X′, , X, , O, Y′, , (iii) We have,, , f ( x ) = x 2 + 2x + 3, , …(i), , f (1) = (1) + 2(1) + 3, 2, , [putting x = 1 in Eq. (i)], =1 + 2 + 3 = 6, f ( 2) = ( 2) 2 + 2( 2) + 3, [putting x = 2 in Eq. (i)], = 4 + 4 + 3 = 11, f ( 3) = ( 3) 2 + 2( 3) + 3, [putting x = 3 in Eq. (i)], = 9 + 6 + 3 = 18, (iv) We have, f (1 + x ) = x 2 + 1, , …(i), , On substituting x = (1 − h ) in Eq. (i), we get, f (1 + 1 − h ) = (1 − h ) 2 + 1, f ( 2 − h ) = 1 + h 2 − 2h + 1 = h 2 − 2h + 2, (v) We know that,, ⇒, ⇒, ⇒, ⇒, , − 1 ≤ sin 3x ≤ 1, − 1 ≤ − sin 3x ≤ 1, 1 ≤ 2 − sin 3x ≤ 3, 1, 1, ≤, ≤1, 3 2 − sin 3x, 1, ≤ f (x) ≤1, 3, , 1, ∴ Range ( f ) =  , 1,  3 

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34, , CBSE New Pattern ~ Mathematics XI (Term I), , 03, Complex Numbers, Quick Revision, Imaginary Numbers, , Equality of Complex Numbers, , The square root of a negative real number is called, imaginary number, e.g. − 2 , − 5 etc., , Two complex numbers z 1 = x 1 + iy 1 and, z 2 = x 2 + iy 2 are said to be equal, iff x 1 = x 2 and, y1 = y 2, i.e. Re (z 1 ) = Re (z 2 ) and Im (z 1 ) = Im (z 2 ), Other relation ‘greater than’ and ‘less than’ are not, defined for complex number., , The quantity − 1 is an imaginary unit and it is, denoted by ‘i ’ called iota., , Integral Power of IOTA ( i ), i = − 1, i 2 = − 1, i 3 = − i , i 4 = 1, So, i 4n + 1 = i , i 4n + 2 = − 1, i 4n + 3 = − i , i 4n = 1, ●, , For any two real numbers a and b, the result, a × b = ab is true only, when atleast one of, the given numbers is either zero or positive., − a × − b ≠ ab So, i 2 = − 1 × − 1 ≠ 1, , ●, ●, , ‘ i ’ is neither positive, zero nor negative., i n + i n +1 + i n + 2 + i n + 3 = 0, , Complex Numbers, A number of the form x + iy , where x and y are, real numbers, is called a complex number. Here, x, is called real part and y is called imaginary part of, the complex number, i.e. Re (z ) = x and Im (z ) = y ., Purely Real and Purely Imaginary, Complex Numbers, A complex number z = x + iy is a purely real if its, imaginary part is 0, i.e. Im (z ) = 0 and purely, imaginary if its real part is 0 i.e. Re (z ) = 0., , Algebra of Complex Numbers, Let z 1 = x 1 + iy 1 and z 2 = x 2 + iy 2 be any two, complex numbers., (i) Addition of Complex Numbers, z 1 + z 2 = ( x 1 + iy 1 ) + ( x 2 + iy 2 ), = ( x 1 + x 2 ) + i ( y1 + y 2 ), Properties of addition, ● Closure z, 1 + z 2 is also a complex number., ●, , Commutative z 1 + z 2 = z 2 + z 1, , ●, , Associative z 1 + (z 2 + z 3 ) = (z 1 + z 2 ) + z 3, , Existence of additive identity, z +0=z =0+z, Here, 0 is additive identity., ● Existence of Additive inverse, z + ( −z ) = 0 = ( − z ) + z, Here, −z is additive inverse., (ii) Subtraction of Complex Numbers, z 1 − z 2 = ( x 1 + iy 1 ) − ( x 2 + iy 2 ), = ( x 1 − x 2 ) + i ( y1 − y 2 ), ●

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35, , CBSE New Pattern ~ Mathematics XI (Term I), , (iii) Multiplication of Complex Numbers, z 1 z 2 = ( x 1 + iy 1 ) ( x 2 + iy 2 ), = ( x 1 x 2 − y1 y 2 ) + i ( x 1 y 2 + x 2 y1 ), Properties of multiplication, ● Closure z z, 1 2 is also a complex number., ●, , Commutative z 1z 2 = z 2z 1, , ●, , Associative z 1 (z 2z 3 ) = (z 1z 2 ) z 3, , ●, , ●, , ●, , Existence of multiplicative identity, z ⋅1 = z = 1 ⋅ z, Here, 1 is multiplicative identity., Existence of multiplicative inverse For every, non-zero complex number z, there exists a, complex number z 1 such that z ⋅ z 1 = 1 = z 1 ⋅ z, Distributive law z 1 (z 2 + z 3 ) = z 1z 2 + z 1 z 3, , (iv) Division of Complex Numbers, z 1 x 1 + iy 1 ( x 1 x 2 + y 1 y 2 ) + i ( x 2 y 1 − x 1 y 2 ), =, =, z 2 x 2 + iy 2, x 22 + y 22, where, z 2 ≠ 0, , Conjugate of a Complex Number, Let z = x + iy , if ‘i ’ is replaced by ( − i ), then it is, said to be conjugate of the complex number z and, denoted by z , i.e. z = x − iy ., Properties of Conjugate, (i) (z ) = z, (ii) z + z = 2 Re (z ), z − z = 2i Im (z ), (iii) z = z , if z is purely real, (iv) z + z = 0 ⇔ z is purely imaginary, (v) z 1 + z 2 = z 1 + z 2, (vi) z 1 − z 2 = z 1 − z 2, (vii) z 1z 2 = z 1 ⋅ z 2, z  z, (viii)  1  = 1 , z 2 ≠ 0, z 2 z 2, (ix) z ⋅ z = {Re (z )} 2 + {Im (z )} 2, (x) z 1z 2 + z 1 z 2 = 2 Re (z 1 z 2 ) = 2 Re (z 1z 2 ), (xi) If z = f (z 1 ), then z = f (z 1 ), (xii) (z )n = (z n ), , Modulus (Absolute Value) of a Complex, Number, Let z = x + iy be a complex number. Then, the, positive square root of the sum of square of real, part and square of imaginary part is called, modulus (absolute values) of z and it is denoted, by | z | i.e. | z | = x 2 + y 2 ., Properties of Modulus, (i) | z | ≥ 0, (ii) If | z | = 0, then z = 0 i.e. Re (z ) = 0 = Im (z ), (iii) − | z | ≤ Re (z ) ≤ | z | and − | z | ≤ IImz ) ≤ | z |, (iv) | z | = | z | = | − z | = | − z |, (v) z ⋅ z = | z |2, (vi) | z 1z 2 | = | z 1 | | z 2 |, (vii), , z1 | z1 |, =, ,z ≠0, z2 |z2 | 2, , (viii) | z 1 + z 2 |2 = | z 1 |2 + | z 2 |2 + 2 Re (z 1z 2 ), (ix) | z 1 − z 2 |2 = | z 1 |2 + | z 2 |2 − 2 Re (z 1 z 2 ), (x) | z 1 + z 2 | ≤ | z 1 | + | z 2 |, (xi) | z 1 − z 2 | ≥ | z 1 | − | z 2 |, , Argand Plane, A complex number z = a + ib can be represented, by a unique point P (a , b ) in the cartesian plane, referred to a pair of rectangular axes., A purely real number a, i.e. (a + 0 i ) is represented, by the point (a , 0 ) on X-axis. Therefore, X-axis is, called real axis., A purely imaginary number ib i.e. (0 + i b ) is, represented by the point (0, b ) on Y-axis., Therefore, Y-axis is called imaginary axis. The, intersection (common) of two axes is called zero, complex number i.e. z = 0 + 0i ., Similarly, the representation of complex numbers, as points in the plane is known as argand, diagram. The plane representing complex, numbers as points, is called complex plane or, argand plane or gaussian plane.

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36, , CBSE New Pattern ~ Mathematics XI (Term I), , If two complex numbers z 1 and z 2 are represented, by the points P and Q in the complex plane, then, | z 1 − z 2 | = PQ = Distance between P and Q, , Quadratic Equation, , Y, Imaginary, axis, , X′, , P (z1 ), , O, (0,0), , Q (z2 ), Real, axis, X, , Y′, , Representation of Conjugate of z, on Argand Plane, Geometrically, the mirror image of the complex, number z = a + ib (represented by the ordered pair, (a , b ) about the X-axis is called conjugate of z, which is represented by the ordered pair (a , − b )., If z = a + ib , then z = a − ib ., Y, , X′, , P (a, b), , X, , O, , Q (a , − b ), , Y′, , Representation of Modulus of z, on Argand Plane, Geometrically, the distance of the complex, number z = a + ib [represented by the ordered pair, (a , b )] from origin, is called the modulus of z., ∴ OP = (a − 0 )2 + (b − 0 )2, = a + b = {Re(z )} + {Im(z )} = | a + ib |, 2, , 2, , 2, , 2, , Y, P (a, b), , X′, , (0, 0) O, , M, , X, , An equation of the form ax 2 + bx + c , a ≠ 0 is, called quadratic equation in variable x, where a , b, and c are numbers (real or complex)., , Nature of Roots of Quadratic Equation, The roots of quadratic equation ax 2 + bx + c = 0,, a ≠ 0 are, −b + b 2 – 4 ac, −b − b 2 – 4 ac, and β =, ., α=, 2a, 2a, Now, if we look at these roots of quadratic, equation ax 2 + bx + c = 0; a ≠ 0, we observe that, the roots depend upon the value of the quantity, b 2 − 4 ac . This quantity is known as the, discriminant of the quadratic equation and, denoted by D., There are following four cases arise :, Case I If b 2 − 4 ac = 0 i.e. D = 0,, b, ., then, α =β = −, 2a, Thus, if b 2 − 4 ac = 0, then the quadratic equation, has real and equal roots and each equal to −b / 2a ., Case II If a , b and c are rational numbers and, b 2 − 4 ac > 0 and it is a perfect square, then, D = b 2 − 4 ac is a rational number and hence α, and β are rational and unequal., Case III If b 2 − 4 ac > 0 and it is not a perfect, square, then roots are irrational and unequal., Case IV If b 2 − 4 ac < 0, then the roots are complex, conjugate of each other., , Quadratic Equations with Real, Coefficients, Let us consider the following quadratic equation, ax 2 + bx + c = 0 with real coefficients a , b , c and, a ≠ 0. Also, let us assume that b 2 − 4 ac < 0. Now,, we can find the square root of negative real, numbers in the set of complex numbers., Therefore, the solutions of the above equation are, available in the set of complex numbers which are, given by, −b ± −(4 ac − b 2 ), −b ± b 2 − 4 ac, =, 2a, 2a, 2, −b ± 4 ac − b i, =, 2a, , x=, Y′

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37, , CBSE New Pattern ~ Mathematics XI (Term I), , Objective Questions, Multiple Choice Questions, 1. If 4x + i ( 3x − y ) = 3 + i ( −6 ), where x and, y are real numbers, then the values of x, and y are, 3, 33, (b) x = , y =, 4, 4, (d) x = 33 , y = 4, , (a) x = 3 , y = 4, (c) x = 4, y = 3, , 2. If (1 − i )x + (1 + i )y = 1 − 3i, then (x , y ) is, equal to, (a) (2, − 1), (c) (−2, − 1), , 3. If i, , 103, , (b) (−2, 1), (d) (2, 1), , = a + ib , then a + b is equal to, (b) −1, (d) 2, , (a) 1, (c) 0, , 4. 1 + i, , 10, , +i, , 20, , +i, , 30, , 5. The value of, , 2, , is equal to, , (b) −1, (d) 0, , 6. If z 1 = 2 + 3i and z 2 = 3 + 2i , then, z 1 + z 2 equals to, (a) 5 + 5 i, (c) 4 + 6 i, , (b) 5 + 10i, (d) 6 + 4 i, , 7. If z 1 = 2 + 3 i and z 2 = 3 − 2 i , then, z 1 − z 2 is equal to, (a) −1 + 5 i, (c) i + 5, , (b) 5 − i, (d) None of these, , −1  3, i  = a + ib then a is equal, 6 , , 8. If ( −i )( 3i ), to, (a) 1, (c) −1, , (b) 0, (d) 2, , 9. If Z 1 = 2 + 3i and Z 2 = 1 − 4i then Z 1 Z 2, is equal to, , 10. If 3(7 + 7i ) + i (7 + 7i ) = a + ib then, , b, is, a, , equal to, (a) 2, (c) 3, , 11. If, , (1 + i ) 2, 2 −i, , (b) 1, (d) −1, , = x + iy then the value of x + y, , is, 1, 5, 3, (c), 5, (a), , 2, 5, 4, (d), 5, (b), , and b are respectively, (b) Complex number, (d) None of these, , i 4 x +1 − i 4 x − 1, , (a) i, (c) −i, , (b) 14 + 5i, (d) −14 − 5i, , 12. If (1 − i )4 = a + ib, then the value of a, , is a, , (a) Real number, (c) Natural number, , (a) 14 − 5i, (c) −14 + 5i, , (a) −4, 0, (c) 4, 0, , (b) 0, − 4, (d) 0, 4, , 13. If z 1 = 6 + 3i and z 2 = 2 − i , then, , z1, z2, , is, , equal to, (a), , 1, (9 + 12 i), 5, , (c) 3 + 2i, , (b) 9 + 12i, (d), , 1, (12 + 9i), 5, , 14. The multiplicative inverse of, , 3 + 5i, 4 − 3i, , is, , equal to, , −3 29i, −, 34 34, 3 29i, (c), −, 34 34, , (a), , 3 29i, +, 34 34, −1 29i, (d), −, 34 34, (b), , 15. If Z 1 = 3 + 3i and Z 2 = 3 + i then, Z , the quadrant in which  1  lies is,  Z2 , , (a) First, (c) Third, , (b) Second, (d) Fourth

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38, , CBSE New Pattern ~ Mathematics XI (Term I), , 16. If Z 1 = 1 + 2i and Z 2 = 2 + 3i , then sum, of Z 1 and additive inverse of Z 2 is, equal to, (a) 1 + 2i, (c) 3 + 5i, , (b) 3 + i, (d) −1 − i, , 17. a + ib form of complex number, 9 −i + 6 + i, , 3, , − 9 + i is given by, 2, , (a) 7 − 2i, (c) −7 − 2i, , (b) 7 + 2i, (d) −7 + 2i, , 18. If Z 1 = 3 + 2i and Z 2 = 2 − i then, Z 1 + Z 2 is given by, , (a) 5 − i, (c) −5 + i, , (b) 5 + i, (d) −5 − i, , 19. If Z 1 = 1 + i , Z 2 = 2 − i and, , Z 1 Z 2 = a + ib , then a + b is equal to, , (a) 2, (c) 3, , (b) 1, (d) 4, , 20. The conjugate of, −2 11, (a), + i, 25 25, 2, 11, (c), + i, 25 25, , 2 −i, (1 − 2i ), , 2, , is equal to, , −2 11, − i, 25 25, 2, 11, (d), − i, 25 25, , (b), ,  Z1 , ,  is equal to,  Z2 , 9 19, + i, 13 13, −9 19, (c), − i, 13 13, , 9 19, − i, 13 13, −9 19, (d), + i, 13 13, (b), , 22. Let Z 1 = 2 − i , Z 2 = − 2 + i and, Z1Z 2, Z1, , = a + ib , then a is equal to, , 2, 5, 11, (c), 5, (a), , 3, 5, −2, (d), 5, , (b), , 23. If Z = − 5i −15 − 6i −8 then Z is equal to, (a) −6 − 5i, (c) 6 − 5i, , 4 + 3i 7 is equal to, , (b) −5, (d) 3, , (a) 5, (c) 2, , 25. If Z 1 = 1 + 3i and Z 2 = 2 + 4i then, | Z 2 − Z 1 | 2 is equal to, (a) 1, (c) 3, , 26. If Z =, , (b) 2, (d) 4, , (1 + i )( 2 + i ), , then| Z | is equal to, (3 + i ), , (a) 1, (c) 2, , (b) 0, (d) 3, , 27. The modulus of the complex number, (1 − i ) −2 + (1 + i ) −2 is equal to, (a) 1, (c) 3, , (b) 2, (d) 0, , 28. If Z 1 = 3 + 2i and Z 2 = 2 − 4i then the, , 21. If Z 1 = 3 + 5i and Z 2 = 2 − 3i , then, , (a), , 24. The modulus of the complex number, , (b) −6 + 5i, (d) 6 + 5i, , value of | z 1 + z 2 | 2 + | z 1 − z 2 | 2 is equal, to, (a) 11, (c) 66, , (b) 22, (d) 55, , 29. Roots of 9x 2 + 16 = 0 is given by, 4 , (a) ±  i, 3 ,  3, (c) ±   i,  2, ,  3, (b) ±   i,  4,  2, (d) ±   i,  3, , 30. Roots of x 2 + 2 = 0 are, (a) ± 2i, (c) 2i, , (b) 2, (d) None of these, , 31. Roots of x 2 + 3x + 9 = 0 are, −3 ± 3 3i, 2, 3 ± 3i, (c), 2, (a), , 3 ± 3 3i, 2, −3 ± 3i, (d), 2, (b), , 32. Roots of x 2 + x + 1 = 0 are, −1 ± 3i, 2, 2 ± 3i, (c), 2, , (a), , 1 ± 3i, 2, −2 ± 3i, (d), 2, (b)

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39, , CBSE New Pattern ~ Mathematics XI (Term I), , 33. Roots of 2x 2 + x + 2 = 0 is given by, −1 ± i 7, 2 2, 1 ± 7i, (c), 2, (a), , (b), , −1 ± i 7, 2, , (d) None of these, , 34. Roots of ( y + 1)( y − 3) + 7 = 0 is given by, (a) −1 ± 3i, (c) 1 ± 2i, , x 2 − px + 8 = 0 is 2, then p is equal to, , Directions (Q. Nos. 36-50) Each of these, questions contains two statements, Assertion (A) and Reason (R). Each of the, questions has four alternative choices, any, one of the which is the correct answer. You, have to select one of the codes (a), (b), (c) and, (d) given below., (a) A is true, R is true; R is a correct, explanation of A., (b) A is true, R is true; R is not a correct, explanation of A., (c) A is true; R is false, (d) A is false; R is true., , 36. Assertion (A) If i = −1, then i, = i, i, , Reason (R), i 4k + i 4k + 1 + i 4k, , = − 1 and i, +2, , + i 4k, , 4k + 3, , +3, , −3, −7, and y =, ., 2, 2, Reason (R) If a + ib = c + id , then a = c, and b = d ., then x =, , Assertion-Reasoning MCQs, , i, , a, = 1., b, , 41. Assertion (A) If (1 + i )(x + iy ) = 2 − 5i,, , (b) ± 2, (d) ± 5, , 4k + 2, , 40. Assertion (A) If (1 + i ) 6 = a + ib, then, Reason(R) If (1 − i ) 3 = a + ib, then, , 35. If difference in roots of the equation, , 4k + 1, , Z 2 = 3 − 2i , then Z 1 − Z 2 = − 1 + 5i ., Reason (R) If Z 1 = a + ib and, Z 2 = c + id , then, Z 1 − Z 2 = (a − c ) + i (b − d ), , b = − 8., , (b) 1 ± 3i, (d) −1 ± 2i, , (a) ± 6, (c) ± 1, , 39. Assertion (A) If Z 1 = 2 + 3i and, , 4k, , 2 − 3i is 2 + 3i., Reason (R) If z = 3 + 4i , then, Z = 3 − 4i ., , 43. Assertion (A), , ( 2 + 3i )[( 3 + 2i ) + ( 2 + i )] = 1 + 21i., Reason (R) z 1 (z 2 + z 3 ) = z 1z 2 + z 1z 3 ., −3 , i  , then z is,  5 , , 44. Assertion (A) If z = 5i , = 1,, , = − i., , = 1., , 37. Assertion (A) Simplest form of i −35, , is −i., Reason (R) Additive inverse of (1 − i ) is, equal to −1 + i., 5 + 2i, 38. Assertion (A) Simplest form of, 1 − 2i, is 1 − 2 2i., Reason (R) The value of (1 + i ) (1 − i ), is 32., 5, , 42. Assertion (A) Multiplicative inverse of, , equal to 3 + 0i., Reason (R) If z 1 = a + ib and z 2 = c + id ,, , then z 1 + z 2 = (a + c ) + i (b + d )., 1 + 2i, 45. Assertion (A) If z =, , then, 1 − 3i, 1, ., |z | =, 2, Reason (R) If z = a + ib , then, |z | = a 2 + b 2 ., , 46. Assertion (A) If x + 4iy = ix + y + 3,, then x = 1 and y = 4., Reason (R) The reciprocal of 3 + 7i is, , 5, , equal to, , 3, 7, −, i., 16 16

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40, , CBSE New Pattern ~ Mathematics XI (Term I), , 47. Assertion (A) If z = i 9 + i 19 , then z is, equal to 0 + 0i., Reason (R) The value of, 1 + i 2 + i 4 + i 6 + .... + i 20 is equal to −1., , 48. Assertion (A) If x 2 + 1 = 0, then, , 100, ,  1 −i , (iii) If , = a + ib, then the values, , 1 + i , of a and b are respectively, (a) 1, 0, (c) 1, 2, , (b) 0, 1, (d) 2, 1, , (1 + i ) 2, = x + iy, then the value of, 2 −i, x + y is, , (iv) If, , solution is ± i ., Reason (R) The value of i −1097 is equal, to i., , 49. Assertion (A) If 3x + 4x + 2 = 0, then, 2, , equation has imaginary roots., Reason (R) In a quadratic equation,, ax 2 + bx + c = 0, if D = b 2 − 4ac is less, than zero, then the equation will have, imaginary roots., , 50. Assertion (A) Roots of quadratic, , equation x + 3x + 5 = 0 is, −3 ± i 11, ., x=, 2, Reason (R) If x 2 − x + 2 = 0 is a, quadratic equation, then its roots are, 1±i 7, ., 2, 2, , Case Based MCQs, 51. Two complex numbers Z 1 = a + ib and, , Z 2 = c + id are said to be equal, if a = c, and b = d ., On the basis of above information,, answer the following questions., (i) If ( 3a − 6 ) + 2ib = − 6b + (6 + a )i , then, the real values of a and b are, respectively, (a) −2, 2, , (b) 2,− 2, , (c) 3, − 3, , (d) 4, 2, , (ii) If ( 2a + 2b ) + i (b − a ) = − 4i , then the, real values of a and b are, respectively., (a) 2, 3, , (b) 2, −2, , (c) 3, 1, , (d) − 2, 2, , (a), , 1, 5, , (b), , 3, 5, , (c), , 4, 5, , (d), , 2, 5, , (v) If (x + y ) + i (x − y ) = 4 + 6 i , then xy is, equal to, (a) 5, , (b) − 5, , (c) 4, , (d) −4, , 52. A complex number z is pure real if and, only if z = z and is pure imaginary if, and only if z = − z ., Based on the above information,, answer the following questions., (i) If (1 + i )z = (1 − i )z , then − iz is, (a) − z, (c) z, , (b) z, (d) z − 1, , (ii) z 1z 2 is, (b) z1 + z2, , (a) z1 z2, (c), , z1, z2, , (d), , 1, z1 z2, , (iii) If x and y are real numbers and the, complex number, ( 2 + i )x − i (1 − i )y + 2i, is pure real,, +, 4 +i, 4i, the relation between x and y is, (a) 8 x − 17y = 16, (c) 17x − 8 y = 16, , (iv) If z =, , (b) 8 x + 17y = 16, (d) 17x − 8 y = − 16, , 3 + 2i sin θ , π,  0 < θ ≤  is pure, 1 − 2i sin θ , 2, , imaginary, then θ is equal to, (a), , π, 4, , (b), , π, 6, , (c), , π, 3, , (d), , π, 12

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41, , CBSE New Pattern ~ Mathematics XI (Term I), , (v) If z 1 and z 2 are complex numbers, z −z2, such that 1, =1, z1 + z 2, z1, is pure real, z2, z, (b) 1 is pure imaginary, z2, , (a) −4, , (a), , higher powers of i as follows, (i) i 2 = − 1, (ii) i 3 = i 2 ⋅ i = ( − 1) ⋅ i = − i, (iii) i 4 = (i 2 ) 2 = ( − 1) 2 = 1, (iv) i 5 = i 4 + 1 = i 4 ⋅ i = 1 ⋅ i = i, (v) i 6 = i 4 + 2 = i 4 ⋅ i 2 = 1 ⋅ i 2 = − 1, M M, M, M, In order to compute i n for n > 4, write, i n = i 4q + r for some q , r ∈ N and, 0 ≤ r ≤ 3. Then, i n = i 4q ⋅ i r, = (i 4 )q ⋅ i r = (1)q ⋅ i r = i r ., In general for any integer k,, i, , 4k + 3, , = i , i 4k, , +2, , = − 1 and, , = − i., , On the basis of above information,, answer the following questions., (i) The value of i 37 is equal to, , (a) 1 + 0i, (c) 0 + 0i, , |z | = a 2 + b 2, Multiplicative inverse of z is, , (a) 0, (c) 2, , (c) z + 3, , 0 + 0i, 1 + 0i, 0+i, 1 + 2i, , (b) 1, (d) 3, , (ii) The value of (z + 3) (z + 3) is, equivalent to, , is equal to, , (a), (b), (c), (d), , . It is, , (i) If (x − iy ) ( 3 + 5i ) is the conjugate of, − 6 − 24i, then the value of x + y is, equal to, , (a) z + 3, , (iii) If z = i 9 +i 19 , then z is equal to, , |z |2, , On the basis of above information,, answer the following questions., , (d) − 1, (b) 1, (d) − i, , z, , also called reciprocal of z ., zz =| z | 2 ., , (c) 1, (a) i, (c) − 1, , (b) 0 + i, (d) 1 + i, , is the complex number, obtained by, changing the sign of imaginary part of, z . It is denoted by z ., The modulus (or absolute value) of a, complex number, z = a + ib is defined, as the non-negative real number, a 2 + b 2 . It is denoted by | z |. i.e., , (b) − i, , (ii) The value of i, , (d) 1, , 54. The conjugate of a complex number z ,, , (a) i, −30, , (c) i, , (v) If z = i , then simplest form of z is, equal to, , 53. We have, i = −1. So, we can write the, , +1, , (b) 4, −39, , (c) z1 is pure real, (d) z1 and z2 are pure imaginary, , i 4k = 1, i 4k, , 2, , 25, ,  1 , (iv) The value of i 19 +    is equal, i  , , , to, , 2, , (b) z − 3, , 2, , (iii) If f (z ) =, , (d) None of these, , 7 −z, , , where z = 1 + 2i ,, 1−z 2, then f (z ) is equal to, z, 2, (c) 2 z, (a), , (b) z, (d) None of these

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42, , CBSE New Pattern ~ Mathematics XI (Term I), , (iv) If z 1 = 1 − 3i and z 2 = − 2 + 4 i , then, | z 1 + z 2 | is equal to, (a), , 2, , (b) 2, , (c), , (d) 1, , 3, , z +z, is equal to, (v) If z = 3 + 4i , then, 2, (a) 1, , (b) 2, , (c) 3, , (d) 4, , 55. An equation of the form ax + bx + c ,, 2, , a ≠ 0 is called quadratic equation in, variable x, where a, b and c are numbers, (real or complex)., The roots of quadratic equation, ax 2 + bx + c = 0, a ≠ 0 are, α=, β=, , −b + b 2 – 4 ac, 2a, −b − b, , 2, , – 4 ac, , ., 2a, Now, if we look at these roots of, quadratic equation, ax 2 + bx + c = 0; a ≠ 0, we observe that, the roots depend upon the value of the, quantity b 2 − 4ac . This quantity is, known as the discriminant of the, quadratic equation and denoted by D., There are following cases :, Case I If b 2 − 4ac = 0 i.e. D = 0, then, b, α =β = − ., 2a, 2, Thus, if b − 4ac = 0, then the quadratic, equation has real and equal roots and, each equal to −b / 2a., Case II If a, b and c are rational, numbers and b 2 − 4ac > 0 and it is a, and, , perfect square, then D = b 2 − 4ac is a, rational number and hence α and β are, rational and unequal., , Case III If b 2 − 4ac > 0 and it is not a, perfect square, then roots are irrational, and unequal., Case IV If b 2 − 4ac < 0, then the roots, are complex conjugate of each other., Based on above information, answer, the following questions, (i) Roots of quadratic equation, 21, 2x 2 − 2 3x +, = 0 are given by, 8, 3 3, ± i, 2, 4, 3, 3, (c) ±, i, 4, 2, , (a), , 3 3, ± i, 2, 4, 3, 3, (d) − ±, i, 4, 2, , (b) −, , (ii) Roots of quadratic equation, 25x 2 − 30x + 11 = 0 are given by, 2 3, ± i, 5, 5, 1, 3, (c) ±, i, 2, 2, , (a), , 3, 2, ±, i, 5, 5, 1, 3, (d) − ±, i, 2 2, , (b), , (iii) Roots of quadratic equation, 2x 2 + x + 1 = 0 are given by, −1 ± 7 i, 4, 3± 3i, (c), 4, (a), , 7±i, 4, 3± 7 i, (d), 4, (b), , (iv) Roots of quadratic equation, − x 2 + x − 2 = 0 are given by, 1± 7 i, 2, −1 ± 7 i, (c), −2, , (a), , 1 ± 5i, 2, 3 ± 2i, (d), 4, (b), , (v) Roots of quadratic equation, 20, 3x 2 − 4x +, = 0 are given by, 3, 2 4, ± i, 3 3, 3 5, (c) ± i, 4 4, , (a), , 4 2, ± i, 3 3, 3 5, (d) − ± i, 4 4, (b)

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43, , CBSE New Pattern ~ Mathematics XI (Term I), , ANSWERS, Multiple Choice Questions, 1. (b), 11. (b), , 2. (a), 12. (a), , 3. (b), 13. (a), , 4. (a), 14. (a), , 5. (a), 15. (a), , 6. (a), 16. (d), , 7. (a), 17. (b), , 8. (b), 18. (a), , 9. (a), 19. (a), , 10. (a), 20. (b), , 21. (c), 31. (a), , 22. (d), 32. (a), , 23. (b), 33. (a), , 24. (a), 34. (b), , 25. (b), 35. (a), , 26. (a), , 27. (d), , 28. (c), , 29. (a), , 30. (a), , 39. (a), 49. (a), , 40. (b), 50. (b), , 41. (a), , 42. (d), , 43. (a), , 44. (b), , 45. (a), , Assertion-Reasoning MCQs, 36. (c), 46. (d), , 37. (d), 47. (c), , 38. (d), 48. (c), , Case Based MCQs, 51. (i) - (a); (ii) - (b); (iii) - (a); (iv) - (d); (v) - (b), 53. (i) - (a); (ii) - (c); (iii) - (a); (iv) - (a); (v) - (b), 55. (i) - (a); (ii) - (b); (iii) - (a); (iv) - (c); (v) - (a), , 52. (i) - (b); (ii) - (a); (iii) - (a); (iv) - (c); (v) - (b), 54. (i) - (a); (ii) - (a); (iii) - (a); (iv) - (a); (v) - (c), , SOLUTIONS, 1. We have, 4 x + i ( 3x − y ) = 3 + i ( −6 ), , i2 −1 − 2, =, 2i, 2i, −1 − i − i, =, = 2 =, =i, i, −1, i, , …(i), , Equating the real and the imaginary parts of, Eq. (i), we get, 4 x = 3 , 3x − y = − 6, which on solving simultaneously, give, 3, 33, x = and y = ., 4, 4, , 2. (1 − i ) x + (1 + i ) y = 1 − 3 i, ( x + y ) + i( y − x ) = 1 − 3 i, Two complex numbers are equal, if their real, and imaginary parts are equal., ∴ x + y = 1 and y − x = − 3, By simplification x = 2 , y = − 1, Here, ( x , y ) is ( 2 , − 1 )., , =, , z 1 + z 2 = ( 2 + 3 i ) + ( 3 + 2i ), = ( 2 + 3) + i ( 3 + 2) = 5 + 5i, , 7. Here, z 1 = 2 + 3 i , z 2 = 3 − 2i , then, z 1 − z 2 = 2 + 3 i − ( 3 − 2i ), = 2 + 3 i − 3 + 2i = − 1 + 5i, 3, , 8. ( − i )( 3i )  − i  = ( − 3i 2 )  −, 1,  6 , , = (1) 25 . ( −i ) = − i = 0 − i, 0 − i = a + ib ⇒ a = 0, b = −1, ∴ a + b = 0 − 1 = −1, , 4. 1 + i 10 + i 20 + i 30 = 1 + ( i 4 ) 2 i 2 + ( i 4 ) 5 + ( i 4 )7 i 2, = 1 − 1 + 1 −1 = 0, i, − i 4x − 1, 5. Consider,, 2, 1, i−, i 4 x ⋅ i − i 4 x ⋅ i −1, i, =, =, 2, 2, , ⇒, , 1 3, i , 216 , 1, , , ( − i ), = ( − 3 × ( − 1))  −, ,  216, , , [Qi 2 = − 1 and i 3 = − i ], 1, =3×, ×i, 216, i, 1, =, =0+, i, 72, 72, a =0, , 9. Z 1 = 2 + 3i , Z 2 = 1 − 4i, [Qi 4 x = 1], , [Q i 2 = − 1], , 6. z 1 + z 2 = ( x1 + x 2 ) + i ( y1 + y 2 ),, , 3. i 103 = i 25 × 4 + 3 = ( i 4 ) 25 . i 3, , 4x + 1, , [Q i 2 = − 1], , ∴ Z 1Z 2 = ( 2 + 3i )(1 − 4i ), = 2 − 8i + 3i + 12, = 14 − 5i

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44, , CBSE New Pattern ~ Mathematics XI (Term I), , 10. We have, 3( 7 + 7i ) + i ( 7 + 7i ), , =, , = 21 + 21i + 7i − 7, = 14 + 28i = a + ib, ⇒ a = 14 and b = 28, b 28, ∴, =, =2, a 14, (1 + i ) 2 1 + i 2 + 2i, 11., =, 2−i, 2−i, 2i, 2 + i 4i − 2 2, =, ×, =, =, 2−i 2+ i, 4 +1 5, , [Q( z 1 + z 2 )( z 1 − z 2 ) = z 12 − z 22 ], =, , [Q ( z 1 − z 2 ) 2 = z 12 − 2z 1 z 2 + z 22 ], [Q i 2 = − 1], , = ( − 2 i )2 = 4 i 2 = − 4 = − 4 + 0 i, which is in the form of a + ib ., ∴ a = −4 and b = 0, , 13. We have, z 1 = 6 + 3 i and z 2 = 2 − i, ∴, , z1, 1, ( 6 + 3i )( 2 + i ), = (6 + 3 i ), =, z2, 2−i, ( 2 − i )( 2 + i ), 1, 2, = (6 + 3 i )  + i , 5, 5, (2 + i), = (6 + 3 i ), 5, 1, = ( 9 + 12i ), 5, , 14. The multiplicative inverse of complex, , 3 + 5i 4 − 3i, =, 4 − 3i 3 + 5i, Q multiplicative inverse of z = 1 , , z , 4 − 3i 3 − 5i, =, ×, 3 + 5i 3 − 5i, [Multiply numerator and denominator by the, conjugate of denominator i.e. ( 3 − 5i )], , quantities, , (12 − 15) + i ( −9 − 20 ), 9 + 25, −3 + i ( −29 ), 3 29i, =, =−, −, 34, 34 34, =, , 15. We have, z 1 = 3 + i 3 and z 2 = 3 + i ., ∴, , z1, =, z2, , 3(1 + i ), 3+i, , 3 −i, 3 −i, , [by rationalising the denominator], 3(1 + i )( 3 − i ), =, ( 3 )2 − (i )2, , 12. (1 − i ) 4 = ((1 − i ) 2 ) 2 = ((1) 2 − 2(1)( i ) + ( i ) 2 ) 2, = (1 − 2 i − 1) 2, , 3(1 + i ), ×, ( 3 + i), , 3( 3 − i + i 3 − i 2 ), 3 − i2, , 3( 3 + i ( 3 − 1) + 1), 3+1, [Q i 2 = − 1], 3, =, (( 3 + 1) + i ( 3 − 1)), 4, 3( 3 + 1) i 3( 3 − 1), =, +, 4, 4, which is represented by a point in first, quadrant., =, , 16. z 1 = 1 + 2i ; z 2 = 2 + 3 i, Additive inverse of z 2 = − 2 − 3 i, z 1 + ( − z 2 ) = 1 + 2i − 2 − 3 i, = −1 − i, , 17. Firstly, write each complex number in, standard form and then find its conjugate., 9 − i + 6 + i3 − 9 + i2, = (9 + i ) + 6 − i − 9 − 1, [Qi 3 = − i and i 2 = − 1], = (9 + i ) + (6 + i ) − 8, = 15 + 2 i − 8, = 7 + 2i, , 18. Given, Z 1 = 3 + 2i , Z 2 = 2 − i, Now, Z 1 + Z 2 = ( 3 + 2i ) + ( 2 − i ) = 5 + i, Z1 + Z 2 = 5 + i = 5 − i, , 19. Given, Z 1 = 1 + i ⇒ Z 1 = 1 − i, Z2 = 2 − i ⇒ Z2 = 2 + i, Now, Z 1 Z 2 = (1 − i )( 2 + i ), = 2 + i − 2i − i 2, = 2−i +1= 3−i, Now, 3 − i = a + ib, ⇒, a = 3, b = −1, ∴, a + b = 3 −1 = 2

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45, , CBSE New Pattern ~ Mathematics XI (Term I), , 2−i, 2−i, =, (1 − 2i ) 2 1 + 4i 2 − 4i, 2−i, 2−i, =, =, 1 − 4 − 4i −3 − 4i, , 23. Let z = − 5 i −15 − 6 i −8, , 20. Given that, z =, , =, , =, , ∴, , [Qi = − 1], 2, , − 9 + 19 i − 9 19, =, =, +, i, 13, 13 13, , z = − 6 + 5i, = 4 + 3(1)( −1)i, [Q i 4 = 1, i 2 = −1], , = 4 − 3i, ∴ Modulus = |4 + 3i 7 | = |4 − 3i |, = 4 2 + ( −3) 2 = 16 + 9 = 25 = 5, , …(i), ,  z 1   −9 19  − 9 19, + i =, − i,   =,  z 2   13 13  13 13, , 22. We have, z 1 = 2 − i and z 2 = − 2 + i, Now,, z 1 z 2 ( 2 − i ) ( −2 + i ) − ( 2 − i ) ( 2 − i ), =, =, z1, 2+i, (2 − i), (4 + i2 − 4i), (4 −1 − 4i), =−, 2+i, 2+i, (3 − 4i) 2 − i, =−, ×, 2+i, 2−i, =−, , [by rationalising the denominator], (6 − 3 i − 8 i + 4 i 2 ), =−, 4 − i2, , ⇒, , 6, −5, [Qi 4 = 1 and i 3 = −i ], −, (1) 3 ⋅ ( −i ) (1) 2, −5, 5, =, −6= −6, i, −i, 5 − 6i (5 − 6i)i, =, =, i, i ⋅i, [by rationalising the denominator], 5i − 6 i 2 5 i + 6, =, =, −1, i2, [Q i 2 = −1], = − 6 − 5i, , =, , 24. We have, 4 + 3i 7 = 4 + 3( i 4 )( i 2 )i, , [by rationalising the denominator], 6 + 9 i + 10 i + 15 i 2 6 + 19 i − 15, =, =, 4+9, 4 − 9i2, , ∴, , [Q i 15 = i 4 × 3 + 3 ], ,  (2−i) (3− 4i) , (2−i), =−, , −( 3 + 4i ), ( 3 + 4 i ) ( 3 − 4 i ) , ,  6 − 8 i − 3i + 4 i 2 , =−, , 9 + 16, , , ( −11 i + 2) −1, =−, =, ( 2 − 11 i ), 25, 25, 1, z =, ( −2 + 11 i ), ⇒, 25, 1, −2 11, ∴, z =, −, ( −2 − 11 i ) =, i, 25, 25 25, 21. Given, z 1 = 3 + 5i and z 2 = 2 − 3i, z1 3 + 5i 3 + 5i 2 + 3i, Now,, =, ×, =, z 2 2 − 3i 2 − 3i 2 + 3i, , 6, −5 6, −5, −, − =, i 15 i 8 ( i 4 ) 3 ⋅ i 3 ( i 4 ) 2, , [Q ( z 1 + z 2 )( z 1 − z 2 ) = z 12 − z 22 ], ( 6 − 11 i − 4 ), =−, [Q i 2 = −1], 5, 2 − 11 i −2 11, =−, =, + i = a + ib, 5, 5, 5, 2, a =−, 5, , 25. Given,, ∴, , Z 1 = 1 + 3i , Z 2 = 2 + 4i, , Z 2 − Z 1 = ( 2 + 4i ) − (1 + 3i ) = 1 + i, , ⇒ | Z 2 − Z 1 |2 = ( 12 + 12 ) 2, =1 + 1 = 2, , 26. Given that,, (1 + i ), , (2 + i), ( 2 + i + 2i + i 2 ), =, (3 + i), (3 + i), =, , 2 + 3i − 1, 3+ i, , =, , 1 + 3i, (1 + 3i )( 3 − i ), =, 3+i, ( 3 + i )( 3 − i ), , =, , 3 + 9i − i − 3i 2, 9 − i2, , =, , 3 + 8i + 3, 6 + 8i, =, 9+1, 10, , 62, 82, +, 100 100, 36 + 64, 100, =, =, =1, 100, 100, , =

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46, , CBSE New Pattern ~ Mathematics XI (Term I), , 27. Let z = (1 − i ) −2 + (1 + i ) −2, =, , 1, 1, (1 + i ) + (1 − i ), +, =, (1 − i ) 2 (1 + i ) 2, (1 − i ) 2 (1 + i ) 2, 2, , 1 + i 2 + 2i + 1 + i 2 − 2i, (1 − i 2 ) 2, 1 −1 + 1 −1 0, =, =, 4, (1 + 1) 2, , 2, , x=, , ⇒, , [Q i = −1], , −b ± ( b 2 − 4ac ), 2a, −1 ± 1 − 4, x=, 2, −1 ± i 3, x=, 2, −1 + i 3 −1 − i 3, ,, x=, 2, 2, , x=, , ⇒, , 28. We have, z 1 = 3 + 2i and z 2 = 2 − 4i, ⇒, , | z 1 + z 2 |2 + | z 1 − z 2 |2, On substituting the values of z 1 and z 2 , we get, |3 + 2i + 2 − 4i |2 + | 3 + 2i − 2 + 4i |, = |5 − 2i |2 + | 1 + 6i |2, 2, , 2, , 2, , [ if z = a + ib , then | z |2 = a 2 + b 2 ], ∴, , = 25 + 4 + 1 + 36, | z 1 + z 2 | 2 + | z 1 − z 2 | 2 = 66, , a = 2, b = 1, c = 2, ∴ D = b − 4ac = (1) 2 − 4 × 2 × 2, , 29. We have, 9 x + 16 = 0, , =1 − 4 × 2 =1 − 8 = − 7 < 0, − 1 ± −7, ⇒ x=, 2× 2, , 9 x 2 = − 16, 16, x2 = −, 9, , ⇒, , 16, x =± −, 9, [taking square root both sides],  16, , x =±, × −1,  9, , , ⇒, , ⇒, , 4 , x = ±  i, 3 , , ∴, , Hence, the roots are, , [Q −1 = i ], , 4, 4, i and − i., 3, 3, , 30. x 2 + 2 = 0, ⇒ x 2 = − 2 ⇒ x = ± −2 = ±, ∴ x = ± 2i, , On comparing the given equation with, ax 2 + bx + c = 0, we get, 2, , 2, , ⇒, , ∴, , 33. Given, 2 x 2 + x + 2 = 0, , = ( 5) + ( −2) + (1) + ( 6 ), 2, , 2i, , ⇒ x=, , a = 1, b = 3, c = 9, Now, D = b 2 − 4ac = ( 3) 2 − 4 × 1 × 9, , [Q − 1 = i ], , ⇒, , y 2 − 2y − 3 + 7 = 0, , ⇒, , y 2 − 2y + 4 = 0, , On comparing with ay 2 + by + c = 0, we get, a = 1, b = − 2 and c = 4, Q, , −b ± b 2 − 4ac, y=, 2a, , ∴, , y=, =, , 31. Given, x + 3x + 9 = 0, On comparing the given equation with, ax 2 + bx + c = 0, we get, , −1 ± i 7, 2 2, , 34. We have, ( y + 1) ( y − 3) + 7 = 0, , 2, , = 9 − 36 = − 27 < 0, , [Q − 1 = i ], , 32. x 2 + x + 1 = 0, Q, , = 0 = 0 + 0i, ∴ |z | = 0 + 0 = 0, , 27, , −3± i 9 × 3 − 3± i 3 3, =, =, 2, 2, , =, , 2, , − 3 ± − 27, −3±i, ,x =, 2 ×1, 2, , 2 ± ( −2) 2 − 4 × 1 × 4 2 ± 4 − 16, =, 2 ×1, 2, 2±, , − 12, 2, , =, , 2±2 3i, =1 ±, 2, , 3i, [Q − 1 = i ], , ∴, , y =1+ 3 i, , or y = 1 − 3 i, Hence, the roots of the given equation are, 1 + 3 i and 1 − i 3.

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47, , CBSE New Pattern ~ Mathematics XI (Term I), , 35. Let α , β be the roots of the equation, , 40. Assertion We have,, (1 + i ) 6 = ((1 + i ) 2 ) 3, , x 2 − px + 8 = 0., Therefore, α + β = p, and, α ⋅β =8, 2 = ± P 2 − 32, , Therefore,, , P 2 − 32 = 4, i.e. P = ± 6, , 36. Assertion We know that, i = −1, k, , Q, , i 4 k = ( i 4 ) = 1k = 1, , ⇒, , i 4k + 1 = i 4k ⋅ i = 1 × i = i, , ⇒, , i 4 k + 2 = i 4 k ⋅ i 2 = 1 × −1 = −1, , ⇒, , i 4k + 3 = i 4k ⋅ i 3 = 1 × − i = − i, , Reason i 4 k + i 4 k + 1 + i 4 k + 2 + i 4 k + 3, = i (1 + i + i + i ), 4k, , 2, , 3, , = i 4 k (1 + i − 1 − i ) = i 4 k 0 = 0, Hence, Assertion is true and Reason is false., , 37. Assertion, i −35 =, , [Q ( z 1 + z 2 ) 2 = z 12 + z 22 + 2z 1 z 2 ], , α − β = ± ( α + β ) 2 − 4αβ, , Now,, ⇒, , = (1 + i 2 + 2 i ) 3, , 1, 1, 1 i, i, =, =, × =, =i, i 35 ( i 2 )17 i −i i −i 2, , Reason Additive inverse of z is −z ., ∴ Additive inverse of (1 − i ) is, −(1 − i ) = − 1 + i, Hence, Assertion is false and Reason is true., , 38. Assertion We have,, 5 + 2i 5 + 2i 1 +, ×, =, 1 − 2i 1 − 2i 1 +, , 2i, 2i, , =, , 5 + 5 2i + 2i − 2, 1 − ( 2i ) 2, , =, , 3 + 6 2i 3 (1 + 2 2i ), =, 1+ 2, 3, , = 1 + 2 2i, Reason (1 + i ) 5 (1 − i ) 5 = (1 − i 2 ) 5, = 25 = 32, Hence Assertion is false and Reason is true., , 39. Assertion Given, Z 1 = 2 + 3i , Z 2 = 3 − 2i, ∴ Z 1 − Z 2 = ( 2 + 3i ) − ( 3 − 2i ), = ( 2 − 3) + i ( 3 − ( −2)) = − 1 + 5i, Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., , = (1 − 1 + 2 i ) 3, ⇒ (1 + i ) 6 = ( 2 i ) 3 = 8 i 3 = − 8 i, , [Q i 2 = − 1], [Q i 3 = − 1], , = a + ib, ∴, b = −8, Reason (1 − i ) 3 = 13 − i 3 − 3(1) 2 i + 3(1)( i ) 2, [Q ( z 1 − z 2 ) 3 = z 13 − 3z 12 z 2 + 3z 1 z 22 − z 23 ], = 1 − ( − i ) − 3i − 3, [Q i 3 = − i and i 2 = − 1], ⇒, , (1 − i ) 3 = − 2 − 2i, , = a + ib, a = − 2 and b = − 2, a −2, ∴, =, =1, b −2, Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion., ⇒, , 41. Assertion We have,, ⇒, , (1 + i )( x + iy ) = 2 − 5i, x + iy + ix + i 2 y = 2 − 5i, , ⇒, , x + i ( y + x ) − y = 2 − 5i, , [Q i 2 = −1], , ⇒ ( x − y ) + i ( x + y ) = 2 − 5i, On equating real and imaginary parts from, both sides, we get, …(i), x −y=2, and, …(ii), x + y = −5, On adding Eqs. (i) and (ii), we get, x −y + x + y =2−5, ⇒, 2x = − 3, −3, ⇒, x=, 2, −3, in Eq. (ii), we get, On substituting x =, 2, −3, + y = −5, 2, 3 −10 + 3 −7, ⇒, y = −5+ =, =, 2, 2, 2, −3, −7, and y =, ∴, x=, 2, 2, Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion.

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48, , CBSE New Pattern ~ Mathematics XI (Term I), , 42. Assertion Let z = 2 − 3i, , =, , Then, z = 2 + 3i and | z | = 2 + ( −3) = 13, 2, , 2, , 2, , Therefore, the multiplicative inverse of 2 − 3i, is, z, 2 + 3i, 2, 3, z −1 =, =, =, + i, 13, 13 13, | z |2, The above working can be reproduced in the, following manner also,, 1, 2 + 3i, z −1 =, =, 2 − 3i ( 2 − 3i ) ( 2 + 3i ), 2 + 3i, 2 + 3i, =, = 2, 13, 2 − ( 3i ) 2, 2, 3, + i, 13 13, Reason If Z = a + ib , then conjugate of Z, =, , i.e., z = a − ib, ∴, z = 3 + 4i, ⇒, z = 3 − 4i, Hence, Assertion is false and Reason is true., , 43. Assertion For any three complex numbers, z 1 , z 2 and z 3 , distributive law is, z 1 ( z 2 + z 3 ) = z 1 z 2 + z 1 z 3 and ( z 1 + z 2 ) z 3, = z 1z 3 + z 2z 3., ∴ ( 2 + 3i ) [( 3 + 2i ) + ( 2 + i )], = ( 2 + 3i ) ( 3 + 2i ) + ( 2 + 3 i ) ( 2 + i ), [ Q z 1 ( z 2 + z 3) = z 1 z 2 + z 1 z 3], = ( 6 − 6 ) + 13 i + ( 4 − 3) + 8 i, = 1 + 21i, Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., −3, −3, 44. Assertion 5i  i  = 5 × i 2,  5 , 5, = − 3( −1) = 3 = 3 + 0i, Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion., 1 + 2i, 45. Assertion Let z =, 1 − 3i, , ∴ z=, , 1 + 2i 1 + 3i 1 + 3i + 2i + 6i 2, ×, =, 1 − 3i 1 + 3i, 12 − ( 3i ) 2, [Q ( a + b ) ( a − b ) = a 2 − b 2 ], , =, , 1 + 5i + 6 ( − 1), 1 − 9i 2, , [Q i 2 = − 1], , ⇒, , 1 + 5i − 6 − 5 + 5i − 1 + i, =, =, 2, 1+9, 10, , z =−, , 1 1, + i, 2 2, 2, ,  1,  1, ∴ | z | = −  +  ,  2,  2, , 2, , [Q | a + ib | = a 2 + b 2 ], 1 1, 2, 1, 1, + =, =, =, 4 4, 4, 2, 2, Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., =, , 46. Assertion x + 4iy = ix + y + 3, ⇒, x =y+3, ⇒, 4y = x, From Eqs. (i) and (ii), we get, 4y = y + 3, ⇒, 3y = 3, ⇒, y =1, From Eq. (i), we get, x =1 + 3 = 4, Reason Let z = 3 + 7i, ∴, , …(i), …(ii), , 1, 1, 3 − 7i, =, ×, z 3 + 7i 3 − 7i, =, , 3 − 7i, 3, 7 $, =, −, i, 9+7, 16 16, , Hence, Assertion is false and Reason is true., , 47. Assertion i 9 + i 19 = i 9 (1 + i 10 ) = i 9 [1 + ( i 2 ) 5 ], = i 9 [1 + ( −1) 5 ] = i 9 (1 − 1) = 0 = 0 + 0i, Reason 1 + i 2 + i 4 + K i 20, =, , 1 [( i 2 )11 − 1] 1( −1 − 1), =, =1, −1 − 1, (i )2 − 1, , Hence, Assertion is true and Reason is false., , 48. Assertion x 2 + 1 = 0, x 2 = −1, Reason i −1097 =, , x = ± −1 ⇒ x = ± i, 1, , i 4 × 274 + 1, 1 1 i, i, i, = = × = 2 =, = −i, −1, i i i i, Hence, Assertion is true and Reason is false.

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49, , CBSE New Pattern ~ Mathematics XI (Term I), , 49. Assertion, 3x + 4 x + 2 = 0, a = 3 , b = 4 , c = 2, 2, , D = b 2 − 4ac, = 16 − 4( 3)( 2), = 16 − 24, = −8, ⇒, D<0, Qb 2 − 4ac < 0, so above equation has, imaginary roots., Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., , 50. Assertion Given, x 2 + 3x + 5 = 0, On comparing the given equation with, ax 2 + bx + c = 0, we get, a = 1, b = 3, c = 5, Now, D = b 2 − 4ac = ( 3) 2 − 4 × 1 × 5, ⇒, , = 9 − 20 = − 11 < 0, − 3 ± − 11, x=, 2 ×1, , − 3 ± i 11, 2, Reason Given, x 2 − x + 2 = 0, , ∴ x=, , [Q − 1 = i ], , On comparing the given equation with, ax 2 + bx + c = 0, we get, a = 1, b = − 1, c = 2, Now, D = b 2 − 4ac = ( − 1) 2 − 4 × 1 × 2, ⇒, , =1 − 8 = − 7 < 0, − ( − 1) ± −7, x=, 2 ×1, , 1±i 7, =, [Q − 1 = i ], 2, Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion., , 51. (i) We have, ( 3a − 6 ) + 2 ib = − 6b + ( 6 + a )i, On equating real and imaginary parts,, we get, …(i), 3a − 6 = − 6b, and, …(ii), 2b = 6 + a, Above equations can be rewritten as, …(iii), 3a + 6b = 6, and, …(iv), a − 2b = − 6, , On multiplying Eq. (iv) by 3 and then, adding with Eq. (iii), we get, 3a + 6b + 3a − 6b = 6 − 18, ⇒, 6a = − 12 ⇒ a = − 2, On substituting a = − 2 in Eq. (iv), we get, − 2 − 2b = − 6, −4, ⇒, − 2b = − 6 + 2 ⇒ b =, =2, −2, ∴ a = − 2 and b = 2., (ii) We have, ( 2a + 2b ) + i ( b − a ) = − 4i , which, can be rewritten as, ( 2a + 2b ) + i ( b − a ) = 0 − 4i, On equating real and imaginary parts, we, get, 2a + 2b = 0, ⇒, a +b =0, [Q 2 ≠ 0 ] …(i), and, …(ii), b −a = − 4, On adding Eqs. (i) and (ii), we get, a + b + b −a = 0− 4, ⇒, 2b = − 4 ⇒ b = − 2, On substituting b = − 2 in Eq. (i), we get, a − 2 = 0 ⇒a = 2, ∴, a = 2 and b = − 2, 1 − i , (iii) Given that, , , 1 + i , , 100, , = a + ib, , 100, ,  (1 − i ) (1 − i ) , ⇒ , ⋅, , (1 + i ) (1 − i ) , ⇒, , 1 + i 2 − 2i , , ,  1−i2 , , 100, ,  −2i , , ,  2 , , ⇒, , = a + ib, = a + ib, 100, , = a + ib, , ⇒, , ( i 4 ) 25 = a + ib, , ⇒, , 1 = a + ib, , [Q i 2 = −1], , [Q i 4 = 1], , On comparing real and imaginary parts, both sides, we get, a = 1 and b = 0, ∴, ( a , b ) = (1, 0 ), (iv) Given that,, ⇒, ⇒, , (1 + i ) 2, = x + iy, 2−i, , (1 + i 2 + 2i ), = x + iy, 2−i, 2i, = x + iy, 2−i

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50, , CBSE New Pattern ~ Mathematics XI (Term I), , ⇒, , 2i ( 2 + i ), = x + iy, (2−i ) (2+ i), , 4i + 2i 2, ⇒, = x + iy, 4 −i2, 4i − 2, = x + iy, ⇒, 4 +1, −2 4i, ⇒, +, = x + iy, 5, 5, On comparing both sides, we get, 2, 4, x = − and y =, 5, 5, −2 4 2, x+y=, + =, ∴, 5 5 5, (v) Given that, ( x + y ) + i ( x − y ) = 4 + 6 i, On comparing both sides, we get, x +y=4, and, x − y =6, Adding Eqs. (i) and (ii), we get, 2x = 10, ⇒, x =5, ∴ From Eq. (i), we get, 5+ y =4, ⇒, y = 4 − 5 = −1, ∴, xy = 5( − 1) = − 5, , Since, z is real ⇒ z = z, ⇒, ⇒, ⇒, (iv), , …(i), …(ii), , z 1 − i 1 − i (1 − i ) 2, =, =, ×, z 1 + i 1 − i 1 − i2, =, , 1 + i 2 − 2i, = −i, 1+1, , z = −i z, , ⇒, , z=, , 3 + 2i sin θ (1 + 2i sin θ ), ×, 1 − 2i sin θ (1 + 2i sin θ ), , =, , ( 3 + 2i sin θ) (1 + 2i sin θ), 1 + 4 sin 2 θ, , =, , ( 3 − 4 sin 2 θ) + i ( 8 sin θ), 1 + 4 sin 2 θ, , Since, z is pure imaginary., , 52. (i) Since, (1 + i ) z = (1 − i ) z, ⇒, , Im z = 0, 2x − 4 y, − =0, 17, 4, 8 x − 16 = 17 y ⇒ 8 x − 17 y = 16, , (ii) Q z 1 z 2 = z 1 z 2, ( 2 + i ) x − i (1 − i ) y + 2i, (iii) Let z =, +, 4i, 4+i, =, , 2x + ( x − 1)i y + ( 2 − y )i i, +, ×, 4+i, 4i, i, , =, , ( 2x + ( x − 1)i ) ( 4 − i ) − iy + ( 2 − y ), +, ( 4 + i )( 4 − i ), 4, , 8 x + x − 1 + i ( 4 x − 4 − 2x ), =, 16 + 1, ( 2 − y ) − iy, +, 4, 9 x − 1 + i ( 2x − 4 ) 2 − y − iy, =, +, 17, 4, , ⇒, , Re( z ) = 0, , ⇒, , 3 − 4 sin 2 θ, =0, 1 + 4 sin 2 θ, 3, 4, π, θ=, 3, , ⇒, , sin 2 θ =, , ⇒, , 3, 2, π, ,  since, 0 < θ ≤ , , 2, ⇒ sin θ = ±, , z1 − z 2, =1, z1 + z 2, , (v) We have,, ⇒, , | z1 − z 2 | = | z1 + z 2 |, , ⇒, , | z1 − z 2 | 2 = | z1 + z 2 | 2, , ⇒ ( z1 − z 2 ) ( z1 − z 2 ) = ( z1 + z 2 ) ( z1 + z 2 ), [Q | z | 2 = z z ], ⇒, 2 z 1z 2 = − 2 z 1z 2, z , z1, z, = − 1 = −  1, z2, z2,  z2, , ⇒, ⇒, , z1, is pure imaginary., z2, , 53. (i) We have, i 37 = ( i ) 36 + 1 = ( i ) 4 × 9 i, = ( i 4 ) 9 ⋅ i = (1) 9 ⋅ i = i, 1, (ii) We have, i −30 = 30, i, , [Q i 4 = 1], , Now, i 30 = ( i ) 4 × 7 + 2, = ( i 4 ×7 ) i 2 = ( i 4 )7 ( −1), = (1) ( − 1) = −1, 7, , ⇒, , i −30 =, , 1, = −1, ( −1), , [Qi 2 = −1], [Qi 4 = 1 ]

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51, , CBSE New Pattern ~ Mathematics XI (Term I), , (ii) Given that, ( z + 3) ( z + 3), , (iii) i 9 + i 19 = i 9 (1 + i 10 ) = i 9 [1 + ( i 2 ) 5 ], 9, , [taking i common], 5, , z = x + iy, , Let, , ⇒ ( z + 3)( z + 3) = ( x + iy + 3)( x + 3 − iy ), , = i [(1 + ( −1) ] = i (1 − 1) = 0, 9, , 9, , [Q i = −1], , = ( x + 3) 2 − ( iy ) 2 = ( x + 3) 2 + y 2, , 2, , = 0 + 0i, , 1, (iv) i 19 +  , i, , , = x + 3 + iy, , 25 2, , 2, , 1 ,  4 ×4 + 3, + 4×6+1,  = i, , , i, , 2, , , , 1 , 1 , = ( i 4 ) 4 ( i ) 3 + 4 6  = (1) 4 ( i ) 3 + 6 , (1) i , (i ) i , , , 2, , 1 , i, , = − i +  = − i + 2, , i , i , 2, , 2, , Now, f ( z ) =, , (v) i −39 =, , =, , 6 − 2i, 6 − 2i, =, 1 −1 − 4i2 − 4i 4 − 4i, , [Q i 3 = − i ], , =, , 3−i, ( 3 − i )( 2 + 2 i ), =, 2 − 2i ( 2 − 2 i )( 2 + 2 i ), , [Q i 2 = − 1], , =, , 6 − 2i + 6 i − 2i 2, 4 − 4i2, , =, , 6 + 4i + 2, 4+4, , 2, , [Q i 2 = − 1], , 8 + 4i, 1, =1 + i, 8, 2, 1, f (z ) =1 + i, 2, =, , 1, , i 39, Multiplying and dividing by i, we get, i, i, i, i, = 40 = 4 10 = 10 = = i, [Q i 4 = 1], 1, (1), i, (i ), =0+i, , 54. (i) We have, ( x − iy ) ( 3 + 5 i ) is the conjugate of, , − 6 − 24 i., ⇒ ( x − iy )( 3 + 5 i ) = − 6 + 24 i, [Q conjugate of − 6 − 24 i = − 6 + 24 i], ⇒ 3x − 3iy + 5 ix − 5i 2 y = − 6 + 24 i, ⇒, , 7−z, 7 − 1 − 2i, =, 1 − z 2 1 − (1 + 2i ) 2, , [Q i 4 = 1], , = ( − i − i ) 2 = ( −2i ) 2 = 4i 2, =−4, , 2, , = 1+ 4 = 5, , z, , 2, , , , i , i , = − i +,  = − i +, , i × i, − 1, , , , = z +3, , z = 1 + 2i, , (iii) Let, ⇒, , 2, , ( 3x + 5y ) + i ( 5x − 3y ) = − 6 + 24 i, [Q i 2 = −1] ...(i), , On equating real and imaginary parts both, sides of Eq. (i), we get, ...(ii), 3x + 5y = − 6, and, ...(iii), 5x − 3y = 24, On multiplying Eq. (i) by 3 and Eq. (ii) by, 5, then adding the result, we get, 9 x + 15y + 25x − 15y = −18 + 120, ⇒, 34 x = 102 ⇒ x = 3, On substituting x = 3 in Eq. (ii), we get, 9 + 5y = −6 ⇒ 5y = −15 ⇒ y = −3, Now, x + y = 3 + ( −3) = 0, , ∴ f (z ) = 1 +, , 1, =, 4, , 4 +1, 4, , z, 5, =, 2, 2, (iv) Given, z 1 = 1 − 3i and z 2 = − 2 + 4 i, ∴ z 1 + z 2 = (1 − 3 i ) + ( − 2 + 4 i ) = − 1 + i, =, , | z 1 + z 2 | = ( −1) 2 + (1) 2 = 1 + 1 = 2, (v) Given, z = 3 + 4 i, ∴, z = 3 − 4i, ⇒ z + z = (3+ 4i) + (3 − 4i) = 6, z +z 6, Now,, = =3, 2, 2, 21, 2, 55. (i) We have, 2x − 2 3 x +, =0, 8, , …(i), , On comparing Eq. (i) with ax 2 + bx + c = 0,, we get, 21, a = 2, b = − 2 3 and c =, 8, Q, , x=, , −b ± b 2 − 4ac, 2a

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52, , CBSE New Pattern ~ Mathematics XI (Term I), , − ( −2 3 ) ± ( −2 3 ) 2 − 4 × 2 ×, , ∴x =, , 2×2, , =, , 21, 8, , [Q − 1 = i ], , [Q − 1 = i ], , 3 3, 3 3, + i and, − i., 2, 4, 2, 4, , Hence, the roots are, , (ii) Given, 25x 2 − 30 x + 11 = 0, , …(i), , On comparing Eq. (i) with ax + bx + c = 0,, we get, a = 25, b = − 30 and c = 11, 2, , Q, , x=, , − b ± b 2 − 4ac, 2a, , ∴, , x=, , 30 ± ( −30 ) 2 − 4 × 25 × 11, 2 × 25, , ⇒, , x=, , 30 ± 900 − 1100, 50, , x=, , ⇒, , ⇒ x=, ⇒, , x=, , 30 ±, , On comparing the given equation with, ax 2 + bx + c = 0, we get, a = − 1, b = 1, c = − 2, Now, D = b 2 − 4ac = (1) 2 − 4( − 1) ( − 2), =1 − 4 ×1 × 2, =1 − 8 = − 7 < 0, x=, , − 1 ± −7 − 1 ± i 7, =, 2 × ( − 1), −2, [Q − 1 = i ], , (v) Given, 3x 2 − 4 x +, , 20, =0, 3, , On comparing the given equation with, ax 2 + bx + c = 0, we get, a = 3, b = − 4, c =, [Q −1 = i ], , ∴, , 20, 3, , D = b 2 − 4ac = ( − 4 ) 2 − 4 × 3 ×, = 16 − 80 = − 64 < 0, , 3, 2, ±, i, 5, 5, , Hence, the roots are, , (iv) Given, − x 2 + x − 2 = 0, , ∴, , − 200, 50, , 30 ± 10 i 2, 50, , =1 − 8 = − 7 < 0, −1 ± − 7 −1 ± i 7, x=, =, 2×2, 4, , ⇒, , 2 3 ± 12 − 21 2 3 ± −9, =, 4, 4, , 2 3 ± 3i, 3 3, =, =, ± i, 4, 2, 4, , D = b 2 − 4ac = (1) 2 − 4 × 2 × 1, , Now,, , 3, 2, 3, 2, +, i and −, i., 5, 5, 5, 5, , (iii) Given, 2x 2 + x + 1 = 0, On comparing the given equation with, ax + bx + c = 0, we get, , ∴, , x=, =, , 4 ± 8i, 2×3, , =, , 2 ( 2 ± 4i ), 2×3, , =, , 2 ± 4i 2 4, = ± i, 3, 3 3, , 2, , a = 2, b = 1, c = 1, , − ( − 4 ) ± − 64, 2×3, , 20, 3

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CBSE New Pattern ~ Mathematics XI (Term I), , 53, , 04, Sequence and Series, Quick Revision, Sequence, A sequence is a succession of numbers or terms, formed according to some rule. e.g. 4, 8, 12, …… ., A sequence is either finite or infinite according as, number of terms in it. A sequence whose range is a, subset of the set of real numbers R, is called a real, sequence., , Series, If a1 , a 2 , a 3 , K , an is a sequence, then the, expression a1 + a 2 + a 3 + a 4 + K + an is called, series., , Progressions, A sequence whose terms follow a certain pattern or, rule, is called progression. e.g. 2, 4 , 6, ..., 100, , Arithmetic Progression (AP), A sequence in which the difference of two, consecutive terms is constant, is called arithmetic, progression (AP). This difference is called the, common difference and denoted by d. Thus,, d = an +1 − an, nth Term of an AP, If a is the first term, d is common difference and l, is the last term of an AP, then, nth term is given by an = a + (n − 1) d ., nth term of an AP from the end is, a ′n = an − (n − 1) d or l − (n − 1) d, Properties of Arithmetic Progression, (i) If a sequence is an AP, then its n th term is a linear, expression in n i.e. its n th term is given by, An + B , where A and B are constants and A is, common difference., , (ii) If a constant is added or subtracted from each, term of an AP, then the resulting sequence is, an AP with same common difference., (iii) If each term of an AP is multiplied or divided, by a non-zero constant, then the resulting, sequence is also an AP., (iv) Any three terms of an AP can be taken as, (a − d ), a , (a + d ) and any four terms of an AP, can be taken as, (a − 3 d ), (a − d ), (a + d ), (a + 3 d )., , Sum of n Terms of an AP, If a is the first term, d is the common difference, and l is the last term, then, n, n, Sum of n terms, S n = [2a + (n − 1 ) d ] = (a + l ), 2, 2, If sum of n terms, S n of an AP is given, then, nth term, an = S n − S n −1 , where a1 = S 1, , Arithmetic Mean, (i) If a, A and b are in AP, then A =, , a +b, is called, 2, , the arithmetic mean of a and b., (ii) If a, A1 , A2 , A3 , K , An , b are in AP and, A1 , A2 , A3 , K , An are n arithmetic mean, between a and b, then, b −a, n (b − a ), and A n = a +, d =, n +1, n +1, (iii) Sum of n arithmetic mean between a and b is, a + b, n, .,  2 , a + b, i.e. A 1 + A2 + A 3 + K + A n = n , ,  2 

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54, , CBSE New Pattern ~ Mathematics XI (Term I), , Geometric Progression (GP), , Sum of n Terms of a GP, , A sequence a1 , a 2 , ..., an is called geometric, a, progression, if it follows the relation k +1 = r, ak, , If a and r are the first term and common ratio of a, GP respectively, then, Sum of n terms of a GP,, , (constant) for all k ∈ N the constant ratio is called, common ratio (r ) of the GP., nth Term of a GP, If a is the first term and r is the common ratio, then, the general term or nth term of GP is, an = ar n − 1 or l = ar n −1, where l is last term., nth term of a GP from the end is, l, an ′ = n − 1, r, where, l is last term, Properties of Geometric Progression, (i) If all the terms of GP are multiplied or, divided by same non-zero constant, then the, resulting sequence is a GP with the same, common ratio., (ii) The reciprocals of the terms of a given GP also, form a GP., (iii) If each term of a GP is raised to some power,, the resulting sequence also forms a GP., (iv) In a finite GP, the product of the terms, equidistant from the beginning and from the, end is always same and is equal to the product, of the first and the last terms., (v) The resulting sequence formed by taking the, product and division of the corresponding, terms of two GP’ s is also a GP., a, (vi) Any three terms can be taken in GP as , a, r, and ar and any four terms can be taken in GP, a a, as 3 , , ar and ar 3 ., r r, , , a, , Sn = , a, , , (1 − r n ), , if | r | < 1, 1−r, (r n − 1) if | r | > 1, ,, r −1, , Sum of an infinite GP,,  a , |r | < 1, , S ∞ = 1 − r,  ∞,, |r | ≥ 1, , , Geometric Mean (GM), (i) If a , G and b are in GP, then G is called the, geometric mean of a and b and is given by, G = ab, (ii) If a , G1 , G 2 , G 3 , K , Gn , b are in GP, then, G1 , G 2 , G 3 , K , Gn are n GM’s between a and b,, 1, , bn + 1, and, then common ratio, r =  , a , n, ,  b  n +1, Gn = a   ., a , , Relation between AM and GM, Let A and G be the AM and GM of two positive, real numbers a and b, respectively., ⇒, ⇒, , ( a − b )2, ≥0, 2, A ≥G, , A −G =

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CBSE New Pattern ~ Mathematics XI (Term I), , 55, , Objective Questions, Multiple Choice Questions, 1. The first five terms of the sequence,, , where a1 = 3, an = 3an −1 + 2 for all n > 1, are, (a) 3, 15, 40, 110, 330, (c) 3, 20, 45, 110, 330, , (b) 3, 11, 35, 107, 323, (d) 3, 11, 40, 107, 323, , 2. The first five terms of the sequence, n −1 n +1, , an = ( −1), (a), (b), (c), (d), , 5, , are, , 3. A man starts repaying a loan as first, instalment of ` 100. If he increases the, instalment by ` 5 every month, then the, amount he will pay in the 30th, instalment is, (b) ` 250, (d) ` 265, , 4. 40 is which term of the sequence 72, 70,, 68 …… ., (a) 18th, (c) 15th, , (b) 17th, (d) 10th, , 5. The number of terms in the AP, 20, 25, 30, …… 100 are, (a) 16, (c) 18, , (b) 17, (d) 19, , 6. In an AP, if mth term is n and the, nth term is m, where m ≠ n, then, pth term is, (a) m + n − p, (c) n − m + p, , (b) m − n + p, (d) m + n + p, , 7. The number of terms in the AP, 7, 13, 19, …… 1205 are, (a) 30, (c) 31, , 75°, 85°, 95° and 105°, 75°, 80°, 90° and 100°, 75°, 85°, 90° and 105°, 70°, 85°, 95° and 105°, , 9. 6th term from the end of the sequence, 9, 12, 15 ……… 20th term is, (a) 50, , (b) 52, , (c) 51, , (d) 55, , 10. If the sum and product of three, , 25, −125, 625, −3125, 15625, 25, 125, 625, 3125, 15625, 25, −125, 625, 3125, 15625, 25, −125, 625, −3125, −15625, , (a) ` 241, (c) ` 245, , (a), (b), (c), (d), , (b) 34, (d) 10, , 8. If the angles of any quadrilateral is in, AP and their common difference is 10,, then the angles are, , numbers of an AP is 24 and 440, respectively, then the common, difference of the AP is, (a) ±1, (c) ±2, , (b) ±3, (d) ±5, , 11. The sum of three consecutive terms of, an AP is 15, and their product is 105,, then the common difference is, (a) ±1, (c) ±3, , (b) ±2, (d) ±4, , 12. If a, b, c are in arithmetic progression,, then the value of, (a + 2b − c ) ( 2b + c − a ) (a + 2b + c ) is, (a) 16abc, (c) 8abc, , (b) 4abc, (d) 3abc, , 13. In an AP, if K th term is 5K + 1. Then,, the sum of first n terms is, n, (5n + 7), 2, n, (c) (n + 5), 2, , (a), , n, (n + 7), 2, n, (d) (7n + 5), 2, (b), , 14. The rth term of an AP sum of whose, first n terms is 2n + 3n 2 is given by, (a) 6 r + 1, (c) 6r, , (b) 6 r − 1, (d) 3r − 1, , 15. If t n denotes the nth term of the series, 2 + 3 + 6 + 11 + 18 + ..., then t 50 is, , [NCERT Exemplar], , (a) 492 − 1, (c) 502 + 1, , (b) 492, (d) 492 + 2

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56, , CBSE New Pattern ~ Mathematics XI (Term I), , 16. If in an AP, first term is 2 and the sum, of first five terms is one-fourth of the, next five terms, then the 20th terms is, (a) −140, (c) −112, , (b) −100, (d) −138, , 17. 6 arithmetic means between 3 and 24, are, (a), (b), (c), (d), , (b) 393206, (d) 303216, , 26. The sum of first three terms of a GP is, , 13, and their product is −1. Then, which, 12, of the following statements is incorrect?, − 3 −4, or, 4, 3, −3, 4, 3, (b) First three terms are , − 1, for r =, 4, 3, 4, −4, 3, 4, (c) First three terms are , − 1, for r =, 3, 4, 3, 3 4, (d) Common ratio is or, 4 3, (a) Common ratio is, , 6, 9, 12, 15, 18 and 21, 6, 9, 10, 15, 18 and 21, 6, 8, 10, 15, 18 and 21, 6, 9, 12, 13, 18 and 21, , 18. The nth term of a GP 5, 25, 125, ..... is, (b) 5 n − 1, (d) 5 n − 2, , (a) 5 n, (c) 5 n + 1, , 27. The sum of first three terms of a GP is, , 19. If the third term of GP is 4, then the, product of its first 5 terms is, [NCERT Exemplar], , (a) 4 3, (c) 4 5, , (b) 4 4, (d) None of these, , 20. In a GP, the 3rd term is 24 and the, 6th term is 192. Then, the 10th term is, (a) 1084, (c) 3072, , (b) 3290, (d) 2340, , in which the third term is greater than, the first term by 9 and the second term, is greater than 4th by 18 is, (b) −2, , (d) −1, , (c) 1, , 13, and their product is −1 then the, 12, common ratio of the GP is, −4 −3, or, 3, 4, 1, −1, (c) or, 4, 4, , 3 4, or, 4 3, 5 −3, (d) or, 3, 5, , (a), , (b), , 1 3 9, …… upto, 2 2 2, , 28. Sum of the sequence , ,, , 21. Common ratio of four numbers of a GP, , (a) 2, , (a) 393216, (c) 313216, , 10 terms is equal to, (a) 14762, (c) 41762, , (b) 14726, (d) 12476, , 29. A person has 2 parents, 4 grandparents,, , common ratio 2, then the 12th term is, , 8 great grandparents and so on. Then,, the number of ancestors during the ten, generations preceding his own is, , (a) 1640, (c) 3072, , (a) 1084, (c) 2250, , 22. If the 8th term of a GP is 192 with the, (b) 2084, (d) 3126, , 23. 5120 is which term of the GP, 5, 10, 20, 40 ……, (a) 11th, (c) 6th, , (b) 10th, (d) 5th, , 24. The 5th term from the end of the, sequence 16, 8, 4, 2, ……, (a) 1, (c) 3, , 1, is, 16, , (b) 2, (d) 4, , 25. 8th term from the end of the sequence, 3, 6, 12, …… 25th term is, , (b) 2046, (d) 1024, , 30. If n terms of GP 3, 3 2 , 3 3 , ..... are, needed to give the sum 120, then the, value of n is, (a) 2, (c) 4, , (b) 3, (d) 5, , 31. The sum of an infinite GP is, common ratio is, , −4, then its first term is, 5, , equal to, (a) 10, (c) 15, , 80, and its, 9, , (b) 14, (d) 16

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CBSE New Pattern ~ Mathematics XI (Term I), , 32. 31/ 2 × 31/4 × 31/ 8 × K upto infinite, terms is equal to, 2, , (b) 3, (d) 3 4, , (a) 3, (c) 33, , 33. The geometric mean of 2 and 8 is, (a) 4, (c) 7, , (b) 6, (d) 5, , 34. 4 gemoteric means between 3 and 96, are, (a) 6, 12, 24, 48, (c) 6, 10, 40, 48, , (b) 6, 10, 24, 48, (d) 48, 24, 10, 5, , 35. The minimum value of, 4 x + 4 1 − x, x ∈ R is, , (a) 2, , (b) 4, , [NCERT Exemplar], , (c) 1, , (d) 0, , Asserion-Reasoning MCQs, Directions (Q. Nos. 36-50) Each of these, questions contains two statements, Assertion (A) and Reason (R). Each of the, questions has four alternative choices, any, one of the which is the correct answer. You, have to select one of the codes (a), (b), (c) and, (d) given below., (a) A is true, R is true; R is a correct, explanation of A., (b) A is true, R is true; R is not a correct, explanation of A., (c) A is true; R is false, (d) A is false; R is true., , 36. If nth term of a sequence is, , 57, , 38. Assertion (A) The first three terms of, 3, 21, the sequence are , x,, whose nth, 2, 2, 2, n(n + 5), 9, term is an =, . Then x =, 4, 2, Reason (R) The third term of the, sequence whose nth term is, an = ( −1)n − 1 5n + 1 is 620., , 39. Assertion (A) If the nth term of a, , sequence is an = 4n − 3. Here, a17 and, a 24 are 65 and 93 respectively., Reason (R) If the nth term of a, sequence is an = ( −1)n − 1 n 3 . Here, 9th, term is 729., , 40. Assertion (A) If the sequence of even, natural number is 2, 4, 6, 8, …, then nth, term of the sequence is an given by, an = 2n, where n ∈ N ., Reason (R) If the sequence of odd, natural numbers is 1, 3, 5, 7, …, then nth, term of the sequence is given by, an = 2n − 1 , where n ∈ N ., , 41. Assertion (A) The fourth term of a GP, is the square of its second term and the, first term is −3, then its 7th term is equal, to 2187., Reason (R) Sum of first 10 terms of the, AP 6, 8, 10, ……… is equal to 150., , an = 2n 2 − n + 1., , 42. Assertion (A) If 5th and 8th term of a, , Assertion (A) First and second terms of, same sequence are 2 and 7 respectively., Reason (R) Third and fourth terms of, same sequence are 16 and 29,, respectively., , 43. Assertion (A) The sum of first 20, , 37. Assertion (A) If nth term of a sequence, n2, , 49, ., 128, 2, Reason (R) If nth term of a sequence is, n(n − 2), 323, ., an =, , then its 20th term is, 22, n+3, is an =, , n, , , then its 7th term is, , GP be 48 and 384 respectively, then, the common ratio of GP is 2., Reason (R) If 18, x, 14 are in AP, then, x = 16., terms of an AP, 4, 8, 12, … is equal to, 840., Reason (R) Sum of first n terms of an, n, AP is given by S n = [2a + (n − 1)d ],, 2, where a = first term and d = common, difference.

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58, , CBSE New Pattern ~ Mathematics XI (Term I), , 44. Assertion (A) The sum of the series, , 3, 4, +, + 5 + …… 25 terms is 75 5., 5, 5, Reason (R) If 27, x, 3 are in GP, then, x = ± 4., , 45. Assertion (A) The sum of first 23, terms of the AP, 16, 11, 6, ...... is − 897., Reason (R) The sum of first 22 terms of, the AP, x + y , x − y , x − 3y , ..... is 22 [x − 20 y ]., , 46. Assertion (A) If the numbers, , −2, −7, are in GP, then k = ±1., ,K,, 7, 2, Reason (R) If a1 , a 2 , a 3 are in GP, then, a2 a3, = ., a1 a 2, , 50. Assertion (A) The sum of first n terms, of the series, 5 + 55 + 555 + … is, , n, , 5 10(10 − 1), − n ., , 9, 9, , , Reason (R) General term of an AP is, Tn = a + (n − 1)d , where a = first term, and d = common difference., , Case Based MCQs, 51. A company produces 500 computers in, the third year and 600 computers in the, seventh year. Assuming that the, production increases uniformly by a, constant number every year., , 47. Assertion (A) The sum of first 6 terms, of the GP 4, 16, 64, … is equal to 5460., Reason (R) Sum of first n terms of the, a(r n − 1), , where, G. P is given by S n =, r −1, a = first term r = common ratio and, | r | > 1., , 48. Assertion (A) If the sum of first two, terms of an infinite GP is 5 and each, term is three times the sum of the, succeeding terms, then the common, 1, ratio is ., 4, Reason (R) In an AP 3, 6, 9, 12………, the 10th term is equal to 30., , 49. Assertion (A) The sum of first n terms, of the series 0.6 + 0.66 + 0.666 + ……. is, 2, 1   1  n , n − 1 −    ., 10  , 3 , 9, Reason (R) General term of a GP is, Tn = ar n − 1 , where a = first term and, r = common ratio., , Based on the above information, answer, the following questions., (i) The value of the fixed number by, which production is increasing every, year is, (a) 25, , (b) 20, , (c) 10, , (d) 30, , (ii) The production in first year is, (a) 400, (c) 450, , (b) 250, (d) 300, , (iii) The total production in 10 years is, (a) 5625, (c) 2655, , (b) 5265, (d) 6525, , (iv) The number of computers produced, in 21st year is, (a) 650, (c) 850, , (b) 700, (d) 950, , (v) The difference in number of, computers produced in 10th year, and 8th year is, (a) 25, , (b) 50, , (c) 100, , (d) 75

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CBSE New Pattern ~ Mathematics XI (Term I), , 52. A sequence whose terms increases or, decreases by a fixed number, is called, an Arithmetic Progression (AP)., In other words, we can say that, a, sequence is called an arithmetic, progression if the difference of a term, and the previous term is always same, i.e. an +1 − an = constant for all n., This constant or same difference is, called the common difference of an AP, and it is denoted by d., In an AP, we usually denote the first, term by a, common difference by d and, the nth term by an or Tn defined as, Tn = an = a + (n − 1) d, Also, l = a + (n − 1)d , where l is the last, term of the sequence., The sum of n terms, S n of this AP is, n, given by S n = [2a + (n − 1)d ]., 2, Also, if l be the last term, then the sum, n, of n terms of this AP is S n = (a + l )., 2, Based on the above information, answer, the following questions., (i) If nth term of an AP is given by, an = 2n 2 + 1, then its 10th term is, equal to, (a), (b), (c), (d), , 200, 301, 400, Given sequence is not an AP, , (ii) 11th term of an AP 11, 18, 25, … is, equal to, (a) 80, (c) 71, , (b) 81, (d) 70, , (iii) If the sum of n terms of an AP is, given by S n = 3n + 2n 2 , then the, common difference of the AP is, (a) 3, (c) 6, , (b) 2, (d) 4, , 59, (iv) If 9 times the 9th term of an AP is, equal to 13 times the 13th term, then, the 22nd term of the AP is, (a) 0, (c) 198, , (b) 22, (d) 220, , (v) Let S n denote the sum of the first n, terms of an AP, if S 2n = 3S n , then, S 3n : S n is equal to, (a) 4, (c) 8, , (b) 6, (d) 10, , 53. A sequence of non-zero numbers is said, to be a geometric progression, if the, ratio of each term, except the first one,, by its preceding term is always constant., Rahul being a plant lover decides to, open a nursery and he bought few, plants with pots. He wants to place pots, in such a way that number of pots in, first row is 2, in second row is 4 and in, third row is 8 and so on… ., , Based on the above information, answer, the following questions., (i) The constant multiple by which the, number of pots is increasing in, every row is, (a) 2, (c) 8, , (b) 4, (d) 1, , (ii) The number of pots in 8th row is, (a) 156, (c) 300, , (b) 256, (d) 456, , (iii) The difference in number of pots, placed in 7th row and 5th row is, (a) 86, (c) 90, , (b) 50, (d) 96

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60, , CBSE New Pattern ~ Mathematics XI (Term I), , (iv) Total number of pots upto 10th row, is, (a) 1046, (c) 1023, , (a) 150, (c) 250, , (b) 2046, (d) 1024, , (v) If Rahul wants to place 510 pots in, total, then the total number of rows, formed in this arrangement is, (a) 7, (c) 9, , (b) 8, (d) 5, , side 10 cm. Another student join the, mid-point of this square to form new, square. Again, the mid-points of the, sides of this new square are joined to, form another square by another, student. This process is continued, indefinitely., B3, , (b) 200, (d) None of these, , (v) The sum of the perimeter of all the, square formed is (in cm), (a) 80 + 40 2, (c) 40, , (b) 40 + 40 2, (d) None of these, , 55. Each side of an equilateral triangle is, , 54. A student of class XI draw a square of, , A4, , (iv) The sum of areas of all the square, formed is (in sq cm), , 24 cm. The mid-point of its sides are, joined to form another triangle. This, process is going continuously infinite., A, , A3, 12 cm, , C3, 10 cm, B4, , A1, , D3, , B, D2, , Based on above information, answer the, following questions., (i) The side of the 5th triangle is (in cm), , C1, , D1, B1, , (c), , A2, , 5, 2, (d) None of these, (b), , 5, , (ii) The area of the fifth square is, (in sq cm), (a), , 25, 2, , (c) 25, , (b) 50, (d), , 25, 4, , (iii) The perimeter of the 7th square is, (in cm), (a) 10, (c) 5, , (b) 20, 5, (d), 2, , C, , P, 24 cm, , B2, , Based on above information, answer the, following questions., (i) The side of fourth square is (in cm), (a) 5, , 12 cm, , C2, , D4, , C4, , 24 cm, , 12 cm, , 24 cm, , (a) 3, (c) 1.5, , (b) 6, (d) 0.75, , (ii) The sum of perimeter of first 6, triangle is (in cm), (a), , 569, 4, , (b), , 567, 4, , (c) 120, , (d) 144, , (iii) The area of all the triangle is, (in sq cm), (a) 576, (c) 144 3, , (b) 192 3, (d) 169 3, , (iv) The sum of perimeter of all triangle, is (in cm), (a) 144, (c) 400, , (b) 169, (d) 625, , (v) The perimeter of 7th triangle is, (in cm), (a), , 7, 8, , (b), , 9, 8, , (c), , 5, 8, , (d), , 3, 4

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CBSE New Pattern ~ Mathematics XI (Term I), , 61, , ANSWERS, Multiple Choice Questions, 1. (b), 11. (b), 21. (b), , 2. (a), 12. (a), 22. (c), , 3. (c), 13. (a), 23. (a), , 4. (b), 14. (b), 24. (a), , 5. (b), 15. (d), 25. (a), , 31. (d), , 32. (b), , 33. (a), , 34. (a), , 35. (b), , 39. (b), 49. (b), , 40. (b), 50. (b), , 6. (a), 16. (c), 26. (d), , 7. (b), 17. (a), 27. (a), , 8. (a), 18. (a), 28. (a), , 9. (c), 19. (c), 29. (b), , 10. (b), 20. (c), , 41. (d), , 42. (b), , 43. (a), , 44. (c), , 45. (b), , 30. (c), , Assertion-Reasoning MCQs, 36. (b), 46. (a), , 37. (c), 47. (a), , 38. (c), 48. (b), , Case Based MCQs, 51. (i) - (a); (ii) - (c); (iii) - (a); (iv) - (d); (v) - (b), 53. (i) - (a); (ii) - (b); (iii) - (d); (iv) - (b); (v) - (b), 55. (i) - (c); (ii) - (b); (iii) - (b); (iv) - (a); (v) - (b), , 52. (i) - (d); (ii) - (b); (iii) - (d); (iv) - (a); (v) - (b), 54. (i) - (d); (ii) - (d); (iii) - (c); (iv) - (b); (v) - (a), , SOLUTIONS, 1. Q a 1 = 3 and a n = 3a n − 1 + 2, Then,, , a2, a3, a4, a5, , = 3a 1, = 3a 2, = 3a 3, = 3a 4, , +, +, +, +, , 2 = 3 ( 3) + 2 = 11, 2 = 3 (11) + 2 = 35, 2 = 3 ( 35) + 2 = 107, 2 = 3 (107 ) + 2 = 323, , 2. We have, a n = ( − 1)n − 1 5n + 1, On putting n = 1, we get, a 1 = ( − 1)1 − 1 51 + 1 = ( − 1) 0 52 = 25, On putting n = 2, we get, a 2 = ( − 1) 2 − 1 52 + 1 = ( − 1)1 53 = − 125, On putting n = 3, we get, a 3 = ( − 1) 3 − 1 53 + 1 = ( − 1) 2 54 = 625, On putting n = 4, we get, a 4 = ( − 1) 4 − 1 54 + 1 = ( − 1) 3 55 = − 3125, On putting n = 5, we get, a 5 = ( − 1) 5 − 1 55 + 1 = ( − 1) 4 56 = 15625, Hence, the first five terms of the given, sequence are 25, − 125, 625, − 3125, 15625., , 3. Given, a = 100, d = 5, Q, ∴, , Tn = a + (n − 1) d, T 30 = 100 + ( 30 − 1) 5 = 100 + 29 × 5, = 100 + 145 = 245, , 4. Given seqeunce is 72, 70, 68, 66, …, Clearly, the successive difference of the terms, is same. So, the above sequence forms an AP, with first term, a = 72 and common difference,, d = 70 − 72 = –2., Let nth term, Tn = 40, and, Tn = a + (n − 1) d, ∴, 40 = 72 + (n − 1) ( − 2), ⇒, 40 = 72 − 2n + 2, ⇒, 2n = 72 − 40 + 2, ⇒, 2n = 34, ⇒, n = 17, Hence, 17th term of the given sequence is 40., , 5. Given AP is 20, 25, 30,..., 100., Here, a = 20, d = 25 − 20 = 5 and l = 100, Q Last term,, l = 100, ⇒ a + (n − 1) d = 100, [Q a = 20 and d = 5], ⇒ 20 + (n − 1) 5 = 100, ⇒, 20 + 5n − 5 = 100, ⇒, 15 + 5n = 100, ⇒, 5n = 100 − 15 = 85 ⇒ n = 17, Hence, there are 17 terms in given AP., , 6. We have,, and, , a m = a + (m − 1) d = n, a n = a + (n − 1) d = m, , …(i), …(ii)

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62, , CBSE New Pattern ~ Mathematics XI (Term I), , Solving Eqs. (i) and (ii), we get, (m − n ) d = n − m, ...(iii), ⇒, d = −1, Now,, ...(iv), a =n + m −1, Therefore,, a p = a + ( p − 1) d, = n + m − 1 + ( p − 1) ( − 1), [using Eqs. (iii) and (iv)], =n + m −1 − p + 1, =n + m − p, , 7. Given, AP is 7, 13, 19 …… 205., ∴ a = 7, d = 13 − 7 = 6, Let it has n terms, ∴ a n = 205, Now, a n = a + (n − 1)d, ∴, 205 = 7 + (n − 1) × 6, 198, = 33, ⇒ n −1 =, 6, ∴, n = 34, Hence, the given AP has 34 terms., , 8. Given, angles of a quadrilateral are in AP, , and common difference ( d ) = 10 °, Let angles of a quadrilateral are a , a + d ,, a + 2d and a + 3d , respectively., We know that, the sum of all interior angles of, a quadrilateral is 360 °., ∴, a + a + d + a + 2d + a + 3d = 360 °, ⇒ a + a + 10 ° + a + 20 ° + a + 30 ° = 360 °, [Q d = 10 °], ⇒, 4a + 60 ° = 360 °, ⇒, 4a = 360 ° − 60 ° = 300 °, 300 °, ⇒, a =, = 75°, 4, ∴, a + d = 75° + 10 ° = 85°, a + 2d = 75° + 20 ° = 95°, and, a + 3d = 75° + 30 ° = 105°, Hence, angles of an quadrilateral are 75°, 85°,, 95° and 105°., , 9. Given sequence is 9, 12, 15, ..., 20th term., Clearly, the successive difference of the terms, is same. So, the above sequence forms an AP, with first term, a = 9 and common difference,, d = 12 − 9 = 3., We know that, if a sequence has n term, then, mth term from end is equal to (n − m + 1)th, term from the beginning., Here, n = 20, m = 6, a = 9 and d = 3, , Now, 6th term from the end of sequence, = ( 20 − 6 + 1), i.e. 15th term from the beginning., ∴ T15 = 9 + (15 − 1) 3, [QTn = a + (n − 1) d ], = 9 + (14 ) 3, = 9 + 42 = 51, Hence, the 6th term from the end of the, sequence is 51., , 10. Let the three number be ( a − d ), a and ( a + d ), According to the question,, ( a − d ) + a + ( a + d ) = 24, ⇒, 3a = 24, ⇒, a =8, and, ( 8 − d )( 8 ) ( 8 + d ) = 440, ⇒, 64 − d 2 = 55, ⇒, d2 =9, ⇒, d =±3, , 11. Let three numbers in AP be, a − d , a and a + d ., According to the question,, Sum of three consecutive terms = 15, ∴, ( a − d ) + a + ( a + d ) = 15, ⇒, 3a = 15, ⇒, a =5, and the product of three consecutive terms, = 105, ∴, ( a − d ) ( a ) ( a + d ) = 105, [put a = 5], ⇒, ( 5 − d ) ( 5) ( 5 + d ) = 105, 2, ⇒, ( 25 − d ) 5 = 105, [Q ( A − B ) ( A + B ) = A 2 − B 2 ], ⇒, , 25 − d 2 = 21, , ⇒, ∴, , [dividing both sides by 5], d2 = 4, d =±2, , 12. Since, 2b = a + c, Now, ( a + 2b − c ) ( 2b + c − a ) ( a + 2b + c ), = ( a + a + c − c ) ( a + c + c − a ) ( 2b + 2b ), = 2a ⋅ 2c ⋅ 4b = 16 abc, , 13. Given, K th term (T K ) = 5K + 1, Putting K = 1, 2, we get, T1 = 5 × 1 + 1 = 6, and, T 2 = 5 × 2 + 1 = 11, ⇒, a = 6, d = 11 − 6 = 5

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CBSE New Pattern ~ Mathematics XI (Term I), , Now,, , n, [ 2a + (n − 1) d ], 2, n, = [ 2 × 6 + (n − 1) 5], 2, n, n, = [12 + 5n − 5] = [ 5n + 7 ], 2, 2, , Sn =, , 14. Given that, sum of n terms of an AP,, S n = 2n + 3n 2, Tn = S n − S n − 1, = ( 2n + 3n 2 ) − [ 2 (n − 1) + 3 (n − 1) 2 ], = ( 2n + 3n 2 ) − [ 2n − 2 + 3 (n 2 + 1 − 2n )], = ( 2n + 3n 2 ) − ( 2n − 2 + 3n 2 + 3 − 6n ), = 2n + 3n 2 − 2n + 2 − 3n 2 − 3 + 6 n, = 6n − 1, ∴ r th term T r = 6 r − 1, , 15. Let S n be sum of the series, , 2 + 3 + 6 + 11 + 18 + K + t 50 ., ∴, S n = 2 + 3 + 6 + 11 + 18 + K + t 50 ...(i), and, S n = 0 + 2 + 3 + 6 + 11 + 18 + K+ t 49 + t 50, ...(ii), On subtracting Eq. (ii) from Eq. (i), we get, 0 = 2 + 1 + 3 + 5 + 7 + L − t 50, ⇒t 50 = 2 + 1 + 3 + 5 + 7 + L upto 49 terms, ∴ t 50 = 2 + [1 + 3 + 5 + 7 + L upto 49 terms], 49, =2+, [ 2 × 1 + 48 × 2], 2, 49, =2+, × [ 2 + 96 ], 2, = 2 + [ 49 + 49 × 48 ], = 2 + 49 × 49, = 2 + ( 49 ) 2, , 16. Let the AP is a , a + d , a + 2d , a + 3 d , ..., Now, given a = 2, According to the given condition,, 1, Sum of first five terms = (Sum of next five, 4, terms), a + ( a + d ) + ( a + 2 d ) + ( a + 3 d ) + a + 4d, 1, = [a + 5 d + a + 6 d + a + 7d, 4, + a + 8d + a + 9d], 1, ⇒ 5a + 10d = [ 5a + 35 d ], 4, , 63, ⇒ 4 [ 5a + 10 d ] = 5 a + 35 d, ⇒ 20a + 40 d = 5 a + 35 d, ⇒, 20a − 5 a = 35 d − 40 d, ⇒, 15 a = − 5 d, ⇒, 15 × 2 = − 5 d, , [Qa = 2], − 30, ⇒, 30 = − 5 d ⇒ d =, = −6, 5, Now, Tn = a + (n − 1) d, ⇒ T 20 = 2 + ( 20 − 1) ( − 6 ) = 2 + 19 ( − 6 ), = 2 − 19 × 6 = 2 − 114 = − 112, , 17. Let A1 , A 2 , A 3 , A 4 , A 5 and A 6 be six, arithmetic means between 3 and 24., Then, 3, A1 , A 2 , A 3 , A 4 , A 5 , A 6 , 24 are in AP, and number of terms is 8., ∴ a = 3, and T 8 = 24, [QTn = a + (n − 1) d ], ⇒, a + ( 8 − 1) d = 24, ⇒, a + 7d = 24, ⇒, 3 + 7d = 24, ⇒, 7d = 21 ⇒ d = 3, Now, A1 = a + d = 3 + 3 = 6, A 2 = a + 2d = 3 + 2( 3) = 9, A 3 = a + 3d = 3 + 3(3) = 12, A 4 = a + 4d = 3 + 4(3) = 15, A 5 = a + 5d = 3 + 5 ( 3) = 18, and A 6 = a + 6d = 3 + 6(3) = 21, Hence, 6 arithmetic means between 3 and 24, are 6, 9, 12, 15, 18 and 21., , 18. Here, a = 5 and r = 5, Thus, a n = ar n − 1 = 5 × ( 5)n − 1 = 5n, , 19. It is given that, T 3 = 4, Let a and r the first term and common ratio,, respectively., Then,, ...(i), ar 2 = 4, Product of first 5 terms = a ⋅ ar ⋅ ar 2 ⋅ ar 3 ⋅ ar 4, = a 5 r 10 = ( ar 2 ) 5 = ( 4 ) 5, [using Eq. (i)], , 20. Here, a 3 = ar = 24, 2, , and, , a 6 = ar 5 = 192, , Dividing Eq. (ii) by Eq. (i), we get, 192, r3 =, ⇒ r3 = 8 ⇒ r = 2, 24, Substituting r = 2 in Eq. (i), we get, a =6, Hence, a 10 = 6( 2) 9 = 3072, , …(i), …(ii)

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64, , CBSE New Pattern ~ Mathematics XI (Term I), , 21. Let the GP is a , ar , ar 2 , ar 3 , …, , 24. Given sequence is 16, 8, 4, 2, …,, , Given, third term = first term + 9, ⇒, T 3 = a + 9 ⇒ ar 2 = a + 9, ⇒, , ar 2 − a = 9, , Again, second term = fourth term + 18, T 2 = T 4 + 18, ⇒, ar = ar 3 + 18, ⇒, , ar − ar = 18, , …(ii), , 22. Given, 8th term,T 8 = 192, , ⇒, , 25. We know that if a sequence has n terms, then, [QTn = ar, , n −1, , ], , a × ( 2) = 192 ⇒ a × 128 = 192, 192, 48 3, a =, =, ⇒ a =, ⇒, 128, 32 2, 3, Now,, T12 = ar 12 − 1 = × ( 2)11, 2, 3, = × 211 = 3 × 210, 2, = 3 × 1024 = 3072, 7, , 23. Given GP is 5, 10, 20, 40, …, 10, =2, 5, Let nth term of given GP = 5120, Here, a = 5 and r =, , Tn = 5120, , i.e., Now,, , Tn, n −1, , ⇒ 5( 2), , = ar, , n −1, , = 5120, , = 5120, 5120, =, = 1024, 5, = 1024, , ⇒, , 2n − 1, , ⇒, , 2n − 1, , ⇒, , 2n − 1 = 210, , 1 / 16, =, (1/ 2) 5 − 1 (1/ 2) 4, 1 / 16, =, =1, 1 / 16, [put the values of l , r and n], Hence, the 5th term from end in GP is 1., , ∴ 5th term from end =, , −1 (1 − r 2 ) 1, 1 1, = ⇒ − = ⇒ r =−2, r (1 − r 2 ) 2, r 2, , and common ratio ( r ) = 2, ⇒, ar 8 − 1 = 192, , l, rn − 1, 1 / 16, =, , On dividing Eq. (i) by Eq. (ii), we get, 9, ar 2 − a, a ( r 2 − 1) 1, =, ⇒, =, 3, 18, ar − ar, ar (1 − r 2 ) 2, ⇒, , 1, ,, 16, 8 1, Common ratio ( r ) =, =, 16 2, We know that, n th term from end in GP, Here, last term ( l ) =, , …(i), , 3, , 1, ., 16, , [Qa = 5, r = 2], , On equating the powers, we get, n − 1 = 10, ⇒, n = 10 + 1 = 11, Hence, 11th term of given GP is 5120., , mth term from end is equal to (n − m + 1)th, term from the beginning., 6, Here, a = 3, r = = 2, m = 8 and n = 25., 3, Now, 8th term from the end of sequence is, equal to the ( 25 − 8 + 1). i.e. 18th term from, , the beginning., Q, ∴, , Tn = ar n − 1, T18 = ar 18 −1 = 3( 2)18 − 1, = 3( 2), , 17, , [Q a = 3, r = 2], , = 3 × 131072 = 393216, , Hence, the 8th term from the end is 393216., a, 26. Let , a , ar be the three terms of the G.P. ., r, a, 13, …(i), Then,, + a + ar =, r, 12, a, and, …(ii), ⋅ a ⋅ ar = −1, r, From Eq. (ii), we get, a 3 = −1, ⇒, , a = −1, [considering only real roots], Substituting a = −1 in Eq. (i), we have, −1, 13, −1 − r =, r, 12, 13r, 2, ⇒ r + r +1+, =0, 12, ⇒ 12r 2 + 25r + 12 = 0

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CBSE New Pattern ~ Mathematics XI (Term I), , 29. Here, a = 2, r = 2 and n = 10, , On solving, we get, −4, −3, or, r=, 3, 4, , Since,, , 4, 3, Thus, the three terms of GP are , −1, for, 3, 4, −3, −4, 3, 4, and , −1, for r =, ., r=, 4, 3, 4, 3, , 27. Let the first three terms of the GP be, a, , a , ar ., r, Now, according to the question,, a, 13, + a + ar =, a, 12, a, and, . r . ar = −1, r, From Eq. (ii),, a 3 = − 1 ⇒ a = −1, , …(i), …(ii), , Put a = −1 in Eq. (i) , we get,  −1,  13, −1 − r  =, ,  r,  12, 13, 1, ⇒, + r +1+ = 0, 12, r, 25, 1, +r + =0, ⇒, r, 12, ⇒, 12r 2 + 25r + 12 = 0, ⇒, , 12r 2 + 16r + 9r + 12 = 0, , ⇒ 4r ( 3r + 4 ) + 3( 3r + 4 ) = 0, ⇒, ( 3r + 4 )( 4r + 3) = 0, −4, −3, or, ⇒, r=, 3, 4, 1 3 9, 28. We have, sequence , , ,..., 2 2 2, Clearly, the successive ratio of the terms is, same., So, the above sequence forms a GP, with first, 1, 3 1, term, a = and common ratio, r = ÷ = 3., 2, 2 2, 1 10, [ 3 − 1] , , a ( r n − 1), ∴ S 10 = 2, as r > 1, Q S n =, 3 −1, r −1, , , 1 ( 310 − 1), 2, 2, 1 10, = [ 3 − 1], 4, = 14762, =, , 65, , Sn =, , a ( r n − 1), r −1, , We have, S 10 =, , 2( 210 − 1), 2 −1, , = 2(1024 − 1), = 2 × 1023, = 2046, Hence, the number of ancestors preceding the, person is 2046., , 30. Here, a = 3, r = 3, S n = 120, Q, , Sn =, , a ( r n − 1), r −1, , ⇒, , 120 =, , 3( 3n − 1), 3 −1, , ⇒, ⇒, ⇒, ⇒, , [Qr > 1 ], , 3( 3n − 1), 2, 120 × 2 = 3 ( 3n − 1), 240, = 3n − 1, 3, 3n − 1 = 80, 120 =, , ⇒, , 3n = 80 + 1, , ⇒, , 3n = 81, , ⇒, , 3n = 34, , On comparing the power of 3 both sides,, we get, n=4, −4, 31. Common ratio i.e r =, 5, 4, 4, −, and, | r | =  = < 1,  5 5, 80, Given, Sum of this in finite GP is S =, 9, Now, Let a be the first term of the given, infinte GP., a, ∴, S =, 1 −r, 80, a, ⇒, =, 4, 9, , 1 + , , 5, 80 5a, =, ⇒, 9, 9, ⇒, a = 16

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66, , CBSE New Pattern ~ Mathematics XI (Term I), , 32. 31 / 2 × 31 / 4 × 31 / 8 K, 1, , = 32, , 1 1, + +K, 4 8, , +, , 1, 1, 1,  1 + +  , 2 , 2  2, , 2, , =3, , =, , 1, +  ,  2, , , , 1  1 , ×, , 1, 2 , 1− , , , 2, 3, 1, , = 32, , ×, , 2, 1, , 3, , , + K, , , 1, , Q a = 1 and r = 2 , , =3, , 33. The geometric mean of 2 and 8 is 16 i.e 4., 34. Let G1 , G 2 , G 3 and G 4 be the required GM’s., Then, 3, G1 , G 2 , G 3 , G 4 , 96 are in GP., Let r be the common ratio. Here, 96 is the 6th, term., ∴, 96 = ar 6 − 1 = 3r 5, ⇒, , 32 = r ⇒ ( 2) = r ⇒ r = 2, , ∴, , G1 = ar = 3 ⋅ 2 = 6, G 2 = ar 2 = 3 ⋅ 22 = 12, , 5, , 5, , 5, , G 3 = ar 3 = 3 ⋅ 23 = 24, and, , G 4 = ar 4 = 3 ⋅ 24 = 48, , 35. We know that,, ⇒, ⇒, ⇒, ⇒, , n2, 2n, 72, 49, Putting n = 7,, a7 = 7 =, 128, 2, n(n − 2), Reason We have, a n =, n+3, , 37. Assertion We have, a n =, , AM ≥ GM, , 4 x + 41 − x, ≥ 4 x ⋅ 41 − x, 2, 4 x + 41 − x ≥ 2 4, 4 x + 41 − x ≥ 2 ⋅ 2, 4 x + 41 − x ≥ 4, , 36. We have, a n = 2n 2 − n + 1, Assertion Putting n = 1, we get, a 1 = 2(1) 2 − 1 + 1 = 2 − 1 + 1 = 2, Putting n = 2, we get, a 2 = 2( 2) 2 − 2 + 1 = 8 − 2 + 1 = 7, Reason Putting n = 3, we get, a 3 = 2( 3) 2 − 3 + 1 = 18 − 3 + 1 = 16, Putting n = 4, we get, a 4 = 2( 4 ) 2 − 4 + 1 = 32 − 4 + 1 = 29, Hence, Assertion and Reason both are true, but Reason is not the correct explanation of, Assertion., , 20( 20 − 2), 20 + 3, 20 × 18, a 20 =, ⇒, 23, 360, =, 23, Hence, Assertion is true and Reason is false., n(n 2 + 5), 38. Assertion We have, a n =, 4, Putting n = 20,, , a 20 =, , Now, we need to find x which is second term, of the sequence, so put n = 2 in a n ., 2( 4 + 5) 18 9, =, =, 4, 4 2, Reason We have, a n = ( −1)n − 1 5n + 1, , ∴ a2 =, , ∴ a 3 = ( −1) 3 − 1 54 = 625, Hence, Assertion is true and Reason is false., , 39. Assertion a n = 4n − 3, Then,, a 17 = 4 (17 ) − 3 = 65, and, a 24 = 4 ( 24 ) − 3 = 93, Reason a n = ( − 1)n −1 ⋅ n 3, Then,, , a 9 = ( − 1) 9 − 1 ⋅ ( 9 ) 3, = ( − 1) 8 ⋅ 729, , = 729, Hence, Assertion and Reason both are true, but Reason is not the correct explanation of, Assertion., , 40. Assertion It is given that, a1 = 2 = 2 × 1, a2 = 4 = 2 × 2, a3 = 6 = 2 × 3, a4 = 8 = 2 × 4, M, Hence, the n th term of this sequence is a n = 2n,, where n ∈ N .

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CBSE New Pattern ~ Mathematics XI (Term I), , Reason It is given that, a1 = 1 = 2 − 1, a2 = 3 = 2 × 2 −1, a3 = 5 = 2 × 3 −1, a4 = 7 = 2 × 4 −1, M, Hence, the n th term of this sequence is, a n = 2n − 1, where n ∈ N ., Hence, Assertion and Reason both are true, but Reason is not the correct explanation of, Assertion., , 41. Assertion Let a be the first term and r be the, common ratio of the given GP., According to the question,, T 4 = (T 2 ) 2 and a = –3, Q, ∴, ⇒, , T 4 = (T 2 ) 2, ar 3 = ( ar ) 2, – 3r 3 = (–3) 2 r 2, , [Qa = –3], , ⇒, r = −3, Now, T7 = ar 6 = – 3(– 3) 6 = – 3 × 729 = – 2187, Reason Given AP is 6, 8, 10, …, ∴ a = 6, d = 8 − 6 = 2, 10, ∴ S 10 = [ 2 × 6 + (10 − 1) × 2], 2, = 512, [ + 18 ], = 5 × 30 = 150, Hence, Assertion is false and Reason is true., , 42. Assertion Let a be the first term and r be the, common ratio of the given GP., According to the question,, T 5 = 48 ⇒ ar 4 = 48, and, , T 8 = 384 ⇒ ar = 384, 7, , 67, 43. Assertion Given AP is 4, 8, 12, …, ∴ a = 4, d = 8 − 4 = 4, 20, Now, S 20 =, [ 2 × 4 + ( 20 − 1) × 4 ], 2, = 10[8 + 76 ], = 840, Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., , 44. Assertion, 3, 4, +, + 5 + … 25th terms, 5, 5, 3, 4, 5, ⇒ Sn =, +, +, + … 25th terms, 5, 5, 5, Clearly, the successive difference of the terms, is same. So, RHS of the above series forms an, 3, and common, AP, with first term, a =, 5, 4, 3, 1, ., difference, d =, –, =, 5, 5, 5, 25 , 3, 1 , 2×, + ( 25 – 1), ∴, S 25 =, 2 , 5, 5 , 12 ,  3, = 25, +, 5, 5 , , Let S n =, , 15, × 5 = 75 5, 5, Reason Given, 27,x , 3 are in GP., x, 3, ∴, =, 27 x, ⇒, x 2 = 81 ⇒ x = ± 9, = 25 ×, , Hence, Assertion is true and Reason is false., …(i), …(ii), , On dividing Eq. (ii) by Eq. (i), we get, ar 7 384, =, 48, ar 4, ⇒, r3 = 8, ⇒, r =2, Reason 18, x, 14 are in AP., ⇒, x − 18 = 14 − x, ⇒, 2x = 32, ⇒, x = 16, Hence, Assertion and Reason both are true, but Reason is not the correct explanation of, Assertion., , 45. Assertion Given AP is 16, 11, 6, …, Here, a = 16, d = 11 − 16 = − 5, n, S n = [ 2a + (n − 1)d ], Q, 2, 23, ∴ S 23 =, [ 2 × 16 + ( 23 − 1)( −5)], 2, 23, 23, =, [ 32 + ( 22)( −5)] =, [ 32 − 110 ], 2, 2, 23, =, [ −78 ] = − 897, 2, Reason Given AP is x + y, x − y, x − 3y, …, Here, a = x + y, d = ( x − y ) − ( x + y ) = − 2y, n, Q, S n = [ 2a + (n − 1) d ], 2

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68, , CBSE New Pattern ~ Mathematics XI (Term I), , 22, [ 2 × ( x + y ) + ( 22 − 1)( −2y )], 2, = 11 [ 2x + 2y + ( 21)( −2y )], = 11 [ 2x + 2y − 42y ], = 11 [ 2x − 40 y ], = 22[ x − 20 y ], Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion., 2, 7, 46. Assertion If − , k , − are in GP., 7, 2, a2 a3, Then,, =, a1 a 2, ∴, , ∴, ⇒, ⇒, ⇒, , S 22 =, , , , a2 a3 a4, Q common ratio ( r ) = a = a = a = ..., , , 1, 2, 3, k, −7 / 2, =, −2 / 7, k, 7, −7 1, k =, ×, −2, 2 k, 7k × 2k = −7 × ( −2), 14k 2 = 14, , ⇒, , k 2 =1 ⇒ k = ± 1, , Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., , 47. Assertion Given GP 4, 16, 64, …, 16, = 4 >1, 4, 4(( 4 ) 6 − 1) 4( 4095), ∴S6 =, =, = 5460, 4 −1, 3, ∴ a = 4, r =, , Reason Given, 3, 6, 9, 12 …, Here, a = 3, d = 6 − 3 = 3, ∴ T10 = a + (10 − 1)d, = 3+ 9×3, = 3 + 27 = 30, Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion., , 49. Assertion Let, S = 0.6 + 0.66 + 0.666 + K upto n terms, S = 6 ( 01, . + 011, . + 0111, ., + K upto n retms), 6, = ( 0.9 + 0.99 + 0.999 + K upto n terms), 9, =, , 29, 99, 999, , +, +, + K upto n terms , 3 10 100 1000, , , =, , 2 , 1 , 1  , 1 , ,  + 1 −, 1 −  + 1 −, 3  10   100   1000 , , , + K upto n terms , , 2, = (1 + 1 + 1 + K upto n terms ), 3, 1, 1, , 1, − +, +, + K upto n terms , , 10 100 1000, , 1, , 2  10, = n −, 3, , , , , , , a (1 − r n ), , r < 1, Q sum of GP =, r, 1, −, , , , Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., , 48. Assertion Let a be the first term and r (| r | < 1), be the common ratio of the GP., ∴ The GP is a , ar , ar 2 , …, According to the question,, T1 + T 2 = 5 ⇒ a + ar = 5 ⇒ a (1 + r ) = 5, and, Tn = 3(Tn + 1 + Tn + 2 + Tn + 3 + ... ), ⇒, ⇒, ⇒, ⇒, ⇒, , n +1, , ar, , = 3( ar + ar, , ar, , n –1, , = 3ar (1 + r + r + ... ), , n, , n, ,  1 , 1 = 3r , , 1 − r , 1 – r = 3r, 1, r=, 4, , + ar, , n+ 2, , n –1, , 2, , + ... ), ,   1  n , 1 −    , 10  , , , 1, 1−, , 10, , , , , 1, , 10, 2, = n −, 3, , , , ,   1  n , 1 −    , 10   2  1, ,  = 3 n − 9, 9, , , 10, , , ,   1  n , 1 −   , 10 , , , Hence, Assertion and Reason both are true, but Reason is not the correct explanation of, Assertion., , 50. Assertion, Let S = 5 + 55 + 555 + K upto n terms, = 5 (1 + 11 + 111 + K upton terms), 5, = ( 9 + 99 + 999 + K upton terms), 9

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CBSE New Pattern ~ Mathematics XI (Term I), , =, , 5, [(10 − 1) + (100 − 1) + (1000 − 1), 9, + K upto n terms ], , 5, [(10 + 100 + 1000 + K upto n terms, 9, − (1 + 1 + 1 + K upto n terms )], , 5 10 (10n − 1), = , − n, 9  10 − 1, , , , a ( r n − 1), , r > 1, Q sum of GP =, r −1, , , and Σ1 = n, , =, , , 5 10 (10n − 1), − n, , 9, 9, , Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion., =, , 51. (i) Since, it is given that, production increases, , (ii), (iii), , (iv), , (v), , uniformly by a constant number, hence, number of production every year forms an, AP., …(i), ∴ a 3 = 500 ⇒ a + 2d = 500, …(ii), a 7 = 600 ⇒ a + 6d = 600, Now, subtracting Eq. (i) from Eq. (ii), we get, 4d = 100 ⇒ d = 25, Put d = 25 in Eq. (i), we get, a + 50 = 500 ⇒ a = 450, The total production in 10 years = S 10, 10, ∴ S 10 = [ 2 × 450 + 9 × 25], 2, = 5 [900 + 225] = 5625, The number of computers produced in 21st, year = a 21, ∴ a 21 = 450 + 20 × 25 = 450 + 500 = 950, a 10 − a 8 = ( a + 9d ) − ( a + 7d ), = 2d = 2 × 25 = 50, , 52. (i) We have, a n = 2n 2 + 1, On replacing n by n + 1, we get, a n+ 1 = 2 (n + 1) 2 + 1 = 2 (n 2 + 1 + 2n ) + 1, = 2n 2 + 2 + 4n + 1 = 2n 2 + 4n + 3, Here, a n + 1 − a n = ( 2n 2 + 4n + 3) − ( 2n 2 + 1), = 2n 2 + 4n + 3 − 2n 2 − 1, = 4n + 2, It is not independent of n. So, given sequence, is not an AP., , 69, (ii) Given, AP 11, 18, 25, …, Here, a = 11, d = 18 − 11 = 7, ∴ a 11 = 11 + 10 × 7 = 11 + 70 = 81, (iii) Given, S n = 3n + 2n 2, First term of an AP,, ∴, , T1 = 3 × 1 + 2 (1) 2, , =3+ 2=5, and T 2 = S 2 − S 1, = [ 3 × 2 + 2 × ( 2) 2 ] − [ 3 × 1 + 2 × (1) 2 ], = [6 + 8 ] − [ 3 + 2], = 14 − 5 = 9, ∴ Common difference ( d ) = T 2 − T1, =9−5= 4, (iv) According to the question,, 9 ⋅ T 9 = 13 ⋅ T13, ⇒ 9 ( a + 8 d ) = 13 ( a + 12 d ), ⇒, , 9 a + 72d = 13 a + 156 d, , ⇒ ( 9 a − 13 a ) = 156 d − 72 d, ⇒, ⇒, ⇒, , − 4 a = 84 d, a = − 21 d, a + 21 d = 0, , ...(i), , ∴ 22nd term i.e. T 22 = [a + 21 d ], T 22 = 0, , [using Eq. (i)], , (v) Let first term be a and common difference, be d., n, ...(i), Then, S n = [ 2a + (n − 1 ) d ], 2, 2n, ∴, S 2n =, [ 2a + ( 2n − 1) d ], 2, …(ii), S 2n = n [ 2a + ( 2n − 1) d ], 3n, ...(iii), S 3n =, [ 2a + ( 3n − 1) d ], 2, According to the question,, S 2n = 3 S n, n, ⇒ n [ 2a + ( 2n − 1) d ] = 3 [ 2a + (n − 1) d ], 2, ⇒ 4 a + ( 4n − 2) d = 6 a + ( 3n − 3) d, ⇒, , − 2a + ( 4n − 2 − 3n + 3) d = 0, , ⇒, , − 2a + (n + 1) d = 0, , ⇒, , d =, , 2a, n +1, , ...(iv)

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70, , CBSE New Pattern ~ Mathematics XI (Term I), , 3n, [ 2a + ( 3n − 1) d ], S 3n, Now,, = 2, n, Sn, [ 2a + (n − 1) d ], 2, 2a, 6 a + ( 9n − 3), n +1, =, 2a, 2a + (n − 1), n +1, =, ⇒, , Then,, , = 52 + 52 = 25 + 25 = 5 2, ∴, , 53. (i) The number of pots in each row is, 2, 4, 8, …. ., ∴ This forms a geometric progression,, 4, where a = 2, r = = 2, 2, Hence, the constant multiple by which the, number of pots is increasing in every row, is 2., (ii) Number of pots in 8th row = a 8, a 8 = ar 8 −1 = 2 ( 2)7 = 28 = 256, , (iii) Number of pots in 7th row,, a 7 = 2 ( 2), , = 2 ⋅ 2 = 2 = 128, , a 5 = 2 ( 2) 5 −1 = 2 ⋅ 24 = 25 = 32, , a ( r 10 − 1) 2 ( 210 − 1), =, r −1, 2 −1, 2 (1024 − 1), =, = 2046, 1, (v) Let there be n number of rows., , ⇒, , 255 = 2n − 1, , 3, ,  1 , (i) Side of fourth square = ar 3 = 10  ,  2, 10, 5, cm, =, =, 2 2, 2, 4, , 6, , ∴ S 10 =, , 510, = 2n − 1, 2, , 25 25, +, = 25 = 5 cm, 2, 2, 5, Similarly, the side of fourth square is, cm., 2, ∴ Sides of each square are, 5 5 5 5, , ,, , , ……… respectively, 10, 5 2, 5,, 2 2 2 2 4, 1, ., which form a GP with a = 10 and r =, 2, =, , 2, , = S 10, , ⇒, , 2, , 25 2,  5, ∴ Area of fifth square =   =, cm,  2, 4, , ∴ Required answer = 128 − 32 = 96, (iv) Total number of pots upto 10th row, , S n = 510 =, ,  5 2,  5 2, = , ,  +,  2 ,  2 , ,  1 , (ii) Side of fifth square = ar 4 = 10  ,  2, 10 5, =, =, 4 2, , 7, , Number of pots in 5th row,, , ∴, , 5 2, 2, 2, , S 3n, =6, Sn, , 6, , C1B 2 = B 2C 2 =, , Similarly, C1C 2 = B 2C 22 + B 2C12, , 6 an + 6 a + 18 an − 6 a 24 an, =, 2an + 2a + 2an − 2a, 4 an, , 7 −1, , B1B 2 = A 2B12 + A 2B 22, , 2 ( 2n − 1), 2 −1, , ⇒, 256 = 2n ⇒ 28 = 2n ⇒ n = 8, 54. Let A1 , A 2 , A 3 , A 4 be the vertices of the first, square with each side equal to 10 cm. Let, B1 , B 2 , B 3 , B 4 be the mid-point of its side., ,  1 , (iii) The side of 7th square = ar 6 = 10  ,  2, 10 5, =, =, 8, 4, 5, ∴ Perimeter of 7th square = × 4 = 5 cm, 4, (iv) Sum of areas of all square formed is, 2, ,  5, 10 2 + ( 5 2 ) 2 + ( 5) 2 +   ....,  2, 25, = 100 + 50 + 25 +, + ……, 2, 50 1, Here, a = 100, r =, = , which is an, 100 2, infinite GP., a, 100, =, =, = 200 cm 2, 1−r 1 − 1, 2

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CBSE New Pattern ~ Mathematics XI (Term I), , 6, ,  1 , 721 −   ,  2  72 × 63 × 2, , =, =, 1, 26, 1−, 2, 567, cm, =, 4, 3, (iii) Area of first triangle is, ( 24 ) 2, 4, 2, 3  24 , Area of second triangle =,  , 4  2, , (v) Sum of perimeter of all square is, 5, , , K, 4 10 + 5 2 + 5 +, , 2 , =4×, , =, , 40 2, 2 +1, 10, =, ×, 1, 2, 1, 2 +1, −, 1−, 2, , 40 2( 2 + 1), = 80 + 40 2, ( 2 − 1), , 55. (i) Side of first triangle is 24., Side of second triangle is, , 24, = 12, 2, , Similarly, side of second triangle is, , =, 12, =6, 2, , 12 1, =, 24 2, ∴ Side of the fifth triangle,, ∴, , a = 24, r =, , 4, , 36, 2, = 18, , Similarly, perimeter of third triangle =, 36 1, =, 72 2, ∴ Sum of perimeter of first 6 triangle, a = 72, r =, , =S6 =, , a (1 − r 6 ), 1−r, , 3, 1, ( 24 ) 2 ×, 4, 4, , 3, 1, ( 24 ) 2 , r =, 4, 4, ∴ Sum of area of all triangles, , ∴, , a =, , a, 3 ( 24 ) 2, =, 1−r, 4 1− 1, 4, 3 × ( 24 ) 2, = 192 3 cm 2, =, 3, =, ,  1, a 5 = ar 4 = 24 ×  ,  2, 24 3, =, = = 1.5 cm, 16 2, (ii) Perimeter of first triangle = 24 × 3 = 72, 72, Perimeter of second triangle =, = 36, 2, , ∴, , 71, , (iv) The sum of perimeter of all triangle, 3( 24 + 12 + 6 + K ) is, , ,  24 , 1, , 3,  = 144 cm, Q a = 24, r = 2 , 1, , 1 − , , 2, 1, (v) Here, a = 72, r =, 2, 6, 1,  , a 7 = ( 72)  ,  2, 72 9, =, = cm, 64 8

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72, , CBSE New Pattern ~ Mathematics XI (Term I), , 05, Straight Lines, Quick Revision, Distance Formula, , ●, , If the points ( x 1 , y 1 ), ( x 2 , y 2 ) and ( x 3 , y 3 ) are, collinear, then, x 1 ( y 2 − y 3 ) + x 2 ( y 3 − y 1 ) + x 3 ( y 1 − y 2 ) = 0., , ●, , The coordinates of centroid of the triangle whose, vertices are ( x 1, y 1 ), ( x 2, y 2 ) and ( x 3, y 3 ), is given,  x + x 2 + x 3 y1 + y 2 + y 3 , by  1, ,, ., , , 3, 3, , The distance between two points A ( x 1 , y 1 ) and, B ( x 2 , y 2 ) is given by, AB = ( x 2 − x 1 )2 + ( y 2 − y 1 )2 ., The distance of a point A ( x , y ) from the origin, O (0, 0 ) is given by OA = x 2 + y 2 ., Three points A, B and C are collinear i.e. in same, straight line, if AB + BC = AC or AC + CB = AB or, BA + AC = BC ., , Section Formulae, ●, , The coordinates of the point which divides the, joining of ( x 1 , y 1 ) and ( x 2 , y 2 ) in the ratio m : n,  mx + nx 1 my 2 + ny 1 , , internally, is  2, ,, m +n ,  m +n,  mx − nx 1 my 2 − ny 1 , ., and externally is  2, ,, m −n ,  m −n, , ●, , The coordinates of the mid-point of the joining,  x + x 2 y1 + y 2 , of ( x 1 , y 1 ) and ( x 2 , y 2 ) is  1, ,, .,  2, 2 , , Area of a Triangle, ●, , If A ( x 1 , y 1 ), B ( x 2 , y 2 ) and C ( x 3 , y 3 ) are the, vertices of a ∆ABC , then area of ∆ABC ., 1, =, [x ( y − y 3 ) + x 2 ( y 3 − y 1 ) + x 3 ( y 1 − y 2 )], 2 1 2, 1, = |( x 1 y 2 + x 2 y 3 + x 3 y 1 ) − ( x 1 y 3 + x 2 y 1 + x 3 y 2 )|, 2, , Locus of a Point, The curve described by a moving point under, given geometrical conditions is called the locus of, that point., , Slope or Gradient of a Line, If θ is the angle of inclination of a line l , then tan θ, is called the slope or gradient of the line l and it is, denoted by m., i.e., m = tan θ, The slope of X -axis is zero and slope of Y -axis is, not defined., , Slope of a Line Joining Two Points, The slope of a line passing through points P ( x 1 , y 1 ), y − y1, and Q ( x 2 , y 2 ) is given by m = tan θ = 2, x 2 − x1, , Angle between Two Lines, The angle θ between two lines having slopes m1 and, m 2 is, tan θ =, , m 2 − m1, 1 + m1 m 2

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73, , CBSE New Pattern ~ Mathematics XI (Term I), , (i) If two lines are parallel, then their slopes are equal, i.e. m1 = m 2 ., (ii) If two lines are perpendicular to each other,, then product of their slopes is − 1, i.e., m1m 2 = − 1., Note, (i) If tanθ is positive, then θ will be an acute angle., (ii) If tanθ is negative, then θ will be an obtuse angle., , Different forms of Ax + By + C = 0 are, −A, C, x − ,B ≠ 0, B, B, y, x, (ii) Intercept form, +, = 1, C ≠ 0, − C /A − C /B, (i) Slope-intercept form y =, , (iii) Normal form x cos α + y sin α = p, A, , where, cos α = ±, , A + B2, 2, , Various Forms of the Equation of a Line, (i) If a line is at a distance a and parallel to, X -axis, then the equation of the line is y = ± a ., (ii) If a line is parallel to Y -axis at a distance b, from Y -axis, then its equation is x = ± b ., (iii) Point-slope form The equation of a line which, passes through the point ( x 1 , y 1 ) and has the, slope m is given by y − y 1 = m ( x − x 1 )., , sin α = ±, , B, A +B, 2, , 2, , and p = ±, , C, A + B2, 2, , Note Proper choice of signs to be made so that p, should be always positive., , Angle Between Two Lines, having, General Equations, , (iv) Two points form The equation of a line, passing through the points ( x 1 , y 1 ) and ( x 2 , y 2 ),  y − y1 ,  ( x − x 1 )., is given by y − y 1 =  2,  x 2 − x1 , , Let general equations of lines be, , (v) Slope-intercept form The equation of line, with slope m and making an intercept c on, Y -axis, is y = mx + c ., (vi) If a line with slope m cuts the X -axis at a, distance d from the origin i.e. makes x-intercept, d . Then, the equation of line is given by, y = m ( x − d )., (vii) Intercept form The equation of a line which, cuts off intercepts a and b respectively on the, x y, X and Y-axes is given by + = 1, a b, y, x, i.e., +, =1, x - intercept y - intercept, , Let θ be the angle between two lines, then,  A2, −, +,  m − m1 ,  B2, =±, tan θ = ±  2, A, 1 + m1m 2 , 1 + 1 ⋅, B1, , , (viii) Normal form The equation of a straight line, upon which the length of the perpendicular, from the origin is p and angle made by this, perpendicular to the X -axis is α, is given by, x cos α + y sin α = p ., , Distance between Two Parallel Lines, , General Equation of a Line, Any equation of the form Ax + By + C = 0, where, A and B are simultaneously not zero, is called the, general equation of a line., , A1 x + B1 y + C 1 = 0 and A2 x + B 2 y + C 2 = 0, then, A, A, slope of given lines are m1 = − 1 and m 2 = − 2 ., B1, B2, A1 , , B1 , A2, , B 2 , , Distance of a Point from a Line, The perpendicular distance d of a point P ( x 1 , y 1 ), from the line Ax + By + C = 0 is given by, d =, , Ax 1 + By 1 + C, A2 + B 2, , The distance d between two parallel lines, y = mx + C 1 and y = mx + C 2 is given by, |C − C 2 |, and if lines are Ax + By + C 1 = 0, d = 1, 1 + m2, and Ax + By + C 2 = 0, then, d =, , |C 1 − C 2 |, A2 + B 2, , .

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74, , CBSE New Pattern ~ Mathematics XI (Term I), , Objective Questions, Multiple Choice Questions, 1. The point ( − 3, 2) is located in the, quadrant, (a) quadrant I, (c) quadrant III, , (b) quadrant II, (d) quadrant IV, , 2. The value of y is, if the distance, between points P ( 2, − 3) and Q (10, y ) is, 10 units., (a) 3, (c) − 3, , (b) 9, (d) None of these, , 3. The point on X-axis which is equidistant, from the points ( 3, 2) and ( −5, − 2) is, (a) (1, 0), (c) (−1, 0), , (b) (2, 0), (d) (−2, 0), , 4. The coordinates of a point which, divides the line segment joining, A (1, − 3) and B ( − 3, 9 ) internally in the, ratio 1 : 3, are given by, (a) (− 2, 6),  − 6 18 , (c) , , ,  4 4, , (b) (0, 0),  10 − 30, (d)  ,, , 4 4 , , 5. The coordinates of a point which, , divides externally the line joining (1, −3), and ( − 3, 9 ) in the ratio 1 : 3 are, (a) (3, − 6), (c) (3, − 9), , (b) (− 6, 3), (d) (− 9, 3), , 6. If the vertices of a triangle are P (1, 3),, , Q ( 2, 5) and R( 3, − 5), then the centroid, of a ∆PQR is, (a) (1, 2), (c) (3, 1), , (b) (1, 3), (d) (2, 1), , 7. The points (1, − 1), (5, 2) and (9, 5), collinear., (a), (b), (c), (d), , Yes, No, Cannot say, Insufficient information, , 8. Area of the triangle whose vertices are, ( 4, 4 ), ( 3, −2) and ( − 3, 16 ), is, (a) 54, (c) 53, , (b) 27, (d) 106, , 9. The slope of a line whose inclination is, 90°, is, (a) 1, (c) − 1, , (b) 0, (d) not defined, , 10. The slope of line, whose inclination is, 60°, is, 1, 3, (c) 3, (a), , (b) 1, (d) Not defined, , 11. The slope of that line, which passes, through the points (at 12 , 2at 1 ) and, (at 22 , 2at 2 ) is, 2, t2 − t1, 1, (c), t2 + t1, (a), , 2, t2 + 2 t1, 2, (d), t2 + t1, (b), , 12. The angle between the lines, y = ( 2 − 3 )( x + 5), and y = ( 2 + 3 )( x − 7 ) is, (a) 30°, (c) 45°, , (b) 90°, (d) 120°, , 13. The angle between the X -axis and, , the line joining the points ( 3, − 1) and, ( 4, − 2) is, (a) 45°, (c) 90°, , (b) 135°, (d) 180°, , 14. The tangent of angle between the lines, whose intercepts on the axes are a, −b, and b, −a respectively, is, a2 − b 2, ab, b 2 − a2, (c), 2ab, , (a), , (b), , b 2 − a2, 2, , (d) None of these

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75, , CBSE New Pattern ~ Mathematics XI (Term I), , 15. The value of y will be, so that the line, through ( 3, y ) and ( 2, 7 ) is parallel to the, line through ( −1, 4 ) and (0, 6 )., (a) 7, (c) 9, , (b) 8, (d) 10, , 16. Using slope of line, till, are A( 4, 4 ),, , B( 3, 5) and C( −1, 1) the vertices of a right, angled triangle., (a), (b), (c), (d), , Yes, No, Cannot say, Insufficient Information, , 17. The points A (x , 4 ), B ( 3, − 2) and, , C ( 4, − 5) are collinear in the value of x, is, (a) 1, (c) − 1, , (b) 2, (d) 0, , 18. The equation of the lines parallel to the, X-axis and passing through the point, ( − 3, 5) is, (a) x = − 3, (c) x = 5, , (b) y = − 3, (d) y = 5, , 19. The equation of the line through ( − 2, 3), with slope − 4 is, (a) x + 4 y − 10 = 0, (c) x + y − 1 = 0, , (b) 4 x + y + 5 = 0, (d) 3 x + 4 y − 6 = 0, , 20. The equation of the line passing, through the point (1, 2) and, perpendicular to the line x + y + 1 = 0 is, (a) y − x + 1 = 0, (c) y − x + 2 = 0, , (b) y − x − 1 = 0, (d) y − x − 2 = 0, , 21. The equation of a line perpendicular to, the line x − 2y + 3 = 0 and passing, through the point (1, – 2) is, (a) y = 2 x, (c) x = − 2 y, , (b) x = 2 y, (d) y = − 2 x, , 22. The equation of line passing through, the points ( −1, 1) and ( 2, − 4 )., , (a), (b), (c), (d), , 5 x + 2y + 2 = 0, 5 x + 3y − 2 = 0, 5 x + 2y + 3 = 0, 5 x + 3y + 2 = 0, , 23. The line passing through, , the points ( − 4, 5) and ( − 5, 7 ) also passes, through the point (l , m ), then 2l + m + 3, is equal to, (a) 1, , (b) −1, , (c) 2, , (d) 0, , 24. A line cutting off intercept −3 from the, Y-axis and the tangent at angle to the, 3, X-axis is , its equation is, 5, (a) 5 y − 3x + 15 = 0, (c) 5 y − 3x − 15 = 0, , (b) 3y − 5 x + 15 = 0, (d) None of these, , 25. The equations of the line which have, slope 1/2 and cuts-off an intercept 4 on, X -axis is, (a) x − 2 y − 4 = 0, (c) x + 2 y + 4 = 0, , (b) x + 2 y − 4 = 0, (d) x − 2 y + 4 = 0, , 26. Slope of a line which cuts off intercepts, of equal lengths on the axes is, (a) −1, , (b) 0, , (c) 2, , (d) 3, , x y, 27. If the line + = 1 passes through the, a b, points (2, −3) and (4, −5), then (a, b) is, (a) (1, 1), (c) (1, −1 ), , (b) (−1, 1), (d) (−1, −1 ), , 28. If the coordinates of the middle point of, the portion of a line intercepted, between the coordinate axes is (3, 2),, then the equation of the line will be, (a) 2 x + 3y = 12, (c) 4 x − 3y = 6, , (b) 3x + 2 y = 12, (d) 5 x − 2 y = 10, , 29. If the normal form of the equation, , 3x + y − 8 = 0 is x cos ω + y sin ω = p,, then p and ω respectively are, , (a) 4, 45°, (c) 3, 45°, , (b) 4, 30°, (d) 3, 30°, , 30. Transform the equation of the line, , 3x + 2y − 7 = 0 to slope intercept form, then the slope and y-intercept will be, 3 7, ,, 2 2, 3 7, (c) − ,, 2 2, (a), , 3 7, (b) − , −, 2 2, , (d) None of these

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76, , CBSE New Pattern ~ Mathematics XI (Term I), , 31. Transform the equation of the line, , 3x + 2y − 7 = 0 to normal form then the, inclination of the perpendicular, segment from the origin on the line, with the axis and its length is,  1, (a) tan− 1   ,,  3,  3, (c) tan− 1   ,,  4, , 7, 13, 7, 13, ,  2, (b) tan− 1   ,,  3,  1, (d) tan− 1   ,,  5, , 7, 13, 7, 13, , 32. The angle between the lines, y − 3x − 5 = 0 and, will not be, (a) 30°, (c) 45°, , 3y − x + 6 = 0, (b) 150°, (d) None of these, , 33. The angle between the lines, x − 2y + 3 = 0 and 3x + y − 1 = 0 is, (a) − tan− 1 (7), ,  1, (b) tan− 1  ,  7, , (c) π − tan− 1 (7), , (d) 2 π − tan− 1 (7), , 34. The equation of line, which passes, through point (4, 3) and parallel to the, line 2x − 3y = 7 is, (a) 2 x − 3y + 1 = 0, (c) 2 x + 3y + 1 = 0, , (b) 2 x − 3y − 1 = 0, (d) 2 x + 3y − 1 = 0, , 35. Lines through the points ( −2, 6 ) and, ( 4, 8 ) is perpendicular to the line, through the points (8, 12) and (x , 24 )., Then, the value of x is, (a) 2, (c) 8, , (b) 6, (d) 4, , 36. The distance of the point ( 3, − 5) from, the line 3x − 4 y − 26 = 0 is, , 3, 7, 7, (c), 5, (a), , 2, 5, 3, (d), 5, , (b), , 37. The perpendicular distance from origin, to the line 5x + 12y − 13 = 0 is, , (a) 10 unit, (c) 2 unit, , (b) 5 unit, (d) 1 unit, , 38. The distance of the point of intersection, of the lines 2x − 3y + 5 = 0 and, 3x + 4 y = 0 from the line 5x − 2y = 0 is, 130, 17 29, 130, (c), 7, (a), , (b), , 13, 7 29, , (d) None of these, , 39. The distance between the parallel lines, 3x − 4 y + 7 = 0 and 3x − 4 y + 5 = 0, is, 3, 7, 2, (c), 5, (a), , 7, 5, 3, (d), 5, (b), , 40. The distance between the lines, 3x + 4 y = 9 and 6x + 8 y = 15 is, , 3, 10, 7, (c), 10, (a), , 2, 25, 3, (d), 25, , (b), , Assertion-Reasoning MCQs, Directions (Q.Nos. 41-54) Each of these, questions contains two statements : Assertion, (A) and Reason (R). Each of these questions, also has four alternative choices, any one of, which is the correct answer. You have to, select one of the codes (a), (b), (c) and (d), given below., (a) A is true, R is true; R is a correct, explanation for A., (b) A is true, R is true; R is not a correct, explanation for A., (c) A is true; R is false., (d) A is false; R is true., , 41. Assertion (A) The point ( 3, 0) is at 3, units distance from the Y -axis measured, along the positive X -axis and has zero, distance from the X -axis., Reason (R) The point ( 3, 0 ) is at 3 units, distance from the X -axis measured, along the positive Y -axis and has zero, distance from the Y -axis.

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77, , CBSE New Pattern ~ Mathematics XI (Term I), , 42. Assertion (A) A point P (h, k ) lies on, , the straight line x + y + 1 = 0 and is at a, distance 5 units from the origin. If k is, negative, then h is equal to − 3., Reason (R) The distance formula is, (x 2 − x 1 ) 2 + ( y 2 − y1 ) 2 ., , 43. Assertion (A) The line x + y = 4, , divides the line joining the points ( −1, 1), and (5, 7 ) in the ratio 1 : 2., Reason (R) Section formula for, internal division is,  m 1x 2 − m 2 x 1 m 1 y 2 − m 2 y1 ,  where, , ,, m1 − m 2 ,  m1 − m 2, p (x , y ) divides the line segment AB,, with A (x 1 , y1 ) and B (x 2 , y 2 ) in the ratio, m1 : m 2 ., , 44. If the vertices of a triangle are (1, a ),, ( 2, b ) and (c 2 , − 3). Then,, Assertion (A) The centroid cannot lie, on the Y-axis., Reason (R) The condition that the, centroid may lie on the X-axis is, a + b = 3., , 45. Assertion (A) Area of the triangle, , whose vertices are ( 4, 4 ), ( 3, −2) and, ( − 3, 16 ), is, Reason (R) Area of triangle whose, vertices are (x 1 , y1 ), (x 2 , y 2 ) and (x 3 , y 3 ),, 1, is | x 1 ( y 2 − y 3 ) + x 2 ( y 3 − y1 ), 2, + x 3 ( y1 − y 2 ) |., , 46. Assertion (A) Slope of X -axis is zero, and slope of Y -axis is not defined., Reason (R) Slope of X -axis is not, defined and slope of Y -axis is zero., , 47. If A ( − 2, − 1), B ( 4, 0 ), C ( 3, 3) and, D ( − 3, 2) are the vertices of a, parallelogram, then, , Assertion (A) Slope of AB = Slope of, BC and Slope of CD = Slope of AD., Reason (R) Mid-point of AC, = Mid-point of BD., , 48. Assertion (A) The angle between the, lines x + 2y − 3 = 0 and 3x + y + 1 = 0 is, tan −1 (1)., , Reason (R) Angle between two lines is,   m 2 − m1  ,  ., given by tan −1  ± ,   1 + m 1m 2  , , 49. Assertion (A) Slope of line, 3, 3x − 4 y + 10 = 0 is ., 4, Reason (R) x-intercept and y-intercept, of 3x − 4 y + 10 = 0 respectively are, − 10, 5, and ., 2, 3, , 50. Assertion (A) If x cos θ + y sin θ = 2 is, , perpendicular to the line x − y = 3, then, one of the value of θ is π /4., Reason (R) If two lines y = m 1x + c 1, and y = m 2 x + c 2 are perpendicular, then m 1 = m 2 ., , 51. Assertion (A) The slope of the line, 1, and y-intercept is 0., 7, Reason (R) The slope of the line, , x + 7 y = 0 is, , 5, ., 3, 52. If the equation of line is x − y = 4, then, 6x + 3y − 5 = 0 is − 2 and y-intercept is, , Assertion (A) The normal form of, same equation is x cos α + y sin α = p,, where α = 315° and p = 2 2., Reason (R) The perpendicular distance, of line from the origin is 3 2.

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78, , CBSE New Pattern ~ Mathematics XI (Term I), , 53. Assertion (A) The distance between, , the lines 4 x + 3y = 11 and 8 x + 6 y = 15, is 7 /10., , Reason (R) The distance between the, lines ax + by = c 1 and ax + by = c 2 is, c1 − c 2, , given by, , a2 + b2, , ., , (v) In which year the population, becomes 110 crores is, (a) 2020, (c) 2021, , 55. Three girls, Rani, Mansi, Sneha are, talking to each other while maintaining, a social distance due to covid-19. They, are standing on vertices of a triangle,, whose coordinates are given., Rani (2, –2), , Case Based MCQs, 54. Population vs Year graph given below., Population, (in Crores), , 102, , C (2010, P), , 97, , B (1995, 97), , 92, , Mansi (1, 1), , A (1985, 92), , 87, O, , 1985 1990 1995 2000 2005 2010, Years, , Based on the above information answer, the following questions., (i) The slope of line AB is, (a) 2, 1, (c), 2, , (b) 1, 1, (d), 3, , (ii) The equation of line AB is, (a), (b), (c), (d), , x + 2 y = 1791, x − 2 y = 1801, x − 2 y = 1791, x − 2 y + 1801 = 0, , (a) 104.5, (c) 109.5, , Sneha (–1, 0), , Based on the above information answer, the following questions., (i) The equation of lines formed by Rani, and Mansi is, (a) 3x − y = 4, (c) x − 3y = 4, , (b) 3x + y = 4, (d) x + 3y = 4, , (ii) Slope of equation of line formed by, Rani and Sneha is, 2, 3, −2, (c), 3, (a), , −3, 2, 1, (d), 3, , (b), , (iii) The equation of median of lines, through Rani is, , (iii) The population in year 2010 is, (in crores), (b) 119.5, (d) None of these, , (iv) The equation of line perpendicular to, line AB and passing through, (1995, 97) is, (a), (b), (c), (d), , (b) 2019, (d) 2022, , 2 x − y = 4087, 2 x + y = 4087, 2 x + y = 1801, None of the above, , (a) 5 x + 4 y = 2, (c) 4 x − 5 y = 1, , (b) 5 x − 4 y = 2, (d) None of these, , (iv) The equation of altitude through, Mansi is, (a) 3x − 2 y = 1, (c) x + 2 y = 3, , (b) 2 x + 3y = 5, (d) None of these, , (v) The equation of line passing through, the Rani and parallel to line formed, by Mansi and Sneha is, (a) x − 2 y = 4, (c) x − 2 y = 6, , (b) x + 2 y = 6, (d) 2 x + y = 4

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79, , CBSE New Pattern ~ Mathematics XI (Term I), , 56. Consider the ∆ABC with vertices A(1, 4 ),, B ( 2, − 3) and C( − 1, − 2) as shown in the, given figure. AD is the median and AM, is the altitude through A., A(1, 4), , 57. Four friends Rishabh, Shubham,, Vikram and Rajkumar are sitting on, vertices of a rectangle, whose, coordinates are given., Rishabh (1, 4), A, , Rajkumar (5, 4), D, , C, Vikram (5, 2), , B, Shubham (1, 2), , Based on the above information answer, the following questions., B(2, – 3), , D, , M, , C(– 1, – 2), , Based on the above information answer, the following questions., (i) Find the distance between A and C, (a) 40 units, (b) 53 units, (d) 29 units, , (ii) Find the slope of BC., 4, 3, 3, (c) −, 2, , 1, 3, 3, (d) −, 4, (b) −, , (iii) Find the equation of median through, A., (a) x − 13y + 9 = 0, (c) 13x − y − 9 = 0, , (b) x + 13y − 9 = 0, (d) 2 x − 13y + 9 = 0, , (iv) Find the equation of the altitude, through A., (a) 3x − y + 1 = 0, (c) x − 3y + 2 = 0, , (b) x + 2 y − 3 = 0, (d) 3x + 2 y − 2 = 0, , (v) Find the equation of right bisector of, side BC., (a) x + 3y − 3 = 0, (c) 3x − y − 4 = 0, , (a), (b), (c), (d), , x + 2y + 3 = 0, x − 2y − 3 = 0, x − 2y + 3 = 0, None of the above, , (ii) The equation formed by Rishabh and, Vikram is, , (c) 41 units, , (a) −, , (i) The equation formed by Shubham, and Rajkumar is, , (b) x − 3y + 3 = 0, (d) 3x + y − 2 = 0, , (a), (b), (c), (d), , x + 2y + 9 = 0, x + 2 y −9 = 0, x − 2 y − 9 =0, None of the above, , (iii) The intersection point of above two, equations is, (a), (b), (c), (d), , (1, 1), (2, 2), (3, 3), (4, 4), , (iv) Slope of equation of line formed by, Rishabh and Rajkumar is, (a) zero, (c) 2, , (b) 1, (d) 3, , (v) Pair for the same slope is, (a) Rishabh-Rajkumar and Shubham-Vikram, (b) Rishabh-Rajkumar and Rajkumar-Vikram, (c) Rishabh-Rajkumar and Rishabh-Shubham, (d) None of the above

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80, , CBSE New Pattern ~ Mathematics XI (Term I), , (ii) The equation of AB is, , 58. If A and B are two persons sitting at the, , positions ( 2, − 3) and (6, − 5). If C is a, third person who is sitting between A, and B such that it divides the line AB in, 1 : 3 ratio., , (a) x + 2 y + 4 = 0, (c) x − 2 y + 4 = 0, , (b) x + 2 y − 4 = 0, (d) None of these, , (iii) Coordinates of point C are,  7, (b)  3, ,  2, 7, , (d)  3, − , , 2, , , 7, (a)  , − 3, , 2, (c) (3, 3), , (iv) Distance between A and C is, , Based on the above information,, answer the following questions., , 5, , (b) 2 5, , (c), , 5, 2, , (d), , 5, 2, , (v) Distance between C and B is, 3 5, 2, 2 5, (c), 3, , (a), , (i) The distance between A and B is, (a) 5, (c) 3 5, , (a), , (b) 2 5, (d) 4 5, , (b) 3 5, (d) None of these, , ANSWERS, Multiple Choice Questions, 1. (b), 11. (d), 21. (d), 31. (b), , 2. (a), 12. (d), 22. (d), 32. (c), , 3. (c), 13. (b), 23. (d), 33. (c), , 4. (b), 14. (c), 24. (a), 34. (a), , 5. (c), 15. (c), 25. (a), 35. (d), , 6. (d), 16. (a), 26. (a), 36. (d), , 7. (a), 17. (a), 27. (d), 37. (d), , 8. (b), 18. (d), 28. (a), 38. (a), , 9. (d), 19. (b), 29. (b), 39. (c), , 10. (c), 20. (b), 30. (c), 40. (a), , 44. (b), , 45. (a), , 46. (c), , 47. (d), , 48. (a), , 49. (b), , 50. (c), , Assertion-Reasoning MCQs, 41. (c), 51. (d), , 42. (d), 52. (c), , 43. (c), 53. (a), , Case Based MCQs, 54. (i) - (c); (ii) - (b); (iii) - (a); (iv) - (b); (v) - (c), 56. (i) - (a); (ii) - (b); (iii) - (c); (iv) - (a); (v) - (c), 58. (i) - (b); (ii) - (a); (iii) - (d); (iv) - (c); (v) - (a), , 55. (i) - (b); (ii) - (c); (iii) - (a); (iv) - (a); (v) - (c), 57. (i) - (c); (ii) - (b); (iii) - (c); (iv) - (a); (v) - (a)

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81, , CBSE New Pattern ~ Mathematics XI (Term I), , SOLUTIONS, 1. Let A = ( − 3, 2), , 5. Let the coordinates of the required point, , Since, x-coordinate of A is negative and its, y-coordinate is positive, therefore A lies in the, second quadrant., , 2. Given, points P ( 2, − 3) and Q (10, y )., According to the question,, PQ = 10, ⇒, , (10 − 2) 2 + ( y + 3) 2 = 10, [Q d = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 ], , ⇒, , ( 8 ) + ( y + 3) = 10, , ⇒, , 64 + y 2 + 9 + 6 y = 10, , 2, , 2, , On squaring both sides, we get, 64 + y 2 + 9 + 6 y = 100, ⇒, , y + 6 y + 73 = 100, , ⇒, , y 2 + 6 y − 27 = 0, , ⇒, , y 2 + 9 y − 3y − 27 = 0, , 2, , ⇒, y( y + 9 ) − 3 ( y + 9 ) = 0, ⇒, ( y + 9 ) ( y − 3) = 0, ⇒, y = 3, − 9, Hence, value of y is 3, − 9., , 3. Let the point on X-axis be P ( x , 0 ), which is, equidistant from (say) A ( 3, 2) and (say), B ( −5, − 2)., Since, P is equidistant from A and B. So,, ∴, PA = PB ⇒ PA 2 = PB 2, ⇒ ( 3 − x ) 2 + ( 2 − 0 ) 2 = ( −5 − x ) 2 + ( −2 − 0 ) 2, ⇒, ⇒, ⇒, , [by distance formula], 9 + x − 6 x + 4 = 25 + x 2 + 10 x + 4, 16 x + 16 = 0, x = −1, 2, , Thus, point on X-axis is ( −1, 0 )., , 4. The coordinates of the point which divides, the line segment joining A (1, − 3) and, B ( − 3, 9 ) internally in the ratio 1 : 3, are, given by, 1 ⋅ ( − 3) + 3 ⋅1, x=, =0, 1+ 3, 1.9 + 3 ⋅ ( − 3), and, y=, =0, 1+ 3, , be P ( x , y )., 1 × ( −3) − 3 × 1, Then, x = , , 1−3, , , and, , 1 × 9 − 3 × ( −3) , y=, , 1−3, , , , i.e. x = 3 and y = − 9, Hence, the required point is ( 3, − 9 )., , 6. We know that, if the vertices of a triangle are, ( x1 , y1 ), ( x 2 , y 2 ) and ( x 3 , y 3 ), then centroid of, a triangle is,  x1 + x 2 + x 3 y1 + y 2 + y 3 , ,, , ., , , 3, 3, Here, P (1, 3) ≡ P ( x1 , y1 );, Q ( 2, 5) ≡ Q ( x 2 , y 2 ), and R( 3, − 5) ≡ R ( x 3 , y 3 ), ∴Centroid of a triangle, 1 + 2 + 3 3 + 5 − 5, ,, =, , , , 3, 3,  6 3, =  ,  = ( 2, 1),  3 3, , 7. Let A = (1, − 1), B = ( 5, 2) and C = ( 9, 5), Now, distance between A and B ,, AB = ( 5 − 1) 2 + ( 2 + 1) 2, [by distance formula], = ( 4 ) 2 + ( 3) 2, = 16 + 9 = 25 = 5, Distance between B and C,, BC = ( 5 − 9 ) 2 + ( 2 − 5) 2, = ( − 4 ) 2 + ( − 3) 2, = 16 + 9, = 25 = 5, Distance between A and C,, AC = (1 − 9 ) 2 + ( − 1 − 5) 2, = ( − 8 )2 + ( − 6 )2, = 64 + 36 = 10, Clearly, AC = AB + BC, Hence, A , B and C are collinear points.

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82, , CBSE New Pattern ~ Mathematics XI (Term I), , 8. Area of the triangle, whose vertices are, (4, 4), (3, –2) and (−3, 16), 1, = |4 ( − 2 − 16 ) + 3 (16 − 4 ) + ( − 3) ( 4 + 2)|, 2, |− 72 + 36 − 18 |, =, 2, | − 54 | 54, =, =, = 27, 2, 2, , 9. The slope of a line whose inclination is 90°,, is not defined., , 10. Let θ be the inclination of a line, then its, , slope = tan θ., At θ = 60 °, slope of a line, m = tan 60 ° = 3, , 11. We know that the slope of a line which, passes through the points ( x1 , y1 ) and ( x 2 , y 2 ), y − y1, = 2, x 2 − x1, Here, ( x1 , y1 ) ≡ ( at12 ,, , 2at1 ), ( x 2 , y 2 ) ≡ ( at 22 ,, , 2at 2 ), 2at 2 − 2at1, ∴ Slope of a line =, at 22 − at12, 2a ( t 2 − t1 ), 2, =, =, a ( t 2 + t1 ) ( t 2 − t1 ) t 2 + t1, , 12. Given lines,, y = ( 2 − 3 )( x + 5), , …(i), , Slope of this line,, m1 = ( 2 − 3), and, , y = (2 +, , Slope of this line,, m2 = (2 +, , 3 )( x − 7 ), , …(ii), , 3), , Let θ be the angle between lines (i) and (ii),, then, m − m 2 , tan θ =  1, , 1 + m 1 m 2 , ⇒, , ( 2 − 3 ) − ( 2 + 3 ) , , tan θ = , 1 + ( 2 − 3 )( 2 + 3 ) , , ⇒, ,  −2 3 , , tan θ = , 1 + 4 − 3 , , ⇒, , tan θ = 3 ⇒ tan θ = tan π/ 3, , ∴, θ = π/ 3 = 60 °, For obtuse angle = π − π/ 3 = 2 π/ 3 = 120 °, Hence, the angle between the lines are, 60° or 120°., , 13. Given, ( x1 , y1 ) = ( 3, − 1), ( x 2 , y 2 ) = ( 4, − 2), y − y1 − 2 + 1, Slope of line m = 2, =, x 2 − x1, 4−3, −1, tan θ =, ⇒ tan θ = − 1, 1, (Q1 = tan 45°), ⇒, tan θ = − tan 45°, ⇒, tan θ = tan (180 ° − 45° ), [Q tan (180° − θ ) = − tan θ ], ⇒, tan θ = tan 135° ⇒ θ = 135°, , and, , 14. Since, intercepts on the axes are a , −b then, equation of the line is, ⇒, ⇒, , x y, − = 1., a b, , y x, = −1, b a, bx, y=, −b, a, , b, So, the slope of this line i.e. m 1 = ., a, Also, for intercepts on the axes as b and − a ,, then equation of the line is, x y, − =1, b a, y x, ⇒, = −1, a b, a, y = x −a, ⇒, b, a, and slope of this line i.e. m 2 =, b, b a, b2 − a2, −, ∴ tan θ = a b = ab, a b, 2, 1+ ⋅, b a, b2 − a2, =, 2ab, , 15. Let A ( 3, y ), B( 2, 7 ), C( −1, 4 ) and D( 0, 6 ) be the, given points., Then, m 1 = Slope of the line, 7−y, AB =, = ( y − 7), 2− 3, and, m 2 = Slope of the line, 6−4, CD =, =2, 0 − ( −1), Since, AB and CD are parallel., ∴ m1 = m 2 ⇒ y − 7 = 2 ⇒ y = 9

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83, , CBSE New Pattern ~ Mathematics XI (Term I), , 16. In ∆ABC, we have, , 4−5, m 1 = Slope of AB =, = −1, 4−3, 5 −1, 4, and m 2 = Slope of BC =, = =1, 3 − ( −1) 4, , Clearly, m 1m 2 = −1, This shows that AB is perpendicular to BC., i.e., , ∠ABC = π / 2, , Hence, the given points are the vertices of a, right angled triangle., Option (a) is correct., , 17. Given points are A ( x , 4 ), B ( 3, − 2) and, C( 4, − 5)., From the condition of collinearity of three, points A , B and C, we should have, Slope of AB = Slope of BC, , y 2 − y1 , −2−4 −5+ 2, i.e., =, Q slope = x − x , 3−x, 4−3, , 2, 1, ⇒, ⇒, ⇒, ⇒, ⇒, , −6, −3, =, (3 − x), 1, 2, 1, =, 3−x 1, 2=3−x, x = 3 − 2 =1, x =1, , 18. Clearly, the equation of a line parallel to the, X -axis and passing through ( − 3, 5) is y = 5., , 19. Here, m = − 4 and given point ( x 0 , y 0 ) is, ( −2, 3)., By slope-point form,, Equation of the given line is, y − 3 = − 4 ( x + 2), or 4 x + y + 5 = 0, which is the required, equation., , 20. Given point is (1, 2) and slope of the required, line is 1., Q x + y + 1 = 0 ⇒ y = − x − 1 ⇒ m 1 = −1, −1, ∴ Slope of the line =, =1, −1, ∴ Equation of required line is, y − 2 = 1 ( x − 1), ⇒, y − 2 = x −1, ⇒ y − x −1 = 0, , 21. Given, line x − 2y + 3 = 0 can be written as, y=, , 1, 3, x+, 2, 2, , …(i), , 1, Slope of the line (i) is m 1 = . Therefore, slope, 2, of the line perpendicular to line (i) is, 1, = − 2., m2 = −, m1, Equation of the line with slope −2 and passing, through the point (1, − 2) is, y − ( −2) = − 2( x − 1) or y = − 2x, which is the required equation., , 22. Let the given points are A ( x1 , y1 ) ≡ A ( −1, 1), and B ( x 2 , y 2 ) ≡ B ( 2, − 4 ), then equation of, line AB is, −4 − 1, y −1 =, ( x + 1), 2+1, , , y 2 − y1, Q y − y1 = x − x ( x − x1 ) , , , 2, 1, −5, y −1 =, ( x + 1), ⇒, 3, ⇒, 3y − 3 = − 5x − 5, ⇒ 5x + 3y + 2 = 0, , 23. Let the given points be A ( − 4, 5) and, B ( − 5, 7 )., We know that, equation of line passing, through two points is, y − y1, y − y1 = 2, ( x − x1 )., x 2 − x1, Then, equation of line passing through the, points A and B is, 7−5, y −5=, ( x + 4), −5+ 4, 2, y −5=, ( x + 4), ⇒, −1, ⇒, ⇒, ⇒, ⇒, ⇒, , y − 5 = − 2 ( x + 4), y − 5 = − 2x − 8, y + 2x = − 8 + 5, y + 2x = − 3, 2x + y + 3 = 0, , …(i), , Now, line (i) passes through the point ( l , m ), so, point ( l , m ) will satisfy Eq. (i)., Put x = l , y = m in Eq. (i),, 2l + m + 3 = 0, Hence, option (d) is correct.

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84, , CBSE New Pattern ~ Mathematics XI (Term I), , 24. Given that,, 3, 5, ∴ Equation of the line is y = mx + c, 3, y= x −3, 5, ⇒, 5y = 3x − 15 ⇒ 5y − 3x + 15 = 0, 1, 25. Given, m = Slope of the line =, 2, Here, d = Intercept of the line on X -axis = 4., Hence, required equation of the line is, 1, y = ( x − 4), 2, ⇒, x − 2y − 4 = 0, [Q y = m ( x − d )], c = − 3 and m =, , 26. Let equation of line be, x y, + =1, a a, ⇒, x + y =a ⇒ y = −x + a, ∴ Required slope = −1, x y, 27. Given, line is + = 1, …(i), a b, Since, the points (2, −3) and (4, −5) lies on this, line., 2 3, …(ii), − =1, ∴, a b, 4 5, and, …(iii), − =1, a b, On multiplying by 2 in Eq. (ii) and then, subtracting Eq. (iii) from Eq. (ii), we get, 6 5, −1, − + =1 ⇒, =1, b b, b, ∴, b = −1, On putting b = −1 in Eq. (ii), we get, 2, 2, + 3 = 1 ⇒ = − 2 ⇒ a = −1, a, a, ( a , b ) = (−1, −1), ∴, , 28. Since, the coordinates of the middle point are, P (3, 2)., Y, B(0, b), 1, P(3, 2), 1, X, , X′, y¢, , (a, 0), , 1⋅ 0 + 1⋅a, 1+1, a, 3 = ⇒a = 6, ⇒, 2, Similarly,, b =4, x y, ∴ Equation of the line is + = 1, 6 4, ⇒, 2x + 3y = 12, , ∴, , 3=, , 29. Given equation is, 3x + y − 8 = 0, , …(i), , Dividing Eq. (i) by ( 3 ) 2 + (1) 2 = 2,, we get, 3, 1, x+ y=4, 2, 2, or cos 30 ° x + sin 30 ° y = 4, …(ii), Comparing Eq. (ii) with x cosω + y sin ω = p,, we get, p = 4 and ω = 30 °, , 30. Given equation is 3x + 2y − 7 = 0., It can be rewritten as, 2y = − 3x + 7, 3, 7, or, ...(i), y=− x +, 2, 2, which is the required slope intercept form of, the given line., On comparing Eq. (i) with y = mx + c , we get, 3, 7, Slope, m = − and y-intercept, c =, 2, 2, , 31. Given equation is 3x + 2y − 7 = 0., , …(i), , On comparing Eq. (i) with Ax + By + C = 0,, we get, A = 3, B = 2 and C = − 7, ∴, , A 2 + B 2 = 32 + 22 = 9 + 4 = 13, , On dividing both sides of Eq. (i) by 13,, we get, 3, 2, 7, x+, y−, =0, 13, 13, 13, 3, 2, 7, …(ii), x+, y=, ⇒, 13, 13, 13, which is the required normal form of the, given line., On comparing Eq. (ii) with, x cos ω + y sin ω = p, we get, 3, 2, 7, and p =, cos ω =, , sin ω =, 13, 13, 13

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85, , CBSE New Pattern ~ Mathematics XI (Term I), , Since, cos ω and sin ω both are positive,, therefore ω is in the first quadrant and is equal, 2, to tan −1 ., 3, Hence, for the given line, we have, 2, 7, ., and p =, ω = tan −1, 3, 13, , Put the value of λ in Eq. (i), we get, 2x − 3y = − 1, ⇒, 2x − 3y + 1 = 0, which is the required equation of line., , 35. Slope of the line through the points (– 2, 6), and (4, 8) is, m1 =, , 32. Given lines are, …(i), y − 3x − 5 = 0 or y = 3x + 5, 1, …(ii), 3y − x + 6 = 0 or y =, x−2 3, 3, Slope of line (i) is m 1 = 3 and slope of, 1, line (ii) is m 2 =, ., 3, The acute angle (say) θ between two lines is, given by, m − m1, …(iii), tan θ = 2, 1 + m 1m 2, Putting the values of m 1 and m 2 in Eq. (iii), we, get, 1, − 3, 1−3, 1, 3, tan θ =, =, =, 1, 2, 3, 3, 1+ 3 ×, 3, which gives θ = 30 °. Hence, angle between, two lines is either 30° or 180 ° − 30 ° = 150 °., , 33. Let m 1 and m 2 be the slopes of the straight, , lines x − 2y + 3 = 0 and 3x + y − 1 = 0., 1, 1, 3, Then, m 1 = −, = and m 2 = − = − 3, −2 2, 1, Let θ be the angle between the given lines., Then,, 1, ,  −3 − ,  m 2 − m1 , 2 = ± 7, tan θ = ± ,  =±, 1 + m 1m 2 ,  1− 3 , , 2 , ⇒, , θ = tan −1 ( 7 ) or π − tan −1 ( 7 ), , 34. Let the equation of line parallel to the given, line is, …(i), 2x − 3y = λ, [Q In two parallel lines, m 1 = m 2 ], Since, the line (i), passes through the point (4,, 3)., So, this point will satisfy the equation of line., ∴, 2×4 −3×3=λ, ⇒, 8 − 9 = λ ⇒ λ = −1, , 8 −6, 2 1, = =, 4 − ( − 2) 6 3, , Slope of the line through the points (8, 12) and, ( x , 24 ) is, 24 − 12, 12, m2 =, =, x −8, x −8, Since, two lines are perpendicular., So,, m 1 m 2 = −1, 1, 12, ⇒, ×, = −1, 3 x −8, ⇒, , 8 − x = 4 or x = 4, , 36. Given line is 3x − 4 y − 26 = 0, , …(i), , On comparing Eq. (i) with general equation of, line Ax + By + C = 0, we get, A = 3, B = − 4 and C = − 26, Given point is ( x1 , y1 ) = ( 3, − 5)., The distance of the given point from given, line is, Ax1 + By1 + C, d =, A2 + B2, =, , 3 ⋅ 3 + ( −4 )( −5) − 26, 3 + ( −4 ), 2, , 2, , =, , 3, 5, , 37. Given equation of line is, …(i), 5x + 12y − 13 = 0, Length of perpendicular from origin to the, line (i) is, − 13, p=, 2, 5 + 122, [Q perpendicular length from origin to the, , c, , line ax + by + c = 0 is p =, a 2 + b 2 , − 13, − 13 13, =, =, =, =1, 25 + 144, 169 13, [Q x = − x , x < 0 ], ∴ Required length of perpendicular is 1 unit.

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86, , CBSE New Pattern ~ Mathematics XI (Term I), , 38. Given equation of lines, and, , 2x − 3y + 5 = 0, 3x + 4 y = 0, , From Eq. (ii), put the value of x =, , 41. Assertion The point (3, 0) is at 3 units, −4y, in, 3, , …(i), …(ii), , Eq. (i), we get,  − 4 y, 2,  − 3y + 5 = 0,  3 , ⇒, , − 8 y − 9 y + 15 = 0 ⇒ y =, , From Eq. (ii), 3x + 4 ⋅, ⇒, , distance from the Y -axis measured along the, positive X -axis and has zero distance from, the X -axis., (3, 0), , Hence, Assertion is true but Reason is false., , 42. Assertion Since, the point ( h , k ) lies on, 15, 17, , 15, =0, 17, − 60 −20, x=, =, 17 ⋅ 3 17, ,  − 20 15, So, the point of intersection is , , .,  17 17 , ∴ Required distance from the line, 5x − 2y = 0 is,, 20,  15, −5 ×, − 2 , 17 , 17, d =, 25 + 4, −100 30, −, 130, 17, 17, =, =, 29, 17 29, [Q distance of a point p( x1 , y1 ) from the line, | ax1 + by1 + c | , , ax + by + c = 0 is d =, a 2 + b 2 , , 39. Here, A = 3, B = − 4, C1 = 7, and C 2 = 5., Therefore, the required distance is, 7−5, 2, =, d =, 2, 2, 5, 3 + ( −4 ), , 40. The equations of lines 3x + 4 y = 9 and, 6 x + 8 y = 15 may be rewritten as, 3x + 4 y − 9 = 0, 15, and, 3x + 4 y −, =0, 2, Since, the slope of these lines are same and, hence they are parallel to each other., Therefore, the distance between them is given, by,  9 − 15 , , 2 = 3, ,  32 + 4 2 ,  10, , , , x + y + 1 = 0., ⇒, h + k +1 = 0, and, h 2 + k 2 = 25, ⇒, ( −1 − k ) 2 + k 2 = 25, ⇒, ⇒, , 2k 2 + 2k − 24 = 0, k 2 + k − 12 = 0 ⇒ k = −4 or k = 3, [k = 3 rejected as k < 0], ∴, h = −1 − ( −4 ) = 3, Hence, Assertion is false and Reason is true., , 43. Assertion Let required ratio be λ : 1., Then, the coordinates of point which divides, the line joining ( −1, 1) and (5, 7) in,  5λ − 1 7 λ + 1, the ratio λ :1, is , ,, .,  λ +1 λ +1 , But it lies on x + y = 4, 5λ − 1 7 λ + 1, ∴, +, =4, λ +1, λ +1, ⇒, 12λ = 4 λ + 4 ⇒ λ = 1 / 2, ∴ Required ratio = 1 : 2, Hence, Assertion is true and Reason is false., , 44. Assertion Centroid of the triangle is, 1 + 2 + c 2 a + b − 3, ,, G ≡, , 3, 3, , , i.e., ,  3 + c 2 a + b − 3, ,, , , 3, ,  3, , Q G will lie on Y -axis, then, 3 + c2, = 0 ⇒ c 2 = − 3 or c ≡ ± i 3, 3, Q Both values of c are imaginary., Hence, G cannot lie on Y -axis., Reason Q G will lies on X -axis, then, a +b −3, =0, 3, ⇒, a + b − 3 = 0 or a + b = 3, Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion.

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87, , CBSE New Pattern ~ Mathematics XI (Term I), , 45. Assertion Area of the triangle, whose, vertices are (4, 4), (3, –2) and (−3, 16), 1, = |4 ( − 2 − 16 ) + 3 (16 − 4 ) + ( − 3) ( 4 + 2)|, 2, |− 72 + 36 − 18 | | − 54 | 54, =, =, =, = 27, 2, 2, 2, Hence, Assertion and Reason both are true, and Reason is the correct explanation of, Assertion., , 46. Assertion Slope of X -axis is zero and slope, of Y -axis is not defined., Hence, Assertion is true but Reason is false., , 47. Assertion, D (–3, 2), , A (–2, –1), , C (3, 3), , B (4, 0), , Q ABCD is a parallelogram., ∴ AB | | CD ⇒ Slope of AB = Slope of CD, and BC | | AD ⇒ Slope of BC = Slope of AD, Reason Mid-point of,  −2 + 3 −1 + 3, AC = , ,, ,  2, 2 ,  1 2  1 , =  ,  =  , 1,  2 2  2 ,  4 − 3 0 + 2, and mid-point of BD = , ,, ,  2,, 2 , 1 , =  , 1, 2 , ⇒ Mid-point of AC = Mid-point of BD, Hence, Assertion is false and Reason is true., , 48. Assertion Let m 1 and m 2 be the slopes of the, straight lines x + 2y − 3 = 0 and 3x + y + 1 = 0., 1, Then, m 1 = − and m 2 = −3, 2, Let θ be the angle between the given lines.,  m −m 1 , Then, tan θ = ±  2, , 1 + m 1m 2 , 1, ,  −3 + , 2  = ±1, =±, 3,  1+ , , 2 , , ⇒, , θ = tan −1 (1) or π − tan −1 (1), , Hence, Assertion and Reason both are true, and Reason is the correct explanation of, Assertion., , 49. Assertion Given equation 3x − 4 y + 10 = 0, can be written as, 3, 5, …(i), x+, 4, 2, Comparing Eq. (i) with y = mx + c , we have, 3, slope of the given line as m = ., 4, Reason Equation 3x − 4 y + 10 = 0 can be, written as, x, y, …(ii), 3x − 4 y = − 10 or, + =1, 10, 5, −, 3, 2, x y, Comparing Eq. (ii) with + = 1, we have, a b, 10, 5, and y-intercept as b = ., x-intercept as a = −, 3, 2, Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion., y=, , 50. Assertion Since, slope of line, x cos θ + y sin θ = 2 is − cot θ and slope of line, x − y = 3 is 1., Also, these lines are perpendicular to each, other., ∴ ( − cot θ ) (1) = − 1, π, π, ⇒, cot θ = 1 = cot, ⇒θ =, 4, 4, Reason Condition of perpendicularity of two, lines is m 1 ⋅m 2 = − 1., Hence, Assertion is true and Reason is false., , 51. Assertion Given equation is x + 7 y = 0, −x, +0, 7, On comparing with y = mx + c , we get, −1, Slope (m ) = , y-intercept = 0, 7, Reason Given equation is 6 x + 3y − 5 = 0, 5, ⇒, y = − 2x +, 3, On comparing with y = mx + c , we get, 5, Slope (m ) = − 2, y-intercept =, 3, Hence, Assertion is false and Reason is true., ⇒, , y=

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88, , CBSE New Pattern ~ Mathematics XI (Term I), , 2y − 184 = x − 1985, , 52. Assertion Given equation of line is, x −y=4, On dividing above equation by, (coefficient of x ) 2 + (coefficient of y ) 2, i.e., , ⇒, , (1) 2 + ( −1) 2 = 2, we get, 1, 1, 4, x−, y=, 2, 2, 2, cos 45° x − sin 45° y = 2 2, , [Q cos x is positive and sin x is negative,, it is possible only in fourth quadrant], ⇒ x cos ( 360 ° − 45° ), + y sin ( 360 ° − 45° ) = 2 2, , Q cos ( 360 ° − θ ) = cos θ,  and sin ( 360 ° − θ ) = − sin θ , , , ⇒, , x cos 315° + y sin 315° = 2 2, , On comparing with x cos α + y sin α = p,, we get, α = 315°, and, p=2 2, Hence, Assertion is true but Reason is false., , 53. Assertion Given lines are 4 x + 3y = 11, and 4 x + 3y = 15/ 2., Distance between them, =, , =, , 15, 2, 16 + 9, , 11 −, , 7, 7, =, 2×5, 10, , Hence, Assertion and Reason both are true, and Reason is the correct explanation of, Assertion., y −y, 97 − 92, 54. (i) Slope of line AB = 2 1 =, x 2 − x1 1995 − 1985, , x − 2y = 1801, (iii) Let the population in year 2010 is P ., Since, A, B, C are collinear, ∴ Slope of AB = slope of BC, 97 − 92, P − 97, =, 1995 − 1985 2010 − 1995, ⇒, , 1 P − 97, =, 2, 15, , ⇒, , 7.5 = P − 97, , ⇒, , P = 97 + 7.5, = 104.5 crores, 1, (iv) Q Slope of AB =, 2, Slope of line perpendicular to AB, −1, =, =−2, 1, 2, ∴ Equation of line perpendicular to AB, passing through (1995, 97) is, y − 97 = − 2( x − 1995), ⇒, y − 97 = − 2x + 3990, ⇒, 2x + y = 4087, (v) Equation of line AB is, x − 2y = 1801, Putting, y = 110, ∴, x = 1801 + 220, ⇒, x = 2021, ∴Population becomes 110 crores in 2021., , 55. Let the point on Rani, Mansi and Sneha, stand on a vertices of triangle be A, B, C., ∴ A( 2, − 2), B(1, 1), C( − 1, 0 ), A (2, –2), , 5 1, =, 10 2, (ii) Equation of line AB is, =, , ∴, , y − y1 = m ( x − x1 ), 1, y − 92 = ( x − 1985), 2, , E, , B (1, 1), , D (0, 1/2), , C (–1, 0)

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89, , CBSE New Pattern ~ Mathematics XI (Term I), , (i) The equation of line AB is, − 2 −1, y −1 =, ( x − 1), 2 −1, , , y 2 − y1, Q y − y1 = x − x ( x − x1 ) , , , 2, 1, ⇒, , y − 1 = − 3x + 3 ⇒ 3x + y = 4, , (ii) Slope of equation of line AC is, 0+2, 2, m =, =, −1 − 2 − 3, (iii) Let D be the mid-point of BC., 1 − 1 0 + 1  1 , Coordinates of D are , ,,  =  0, ,  2, 2   2, 1, +2, ∴ Equation of AD is y + 2 = 2, ( x − 2), 0−2, −5, y + 2=, ( x − 2), ⇒, 4, ⇒, 4 y + 8 = − 5x + 10, ⇒, 5x + 4 y = 2, −2, (iv) Slope of AC =, 3, 3, ∴ Slope of BE =, [Q BE ⊥ AC ], 2, Equation of altitude through B is, 3, y − 1 = ( x − 1) ⇒ 3x − 2y = 1, 2, 0 −1, 1, (v) Slope of line BC =, =, −1 −1 2, Equation of line passing through A and, parallel to BC is, 1, y + 2 = ( x − 2), 2, ⇒, 2y + 4 = x − 2, ⇒, x − 2y = 6, , 56. (i) AC = ( − 1 − 1) 2 + ( − 2 − 4 ) 2, [using distance formula], = ( − 2) + ( − 6 ) 2, 2, , = 4 + 36 = 40 units, (ii) Slope of BC =, , y 2 − y1 − 2 − ( − 3), =, x 2 − x1, −1 − 2, , −2+ 3, 1, =, =−, −3, 3, , (iii) Since D is the mid-point of BC.,  2 − 1 − 3 − 2, ,, ∴ Coordinates of D are , ,  2, 2 , 5, 1, = ,− , 2, 2, 5, 13, − −4 −, 2, ∴ Slope of AD =, = 2 = 13, 1, 1, −1, −, 2, 2, ∴ Equation of the median AD is, y − 4 = 13( x − 1), ⇒, 13x − y − 9 = 0, (iv) Since AM is the altitude through A., 1, 1, =−, =3, ∴ Slope of AM = −, 1, slope of BC, −, 3, ∴ Equation of the altitude through A is, given by, y − 4 = 3( x − 1), ⇒, y − 4 = 3x − 3 ⇒ 3x − y + 1 = 0, (v) Equation of the right bisector of BC is a line, which passes through D and having slope, is 3., 1,  5, , y −  −  = 3 x − , ∴,  2, , 2, ⇒, , y+, , 5, 3, = 3x − ⇒ 3x − y − 4 = 0, 2, 2, , 57. (i) We have the positions,, Shubham B (1, 2) and Rajkumar D (5, 4), 4−2 2 1, Slope, m 1 =, = =, 5 −1 4 2, 1, 2, Equation of line joining points B and D is, , Taking point (1, 2) = ( x1 , y1 ) and m 1 =, ( y − y1 ) = m 1 ( x − x1 ), 1, ( x − 1), 2, ⇒, 2y − 4 = x − 1, ⇒ x − 2y + 3 = 0, (ii) We have the positions,, ⇒, , ( y − 2) =, , Rishabh A (1, 4) and Vikram C (5, 2), Slope, m 2 =, , 2 − 4 −2, 1, =, =−, 5 −1, 4, 2

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90, , CBSE New Pattern ~ Mathematics XI (Term I), , Taking point ( x1 , y1 ) = (1, 4) and m 2 = −, , 1, 2, , Equation of line is, 1, ( y − 4 ) = − ( x − 1), 2, ⇒, 2y − 8 = − x + 1 ⇒ x + 2y − 9 = 0, (iii) We have, x − 2y + 3 = 0, …(i), and, …(ii), x + 2y − 9 = 0, On adding Eqs. (i) and (ii), we get, 2x − 6 = 0 ⇒ x = 3, [From Eq. (ii)], ∴, 3 + 2y − 9 = 0, ⇒, 2y = 6 ⇒ y = 3, Hence, point of intersection is (3, 3)., (iv) We have positions,, Rishabh A (1, 4) and Rajkumar D (5, 4), 4−4, Slope AD =, =0, 5 −1, Hence, slope is zero., (v) The line formed by Rishabh-Rajkumar is, opposite and parallel to the line formed by, Shubham and Vikram. Hence, first pair, have same slope., , 58. (i) Given positions of person A and B are as, follows, A( 2, − 3) and B( 6, − 5), d = ( 6 − 2) 2 + ( − 5 + 3) 2, [using distance formula], = ( 4 ) 2 + ( −2) 2 = 16 + 4 = 20 = 2 5, (ii) We have, A( 2, − 3) and B( 6, − 5), − 5 − ( − 3) − 5 + 3 −2, 1, Slope, m =, =, =−, =, 4, 2, 6−2, 4, , Taking point A ( 2, − 3) = ( x1 , y1 ) and m = −, , 1, 2, , Equation of line AB is, 1, ( y − ( − 3)) = − ( x − 2), 2, ⇒, 2( y + 3) = − ( x − 2), ⇒, 2y + 6 = − x + 2 ⇒ x + 2y + 4 = 0, (iii) Let point C divides AB in the ratio m 1 and, m 2., A, (2, –3), , 1:3, C, (x, y), , B, (6, –5), , m x + m 2 x1 m 1 y 2 + m 2 y1 , Then, ( x , y ) =  1 2, ,, , m1 + m 2 ,  m1 + m 2, 1 × 6 + 3 × 2 1 × ( − 5) + 3 ( − 3) , ,, =, , 1+ 3, 1+ 3, , , 7, 12 −14  , = ,,  =  3, − , 4 4  , 2, 7, , (iv) We have, A( 2, − 3) and C  3, − , , 2, 2, ,  7, , AC = ( 3 − 2) 2 +  − + 3,  2, ,  1, = 12 +  − ,  2, , 2, , = 1+, , 1, 5, =, 4, 2, , 7, , (v) We have, C  3, −  and B ( 6, − 5), , 2, 7, , CB = ( 6 − 3) 2 +  − 5 + , , 2,  3, = 32 +  − ,  2, , 2, , = 9+, , 2, , 9, =, 4, , 45 3 5, =, 4, 2

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CBSE New Pattern ~ Mathematics XI (Term I), , 91, , 06, Limits, Quick Revision, If x approches a i.e. x → a , then f ( x ) approaches l, i.e. f ( x ) → l , where l is a real number, then l is, called limit of the function f ( x ). In symbolic form,, it can be written as lim f ( x ) = l ., x→a, , Left Hand and Right Hand Limits, If values of the function at the point which are very, near to a on the left tends to a definite unique, number as x tends to a, then the unique number so, obtained is called the Left Hand Limit (LHL) of, f ( x ) at x = a , we write it as, f (a − 0 ) = lim f ( x ) = lim f (a − h ), x →a−, , h→0, , Limits of a Polynomial Function, A function f is said to be a polynomial function if, f ( x ) is zero function, or if f ( x ) = a 0 + a1 x + a 2 x 2 + … + an x n ,, where ai ’s are real number such that an ≠ 0., Then, limit of polynomial functions is, f ( x ) = lim f ( x ) = lim [a 0 + a1 x + a 2 x 2 + ... + an x n ], x→a, , x→a, , = a 0 + a1a + a 2a 2 + ... + an a n = f (a ), , Limits of Rational Functions, , Similarly, Right Hand Limit (RHL) is, f (a + 0 ) = lim f ( x ) = lim f (a + h ), , A function f is said to be a rational function, if, g (x ), , where g ( x ) and h ( x ) are polynomial, f (x ) =, h (x ), , Existence of Limit, , functions such that h ( x ) ≠ 0., , x →a+, , h→0, , If the right hand limit and left hand limit coincide, (i.e. same), then we say that limit exists and their, common value is called the limit of f ( x ) at x = a, and denoted it by lim f ( x )., x →a, , Algebra of Limits, Let f and g be two functions such that both, lim f ( x ) and lim g ( x ) exist, then, x→a, , x→a, , (i) lim [ f ( x ) ± g ( x )] = lim f ( x ) ± lim g ( x ), x→a, , x→a, , x→a, , (ii) lim kf ( x ) = k lim f ( x ), x→a, , x→a, , (iii) lim f ( x ) ⋅ g ( x ) = lim f ( x ) × lim g ( x ), x→a, , (iv) lim, , x→a, , x→a, , x→a, , lim f ( x ), f (x ) x → a, , where lim g ( x ) ≠ 0, =, x→a, g (x ), lim g ( x ), x→a, , Then, lim f ( x ) = lim, x→a, , x→a, , lim g ( x ), g (x ) x → a, g (a ), =, =, h( x ), lim h ( x ) h (a ), x→a, , However, if h (a ) = 0, then there are two cases arise,, (i) g (a ) ≠ 0, (ii) g (a ) = 0., In the first case, we say that the limit does not exist., In the second case, we can find limit., Limit of a rational function can be find with the help of, following methods, 1. Direct Substitution Method In this method,, we substitute the point, to which the variable, tends to in the given limit. If it give us a real, number, then the number so obtained is the, limit of the function and if it does not give us a, real number, then use other methods.

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92, , CBSE New Pattern ~ Mathematics XI (Term I), , 2. Factorisation Method Let lim, , x→a, , f (x ), reduces, g (x ), , 0, to the form , when we substitute x = a . Then,, 0, we factorise f ( x ) and g ( x ) and then cancel, out the common factor to evaluate the limit., 0, 3. Rationalisation Method If we get form and, 0, numerator or denominator or both have, radical sign, then we rationalise the numerator, or denominator or both by multiplying their, 0, conjugate to remove form and then find, 0, limit by direct substitution method., , Some Standard Limits, xn − an, (i) lim, = na n − 1, x→a x − a, (iii) lim, , x→ 0, , 1 – cos x, =0, x, , ax −1, = loge a, x→0, x, , (v) lim, (vii) lim, , x→0, , (ii) lim, , x→0, , (iv) lim, , x→0, , sin x, =1, x, , tan x, =1, x, , ex −1, =1, x→0, x, , (vi) lim, , log (1 + x ), =1, x, , log ( 1 − x ), =1, x→0, −x, , (viii) lim, , Objective Questions, Multiple Choice Questions, 1. The value of lim ( 4x 3 − 2x 2 − x + 1) is, x→ 3, , equal to, (a) 40, (c) 38, , (b) 20, (d) 88, , 1 + x , if 0 ≤ x ≤ 1, , then right, 2, 2 − x , if x > 1, 2, , 2. If f (x ) = , , hand limit of f (x ) at x = 1 is equal to, (a) 1, (c) 3, , (b) 2, (d) 4, , 2x + 3, if x ≤ 2, , then the left, 3. If f (x ) = , x + 5, if x > 2, hand limit of f (x ) at x = 2 is equal to, (a) 6, (c) 9, , (b) 7, (d) 8, , | x − 3 |, , , x≠3, 4. If f (x ) =  x − 3, , then left,  0,, x=3, hand limit of f (x ) at x = 3 is equal to, (a) 1, (c) 2, , (b) −1, (d) 0, , 5. The value of lim, , x→ 0, , 2+x + 2−x, is, 2+x, , equal to, (a) 2, (c) 2 2, , (b), (d), , 6. The value of lim, , x→1, , x −4, is equal to, 3 − 13 − x, , (a) 3 + 2 3, (c) 2 + 3, , 7. lim, , (b) 3 − 2 3, (d) 2 − 3, , ( x − 1)( 2x − 3), , x→1, , 2, 3, , 2x 2 + x − 3, , is equal to, [NCERT Exemplar], , −1, 10, (d) None of these, , 1, (a), 10, (c) 1, , (b), ,  4 x 2 − 1,  is equal to,  2x − 1 , , 8. The value of lim , x → 1/ 2, (a) 1, (c) 3, , (b) 2, (d) 4, , 9. The value of lim, , x→ 2, , (a) 10, (c) 12, , x3 −8, is equal to, (x − 2), (b) 11, (d) 13

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CBSE New Pattern ~ Mathematics XI (Term I), , 1 ,  2, +,  is, x → 1 1 − x 2, x − 1, , sin 7x, is equal to, x → 0 tan 5x, , 10. The value of lim , , 17. The value of lim, 5, 7, 2, (c), 7, , equal to, 1, (b), 3, , 11. The value of lim, , 1, x→, 2, , 4x − 1, is equal to, 2x − 1, , (a) 1, (c) 3, , (a) 0, (c) 2, , 2, 3, 3, (d), 4, (b), , x −2, n, , x→ 2, , n, , x−2, , ( x + 2)1/ 3 − 21/ 3, , x→ 0, , x, , is, , 1, (b), 3 (2)2 / 3, 1, (d), 3 3, ,  1 + x − 1,  is equal, x→ 0, x, , , , 16. The value of lim , to, , 1, 2, 1, (d), 5, (b), , x, , is equal to, , (a) 0, (c) 2, , (b) 1, (d) 3, , sin x, is equal to, x → 0 x (1 + cos x ), , 21. lim, , 1, 2, (d) −1, , (b), , 22. lim, , sin( 2 + x ) − sin( 2 − x ), , is equal to p, x, cos q, where p and q are respectively, x→ 0, , (a) 1 , 2, (c) 1, 1, , equal to, 1, (a), 3 (2) 3/2, 1, (c), 2 (3)2 / 3, , 1 − cos 4x, , (b) 3, (d) −1, , (c) 1, , (b) 3, (d) 7, , 15. The value of lim, , (a) 1, (c) 2, , (a) 0, , = 80, then n is equal to, , (a) 1, (c) 5, , (b) 2, (d) 4, , x→ 0, ,  x 15 − 1, 13. The value of lim  10,  is equal to, x → 1 x, − 1, , 3, 2, 4, (c), 3, , sin 4x, is equal to, sin 2x, , (a) 1, (c) 3, , 20. lim, , (b) 1, (d) 3, , (a), , x→0, , 19. lim, , x, is equal to, 1+ x +1, , x→ 0, , 18. The value of lim, , tan x °, is equal to, x→ 0 x °, , (b) 2, (d) 4, , 12. The value of lim, , 1, 3, 1, (c), 4, , (b), , (d) 1, 2, , (a), , 7, 5, 7, (d), 2, , (a), , 1, (a), 2, 1, (c), 4, , 14. If lim, , 93, , 23. lim, , x→ 0, , (b) 2, 1, (d) 2, 2, , tan x − sin x, sin 3 x, , is equal to, , 1, 2, (c) 1, , (b) 0, , (a), , 24. lim, , x→ 0, , (a) 2, , (d) Not defined, , tan 2x − x, is equal to, 3x − sin x, [NCERT Exemplar], (b), , 1, 2, , (c), , 25. The value of limπ, x→, , (a) 1, (c) 4, , 2, , −1, 2, , (d), , 1, 4, , tan 2x, is equal to, π, x−, 2, (b) 3, (d) 2

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94, , CBSE New Pattern ~ Mathematics XI (Term I), , 26. lim, , x → π /4, , sec 2 x − 2, is, tan x − 1, , (a) 3, (c) 0, x→ 0, , (a), , [NCERT Exemplar], , (b) 1, (d) 2, , cosec x − cot x, , 27. lim, , 35. lim, , x, , −1, 2, , is equal to, (c), , 1, 2, , (d) 1, , x 2 cos x, is equal to, x → 0 1 − cos x, [NCERT Exemplar], , 28. lim, , (a) 2, (c), , (b), , −3, 2, , 3, 2, , (d) 1, , e 3x − 1, is equal to, 29. lim, x→ 0, x, (a) 1, (c) 3, , (b) 2, (d) 4, , ex −e3, is equal to, x−3, , 30. lim, , x→ 3, , (b) e2, (d) e 4, , (a) e, (c) e 3, , 31. lim, , x→ 0, , e sin x − 1, is equal to, x, , (a) 1, (c) 3, , 32. lim, , (b) 2, (d) 4, , e x + e −x − 2, x2, , x→ 0, , (a) 1, (c) 3, , 33. lim, , is equal to, (b) 2, (d) 4, , 3x − 2x, , x→ 0, , 3, (a) log, 2, 1, (c) log, 2, , x, , is equal to, 2, (b) log, 3, 1, (d) log, 3, , 2x − 1, 34. lim, is equal to, x→ 0 1 + x − 1, (a) log2, (c) 3log 2, , (b) 2 log 2, (d) 4 log 2, , x→ 0, , (a) 1, , x, , is equal to, , (b) 2, , (c) 3, , (d) 4, , Assertion-Reasoning MCQs, , [NCERT Exemplar], , (b) 1, , log e (1 + 2x ), , Directions (Q. Nos. 36-50) Each of these, questions contains two statements, Assertion (A) and Reason (R). Each of the, questions has four alternative choices, any, one of the which is the correct answer. You, have to select one of the codes (a), (b), (c) and, (d) given below., (a) A is true, R is true; R is a correct, explanation of A., (b) A is true, R is true; R is not a correct, explanation of A., (c) A is true; R is false, (d) A is false; R is true., , 36. Assertion (A) lim, , ax 2 + bx + c, , is, cx 2 + bx + a, equal to 1, where a + b + c ≠ 0., 1 1, +, x, 2 is equal to 1 ., Reason (R) lim, x → −2 x + 2, 4, x →1, , sin ax, a, is equal to ., x→0, bx, b, sin x, Reason (R) lim, = 1., x→0, x, sin ax + bx, 38. Assertion (A) lim, is equal, x → 0 ax + sin bx, to −2., Reason (R) lim (5x 3 + 5x + 1) is equal, , 37. Assertion (A) lim, , to 11., , x →1, , 39. Assertion (A) lim, to π., , x→π, , Reason (R) lim, , x→0, , sin( π − x ), π(π − x ), , is equal, , cos x, 1, is equal to ., π, π −x, , a, sin ax, is equal to ., b, sin bx, sin x, Reason (R) lim, =1., x→0, x, , 40. Assertion (A) lim, , x→0

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CBSE New Pattern ~ Mathematics XI (Term I), , 41. Assertion (A) lim, , x→0, , cos 2x − 1, cos x − 1, , is equal, , to 4., tan x, = 1., x→0 x, ax + x cos x, 42. Assertion (A) lim, is equal, x→0, b sin x, a +1, ., to, b, Reason (R) lim x sec x is equal to 1., Reason (R) lim, , x→0, , e 3 + x − sin x − e 3, is, x→0, x, , 43. Assertion (A) lim, equal to e 3 + 1., , tan 4x, is equal to 2., x → 0 sin 2x, , Reason (R) lim, , e x − e −x, is equal, x→0, x, , 44. Assertion (A) lim, to 2., , e x −1, = 1., x→0, x, , Reason (R) lim, , e tan x − 1, 45. Assertion (A) lim, is equal, x→0, x, to 1.,  e 4 x − 1,  is equal to 2., Reason (R) lim , x→0, x , , 46. Assertion (A) lim, to −1., , x→0, , e, , −e, is equal, x − sin x, x, , sin x, , 95, 3x − 2x, is equal to, x → 0 tan x, , 48. Assertion (A) lim,  3, log  .,  2, , Reason (R) lim, , log (1 + x ), , is equal to 2., tan x, x −1, 49. Assertion (A) lim, is equal to 1., x → 1 log e x, x→0, , Reason (R) lim, , log (sin x + 1), , x→0, , x, , is equal, , to 0., 32 +x − 9, is equal to, x→0, x, , 50. Assertion (A) lim, 9 log 2., , a sin x − 1, is equal to, x → 0 sin x, , Reason (R) lim, log a., , Case Based MCQs, 51. Raj was learning limit of a polynomial, function from his tutor Rajesh., His tutor told that a function f is said to, be a polynomial function, if f (x ) is zero, function., , Limit of a, Polynomial Function, ,  32 x − 23 x ,  is equal to, Reason (R) lim , x→0, x, , 9, log  ., 8, , 47. Assertion (A) lim, 1, ., to, 2, , x→0, , x −1, is equal to 1., x −1, 2, , Reason (R) lim, , x →1, , e x −1, is equal, 1 − cos 2x, , Now, let, f (x ) = a 0 + a1x + a 2 x 2 + ... + an x n be a, polynomial function, where ai′ s are real, numbers and an ≠ 0 .

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96, , CBSE New Pattern ~ Mathematics XI (Term I), , Then, limit of a polynomial function f (x ), = lim f (x ), x→a, , = lim [a 0 + a1x + a 2 x 2 + ... + an x n ], x→a, , = lim a 0 + lim a1 x + lim a 2 x 2, x→a, , x→a, , x→a, , + ... + lim an x n, x→a, , = a 0 + a1 lim x + a 2 lim x 2, x→a, , x→a, , + ... + an lim x n, x→a, , = a 0 + a1a + a 2 a 2 + ...+ an a n = f (a ), , Based on above information, answer the, following questions., (i) lim (1 + x + x 2 + ... x 9 ) is equal to, x → −1, , (a) 0, (c) 2, , (b) 1, (d) 3, , (iii) lim (x 3 + x 2 + x − 1) is equal to, x→ 2, , (b) 11, (d) 13, , x → −3, , (b) −28, (d) −15, , (b) 180, (d) 165, , 52. A function f is said to be a rational, g (x ), , where g (x ), h (x ), and h (x ) are polynomial functions such, that h (x ) ≠ 0., g (x ), Then, lim f (x ) = lim, x→a, x → a h (x ), function, if f (x ) =, , lim g (x ), lim h (x ), , x→a, , =, , g (a ), h (a ), , is equal to, , 6, 5, 3, (d), 4, , (b), , , , x2 − 4, (iii) The value of lim  3,  is, x → 2 x − 4x 2 + 4x, , , (b) 1, (d) Does not exist, , x 7 − 2x 5 + 1, − 3x 2 + 2, , (a) 0, (c) 2, , x→4, , =, , (x 4 + 1) 2, , x→1x 3, , (v) lim (x − x ) is equal to, , x→a, , (x −1) 2 + 3x 2, , 7, 4, 4, (c), 7, , (iv) lim, , 3, , (a) 192, (c) 50, , (c) 2, , (a) 0, (c) 2, , (iv) lim (x 3 + x + 2) is equal to, , 4, , (b), , (a), , (b) 100, (d) 125, , (a) 28, (c) 30, , −1, 2, 3, (d), 2, , 1, 2, , x → −1, , x→ 5, , (a) 9, (c) 10, , (a), , (ii) lim, , (ii) lim [x 2 (x − 1)] is equal to, (a) 10, (c) 25, , However, if h (a ) = 0, then there are two, cases arise,, (i) g (a ) ≠ 0, (ii) g (a ) = 0., In the first case, we say that the limit, does not exist., In the second case, we can find limit., Based on above information, answer the, following questions.,  x 10 + x 5 + 1,  is equal to, (i) lim , , x → −1, x −1, , , (v) lim, , x→ 0, , (a) 1, (c) −1, , is equal to, (b) 1, (d) 3, , 1 + x 3 − 1− x 3, x2, , is equal to, , (b) 0, (d) 2, , 53. The great Swiss Mathematician, Leonhard Euler (1707-1783) introduced, the number e, whose value lies between, 2 and 3. This number is useful in, defining exponential function., A function of the form of f (x ) = e x is, called exponential function.

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CBSE New Pattern ~ Mathematics XI (Term I), , The graph of the function is given below, Y, , X′, , f (x)=e x, X, , O, Y′, , (i) Domain of f (x ) = ( −∞, ∞ ), (ii) Range of f (x ) = (0, ∞ ), To find the limit of a function involving, exponential function, we use the, following theorem, Theorem lim, , e x −1, , x→ 0, , x, , 97, , 54. To find the limits of trigonometric, functions, we use the following, theorems, Theorem 1 Let f and g be two real, valued functions with the same domain, such that f (x ) ≤ g (x ) for all x in the, domain of definition. For some real, number a, if both lim f (x ) and lim g (x ), x →a, x →a, exist, then, lim f (x ) ≤ lim g (x )., , x →a, , x →a, , This is shown in the figure, Y, , =1, , y=g(x), y=f(x), , Based on above information, answer the, following questions., (i) lim, , e x −e 4, x−4, , x→4, , O, , is equal to, 2, , (a) e, (c) e 3, , (ii) lim, , (b) e, (d) e 4, , e kx − 1, , x→ 0, , x, , is equal to, , k, 2, (c) − k, , (a), , (b) k, (d) 1, ,  e − x − 1,  is equal to, (iii) lim , x→ 0, x , (a) 1, (c) 0, ,  e 5x − e 4x, (iv) lim , x → 0, x, (a) 1, (c) 3, , X, , Theorem 2 (Sandwich theorem) Let, f , g and h be real functions such that, f (x ) ≤ g (x ) ≤ h(x ) for all x in the, common domain of definition. For some, real number a, if, lim f (x ) = l = lim h(x ), then, , x →a, , lim g (x ) = l ., , x →a, , x →a, , This is shown in the figure, , (b) −1, (d) 2, , ,  is equal to, , , (b) 2, (d) 4, ,  2e x − 3x − 2 ,  is equal to, (v) lim , x→ 0, x, , (a) −1, (c) 1, , a, , (b) 0, (d) 2, , Theorem 3 Three important limits are, sin x, (i) lim, =1, x→ 0, x, 1 − cos x, (ii) lim, =0, x→ 0, x, tan x, (iii) lim, =1, x→ 0, x

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98, , CBSE New Pattern ~ Mathematics XI (Term I), , Based on above information, answer the, following questions., sin 3x, (i) lim, is equal to, x → 0 5x, 1, 5, 3, (c), 5, , 2, 5, 4, (d), 5, (b), , (a), , (ii) lim, , tan(θ − b ), θ −b, , θ→b, , (a) 0, (c) 2, , (iii) lim, , (b) 1, (d) 3, , tan 2x − sin 2x, , x→ 0, , x, , 3, , (a) 4, (c) 2, , (iv) lim, , 2 sin x − sin 2x, x3, , (a) 0, (c) 2, π, x→, 4, , is equal to, , (b) 3, (d) 1, , x→ 0, , (v) lim, , is equal to, , is equal to, , (b) 1, (d) 3, , sin x − cos x, is equal to, π, x−, 4, (b) 3, (d) 3, , (a) 2, (c) 1, , 55. The logarithmic function expressed as, , log e R + → R and given by log e x = y, iff e y = x ., The graph of the function is given below, Y, f (x)=loge x, X′, , O, , Y′, , (1, 0), , X, , (i) Domain of f (x ) = (0, ∞ ) or R +, (ii) Range of f (x ) = ( −∞, ∞ ) or R, To find the limit of functions involving, logarithmic function, we use the, following theorem, log e (1 + x ), Theorem lim, =1, x→ 0, x, Based on above information, answer the, following questions., log e (1 + 5x ), is equal to, (i) lim, x→ 0, x, (a) 5, (c) 3, , (b) 4, (d) 1, , log e (1 + 6x ) − 5x 2, (ii) lim, is equal to, x→ 0, x, (a) 1, (c) 3, , (iii) lim, , x→ 0, , (b) 2, (d) 6, , 1+ x −1, log(1 + x ), , is equal to, 1, 2, 3, (d), 2, , (a) 1, (c), , (b), , 1, 3, , (iv) lim, , x→ 5, , 1, 5, 1, (c), 4, , (a), , log x − log 5, x −5, , is equal to, 3, 5, 2, (d), 3, (b), , log(5 + x ) − log(5 − x ), is equal to, x→ 0, x, , (v) lim, , 1, 5, 3, (c), 5, (a), , 2, 5, 4, (d), 5, (b)

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CBSE New Pattern ~ Mathematics XI (Term I), , 99, , ANSWERS, Multiple Choice Questions, 1. (d), 11. (b), 21. (b), , 2. (a), 12. (a), 22. (d), , 3. (b), 13. (a), 23. (a), , 4. (b), 14. (c), 24. (b), , 5. (b), 15. (b), 25. (d), , 31. (a), , 32. (a), , 33. (a), , 34. (b), , 35. (b), , 39. (d), 49. (c), , 40. (a), 50. (d), , 6. (a), 16. (b), 26. (d), , 7. (b), 17. (b), 27. (c), , 8. (b), 18. (b), 28. (a), , 9. (c), 19. (a), 29. (c), , 10. (a), 20. (a), , 41. (b), , 42. (c), , 43. (d), , 44. (a), , 45. (c), , 30. (c), , Assertion-Reasoning MCQs, 36. (c), 46. (d), , 37. (a), 47. (c), , 38. (d), 48. (c), , Case Based MCQs, 51. (i) - (a); (ii) - (b); (iii) - (d); (iv) - (b); (v) - (a), 53. (i) - (d); (ii) - (b); (iii) - (b); (iv) - (a); (v) - (a), 55. (i) - (a); (ii) - (d); (iii) - (b); (iv) - (a); (v) - (b), , 52. (i) - (b); (ii) - (a); (iii) - (d); (iv) - (b); (v) - (b), 54. (i) - (c); (ii) - (b); (iii) - (a); (iv) - (b); (v) - (a), , SOLUTIONS, | x − 3 |, , 1. lim ( 4 x 3 − 2x 2 − x + 1), x→3, , = 4 lim x − 2 lim x − lim x + lim 1, 3, , x→3, , 2, , x→3, , x→3, , x→3, , = 4 ( 3) 3 − 2 ( 3) 2 − 3 + 1, = 108 − 18 − 2 = 88, 1 + x 2 , if 0 ≤ x ≤ 1, 2. We have, f ( x ) = , 2, 2 − x , if x > 1, RHL = lim f ( x ) = lim ( 2 − x 2 ), x →1+, , x →1+, , [Q f ( x ) = 2 − x 2 , if x > 1], = lim [ 2 − (1 + h ) ], 2, , h→ 0, , [putting x = 1 + h and when x → 1+ ,, then h → 0], = 2 −1 =1, 2x + 3, if x ≤ 2, 3. Given, f ( x ) = , x + 5, if x > 2, LHL = lim f ( x ) = lim 2x + 3, x→2, , −, , x→2, , −, , [Q f ( x ) = 2x + 3, if x ≤ 2], = lim [ 2 ( 2 − h ) + 3] = 2 ( 2 − 0 ) + 3, h→ 0, , [ putting x = 2 − h and when x → 2−,, then h → 0 ], = 4 + 3=7, , 4. Given, f ( x ) =  x − 3, , ,, , x ≠3, ,  0,, x =3, ∴ Left hand limit at x = 3 is, | x − 3|, lim f ( x ) = lim, x → 3−, x → 3− x − 3, , ...(i), , On putting x = 3 − h and changing the limit, x → 3− by h → 0 in Eq. (i), we get, | x − 3|, | −h |, lim f ( x ) = lim, = lim, −, −, h, →, 0, x −3, −h, x→ 3, x→3, h, [ Q| x | = x ], ⇒, lim f ( x ) = lim, h → 0 ( −h ), x → 3−, = −1, 2+ x + 2− x, 5. lim, =, x→ 0, 2+ x, =, , 6. lim, , x→1, , 2+ 0 + 2− 0, 2+ 0, 2+ 2 2 2, =, = 2, 2, 2, , −3, x −4, =, 3 − 13 − x 3 − 12, −3, −3 ( 3 + 2 3 ), =, =, 3 − 2 3 ( 3 − 2 3 )( 3 + 2 3 ), −3 ( 3 + 2 3 ), 9 − 12, −3 ( 3 + 2 3 ), = 3+ 2 3, =, −3, =

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100, , CBSE New Pattern ~ Mathematics XI (Term I), , ( x − 1)( 2x − 3), x→1, 2x 2 + x − 3, = lim, , x→1, , 2 − (1 + x ), 0, , 2,  0 form , 1− x, 1− x, 1, 1, = lim, = ., = lim, x →11− x2, x →11+ x, 2, 0, 1, 11. On putting x = , we get the form ., 0, 2, So, let us first factorise it., Consider,, ( 2x + 1) ( 2x − 1), 4x 2 − 1, lim, = lim, 1 2x − 1, 1, ( 2x − 1), x→, x→, = lim, , 7. Given, lim, , x→1, , ( x − 1)( 2x − 3), ( 2x + 3) ( x − 1), , ( x − 1)( 2x − 3), x → 1 ( 2x + 3)( x − 1)( x + 1), 2x − 3, −1, −1, = lim, =, =, x → 1 ( 2x + 3)( x + 1), 5 × 2 10, = lim, , 4 x2 − 1, x → 1 / 2 2x − 1, , 8. Given, lim, , 2, , 2, , [using factorisation method], = lim ( 2x + 1), , ( 2x ) 2 − (1) 2, x → 1/2, 2x − 1, ( 2x + 1) ( 2x − 1), = lim, x → 1/2, ( 2x − 1), = lim, , x→, ,  1, = 2  +1= 2,  2, , = lim ( 2x + 1), x → 1/2, , 12. lim, , 1, = 2 × + 1 =1 + 1 = 2, 2, x3 −8, 9. We have, L = lim, ., x→ 2 x − 2, 3, Let f ( x ) = x − 8 and g ( x ) = x − 2, , x→0, , x→ 2, , = lim, , x→ 2, , lim g ( x ) = lim x − 2 = 2 − 2 = 0, , and, , x→ 2, , x→ 2, , 0, Thus, we get form., 0, Now, factorise f ( x ) and g ( x ) such that ( x − 2), is a common factor., Here, f ( x ) = x 3 − 8 = ( x 3 − 23 ), = ( x − 2) ( x 2 + 4 + 2x ), and, , 1 + x −1, x, x, = lim, ×, 1+ x +1 x→0 1+ x +1, 1 + x −1, , [multiplying numerator and denominator by, 1 + x − x], , lim f ( x ) = lim x 3 − 8 = 23 − 8 = 0, , Here,, , x→0, , x ( 1 + x − 1), (1 + x ) − (1) 2, , x ( 1 + x − 1), = lim ( 1 + x − 1), x→0, x, [ cancel out x from numerator and denominator ], [put x = 0], = 1 + 0 −1 =1 −1 = 0, x→0, ,  x 15 − 1 x 10 − 1 , x 15 − 1, = lim , ÷, , 10, x→1 x, x −1 , − 1 x→1  x − 1, , 13. Given, lim, ,  x 10 − 1 ,  x 15 − 1 , ÷ lim , = lim , , , x→1,  x − 1  x→1  x − 1 , , ( x − 2) ( x 2 + 4 + 2x ), x→2, ( x − 2), , L = lim, , On cancelling the common factor ( x − 2),, we get, L = lim ( x 2 + 4 + 2x ) = ( 2) 2 + 4 + 2 ( 2), x→2, , = 4 + 4 + 4 = 12, x3 −8, Hence, lim, = 12, x→ 2 x − 2, , [Q ( a + b ) ( a − b ) = a 2 − b 2 ], , = lim, , g(x) = x − 2, , ∴, , 1, 2, , = 15(1)14 ÷ 10(1) 9 = 15 ÷ 10 =, , 14. Given,, ⇒, , 3, 2, , x n − 2n, = 80, x→2 x − 2, lim, , n ( 2)n − 1 = 80, , , xn − a n, = na n −1 , Q xlim, →a x −a, , , , 10. We have,, 1 , 1 ,  2,  2, lim , +, −,  = lim , , x → 1 1 − x 2, x − 1 x → 1 1 − x 2 1 − x , [ ∞ − ∞ form ], , ⇒, , n ( 2)n − 1 = 5 × 16, , ⇒, , n × 2n − 1 = 5 × ( 2) 4, , ⇒, , n × 2 n − 1 = 5 × ( 2) 5 − 1, , ∴, , n=5

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CBSE New Pattern ~ Mathematics XI (Term I), , ( x + 2)1 / 3 − 21 / 3, x→ 0, x, , 15. Given, lim, , ( x + 2)1 / 3 − 21 / 3, x→ 0, ( x + 2) − 2, 1, , −1, 1, × 23, 3, 1, 1, = × ( 2) −2 / 3 =, 3, 3 ( 2) 2 / 3, , =, , 16. Put y = 1 + x , so that y → 1 as x → 0., x→ 0, , y −1, 1 + x −1, = lim, y →1 y − 1, x, 1, , 1, , y 2 − 12, = lim, y →1 y − 1, 1, , 1 2 −1 1, =, (1), 2, 2,  sin 7 x , sin 7 x, 7x , ,  7 x  7 xlim, sin 7 x, →0 7x, 17. lim, =, = lim, x→ 0 tan 5x, x→ 0,  tan 5x  5 lim tan 5x, 5x , , x→ 0,  5x , 5x, =, , 7 1 7, × =, 5 1 5, sin θ, tan θ, , , lim, = 1 and lim, = 1, θ→ 0 θ, , θ→ 0 θ, sin 4 x, 2x,  sin 4 x, , 18. lim, = lim, ⋅, ⋅2, x→ 0  4 x, x→0 sin 2x, sin 2x , =, , ,  sin 4 x   sin 2x , = 2 ⋅  lim , ÷, ,  x→ 0  4 x   2x , ,  sin 4 x ,  sin 2x , = 2 ⋅  lim , ÷ lim, ,  4 x→ 0  4 x  2 x→ 0  2x , [as x → 0, 4 x → 0 and 2x → 0], = 2 (1 ÷ 1) = 2, πx, tan, tan x °, π, , , 180, 19. lim, rad , = lim, = 1 Q 1° =, x, π, x→ 0, x, →, 0, x°, 180, , , 180, 1 − cos 4 x, 2 sin 2 2x x, 20. lim, = lim, ×, x→ 0, x→ 0, x, x, x, [Q1 − cos 2θ = 2 sin 2 θ], 2, , x, x, cos, 2, 2, x, , x  2 cos 2 , , 2, x, tan, 1, 2 =1, = lim, 2 x→ 0 x, 2, 2, sin( 2 + x ) − sin( 2 − x ), 22. We have, lim, x→ 0, x, (2 + x + 2 − x), (2 + x − 2 + x), sin, 2 cos, 2, 2, = lim, x→ 0, x, 2 cos 2 sin x, = lim, x→ 0, x, C+D, C − D, , Q sin C − sin D = 2 cos, sin, 2, 2 , , sin x, 21. We have, lim, = lim, x→ 0 x (1 + cos x ), x→ 0, , = lim, , Then, lim, , 101, ,  sin 2x , = lim 2 ,  × 4x, x→ 0  2x , sin x, , , = 2 ×1 × 0 = 0, Q lim, = 1,  x→ 0 x, , , 2 sin, , sin x, sin x, , , = 2 cos 2 Q lim, = 1, x, →, 0, x, x, , , Hence, p = 2 and q = 2, tan x − sin x, 23. We have, lim, x→ 0, sin 3 x, ,  1, − 1, sin x ,  cos x , = lim, x→ 0, sin 3 x, 1 − cos x, = lim, x→ 0 cos x sin 2 x, x, 2 sin 2, 1, 2, =, = lim, x→ 0, x, x,  2, , cos x  4 sin 2 ⋅ cos 2 , , 2, 2, tan 2x − x, 24. Given, lim, x → 0 3x − sin x,  tan 2x, , − 1, x, x, , , = lim, x→0, sin x , , x 3 −, x , , tan 2x, lim 2 ×, −1, 2 −1 1, x→0, 2x, =, =, =, sin x, 3 −1 2, 3 − lim, x→0 x, tan 2x, 25. Given, lim, π, π, x→, x−, 2, 2, π, Let x − = h ,, 2, π, when x → , then h → 0, 2, = 2 cos 2 lim, x→ 0

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102, , CBSE New Pattern ~ Mathematics XI (Term I), π, , tan 2  + h , 2, , Therefore, given limit = lim, h→ 0, h, tan ( π + 2h ), = lim, h→ 0, h, tan 2h, = lim, [Q tan ( π + θ ) = tan θ ], h→ 0, h, 2 tan 2h, = lim, h→ 0, 2h, tan x, , , = 2 ×1 = 2, = 1, Q xlim, →0, x, , , 26. Given,, , lim, , x→ π /4, , sec 2 x − 2, tan x − 1, , 1 + tan 2 x − 2, x→ π /4, tan x − 1, , = lim, , tan 2 x − 1, x → π / 4 tan x − 1, (tan x + 1)(tan x − 1), = lim, x→ π /4, (tan x − 1), , e 3x − 1, e 3x − 1 3, = lim, ×, x→ 0, x→ 0, x, x, 3, [multiplying numerator and, denominator by 3], 3x, e −1, ...(i), = 3 lim, x→ 0, 3x, Let h = 3x ., Then, x → 0 ⇒ h → 0, Now, from Eq. (i), we get, e 3 x −1, eh −1, lim, = 3 lim, = 3 (1), x→ 0, h→ 0, x, h, , eθ −1 , = 1, Q θlim, →0, θ, , , =3, x, 3, e −e, 30. We have, lim, x→3 x − 3, , 29. lim, , On putting h = x − 3 we get, ex − e3, eh + 3 − e3, lim, = lim, x→3 x − 3, h→ 0, h, , = lim, , [Q x → 3 ⇒ h → 0 ], e he 3 − e 3, e h −1, = lim, = e 3 lim, h→ 0, h→ 0, h, h, θ, , e −1 , = e 3 × 1 = e 3 Q lim, = 1, θ→ 0, θ, , , , = lim (tan x + 1), x→ π /4, , =2, , cosec x − cot x, x, cos x, 1, −, sin, x, sin x = lim 1 − cos x, = lim, x→0, x → 0 x ⋅ sin x, x, x, x, 2 sin 2, tan, 2, 2, = lim, = lim, x, x x→0 x, x→0, x ⋅ 2 sin cos, 2, 2, x, tan, 2 ⋅ 1 = 1 Q lim tan θ = 1, = lim, , x, x→0, 2 2  θ → 0 θ, 2, x 2 cos x, x 2 cos x, 28. Given, lim, = lim, x, x→0, x → 0 1 − cos x, 2 sin 2, 2, , 2 x, Q1 − cos x = 2 sin 2 , , 27. Given, lim, , x→0, , = 2 lim, , x→0, ,  x,  ,  2, sin 2, , = 2 ⋅1 = 2, , 2, , x, 2, , ⋅ limcos x, x→0, , e sin x − 1, e sin x − 1 sin x, = lim, ×, x→0, x→0, sin x, x, x, , 31. lim, , [multiplying numerator and, denominator by sin x], e sin x − 1 sin x , = lim , ×, , x→0, x ,  sin x, sin x, e sin x − 1, × lim, x → 0 sin x, x→0, x, , = lim, , =1 ×1 =1, , , eθ −1, sin θ, = 1 and lim, = 1, Q θlim, θ→ 0, θ, ,  →0 θ, e 2 x + 1 − 2e x, e x + e −x − 2, = lim, 2, x→ 0, x→ 0, x 2e x, x, , 32. lim, , 2, ,  e x − 1, = lim,  × e −x, x→ 0  x , 2, ,  e x − 1, = lim,  × lim e − x, x→ 0  x , x→ 0, = (1) 2 × e 0 = 1

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CBSE New Pattern ~ Mathematics XI (Term I),  3x − 2x , ,  x ,  3x − 1,  2x − 1, = lim ,  − lim , , x→0 , x  x→0  x , = log 3 − log 2 = log ( 3 / 2), , 34. lim, x→ 0, , sin ax, , , = 1, Q xlim, → 0 ax, , Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., sin ax + bx, 38. Assertion lim, x → 0 ax + sin bx, =, , 33. lim , x→0, , 103, , 2x − 1, ( 1 + x + 1), 2x − 1, ×, = lim, 1 + x − 1 x→ 0 1 + x − 1 ( 1 + x + 1), 2x − 1, × { 1 + x + 1}, x→ 0, x, 2x − 1, = lim, × lim ( 1 + x + 1), x→0, x→0, x, = (log 2) 2 = 2 log 2, = lim, , loge (1 + 2x ) 2, 35. We have, lim, ×, x→ 0, x, 2, [multiplying numerator and, denominator by 2], loge (1 + 2x ), = 2 lim, x→ 0, 2x, On putting h = 2x , we get, loge (1 + 2x ), loge (1 + h ), = 2 lim, h, →, 0, x, h, [Q x → 0 ⇒ h → 0 ], loge (1 + x ), , , = 1, = 2 (1) Q lim, x, →, 0, x, , , =2, ax 2 + bx + c, 36. Assertion Given, lim 2, x → 1 cx + bx + a, lim, , x→ 0, , =, , a × (1) 2 + b × 1 + c, c × (1) 2 + b × 1 + a, , =, , a +b +c, =1, c +b +a, , 1 1, +, Reason lim x 2, x → −2 x + 2, (2 + x), 1, = lim, = lim, x → − 2 2x ( x + 2), x → − 2 2x, 1, 1, =, =−, 2( − 2), 4, Hence, Assertion is true and Reason is false., sin ax, ( a ) sin ax, 37. Assertion Given, lim, = lim, x→0, x→0, bx, b ( ax ), [dividing and multiplying by a], , a, a, ×1 =, b, b, , Dividing each term by x, we get, a sin ax, sin ax bx, +b, +, ax, x, x, = lim, = lim, x → 0 ax, b sin bx, sin bx x → 0, +, a +, x, x, bx, sin x, a ×1 + b a + b, , , =, =, = 1, = 1 Q lim, a + b ×1 a + b,  x→0 x, , Reason lim ( 5x 3 + 5x + 1), x→1, , = 5 (1) 3 + 5 (1) + 1 = 5 + 5 + 1 = 11, Hence, Assertion is false and Reason is true., , 39. Assertion Given, lim, , x→π, , sin( π − x ), π (π − x), , Let π − x = h , As x → π, then h → 0, sin( π − x ), sin h, = lim, π ( π − x ) h → 0 πh, 1 sin h, = lim, ×, h→ 0 π, h, 1, 1 , sin h, , = × 1 = Q lim, = 1, π, π  h→ 0 h, , cos x, Reason Given, lim, x→0 π − x, , ∴ lim, , x→π, , Put the limit directly, we get, cos 0 1, =, π−0 π, Hence, Assertion is false and Reason is true., sin ax, 40. Assertion Given, lim, x → 0 sin bx, Multiplying and dividing by ( ax ) and ( bx ),, we get, sin ax, bx, ax, = lim, ×, ×, x → 0 ax, sin bx bx, a a, =1 ×1 × =, b b, sin ax, bx, , , = lim, = 1, Q xlim, x → 0 sin bx, →0, ax, , , Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion.

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104, , CBSE New Pattern ~ Mathematics XI (Term I), cos 2x − 1, cos x − 1, 1 − cos 2x, 2 sin 2 x, = lim, = lim, x, x→0, x → 0 1 − cos x, 2 sin 2, 2, Q1 − cos 2x = 2 sin 2 x , , , and 1 − cos x = 2 sin 2 x , , 2 , , 41. Assertion Given, lim, , x→0, , Multiplying and dividing by x 2 and then, 4, multiplying by in the numerater. we get, 4, x2, 4×, 2, sin x, 4, = lim, ×, x→0, 2 x, x2, sin, 2, 2,  x , 2, , ,  sin x , = lim ,  × 2  ×4, x→0  x ,  sin x , , 2, =1 ×1 × 4 = 4, Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion., ax + x cos x, 42. Assertion Given, lim, x→0, b sin x, Dividing each term by x, we get, ax x cos x, +, a + cos x, x, = lim x, = lim, x→0, x→0, b sin x,  sin x , b, ,  x , x, =, , a + cos 0 a + 1, =, b ×1, b, , sin x, , , = 1, Q xlim, →0, x, , , Reason lim x sec x = 0 × sec 0 = 0 × 1 = 0, x→0, , Hence Assertion is true and Reason is false., e 3 + x − sin x − e 3, 43. Assertion Given, limit = lim, x→0, x, e 3 ( e x − 1) sin x , = lim , −, , x→0, x, x , , = e 3 −1, tan 4 x, sin 2x,  tan 4 x   2x , = lim , , , x → 0  4 x   sin 2x , , Reason Given limit = lim, , x→0, ,  4x ,  ,  2x , , 1, tan 4 x, 4x, ×, ×, sin 2x 2x, 4x, lim, 2 x → 0 2x, = 1 ×1 × 2 = 2, Hence, Assertion is false and Reason is true., e x − e −x, 44. Assertion Given, limit = lim, x→0, x, e 2 x −1, = lim, x → 0 xe x, e 2 x −1, 2, = lim, × lim x, 2x → 0, x→0e, 2x, 2, =1× = 2, 1, Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., e tan x − 1 tan x, 45. Assertion Given limit = lim, ⋅, x→ 0 tan x, x, = lim, , 4x → 0, , = 1 ⋅1 = 1,  e 4 x − 1, , Reason Given limit = lim , x→ 0 , x ,  e 4 x − 1,  =4, = lim 4 , 4 x→ 0  4 x , Hence, Assertion is true and Reason is false.,  e x − e sin x , , x→ 0  x − sin x , , 46. Assertion Given limit = lim , ,  e x − sin x − 1, = lim e sin x ,  = e sin 0 × 1 = 1, x→ 0,  x − sin x ,  32 x − 23 x , Reason Given limit = lim , , x→ 0 , x, ,  ( 32 x − 1) − ( 23 x − 1) , = lim , , x→ 0 , x, ,  23 x − 1,  32 x − 1, = 2 ⋅ lim , ,  − 3 ⋅ lim , x→ 0  2x , x→ 0  3x , = 2 log 3 − 3 log 2, = log 9 − log 8,  9, = log  ,  8, Hence Assertion is false and Reason is true.

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CBSE New Pattern ~ Mathematics XI (Term I), , 47. Assertion Given, limit = lim, x→ 0, , = lim, , e x −1, 1 − cos 2x, e x −1, , x→ 0, , 2, , 2 sin x, e x −1, x, 1, =, lim, ×, x, →, 0, x, sin x, 2, 1, 1, =, ×1 ×1 =, 2, 2, ( x 2 − 1), Reason Given, limit = lim, x→1 ( x − 1), ( x − 1) ( x + 1), = lim, x→1, ( x − 1), = lim( x + 1) = 2, , 105, a sin x − 1, x→ 0 sin x, , Reason Given, limit = lim, , Let y = sin x, Then, y → 0 as x → 0, a sin x − 1, a y −1, ∴ lim, = lim, = log a, x→ 0 sin x, y→0, y, Hence, Assertion is false and Reason is true., , 51. (i) Given, limit, = lim (1 + x + x 2 + x 3 + x 4 + x 5 + x 6, x → −1, , + x7 + x 8 + x 9 ), , = 1 − 1 + 1 −1 + 1 − 1 + 1 − 1 + 1 − 1 = 0, limit = lim [ x 2 ( x − 1)], , (ii) Given,, , x→5, , x→1, , Hence Assertion is true and Reason is false., 3x − 2x, 48. Assertion Given limit = lim, x→ 0 tan x, x, ( 3x − 1) − ( 2x − 1), = lim, ×, x→ 0, x, tan x, , x, 3x − 1, 2x − 1, =  lim, − lim,  × lim, x→ 0, x  x→ 0 tan x,  x→ 0 x, , x→5, , x→5, , = ( 5) 3 − ( 5) 2, = 125 − 25 = 100, (iii) Given, limit = lim ( x 3 + x 2 + x − 1), x→2, , = lim x 3 + lim x 2 + lim x + lim ( − 1), x→2, , x→2, , x→2, , (iv) Given, limit = lim ( x 3 + x + 2), , log (1 + x ), log (1 + x ), x, = lim, ×, x→ 0, x→ 0, tan x, tan x, x, = 1 ×1 = 1, Hence Assertion is true and Reason is false., x −1, 49. Assertion Given limit = lim, x→1 loge x, Put x = 1 + h as x → 1, h → 0, 1, 1 + h −1, ∴ lim, =, =1, log (1 + h ), h→ 0 loge (1 + h ), lim, h→ 0, h, log (sin x + 1), Reason Given, limit = lim, x→ 0, x, log (sin x + 1) sin x, = lim, ×, =1, x→ 0, sin x, x, Hence Assertion is true and Reason is false., 2+ x, , 3, , −9, , x, , 32 ( 3x − 1), = 9 log 3, x→ 0, x, , = lim, , = lim x 3 − lim x 2, , = ( 2) 3 + ( 2) 2 + ( 2) − 1 = 8 + 4 + 2 − 1 = 13, , Reason lim, , x→ 0, , x→5, , x→2, , = (log 3 − log 2) × 1,  3, = log  ,  2, , 50. Assertion Given, limit = lim, , = lim [ x 3 − x 2 ], , x → −3, , = lim x 3 + lim x + lim 2, x → −3, , x → −3, , x → −3, , = ( −3) + ( − 3) + 2 = − 27 − 3 + 2, 3, , = − 30 + 2 = − 28, (v) Given, limit = lim ( x 4 − x 3 ), x→4, , = lim x 4 − lim x 3 = ( 4 ) 4 − ( 4 ) 3, x→4, , x→4, , = 256 − 64 = 192, x 10 + x 5 + 1 ( −1)10 + ( −1) 5 + 1, 52. (i) lim, =, x→−1, x −1, −1 − 1, =, (ii) lim, , x→−1, , ( x − 1) 2 + 3x 2, ( x 4 + 1) 2, , 1 − 1 + 1 −1, =, −2, 2, , =, , ( −1 − 1) 2 + 3 ( − 1) 2, (( −1) 4 + 1) 2, , =, , ( − 2) 2 + 3 (1), (1 + 1) 2, , =, , 4+3 7, =, 4, 22

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106, , CBSE New Pattern ~ Mathematics XI (Term I), , x2 − 4, x − 4x 2 + 4x, On putting x = 2, we get, 4−4, 0, f ( 2) =, =, 8 − 16 + 8 0, 0, i.e. it is the form ., 0, So, let us first factorise it., , (iii) Consider f ( x ) =, , (v) Given, lim, , 3, , x→2, , = lim, , = lim, , x→0, , = lim, , ( x + 2) ( x − 2), = lim, x→2, x ( x − 2) 2, , =, , x→0, , ( x + 2), x ( x − 2), , = lim, , x→0, , 2+ 2, 4, =, 2 ( 2 − 2) 0, , = lim, , x→0, , , , x2 − 4, ∴ lim  3,  does not exist., x → 2 x − 4x 2 + 4x, , , x→1, , x 7 − 2x 5 + 1, x 3 − 3x 2 + 2, , , 0,  0 form, , x7 − x 5 − x 5 + 1, x 3 − x 2 − 2x 2 + 2, , = lim, , x→1, , x ( x − 1) 1 ( x − 1), −, ( x − 1), ( x − 1), = lim, x → 1 x 2 ( x − 1), 2( x 2 − 1), −, ( x − 1), ( x − 1), 2, , 5, ,  x 5 − 1, lim x 5 ( x + 1) − lim , , x→1, x → 1  x −1 , lim x 2 − lim 2( x + 1), x→1, , 1 × 2 − 5 × (1), 2− 5, =, 1− 2× 2, 1− 4, , , xn − a n, = na n −1 , Q xlim, →a x − a, , , −3, =, =1, −3, =, , x 2( 1 + x 3 + 1 − x 3 ), 1 + x3 −1 + x3, x 2( 1 + x 3 + 1 − x 3 ), 2x 3, x ( 1 + x3 + 1 − x3 ), 2, , 2x, ( 1 + x + 1 − x3 ), 3, , Put, h = x − 4, ex − e4, eh + 4 − e4, = lim, x→4 x −4, h→ 0, h, , ∴ lim, , e he 4 − e 4, h→ 0, h, , = lim, , e h −1, h→ 0, h, , = e 4 lim, , On dividing numerator and denominator, by ( x −1), then, , x→1, , (1 + x 3 ) − (1 − x 3 ), , ex − e4, x→4 x −4, , x 5 ( x 2 − 1) − 1( x 5 − 1), x → 1 x 2 ( x − 1) − 2( x 2 − 1), , =, , 1 + x3 + 1 − x3, , 53. (i) We have, lim, , = lim, , 5, , 1 + x3 + 1 − x3, , =0, , which is not defined., , (iv) Given, lim, , x2, ⋅, , x −4, x − 4x 2 + 4x, , x→2, , 1 + x3 − 1 − x3, , x→0, , 3, , = lim, , x2, , x→0, , 2, , Consider, lim, , 1 + x3 − 1 − x3, , 4, , = e4 ×1 = e4, e kx − 1, x→0, x, , (ii) We have, lim, ,  e kx − 1, = lim ,  × (k ), kx → 0  kx , = 1 × k = k [Q x → 0 ⇒ kx → 0 ],  e − x − 1, (iii) We have, lim , , x → 0, x , Put, − x = y, as x → 0 ⇒ y → 0,  e y − 1,  e − x − 1, ∴ lim , ,  = lim , x → 0, x  y → 0  −y ,  e y − 1, = − lim ,  = −1, y → 0, y 

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108, , CBSE New Pattern ~ Mathematics XI (Term I), , = lim, , x→ π /4, , 1, 1, lim, ( 1 + 0 + 1) x→ 0 log(1 + x ), x, 1, 1, =, ×, 1 + 1 lim log (1 + x ), x→ 0, x, 1, 1, =, ×1 =, 1+1, 2, , π ,  , 2 sin  x −  , , 4 , , π, , x − , , 4, , =, , [Q sin A cos B − cos A sin B = sin ( A − B )], π, , sin  x − , , 4, = 2 lim, = 2, π, π, x− → 0 x − , , , 4, , 4, π, π, sin x, , , , = 1, Q x → 4 ⇒  x − 4  → 0 and xlim, →0 x, , , loge (1 + 5x ), 55. (i) We have, lim, x→0, x, loge (1 + 5x ), = 5 lim, = 5 ×1 = 5, 5x → 0, 5x, [Q x → 0 ⇒ 5x → 0 ], loge (1 + 6 x ) − 5x 2, x→0, x, loge (1 + 6 x ), = 6 lim, − 5 lim x, 6x → 0, x→0, 6x, [Q x → 0 ⇒ 6 x → 0 ], = 6 × (1) − 5 × ( 0 ) = 6, , (ii) We have, lim, , (iii) lim, x→ 0, , 1 + x −1, log(1 + x ), , On multiplying numerator and, denominator by 1 + x + 1, we get, 1 + x −1, 1+ x +1, ×, lim, x→ 0 log(1 + x ), ( 1 + x + 1), = lim, , 1 + x −1, ( 1 + x + 1) log(1 + x ), , = lim, , x, ( 1 + x + 1) log(1 + x ), , x→ 0, , x→ 0, , (iv) Put x − 5 = h and as x → 5, then h → 0, log( h + 5) − log 5, ∴ lim, h→ 0, h, h, , log 1 + , , 5 1, =, = lim, h, h, 5, →0, ×5, 5, 5, m , , Q log m − log n = log n , , , , h, , h → 0 ⇒ → 0, , , 5, x , x ,  ,  , log 51 +   − log 51 −  , , , , 5, 5 , , , , (v) lim, x→ 0, x, x  , ,  x , , log 5 + log 1 +   − log 5 + log 1 −  , 5, 5 ,  , = lim , x→ 0, x, x, x, , , log 1 + , log 1 − , , , 5, 5 1, ⋅, − lim, x, x, x, ( −5), →0, −, 5, 5, 5, 5, x, , Q x → 0 ⇒ 5 →, 1, 1, 2, = × (1) + × (1) =, 5, 5, 5, 1, = lim, x, →0 5, , , 0, 

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109, , CBSE New Pattern ~ Mathematics XI (Term I), , 07, Statistics, Quick Revision, Measures of Dispersion, The dispersion is the measure of variations in the values of, the variable. It measures the degree of scatterdness of the, observation in a distribution around the central value., , Range, Range is defined as the difference between two extreme, observations of the distribution., Range of distribution = Maximum value of observation, − Minimum value of observation, , Mean Deviation, Mean deviation is defined as the arithmetic mean of the, absolute deviations of all the values taken about any, central value ‘a’ (mean or median). The mean deviation, from ‘a’ is denoted as MD(a)., Sum of absolute values of deviations from ‘ a ’, ∴ MD(a ) =, Number of observations, (i) Mean deviation for ungrouped data Let n, observations be x 1 , x 2 , x 3 , K , x n , then mean, deviation about their mean or median is given by, Σ | xi − A |, MD =, n, where, A = mean or median, (ii) Mean deviation for discrete frequency, distribution, Let the given data consist of discrete observations, x 1 , x 2 , x 3 , K , x n occurring with, frequencies f 1 , f 2 , f 3 , K , f n respectively, then, Σf | x − A | Σf i | x i − A |, MD = i i, =, Σf i, N, where, A = mean or median, , (iii) Mean deviation for continuous, frequency distribution Here, the, procedure is same as for a discrete, frequency distribution. The only, difference is that here we have to, obtain the mid-points of the various, classes and take the deviations of these, mid-points from the given central, value., N, − cf, Note Median = l + 2, ×h, f, where, l = lower limit, f = frequency, h = width of median class, and cf = cumulative frequency of class, just preceding the median, class., , Variance, Variance is the arithmetic mean of the, square of the deviations about mean x., Let x 1 , x 2 , K , x n be n observations with x, as the mean, then the variance denoted by, Σ( x i − x )2, ., σ 2 , is given by σ 2 =, n, , Standard Deviation, If σ 2 is the variance, then σ is called the, standard deviation which is given by, σ=, , Σ( x i − x )2, ., n, , Thus, standard deviation (SD), = Variance

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110, , CBSE New Pattern ~ Mathematics XI (Term I), , Also, by shortcut method,, , (i) Standard deviation for ungrouped data, SD of n observations x1 , x 2 , … , xn is given by, σ=, , Σ( x1 − x ), ., n, , (ii) Standard deviation of a discrete frequency, distribution Let the discrete frequency, distribution be xi : x1 , x 2 , … , xn and, f i : f 1 , f 2 , … , f n , then, σ=, , Σf i ( xi − x ) 2, 1, or σ =, N, N, , NΣ f i xi2 − ( Σ f i xi ) 2, , where, f i ’s are the frequency of xi ’s and N =, , n, , ∑ fi., , 1, N, , σ=, , 2, , n, , ∑, i =1, ,  Σf d , f i di 2 −  i i ,  N , , 2, , where, di = xi − a , a = assumed mean, , Standard Deviation of a continuous, frequency distribution, If there is a frequency distribution of n classes, and each class defined by its mid-point x i , with, corresponding frequency f i , then, 1, σ=, NΣf i x i2 − ( Σf i x i )2, N, , i =1, , Objective Questions, Multiple Choice Questions, 1. The mean deviation from the mean of, the set of observations −1, 0 and 4 is, , (a) 3, (c) −2, , (b) 1, (d) 2, , Then, the mean deviation about the, median for the data is, (a) 5, (c) 5.1, , 6. Consider the following data, , 2. When tested, the lives (in hours) of, 5 bulbs were noted as follows, 1357, 1090, 1666, 1494, 1623, The mean deviations (in hours) from, their mean is, (a) 178, (c) 220, , (b) 179, (d) 356, , 3. Mean deviation about the median for, the data, 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21 is, (a) 4.27, (c) 5.27, , (b) 5.24, (d) 4.24, , xi, , 5, , 7, , 9, , 10, , 12, , 15, , fi, , 8, , 6, , 2, , 2, , 2, , 6, , Then, the mean deviation about the, median for the data is, (a) 3.15, (c) 3.21, , (b) 3.23, (d) 3.17, , 7. The mean deviation of the data 3, 10,, 10, 4, 7, 10, 5 from the mean is, (a) 2, (c) 3, , (b) 2.57, (d) 3.75, , 8. Following are the marks obtained by, , 4. The mean deviation about the median, for the data, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50 is, (a) 4.5, (c) 5.5, , (b) 5.3, (d) 5.2, , (b) 4, (d) 5, , 9 students in a mathematics test, 50, 69, 20, 33, 53, 39, 40, 65, 59, The mean deviation from the median is, (a) 9, (c) 12.67, , (b) 10.5, (d) 14.76, , 9. The mean deviation about the median, , 5. Consider the following data, xi, , 15, , 21, , 27, , 30, , 35, , fi, , 3, , 5, , 6, , 7, , 8, , for the data 34, 66, 30, 38, 44, 50, 40,, 60, 42, 51 is, (a) 8.7, (c) 87, , (b) 7.7, (d) 77

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111, , CBSE New Pattern ~ Mathematics XI (Term I), , 16. Consider the following data, , 10. Consider the following data, Marks, 10-20 20-30 30-40 40-50 50-60 60-70 70-80, obtained, Number of, students, , 2, , 3, , 8, , 14, , 8, , 3, , 2, , Then, the mean deviation about the, mean is, (a) 20, (c) 30, , (b) 10, (d) 15, , are 48, 80, 58, 44, 52, 65, 73, 56, 64,, 54, then the mean deviation from the, median is, (b) 8.6, (d) 10.1, , (b) 2.1, (d) 5.1, , The mean deviation about the mean for, the given data is, (b) 7.4, (d) 4, , Age (in years), , 10, , 12, , 15, , 18, , 21, , 23, , Frequency, , 3, , 5, , 4, , 10, , 8, , 4, , The mean deviation about the median, of the given frequency distribution, is (in years), (b) 2.24, (d) 7.2, , 15., xi, , 2, , 5, , 6, , 8, , 10, , 12, , fi, , 2, , 8, , 10, , 7, , 8, , 5, , The mean deviation about the mean for, the given data is, (b) 2.2, (d) 2.4, , (b) 2.33, (d) 1.33, , (a) 23.33, (c) 46.66, , (b) 25.33, (d) 48.66, , 8, 9, and 10. If 1 is added to each, number the variance of the numbers,, so obtained is, (a) 6.5, (c) 3.87, , (b) 2.87, (d) 8.25, , 20. Consider the following data, , 14., , (a) 2.1, (c) 2.3, , (a) 2.44, (c) 1.44, , 19. Consider the numbers 1, 2, 3, 4, 5, 6, 7,, , 13. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44., , (a) 3.24, (c) 8.1, , the data, 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17, is, , 23.33. Then, variance of 4, 8, 10, 12, 16,, 34 will be, , The mean deviation from the mean for, the given data is, , (a) 8.4, (c) 6.3, , 6, 8, 7, None of the above, , 18. Variance of the data 2, 4, 5, 6, 8, 17 is, , 12. 6, 5, 5.25, 5.5, 4.75, 4.5, 6.25, 7.75, 9., (a) 1.1, (c) 4.1, , (a), (b), (c), (d), , 17. Mean deviation about the median for, , 11. The scores of a batsman in 10 innings, , (a) 7.6, (c) 9.6, , 36, 72, 46, 42, 60, 45, 53, 46, 51, 49, Then, the mean deviation about the, median for the data is, , 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, If 2 is added to each number, then, variance of the numbers so obtained is, (a) 6.5, (c) 3.87, , (b) 2.87, (d) 8.25, , 21. Find the variance of the following data, Class interval, , 4-8, , 8-12, , 12-16, , 16-20, , Frequency, , 3, , 6, , 4, , 7, , (a) 13, (c) 19, , (b) 18, (d) 20, , 22. The mean and variance for the data, 6, 7, 10, 12, 13, 4, 8, 12 respectively are, (a), (b), (c), (d), , 9, 9.50, 8, 8.50, 9, 9.25, 8, 8.25

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112, , CBSE New Pattern ~ Mathematics XI (Term I), , 23. Find the variance of the following data, Class interval, , 4-8, , 8-12, , 12-16, , 16-20, , Frequency, , 3, , 6, , 4, , 7, , (a) 13, , (b) 18, , (c) 19, , (d) 20, , 24. Following are the marks obtained by, 9 students in a mathematics test, 50, 69, 20, 33, 53, 39, 40, 65, 59, The mean deviation from the median is, (a) 9, (c)12.67, , (b) 10.5, (d) 14.76, , 25. The standard deviation of data 6, 5, 9,, 13, 12, 8 and 10 is, (a), , 52, 7, , (b), , 52, 7, , (d) 6, , (c) 6, , 26. The standard deviation for the data, (b), (d), , 8.25, 9.25, , 7.25, 1025, ., , dispersion of values of x from mean x ,, we take absolute measure of dispersion., , Reason (R) The mean deviation about, the mean for the data 38, 70, 48, 40, 42,, 55, 63, 46, 54, 44 is 8.5., , (b) 5000, 256100, (d) 4000, 255600, , 33. Assertion (A) Consider the following, , 28. 6, 7, 10, 12, 13, 4, 8, 12, , data., , The variance for the given data is, (b) 9.25, (d) 8.9, , 29. 45, 60, 62, 60, 50, 65, 58, 68, 44, 48, , xi, , 5, , 10, , 15, , 20, , 25, , fi, , 7, , 4, , 6, , 3, , 5, , Then, the mean deviation about the, mean is 6.32., , The variance for the given data is, (a) 56.2, (c) 65.2, , 31. Assertion (A) In order to find the, , about the mean for the data 4, 7, 8, 9,, 10, 12, 13, 17 is 3., , 100 items are 50 and 4 respectively,, then the sum of all the item and the, sum of the squares of item is, , (a) 8.25, (c) 10.5, , (a) A is true, R is true; R is a correct, explanation of A., (b) A is true, R is true; R is not a correct, explanation of A., (c) A is true; R is false, (d) A is false; R is true., , 32. Assertion (A) The mean deviation, , 27. If mean and standard deviation of, , (a) 5000, 251600, (c) 4000, 215600, , Directions (Q. Nos. 31-45) Each of these, questions contains two statements, Assertion (A) and Reason (R). Each of the, questions has four alternative choices, any, one of the which is the correct answer. You, have to select one of the codes (a), (b), (c) and, (d) given below., , Reason (R) Sum of the deviations from, mean (x ) is zero., , 6, 7, 10, 12, 13, 4, 8, 12 is, (a), (c), , Assertion-Reasoning MCQs, , (b) 66.2, (d) 55, , Reason (R) Consider the following, data., , 30., xi, , 10, , 15, , 18, , 20, , 25, , xi, , 10, , 30, , 50, , 70, , 90, , fi, , 3, , 2, , 5, , 8, , 2, , fi, , 4, , 24, , 28, , 16, , 8, , The variance for the data given, distribution is, (a) 16, , (b) 15, , (c) 11, , (d) 17, , Then, the mean deviation about the, mean is 15.

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113, , CBSE New Pattern ~ Mathematics XI (Term I), , Reason (R) The units of individual, observations x i and the unit of their, mean are different that of variance., Since, variance involves sum of squares, of (x − x )., , 34. Assertion (A) The mean deviation, about median calculated for series,, where variability is very high, cannot, be fully relied., Reason (R) The median is not a, representative of central tendency for, the series where degree of variability is, very high., , 35. Assertion (A) The mean deviation, about the mean to find measure of, dispersion has certain limitations., Reason (R) The sum of deviations from, the mean is more than the sum of, deviations from median. Therefore, the, mean deviation about the mean is not, very scientific, where degree of, variability is very high., , 36. Assertion (A) The average marks of, boys in a class is 52 and that of girls is, 42. The average marks of boys and girls, combined is 50. The percentage of boys, in the class is 80%., Reason (R) Mean marks scored by the, students of a class is 53. The mean, marks of the girls is 55 and the mean, marks of the boys is 50. The percentage, of girls in the class is 64%., , 37. Assertion (A) The weights (in kg) of, 15 students are as follows, 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36,, 44, 45, 42, 30, If the weight 44 kg is replaced by 46 kg, and 27 kg is by 25 kg, then new median, is 35., Reason (R) The mean deviation from, the median of the weights (in kg) 54,, 50, 40, 42, 51, 45, 47, 57 is 4.78., , 38. Assertion (A) The proper measure of, dispersion about the mean of a set of, observations i.e. standard deviation, is expressed as positive square root of, the variance., , 39. Consider the following data, xi, , 4, , 8, , 11, , 17, , 20, , 24, , 32, , fi, , 3, , 5, , 9, , 5, , 4, , 3, , 1, , Assertion (A) The variance of the data, is 45.8., Reason (R) The standard deviation of, the data is 6.77., , 40. Consider the following data, xi, , 6, , 10, , 14, , 18, , 24, , 28, , 30, , fi, , 2, , 4, , 7, , 12, , 8, , 4, , 3, , Assertion (A) The mean of the data, is 19., Reason (R) The variance of the data, is 43.4., , 41. Consider the following data, xi, , 60, , 61, , 62, , 63, , 64, , 65, , 66, , 67, , 68, , fi, , 2, , 1, , 12, , 29, , 25, , 12, , 10, , 4, , 5, , Assertion (A) The mean of the data, using shortcut method is 32., Reason (R) The standard deviation of, the data using shortcut method is 1.69., , 42. Assertion (A) If each of the, observations x 1 , x 2 , …, x n is increased, by a, where a is a negative or positive, number, then the variance remains, unchanged., Reason (R) Adding or subtracting, a positive or negative number to, (or from) each observation of a group, does not affect the variance.

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114, , CBSE New Pattern ~ Mathematics XI (Term I), , 43. If for a distribution Σ(x − 5) = 3,, Σ(x − 5) = 43 and the total number of, items is 18., Assertion (A) Mean of the distribution, is 4.1666., Reason (R) Standard deviation of the, distribution is 1.54., 2, , 44. Assertion (A) The variance of first n, n2 −1, ., even natural numbers is, 4, Reason (R) The sum of first n natural, n(n + 1), numbers is, and the sum of, 2, squares of first n natural numbers is, n(n + 1)( 2n + 1), ., 6, , 45. Assertion (A) If the mean of n, observations 12 , 2 2 , 3 2 , …, n 2 is, 46n, , then n is equal to 11., 11, Reason (R) For two data sets each of, size 5, the variances are given to be, 4 and 5 and the corresponding means, are given to be 2 and 4, respectively., 11, The variance of combined data set is ., 2, , Case Based MCQs, 46. For a group of 200 candidates, the, mean and the standard deviation of, scores were found to be 40 and 15,, respectively. Later on it was discovered, that the scores of 43 and 35 were, misread as 34 and 53, respectively., Student, Ramu, Rajitha, Komala, Patil, Pursi, Gayathri, , Eng, 39, 79, 41, 77, 72, 46, , Hind, 59, 92, 60, 77, 65, 96, , Social Science Maths, 84, 80, 41, 68, 38, 75, 38, 71, 82, 87, 75, 42, 69, 83, 67, 53, 71, 39, , Answer the following questions on the, basis of above information., (i) Find the sum of correct scores., (a) 7991, (c) 8550, , (b) 8000, (d) 6572, , (ii) Find the correct mean., (a) 42.924, (c) 38.423, , (b) 39.955, (d) 41.621, , (iii) The formula of variance is, n, , ∑ (x i − x) 2, (a), , i =1, , n, , (b), , n, , i =1, , n, , ∑ (x i − x), (c), , ∑ (x i − x) 2, , 2, , i =1, , Σfi, , n, , (d), , ∑ fi (xi − x)2, i =1, , (iv) Find the correct variance., (a) 280.3, (c) 224.143, , (b) 235.6, (d) 226.521, , (v) Find the correct standard deviation., (a) 14.971, (c) 16.441, , (b) 11.321, (d) 12.824, , 47. You are given some observations as 34,, 66, 30, 38, 44, 50, 40, 60, 42, 51., Based on these observations, answer the, following questions., (i) The mean of the given data is, (a) 40.5, (c) 45.5, , (b) 45.0, (d) 50.5, , (ii) The mean deviation about the mean, is, (a) 10.0, (c) 9.1, , (b) 9.5, (d) 9.0, , (iii) The median of the given data is, (a) 41, (c) 43, , (b) 42, (d) 44, , (iv) The mean deviation about the, median is, (a) 8.0, , (b) 8.3, , (c) 8.5, , (d) 8.7, , (v) The difference between mean, deviation about the mean and mean, deviation about the median is, (a) 0.1, , (b) 0.2, , (c) 0.3, , (d) 0.4

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115, , CBSE New Pattern ~ Mathematics XI (Term I), , (iv) Variance of the given data is, , 48. You are given the following grouped, data., xi, , 2, , 5, , 6, , 8, , 10, , 12, , fi, , 2, , 8, , 10, , 7, , 8, , 5, , (ii) Mean deviation about the mean is, (b) 2.2, (d) 2.4, , (iii) The value of median is, (a) 5, (c) 7, , (b) 6, (d) 8, , (iv) Mean deviation about the median is, (a) 1.9, (c) 2.2, , (b) 2.0, (d) 2.3, , (v) The difference between mean and, median is, (a) 0.9, (c) 0.5, , (b) 0.7, (d) 0.3, , 49. Consider the data, xi, , 4, , 8, , 11, , 17, , 20, , 24, , 32, , fi, , 3, , 5, , 9, , 5, , 4, , 3, , 1, , Based on above information answer the, following questions., (i) Mean is calculated by using the, formula, Σfi xi, N, Σfi x2i, (c) x =, N, (a) x =, , (b) x = ∑ fi xi, (d) None of these, , (ii) Variance is calculated by using the, formula, 1, 1, Σfi (xi − x)2 (b) σ2 = Σfi (xi + x)2, N, N, 1, (c) σ2 = Σ (xi − x) (d) None of these, N, (a) σ2 =, , (iii) Mean of the given data is, (a) 10, , (b) 12, , (c) 14, , (d) 15, , (b) 5, (d) 3.19, , 50. Consider the data, , (b) 7.5, (d) 8.5, , (a) 2.1, (c) 2.3, , (b) 45.8, (d) 39.8, , (v) Standard deviation of the given data, is, (a) 6.77, (c) 4.8, , Based on these data, answer the, following questions., (i) Mean of the grouped data is, (a) 7.0, (c) 8.0, , (a) 40, (c) 41.5, , Class, , Frequency, , 0-10, , 6, , 10-20, , 7, , 20-30, , 15, , 30-40, , 16, , 40-50, , 4, , 50-60, , 2, , Based above information answer the, following questions., (i) Median is calculated by using the, formula, N, N, − cf, + cf, 2, (a) M = l +, × h (b) M = l + 2, ×h, f, f, N, − cf, (c) M = l − 2, × h (d) None of these, f, , (ii) Mean deviation about median is, calculated by using the formula, Σfi | xi + M |, N, Σ | xi − M |, (c) MD =, N, , (a) MD =, , (b) MD =, , Σfi | xi − M |, N, , (d) None of these, , (iii) Total frequency of the given data is, (a) 10, (c) 50, , (b) 20, (d) 60, , (iv) Median of the given data is, (a) 28, (c) 18, , (b) 20, (d) 8, , (v) Mean deviation about median is, (a) 10.16, (c) 9.16, , (b) 15, (d) 8.5

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116, , CBSE New Pattern ~ Mathematics XI (Term I), , ANSWERS, Multiple Choice Questions, 1. (d), 11. (b), 21. (c), , 2. (a), 12. (a), 22. (c), , 3. (c), 13. (a), 23. (c), , 4. (d), 14. (a), 24. (c), , 5. (c), 15. (c), 25. (a), , 6. (b), 16. (c), 26. (c), , 7. (b), 17. (b), 27. (a), , 8. (c), 18. (c), 28. (b), , 9. (a), 19. (d), 29. (b), , 10. (b), 20. (d), 30. (d), , 34. (a), 44. (d), , 35. (a), 45. (b), , 36. (c), , 37. (b), , 38. (a), , 39. (b), , 40. (b), , Assertion-Reasoning MCQs, 31. (a), 41. (d), , 32. (c), 42. (a), , 33. (c), 43. (d), , Case Based MCQs, 46. (i) - (a); (ii) - (b); (iii) - (a); (iv) - (c); (v) - (a), 48. (i) - (b); (ii) - (c); (iii) - (c); (iv) - (d); (v) - (c), 50. (i) - (a); (ii) - (b); (iii) - (c); (iv) - (a); (v) - (a), , 47. (i) - (c); (ii) - (d); (iii) - (c); (iv) - (d); (v) - (c), 49. (i) - (a); (ii) - (a); (iii) - (c); (iv) - (b); (v) - (a), , SOLUTIONS, −1 + 0 + 4, =1, 3, Σ| xi − x |, ∴MD ( x ) =, n, |− 1 − 1 | + |0 − 1 | + |4 − 1 |, =, =2, 3, , 1. Mean ( x ) =, , 11, , Therefore,, , ∑ | xi, , and MD (M ) =, , 1 11, ∑ | xi − M |, 11 i = 1, , =, , 1, × 58 = 5. 27, 11, , 2. Since, the lives of 5 bulbs are, 1357, 1090, 1666, 1494 and 1623., 1357 + 1090 + 1666 + 1494 + 1623, ∴ Mean =, 5, 7230, x =, ⇒, 5, ⇒, x =1446, , 4. The given data, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, are in ascending order., Here, total number of observations are 10, i.e. n = 10, which is even., , |1357 − 1446 | + |1090 − 1446 |, + |1666 − 1446 | + |1494 − 1446 |, + |1623 − 1446 |, ∴ MD ( x ) =, 5, 89 + 356 + 220 + 48 + 177, =, 5, = 178, , 3. Arranging the data into ascending order,, we have, 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21., 11 + 1, Now, median = ,  th observation = 9,  2 , Now, | xi − M| are 6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12., , − M| = 58, , i =1, , ∴ Median, ( M ) =, , n ,   th observation,  2, n , +  + 1 th observation, 2 , , 2, 5th observation + 6 th observation, =, 2, 40 + 42 82, =, =, = 41, 2, 2, | xi − M | are 9, 7,5, 3, 1, 1, 3, 5, 7, 9., 10, , Therefore, ∑ | xi − M | = 50, i =1, , and MD ( M ) =, , 1, 10, , 10, , 50, , ∑ | xi − M | = 10, i =1, , =5

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117, , CBSE New Pattern ~ Mathematics XI (Term I), ,  26 ,   th observation,  2, , 5., | xi − M |, , f i | xi − M |, , xi, , fi, , cf, , 15, , 3, , 3 |15 − 30| = 15, , 45, , 21, , 5, , 8 | 21 − 30 | = 9, , 45, , 27, , 6, , 14 | 27 − 30 | = 3, , 18, , 30, , 7, , 21 | 30 − 30 | = 0, , 0, , 35, , 8, , 29 | 35 − 30 | = 5, , 40, , Total, , Σf i = 29, , Σf i | x i − M |, = 148, , N = Σf = 29 (odd),  N + 1, ∴ Median M = ,  th observation,  2 ,  29 + 1, =,  th observation,  2 , Here,, , = 15th observation, ⇒, M = 30, ∴ Mean deviation about median, Σf | x − M | 148, = i i, =, = 51, ., Σf i, 29, , 6., ,  26, , +, + 1 th observation,  2, , , =, , 2, 13th observation + 14 th observation, =, 2, 7 + 7 14, =, =, =7, 2, 2, ∴ Mean deviation about median, Σf | x − M |, = i i, Σf i, 84, =, = 3.23, 26, , 7. Given, observations are 3, 10, 10, 4, 7, 10, and 5., ∴, , 3 + 10 + 10 + 4 + 7 + 10 + 5, 7, 49, =, =7, 7, , x =, , d i = xi − x, , xi, 3, , 4, , 10, , 3, 3, , xi, , fi, , cf, , | xi − M |, , f i | xi − M |, , 10, , 5, , 8, , 8, , | 5 − 7| = 2, , 16, , 4, , 3, , 7, , 6, , 14, , |7 − 7| = 0, , 00, , 7, , 0, 3, , 2, , 16, , |9 − 7| = 2, , 10, , 9, , 04, , 5, , 2, , 10, , 2, , 18, , |10 − 7 | = 3, , 06, , Total, , Σd i = 18, , 12, , 2, , 20, , |12 − 7 | = 5, , 10, , 15, , 6, , 26, , |15 − 7 | = 8, , 48, , Total, , Σf i | x i − M |, = 84, , Σf i = 26, Here, N = Σf i = 26 (even), , N ,   th observation,  2, Median ( M ) =, , , N, +  + 1 th observation, ,  2, 2, , Σdi, N, 18, =, = 2.57, 7, , Now, MD =, , 8. Since, marks obtained by 9 students in, Mathematics are, 50, 69, 20, 33, 53, 39, 40, 65 and 59., Rewrite the given data in ascending order., 20,33, 39, 40, 50, 53, 59, 65, 69, Here, n = 9, [odd],  9 + 1, ∴ Median = ,  term = 5 th term,  2 , Median = 50

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118, , CBSE New Pattern ~ Mathematics XI (Term I), , ∴, , xi, , d i = xi − Me, , 20, 33, 39, 40, 50, 53, 59, 65, 69, , 30, 17, 11, 10, 0, 3, 9, 15, 19, , N =2, , Σ d i = 114, , MD =, , 114, = 1267, ., 9, , 9. The given data can be arranged in ascending, order as 30, 34, 38, 40, 42, 44, 50, 51, 60, 66., Here, total number of observations are 10., i.e. n = 10, which is even., ∴ Median,, , n, n ,   th observation +  + 1 th observation, 2 ,  2, (M ) =, 2, 10 , 10, ,   th observation +  + 1 th observation,  2,  2, , =, 2, (5th observation + 6th observation), =, 2, 42 + 44 86, =, =, = 43, 2, 2, Let us make the table for absolute deviation, , xi, , | xi − M |, , 30, 34, 38, 40, 42, 44, 50, 51, 60, 66, , |30 − 43 | = 13, |34 − 43 | = 9, |38 − 43 | = 5, |40 − 43 | = 3, |42 − 43 | = 1, |44 − 43 | = 1, |50 − 43 | = 7, |51 − 43 | = 8, |60 − 43 | = 17, |66 − 43 | = 23, 10, , Total, , ∑ | x i − M | = 87, , i =1, , Now, mean deviation about median,, 10, , MD =, , ∑, i =1, , | xi − M |, 10, , 87, =, 10, = 8.7, , 10. Take the assumed mean a = 45 and h = 10,, and form the following table, No. of Mid di =, Marks, |x − x|, students value xi − 45 f i di i, f i | xi − x |, obtained, (f i ), ( xi ), 10, 10-20, , 2, , 15, , −3, , −6, , 30, , 60, , 20-30, , 3, , 25, , −2, , −6, , 20, , 60, , 30-40, , 8, , 35, , −1, , −8, , 10, , 80, , 40-50, , 14, , 45, , 0, , 0, , 0, , 0, , 50-60, , 8, , 55, , 1, , 8, , 10, , 80, , 60-70, , 3, , 65, , 2, , 6, , 20, , 60, , 70-80, , 2, , 75, , 3, , 6, , 30, , 60, , 40, , 0, , 400, , 7, , Therefore, x = a +, , ∑ f i di, , i =1, , ×h, N, 0, = 45 +, × 10, 40, = 45, 1 7, and MD ( x ) =, ∑ f i | xi − x |, N i =1, =, , 400, = 10, 40, , 11. Arranging the data in ascending order,, we have, 44, 48, 52, 54, 56, 58, 64, 65, 73, 80, Here, n = 10., So, median is the mean of 5th and 6th terms.,  56 + 58 , ∴ Median (M ) = ,  = 57, , 2 

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119, , CBSE New Pattern ~ Mathematics XI (Term I), , We make the table from the given data., , 13. Given observations are, , Scores, ( xi ), , Deviation from median, ( xi − M ), , | xi − M |, , 44, 48, 52, 54, 56, 58, 64, 65, 73, 80, , 44 − 57 = − 13, 48 − 57 = − 9, 52 − 57 = − 5, 54 − 57 = − 3, 56 − 57 = − 1, 58 − 57 = 1, 64 − 57 = 7, 65 − 57 = 8, 73 − 57 = 16, 80 − 57 = 23, , 13, 9, 5, 3, 1, 1, 7, 8, 16, 23, , Total, , 86, , ∑ | xi − M |, , 86, =, = 8.6, 10, n, Hence, the mean deviation from the median is, 8.6., , ∴Mean deviation =, , 12. Let x be the mean of given data., 6.5 + 5 + 5.25 + 5.5 + 4.75 + 4.5, + 6.25 + 7.75 + 8.5, 9, , Then, x =, , 54, =6, 9, Let us make the table for deviation and, absolute deviation., =, , xi, , xi − x, , xi − x, , 6.5, 5.0, 5.25, 5.5, 4.75, , 0.5, −1, − 0.75, − 0.5, − 1.25, , 0.50, 1.00, 0.75, 0.50, 1.25, , 4.5, 6.25, 7.75, 8.5, , − 1.50, 0.25, 1.75, 2.5, , 1.50, 0.25, 1.75, 2.50, 9, , ∑ | x i − x | = 10.00, , Total, , i =1, , ∴ Mean deviation about mean,, 9, , ∑ | xi, , − x|, , 10, =11, ., =, 9, 9, Hence, the mean deviation about mean is 11, .., MD ( x ) =, , i =1, , 38, 70, 48, 40, 42, 55, 63, 46, 54 and 44, Here, number of observations, n = 10, ( 38 + 70 + 48 + 40 + 42 + 55, + 63 + 46 + 54 + 44 ), 10, , ∴ Mean, x =, , 500, = 50, 10, Let us make the table for deviation and absolute, deviation, =, , | xi − x |, , 38, , xi − x, 38 − 50 = − 12, , 70, , 70 − 50 = 20, , 48, , 48 − 50 = − 2, , 2, , 40, , 40 − 50 = − 10, , 10, , 42, , 42 − 50 = − 8, , 8, , 55, , 55 − 50 = 5, , 5, , 63, , 63 − 50 = 13, , 13, , 46, , 46 − 50 = − 4, , 4, , 54, , 54 − 50 = 4, , 4, , 44, , 44 − 50 = − 6, , xi, , 12, 20, , 6, 10, , ∑ | x i − x | = 84, , Total, , i =1, 10, , Now, MD =, , ∑ | xi, , −x|, , i =1, , 10, 84, =, = 8.4, 10, , 14. The given observations are already in, ascending order., Now, let us make the cumulative frequency., , Age (x i ), , Frequency (f i ), , cf, , 10, 12, 15, 18, 21, 23, , 3, 5, 4, 10, 8, 4, , 3, 8, 12, 22, 30, 34, , Total, , N = 34, , Here, Σf i = N = 34, which is even.

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120, , CBSE New Pattern ~ Mathematics XI (Term I), , ∴Median,  34, Value of   th observation,  2, =, , 16. The given data is 36, 72, 46, 42, 60, 45, 53,, ,  34 , + Value of  + 1 th observation, , 2, 2, Value of 17th observation, + Value of 18th observation, 2, , =, , 18 + 18, = 18, 2, [Q both of these observation lies in the, cumulative frequency 22 and its, corresponding observation is 18.], Now, let us make the following table from the, given data., =, , | x i − 18 |, , 8, , 6, , 3, , 0, , 3, , 5, , Total, , f i | x i − 18 |, , 24, , 30, , 12, , 0, , 24, , 20, , 110, , Σ f i | xi − M |, Σ fi, 110, =, = 3.24 yr, 34, , =, , 15. Let us make the following table from the, given data., , f i xi, , | xi − x | f i | xi − x |, , 46, 51, 49, Arranging the data in ascending order,, 36, 42, 45, 46, 46, 49, 51, 53, 60, 72, Number of observations = 10 (even), N ,   th observation,  2, , N, +  + 1 th observation, ,  2, Median M =, 2, 10 ,   th observation,  2, 10, , +  + 1 th observation,  2, , =, 2, 5th observation + 6th observation, =, 2, 46 + 49, =, = 47.5, 2, , xi, , | xi − M |, , 36, , | 36 − 47.5 | = 11.5, , 42, , | 42 − 47.5 | = 5.5, , 45, , | 45 − 47.5 | = 2.5, , 46, , | 46 − 47.5 | = 1.5, , 46, , | 46 − 47.5 | = 1.5, , xi, , fi, , 2, , 2, , 4, , 5.5, , 11, , 49, , |49 − 47.5 | = 1.5, , 5, , 8, , 40, , 2.5, , 20, , 51, , | 51 − 47.5 | = 3.5, , 6, , 10, , 60, , 1.5, , 15, , 53, , | 53 − 47.5 | = 5.5, , 8, , 7, , 56, , 0.5, , 3.5, , 60, , | 60 − 47.5 | = 12.5, , 10, , 8, , 80, , 2.5, , 20, , 72, , | 72 − 47.5 | = 24.5, , 12, , 5, , 60, , 4.5, , 22.5, , Total, , 40, , 300, , 92, , Here, N = Σ f i = 40, Σ f i x i = 300, 1, 1, Now, mean ( x ) =, × 300 = 7.5, Σ f i xi =, N, 40, ∴Mean deviation about the mean,, 1, MD ( x ) =, Σ f i | xi − x |, N, 1, =, × 92 = 2.3, 40, Hence, the mean deviation about mean, is 2.3., , Σ | x i − M | = 70, ∴ Mean deviation about median, Σ | xi − M |, =, n, 70, =, =7, 10, , 17. The given data is 13, 17, 16, 14, 11, 13, 10, 16,, 11, 18, 12, 17, Arranging in ascending order,, 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18, Number of observations = 12 (even)

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121, , CBSE New Pattern ~ Mathematics XI (Term I), , N ,   th observation,  2, Median M =, , 19. Given numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10., If 1 is added to each number, then, observations will be 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11., ∴, Σxi = 2 + 3 + 4 + K + 11, 10, = [ 2 × 2 + 9 × 1], 2, = 5 [ 4 + 9 ] = 65, , N, , +  + 1 th observation,  2, , , 2, 12,   th observation,  2, , ⇒, , 12 , +  + 1 th observation, 2, , =, 2, 6th observation + 7th observation, =, 2, 13 + 14 27, =, =, 2, 2, M = 13.5, , xi, , | xi − M |, , 10, , | 10 − 13.5 | = 3.5, , 11, , | 11 − 13.5 | = 2.5, , 11, , | 11 − 13.5 | = 2.5, , 12, , | 12 − 13.5 | = 1.5, , 13, , | 13 − 13.5 | = 0.5, , 13, , | 13 − 13.5 | = 0.5, , 14, , | 14 − 13.5 | = 0.5, , 16, , | 16 − 13.5 | = 2.5, , 16, , | 16 − 13.5 | = 2.5, , 17, , | 17 − 13.5 | = 3.5, , 17, , | 17 − 13.5 | = 3.5, , 18, , and, , Σxi2 = 22 + 32 + 4 2 + 52 + ... + 112, = (12 + 22 + 32 + K + 112 ) − (12 ), 11 × 12 × 23, =, −1, 6, 11 × 12 × 23 − 6, =, 6, = 505, , Variance ( σ 2 ) =, =, , then variance is also multiplied by 2., We are given, 2, 4, 5, 6, 8, 17., When each observation multiplied by 2,, we get 4, 8, 10, 12, 16, 34., ∴Variance of new series, = 2 × Variance of given data, = 2 × 23.33 = 46.66, , 2, , 20. We have the following numbers, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, If 2 is added to each number, we get, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, Sum of these numbers,, Σ xi = 3 + K + 11 + 12 = 75, Sum of squares of these numbers,, Σxi2 = 32 + K + 112 + 122 = 645, Σxi2  Σxi , −, ,  n , n, 645, =, − ( 7.5) 2, 10, = 64.5 − 56.25, = 8.25, , Variance ( σ 2 ) =, , Σ | x i − M | = 28, , 18. When each observation is multiplied by 2,, , 505  65, − , 10 10 , , 2, , = 50.5 − 42.25 = 8.25, , | 18 − 13.5 | = 4.5, , ∴Mean deviation about median, Σ | xi − M | 28, =, =, = 2.33, n, 12, , ∑ xi2  ∑ xi , −, ,  n , n, , 2, , 21., Class interval, , Mid-value ( x i ), , fi, , 4-8, , 6, , 3, , 8-12, , 10, , 6, , 12-16, , 14, , 4, , 16-20, , 18, , 7

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122, , CBSE New Pattern ~ Mathematics XI (Term I), , Σf i xi, Σf i, 3 × 6 + 6 × 10 + 4 × 14 + 7 × 18, =, = 13, 20, Σf ( x − x ) 2, Variance ( σ 2 ) = i i, Σf i, , Mean ( x ) =, , =, , 3 ( −7 ) 2 + 6 ( −3) 2 + 4(1) 2 + 7( 5) 2, = 19, 20, , 22., xi, , x i2, , 6, , 36, , 7, , 49, , 10, , 100, , 12, , 144, , 13, , 169, , 4, , 16, , 8, , 64, , 12, , 144, , Total = 72, , 722, , Variance ( σ 2 ) =, , 3 ( −7 ) 2 + 6 ( −3) 2 + 4(1) 2 + 7( 5) 2, 20, 147 + 54 + 4 + 175, =, = 19, 20, , =, , 24. Since, marks obtained by 9 students in, Mathematics are 50, 69, 20, 33, 53, 39, 40,, 65 and 59., Rewrite the given data in ascending order., 20,33, 39, 40, 50, 53, 59, 65, 69,, Here,, [odd], n =9,  9 + 1, ∴ Median = ,  term = 5 th term,  2 , Me = 50, , xi, , Σx, 72, Mean = i =, =9, n, 8, Variance =, =, , Σxi2  Σx i , −, ,  n , n, 722  72, − , 8, 8, , Σf i ( xi − x ) 2, Σf i, , 2, , 2, , 20, , 30, , 33, , 17, , 39, , 11, , 40, , 10, , 50, , 0, , 53, , 3, , 59, , 9, , 65, , 15, , 69, , 19, , N =2, , = 90.25 − 81, = 9.25, , ∴, , 23., Class interval, , Mid value ( x i ), , fi, , 4-8, , 6, , 3, , 8-12, , 10, , 6, , 12-16, , 14, , 4, , 16-20, , 18, , 7, , Σf x, Mean ( x ) = i i, Σf i, 3 × 6 + 6 × 10 + 4 × 14 + 7 × 18, =, 20, = 13, , d i = xi − Me, , Σ d i = 114, MD =, , 114, = 1267, ., 9, , 25. Given, data are 6, 5, 9, 13, 12, 8, and 10., xi, , xi2, , 6, , 36, , 5, , 25, , 9, , 81, , 13, , 169, , 12, , 144, , 8, , 64, , 10, Σ xi = 63, , 100, Σxi2, , = 619

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123, , CBSE New Pattern ~ Mathematics XI (Term I), , Σ xi2  Σ xi , −, ,  N , N, , ∴ SD = σ =, , 2, , 28. Given observations are 6, 7, 10, 12, 13, 4, 8, 12., , 2, , =, , 619  63, 7 × 619 − 3969, −  =, 7, 7, 49, , =, , 4333 − 3969, 364, 52, =, =, 49, 49, 7, , 26., , ∴ SD, σ =, =, , 6, , 36, , 7, , 49, , 10, , 1, , 10, , 100, , 12, , 3, , 12, , 144, , Total, , 74, , Total, , 13, , 169, , 4, , 16, , 8, , 64, , 12, , 144, ∑, , ∑ xi2  ∑ xi , −, ,  N , N, 722  72, − , 8, 8, , x i2, , = 722, , 2, , 2, , 27. Here, x = 50, n =100 and σ = 4, , and, ⇒, ⇒, ⇒, ∴, , xi, , xi, , = 90.25 − 81, = 9.25, , ⇒, , xi − x ( xi − x ) 2, , xi, , x i2, , ∑ x i = 72, , ∴, , Number of observations = 8, 6 + 7 + 10 + 12 + 13 + 4 + 8 + 12, ∴Mean, ( x ) =, 8, 72, =, =9, 8, Now, let us make the following table for, deviation., , Σxi, = 50, 100, Σxi = 5000, σ2 =, , Σxi2  Σxi , −, ,  n , n, , 2, , Σxi2, − ( 50 ) 2, 100, Σ xi2, 16 =, − 2500, 100, Σ xi2, = 16 + 2500 = 2516, 100, Σ xi2 = 251600, ( 4 )2 =, , Hence, required answer is 5000, 251600., , xi − x ( xi − x ) 2, , 6, , −3, , 9, , 13, , 4, , 16, , 7, , −2, , 4, , 4, , −5, , 25, , 1, , 8, , −1, , 1, , 9, , 12, , 3, , 9, 74, , ∴ Sum of squares of deviations, 8, , =, , Σ ( xi, i =1, , − x ) 2 = 74, , 8, , Σ ( xi − x )2, i =1, , 74, =, = 9.25, n, 8, 29. Let x be the mean of the given set of, observations., Number of observations = 10, 45 + 60 + 62 + 60 + 50 + 65 + 58 , , + 68 + 44 + 48 , ∴ x =, 10, 560, =, = 56, 10, Make a table from the given data., Variance, σ 2 =, , xi, , xi − x, , ( x i − x )2, , 45, , 45 − 56 = − 11, , 121, , 60, 62, 60, 50, 65, 58, 68, 44, 48, , 60 − 56 = 4, 62 − 56 = 6, 60 − 56 = 4, 50 − 56 = − 6, 65 − 56 = 9, 58 − 56 = 2, 68 − 56 = 12, 44 − 56 = − 12, 48 − 56 = − 8, , 16, 36, 16, 36, 81, 4, 144, 144, 64, , Total, , 662, , We have, n = 10 and Σ ( xi − x ) = 662, 1, 662, = 66.2, ∴Variance, σ 2 = Σ( xi − x ) 2 =, n, 10, 2

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124, , CBSE New Pattern ~ Mathematics XI (Term I), , 30., xi, 10, 15, 18, 20, 25, Total, , fi, , f i xi, , 3, 30, 2, 30, 5, 90, 8, 160, 2, 50, N = Σf i Σf i xi, , xi − x, 2, 2, = ( xi − 18 ) ( xi − x ) f i ( xi − x ), −8, −3, 0, 2, 7, , 64, 9, 0, 4, 49, , 192, 18, 0, 32, 98, Σf i ( xi − x ) 2, = 340, , = 20 = 360, , N = 20 and Σf i xi = 360, 1, 360, Mean, ( x ) =, = 18, ∴, Σf i xi =, N, 20, 1, Now, variance, ( σ 2 ) =, Σf i ( xi − x ) 2, N, 1, =, × 340 = 17, 20, Here,, , 31. Assertion The deviation of an observation x, , from a fixed value ‘a ’ is the difference ( x − a )., In order to find the dispersion of values of x, from a central value a , we find the deviations, about a . An absolute measure of dispersion is, the mean of these deviations., Reason To find the mean, we must obtain, the sum of the deviations. But, we know that, a measure of central tendency lies between, the maximum and the minimum values of, the set of observations., Therefore, some of the deviations will be, negative and some positive. Thus, the sum of, deviations may vanish. Moreover, the sum of, the deviations from mean ( x ) is zero. Also,, Mean of deviations, Sum of deviations, 0, =, = =0, Number of observations n, Thus, finding the mean of deviations about, mean is not of any use for us, as far as the, measure of dispersion is concerned., Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., , 32. Assertion Mean of the given series, , Σ xi, n, 4 + 7 + 8 + 9 + 10 + 12 + 13 + 17, =, = 10, 8, , x =, , Sum of terms, , Number of terms, , =, , xi, , | xi − x |, , 4, , | 4 − 10 | = 6, , 7, , | 7 − 10 | = 3, , 8, , | 8 − 10 | = 2, , 9, , | 9 − 10 | = 1, , 10, , |10 − 10 | = 0, , 12, , |12 − 10 | = 2, , 13, , |13 − 10 | = 3, , 17, Σx i = 80, , |17 − 10 | = 7, Σ| x i − x | = 24, , ∴ Mean deviation about mean, Σ| xi − x | 24, =, =, =3, n, 8, Reason Mean of the given series, Sum of terms, Σx, x =, = i, n, Number of terms, 38 + 70 + 48 + 40 + 42 + 55, + 63 + 46 + 54 + 44, =, = 50, 10, , xi, , | xi − x |, , 38, , | 38 − 50 | = 12, , 70, , | 70 − 50 | = 20, , 48, , | 48 − 50 | = 02, , 40, , | 40 − 50 | = 10, , 42, , | 42 − 50 | = 08, , 55, , | 55 − 50 | = 05, , 63, , | 63 − 50 | = 13, , 46, , | 46 − 50 | = 04, , 54, , | 54 − 50 | = 04, , 44, , | 44 − 50 | = 06, , Σx i = 500, , Σ | x i − x | = 84, , ∴Mean deviation about mean, Σ| xi − x |, =, n, 84, =, = 8.4, 10, Hence, Assertion is true and Reason is false.

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125, , CBSE New Pattern ~ Mathematics XI (Term I), , 33. Assertion, xi, , fi, , f i xi, , | xi − x |, , f i | xi − x |, , 5, , 7, , 35, , | 5 − 14 | = 9, , 63, , 10, , 4, , 40, , |10 − 14 | = 4, , 16, , 15, , 6, , 90, , |15 − 14 | = 1, , 06, , 20, , 3, , 60, , | 20 − 14 | = 6, , 18, , 25, , 5, , 125, , | 25 − 14 | = 11, , 55, , Total Σf i = 25 350, Mean ( x ) =, , 158, , Σf i xi 350, =, = 14, Σf i, 25, , ∴ Mean deviation about mean, Σf | x − x | 158, = i i, =, = 6.32, Σf i, 25, , fi, , xi, , fi, , f i xi, , | xi − x |, , | xi − x |, , 10, , 4, , 40, , |10 − 50 | = 40, , 160, , 30, , 24, , 720, , | 30 − 50 | = 20, , 480, , 50, , 28, , 1400, , | 50 − 50 | = 00, , 000, , 70, , 16, , 1120, , | 70 − 50 | = 20, , 320, , 90, , 8, , 720, , | 90 − 50 | = 40, , 320, , Total, , Σf i = 80, , Mean =, , Σf i x i, , 35. Assertion The sum of the deviations from, the mean (minus signs ignored) is more than, the sum of the deviations from median., Therefore, the mean deviation about the, mean is not very scientific. Thus, in many, cases, mean deviation may give unsatisfactory, results. Also mean deviation is calculated on, the basis of absolute values of the deviations, and therefore, cannot be subjected to further, algebraic treatment. This implies that we, must have some other measure of dispersion., Standard deviation is such a measure of, dispersion., Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., , 36. Assertion Let the number of boys and girls, , Reason, , = 4000, , Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., , 1280, , Σf i xi 4000, =, = 50, Σf i, 80, , ∴ Mean deviation about mean, Σf | x − x |, = i i, Σf i, 1280, =, = 16, 80, Hence, Assertion is true and Reason is false., , 34. Assertion In a series, where the degree of, variability is very high, the median is not a, representative central tendency. Thus, the, mean deviation about median calculated for, such series can not be fully relied., , be x and y., ∴, 52x + 42y = 50( x + y ), ⇒, 2x = 8 y, ⇒, x = 4y, ∴Total number of students in the class, = x + y = 5y, ∴ Required percentage of boys, 4y, =, × 100%, 5y, = 80%, Reason Let the number of boys be x and, number of girls be y., ∴, 53 ( x + y ) = 55y + 50 x, ⇒, 3x = 2y, 2y, ⇒, x=, 3, ∴ Total number of students, 2y, 5, =x +y=, +y= y, 3, 3, Hence, required percentage, y, =, × 100%, 5y / 3, 3, = × 100% = 60%, 5, Hence, Assertion is true and Reason is false., , 37. Assertion Since, 44 kg is replaced by 46 kg, and 27 kg is replaced by 25 kg, then the, given series becomes 31, 35, 25, 29, 32, 43,, 37, 41, 34, 28, 36, 46, 45, 42, 30.

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126, , CBSE New Pattern ~ Mathematics XI (Term I), , On arranging this series in ascending order,, we get, 25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42,, 43, 45, 46., Total number of students are 15, therefore, middle term is 8th whose corresponding value, is 35., Reason On arranging the terms in increasing, order of magnitude, 40, 42, 45, 47, 50, 51, 54, 55, 57, , 39. Assertion Presenting the data in tabular, , Number of terms, N = 9,  9 + 1, ∴ Median = ,  th term = 5th term = 50 kg,  2 , , Weight, (in kg), , Deviation from, median (d ), , |d |, , 40, , −10, , 10, , 42, , −8, , 8, , 45, , −5, , 5, , 47, , −3, , 3, , 50, , 0, , 0, , 51, , 1, , 1, , 54, , 4, , 4, , 55, , 5, , 5, , 57, , 7, , 7, | d | = 43, , 43, = 4.78 kg, 9, Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion., , MD from median =, , 38. Assertion In the calculation of variance, we, find that the units of individual observations, xi and the unit of their mean x are different, from that of variance, since variance involves, the sum of squares of ( xi − x )., For this reason, the proper measure of, dispersion about the mean of a set of, observations is expressed as positive, square-root of the variance and is called, standard deviation., Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., , form, we get, , xi − x ( xi − x ) 2 f i ( xi − x ) 2, , xi, , fi, , f i xi, , 4, , 3, , 12, , −10, , 100, , 300, , 8, , 5, , 40, , −6, , 36, , 180, , 11, , 9, , 99, , −3, , 9, , 81, , 17, , 5, , 85, , 3, , 9, , 45, , 20, , 4, , 80, , 6, , 36, , 144, , 24, , 3, , 72, , 10, , 100, , 300, , 32, , 1, , 32, , 18, , 324, , 324, , 30, , 420, , N = 30,, , 7, , 1374, , ∑ f i xi, , 7, , = 420,, , ∑ f i ( xi, , − x ) 2 = 1374, , i =1, , i =1, , 7, , Therefore,, , x =, , ∴Variance ( σ 2 ) =, , ∑ f i xi, , i =1, , N, 1, N, , 7, , =, , 1, × 420 = 14, 30, , ∑ f i ( xi, , − x )2, , i =1, , 1, × 1374 = 458, ., 30, Reason Standard deviation, =, , ( σ ) = 458, . = 6.77, Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion., , 40. Assertion, xi, , fi, , x i2, , f i xi, , f i x i2, , 6, , 2, , 36, , 12, , 72, , 10, , 4, , 100, , 40, , 400, , 14, , 7, , 196, , 98, , 1372, , 18, , 12, , 324, , 216, , 3888, , 24, , 8, , 576, , 192, , 4608, , 28, , 4, , 784, , 112, , 3136, , 30, , 3, , 900, , 90, , 2700, , Total = 130, , 40, , 760, , 16176

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127, , CBSE New Pattern ~ Mathematics XI (Term I), , Mean ( x ) =, , Σf i x i 760, =, = 19, Σf i, 40, , Reason Variance =, , Σf i xi2  Σf i x i , −, ,  Σf i , Σf i, , 16176  760 , −, ,  40 , 40, = 404.4 − (19 ) 2, =, , If a is added to each observation, the new, observations will be, …(i), yi = xi + a, Let the mean of the new observations be y., Then,, 1 n, 1 n, y = ∑ yi = ∑ ( xi + a ), n i =1, n i =1, , 2, , 2, , = 404.4 − 361 = 43.4, Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion., , 41. Assertion, Mid, value, (xi ), , i.e., , Deviation, Frequency from mean, (f i ), di = xi − A,, A = 64, , di 2, , f i di, , f i di 2, , 60, , 2, , −4, , 16, , −8, , 32, , 61, , 1, , −3, , 9, , −3, , 9, , 62, , 12, , −2, , 4, , − 24, , 48, , 63, , 29, , −1, , 1, , − 29, , 29, , 64, , 25, , 0, , 0, , 0, , 0, , 65, , 12, , 1, , 1, , 12, , 12, , 66, , 10, , 2, , 4, , 20, , 40, , 67, , 4, , 3, , 9, , 12, , 36, , 68, , 5, , 4, , 16, , 20, , 80, , Total, , 100, , 0, , 0, , 286, , Σf d, Mean ( x ) = A + i i, Σf i, 0, ( x ) = 64 +, = 64, 100, Reason Standard deviation ( σ ), =, , Σf idi2  Σf id i , −, ,  Σf i , Σf i, , 2, , 2, , 286  0 , . = 169, ., =, −,  = 286, 100 100 , Hence Assertion is false and Reason is true., , 42. Assertion Let x be the mean of x1 , x 2 ..., xn ., Then, variance is given by, 1 n, σ12 = ∑ ( xi − x ) 2, n i =1, , =, , n, , 1 n,  ∑ xi + ∑ a , n i = 1, i =1 , , , , =, , na, 1 n, xi +, =x +a, n i∑, n, =1, , y =x +a, , ...(ii), , Thus, the variance of the new observations is, 1 n, 1 n, σ 22 = ∑ ( yi − y ) 2 = ∑ ( xi + a − x − a ) 2, n i =1, n i =1, [using Eqs. (i) and (ii) ], 1 n, = ∑ ( xi − x ) 2 = σ12, n i =1, Thus, the variance of the new observations is, same as that of the original observations., Reason We may note that adding, (or subtracting) a positive number to, (or from) each observation of a group does, not affect the variance., Hence, Assertion and Reason both are true and, Reason is the correct explanation of Assertion., , 43. Assertion Given,, Σ ( x − 5) = 3, Σx − Σ 5 = 3, Σx − 5 × 18 = 3, Σx = 3 + 90, Σx = 93, Σ ( x − 5) 2 = 43, , ∴, ⇒, ⇒, ⇒, Now,, ⇒, , Σ( x 2 + 25 − 10 x ) = 43, , ⇒, , Σx 2 + Σ 25 − 10 Σx = 43, , [Qn = 18 ], , ⇒ Σx 2 + 25 × 18 − 10 × 93 = 43, ⇒, , Σx 2 = 43 + 930 − 450, , ⇒, , Σx 2 = 973 − 450, , ⇒, , Σx 2 = 523, , Now, mean =, , Σx 93, =, = 5.16, n, 18

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128, , CBSE New Pattern ~ Mathematics XI (Term I), , Reason SD ( σ) =, , Σx 2  Σx , − ,  n , n, , 2, , =, , 523  93, − , 18 , 18, , 2, , =, , 523 × 18 − 93 × 93, 18 × 18, , ⇒, Since,, ⇒, , Similarly, Σyi2 = 105, , 1, 9414 − 8649, 18, 1, 27. 66, =, = 1.54, 765 =, 18, 18, Hence Assertion is false and Reason is true., =, , 44. Assertion Sum of n even natural numbers, = n(n + 1), n(n + 1), Mean ( x ) =, =n + 1, n, 1, , Variance =, Σ ( xi ) 2 − ( x ) 2, n, , 1, = [ 22 + 4 2 + ... + ( 2n ) 2 ] − (n + 1) 2, n, 1, = 22 [12 + 22 + ... + n 2 ] − (n + 1) 2, n, 4 n (n + 1) ( 2n + 1), =, − (n + 1) 2, n, 6, (n + 1) [ 2 ( 2n + 1) − 3 (n + 1)], =, 3, (n + 1) (n − 1), =, 3, n2 − 1, =, 3, Hence Assertion is false and Reason is true., , 45. Assertion Mean of 12 , 22 , 32 , ...,n 2 is, , ∴, , 2, , 2, , 2, , 22n 2 + 33n + 11 − 276 n = 0, , ⇒, , (n − 11) ( 22n − 1) = 0, , ⇒, , n = 11 and n ≠, , Reason Q σ 2x = 4 and σ 2y = 5, Also,, Now,, , —, , x = 2 and y = 4, Σxi, = 2 ⇒ Σxi = 10, 5, , 46. (i) We have, n = 200, incorrect mean = 40, and incorrect standard deviation = 15, Now, incorrect mean = 40, Incorrect Σxi, = 40, ⇒, 200, Incorrect Σxi = 8000, Correct Σxi = 8000 − ( 34 + 53) + ( 43 + 35), = 8000 − 87 + 78 = 7991, (ii) Correct mean =, n, , ∑ ( xi, , (iii), , 7991, = 39.955, 200, , − x )2, , i =1, , n, (iv) Incorrect SD = 15, , 2, , ⇒, , 2, , 1, x + y, ( Σxi2 + Σyi2 ) − , ,  2 , 10, 1, = ( 40 + 105) − 9, 10, 55 11, =, =, 10, 2, Hence, Assertion and Reason both are true, and Reason is not the correct explanation of, Assertion., σ z2 =, , ∴, , ⇒ Incorrect variance = (15) 2 = 225, , 1 + 2 + 3 + ... + n, Σn, =, n, n, 46 n n(n + 1) ( 2n + 1), =, 11, 6n, 2, , Σyi, =4, 5, Σyi = 20, 1, σ 2x = ( Σxi2 ) − ( x ) 2, 5, Σxi2 = 40, , ⇒, , Incorrect Σxi2, − (Incorrect mean) 2 = 225, 200, , ⇒, , Incorrect Σxi2, − ( 40 ) 2 = 225, 200, , ⇒ Incorrect Σxi2 = 200(1600 + 225), 1, 22, , = 200 × 1825, = 365000, Now, Correct Σxi2 = Incorrect Σxi2, − ( 34 2 + 532 ) + ( 432 + 352 ), = 365000 − 3965 + 3074, = 364109

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129, , CBSE New Pattern ~ Mathematics XI (Term I), , So, correct variance, 1, =, (correct Σxi2 ) − (correct mean )2, 200, =, , 1,  7991, ( 364109 ) − , ,  200 , 200, , 2, , = 1820.545 − 1596 . 402 = 224 .143, (v) Correct standard deviation, = correct variance, = 224 .143, , [using part (iv)], , = 14.971, 34, 66, 30, 38, 44, 50, 40, 60, 42, 51., Here, number of observations, n = 10, 34 + 66 + 30 + 38 + 44 + 50, + 40 + 60 + 42 + 51, ∴ Mean, x =, 10, 455, x =, = 45.5, ⇒, 10, (ii) Let us make the table for absolute deviation, , xi, , xi − x, , | xi − x |, , 34, , 34 − 45.5 = − 11.5, , 11.5, , 66, , 66 − 45.5 = 20.5, , 20.5, , 30, , 30 − 45.5 = − 15.5, , 15.5, , 38, , 38 − 45.5 = − 7.5, , 7.5, , 44, , 44 − 45.5 = − 1.5, , 1.5, , 50, , 50 − 45.5 = 4 .5, , 4.5, , 40, , 40 − 45.5 = − 5.5, , 5.5, , 60, , 60 − 45.5 = 14 .5, , 14.5, , 42, , 42 − 45.5 = − 3.5, , 3.5, , 51, , 51 − 45.5 = − 5.5, , 5.5, , 10, , Now, MD =, , i =1, , 10, , =, , ∴ Median ( M ) =, , , n, +  + 1 th observation, 2 , , 2, 10 , 10, ,   th observation +  + 1 th observation,  2,  2, , =, 2, ( 5th observation + 6th observation ), 2, 42 + 44 86, =, =, = 43, 2, 2, (iv) Let us make the table for absolute deviation, =, , 47. (i) Given, observations are, , Σ | xi − x |, , Here, total number of observations are 10., i.e. n = 10, which is even., n ,   th observation,  2, , 90, = 9.0, 10, , (iii) The given data can be arranged in, ascending order as 30, 34, 38, 40, 42, 44,, 50, 51, 60, 66., , | xi − M |, , xi, 30, 34, 38, 40, 42, 44, 50, 51, 60, 66, , |30 − 43 | = 13, |34 − 43 | = 9, |38 − 43 | = 5, |40 − 43 | = 3, |42 − 43 | = 1, |44 − 43 | = 1, |50 − 43 | = 7, |51 − 43 | = 8, |60 − 43 | = 17, |66 − 43 | = 23, , Total, , ∑ | x i − M | = 87, , 10, , i =1, , Now, mean deviation about median,, 10, , MD =, , ∑, i =1, , | xi − M |, , 10, 87, =, = 8.7, 10, (v) The difference between mean deviation, about the mean and mean deviation about, the median = 9.0 − 8.7, = 0.3

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130, , CBSE New Pattern ~ Mathematics XI (Term I), , 6 + 8 14, =, =7, 2, 2, [Q 20th observation lies in the cumulative, frequency 20 and its corresponding, observation is 6 and the 21st observation lies, in the cumulative frequency 27 where the, corresponding observation is 8], , 48. (i) Let us make the following table from the, , =, , given data., , xi, , fi, , f i xi, , | xi − x |, , f i | xi − x |, , 2, , 2, , 4, , 5.5, , 11, , 5, , 8, , 40, , 2.5, , 20, , 6, , 10, , 60, , 1.5, , 15, , 8, , 7, , 56, , 0.5, , 3.5, , 10, , 8, , 80, , 2.5, , 20, , 12, , 5, , 60, , 4.5, , Total, , 40, , 300, , (iv), , xi, , fi, , | x i − 7|, , f i | x i − 7|, , 22.5, , 2, , 2, , 5, , 10, , 92, , 5, , 8, , 2, , 16, , 6, , 10, , 1, , 10, , 8, , 7, , 1, , 7, , 10, , 8, , 3, , 24, , 12, , 5, , 5, , 25, , Here, N = Σ f i = 40, Σ f i x i = 300, 1, 1, × 300 = 7.5, Σ f i xi =, N, 40, (ii) Mean deviation about the mean,, 1, 1, MD ( x ) =, Σ f i | xi − x | =, × 92 = 2.3, N, 40, Hence, the mean deviation about mean, is 2.3., (iii) The given observations are already in, ascending order. Now, let us make the, cumulative frequency., Now, mean ( x ) =, , Here, Σf i | xi − M | = 92, and, Σf i = 40, ∴The required mean deviation, Σf | x − M | 92, = i i, =, = 2.3, Σf i, 40, , xi, , fi, , c f, , 2, , 2, , 2, , 5, , 8, , 10, , 6, , 10, , 20, , 8, , 7, , 27, , xi, , fi, , f i xi, , 10, , 8, , 35, , 4, , 3, , 12, , 12, , 5, , 40, , Total, , N = 40, , Here, Σf i = 40, which is even., ∴ Median,,  40 , Value of   th observation,  2, ,  40, + Value of , + 1 the observation,  2, , =, 2, Value of 20th observation, =, , 92, , Total, , + Value of 21st observation, 2, , (v) The difference between mean and median, = 7.5 − 7 = 0.5, , 49. Let us make the following table from the, given data., , xi − x, ( x i − x )2 f i ( x i − x )2, = x i − 14, − 10, , 100, , 300, , 8, , 5, , 40, , −6, , 36, , 180, , 11, , 9, , 99, , −3, , 9, , 81, , 17, , 5, , 85, , 3, , 9, , 45, , 20, , 4, , 80, , 6, , 36, , 144, , 24, , 3, , 72, , 10, , 100, , 300, , 18, , 324, , 32, , 1, , 32, , Total, , 30, , 420, , 324, 1374, , Here, we have, N = Σf i = 30, Σf i xi = 420, 7, , x =, , Σ f i xi, , i =1, , N, , =, , 420, = 14, 30

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131, , CBSE New Pattern ~ Mathematics XI (Term I), , Σf i xi, N, 1, (ii) σ 2 =, Σf i ( xi − x ) 2, N, (iii) x = 14, 1 7, (iv) Variance ( σ 2 ) =, Σ f i ( xi − x ) 2, N i =1, , 1  n, , 2, f i ( xi − x ) 2 , Q σ = N i Σ, =, 1, , , , 1, =, × 1374 = 458, ., 30, (v) Standard deviation,, (i) x =, , σ = σ 2 = 458, . = 6.77, , 50. Let us make the following table from the, given data., , Class f i, 0-10, , cf, , Mid-point x i − M ,, ( xi ), M = 28, 5, 23, , f i xi − M, , 6, , 6, , 10-20 7, , 13, , 15, , 13, , 91, , 20-30 15 28, , 25, , 3, , 45, , 30-40 16 44, , 35, , 7, , 112, , 40-50 4, , 48, , 45, , 17, , 68, , 50-60 2, , 50, , 55, , 27, , Total 50, , 138, , 54, 508, , N, 50, =, = 25, 2, 2, which item lies in the cumulative frequency, 28. Therefore, 20-30 is the median class., So, we have, l = 20, cf =13, f = 15, h = 10 and, N = 50, N, − cf, Now, Median, M = l + 2, ×h, f, 25 − 13, × 10, = 20 +, 15, = 20 + 8 = 28, N, − cf, (i) Median, M = l + 2, ×h, f, Here,, , (ii) Mean deviation, MD =, , Σf i | xi − M |, N, , (iii) N = ∑ f i = 50, (iv) Median = 28, (v) The mean deviation about median is given, by, MD(M) =, , 1, N, , 6, , ∑ fi, , i =1, , xi − M, , 1, × 508 = 1016, ., 50, Hence, the mean deviation about median, is 10.16., =

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132, , CBSE New Pattern ~ Mathematics XI (Term I), , PRACTICE PAPER 1, Mathematics Class 11th (Term I), Instructions, 1. This paper has 40 questions., 2. All questions are compulsory., 3. Each question carry 1 mark., 4. Answer the questions as per given instructions., Time : 90 Minutes, , Max. Marks : 40, , Multiple Choice Questions, 1. Suppose the function is defined by, , | x − 3|, x≠3, , , then the left hand, f (x ) =  x − 3,  0,, x=3, limit of f (x ) at x = 3 is, (a) 0, (c) − 2, , (b) 1, (d) − 1, , 2. Which of the following relations are, functions?, R1 = {( 2, 1), (5, 1), (8, 1), (11, 1),, (14, 1), (17, 1)}, R 2 = {( 2, 1), ( 4, 2), (6, 3), (8, 4 ), (10, 5),, (12, 6 ), (14, 7 )}, R 3 = {(1, 3), (1, 5), ( 2, 5)}, (a) R1 and R2, (c) R1, R2 and R3, , (b) R2 and R3, (d) None of these, , 3. The income of a person is ` 300000 in, the first year and he receives and, increment of ` 10000 to his income per, year for the next 19 yr. Then, the total, amount he received in 20 yr, is, (a), (b), (c), (d), , 7900000, 6000000, 8000000, 6900000, , 1 − sin, , 4. lim, , x→ π, , x, 2, , x, x, x, cos  cos − sin , 2, 4, 4, , (a) 1, , (b), , 1, (c), 2, , (d) −, , equals to, , 2, 1, 2, , 5. Let A = {x , y , z } and B = {1, 2}. Then, the, number of relations from A to B is, (a) 32, (c) 128, , (b) 64, (d) 8, , 6. The tangent of an angle between the, 2m, x y, x y, ,, + = 1 and − = 1 is, n, a b, a b, where m and n respectively are, lines, , (a) a2 − b 2 , ab, (c) a2 + b 2 , ab, , (b) ab , a2 − b 2, (d) ab , a2 + b 2, , 7. The six numbers which will be inserted, between 3 and 24 such that resulting, sequence become an AP, are, (a) 6, 9, 12, 15, 19, 22, (c) 6, 8, 10, 12, 14, 16, , (b) 4, 6, 8, 10, 12, 14, (d) 6, 9, 12, 15, 18, 21, , 8. The coordinates of the foot of the, perpendicular from the point (2, 3) on, the line x + y − 11 = 0 are, (a) (−6, 5), (c) (−5, 6), , (b) (5, 6), (d) (6, 5)

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133, , CBSE New Pattern ~ Mathematics XI (Term I), , 9. lim, , x→ 0, , x (e x − 1), 1 − cos x, , is equals to, , (a) 4, (c) 2, , (b) 3, (d) 1, , 10. The equations of the side AC of a, triangle whose vertices are, A( −1, 8 ), B ( 4, − 2) and C( −5, − 3) is, (a), (b), (c), (d), , 11x + 4 y + 43 = 0, 11x − 4 y + 43 = 0, 11x − 4 y − 43 = 0, None of the above, , (b) 492, (d) 492 + 2, , 12. Let A = {1, 2, 3, 5} and B = {4, 6, 9} and a, relation R from A to B is defined by, R = {(x , y ) the difference between x and, y is odd; x ∈ A and y ∈ B }. Then, Roster, form of R is …K… Here, K refers to, {(1, 4), (1, 6), (1, 9)}, {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 9)}, {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5,6)}, None of the above, , 1 − cos 4θ, is equal to, θ → 0 1 − cos 6θ, , 13. lim, , 4, 9, −1, (c), 2, , (a), , (b), , 1, 2, , (d) −1, , 14. Let A ={1, 2, 3} and {−1, 0, 1}. Then,, which of the following is not the, relation from A to B?, (a), (b), (c), (d), , R = {(1, − 1), (2, 0), (3, 1)}, R = {(1, − 1), (1, 0), (2, 0), (2, 1)}, R = {(2, 0), (2, 1), (1 − 1), (3, 1), (3, 0)}, R = {(1, − 1), (−1, 1), (2, 0), (0, 2), (3, 1),(1, 3)}, , x 2 − x log x + 2 log x − 4, 15. lim, is equals, x→ 2, x−2, to, (a) 4 − log 2, (c) −4 − log 2, , (a) {0, 3, 4, 5 }, (c) {0, 3, 4 }, , (b) {5, 4, 3, 0}, (d) {5, 4, 3}, , 17. If the sum of n terms of an AP is given, by S n = 3n + 2n 2 , then the common, difference of the AP is, (b) 2, (d) 4, , 18. If R2 = {x , y )| x and y are integers and, , 2 + 3 + 6 +11 + 18 + ..., then t 50 is, , (a), (b), (c), (d), , then the range of R is, , (a) 3, (c) 6, , 11. If t n denotes the nth term of the series, (a) 492 − 1, (c) 502 + 1, , 16. If R = {(x , y ): x , y ∈W , x 2 + y 2 = 25},, , (b) 4 + log 2, (d) log 4, , x 2 + y 2 = 64} is a relation, then the, value of R 2 is, (a), (b), (c), (d), , {(0, 0), (0, 8), (8, 0), (8, 8)}, {(0, 0), (0, − 8), (−8, 0), (8, 8)}, {(0, 8), (0, − 8), (8, 0)}, {(0, 8), (0, − 8), (8, 0), (−8, 0)}, , 19. The rth term of an AP sum of whose, first n terms is 2n + 3n 2 , is, (a) 6 r − 1, (c) 1 − 6 r, , (b) 6 r + 1, (d) 1 + r, , 20. The equation of the line upon which, the length of perpendicular p from, origin and angle α made by this, perpendicular with the positive, direction of X -axis are p = 4, α = 120 °,, is, (a) x + 3y = 8, (c) − x + 3y = 8, , (b) x − 3y = 8, (d) − x − 3y = 8, , Assertion-Reasoning MCQs, For question numbers 21 to 25, two, statements are given-one labelled, Assertion (A) and the other labelled, Reason (R). Select the correct answer to these, questions from the codes (a), (b), (c) and (d) as, given below., (a) Both A and R are true and R is the correct, explanation of A., (b) Both A and R are true but R is not the, correct explanation of A., (c) A is true but R is false., (d) A is false and R is true.

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134, , CBSE New Pattern ~ Mathematics XI (Term I), , 21. Let A = {1, 2, 3, 4, 6}. If R is the relation, on A defined by {(a, b ) : a, b ∈ A, b is, exactly divisible by a}., Assertion (A) The relation R in roster, form is {(6, 3), (6, 2), ( 4, 2)}., Reason (R) The domain and range of R, is {1, 2, 3, 4, 6}., , 22. If the sequence is 54, 51, 48, ……, then, the sum of first, Assertion (A) 18 terms is 513, Reason (R) 19 terms is 513, , 23. Assertion (A), a × b = ab ∀a, b ∈, (0, ∞ )., Reason (R) a × b = ab , when, a > 0, b < 0 or a < 0, b > 0., x, 2, 5 1, 24. Assertion (A) If  + 1, y −  =  ,  ,, 3, 3  3 3, then the values of x and y are 2 and 1,, respectively., Reason (R) If the set A has 3 elements, and the set B = {3, 4, 5}, then the, number of elements is ( A × B ) is 6., , 25. Assertion (A) lim (cosec x − cot x ) = 0, x→ 0, , Reason (R) lim, x→, , π, 2, , tan 2x, =1, π, x−, 2, , Case Based MCQs, Direction (Q. Nos. 26-30) Answer the, questions from based on the following case., Operations on Sets, There are some operations which when, performed on two sets give rise to another set., Here, we will define certain operations on set, and examine their properties., Union of Sets Let A and B be any two sets., The union of A and B is the set of all those, , elements which belong to either A or B or, both. It is denoted by A ∪ B and read as, A union B. The symbol ‘∪’ is used to denote, the union., ∴ A ∪ B = {x : x ∈ A or x ∈ B ), Intersection of sets Let A and B be any two, sets. The intersection of A and B is the set of, all those elements which belong to both A and, B. It is denoted by A ∩ B and read as A, intersection B. The symbol ‘∩’ is used to, denote the intersection., ∴ A ∩ B = {x : x ∈ A or x ∈ B ), Based on the above information answer the, following questions., , 26. If A = {a, e , i , o, u } and B ={a, c , d }, then, A ∪ B is, (a) {a, c, e,i,o, u }, (c) {a,c,d, e,i,o, u }, , (b) {a,c,d, e,i,o }, (d) {a,c,d,i,o, u }, , 27. If A = {x : x is a natural number and, 1 < x ≤ 5}, and B = {x : x is a natural number and, 5 < x ≤ 10},, then union of A and B is, (a), (b), (c), (d), , { 2, 3, 4, 5, 6, 7, 9, 10}, {2, 3, 4, 5, 6, 7, 8, 9, 10}, {2, 3, 4, 6, 7, 8, 9, 10}, {6, 7, 8, 9, 10}, , 28. The smallest set A such that, A ∪ {1, 2} = {1, 2, 3, 5, 9} is, (a) {1, 2, 3}, (c) {1, 2}, , (b) {2, 3, 5 }, (d) {3, 5, 9}, , 29. A = {e , f , g }, B = φ, then intersection of, A and B is, (a) { e, f, g }, (c) { e, f }, , (b) φ, (d) { f, g }, , 30. If X = {a, b , c , d } and y = { f , b , d , g }, then, X ∩ Y is, (a) {a,c, f, g }, (c) {b , d }, , (b) {a,b ,d, g }, (d) {c,d, f,b }

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135, , CBSE New Pattern ~ Mathematics XI (Term I), , Direction (Q. Nos. 31-35) Answer the, questions from based on the following case., We already know that, a × b = ab for all, positive real numbers a and b. This result also, holds true when either a > 0, b < 0 or a < 0,, b > 0. But above result is not true for a < 0,, b < 0, which can be explained as follows., Let us consider, i 2 = −1 −1 = ( −1) ( −1), [by assuming a × b = ab, for all real numbers], = 1 =1, which is a contradiction to the fact that, i 2 = −1., Therefore, a × b ≠ ab , if both a and b are, negative real numbers., Modulus (Absolute Value) of Complex, Numbers The modulus (or absolute value) of, a complex number, z = a + ib is defined as the, non-negative real number a 2 + b 2 . It is, denoted by | z |. i.e. | z | = a, , 2, , +b ., 2, , 31. The value of −25 + 3 −4 + 2 −9 is, , (c), , 19, 37, , (b), (d), , 34. The modulus of the conjugate of the, complex number − 3i is, , (b) −3, (d) None of these, , (a) 9, (c) 3, , 35. If z 1 = 3 + i and Z 2 = 1 + 4i , then, | z 1 − z 2 | is, (a), (c), , 10, 5, , (b), (d), , 11, 13, , Direction (Q. Nos. 36-40) Answer the, questions from based on the following case., The statistics data is as following, 6, 7, 10, 12, 13, 4, 8, 12, Based on the above data, answer the, following questions., , 36. The mean of the given data is, (a) 7, (c) 9, , (b) 8, (d) 10, , 37. The mean deviation about the mean is, (b) 2.75, (d) 3.25, , 38. The median is, (a) 9, (c) 11, , (b) 10, (d) 8, , (a) 9.00, (c) 9.20, , (b) 9.15, (d) 9.25, , 40. The standard deviation of the data is, , 19 i, 19 i, , (b) 3, (d) 5, , 39. Variance of the data is, , (b) 16 i, (d) 18 i, , 32. The imaginary part of 37 + − 19 is, (a), , (a) 2, (c) 4, , (a) 2.50, (c) 2.85, , Based on the above information, answer the, following questions., (a) 15 i, (c) 17 i, , 33. The modulus of 4 + 3i 7 is, , 2, , (a) 2.04, (c) 5.04, , (b) 4.04, (d) 3.04

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136, , CBSE New Pattern ~ Mathematics XI (Term I), , PRACTICE PAPER 1, OMRSHEET, Instructions, Use black or blue ball point pens and avoid gel pens and fountain, pens for filling the sheets, Darken the bubbles completely. Don't put a tick mark or a cross, mark half-filled or over-filled bubbles will not be read by the software., , ü, , û, , Incorrect, , Incorrect, , Incorrect, , Correct, , Do not write anything on the OMR Sheet, Multiple markings are invalid, 1, , a, , b, , c, , d, , 21, , a, , b, , c, , d, , 2, , a, , b, , c, , d, , 22, , a, , b, , c, , d, , 3, , a, , b, , c, , d, , 23, , a, , b, , c, , d, , 4, , a, , b, , c, , d, , 24, , a, , b, , c, , d, , 5, , a, , b, , c, , d, , 25, , a, , b, , c, , d, , 6, , a, , b, , c, , d, , 26, , a, , b, , c, , d, , 7, , a, , b, , c, , d, , 27, , a, , b, , c, , d, , 8, , a, , b, , c, , d, , 28, , a, , b, , c, , d, , 9, , a, , b, , c, , d, , 29, , a, , b, , c, , d, , 10, , a, , b, , c, , d, , 30, , a, , b, , c, , d, , 11, , a, , b, , c, , d, , 31, , a, , b, , c, , d, , 12, , a, , b, , c, , d, , 32, , a, , b, , c, , d, , 13, , a, , b, , c, , d, , 33, , a, , b, , c, , d, , 14, , a, , b, , c, , d, , 34, , a, , b, , c, , d, , 15, , a, , b, , c, , d, , 35, , a, , b, , c, , d, , 16, , a, , b, , c, , d, , 36, , a, , b, , c, , d, , 17, , a, , b, , c, , d, , 37, , a, , b, , c, , d, , 18, , a, , b, , c, , d, , 38, , a, , b, , c, , d, , 19, , a, , b, , c, , d, , 39, , a, , b, , c, , d, , 20, , a, , b, , c, , d, , 40, , a, , b, , c, , d, , Students should not write anything below this line, , SIGNATURE OF EXAMINER WITH DATE, , MARKS SCORED

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137, , CBSE New Pattern ~ Mathematics XI (Term I), , PRACTICE PAPER 2, Mathematics Class 11th (Term I), Instructions, 1. This paper has 40 questions., 2. All questions are compulsory., 3. Each question carry 1 mark., 4. Answer the questions as per given instructions., Time : 90 Minutes, , Max. Marks : 40, , Multiple Choice Questions, a, 1. If  , a − 2b  = (0, 6 + b ), then the, , 4, value of a and b is, (a) 0, −2, (c) 0, 4, , (b) 0, 2, (d) 0, 3, , 2. Given set in the roaster form is, B = {x | x 2 = x , x ∈ R}, (a) {0}, (c) {1, 2}, , (b) {1}, (d) {0, 1}, , 3. The mean deviation from the mean of, the set of observations −1, 0 and 4 is, (a) 3, (c) −2, , (b) 1, (d) 2, , 4. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10},, A ={1, 2, 3, 5}, B ={ 2, 4 , 6 , 7} and, C ={ 2, 3, 4 , 8}, then B ∪ C is, (a), (b), (c), (d), , {2, 4, 6, 7}, {2, 3, 4, 6, 7, 8}, {2, 3, 4, 8}, None of the above, , 5. If A = {−1, 2, 3} and B = {1, 3}, then, B × A consists, (a) (−1, 1), (c) (3, − 1), , (b) (2, − 1), (d) (3, 1), , 6. If z 1 = 3 + i and z 2 = 1 + 4i , then, | z 1 + z 2 | is less than, (a) 6.38, (c) 5.40, , (b) 6.10, (d) 7.40, , 7. Mean deviation about the median for, the data 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21, is, (a) 4.27, (c) 5.27, , (b) 5.24, (d) 4.24, , 8. Let A = {x : x ∈ Z and x 2 ≤ 4} and, B = {x : x ∈ R and x 2 − 3x + 2 = 0}., Then,, (a) A = B, (c) A ∈B, , (b) A ≠ B, (d) A ∉B, , 9. The domain of the function f given by, f (x) =, , x 2 + 2x + 1, x2 − x −6, , ., , (a) R − { 3, − 2 }, , (b) R − { − 3, 2 }, , (c) R − [3, − 3], , (d) R − (3, − 4), , 10. The standard deviation of data 6, 5, 9,, 13, 12, 8 and 10 is, (a), (c), , 52, 7, 6, , 52, 7, (d) 6, , (b)

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138, , CBSE New Pattern ~ Mathematics XI (Term I), , 11. Identify the null set from the following, I. {x : x is a natural number, x < 5 and, x > 7}., II. {y : y is a point common to any two, parallel lines}, (a) Only I, (c) Both I and II, , (b) Only II, (d) None of these, , 12. Let R be a relation in N defined by, R = (1 + x , 1 + x 2 ) : x ≤ 5, x ∈ N )., Which of the following is false?, (a), (b), (c), (d), , R = (2, 2), (3, 5), (4, 10), (5, 17), (6, 25), Domain of R = {2, 3, 4, 5, 6 }, Range of R = {2, 5, 10, 17, 26 }, None of the above, , and their standard deviation is 5 than, the sum of all squares of all the, obsevations is, (b) 250000, (d) 255000, , 14. The modulus of the complex number, 3 + 2i, 2 − 5i, , +, , 6, 29, 10, (c), 29, (a), , 3 − 2i, 2 + 5i, , is, 8, 29, 12, (d), 29, (b), , 15. If X and Y are two sets such that, , n(X) = 17, n(Y) = 23 and n( X ∪ Y ) = 38,, then n( X ∩ Y ) is, , (a) 2, (c) 4, , (b) 3, (d) 0, , 16. The domain of f (x ) =, (a) (−∞, ∞), (c) (−∞, 2), , 1, x +| x |, , is, , (b) (2, ∞), (d) (0, ∞), , 17. The standard deviation of some, temperature data in °C is 5. If the data, were converted into °F, then the, variance would be, (a) 81, (c) 36, , (b) 57, (d) 25, , 3i 3 − 2ai 2 + (1 − a )i + 5 is real, is, (a) −4, (c) −2, , (b) −3, (d) −1, , z −1, is a purely imaginary number, z +1, (z ≠ −1), then the value of | z | is, , 19. If, , (a) 0, (c) 3, , (b) 4, (d) 1, , 20. If (x + iy )1/ 3 = a + ib , where, x , y , a, b ∈ R, then, , 13. If the mean of 100 observations is 50, , (a) 50000, (c) 252500, , 18. The real value of a for which, , (a), (b), (c), (d), , x y, − equal to, a b, , (a2 + b 2 ), − (a2 + b 2 ), 2 (a2 + b 2 ), −2 (a2 + b 2 ), , Assertion-Reasoning MCQs, For question numbers 21 to 25, two, statements are given-one labelled, Assertion (A) and the other labelled, Reason (R). Select the correct answer to these, questions from the codes (a), (b), (c) and (d) as, given below., (a) Both A and R are true and R is the correct, explanation of A., (b) Both A and R are true but R is not the, correct explanation of A., (c) A is true but R is false., (d) A is false and R is true., , 21. Assertion (A) The domain of the, , relation R = (a, b ) : a ∈ N , a < 5, b = 4 is, {1, 2, 3, 4}., Reason (R) The range of the relation, S = {(a, b ) : b = | a − 1|, a ∈ Z and | a| ≤ 3}, is {1, 2, 3, 4}., , 22. Assertion (A) Three sets A, B , C are, such that A = B ∩ C and B = C ∩ A,, then A = B ., Reason (R) If A = (x , y ), then, A ∩ P ( A ) = A.

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139, , CBSE New Pattern ~ Mathematics XI (Term I), , 23. Assertion (A) Conjugate of complex, number Z = x + iy is Z = x − iy ., Reason (R) Geometrically the point, (x , − y ) is the mirror image of the point, (x , y ) on the real axis., , 24. If the equation of line is x − 3y + 8 = 0,, then, Assertion (A) The normal form of the, equation is x cos α + y sin α = p where, α = 120 ° and p = 4., Reason (R) The perpendicular distance, of the line from the origin is 8., , 25. Consider the following data, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, Assertion (A) The variance of the data, is 33., Reason (R) The standard deviation of, the data is 4.74., , Case Based MCQs, Direction (Q. Nos. 26-30) Answer the, questions from based on the following case., A function f is said to be a rational function,, g (x ), if f (x ) =, , where g (x ) and h(x ) are, h(x ), polynomial function such that h(x ) ≠ 0., g (x ), Then, lim f (x ) = lim, x →a, x → a h (x ), =, , lim g (x ), , x →a, , lim h(x ), , x →a, , =, , g (a ), h(a ), , however, if h(a ) = 0, then there are two cases, arise,, Case I g (a ) ≠ 0, Case II g (a ) = 0, In the first case, we say that the limit does not, exist., In the second case, we can find limit., , Based on above information, answer the, following questions.,  x 2 − 9x + 18 ,  is equal to, 26. lim , x→ 3 , x−3, , (a) −3, (c) 1, , 27. lim, , (b) 3, (d) −1, , (x + 1) 2 + 5x 2, , x → −1, , (x 2 + 1), , is equal to, , 1, 2, 3, (c), 2, , 5, 2, 6, (d), 7, (b), , (a), ,  x 2 − 13x + 36 ,  is equal to, x →4 , x−4, , , 28. lim , , (b) −5, (d) −1, , (a) 5, (c) 1, ,  x 8 − 2x 6 + 1 ,  is equal to, x → 1 x 4 − 3x 2 + 4 , , , , 29. lim, (a) 0, (c) 2, , (b) 1, (d) 3, ,  x 2 + 3x + 2 ,  is equal to, x → −2 , x+2 , , 30. lim , (a) 3, (c) 1, , (b) 0, (d) −1, , Direction (Q. Nos. 31-35) Answer the, questions from based on the following case., Internal Division If point p(x , y ) divides the, line segment AB, obtained by joining, A(x 1 , y1 ) and B (x 2 , y 2 ) internally in the ratio, m 1 : m 2 , then coordinates of p are, m x 2 + m 2 x1, ,, x= 1, m1 + m 2, m y 2 + m 2 y1, y= 1, m1 + m 2, (x1 , y1 ) A, , m1 : m 2, P (x, y), , B (x2 , y2 )

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140, , CBSE New Pattern ~ Mathematics XI (Term I), , External Division If point P (x , y ) divides, the line segment AB, obtained by joining, A(x 1 , y1 ) and B (x 2 , y 2 ), externally in the ratio, m 1 : m 2 , then coordinates of p are, m x 2 − m 2 x1, m y 2 − m 2 y1, ,y = 1, x= 1, m1 − m 2, m1 − m 2, m1 : m2, A (x1 , y1 ), , P (x , y), , B(x2 , y2 ), , Mid-point The coordinates of mid points of, line segment obtained by joining A(x 1 , y1 ),  x + x 2 y1 + y 2 , and B (x 2 , y 2 ) are  1, ,, ., , 2, 2 , , the axis is divided internally in the ratio, 5 : 3 by this point is, (a) 9 x − 20y + 80 = 0, (c) 8 x − 20y + 96 = 0, , (b) 9 x − 20y + 96 = 0, (d) None of these, , Direction (Q. Nos. 36-40) Answer the, questions from based on the following case., A leather company produces 8000 leather, bags in third year and 16000 Leather bags in, the seventh year. Assuming that the, production increases uniformly by a constant, number every year., , Based on above information answer the, following question, , 31. The coordinates of mid points of line, segment joining the points (1, 8) and, (7, 4) is, (a) (4, 6), (c) (4, 0), , (b) (1, 2), (d) (6, 1), , 32. The coordinates of the point which, , divides the join of (−1, 7) and (4, −3) in, the ratio 2 : 3 internally is, (a) (3, 1), (c) (1, 3), , (b) (2, 1), (d) (4, 0), , 33. The coordiantes of the point which, , divides the join of (−1, 7) and (4, −3 ) in, the ratio 2 : 3 externally is, (a) (1, 2), (c) (27, −11), , (b) (−11, 27), (d) (3, 8), , 34. If the intercept of a line between the, coordinate axis is divided by the point, ( −5, 4 ) in the ratio 1 : 2, then the, equation of line is, (a) 8 x − 5 y + 60 = 0, (c) 8 x − 5 y − 60 = 0, , (b) 8 x + 5 y − 60 = 0, (d) None of these, , 35. The equation of line which passes, , through the point ( −4, 3) and the, portion of the line intercepted between, , Based on above information answer the, following questions., , 36. The value of the fixed number by, which production is increasing every, year., (a) 2000, (c) 8000, , (b) 4000, (d) 6000, , 37. The production in first year is, (a) 2000, (c) 4000, , (b) 15000, (d) 5000, , 38. The total production in 6 yr is, (a) 50000, (c) 55000, , (b) 54000, (d) 60000, , 39. The number of leather bags produced, in tenth year is, (a) 22000, (c) 20000, , (b) 23000, (d) 21000, , 40. The difference in number of leather, bags produced in 10th year and 5th, year is, (a) 10000, (c) 21000, , (b) 30000, (d) 22000

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141, , CBSE New Pattern ~ Mathematics XI (Term I), , PRACTICE, PAPER 22, PRACTICE PAPER, OMRSHEET, OMRSHEET, , Instructions, Instructions, Use black or blue ball point pens and avoid gel pens and fountain, , Use black, or blue, point pens and avoid gel pens and fountain, pens, for filling, theball, sheets, pens for filling the sheets, Darken the bubbles completely. Don't put a tick mark or a cross, Darken, the bubbles, completely., Don't put, tick, cross, mark, half-filled, or over-filled, bubbles, will anot, bemark, readorbya the, software., mark half-filled or over-filled bubbles will not be read by the software., û, ü, û, ü, Incorrect, Incorrect Correct, Incorrect, Incorrect, , Incorrect, , Incorrect, , Do not write anything on the OMR Sheet, Do not write anything on the OMR Sheet, Multiple markings are invalid, Multiple markings are invalid, 1, 1, 2, 2, 33, , a, , b, , c, , d, , a, , b, , c, , d, , a, , b, , c, , d, , 44, 55, , a, , b, , c, , d, , a, , b, , c, , d, , 66, , a, , b, , c, , Correct, , 21, 21, 22, 22, 23, 23, 24, 24, , a, , b, , c, , d, , a, , b, , c, , d, , a, , b, , c, , d, , a, , b, , c, , d, , a, , b, , c, , d, , d, , 25, 25, 26, 26, , a, , b, , c, , d, , a, , b, , c, , d, , 77, , a, , b, , c, , d, , 27, 27, , 88, , a, , b, , c, , d, , 28, 28, , a, , b, , c, , d, , 99, , a, , b, , c, , d, , 29, 29, , a, , b, , c, , d, , 10, 10, , a, , b, , c, , d, , 30, 30, , a, , b, , c, , d, , 11, 11, , a, , b, , c, , d, , 31, 31, , a, , b, , c, , d, , 12, 12, , a, , b, , c, , d, , 32, 32, , a, , b, , c, , d, , 13, 13, , a, , b, , c, , d, , 33, 33, , a, , b, , c, , d, , 14, 14, , a, , b, , c, , d, , 34, 34, , a, , b, , c, , d, , 15, 15, , a, , b, , c, , d, , 35, 35, , a, , b, , c, , d, , 16, 16, , a, , b, , c, , d, , 36, 36, , a, , b, , c, , d, , 17, 17, , a, , b, , c, , d, , 37, 37, , a, , b, , c, , d, , 18, 18, 19, 19, , a, , b, , c, , d, , 38, 38, , a, , b, , c, , d, , a, , b, , c, , d, , a, , b, , c, , d, , 20, 20, , a, , b, , c, , d, , 39, 39, 40, 40, , a, , b, , c, , d, , Students should not write anything below this line, Students, should not write anything below this line, , SIGNATURE OF EXAMINER WITH DATE, SIGNATURE OF EXAMINER WITH DATE, , MARKS SCORED, MARKS SCORED

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142, , CBSE New Pattern ~ Mathematics XI (Term I), , PRACTICE PAPER 3, Mathematics Class 11th (Term I), Instructions, 1. This paper has 40 questions., 2. All questions are compulsory., 3. Each question carry 1 mark., 4. Answer the questions as per given instructions., Time : 90 Minutes, , Max. Marks : 40, , Multiple Choice Questions, 1. If X = {8 n − 7n − 1| n ∈ N } and, Y = {49n − 49| n ∈ N }, then, (a) X ⊂ Y, (c) X = Y, , (b) Y ⊂ X, (d) X ∩ Y = φ, , 2. If X and Y are two sets such that, , n( X ) =17, n(Y ) = 23 and n( X ∪ Y ) = 38,, then n( X ∩ Y ) is, (a) 1, (c) 3, , (b) 2, (d) 4, , 3. The set of all letters in the word, ‘BETTER’ in roster form is, (a) {B, E, T, R}, (c) {B, E, R}, , (b) {B, T, R}, (d) {B, R}, , 4. Let A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13},, , C = {11, 13, 15} and D = {15, 17}, then the, set ( A ∪ D ) ∩ (B ∪ C ) is, (a) { 7, 9, 11}, (c) { 7, 9, 11, 15 }, , (b) { 7, 9, 11, 13}, (d) { 7, 11, 13, 15 }, , 5. The number of non-empty subsets of, the set {1, 2, 3, 4} is, (a) 15, (c) 16, , (b) 14, (d) 17, , 6. If P = {a, b , c } and Q = {r }, then, (a) P × Q = Q × P, (c) P × Q ⊂ Q × P, , (b) P × Q ≠ Q × P, (d) None of these, , 7. Let A = {3, 5}, B = {7, 11} and, , R = {(a, b ) : a ∈ A, b ∈ B , a + b is odd}., Then, R in Roster form, is, (a) {(3, 7), (5, 11)}, (c) {}, , (b) {(3, 11), (5, 7)}, (d) None of these, , 8. The Domain of the function, f (x ) = 16 − x 2 is, (a) x ∈ [−4, 4], (c) x ∈[0, 1], , (b) x ∈[0, 4], (d) x ∈ [4, ∞), , 9. The range of the function f (x ) =, , 1, x −5, , is, (a) (0, ∞), (c) (5, ∞), , (b) (−∞, 5), (d) (1, 5), , 10. The range of the function f (x ) =, is, , 3, , (a)  −∞,, , 2 , (c) [2, ∞), , 3, 2 −x2, , 3 , ,∞, 2 , (d) [1, 2], (b), , 11. The equation of the straight line which, , passes through the point (1, −2 ) and cut, off equal intercepts from axes is given, by, (a) x + y − 1 = 0, (c) x + y + 1 = 0, , (b) x − y + 1 = 0, (d) x − y − 1 = 0

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143, , CBSE New Pattern ~ Mathematics XI (Term I), , 12. The angle between the X -axis and the, , line joining the points ( 3, − 1) and ( 4, − 2), is, (a) 45°, (c) 180°, , (b) 135°, (d) 90°, , 13. If the vertices of a triangle are P (1, 3),, , Q ( 2, 5) and R( 3, − 5), then the centroid, of a ∆PQR is., (a) (2, 1), (c) (6, 3), , (b) (1, 2), (d) (3, 6), , 14. The equation of the right bisector of the, line segment joining the points (3, 4), and (−1, 2) is, (a) 2 x − y + 1 = 0, (c) 2 x − y − 1 = 0, , (b) 2 x + y − 5 = 0, (d) 2 x − y − 5 = 0, , 15. The distance between the points (6, − 4 ), and ( 3, 0 ) is, (a) 7 units, (c), , 3, , (b) 25 units, , 25 units, , (d) 5 units, , cot x − 3, is equal to, cosec x − 2, 2, , 16., , lim, , x → π /6, , (a) 1, (c) 3, , 17. lim, , (b) 2, (d) 4, , x 10 − 1024, x−2, , x→ 2, , is equal to, , (a) 5120, (c) 5210, , 18. lim, x→, , 2, 5, 3, (c), 5, (a), , (b) 512, (d) 521, , x, 3, , 2, , −3, , x 2 + 3 3x − 12, , is equal to, , 1, 5, 4, (d), 5, (b), , 1, 19. lim x sin is equal to, x→ 0, x, (a) 0, 1, (c), 2, , (b) 1, (d) Does not exist, , 20. limπ (sec x − tan x ) is equal to, x→, , 2, , (a) 1, (c) 0, , (b) 2, (d) Not defined, , Assertion-Reasoning MCQs, For question numbers 21 to 25, two, statements are given-one labelled, Assertion (A) and the other labelled, Reason (R). Select the correct answer to these, questions from the codes (a), (b), (c) and (d) as, given below., (a) Both A and R are true and R is the correct, explanation of A., (b) Both A and R are true but R is not the, correct explanation of A., (c) A is true but R is false., (d) A is false and R is true., , 21. Assertion (A) The set of letters needed, to spell ‘CATARACT’ and the set of, letters needed to spell ‘TRACT’ are, equal., Reason (R) All the subsets of the set, {−1, 0,1} are φ, {−1}, {0}, {1}, {−1, 0}, {−1, 1}, and {0, 1}., , 22. Assertion (A) If A ⊆ B and C ⊆ D , then, A × C ⊆ B × D., Reason (R) If A = {2, 3}, B = {4, 5} and, C = {5, 6}, then A × (B ∪ C ) = A × (B ∩ C )., , 23. Assertion (A) lim (cosec x − cot x ) is, x→ 0, , equal to 0., Reason (R) lim, , x→ 0, , e 8x − 1, x, , is equal to 8., , 24. Assertion (A) If the perpendicular, distance between the line and origin is, 5 units and the angle made by the, perpendicular with the positive x-axis is, 30°, then the equation of line is, 3x + y = 10., Reason (R) If three points (h, 0 ), (a, b ), a b, and (0, k ) lie on a line, then + = 1., h k

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144, , CBSE New Pattern ~ Mathematics XI (Term I), , 25. Assertion (A) Coordinates of the mid, point of the line segment joining the, points (1, 2) and (3, 8) is (2, 5)., Reason (R) Coordinates of the mid, point of the line segment joining the, points (x 1 , y1 ) and (x 2 , y 2 ) is,  x 1 + x 2 y1 + y 2 , ,, , .,  2, 2 , , Case Based MCQs, Direction (Q. Nos. 26-30) Answer the, questions from based on the following case., As, we have, i = −1. So, we can write the, higher powers of i as follows, , 29. i 4 + i 8 + i 12 +i 16 is equal to, (a) 1, , 30. i, , 107, , (b) 2, , +i, , 112, , +i, , 117, , (a) 1, (c) i, , (c) 3, , +i, , 122, , (d) 4, , is equal to, , (b) −1, (d) 0, , Direction (Q. Nos. 31-35) Answer the, questions from based on the following case., A firm produces 600 smart phones in, 3rd year, and 700 smart phones in the 7th, year. Assuming that the production increases, uniformly by a constant number every year., , (i) i 2 = − 1, (ii) i 3 = i 2 ⋅ i = ( − 1) ⋅ i = − i, (iii) i 4 = (i 2 ) 2 = ( − 1) 2 = 1, (iv) i 5 = i 4 + 1 = i 4 ⋅ i = 1 ⋅ i = i, (v) i 6 = i 4 + 2 = i 4 ⋅ i 2 = 1 ⋅ i 2 = − 1, M, M, M, M, While evaluating i n for n > 4, we are writing n, as 4 q + r for some q , r ∈ N and 0 ≤ r ≤ 3. So,, in order to compute i n for n > 4, write, i n = i 4q + r for some q , r ∈ N and 0 ≤ r ≤ 3., Then, i n = i 4q ⋅ i r = (i 4 )q ⋅ i r = (1)q ⋅ i r = i r, In general for any integer k, i 4k = 1, i 4k + 1 = i ,, i 4k, , +2, , = − 1 and i 4k, , +3, , = −i, , Based on the above information, answer the, following questions., , 26. i 80 is equal to, (a) 0, (c) 2, , (b) 1, (d) 3, , 27. 2i 2 + 6i 3 + 3i 16 − 6i 19 + 4i 25 is equal to, (a) 1 + 4i, (c) 1, , (b) 1 − 4i, (d) 4i, , 28. i 12 + i 13 + i 14 + i 15 is equal to, (a) 0, (c) −1, , (b) 1, (d) 2, , Based on above information, answer the, following questions., , 31. The value of the fixed number by, which production is increasing every, year is, (a) 20, (c) 30, , (b) 25, (d) 35, , 32. The production in Ist year is, (a) 400, (c) 500, , (b) 350, (d) 550, , 33. The total production in 6 yr is, (a) 3600, (c) 3675, , (b) 3650, (d) 3725, , 34. The number of smart phone production, in 10th year is, (a) 775, (c) 800, , (b) 770, (d) 850, , 35. The difference in number of smart, phones produced in 10th year and, 8th year is, (a) 50, (c) 75, , (b) 25, (d) 100

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145, , CBSE New Pattern ~ Mathematics XI (Term I), , Direction (Q. Nos. 36-40) Answer the, questions from 36-40 based on the following, case., Consider the following data, Class, 30-40, , Frequency, 3, , 40-50, , 7, , 50-60, , 12, , 60-70, , 15, , 70-80, , 8, , 80-90, , 3, , 90-100, , 2, , Based on above information answer the, following question., , 36. Mean is calculated by using the, formula, , Σfixi, N, Σxi, (c) x =, Σfi, (a) x =, , (b) x =, , Σfi, N, , (d) None of these, , 37. Variance is calculated by using the, formula, 1, Σfi (xi + x)2, N, (d) σ2 = Σfi (xi + x)2, (a) σ2 =, , 1, Σfi (xi − x)2, N, (d) None of these, (b) σ2 =, , 38. Mean is equal to, (a) 62, (c) 50, , (b) 60, (d) 55, , 39. Variance is equal to, (a) 200, (c) 202, , (b) 201, (d) 203, , 40. Standard deviation is equal to, (a) 14.18, (c) 10.8, , (b) 19, (d) 6.5

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146, , CBSE New Pattern ~ Mathematics XI (Term I), PRACTICE, PAPER 3, PRACTICE PAPER 3, , OMRSHEET, OMRSHEET, Instructions, Instructions, Use black or blue ball point pens and avoid gel pens and fountain, Use black, or blue, point pens and avoid gel pens and fountain, pens, for filling, theball, sheets, pens for filling the sheets, Darken the bubbles completely. Don't put a tick mark or a cross, Darken, the bubbles, completely., Don't will, put anot, tick, cross, mark, half-filled, or over-filled, bubbles, bemark, readorbya the, software., mark half-filled or over-filled bubbles will not be read by the software., û, ü, û, ü, Incorrect, Incorrect Correct, Incorrect, Incorrect, , Incorrect, , Incorrect, , Do not write anything on the OMR Sheet, Do not write anything on the OMR Sheet, Multiple markings are invalid, Multiple markings are invalid, 1, 1, 2, 2, 3, 3, 44, 55, 66, , a, , b, , c, , d, , a, , b, , c, , d, , a, , b, , c, , d, , a, , b, , c, , d, , Correct, , 21, 21, 22, 22, 23, 23, 24, 24, , a, , b, , c, , d, , a, , b, , c, , d, , a, , b, , c, , d, , a, , b, , c, , d, , a, , b, , c, , d, , a, , b, , c, , d, , a, , b, , c, , d, , 25, 25, 26, 26, , a, , b, , c, , d, , a, , b, , c, , d, , 77, , a, , b, , c, , d, , 27, 27, , 88, , a, , b, , c, , d, , 28, 28, , a, , b, , c, , d, , 99, , a, , b, , c, , d, , 29, 29, , a, , b, , c, , d, , 10, 10, , a, , b, , c, , d, , 30, 30, , a, , b, , c, , d, , 11, 11, , a, , b, , c, , d, , 31, 31, , a, , b, , c, , d, , 12, 12, , a, , b, , c, , d, , 32, 32, , a, , b, , c, , d, , 13, 13, , a, , b, , c, , d, , 33, 33, , a, , b, , c, , d, , 14, 14, , a, , b, , c, , d, , 34, 34, , a, , b, , c, , d, , 15, 15, , a, , b, , c, , d, , 35, 35, , a, , b, , c, , d, , 16, 16, , a, , b, , c, , d, , 36, 36, , a, , b, , c, , d, , 17, 17, , a, , b, , c, , d, , 37, 37, , a, , b, , c, , d, , 18, 18, , a, , b, , c, , d, , 38, 38, , a, , b, , c, , d, , 19, 19, 20, 20, , a, , b, , c, , d, , 39, 39, , a, , b, , c, , d, , a, , b, , c, , d, , 40, 40, , a, , b, , c, , d, , Students should, Students, should not, not write, write anything, anythingbelow, belowthis, thisline, line, , SIGNATURE OF EXAMINER WITH DATE, SIGNATURE OF EXAMINER WITH DATE, , MARKS SCORED, MARKS SCORED

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147, , CBSE New Pattern ~ Mathematics XI (Term I), , ANSWERS, Practice Set 1, 1. (d), 11. (d), 21. (d), 31. (c), , 2. (a), 12. (c), 22. (b), 32. (a), , 3. (a), 13. (a), 23. (b), 33. (d), , 4. (c), 14. (d), 24. (c), 34. (c), , 5. (b), 15. (a), 25. (c), 35. (d), , 6. (b), 16. (b), 26. (c), 36. (c), , 7. (d), 17. (d), 27. (b), 37. (b), , 8. (b), 18. (d), 28. (d), 38. (a), , 9. (c), 19. (a), 29. (b), 39. (d), , 10. (b), 20. (c), 30. (c), 40. (d), , Practice Set 2, 1. (a), 11. (c), 21. (c), , 2. (d), 12. (a), 22. (c), , 3. (d), 13. (c), 23. (a), , 4. (b), 14. (b), 24. (c), , 5. (c), 15. (a), 25. (c), , 6. (d), 16. (d), 26. (a), , 7. (c), 17. (a), 27. (b), , 8. (b), 18. (c), 28. (b), , 9. (a), 19. (d), 29. (a), , 10. (a), 20. (d), 30. (d), , 31. (a), , 32. (c), , 33. (b), , 34. (a), , 35. (b), , 36. (a), , 37. (c), , 38. (b), , 39. (a), , 40. (a), , Practice Set 3, 1. (a), 11. (c), 21. (c), , 2. (b), 12. (b), 22. (c), , 3. (a), 13. (a), 23. (b), , 4. (c), 14. (b), 24. (b), , 5. (a), 15. (d), 25. (a), , 6. (b), 16. (d), 26. (b), , 7. (c), 17. (a), 27. (a), , 8. (a), 18. (a), 28. (a), , 9. (a), 19. (a), 29. (d), , 10. (b), 20. (c), 30. (d), , 31. (b), , 32. (d), , 33. (c), , 34. (a), , 35. (a), , 36. (a), , 37. (b), , 38. (a), , 39. (b), , 40. (a)