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Optional Mathematics, Grade 10, , Government of Nepal, , Ministry of Education, Science and Technology, , Curriculum Development Centre, Sanothimi, Bhaktpur, Nepal
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Optional Mathematics, Grade – 10, , Government of Nepal, Ministry of Education, Science and Technology, , Curriculum Development Centre, Sanothimi, Bhaktapur, 2076
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Publisher:, , Government of Nepal, Ministry of Education, Science and Technology, Curriculum Development Centre, Sanothimi, Bhaktapur, , © Publisher, , First Edition: 2076 B.S.
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Preface, The curriculum and curricular materials have been developed and revised on a regular, basis with the aim of making education objective-oriented, practical, relevant and job, oriented. It is necessary to instill the feelings of nationalism, national integrity and, democratic spirit in students and equip them with morality, discipline and selfreliance, creativity and thoughtfulness. It is essential to develop in them the linguistic, and mathematical skills, knowledge of science, information and communication, technology, environment, health and population and life skills. it is also necessary to, bring in them the feeling of preserving and promoting arts and aesthetics, humanistic, norms, values and ideals. It has become the need of the present time to make them, aware of respect for ethnicity, gender, disabilities, languages, religions, cultures,, regional diversity, human rights and social values so as to make them capable of, playing the role of responsible citizens. This textbook for grade nine students as an, optional mathematics has been developed in line with the Secondary Level Optional, Mathematics Curriculum, 2074 so as to strengthen mathematical knowledge, skill and, thinking on the students. It is finalized by incorporating recommendations and, feedback obtained through workshops, seminars and interaction programmes., The textbook is written by Prof. Dr. Siddhi Prasad Koirala, Mr. Ramesh Prasad Awasthi, and Mr. Shakti Prasad Acharya. In Bringing out the textbook in this form, the, contribution of the Director General of CDC Dr. Lekha Nath Poudel is highly, acknowledged. Similarly, the contribution of Prof. Dr. Ram Man Shrestha, Mr. Laxmi, Narayan Yadav, Mr. Baikuntha Prasad Khanal, Mr. Krishna Prasad Pokharel, Mr., Anirudra Prasad Neupane, Ms. Goma Shrestha, Mr. Rajkumar Mathema is also, remarkable. The subject matter of the book was edited by Dr. Dipendra Gurung,, Mr. Jagannath Adhikari and Mr. Nara Hari Acharya. The language of the book was, edited by Mr. Nabin Kumar Khadka. The layout of this book was designed by, Mr. Jayaram Kuikel. CDC extends sincere thanks to all those who have contributed to, developing this textbook., This book contains various mathematical concepts and exercises which will help the, learners to achieve the competency and learning outcomes set in the curriculum., Efforts have been made to make this textbook as activity-oriented, interesting and, learner centered as possible. The teachers, students and all other stakeholders are, expected to make constructive comments and suggestions to make it a more useful, textbook., 2076, , Curriculum Development Centre, Sanothimi, Bhaktapur
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Table of Content, Unit 1, , Algebra, , 1, , 1.1 Function, , 1, , 1.2 Polynomials, , 16, , 1.3 Sequence and Series, , 29, , 1.4 Linear Programming, , 68, , 1.5 Quadratic Equations and Graphs, , 81, , Unit 2, , Continuity, , 92, , Unit 3, , Matrix, , 103, , Unit 4, , Coordinate Geometry, , 121, , Unit 5, , Trigonometry, , 148, , Unit 6, , Vector, , 207, , Unit 7, , Transformation, , 237, , Unit 8, , Statistics, , 253, , Answer, , 276
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Unit 1, , Algebra, , 1.1, , Function, , 1.1.0, , Review, , Study the square given and answer the following questions:, (a) What is the area function 'f' in terms of side (l)?, (b) What is the perimeter function ‘g’ in terms of side (l)?, (c) Are functions 'f' and ‘g’ one to one?, (d) Are functions 'f' and 'g' onto?, Study the following information and relate them with function., (a) To each person, there corresponds an annual income., (b) To each square, there corresponds an area., (c) To each number, there corresponds its cube., Each (a), (b) and (c) are examples of function., Take an example of function and represent it in different forms. Discuss about the graph, among your friends., Graphs provide a means of displaying, interpreting and analyzing data in a visual form. To, graph an equation is to make a drawing that represents the solutions of that equation., When we draw the graph of an equation, we must remember the following points:, (a) Calculate solutions and list the ordered pairs in a table., (b) Use graph paper and scale the axes., (c) Plot the ordered pairs, look for patterns and complete the graph with the, equation being graphed., 1.1.1.(a): Algebraic function and it's graph, Study the following functions and think about their graphical representation:, f(x) = x + 5, f(x) = x2, f(x) = x3, The algebraic equation consisting of the least one variable is called the algebraic, function., A function f is linear function if it can be written as f(x) = mx + c, where m and c are, constant., If m = 0, the function is constant function and written as f(x) = c. If m = 1 and c = 0, the, function is the identity function written as f(x) = x., 1
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Linear Fuction, , Identity function, , Constant function, , functions like f(x) = ax2 + bx + c, a ≠ 0 and f(x) = ax3 + bx2 + cx + d, a ≠ 0 are examples of, Non-linear functions., f(x) = ax2 + bx + c, a ≠ 0 is quadratic function where a, b, and c are real numbers. The, graph of quadratic function is a parabola., The vertex form of f(x) = ax2 + bx + c, a ≠ 0 is f(x) = a(x-h)2 + k, For example, f(x) = x2 + 2x + 2 = (x-(-1))2 + 1. The vertex of the parabola is (h, k) =, (-1, 1). When, x = 0, the curve meets y-axis at (0, 2)., x = h is the axis of parabola about which the curve is symmetric. If a > 0, the curve turns, upward from vertex and it turns downwards for a < 0., , 2
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The function defined as, f(x) = ax3 + bx2 + cx + d, a ≠ 0 is, called cubic function, where a, b, c and d are real numbers., The nature of curve in graph is as shown at the right., When b, c, d are non-zero, the curve meets X-axis at, three different points., , o, , o, , o, , If each of b, c, d is zero, then the curve passes through the origin., Example 1, (a) When we draw the graph of y = 2x2, we can take different values of x and find their, corresponding y-values. Representing the ordered pairs (x, y) in graph, we can find, the shape of curve in graph., , (b) When we draw the graph of y = -2x2, we follow the same steps as we do in (a) and, give the shape of curve in graph., , 3
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Exercise: 1.1.1 (a), 1., , 2., , (a), , Define linear function with example., , (b), , What is the coordinates of vertex of f(x) = ax2 + bx + c, a ≠ 0, , (c), , Identify the identity function: f(x) = 5 and f(x) = x., , Study the following graphs and identify their nature as identity, constant,, quadratic and cubic function., , 4
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3., , Draw the graph of the following function, (a) y = x + 2, , (c) y = x2, , (b) y = 6, , (d) y = -x2, , (e) y = x3, 4., , Pemba estimates the minimum ideal weight of a woman, in pounds is to multiply her, height, in inches by 4 and subtract 130. Let y = minimum ideal weight and x = height., (a) Express y as a linear function of x., (b) Find the minimum ideal weight of a woman whose height is 62 inches., (c) Draw the graph of height and weight., , 5., , Investigate the nature of graph showing linear, quadratic and cubic function in our, daily life. Make a report and present it in classroom., , 1.1.1 (b): Trigonometric function, Discuss about trigonometric ratios of any angle in your classroom. Also, tell any nine, relations among trigonometric ratios of any magnitude., The function which relates angles and their measurement to the real number is called, trigonometric function. It associates an angle with the definite real number. We know sin, (x + 2) = sinx, cos (x + 2) = cosx and tan(x + ) = tanx. So, if a function f(x + k) = f(x) for, positive value of k. k is called period for f(x)., In case of sinx and cosx, the period is taken as 2, but for tanx, it is taken as (Why?), Trigonometric functions are said to be non-algebraic or transcendental functions., (i) Graph of sinx or sine – graph: Let us take the values of x differing 90° and, corresponding values of y for y = sinx. The maximum and minimum values of sinx are, 1 and -1 respectively., x:, , -360°, , -270°, , -180°, , -90°, , 0°, , 90°, , 180°, , 270°, , 360°, , sinx:, , 0, , 1, , 0, , -1, , 0, , 1, , 0, , -1, , 0, , y = sinx, Domain:, , -360° < x° < 360°, -2 < x < 2, , Range:, , -1 < y < 1, , Period:, , 2, , 5
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(ii) Graph of cosx or cosine – graph, Let us take the values of x differing 90° and corresponding values of y for y = cosx., The maximum and minimum values of cosx are 1 and -1 respectively., x, , -360°, , -270°, , -180°, , -90°, , 0°, , 90°, , 180°, , 270°, , 360°, , cosx, , 1, , 0, , -1, , 0, , 1, , 0, , -1, , 0, , 1, , y = cosx, Domain:, , -360° < x° < 360°, -2 < x < 2, , Range:, , -1 < y < 1, , Period:, , 2, , , 6
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(iii) Graph of tanx or tangent - graph, Let us take the values of x differing 90° and the corresponding values of y., For y = tanx., As x passes through – 360° to 360°, the values of tanx suddenly changes from very, large positive and negative values. The line parallel to the y-axis corresponding to the, odd multiples of 90° are continuously approached by the graph on either side, but, never actually meet. Such lines are called asymptote to the curve., , x, tanx, , -360°, , -270°, , -180°, , -90°, , 0°, , 90°, , 180°, , 270°, , 360°, , 0, , Not, defined, , 0, , Not, defined, , 0, , Not, defined, , 0, , No, defined, , 0, , y = tanx, range:, , -<x<, , domain:, , - < x < , , period:, , , , O, , Note: For least value of x = (2), sin(x + 2 = sinx and cos(x + 2 = cosx., So, period of sine and cosine is 2But tan(x + ) = tanx, so period of tangent is ., , Exercise: 1.1.1. (b), 1., , Write the maximum and minimum values of the following:, (a) f(x) = sinx, , (b) f(x) = cosx, , (c) f(x) = tanx, , 7
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2., , Write the period of the following:, (a) f(x) = sinx, , 3., , 4., , (b) f(x) = cosx, , (c) f(x) = tanx, , Draw the graph of, (a) f(x) = sinx (- < x < ), , (b) f(x) = cosx (o < x < 2), , (c) f(x) = tanx (0 < x < ), , (d) f(x) – 2sinx ( -/2 < x < /2), , Study the topic 'sound' in physics of your science book and find the nature of, longitudinal wave. Relate this concept with trigonometric function., , 1.1.2, , Composition of function, , Study the following situations:, (i) When paddy is kept into the rice mill,, rice comes out., (ii) When the rice is kept in grinding machine,, flour comes out from grinding machine., (iii) The composite machine produces flour from paddy., How can we link this situation in mathematical form? Discuss., Again, Let us suppose two real-valued function, f = {(1,5), (2, 6), (3, 7), (4, 8)} and, g = {(5, 25), (6, 36), (8, 64), (7, 49)}. Now,, Answer the following questions:, (i) What is the domain set of function g?, , (ii) What is the range set of function f?, , (iii) What is the range set of function g?, , (iv) What is the domain set of function f?, , (v) What is the relation between range of 'f' and domain of 'g'?, (vi) Can we show this situation in combined form given as below?, Let, f and g are two real-valued functions. The, composition of f and g, is defined as, -, , (fg)(x) = f(g(x)): Where x is in the domain, of g and g(x) is in the domain of f., , -, , (gf) (x) = g(f(x)): Where x is in the domain, of f and f(x) is in the domain of g., , 8
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Note: (i) 'gf' is read as composite of f and g., (ii) 'fg' is read as composite of g and f., (iii) 'gf' and 'fg' cannot be defined when (domain of g) (Range of f) = and, (domain of f) (range of g) = , (iv) for 'gf' range of f is subset of domain of 'g' and domain of 'f' = domain of gf., (v) for 'fg', range of g is subset of domain of 'f' and domain of 'g' = domain of fg., , Example 1, In the adjoining mapping diagram given at right,, (i) Write the range of function f., (ii) Write the domain of function g., (iii) Write 'gf' in ordered pair form., (iv) Does 'fg' exist? Give reason., Solution:, The given mapping diagram helps us in visualizing the composite function g o f., (i) range of f = {3, 4, 5} domain of g., (ii) domain of g = {3, 4, 5, 7}, (iii) g o f = {(1, 9), (2, 16), (3, 25)}, (iv) f o g does not exist because (range of g) (domain of f) = ., Example 2, Given f and g are two real valued functions defined as f(x) = √𝑥 and g(x) = x – 1. Answer, the following questions from given information:, (i) domain of f, , (ii) (gf)(x), , (iii) domain of g, , (iv) (fg)(x), , Solution: Here,, f(x) = √𝑥. the negative real number does not satisfy the function. So,, (i) domain of f is 0 < x < [i.e. real numbers greater than or equal to 0.), Again, (ii) (gf)(x) = g(f(x)) = g(√𝑥) = √𝑥 − 1, , 9
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(iii) Domain of g is the set of all real numbers each of the real number satisfy g(x)., So, domain of g = -∞ < x < ∞., (iv) (fg)(x) = f(g(x)) = f(x-1) = √𝑥 − 1, Example 3, Let f(x) = 3 + 2x for all x R and g(x) = x2 + 1 for all x R, then find the formula for the, following functions:, (a) (ff)(x), , (b) (gf)(x), , (c) (fg)(x), , (d) (gg)(x), , Solution: Here,, f(x) = 3 + 2x for all x R, g(x) = x2 + 1 for all x R, Now, (a) (ff)(x) = f(f(x) = f(3 + 2x) = 3 + 2(3 + 2x) = 3 + 6 + 4x = 4x + 9, (b) (gf)(x) = g(f(x)) = g(3 + 2x) = (3 + 2x)2 + 1 = 9 + 12x + 4x2 + 1 = 4x2 + 12x + 10, (c) (fg)(x) = f(g(x)) = f(x2 + 1) = 3 + 2(x2 + 1) = 3 + 2x2 + 2 = 2x2 + 5, (d) (gg)(x) = g(g(x)) = g(x2 + 1) = (x2 + 1)2 + 1 = x4 + 2x2 + 1 + 1 = x4 + 2x2 + 2, Example 4, If f(x) = x2 + 1 and g(x) = x – 3, find, (a) (fg)(x), , (b) g(f(x)), , (c) f(g(2)), , (d) (gf)(3), , Solution: Here,, f(x) = x2 + 1, g(x) = x – 3,, (a) (fg)(x) = f(g(x)) = f(x – 3) = (x – 3)2 + 1 = x2 – 6x + 9 + 1 = x2 – 6x + 10, (b) g(f(x)) = g(x2 + 1) = x2 + 1 – 3 = x2 – 2, (c) f(g(2)) = 22 – 6 x 2 + 10 = 4 – 12 + 10 = 2, (d) g(f(3)) = 32 – 2 = 9 – 2 = 7, Exercise 1.1.2, 1., , 2., , (a), , Let f: A B such that f(x) = y and g: B C such that g(y) = Z. Name the, function defined from A to C., , (b), , Write the difference between (fg) (x) and (gf) (x)., , (c), , Define composition function of f and g., , Given f and g are two real-valued functions defined as below. Find (i) domain of f (ii), domain of g (iii) domain of (fg) (iv) range of (gf) in each of the following if exist., , 10
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(a) f = {(3, 4), (5, 6), (9, 10)} and g = {(4, 16), (6, 36), (10, 100)}, (b) f = {(1, 2), (2, 3), (3, 4)} and g = {(2, 3), (3, 4), (4, 5)}, 3., , 4., , Given that f(x) = 2x – 5 and g(x) = x2 – 2x + 6, calculate:, (a) (fg) (x) and (gf) (x), , (b) (fg) (5) and (gf) (4), , (c) (gg) (2) and (ff) (9), , (d) (fg) (-4) and (gf) (-4), , (a), , If f(x) = x, g(x) = x + 1 and h(x) = 2x – 1 then find f(gh)(x)) and (g(hf)(x)), , (b), , Given that f(x) = 2x – 3, g(x) = x3 + 2 and h(x) = x2 – 2x + 3, find (f(gh)(x)), and ((hf)g(x)). (Taking composition of two functions as a single function)., , 5., , If f and g are linear functions, what can you say about the domain of (fg) and, (gf)? Explain., , 6., , Dolma determines the domain of fg by examining only the formula for (fg)(x). Is her, approach valid? Why or why not?, , 7., , Write yourself any two real-valued function. Find their composition., , 8., , A stone is thrown into a pond, creating a circular ripple that spreads over the pond in, such a way that the radius is increasing at the rate of 3 ft/sec., (a) Find a function r(t) for the radius in terms of t., (b) Find a function A(r) for the area of the ripple in terms of the radius r., (c) Find (Ar) (t). Explain the meaning of this function., , 1.1.3, , Inverse of a function, , Study the function 'f' given in mapping diagram and answer the following:, (a) What is the domain of f?, , (b) What is the range of f?, , (c) Is the functions 'f' one to one and onto (d) What is the formula for 'f'?, Let, f: A B be a one to one and onto function then a function g: B A such that for each, x B, g(x) A is called inverse of f. It is denoted by g = f-1., , 11
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In the above mapping diagram, f-1 exists because it satisfies the following conditions., If f-1 exists, the domain and range of f are range and domain of f-1 respectively. If y is the, image of x under f then x is image of y under f-1. In the above mapping diagram,, f(-1) = -1 if and only if f-1(-1) = -1, f(-2) = -8 if and only if f-1(-8) = -2, f(1) = 1 if and only if f-1 (1) = 1, f(2) = 8 if and only if f-1(8) = 2, 3, , Again, f(x) = x3 if and only if f-1(x) = √𝑥, f = {(-1, -1), (-2, -8), (1, 1), (2, 8)}, f-1 = {(-1, -1), (-8, -2), (1, 1), (8, 2)}, Given function f: A B, the inverse image of an element y B with respect to f is denoted, by f-1 (y). That is f(x) = y if and only if x = f-1(y). It is read as 'f inverse of y'., f: A B such that f(x) = {y: x A}, f-1 : B A such that f-1(y) = {x A : y = f(x)}, Example 1, Let f: R R is a function such that f(1) = 2, f(5) = 10, f(-4) = -8 and f(-3) = -6, (i) Write function 'f' in ordered pair form., (ii) Represent 'f' in mapping diagram., (iii) Represent 'f-1' in ordered pair form., (iv) Represent 'f-1' in mapping diagram., Solution: Here,, f is function, defined from set of real numbers to set of real number. Certain elements of, real numbers are given, So,, (i) f = {(1, 2), (5, 10), (-4, -8), (-3, -6)}, , (ii), , -1, , (iii) f = {(2, 1), (10, 5), (-8, -4), (-6, -3)}, (iv), , 12
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Example 2, Let f: R R be defined by f(x) = 2x + 3 then find, (i) f-1(x), , 1, , (ii) f-1(-4), , (iii) f-1(4), , Solution:, Let, f: R R such that f(x) = y, (x, y) R x R, , (i), , f(x) = y if and only if f-1(y) = x, , Alternatively, , Now, (i) y = f(x), , f(x) = y = 2x + 3, , or, y = 2x + 3, , y = 2x + 3, , or, y – 3 = 2x, , Interchanging the places of x and y we get, , y−3, or, 2 =, , or, x = 2y + 3, , or,, , x, , or, x – 3 = 2y, , y−3, = f-1(y), 2, , x−3, or, 2, , or, y =, , -1, , = f (x), , f (x) =, -1, , (x−3), 2, , f-1(x) =, , x−3, 2, , (x−3), 2, , 1, , (ii) f-1(-4) = 2(-4-3), 1, , = 2 (-7), =–, , 7, 2, , 1, , 1 1, , (iii) f-1(4) = 2 (4 − 3), 1 1−12, ), 4, , = 2(, , 1 −11, ), 4, , = 2(, =, , −11, 8, , Example 3, Let f: R –{3} R is defined as f(x) =, , 2𝑥+5, ., 𝑥−3, , Find (i) f-1(x), , Solution, Here, domain of f is set of real numbers except 3., (i), , Let, y = f(x), , i.e. x = f–1(y), , 13, , (ii) (f-1of)(x)
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or,, , 𝑦1 −9, = f-1(y1), 𝑎, , or,, , 𝑥−9, = f-1(x), 𝑎, , Again,, [g(x) = y2 i.e. x = g-1(y2), , y2 = g(x), or, y2 = 3x + 8, or, y2 – 8 = 3x, or,, , 𝑦2 −8, 3, , =𝑥, , or, y2 - 8 = g-1(y2), or, x – 8 = g-1(x), Now, f-1(10) =, , 10−9, 𝑎, , 1, , =𝑎, , g-1(11) = 11– 8 = 3, but, f-1(10) = g-1(11), or,, , 1, 𝑎, , = 3, 1, , or, a = 3, 1, , a = 3, Exercise 1.1.3, 1., 2., , (a), , Define inverse of function, f: R R., , (b), , What is the relation between composition of a function and its inverse., , Represent the following functions in mapping diagram and find their inverse., (a) f = {(1, 2), (2, 3), (4, 5)}, , (b) g = {(1, 4), (2, 5), (3, 6)}, , (c) h = {(-2, 4), (-3, 9), (-6, 36)}, 3., , If f is the real-valued function, find, (a) f-1(x), (d), , (b) f-1(6), , 1, , (c) f-1(4), , f-1(-2) in each of the following:, (i) f(x) = 3x + 1, (iii) f(x) =, , 𝑥+1, 2, , (ii) f(x) = 2x – 5, 𝑥−2, , (iv) f = {(x, 𝑥+2), x ≠ -2}, , 15
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4., , If f(x) = x + 1, g(x) = 2x, find, (a) (fg-1)(x), , 5., , 6., , (b) (gf-1)(x). f and g are real-valued functions., 𝑥+2, , (a), , If f(x) = 3x – 7, g(x) = 5 and (g-1f)(x) = f(x), find the value of x. f and g, are real-valued functions., , (b), , f is a real-valued function defined as f(x) = 3x + a. If (ff)(6) = 10 then find, the values of a and f-1(4)., , Write the formula of volume and surface area of sphere in terms of radius. Find the, functional relation and write their inverse., , 1. 2. Polynomials, 1.2.0 Review, Study the following algebraic expressions and answer the following questions in group., 1, , (i) 4x + 5y (ii) 3y3 – 2y2 + 5y – 7, , (ii) 7y5 – 2y3/2 + 7√𝑦 – 6, , (iii) 3x3 + 4x2y + 7xy2 + 9y3, , (iv) 15xy3/2 – 12x2y1/2, , (v) x3 + x2 – x + 6, , -, , Which are examples of polynomials and why?, , -, , Which one is polynomial in one variable?, , -, , What are the main characteristics of polynomial that you study in previous grades?, , A polynomial in one variable is any expression of the type anxn + an-1 xn-1 + … + a2x2 + a1x+a0,, where n is non-negative integers and an, an-1 … a0 are real numbers, called coefficients anxn, is called the leading term of the polynomial. 'n' is degree of the polynomial. If an 0., In the above examples (i), (ii), (iv), (v) are examples of polynomials because they have nonnegative integer in power of each term. (vi) is polynomial in one variable., 1.2.1, , Division of Polynomials, , Let us take two examples:, x2 – 3x + 2 = (x – 1) (x – 2), or, (x2 – 3x + 2) ÷ (x – 1) = (x – 2), Similarly,, x5 + 2x3 + 4x2 + 5 = (x2 + 2) (x3 + 4) – 3, or, (x5 + 2x3 + 4x2 + 5) ÷ (x2 + 2) = x3 + 4 -, , 3, 𝑥 2 +2, , 16
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In the above examples, we divide one polynomial by another, we obtain a quotient and a, remainder. If the remainder is 0 as in first example, then the divisor is a factor of the, dividend. If the remainder is not 0, as in second examples, then the divisor is not a factor, of the dividend., When we divide a polynomial P(x) by a divisor D(x), we get quotient Q(x) and remainder, R(x). The quotient Q(x) must have degree less than that of the dividend P(x)., Remainder R(x) must either be 0 or have degree less than that of the divisor D(x)., We can check the division by algorithm P(x) = D(x). Q(x) + R(x)., , Example 1, Divide x2 – 5x + 6 by x – 2 and find quotient and remainder., Solution: Here,, We know, x2 – 5x + 6, , = x2 – 3x – 2x + 6, = x(x – 3) -2(x – 3), = (x – 3) (x – 2), , Now, (x2 – 5x + 6) ÷ (x – 2), , =, , (𝑥−3)(𝑥−2), (𝑥−2), , = (x – 3), , Hence the quotient is (x-3) and remainder is 0., Alternatively:, 𝑥−2, , , , , 𝑥 2 − 5𝑥 + 6 x – 3, 𝑥 2 − 2𝑥, −3𝑥 + 6, −3𝑥 + 6, , , , 0, , Quotient = (x – 3) and remainder = 0., Steps for division:, - Arrange divisor and dividend in division form., - First term of dividend is product of x and first term of divisor., - Since division is repeated subtraction, we change the sign in second row., Example 2, Divide x3 + 6x2 – 25x + 18 by (x + 9) using long division method. Also write the quotient, and remainder., Solution: Here,, Writing the dividend and divisor in long division method, we get:, 17
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(c) x3 – 8 ÷ (x – 2), 3., , 4., , Divide using long division method and find quotient and remainder., (a) x3 + 2x2 – 5x – 7 ÷ (x + 1), , (b) x3 – 10x2 + 16x + 26 ÷ (x – 5), , (c) 2x4 + 5x2 – 3x – 7 ÷ (2x – 1), , (d) y5 + y3 – y ÷ (3-y), , For the function f(y) = y3 – y2 – 17y – 15, use long division to determine whether each, of the following is a factor of f(y) or not., (a) y + 1, , 5., , (d) x3 + 9x2 + 27x + 27 ÷ (x + 3), , (b) y + 3, , (c) y + 5, , (d) y – 1, , (e) y – 5, , For the polynomial p(x) = x4 – 6x3 + x – 2 and divisor d(x) = x – 1, use long division to, find the quotient Q(x) and the remainder R(x) when P(x) is divided by d(x). Express, p(x). In the form of d(x). Q(x) + R(x). Write your finding., , Synthetic division, When the divisor is in the form of x – a, we can simplify using addition rather than, subtraction. When the procedure is finished, the entire algorithm is known as synthetic, division., Example 1, Divide x4 – x3 – 3x2 – 2x + 5 by x – 2 by synthetic division., Solution:, Here, the devisor is x – 2, thus we use '2' in synthetic division., 2, , -, , We 'bring down' the 1 [coefficient of x4]. Then multiply it by 2 to get 2 and add to, get 1., , -, , We then multiply 1 by 2 to get 2, add, and so on., , 19
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-, , The number -3 is the remainder. The others numbers 1, 1, -1, -4 are coefficients of, the quotient, x3 + x2 – x – 4. In this case, the degree of the quotient is 1 less than the, degree of the dividend and the degree of the divisor is 1., , Example 2, Divide 8x3 – 1 by 2x – 1, using synthetic divisions method., Solution:, When the divisor is not in the form of x – a, we first make it in the form of x – a, taking, 1, , coefficient of x common. i.e. 2x – 1 = 2(𝑥 − 2)., We write coefficient as zero for missing term in the dividend. Now, dividing by synthetic, division method., , So, quotient, , 1, , = 2 (8x2 + 4x + 2), = 4x2 + 2x + 1, , Remainder = 0, 1, 2, , 1, 2, , Since, divisor is 2 (𝑥 − ), so quotient is (8x2 + 4x + 2) (why?), , Exercise 1.2.2, 1., , 2., , (a), , What is the degree of quotient when the degree of polynomial is 'n' in, synthetic division?, , (b), , What should be the expression of division in synthetic division?, , Use synthetic division and divide in each of the following:, (a) x3 + 8 by x – 2, , (b) 2x4 + 7x3 + x – 12 by (x + 3), , (c) 4x3 – 3x2 + x + 9 by x – 2, , (d) 2x3 + 7x2 – 8 by (x + 3), , (e) 8x3 + 4x2 + 6x – 7 by 2x – 1, , 20
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1.2.3 (A) Some theorems on polynomials, Does x – 2 exactly divide x2 + 4x + 4?, Divide and write the conclusion., We can use 'Remainder theorem' for finding remainder and 'factor theorem' for finding, factors of the polynomial., (a) Remainder theorem: If a number 'a' is substituted for x in the polynomials f(x), the, result f(a) is the remainder that would be obtained by dividing f(x) by x –a. That is f(x), = (x – a) Q(x) + R(a), Proof: We know, f(x) = d(x) x Q(x) + R(x), or, f(x) = (x – a) Q(x) + R(x), or, f(a) = (a – a) Q(a) + R(a), or, f(a) = R(a), , , Remainder is f(a), 𝑏, , If the divisor is in the form of ax ± b, the remainder is 𝑓 (± 𝑎)., Example 1, If f(x) = 2x3 – 3x2 + 4x + 7 is divided by x – 1, find the remainder using remainder theorem., Solution, Here, polynomial, f(x) = 2x3 – 3x2 + 4x + 7, divisor (x – a) = x – 1, by reminder theorem, remainder = f(a) = f(1), So, f(1), , = 2 x 13 – 3 x 12 + 4 x 1 + 7, =2x1–3x1+4x1+7, =2–3+4+7, = 13 – 3 = 10., , Remainder (R) = 10, Example 2, Use remainder theorem and find the remainder: (4x3 + 2x2 – 4x + 3) ÷ (2x + 3), Solution:, Here, let polynomial f(x) = 4x3 + 2x2 – 4x + 3, , 21
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3, , −3, , Divisor d(x) = 2x + 3 = 2 (𝑥 + 2) = 2 (𝑥 − ( 2 )), By remainder theorem;, Remainder = f(a) = f(-3/2), Now, f(-3/2), , −3 3, , −3 2, , −3, , = 4 ×( 2 ) + 2 × ( 2 ) −4( 2 ) + 3, −27, )+, 8, , =4×(, =, , −27, 9, +2, 2, , =, , −27+9+12+6, 2, , +, , 12, 2, , 9, , 3, , 2×4+4×2+3, +3, , 0, 2, , = =0, Remainder (R) = 0, Example 3, If 2x4 + 2x2 – 2x + k is divided by x – 2, the remainder is 5, find the value of k., Solution:, Here, let f(x) = 2x4 + 3x2 – 2x + k, Divisor (x – a) = x – 2, By remainder theorem,, Remainder = f(a), = f(2), = 2 x (2)4 + 3 x (2)2 – 2 x 2 + k, = 32 + 12 – 4 + k, = 40 + k, By the question, Remainder (R) = f(2) = 5, or, 40 + k = 5, or, k = 5 - 40, k = –35, Hence, k = -35, , 22
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Exercise 1.2.3 (A), 1., 2., , (a), , State remainder theorem, , (b), , If ax + b (a ≠ 0) divides f(x), what is the remainder?, , Use remainder theorem and find the remainder in each of the following:, (a), , (2x3 – 5x2 + x – 5) ÷ (x – 2), , (b), , (4x3 + 7x2 – 3x + 2) ÷ (x + 2), , (c), , (x4 – 3x2 + 15) ÷ (x – 1), , (d), , (x5 + x3 + 20) ÷ (2x – 1), , (e) 7x4 - 6x3 + 8x2 – 10x + 9 ÷ (3x – 9), (f) 6x4 – 4x3 + 6x2 + 8x + 10 ÷ (2x + 3), (g) 9x5 – 7x2 + 12x + 10 ÷ (3x + 1), 3., , 4., , (a), , If x4 + 2x2 – 4x + k is divided by x – 2, the remainder is 4, find the value of k,, using remainder theorem., , (b), , If x3 – 9x2 + (k + 1)x – 8 is divided by x – 5, the remainder is 6, find the value of, k, using remainder theorem., , (c), , If x3 – ax2 + 8x + 11 and 3x3 – ax2 + 7ax + 13, both are divided by (x – 1),, remainder is same, find the value of a., , (d), , If (x – 2) divides the polynomials 4x3 + 2x2 + kx + 5 and kx2 + 5x + 4 to get the, same remainder, find the value of k., , Take a polynomial function. Take any three linear divisors in the form of, (x + a), (ax + b) and (ax – b). Use remainder theorem and find the remainder., , 1.2.3 (B) Factor theorem, Let f(x) be a polynomial of degree n(n > 1) such that f(a) = 0, then (x – a) is a factor of f(x)., Proof: If we divide f(x) by x – a, we obtain a quotient and remainder using algorithm,, f(x) = (x – a). Q(x) + f(a), If f(a) = 0, we have f(x) = (x – a). Q(x), So, (x – a) is factor of f(x)., Conversely: Let (x – a) is a factor of f(x) then remainder f(a) = 0, where f(x) is a polynomial, of degree n., The factor theorem is used in factoring polynomials and hence, is solving polynomial and, finding factors or zeros of polynomial function., , 23
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Example 1, Show that (x – 3) is a factor of x3 – 6x2 + 11x – 6., Solution: Here,, Let, f(x) = x3 – 6x2 + 11x – 6 be the given polynomial. By factor theorem, (x – a) is factor of, a polynomial f(x) if f(a) = 0., Therefore, in order to prove that x – 3 is a factor of f(x), it is sufficient to show that, f(3) = 0, Now, f(x) = x3 – 6x2 – 11x – 6, f(3) = 33 – 6 x 32 + 11 x 3 – 6, = 27 – 54 + 33 – 6, = 60 – 60, =0, Since, f(3) = 0, Hence, this shows that (x-3) is a factor of f(x)., Example 2, Find the value of a such that (x – 4) is a factor of 5x2 – 7x2 – ax – 28., Solution, Let, f(x) = 5x3 – 7x2 – ax – 28 be the given polynomial. If (x – 4) is a factor of f(x), then, f(4) = 0, or, 5 x 43 – 7 x 42 – a x 4 – 28 = 0, or, 5 x 64 – 7 x 16 – 4a – 28 = 0, or, 320 – 112 – 4a – 28 = 0, or, 180 – 4a = 0, or, 4a = 180, or, a =, , 180, 4, , = 45, , Hence, a = 45, Example 3, What must be added in f(x) = 3x3 + 2x2 + 5x + 6, so that the result is exactly divided by, x – 1?, , 24
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Solution, Here, f(x) = 3x3 + 2x2 + 5x + 6, When f(x) is divided by x – 1, the remainder f(1) = 0. It is only possible when we add certain, number in f(x)., Let us add k in f(x), such that 3x3 + 2x2 + 5x + 6 + k is exactly divided by x – 1. Now, by, factor theorem, 3 x 12 + 2 x 12 + 5 x 1 + 6 + k = 0, or, 3 + 2 + 5 + 6 + k = 0, or, 16 + k = 0, or, k = -16, -16 is added in f(x)., Example 4, Using factor theorem, factorize the polynomial x3 + 6x2 + 11x + 6., Solution, Let f(x) = x3 + 6x2 + 11x + 6. The constant term f(x) is equal to 6, and possible factors of 6, are ±1, ±2, ±3, ±6 putting x = 1 in f(x), w have, f(1) = 13 + 6 x 12 + 11 x 1 + 6 = 1 + 6 + 11 + 6 = 24, Since, the remainder is non-zero so, (x-1) is not a factor of f(x)., Again, putting x = -1 in f(x), we have, f(-1) = (-1)3 + 6 x (-1)2 + 11 x (-1) + 6, = -1 + 6 – 11 + 6, =0, x + 1 is a factor of f(x), Now using synthetic division, we have, , Quotient = x2 + 5x + 6, = x2 + 3x + 2x + 6, = x(x+3) + 2(x+3) = (x+3) (x+2), , 25
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Hence the factors of f(x) are (x+1) x Q(x) = (x + 1) (x + 2) (x + 3), Example 5, Using factor theorem, factorize the polynomial, f(x) = (x + 5) (x – 2) – 3(x + 18), Solution, Here, f(x) = (x + 5) (x – 2) – 3(x + 18), = x2 + 5x – 2x – 10 – 3x – 54, = x2 – 64, The possible factors of -64 are ±1, ±2, ±4, ±8, ±16, ±32, ±64, Putting x = 1 in f(x), f(1) = 12 – 64 = – 63 ≠ 0, (x – 1) is not factor of x2 – 64, Putting x = 8 in f(x), we have, f(8) = 82 – 64 = 64 – 64 = 0, So, (x – 8) is factor of f(x), Using synthetic division, , Quotient = x + 8, Now, f(x) = (x – 8) Q(x), = (x – 8) (x + 8), Hence, the factor of f(x) are (x – 8) and (x + 8), Example 6, Solve for x, using factor theorem 6x3 – 13x2 + x + 2 = 0, Solution, Let, the polynomial 6x3 – 13x2 + x + 2 be f(x), Now, f(x) = 0, The possible factors of constant term in f(x) are ±1, ±2., When x = 1, f(1) = 6 x 13 – 13 x 12 + 1 + 2 = 6 – 13 + 3 ≠ 0, , 26
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when x = 2, f(2) = 6 x 23 – 13 x 22 + 2 + 2 = 48 – 52 + 4 = 0, So, (x – 2) is a factor of f(x)., Now, by using synthetic division, we have, , = 6x2 – x – 1, , Now, quotient, Q(x), , = 6x2 – 3x + 2x – 1, = 3x(2x – 1) + 1(2x – 1), = (2x – 1) (3x + 1), Hence, f(x), , = (x – 2) x Q(x), = (x – 2) (2x – 1) (3x + 1), , But, f(x) = 0, or, (x – 2) (2x – 1) (3x + 1) = 0, Either x – 2 = 0 or, 2x – 1 = 0 or, 3x + 1 = 0, or, x = 2, , or, x =, , 1, 2, , or, x = –, , 1, 3, 1 1, , Hence, the roots/zeros of 6x3 – 13x2 + x + 2 are – 3, 2 and 2., Example 7, Observe the given graph of f(x) and find the values of x, when the graph meet X – axis and write f(x)., Solution:, Here, the graph of f(x) meet X- axis at -3, -1 and 2., So, the polynomial has roots x = -3, x = -1 and x = 2, Now f(x) = (x + 3) (x + 1) (x – 2) = x3 + 2x2 – 5x – 6, , 27
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(c), , (d), , 7., , Factorize and solve f(x = x3 – 3x2 – 6x + 8. Give rough sketch of f(x) in graph., , 8., , Investigate the use of polynomial in our daily life. Write the relation between roots, (zeros) of polynomial and polynomial function, with suitable example., , 1.3 Sequence and series, 1.3.0 Review, Observe the following pattern of the figures and discuss on the following questions., (i), , (ii), , (a) Can you draw two more figure in the same pattern of fig (i) and (ii)?, (b) Count the blocks and write the numbers in each figure., (c) Observe the numbers and discuss, how they are increasing., (d) Can you find the relation how they are increasing?, Sequences: A set of numbers which is formed under a definite mathematical rule is called, a sequence. From above patterns, 1, 3, 5, …, …, and, 2, 4, 8, ….. are the example of sequence., , 29
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Series: The sum of the terms of a sequence is called a series. The sum of the terms of the, series is denoted by Sn. For example, (i) 1 + 3 + 5 + …n, (ii) 2 + 4 + 8 + … n, 1.3.1 Arithmetic Sequence, Consider the following sequence of numbers, (i) 1, 2, 3, 4, …,…,…, (ii) 100, 70, 40, 10, …,…,…, (iii) -3, -2, -1, 0, …,…,…, Each of the number in the sequence is called a term. Can you write the next term in each, of the above sequence. If so how will you write it? Let us observe and write the rule., In (i) each term is 1 more than the term preceding it., In (ii) each term is 30 less than the preceding term., In (iii) each term is obtained by adding 1 to the term preceding it., In all the sequence above, we see that the successive terms are obtained by adding a fixed, number to the preceding terms., Definition, A sequence is said to be an arithmetic sequence or arithmetic progression if the difference, between a term and its preceding term is constant throughout the whole sequence. It is, denoted by AP., The constant difference obtained by subtracting a term from its succeeding term and is, called the common difference. In above example (i) 1, 2, 3, 4, ………………, the common, difference = 2 – 1 = 3 – 2 = 4 – 3 = 1, Similarly, the common difference of example (ii) and (iii) are – 30, 1 respectively. The, common difference of an arithmetic sequence is denoted by d., A series corresponding to any arithmetic sequence is known as the arithmetic series, associated with the given arithmetic sequence. Hence, 1 + 2 + 3 + ….. is an arithmetic, series associated with the arithmetic sequence 1, 2, 3, 4 ……, Note: Common difference can be positive or negative., 1.3.2. General term or nth term of AP, If a = first term and d = common difference of an arithmetic progression (AP), then the, terms of progression are, a, a + d, a + 2d, a + 3d, …, , 30
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If t1, t2, t3, t4, …,…,…., tn be the first, second, third, fourth, …….. nth term of an AP, then, t1 = a = a + 0 = a + (1 – 1)d, t2 = a + d = a + (2 – 1)d, t3 = a + 2d = a + (3 – 1)d, t4 = a + 3d = a + (4 – 1)d, ……………………, ……………………, tn = a + (n – 1)d, Now, to know about an AP., What is the minimum information that you need?, Is it enough to know the first term? or is it enough to know only the common difference?, We need to know both the first term (a) and the common difference (d)., List of formulae, 1. General term or nth term (tn) = a + (n – 1)d, 2. Common difference (d) = t2 – t1 = t3 – t2 = tn – tn-1, Example 1, Write the common difference of the given arithmetic sequence 20, 25, 30, 35, ……………, Solution: Here,, The given sequence is, 20, 25, 30, 35, …………….., Common difference (d) = t2 – t1, = 25 – 20, d=5, Common difference (d) = 5, Example 2, Find the 20th term of the AP 2, 7, 12, …., Solution, Here, The given terms of AP are 2, 7, 12, First term (a) = 2, Common difference (d) = t2 – t1 = 7 – 2 = 5, , 31
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To find: 20th terms (t20), By formula, tn = a + (n – 1)d, t5, , = 50 + (5 – 1)d, , t20, , = 2 + (20 – 1)5, = 2 + 19 x 5, = 97, , The 20th term of the given AP is 97., Example 3, If the first and fifth term of an AP are 50 and 30 respectively, find its common difference., Solution: Here,, First term (a) = 50, Fifth term (t5) = 30, No. of term (n) = 5, To find: common difference (d), By formula, tn = a+ (n – 1)d, or, 30 = 50 + (5 – 1)d, or, 30 – 50 = 4d, or,, , −20, 4, , =d, , or, d = -5, The common difference (d) = -5, Example 4, If 21, 18, 15, … is an AS. Find value of n for tn = -81 and tn = 0 and justify., Solution: Here,, The given AP is 21, 18, 15, …, -81, First term (a) = 21, Common difference (d) = t2 – t1 = 18 – 21 = -3, Last term (tn) or (l) = -81, To find: number of terms (n), By formula, tn = a + (n – 1)d, , 32
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or, -81 = 21 + (n – 1) (-3), or, -81 – 21 = (n – 1) (-3), or,, , −102, =, −3, , n–1, , or, 34 + 1 = n, or, n = 35, Therefore, the 35th term of the given AP is -81., Again, If there is any n for which tn = 0 then tn = a + (n – 1)d, or, 0 = 21 + (n – 1) (-3), or,, , −21, −3, , =n–1, , n=7+1=8, So, eight term is 0., Since, the sequence is in decreasing order with common difference-3, so its eighth term, is zero., Example 5, Determine the AP whose 3rd term is 5 and the 7th term is 9., Solution: Here,, Third term (t3) = 5, Seventh term (t7) = 9, To find: An AP, By formula,, tn = a + (n – 1)d, or, t3 = a + (3 – 1)d, 5 = a + 2d, or, a = 5 – 2d ………………. (i), Similarly, t7 = a + 6d, or, 9 = 5 – 2d + 6d [ From equation (i)], or, 9 – 5 = 4d, 4, , or, 4 = d, or, d = 1, , 33
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Substituting d = 1 in equation (i), we get, a = 5 – 2d = 5 – 2 x 1 = 3, Second term (t2) = a + d = 3 + 1 = 4, Third term (t3) = a + 2d = 3 + 2 x 1 = 5, The required A.P is 3, 4, 5, …., Example 6, Check whether 301 is a term of the arithmetic sequence 5, 11, 17, 23, … or not., Solution: Here,, The given sequence is 5, 11, 17, 23, …., First term (a) = 5, Common difference (d) = t2 – t1 = 11 – 5 = 6, Last term (tn) = 301, By formula, tn = a + (n – 1)d, or, 301 = 5 + (n – 1)6, or, 301 = 5 + 6n – 6, or, 301 + 1 = 6n, n=, , 302, 6, , =, , 151, 3, , But n should be a positive integer (why?). So 301 is not a term of the given sequence., Example 7, If k + 1, K + 5 and 3k + 1 are in AP, find the value of K and its three terms., Solution: Here,, K + 1, K + 5 and 3k + 1 are in AP, To find:, (i) The value of K, , (ii) Three terms, , Now, K + 5 – (K + 1) = 3k + 1 – (K + 5), or, K + 5 – K – 1 = 3k + 1 – k – 5, or, 4 = 2k – 4, or, k = 8/ 2 = 4, k =4, , 34
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Again, three terms are, k+1=4+1=5, k+5=4+5=9, 3k + 1 = 3 x 4 + 1 = 13, , Exercise 1.3.1, 1., , 2., , a., , Define sequence and series., , b., , What do you mean by arithmetic sequence?, , c., , Define common difference in arithmetic sequence., , Determine with reason, which of the following are in arithmetic progression, (a) 3, 7, 11, 15, …., , (b) 10, 6, 2, -2, …, , (c) 0, -4, -8, -12, …, , (d) 5 + 11 + 15 + 23 + …, , (e), 3., , 4., , 5., , −3, −5, , 2, 2 , −3, …, 2, , (f) a,, , a+b, 3b−a, , b, 2, 2, , From the following arithmetic sequences, find, (i) First term, , (ii) Common difference, , (iii) the general term (tn), , (iv) the next two terms., , (a) -1, -3, -5, -7, …, , (b) 2, 6, 10, 14, …, , (c) 7, 17, 27, 37, , (d) , 1, ,, , 5, 4, , 3 2, 4 4, , (a), , Write the formula for nth terms of an AP, , (b), , Is 7, 12, 17, 22, … an arithmetic sequence? Write with reasons., , (c), , In tn = a + (n – 1)d, what does 'a' represent?, , (a), , In an arithmetic sequence, 2nd term is 8 and the common difference is 5,, what is the third term? Write it., , (b), , In an arithmetic sequence, the tenth term is 120 and the common, difference is 8. What is the ninth term?, , 6., , (a), , In an arithmetic sequence, 14th term is 150 and the 13th term is 139 then, find the common difference., , (b), , In an AP 6th term is -20 and the 7th term is -25 then find the common, difference., , 35
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(a), , In an AP, 2nd term and 3rd term are 10 and 14 respectively. Find the value, of 1st term., , (b), , In an arithmetic sequence, common difference is 7 and the 2nd term is 5,, what is the first term?, , (a), , The nth term of an AP is 6n + 11. Find the common difference., , (b), , The nth term of an AP is 5n – 2. Find the common difference., , (a), , If the first term and the common difference of an AP are 3 and 5 respectively,, find the 5th term., , (b), , Find the 11th term of arithmetic sequence when the first term and the common, difference are 20 and 5 respectively., , 10. (a), , Determine the first term of an AP whose common difference is -8 and 10th term, is 240., , (b), , What is the first term of an arithmetic sequence whose common difference is, 4 and tenth term is 40., , 11. (a), , What is the common difference of an arithmetic sequence whose first term is, 150 and 12th term is 40?, , 7., , 8., 9., , (b), , Find the common difference of an arithmetic sequence when first term and, fifth term are 9 and 17 respectively., , 12. Find the number of terms of following AP's, (a) 7, 13, 19, ….., 205, , 1, 2, , (b) 18, 15 , 13, … , −47, , (c) First term = 15, common difference = 10, last term = 115, 13. (a), (b), 14. (a), , Is 44 a term of the arithmetic sequence 11, 14, 17 … ?, Does the Arithmetic sequence 11, 8, 5, 2, …., contain -150?, If 2x + 3, x + 11 and 8x + 3 are in AP, find the value of x., , (b), , If 8b + 4, 6b – 2 and 2b + 7 are first three terms of an AP, find the value of b., Also find first three terms., , 15. (a), , If the 11th term and 16th term of an arithmetic sequence are 38 and 73, respectively, find, (i) first term and common difference, , (ii) 31st term, , (iii) An arithmetic sequence, (b), , If the 3rd and 9th terms of an AP are 4 and -8 respectively find, (i) first term and common difference., , 36
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(ii) Which term of an AP is zero., (iii) Is 10 a term of an AP? Write with reason, 16. (a), , An AP consists of 47 terms whose fifth term is 16 and the last term is 100., Find the 20th term., , (b), , An AP consists of 25 terms whose 3rd term is 18 and the last term value is 128., Find the 15th terms., , 17. (a), , In an AP whose third term is 16 and 7th term exceeds the 5th term by 12, then, show that 11th term is 64., , (b), , If 7 times the 7th term of an AP is equal to 11 times its 11th term then show that, the 18th term is zero., , 18. (a), (b), , For what value of n, are the nth terms of two AP's. 63, 65, 67, … and 3,10, 17,, … equal?, If the nth term of the AP 1, 5, 9, 13, … and nth term of an AP 43, 46, 49, … are, equal then, find the value of 'n'., , 19. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th term is, 44. Find first three terms of the AP., 20. (a), (b), , 1.3.2, , A man takes a job in 2019 at an annual salary of Rs. 50,000. He receives an, annual increase of Rs. 2000. In which year will his income reach Rs . 70,000?, After knee surgery, your trainer tell you to return to your jogging program, slowly. He suggests for 12 minutes each for the first week. Each week, thereafter, he suggests you increase that time by 6 minutes. How many weeks, will it be before you are up to jogging 60 minutes per day. Find it., , Arithmetic mean, , The arithmetic sequences are, (i) 5, 10, 15, , (ii) 4, 8, 12, 16, 20, , Discuss in groups about the terms of each sequence., In (i) 10 is arithmetic mean and, In (ii) 8, 12 and 16 are called arithmetic means. Thus, the terms between first term, and last term of an arithmetic progression are called arithmetic means. It is denoted, by AM., , 37
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Arithmetic mean between two numbers, Let, m be the arithmetic mean between two numbers a and b, then, a m, b are in AP then, or, m – a = b – m, or, m + m = a + b, or, 2m = a + b, , , m=, , [ t2 – t1 = t3 – t2], , a+b, 2, , Hence, one AM between a and b is, , a+b, 2, , n arithmetic mean between the two numbers a and b, Let, m1, m2, m3, …, mn be the n arithmetic means between a and b. Then a, m1, m2, m3, …, mn, b be an arithmetic sequence., If d be the common difference, then number of terms = number of means + 2 = n + 2, Last term = nth term (tn) = b, First term (a) = a, By formula, tn = a + (n – 1) d, or, b = a + (n + 2 – 1) d, or, b – a = (n + 1)d, or, d =, , b−a, n+1, , [where n = no. of means], , The arithmetic means are, b−a, , First mean (m1) = t2 = a + d = a + n+1, b−a, , 2nd mean (m2) = t3 = a + 2d = a + 2 (n+1), b−a, , 3rd mean (m3) = t4 = a + 3d = a + 3. (n+1), ……………………………………., …………………………………………, b−a, ), n+1, , nth mean (mn) = tn+1 = a + nd = a + n(, List of important formulae, a+b, 2, b−a, = n+1, , 1., , An arithmetic mean (AM) =, , 2., , Common difference (d), , 3., , First mean (m1) = a + d, , 4., , 2nd mean (m2) = a + 2d and so on., , 38
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Example 1, Find the arithmetic mean between (m – n)2 and (m + n)2., Solution: Here,, First term (a) = (m – n)2, Last term (b) = (m + n)2, To find: An AM, By formula, A.M =, , a+b, 2, , (m−n)2 +(m+n)2, , =, , 2, , =, =, , m2 −2mn+n2 +m2 +2mn+n2, 2, , 2m2 +2n2, 2, 2, , Hence, A.M, , =, , 2(m2 +n2 ), 2, , 2, , =m +n, , Example 2, Find the 13th terms of an AP whose 12th term and 14th terms are -7 and 23 respectively., Solution: Here,, 12th term (t12) = -7, 14th term (t14) = 23, To find: 13th term (t13), We know that, 13th term is arithmetic mean of 12th term and 14th term, By formula, 13th term =, =, , −7+23, 2, , =, , 12th term+14th term, 2, 16, 2, , =8, 13th term (t12) = 8, , Example 3, Find the value of p, q and r if 3, p, q, r, 27 are in A.P, Solution: Here,, The given AP is 3, p, q, r, 27, To find: The value of p, q and r, First term (a) = 3, , 39
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Last term (b) = 27, number of mean (n) = 3, 𝑏−𝑎, , By formula, common difference (d) = 𝑛+1 =, , 27−3, 3+1, , =, , 24, 4, , =6, , then, p = m1 = a + d = 3 + 6 = 9, q = m2 = a + 2d = 3 + 2 x 6 = 15, r = m3 = a + 3d = 3 + 3 x 6 = 21, Hence, p = 9, q = 15 and r = 21, , Example 4, Insert 5 AM's between -6 and 54, Solution, Let, m1, m2, m3, m4 and m5 be 5 AM's between -6 and 54., -6, m1, m2, m3, m4, m5, 54 are in AP, Now, First term (a) = -6, Last term (b) = 54, No. of mean (n) = 5, To find: 5 AM's (i.e. m1, m2, m3, m4 and m5), b−a, , 54+6 60, = 6, 5+1, , By formula, common difference (d) = n+1 =, , = 10, , Again,, First mean (m1), , = a + d = -6 + 10 = 4, , Second mean (m2), , = a + 2d = -6 + 2 x 10 = 14, , Third mean (m3), , = a + 3d = -6 + 3 x 10 = 24, , Fourth mean (m4), , = a + 4d = -6 + 4 x 10 = 34, , Fifth mean (m5), , = a + 5d = -6 + 5 x 10 = 44, , Example 5, If n arithmetic means are inserted between 20 and 5. If the ratio of the third mean and, the last mean is 31:13, find the value of n., Solution:, Let, m1, m2, m3 … mn be the n AM's between 20 and 5., Frist term (a) = 20, , 40
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2., 3., 4., , 𝑏−𝑎, 𝑛+1, , (c), , Write the meaning of 'b' in formula d =, , ., , (a), , State an arithmetic mean between -20 and 20., , (b), , Find the arithmetic mean between (a + b) and (a – b)., , (a), , If the arithmetic mean between x and 2x is 3, find the value of x., , (b), , If the arithmetic mean between y – 1 and 2y is 7, find the value of y., , (a), , In an AP, 4th mean is 12 and the ratio of first mean to the 4th mean is, 2, . Find the first mean., 3, , 5., , 6., , 7., , 8., 9., , (b), , Write the number of terms which is equal to 5th mean., , (a), , In an AP, 10th and 12th terms are 100 and 120 respectively, find the 11th, term., , (b), , If 10, x, 20 are in AP, find x., , (a), , Find the value of x and y if 10, x, y, 25 are in AP., , (b), , If -13, p, q, r, 7 are in AP, find the values of p, q and r., , (a), , If l, m and n are in AP, show that m =, , (b), , If the arithmetic mean between two number is 50 and the second, number is 60, find the first number., , (c), , Two numbers are in the ratio of 2:3. If their AM is 20, find the numbers., , (a), , Insert two arithmetic means between 4 and 22., , (b), , Insert two arithmetic means between 100 and 10., , (a), , Insert 5 arithmetic means between 10 and 70., , (b), , Insert 6 arithmetic means between 11 and 39., , 𝑙+𝑛, 2, , ., , 10. (a), , There are 'n' arithmetic means between 15 and 45. If the third mean is, 30, find the value of n. Also, find the ratio of 3rd mean to fifth mean., , (b), , Some arithmetic means are inserted between 9 and 33. If the third mean is 21,, find the number of means and the values of the remaining means., , 11. (a), (b), , If 'n' arithmetic means are inserted between 5 and 35. The ratio of, second and last mean is 1:4, find the value of n., There are 6 AM's between a and b. If the 2nd mean and 5th means are 11 and, 23 respectively, then find the values of a and b., , 42
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1.3.3, , Sum of first n terms of an arithmetic series, , Let, an arithmetic sequence be, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and its corresponding series is, 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 and the sum of series is 55., Hence, the result obtained by adding the terms of an AP is known as sum of arithmetic, series. Sum of n terms is denoted by Sn., Now, S10 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 ……….. (i) and in reverse order is, S10 = 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 …………… (ii), Adding (i) and (ii) we get 2S10 = 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11, 2S10 = 11 x 10, , S10 =, , 11×10, 2, , = 55, , S10 = 55, Similarly, a be the first term, d be the common difference, n be the number of terms, l be, the last terms and Sn be the sum of the n terms of an arithmetic series (AS), then, Sn = a + (a + d) + (a + 2d) + …………. + (l – 2d) + (l – d) + l …………… (iii), Writing in the reverse order, we have, Sn = l + (l – d) + (l – 2d) + ………. + (a + 2d) + (a + d) + a ………………. (iv), Adding equation (iii) and (iv) we get, 2Sn = (a + l) + (a + l) + (a + l) + …………. + (a + l) + (a + l) + (a + l), or, 2Sn = n(a+l), 𝑛, 2, , or, Sn = (𝑎 + 𝑙), We know that, Last term (l) = a + (n – 1)d, 𝑛, , Sn = 2 [a + a + (n – 1)d], 𝑛, , Sn = 2 [2a + (n – 1)d], Important formulae, 𝑛, , (i) Sn = 2 (a + l), 𝑛, , (ii) Sn = 2 (2a + (n – 1)d], Note: To solve the problem easily, we can consider, (i) Three terms of an A.P be a – d, a, a + d, 43
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(ii) Four terms in an AP be a – 3d, a – d, a + d, a + 3d, Sum of first 'n' natural numbers, Let, 1 2, 3, 4, ……………, n be the set of 'n' natural numbers. It is an arithmetic progression, with first term (a) = 1 and the common difference (d) = 1, We have,, 𝑛, , = 2 [2𝑎 + (𝑛 − 1)𝑑], , Sn, , 𝑛, , = 2 [2 x 1 + (n – 1).1], 𝑛, , = 2 (2 + n – 1), 𝑛, 2, , = (n + 1), 𝑛, , Therefore, the sum of first 'n' natural numbers (Sn) = 2 (n + 1), Example 1, Find the sum of 1 + 2 + 3 + ……… + 20, Solution: Here,, The given series is 1 + 2 + 4 + ………… + 20, First term (a) = 1,, common difference (d) = 2 – 1 = 1, no. of terms (n) = 20, 𝑛, , By formula, Sn = 2 [2a + (n – 1)d], S20, , =, , 20, 2, , [2 x 1 + (20 – 1).1], , = 10 (2 + 10), = 10 x 21 = 210, S20 = 210, Sum of First n Even Natural Number, The first n even numbers are 2, 4, 6, … 2n, Number Sn = 2 + 4 + 6 + …. + 2n, First term (a) = 2, d = 2, Sum of first n terms be Sn, , 44
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By formula,, 𝑛, 2, 𝑛, 2, 𝑛, 2, 𝑛, 2, 𝑛, 2, , Sn = [2a + (n – 1)d], = [2 x 2 + (n – 1)2], = (4 + 2n – 2), = (2n + 2), = x 2(n + 1), Sn = n(n + 1) = n2 + n, Sum of first 'n' odd natural numbers, The first n odd natural numbers are 1, 3, 5, ..., (2n – 1), Let, Sn = 1 + 3 + 5 + ………………….. + (2n – 1), First terms (a) = 1, Common difference (d) = 3 – 1 = 2, Number of terms (n) = n, Sum of 'n' terms be Sn, By formula,, Sn, , 𝑛, 2, 𝑛, [2 x 1 + (n – 1)2], 2, 𝑛, (2 + 2n – 2), 2, 𝑛, x 2n = n2, 2, , = [2a + (n – 1)d], =, =, =, , Example 2, Find the sum of the first 10 terms of the AP: 2, 7, 12, …, Solution: Here,, The given AP is 2, 7, 12, …, To find: The sum of first 10 terms (S10), First term (a) = 2, Common difference d) = 7 – 2 = 5, no. of terms (n) = 10, By formula,, 45
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Sn, , 𝑛, 2, , = [2a + (n – 1)d], =, , 10, 2, , [2 x 2 + (10 – 1)5], , = 5[4 + 45], = 5 x 49 = 245, Hence, Sum of first 10 terms is 245., Example 3, Find the sum of the series: 34 + 32 + 30 + …… + 10, Solution: Here,, The given series is 34 + 32 + 30 + …………. + 10, First term (a) = 34, Last term (l) = 10, Common difference (d) = t2 – t1 = t3 – t2, = 32 – 34 = 30 – 32, = -2 = -2, It is an arithmetic series., To find: Sum of series (Sn), Now, tn = l = a + (n – 1)d, or, 10 = 34 + (n – 1) (-2), or, 10 – 34 = -2n + 2, or, -24 = -2n + 2, or, 2n = 24 + 2, or, n =, , 26, 2, , = 13, , n = 12, Again, by formula,, 𝑛, , Sn = 2 [a + l], =, , 13, [34, 2, , =, , 13, 2, , + 10], , x 44 = 286, , S13 = 286, , 46
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Example 4, How many terms of the AP 24, 21, 81, … must be taken so that their sum is 78?, Solution: The given AP is 24, 21, 18, …, Sum of n terms (Sn) = 78, To find: number of terms (n), First terms (a) = 24, Common difference (d) = t2 – t1 = 21 – 24 = -3, By formula,, 𝑛, 2, , Sn = [2 x 24 + (n – 1) (-3)], 𝑛, , or, 78 = 2 [48 – 3n + 3], 𝑛, , or, 78 = 2 [51 – 3n), or, 156 = 51n – 3n2, or, 3n2 – 51n + 156 = 0, or, -3(n2 – 17n + 52) = 0, or, n2 – (13 + 4)n + 52 = 0, or, n2 – 13n – 4n + 52 = 0, or, n(n – 13) –4(n – 13) = 0, or, (n – 13) (n – 4) = 0, Either, n – 13 = 0, n = 13, , or, n – 4 = 0, n=4, , Since, both values of n are positive. So the number of terms is either 4 or 13., Note: (i) In this case the sum of the 1st four terms = the sum of the first 13 terms = 78., Example 5, Find the common difference of an arithmetic series whose first term is 2 and the sum of, the 10 terms is 120, Solution: Here, First term (a) = 2, Sum of the first 10 terms (S10) = 120, To find: common difference (d), , 47
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By formula,, 𝑛, 2, , Sn = [2a + (n – 1)d], or, S10 =, , 10, [2, 2, , x 2 + (10 – 1)d], , or, 120 = 5 (4 + 9d), or,, , 120, 5, , = 4 + 9d, , or, 24 – 4 = 9d, or d =, , 20, 9, , Hence, common difference (d) =, , 20, ., 9, , Example 6, Find the sum of all natural numbers less than 100 which are exactly divisible by 6., Solution: Here,, The natural numbers less than 100 and exactly divisible by 6 are 6, 12, 18, ……. 96, Now, first term (a) = 6, Common difference, (d) = 6, Last term (l) = 96, To find: Sum (Sn), By formula, l = a + (n – 1)d, or, 96 = 6 + (n – 1)6, or, 96 – 6 = (n – 1)6, or,, , 90, 6, , =n–1, , or, 15 + 1 = n, n = 16, Again, by formula, 𝑛, , Sn = 2 (a + l), or, S16 =, , 16, (6, 2, , + 96), , or, S16 = 8 x 102, or, S16 = 816, Hence, the required sum is 816., , 48
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Example 7, If the sum of first seven term of an arithmetic series is 14 and the sum of the first ten, terms is 125, then find the fourth term of the series., Solution: Here,, Sum of the 1st seven terms of the series in A.P. (S7) = 14, Sum of the 1st ten terms (S10) = 125, To find: Fourth term (t4) ., By formula,, 𝑛, , Sn = 2 [2a + (n – 1)d], 7, , or, S7 = 2 [2a + (7 – 1)d, 7, , or, 14 = 2 x 2 (a + 3d), a = 2 – 3d …………. (i), Similarly, S10 = 125, or,, , 10, 2, , [2a + (10 – 1)d] = 125, , or, 5[2a + 9d] = 125, or, 2a + 9d = 25, a=, , 25−9d, 2, , ……….. (ii), , Equating equation (i) and (ii), we get, 2 – 3d =, , 25−9𝑑, 2, , or, 4 – 6d = 25 – 9d, or, 9d – 6d = 25 – 4, or, 3d = 21, d=7, From equation (i) a = 2 – 3d = 2 – 3 x 7 = -19, Again, fourth term (t4) = a + 3d = -19 + 3 x 7 = 2, Fourth term (t4) = 2, , 49
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Example 8, The sum of three consecutive terms in an arithmetic series is 18 and their product is 192,, find these three terms., Solution: Here, Let, three consecutive terms in AP be a – d, a, a + d, To find: Three terms in AP., From first condition, a – d + a + a + d = 18, or, 3a = 18, or, a = 6, From 2nd condition: (a – d) x a x (a + d) = 192, or, (6 – d) x 6 x (6 + d) = 192, or, 36 – d2 =, , 192, 6, , or, 36 – d2 = 32, or, -d2 = 32 – 36, or, -d2 = -4, d = ±√4 = ± 2, (i), , When a = 6 and d = 2 then three terms are, a–d=6–2=4, a=6, a+d=6+2=8, , (ii), , When a = 6 and d = -2 then three terms are, a–d=6+2=8, a=6, a + d = 6 + (-2) = 4, , Hence, three terms of an AP are 4, 6, 8 or 8, 6, 4., , Exercise 1.3.3, 1., , (a), , Write the formula for finding the sum of first 'n' terms in arithmetic series, whose first and the last terms are given., , (b), , If sum of first 10 terms of arithmetic series is 80 and the sum of 1st 9 terms of, the same series is 72. Find the 10th terms., , 50
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2., 3., , (a), , What is the sum of first 5 odd natural numbers?, , (b), , What is the sum of first 5 even natural numbers?, , Find the sum of the following series:, (a) -37 – 33 - 29, ………….. to 12 terms, (b) 5 + 8 + 11 + 14 + ………. to 20 terms, 1, , (c) 7 + 10 2 + 14 + …………. + 84, (d) -5 + (-8) + (-11) + ………… + (-230), 4., , Find the sum of the series:, (a) ∑10, 𝑛=1(2𝑛 + 1), , 5., , 6., , 7., , 8., , 9., , (b) ∑15, 𝑛=2(𝑛 + 2), , (a), , The first terms and the common difference of an arithmetic series are 2, and 8 respectively. Find the sum of first 10 terms., , (b), , If the common difference and the sum of 1st 9 terms of an AP are 5 and 75, respectively, find the first terms., , (a), , Find the common difference of an AS whose first terms is 3 and the sum of first, 8 terms is 192., , (b), , How many terms of the series 9 + 6 + 3 + … + … must be taken so that the sum, of the series is zero?, , (a), , Find the sum of all numbers from 50 to 150 which are exactly divisible, by 9., , (b), , Find the sum of all two digit numbers in AP which are multiple of 5., , (c), , Find the sum of the odd numbers between 0 and 50., , (a), , The first term of an AP is 5, the last term is 45 and the sum is 400. Find, the numbers of terms., , (b), , The first term of an AP is -9, the last term is 47 and the sum is 152. Find, the number of terms., , (a), , How many terms of the series 20 + 18 + 16 + …… must be taken so that, the sum of the series may be 110? Explain the double answer., , (b), , If the 3rd term and 11th term of an AP are 18 and 50 respectively, find, (i) First terms and the common difference, (ii) Arithmetic series (iii) Sum of first 20 terms., , 10. (a), , If the 2nd term and 12th term of an AP are 20 and 50 respectively, find, (i) First terms and the common difference, (ii) Sum of first 25th terms., , 51
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(b), , If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289:, (i) Find the first term and the common difference, (ii) Find the arithmetic series, (iii) Find the sum of the first 20 terms., , 11. (a), (b), , The sum of three terms of an AP is 30 and their product is 840. Find the, three terms., The sum of three terms of an AP is 12 and the sum of their squares is 56., Find the three terms., , 12. A sum of Rs. 700 is to be used to give seven cash prizes to students of a school for, their overall academic performance. If each prize is Rs. 20 less than its preceding, prize, find?, (i) The first prize, , (ii) The 2nd prize, , (iii) The 3rd prize, , (iv) Seventh prize, , 52
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Geometric sequence (geometric progression) (GP), Let us consider the following sequence:, a), , i) 2, 4, 8, 16, 32, …, ii) 81, 27, 9, 3, …, How are the numbers arranged in both sequences?, , b) Is the difference between a term and its preceding term constant in both, sequences?, c) What relation do you see in any two consecutive of terms in each of the above, sequence?, d) What characteristics are found in both sequences?, In (i) sequence the numbers are increased in the multiple of 2, similarly, in (ii) the, 1, , numbers are decreased by multiple of 3., Therefore, from above two sequences, the numbers are increased or decreased by a, constant number called common ratio and such sequences are called geometric, sequences or geometric progression. (GP), Hence, a sequence or series of numbers that increased or decreased with a constant ratio, is called a geometric sequence or series. The constant ratio and it is denoted by r., Common ratio (r) =, , a term, preceding term, , Or, Common ratio (r) =, , t2, t1, , t, , t, , = t3 = t4, 2, , 3, , 1, 3, , From the above sequences (I) and (ii), the common ratios are 2 and respectively., 1.3.4, , General term (nth term) of a geometric sequence, , If ‘a’ be the first term and r be the common ratio of a geometric sequence, then the, terms of the sequences are, a, ar, ar2, ar3, …, If, t1, t2, t3, … tn be the first, second, third, …, nth terms of a geometric sequence, respectively, then, First term (t1) = a = ar1-1, Second term (t2) = ar = ar2-1, Third term (t3) = ar2 = ar3-1, ………………….., …………………, nth term (tn) = arn-1, Note: If we have first term and common ratio then we can find any term of a, , geometric progression., 53
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Example 1, Find the 10th term of a sequence: 3, 9, 27, …, , Solution: Here,, the given sequence is 3, 9, 27, …, To find: 10th term (t10), Now,, , t2, t1, , =, , 9, 3, , =, , 27, 3, , or,, , t3, t2, , or, 3 = 3, So, the given sequence forms a GP, First term (a) = 3, common ratio (r) = 3, By formula, tn = arn-1, t10 = 3 x 310-1, t10 = 59,049, Hence, 10th term t10 = 59,049, Example 2, What is the common ratio of a geometric sequence whose first term is 48 and 4 th term is 6?, , Solution: Here,, First term (a) = 48, 4th term (t4) = 6, Common ratio (r) = ?, By formula,, 𝑡𝑛 = 𝑎𝑟 𝑛−1, 𝑜𝑟, 𝑡4 = 48 × 𝑟 4−1, 𝑜𝑟,, , 6, 48, , = 𝑟3, , 1, , 𝑜𝑟, 8 = 𝑟 3, 1 3, , 𝑜𝑟, (2) = (𝑟)3, 1, , 𝑟=2, 1, , Hence, Common ratio (r) = 2, , 54
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Example 3, The 3rd term and 7th term of GP are 8 and 128 respectively. Find, i) Common ratio and first term., ii) Geometric sequence, iii) 15th term, Solution: Here,, 3rd term (𝑡3 ) = 8, 7th term (𝑡7 ) = 128, To find:, (i) Common ratio (r) and first term (a), By formula,, , (ii) Geometric sequence, , (iii) 15th term (𝑡15 ), , 𝑡𝑛 = 𝑎𝑟 𝑛−1, Now, 𝑡3 = 8,, , t3 = ar3-1, , 𝑜𝑟, 𝑎𝑟 3−1 = 8, 𝑜𝑟, 𝑎𝑟 2 = 8 ………………….. (i), Similarly, 𝑎𝑟 7−1 = 128, 𝑎𝑟 6 = 128 …………………. (ii), Dividing equation (ii) by (i), we get, 𝑎𝑟 6, 𝑎𝑟 2, , =, , 128, 8, , 𝑜𝑟, 𝑟 4 = 16, 𝑜𝑟, 𝑟 4 = (±2)4, 𝑟 = ±2, Taking r = 2 then from equation (i) [Since, GP is in increasing order.], 𝑎(2)2 = 8, 8, , 𝑎=4=2, First term (a) = 2, Second term (t2) = ar = 2 x 2 = 4, GP is 2, 4, 8, ………. and 15th term (t15) = 𝑎𝑟14 = 2 × (2)14 = 32,768, , 55
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Exercise 1.3.4, 1. a), b), , Define geometric sequence with an example., If a, ar, ar2, … be the first, second and third term of a geometric sequence,, write the sixth term., , 2. Which of the following are geometric sequences?, a) -1, 6, -36, 316, …, , b) -1, 1, 4, 8, …, , c) 4, 16, 36, 64, …, , d) -3, -15, -75, -375, …, , e) -2, -4, -8, -16, …, , d) 1, -5, 25, -125, …, , 3. From the following geometric sequences, find the common ratio and next two terms., 1, 2, , a) 2, 1, , …, 4. a), b), , b) 5, -10, 20, …, , c) 4, 12, 36, …, , d) a, ab, ab2 …, , Find the first term of a geometric sequence, whose sixth term is 729 and the, common ratio 3., Determine the first term of geometric sequence, whose 4th term is -8 and the, 1, , common ratio is − 2., 5. a), , If the first term of a GP is 4 and the 5th term is 64, find the common ratio., , b), , How many terms are there in the geometric sequence 3, 12, 48, …, 192?, , c), , Find the number of terms in a sequence, 1, 5, 25, ..., 3125., , 6. a), b), 7. a), , Find the value of x, if x, x + 4, and x + 6 are consecutive terms of geometric, sequence., Find the value of p, if 2p, 2p + 3, and 2p + 9 are three consecutive terms of a, geometric sequence. Also find the common ratio., The third term of a geometric progression is -108 and the sixth term is 32. Find, i) The common ratio and the first term., iii) 10th term, , ii) Geometric sequence, b), , The second and fifth term of a geometric sequence are 750 and -6 respectively,, Find, i) Common ratio and the first term, iii) 8th term, , ii) Geometric Sequence, 8. a), , The fourth term of G.P is square of its second term and the first term is -3., Determine the 8th term of GP., , 56
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b), , In a geometric sequence, 2nd term is 9 and 27 times of 8th term is equal to 5th, term, then find the 12th term., , c), , The first term of a geometric sequence is 1. The ninth term exceeds the fifth, term by 240. Find the possible values for the eighth term., , 9. The first three terms of a geometric sequence are (k + 4), k and (2k – 15) respectively,, where k is positive constant. (i) show that k = 12 (ii) Find the common ratio, (iii) 10th term., 10. Suppose, your parents income is Rs. 50, 0000 per month. They save Rs. 2,000 in the, first week of the new year, if they double the amount they save every week, after, that, how much will they save in the 2nd, 3rd, 4th and 5th week of the year?, Complete the following table and explain in classroom with reason., 1st week, , Week No., Amount, Saved, , 1.3.5, , 2nd week, , 3rd week, , 4th week, , 5th week, , Geometric mean, , Consider a geometric sequence 2, 6, 18, where 6 is called geometric mean. We can write, it, geometric mean of 2 and 18 = √2 × 18 = 6, Also, it is like the area which is same, , 6, , 18, 2 =, , 6, , Area (A) = 2 x 18, , Area = 6 x 6, , Thus, the term between the first and the last term of a GP are known as the geometric, means., Geometric mean between two numbers a and b, Let, GM be the Geometric Mean between a and b then a, GM, b from a GP, then, by definition,, GM, a, , b, , = GM, 57
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AM =, , a+b, 2, , And GM = √ab, Now,, AM − GM =, =, , a+b, −, 2, , √ab, , (a+b−2√ab), 2, 2, , 2, , =, , (√a) +(√b) −2√a √b, 2, , =, , (√𝑎−√𝑏), 2, , 2, , or, AM – GM ≥ 0 [ Square of any two quantities is greater than or equal to zero.], Hence, AM ≥ GM, Case I – When a = b then, AM = GM, Case II – When a ≥ b or b ≥ a then AM ≥ GM, Example 1, Find the geometric mean between 5 and 20, Solution: Here,, First term (a) = 5, Last term (b) = 20, To find: GM, By formula, GM = √ab = √5 × 20 = √100 = 10, Hence, G.M = 10, Example 2, 1, , Find the value of p and q if 8, p, q, 8 are in G.P, Solution: Here,, 1, , The given GP is 8, p, q, 8, To find: The value of p and q, 1, , Now, First term (a) = 8, Last term (b) = 8, no. of means (n) = 2, 59
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By formula, common ratio (r), 1, 2+1, , 1, , =, , 𝑏 𝑛+1, (𝑎), , 8, , = (1), , 1, , 1, , = (64)3 = (43 )3 = 4, , 8, , 1, 8, , 1st mean (m1) = 𝑝 = 𝑎𝑟 = × 4 =, 1, , 1, 2, , 2nd mean (m2) = 𝑞 = 𝑎𝑟 2 = 8 (4)2 = 2, Example 3, , Find the number of geometric means inserted between 1 and 64 in which the ratio of, first mean to the last mean is 1:16, Solution: Here,, Let, m1, m2, m3 … mn be the n GM’s between 1 and 64, 1, m1, m2, m3, … mn, 64 form a GP, Since,, , 1st mean(m1 ), last mean (mn ), , 1, , = 16, , To find: number of GM (n), Now, First term (a) = 1, Last term (b) = 64, 1, , By formula, common ratio (r) =, , 𝑏 1+1, ( ), 𝑎, , 1, , =, , 64 𝑛+1, ( ), 1, , 1, , 𝑜𝑟, 𝑟 = (64)𝑛+1 ……………. (i), 𝑚1, 1, = 16, 𝑚𝑛, 𝑎𝑟, 1, 𝑜𝑟, 𝑎𝑟𝑛 = 16, 1, 𝑟1−𝑛 =, 16, , and,, , 1, , Raising power 1−𝑛 on both sides we get, 1, , 1 1−𝑛, , 𝑟 = (16), , … (ii), , Equating equation (i) and (ii) we get, 1, 𝑛+1, , (64), , 1, , =, , 3, n+1, , or, (4), or, 4, , 3, 𝑛+1, , 1 1−𝑛, (16), 16 −, , = (1), , 1, −(n−1), , 2, , = 4𝑛−1, , 60, , 1, , = (64)𝑛+1
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or,, , 3, n+1, , =, , 2, n−1, , or, 3n − 3 = 2n + 2, n=5, Therefore, there are 5 GM’s between 1 and 16., Exercise 1.3.5, 1. a), b), c), , Define geometric mean with an example., Determine a geometric mean between p and q., Find a geometric mean between, , 1, 10, , and 10., 1, , 2. a), , Find 10th term of a GP whose 9th term is 5 and 11th term is 125., , b), , The Geometric mean between 9 and b is 2. Find the value of b., , c), , There are 3 GM’s in a GP in which first term is 10 and common ratio is 2 then, , 3. a), b), 4. a), b), c), 5. a), , 1, , 1, , find first mean only., Find AM and GM between 4 and 16. Also write their difference., The arithmetic mean of a and 24 is 15. Find the Geometric Mean (GM)., Insert 2 GM’s between 6 and 48, Insert 3 GM’s between 5 and 80, 1, , Find the values of x and y when 2, x, 2, y are in GP, Insert 7 GM’s between, , 1, 16, , and 16 and find the difference between 5th mean, , and first mean., b), , 1, , Insert 6 GM’s between 2 and 64 and compare 1st mean and 6th mean., , 7. a), b), , There are 4 geometric means between 4 and q. If 2nd mean is 36, find the, value of q and the remaining other means., Some geometric means are inserted between 5 and 80. Find the number of, means between two numbers if the third mean is 40. Also find the remaining, means., The AM between two positive numbers is 50 and GM is 40. Find the numbers., The AM between two natural numbers is 45 and GM is 27. Find the numbers., , 8. a), , There are n geometric means between 81 and 81. If the ratio of 3rd mean to, , 6. a), b), , 1, , the last mean is 1:81. Find the value of n., b), , 27, , 8, , There are n geometric means between 16 and 81. If the ratio of (n-1)th mean, to the 2nd mean is 8:27. Find the value of n., , 61
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Note: to solve the problem easily, we can suppose, 𝑎, i) Three terms in GP be , a, ar, 𝑟, 𝑎 𝑎, , , ar, ar3, 𝑟3 𝑟, 𝑎 𝑎, be 2 , , a, ar, ar2, 𝑟 𝑟, , ii) Four terms in GP be, iii) Five terms in GP, Example 1, , Find the sum of the following series: 36 + 12 + 4 + …… up to 6 terms., Solution: Here,, The given series is 36 + 12 + 4 + … to 6 terms, First term (a) = 36, 12, , 1, , Common ratio (r) = 36 = 3, It is geometric series, no. of terms (n) = 6, Sum of 6 terms (s6) = ?, By formula,, 𝑠𝑛 =, =, , 𝑎(1−𝑟 𝑛 ), 1−𝑟, 1 6, 3, 1, 1−, 3, , 36{1−( ) }, , r<1, =, , 1, ), 729, 3−1, 3, , 36(1−, , =, , 36×728×3, 729×2, , =, , 1456, =, 27, , 25, , 53 27, , Example 2, If the first and last term of a GP are 7 and 448 respectively and the sum of the series is, 889, find the common ratio., Solution : Here,, First term (a) = 7, Last term (l) = 448, Sum of n terms (sn) = 889, To find: Common ratio (r), By formula,, sn =, , lr−a, r−1, , or, 889 =, , 448r−7, r−1, , or, 889r – 889 = 448r – 7, or, 889r – 448r = 889 – 7, 63
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𝑜𝑟, 3(1 + 𝑟 + 𝑟 2 ) = 13𝑟, 𝑜𝑟, 3 + 3𝑟 + 3𝑟 2 − 13𝑟 = 0, 𝑜𝑟, 3𝑟 2 − 10𝑟 + 3 = 0, 𝑜𝑟, 3𝑟 2 − (9 + 1)𝑟 + 3 = 0, 𝑜𝑟, 3𝑟 2 − 9𝑟 − 𝑟 + 3 = 0, 𝑜𝑟, 3𝑟(𝑟 − 3) − 2(𝑟 − 3) = 0, 𝑜𝑟, (𝑟 − 3)(3𝑟 − 1) = 0, Either, 𝑟 − 3 = 0 𝑟 = 3, or, 3𝑟 − 1 = 0 𝑟 =, , 1, 3, , Case I: When a = 12 and r = 3 then three numbers are, 𝑎, 𝑟, , =, , 12, 3, , =4, , 𝑎 = 12, 𝑎𝑟 = 12 × 3 = 36, 1, 3, , Case II When a = 12 and r = then three numbers are, 𝑎, 𝑟, , =, , 12, 1, , 3, , = 12 × 3 = 36, , 𝑎 = 12, 1, 3, , 𝑎𝑟 = 12 × = 4, Hence, the required numbers are 4, 12, 36 or 36, 12, 4, Exercise 1.3.6, 1. a), b), 2. a), b), c), 3., , a), , If lr – a = 90 and r – 1 = 9, find sn., Write the sum of first 5 terms of the geometric series if a = 1 and r = 2., 1893×3−𝑎, If 500 =, , Find the value of a., 3−1, In a geometric series 2 + 4 + 8 + 16 + 32 + 64 + 128. What is the common, ratio?, 1, Write the first term and last term of the given GP: 81 + 27 + 9 + 3 + 1 + 3, Find the sum of the following geometric series:, i) 24 + 12 + 6 + … to + 10 terms, ii) 3 – 6 + 12 - … to 7 terms, 1, 1, iii) 27 + 18 + 12 + … to 6 terms, iv) √2 + 2 + 2 2 +… to 8 terms, √, , (v) 3 + 6 + 18 … + 4374, , √, , vi) 3 + 6 + 12 + … + 768, , 66
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4., , a), b), , 5. a), b), a), 6. a), b), 7. a), b), , 8., , a), , The first term of a GP is 1, the sum of the third and fifth term is 90. Find the, common ratio., If the sum of first two terms is 10 and the first term is 2. Find the common, ratio., The sum of 4th terms of a series in GP whose common ratio is 2 is 255. Find, the first term of a GP., The sum of a series in GP whose common ratio 4 is 1364, and the last term is, 1024. Find the first term., 1, If the first term and common ratio of a geometric series are 3 and 3, respectively. Find the sum of 1st 6 terms., How many terms are in a GP where a = 2, r = 3 and the sum is 728?, How many terms of the series 5 + 10 + 20 + … must be taken so that the sum, becomes 315?, Find the sum of first 8 terms of a GP whose 3rd and 7th terms are 8 and 128, respectively., In a GP, 4th term and 7th terms are 27 and 729 respectively. Find, i) common ratio and first term, ii) Geometric series, iii) Sum of first eight terms, In a GP the sum of 1st two terms is 18 and the sum of first 4 terms is 90. Find, i) Common ratio and first term, ii) Geometric Series, iii) Sum of first 6 terms, 7, , 63, , In a GP the sum of 1st three terms is and the sum of 1st six terms is . Find, 4, 4, the sum of 1st 8th terms, 9. a), In a GP the sum of three numbers is 28 and their product is 512. Find the, numbers., b), The sum of three consecutive terms of GP 13 and their product is 27. Find, the three consecutive terms., 10. a), The sum of three numbers in GP is 56. If 1, 7, 21 are subtracted from the, numbers respectively which form the consecutive term of an A.P. Find the, original numbers., b), The sum of three numbers in AP is 15. If 1 and 5 are respectively added to 2nd, and 3rd numbers then the first number together with these two are in GP find, the original numbers., 11. The IQ scores of 10 students in a test are as the rule that the score of second student, is the double of score of first student. The score of third student is double the score, of second and so on. If the score of the first student is 2, find the score of 10th students., Also find total score of all 10 students., b), , 67
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1.4 Linear programming, 1.4.0 Review, The general form of a linear equation of two variables x and y is ax + by + c = 0. A relation, represented by ax + by + c > 0, ax + by + c < 0, or ax + by + c ≥ 0 or ax + by + c ≤ 0 is known, as the linear inequality in two variables x and y., Let us consider one example; x ≥ 3., i) How many variable are there in the above inequality?, ii), , What do you mean by sign >?, , The associated equation of x ≥ 3 is x = 3 is a straight line (say MN) parallel to y-axis at a, distance of 3 units from x-axis., From the graph, we observe that the line x = 3 i.e. MN divides the whole plane into two, parts, one on the right side of MN each point of which x-coordinate is greater than 3 and, other on the left of MN each point of which x-coordinate less than 3. Each point on MN, has x-coordinate 3. The line x = 3 is said to be the boundary line., , Hence, the graph of x ≥ 3 is shown in the plane region on the right of MN including MN., Hence, MN must be drawn by the solid line because the line MN is also included on the, graph., Let’s take another inequality, x<3, The graph of x < 3 will be the plane region on the left of PQ (not containing PQ as the, inequality does not contain equality sign as well. Here, we have drawn broken line to, indicate that the line PQ is not included on the graph., , 68
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When we put x = 0, y = 0 in x < 3, we have 0 < 3 which is true, so its graph is a plane, region containing the origin (0, 0)., 1.4.1. Graph of inequalities in two variables, Let us consider 2x + 3y ≥ 6, What are the two variables? Let’s discuss., Now,, i), ii), , The corresponding equation of the given inequality is 2x + 3y = 6 which is the, boundary line of the given inequality., The table from equation is, x, y, , 0, , 3, , 6, , 2, , 0, , -2, , The boundary line passes through the points (0, 2), (3, 0) and (6, -2). It divides the, plane region into two parts, iii) Plot the points (0, 2), (3, 0) and (6, -2). Join the points by the solid line (not by, dotted line)., (iv) For the graph of 2x + 3y ≥ 6, we use the test point (0, 0), put x = 0, y = 0 in the given, in equation and if (0, 0) satisfies the given in equation, then the graph of the given in, equation is the plane region containing the origin. But if (0, 0) does not satisfy the, given in equation, the graph is the plane region not containing the origin., 69
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Taking testing point (0, 0), put x = 0, y = 0 in 2x + 3y ≤ 6, i.e. 0 + 0 ≥ 6, , 0 ≥ 6 which is false, , The graph of 2x + 3y ≥ 6 is the plane region not containing the origin., , Example 1, Draw the graph of 2x - 3y < 6, Solution: Here,, The corresponding equation of given inequality is 2x – 3y = 6 which is the boundary line, or, 2𝑥 = 6 + 3y, or, 𝑥 =, , 6+3𝑦, 2, , ……………. (i), , Table from the equation (i), x, , 3, , 0, , -3, , y, , 0, , -2, , -4, , 70
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The boundary line passes through the points (3, 0), (0,-2) and (-3,-4). Plot the points, (3, 0), (0, -2) and (-3, -4). Join the points by a dotted line., For the graph, taking testing point (0, 0), put x = 0, y = 0 in 2x – 3y < 6, I.e. 0 – 0 < 6, , or,, , 0 < 6 which is true., , Hence, the graph of 2x – 3y < 6 is the plane region containing the origin but not, boundary line., Note: If the corresponding equation of a given inequation passes through the origin,, then the test point should be different from (0, 0) i.e. (1, 0) or (0, 1) etc., 1.4.2 System of Linear Inequalities, A set of two or more linear inequalities having a common solution region (set) is said to, be the system of linear inequalities. In this system, the plane regions determined by the, set of inequalities are shown in the same graph., Example 2, Draw the graph of x - 2y ≥ 4 and 2x + y ≤ 8, Solution, The given in equations are, x – 2y ≥ 4 and 2x + y ≤ 8, The corresponding equation of given inequations are, x – 2y = 4 …………………. (i), 2x + y = 8 ………………… (ii), , 71
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From equation (i), x = 4 + 2y, table from equation (i), x, , 4, , 0, , 2, , y, , 0, , -2, , -1, , Similarly from equation (ii), y = 8 - 2x, Table from equation (ii), x, , 4, , 3, , 2, , y, , 0, , 2, , 4, , One boundary line passes through the points (4, 0), (0, -2) and (2, -1). Plot the points and, join these points by solid line. Similarly, another boundary line passes through the points, (4, 0) (3, 2) and (2, 4). Plot these points and join these points by another solid line., Taking testing point (0, 0) in x – 2y ≥ 4, put x = 0 and y = 0, 0 – 2 x 0 > 4 0 > 4 which is, false., Hence, the graph of x-2y ≥ 4 is the plain region does not contain the origin. Similarly, taking, testing point (0, 0) in 2x+y ≤ 8, Put x = 0, y = 0, then 2 x 0 + 0 ≤ 8 = 0 ≤ 8 which is true., Hence, the graph of 2x+y ≤ 8 is the plane region containing the origin., The intersection part of the shaded region gives the required solution set of the given, system of inequalities., 1.4.3 Linear programming, Most of the business and economic activities may have various problem of planning due, to limited resources. In order to achieve the business goal i.e. minimizing the cost of, production and maximize the profit from the optimum use of available limited resources,, linear programming is used., Linear programming is a mathematical technique of finding the maximum or minimum, value of the objective function satisfying the given condition. The problem which has, object of finding maximum or minimum value satisfying all the given condition is called, linear programming (L.P) problem. To define linear programming, we need some basic, definitions:, (i) Decision variables: The non-negative independent variables involving in the L.P, problem are called decision variables. For example: in 2x + 3y = 7, x and y are decision, variable., (ii) Objective function: The linear function whose value is to be maximized or minimized, (optimized) is called an objective function., , 72
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(iii) Constraints: - The conditions satisfied by the decision variables are called constraints., For example: if x and y be the number of first two kinds of articles produced, then, x + y ≥ 1000; x ≥ 0 and y ≥ 0 are the constraints, (iv) Feasible region (convex polygonal region): A closed plane region bounded by the, intersection of finite number of boundary lines is known as feasible region., (v) Feasible solution: The values of decision variables x and y involved in objective, function satisfying all the given condition is known as feasible solution., Maximum or the minimum value of an objective function will always occur at the vertex, of feasible region., Example 3, Find the maximum and the minimum values of the objective function (F) = 4x – y subject, to 2x + 3y ≥ 6, 2x – 3y ≤ 6 and y ≤ 2, Solution: - Here,, The given constraints are, 2x + 3y ≥ 6, 2x – 3y ≤ 6 and y ≤ 2, The objective function (F) = 4x – y, To find: The maximum and the minimum value., The corresponding equation of given constraints are, 2x+3y = 6 …………….. (i), 2x-3y = 6 ……………… (ii), and y = 2 ……………... (iii), From equation (i),, 3y = 6 – 2x, or, y =, , 6−2𝑥, 3, , Table from equation (i), x, 0, 3, 6, y, , 2, , 0, , -2, , Similarly, from equation (ii), 2x = 6+3y, or, x =, , 6+2𝑦, 3, , 73
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Table form equation (ii), x, , 3, , 0, , -3, , y, , 0, , -2, , -4, , The Boundary line (i) passes through the point (0, 2), (3, 0) and (6, -2). Draw a solid, line through these points in graph. Similarly, another boundary line (ii) passes through, the points (3, 0), (0, -2) and (-3, -4). Draw a solid line through these points in graph., Taking common testing point (0, 0) in both inequality 2x + 3y ≥ 6 and 2x – 3y ≤ 6, Now, x = 0, and y = 0 then 2x + 3y ≥ 6 = 2 x 0 + 3 x 0 ≥ 6 = 0 ≥ 6 which is false., So, the graph of 2x + 3y ≥ 6 is the plane region does not contain the origin., Similarly, taking x = 0 and y = 0 in (ii) then 2x – 3y ≤ 6 2 x 0 – 3 x 0 ≤ 6 = 0 < 6, which is true. So the graph of 2x-3y ≤ 6, is the plain region containing the, origin., From equation (iii), y = 2 is a straight, line parallel to x-axis lying at a, distance of 2 units above from x-axis., The graph of y ≤ 2 is the lower half, plane from y = 2 including the, boundary line., , B(6, 2, , C, A, , From the graph, ∆ABC is the feasible, region. The vertices of the feasible, region are A(3, 0), B(6, 2) and C(0, 2), Now, the feasible solution is, , , , Vertices, , Objective function, F = 4x – y, , A (3, 0), B (6, 2), C (0, 2), , F = 4 x 3 – 0 = 12, F = 4 x 6 – 2 = 22, F = 4 x 0 – 2 = –2, , Remarks, , 22 (max), -2 (min), , Maximum value of F = 22 at the vertex B(6, 2) i.e. when x = 6 and y = 2 and the, minimum value of F = -2 at the vertex C(0, 2) i.e. when x = 0 and y = 2, , 74
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Example 4, Maximize and minimize Z = 2x + y under the constraints, x + y ≤ 6, x – y ≤ 4, x ≥ 0, y ≥ 0, Solution: Here,, The given constraints are,, x + y ≤ 6, x-y ≤ 4, x ≥ 0, and y ≥ 0, The objective function (Z) = 2x+y, To find: maximum and minimum value, The corresponding equation of the given constraints are:, x + y = 6 …………… (i), x – y = 4 ……………. (ii), x = 0 ……………… (iii), y = 0 ………………. (iv), From the equation (i) y = 6 – x, Table from equation (i), x, , 0, , 6, , 3, , y, , 6, , 0, , 3, , From equation (ii) y = x – 4, Table from equation (ii), x, , 0, , 4, , 2, , y, , -4, , 0, , -2, , The boundary line (i) passes through (0, 6), (6, 0) and (3, 3). Plot the points and join, them by the solid line. Similarly, the boundary line (ii) passes through (0, -4), (4, 0), and (2, -2). Plot the points and join them by the solid line., Taking common testing point (1, 1) in the given inequalities, Now, x = 1 and y = 1 in x + y < 6, 1 + 1 ≤ 6 = 2 ≤ 6 which is true. So the graph of x + y ≤ 6 in the plane region, containing the testing point (1, 1). Similarly, taking x = 1 and y = 1 in x – y ≤ 4, , 75
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1-1 ≤ 4 = 0 ≤ 4 which is true. So the graph of x – y ≤ 4 in the plane region containing, the origin., From the equation (iii) and (iv) x = 0 and, y = 0 are the y-axis and x-axis, respectively. x ≥ 0 is the right half plane, containing the y-axis and y ≥ 0 is the, upper half plane containing the x-axis., From the graph quadrilateral OABC is, the feasible region. The vertices of the, feasible region are O(0, 0), A(4, 0),, B(5, 1) and C(0, 6)., Hence the feasible solution is, Vertices, , Objective function, Z = 2x + y, , Remarks, , O (0, 0), , Z=2x0+0=0, , 0 (min), , A (4, 0), , Z=2x4+0=8, , B (5, 1), , Z = 2 x 5 + 1 = 11, , C (0, 6), , Z=2x0+6=6, , 11 (Max), , Maximum value of Z = 11 at the vertex B(5, 1) and minimum value of Z = 0 at the, vertex O (0, 0), Example 5, In the given diagram the coordinates of A, B, and C are (2, 0), (6, 0) and (1, 4) respectively., The shaded region inside the ∆ABC is represented by inequalities. Write down the, equations of these inequalities and also calculate the minimum value of 2x+3y from the, values which satisfy all the three inequalities., Solution: Here,, ∆ABC is a feasible region and their coordinates are A(2, 0), B(6, 0) and C(1, 4), The objective function F = 2x + 3y, To find: (i) Equation of inequalities, , (ii) The minimum value, , 76
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For the line BC,, The equation of line BC passing, through B(6, 0) and C(1, 4), , 𝑦 − 𝑦1 =, , 𝑦2 −𝑦1, 𝑥2 −𝑥1, 4−0, , Or, 𝑦 − 0 =, Or, 𝑦 =, , 4, −5, , 1−6, , (𝑥 − 𝑥1 ), (𝑥 − 6), , (𝑥 − 6), , Or, 4𝑥 − 24 = −5𝑦, Or, 4𝑥 + 5𝑦 = 24, Since, the half plane with the, boundary line BC contains origin. So, the inequality of BC is 4x + 5y ≤ 24, Similarly, for AC, The equation of line AC Passing through A(2, 0) and C(1,4) is, 𝑦 −𝑦, , 𝑦 − 𝑦1 = 𝑥2−𝑥1 (𝑥 − 𝑥1 ), 2, , 1, , 4−0, , Or, 𝑦 − 0, , = 1−2 (𝑥 − 2), , Or, 𝑦 − 0, , = −1 (𝑥 − 2), , 4, , Or, 𝑦 − 0 = −4𝑥 + 8, Or, 4𝑥 + 𝑦 = 8, , Since the half plane with the boundary line AC doesn’t contain the origin, the, inequality of AC is 4x + y ≥ 8., The equation of AB means the equation of x-axis is y = 0 and the equation of y-axis, is x = 0., The shaded region lies in the first quadrant only. So the inequality of AB is y > 0 and, inequality of y-axis is x ≥ 0, The inequality equations are, 4x + 5y ≤ 24, 4x + y > 8, x > 0, y > 0, , 77
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Again for the minimum value,, Vertices, , Objective function, (F) = 2x+3y, , Remarks, , A (2, 0), , F= 2 x 2 +3 x 0 = 4, , 4 (Min), , B (6, 0), , F = 2 x 6 + 3 x 0 = 12, , C (1, 4), , F = 2 x 1 + 3 x 4 = 14, , Minimum value of F = 4 at the vertex A(2, 0), Exercise 1.4.1, 1., , a), b), , Define boundary line with an example., In which condition the boundary line is dotted line?, , 2., , a), b), , Define constraints with an example., What do you mean by objective function? Also write an example., , 3., , a), b), , In an objective function (F) = 5x – 2y, one vertex of feasible region is (5, 2), then find the value of objective function., From the given inequality 4x + 3y ≥ 10, Write the boundary line equation., , a), , From the adjoining figure,, , 4., , i) What is called the shaded region, ∆ABC?, ii) Find the equation of AB., c), , Draw the graph of x ≥ 0., , d), , Draw the graph of y ≤ 0., , Y', , 5., , Where does the solution set lies for the inequalities x ≥ 0 and y ≥ 0?, , 6., , a), , Draw the graph of the following inequalities., i) x ≥ y, , ii) x + 2y ≤ 8, , iii) x ≥ -5, , 78, , iv) y ≥ 2x
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b), , 7., , Draw the graph of the following inequalities and shade the common solution, set., i) 2x + 2y ≥ 6 and y ≥ 0, , ii) 2x + y ≥ 6 and x ≥ 2, , iii) x + y ≤ 2 and x ≤ 0, , iv) x – 3y ≤ 6 and y ≤ 3, , a), , Find the maximum and minimum value of the function Z = 3x + 5y for each of, the following feasible region,, , i), , ii), , Y', , 8., , Draw the graphs of the following inequalities and find the feasible region. Also find, the vertices of the feasible region., i) x + y ≤ 3, x≥2, y≤1, , 9., , Y', , ii) x – 2y > 4, 2x + y ≤ 8, y ≥ -1, , iii) 2x + y ≥ 4, 3x + 4y ≤ 12, x ≥ 0, y ≥ 0, , iv) 2y ≥ x – 1, x+y≤4, x ≥ 0, y ≥ 0, , Find the maximum values of the following objective functions with the given, constraints., i) p = 14x + 16y subject to, 3x + 2y ≤ 12, 7x + 5y ≤ 28, x ≥ 0, y ≥ 0, , ii) z = 6x + 10y + 20 subject to, 3x +5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0 and y ≥ 0, , iii) F = 6x + 5y subject to, x+y≤6, x – y ≥ -2, x ≥ 0, y ≥ 0, , iv) Q = 3x + 2y subject to, x+y≥0, x–y≤0, y ≤ 2, x ≥ -1, , 79
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10. Find the minimum values of the following objective functions with the given, constraints., i) 2x + 4y < 8, 3x + y ≤ 3, x ≥ 0, y ≥ 0, objective function (F) = 5x + 4y, , ii) Objective function (M) = x + y subject to, 3x + 4y ≤ 21, 2x + y ≥ 4, x ≥ 0, y ≥ 0, , iii) Objective function (L) = 3x + 5y, subject to 2x + y ≤ 6, x+y≥3, x ≥ 0, y ≥ 0, , iv) Objective function (F) = 6x + 9y, Subject to, 2x + y ≤ 9, y ≥ x, x ≥ 1, , 11. In the adjoining figure, the coordinates, of A, B, C are (-3, 0), (2, 0) and (0, 2), respectively. Find the inequalities, represented by the shaded region and, also calculate the maximum value of 3x, + 4y which satisfy all three inequalities., , Y', , 12. Study and discuss, in the adjoining, figure, find the inequalities which, represent the boundaries of, shaded region as solution set and, find the maximum and minimum, value of Z = 3x + 5y. Prepare a, report and present it in your class, room., , Y', , 80
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1.5 Quadratic equations and graph, 1.5.0 Review, Let us consider, an equation and discuss on it, y = 2x + 3, (i) What are x and y called?, (ii) What are the maximum degree of x and y?, (iii) What has the degree of constant term?, (iv) Write the name of given equation., Can you draw the graph of the given equation? If it is, let’s discuss on its shape., Again, let us consider another equation, x2 + 3x + 2 = 0, (i) Write the degree of the equation., (ii) What is the variable in that equation?, (iii) Can you write the roots of the equation x2 + 3x + 2 = 0?, Discuss above questions in different groups and write the conclusion, 1.5.1 Graph of quadratic function, The general form of quadratic function is y = ax2 + bx + c where, a, b and c are called, coefficient of x2, coefficient of x and constant term respectively. The graph of quadratic, function is called parabola., a), i), , Graph of the quadratic function (y = ax2), y = ax2 where a = 1, to draw the graph, let’s find some values of x and y, x, , 0, , ±1, , ±2, , ±3, , ±4, , y, , 0, , 1, , 4, , 9, , 16, , From the table, plotting the pair of points in a graph and joined them freely, ii), , y = x2 where a = -1, To draw the graph, let’s find some values of x and y, x, , 0, , ±1, , ±2, , ±3, , ±4, , y, , 0, , -1, , -4, , -9, , -16, , Plotting the pair of points in a graph and joined them freely., , 81
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Which is in the form of y = a (x - h)2 + k Where h =, The vertex of parabola = (h, k) = (−, , −b, 2a, , ,k=, , 4ac−b2, 4a, , 2, , b 4ac−b, , 4a, 2a, , ), , When k = 0, the equation of parabola takes the form y = a(x – h)2 where the vertex, = (h, 0)., Conclusion:, i), The graph of y = ax2 + bx + c is called parabola is symmetrical to a line, parallel to y-axis., The equation of line of symmetry is x = −, , ii), , 𝑏, 2𝑎, , Example 1, Draw the graph of y = x2 +2x – 8. Also find the equation of line of symmetry., Solution: here,, P, , The given quadratic equation is, y = x2 +2x -8 …………. (i), Now, comparing equation (i) with y = ax2 + bx +, c we get, a = 1, b = 2 and c = -8, Now, x - coordinate of the vertex of parabola, (x) = −, , 𝑏, 2𝑎, , =, , −(2𝑎), 2𝑥𝑎, , = −1, , y-coordinate of the vertex of parabola, (y) = (-1)2 + 2(-1) – 8, = 1 -2 - 8, = -9, Vertex or turning point of the parabola (h, k) = (-1, -9), , Q, , Table from equation (i), x, Y, , -1, -9, , 0, -8, , 1, -5, , 2, 0, , -2, -8, , 3, 7, , -3, -5, , -4, 0, , -5, 7, , Plotting the pair of points, (-1, -9), (0, -8), (1, -5), (2, 0), (-2, -8), (3, 7), (-3,-5), (-4,0), (-5,7). Join these points freely., The Parabola meets the x-axis at two points (2, 0) and (-4, 0)., Again, PQ is the line of symmetry and its equation is x = -1., Note: For the quadratic function y = ax2 + bx + c the turning point not at the origin., 83
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1.5.2 Graph of a cubic function, The general form of cubic function is defined by y = ax3 + bx2 + cx + d where, a, b, c and d, are constants and a ≠ 0. The simplest form of a, cubic function passing through origin is y = ax3., Different values of ‘a’ will give different curves, passing through the origin with similar nature., Now y = ax3 when a = 1 then discuss on its, graph, y = x3 …………… (i), Table from equation (i), x, , 0, , 1, , 2, , -1, , -2, , y, , 0, , 1, , 8, , -1, , -8, , Plotting the pair of points (0, 0), (1, 1), (2, 8),, (-1, -1), (-2, -8) in a graph and joined them freely., The curve line passes through the points of 1st quadrant, origin and 3rd quadrant points, when coefficient of x3 (a) is positive., Example 2, Draw the graph of y = -x3 where a = -1 also, write the nature of graph., Solution: Here,, The given cubic function is, y = -x3, Table from equation (i), x, y, , 0, 0, , 1, -1, , -1, 1, , 2, -8, , -2, 8, , Plotting the pair of points (0, 0), (1, -1), (-1,, 1), (2, -8) and (-2, 8) in a graph and joined, them freely, Again, the curve line passes through the, points of 2nd quadrant, origin and 4th, quadrant points when coefficient of x3 (a) is, negative, , 84
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Example 3, Draw the graph of y = 2x3, also write the value of ‘a’., Solution: Here,, The given cubic equation is, y = 2x3 ………………….. (i), Table from the equation, x, , 0, , 1, , -1, , 2, , -2, , y, , 0, , 2, , -2, , 16, , -16, , Plotting the pair of points, (0, 0), (1, 2), (-1, -2), (2, 16), (-2, -16) in a graph, and joined them freely, The value of a = 2, Let us discuss on the following equations and, draw the graph of, (i) 𝑦 = −2𝑥 3, (ii) 𝑦 =, , 1, 2, , 𝑥3, 1, , (iii) 𝑦 = − 2 𝑥 3, 1.5.3 Solution of quadratic equation and linear equation, Let us consider two equations:, y = x2 – 5x + 6 and y = 2, Let us discuss on the nature of the graph of both equations and draw the graph, Now, y = x2 – 5x + 6 ………… (i), y = 2 ………………….. (ii), Comparing equation (i) with y = ax2 + bx + c, we get a = 1, b = -5, c = 6, X-coordinate of the vertex of parabola (x) =, , −b, 2a, , =, , (−5), 2x1, , =, , 5, 2, , = 2.5, , Y-coordinate of the vertex of parabola (y), 5 2, , 5, , 2, , 2, , = ( ) − 5. + 6, , =, , 25, 4, , −, , 25, 2, , +6 =, , 25−50+24, 4, 5 −1, ), 4, , The vertex of parabola (h, k) = ( 2.5, - 0.25) = (2 ,, , 85, , =, , −1, 4, , = – 0.25
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Table from equation (i), x, , 2.5, , 0 1 2 3 4 5, , y -0.25 6 2 0 0 2 6, Plotting the pair of points, (2.5, -0.25), (0, 6), (1, 2), (2, 0), (3, 0), (4,, 2), (5, 6) in a graph and joined them, freely, Again, From equation (ii) y = 2, represent a straight line parallel to yaxis which is 2 units above the X-axis., From the graph, the intersection points, of the parabola and the straight lines, are (1, 2) and (4, 2), The solutions are x = 1, y = 2 and x = 4, y = 2, Hence, x = 1, 4 and y = 2, By substitution method, the given equations are, y = x2 – 5x + 6 ………………….. (i), and y = 2 ……………………….. (ii), Substituting y = 2 in equation (i), we get, or, 2 = x2 – 5x + 6, or, x2 – 5x + 6 – 2 = 0, or, x2 – 5x + 4 = 0, or, x2 – (4 + 1)x + 4 = 0, or, x2 – 4x – x + 4 = 0, or, x (x - 4) – 1 (x - 4) = 0, or, (x – 4) (x – 1) = 0, Either, x – 4 = 0, x=4, or, x – 1 = 0, x=1, x = 4, 1 and y = 2, Note: When vertex of parabola (x-coordinate or y coordinate) is in fraction especially their, denominator is 4 then take a scale as 8 small boxes = 1 unit so that it is easy to draw the, graph., , 86
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Example 1, Solve: y = x2 – 3x + 5 and y = 2x + 1 by graphical method and substitution method., Solution: Here,, The given equations are:, y = x2 – 3x + 5 …………………. (i), y = 2x + 1 ………………………. (ii), Comparing equation (i) with y = ax2 +bx + c, we get, a = 1, b = -3, c = 5, Now, x-coordinate of the vertex of parabola (x) =, , −b, 2a, , =−, , (−3), , 3, , = = 1.5, , 2, , 2, , 3 2, , 3, , 2, , 2, , And y-coordinate of the vertex of parabola (y) = ( ) − 3 ( ) +5, 9, , 9, , 9−18+20, , 4, , 2, , 4, , = − +5 =, 3 11, , The vertex of parabola (h, k) = ( ,, 2, , 4, , =, , 11, 4, , = 2.75, , ) = (1.5, 2. 75), , Table from equation (i), x, , 1.5, , 0, , 1, , -1, , 2, , 3, , 4, , y, , 2.75, , 5, , 3, , 9, , 3, , 5, , 9, , Plotting the pair of points, (1.5, 2.75), (0, 5), (1, 3), (-1, 9), (2, 3), (3, 5) (4, 9) in, a graph and joined them freely, Again,, From equation (ii), y = 2x + 1, Table for equation (ii), x, , 0, , -1, , 1, , y, , 1, , -1, , 3, , Plotting the pair of points (0, 1), (-1, -1), (1, 3) in, the same graph and draw the straight line, The intersection points of the parabola and a straight line are (1, 3) and (4, 9), , 87
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The solutions are x = 1, y = 3, and x = 4, y = 9, Hence, (1, 3) and (4, 9) are solution set, By substitution method, The given equations are, y = x2 – 3x + 5 …………………. (i), y = 2x + 1 ………………………. (ii), Substituting y = 2x + 1 from equation (ii) in equation (i), we get, or, 2x + 1 = x2 – 3x + 5, or, x2 - 3x + 5 – 2x -1 = 0, or, x2 – 5x + 4 = 0, or, x2 – (4 + 1)x + 4 = 0, or, x2 – 4x – x + 4 = 0, or, x(x – 4) - 1(x – 4) = 0, or, (x – 4) (x – 1) = 0, Either, x – 4 = 0, x=4, or,, , x–1=0, x=1, , When x = 4 then y = 2 x 4 + 1 = 9, When x = 1 then y = 2 x 1 + 1 = 3, Hence, the solutions are x = 4, y = 9, and x = 1, y = 3, Example 2, Solve the equation graphically: x2 + 2x – 3 = 0., Solution: Here,, The given equation is, x2 + 2x – 3 = 0, or, x2 = 3 – 2x, Let, y = x2 = 3 – 2x, Taking, y = x2 …………………. (i), And y = 3 – 2x ……………… (ii), From equation (i) y = x2 is in the form of y = ax2 where (a = 1) so its turning point (vertex), is always origin (0, 0)., , 88
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Table from equation (i), x, , 0, , 1, , -1, , 2, , -2, , 3, , -3, , y, , 0, , 1, , 1, , 4, , 4, , 9, , 9, , Plot the pair of points (0, 0), (1, 1), (-1, 1), (2, 4), (-2, 4), (3, 9) (-3, 9) in a graph and, joined them freely, Again, from equation (ii), y = 3 – 2x, Table from equation (ii), X, , 0, , 1, , 2, , Y, , 3, , 1, , -1, , Plotting the pair of points (0, 3), (1, 1), (2, -1), in the same graph and draw the straight line., From the graph, the intersection points of, parabola and the straight line are, (1, 1) and (-3, 9), x = 1, -3, , Exercise 1.5.1, 1., , a), b), , Define vertex of parabola., In a quadratic equation ax2 + bx + c = 0, what a, b and c are called?, , 2., , a), , Define line of symmetry in parabolic curve., , b), , Write the equation of line of symmetry in the equation y = x2., , c), , Write the equation of line of symmetry in the equation y = ax2 + bx + c., , 89
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3., , Write the vertex and the equation of line of symmetry of the following graph:, Scale: 5 small boxes = 1 unit, i), , ii), , y, , x’, , y, , y’, , y’, , (iii), , y, , x’, , x, , O, , y’, , 90, , x, , O, , x’, , x, , O
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c. Consider the function 𝑓(𝑥) =, , 𝑥 2 −1, ., 𝑥−1, , Finding the limiting value of f(x) as 𝑥 → 1. Here,, 0, 0, , x = 1 makes f(x) undefined, since𝑓(𝑥) = . Therefore, we try to find the limiting value, of f(x) when x is sufficiently close to 1 but not equal to 1., putting x = 0.99, f(0.99) =, , =, , (0.99)2 −1, 0.99−1, −0.0199, −0.01, , = 1.99, 2, putting x = 1.01, f(1.01) =, , (1.01)2 −1, 1.01−1, , = 2.01, 2, 𝑥 2 −1, =, 𝑥→1 𝑥−1, , So, 𝑙𝑖𝑚, , 2, , 2.1 Investigation of continuity in different sets of numbers, We know, some set of numbers are natural numbers, whole numbers integers, fractions,, rational numbers, irrational numbers and real numbers. Here we shall discuss about, continuity and discontinuity in their order in real number line. In real number line,, numbers are extended in a line from – ∞ to ∞., a. Observe the number line with Natural number below., , It is obvious that we can find natural numbers between 2 and 6. These are 3, 4, 5., Similarly, we can find natural numbers between 10 and 20. Obviously there are 11,, 12, 13, 14, 15, 16, 17, 18, 19. But in number line there is no natural number between, 4 and 5. 4.5 is also the number in number line but not natural number. Same property, exists in case of integers also. There is no continuity in the set of integers. The set of, integers does not hold a property of continuity., b. The set of whole numbers also holds the same property of discontinuity as that of the, set of natural numbers. The set of whole numbers is discontinuous as can be seen, below:, 93
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c. Consider the set of rational numbers Q, which have no common factors (or in lowest, p, form). 𝑄 = {𝑥|𝑥 = q where p and q are integers but q ≠ 0} All the numbers in, number line are not only rational numbers. There are other numbers which are, irrationals. Rational numbers and irrational numbers form the set of real numbers., Graphically:, , Therefore, the set of rational numbers Q holds the property of discontinuity in, increasing or decreasing order., d. Continuity in set of real numbers, The set of real numbers is represented by the points in the real number line. There is, one-one correspondence between the real numbers and the points on the number, line. Between any two real numbers on the line there correspond real numbers, represented by the points between given two points. This establishes the fact that, the set of real numbers is dense. Hence, the real number line is continuous., Exercise 2.1, 1., , 2., , Write the following sets in set builder or tabular form., (a) natural number, , (b) whole number, , (c) integers, , (d) rational numbers, , (e) irrational numbers, , (f) real numbers, , (a), , Show the first five natural number in a number line., , (b) Write the natural numbers from 20 to 30., (c) Write the rational numbers from -5 to 5., (d) Form a set of whole numbers with first ten elements. Present them in a number line., , 3., , Study the following number line and examine the continuity or discontinuity in the, set of numbers., (a), , 94
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(i) Set of natural numbers, , (ii) set of integers, , (iii) set of real numbers, , (b), , 4., , (i) set of integers, , (ii) set of rational numbers, , (iii) set of whole number, , (iv) set of irrational numbers, , Observe the following and find the continuity or discontinuity in their situation:, (a) The movement of frog from one place to other, (b) The movement of crocodile from one place to other, (c) The height of a plant from first Sunday to second Sunday of the same month, (d) The weight of a person, (e) The flow of water in a river, (f) The number of presence of students in your class for a week, , 5., , Collect any three examples of continuity and three examples of discontinuity which, can be shown in scale. Present your findings in classroom, , 2.2 Investigation of continuity and discontinuity in graphs, Let us observe the following graphs for inequalities:, (a), , (b), , What is the inequality?, , What is the inequality?, , (c), , (d), , What is the inequality?, , What is the inequality?, 95
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Here,, (a) The inequality for solution region is -5 < x < 5. -5 and 5 both do not belong to, the solution region. It is written as (-5, 5) = {x: -5 < x < 5}, (b) The inequality for solution region is -5 < x < 5. The end point -5 belongs to the, solution region but 5 does not belong to the region., (c) The inequality for solution region is -5 < x < 5. The end point -5 does not belong, to the region but 5 belongs to the region., (d) The inequality for solution region is -5 and 5 both belong to the region., Again observe the following graphs from -5 to +5., (a), , (b), , (c), , (d), , 96
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If we discuss about continuity and discontinuity of the above function at x = 0, we get the, following results:, (a) The function is defined from -5 to +5 in X-axis. At x = 0 there is no break, jump, gap, or hole. So the function is continuous at x = 0., (b) The function is defined from -5 to +5 in X-axis At x = 0, the curve has no break or no, gap, so the function representing by the curve is continuous at x = 0., (c) The function representing by the curve is defined from x = -5 to x = 5. At x = 0 the, graph has a break so the function representing the curve is discontinuous at x = 0., (d) The function representing by the straight line is defined from x = –5 to x = 5. The, graph has break/gap at x = 0, so it is discontinuous at x = 0., Let y = f(x) be a function defined from x = a to x = b. The function f(x) is said to be, continuous at x = c if its graph has no 'break', 'jump', 'gap' or 'hole' at x = c otherwise, a, function y = f(x) is said to be discontinuous at x = c., Example 1, Draw the graph of y = sinx when -180° < x < 360° and discuss about its continuity at, x = -90° and x = 180°., Solution, To draw the sine graph as in chapter one, list the value of sinx corresponding to x as, following in the difference of 90°. Let 10 small division along horizontal axis represents, 90° and along vertical axis 10 small divisions represent 1 unit. We get (-180° and along, vertical axis small 10 small divisions represent unit. We get (-180°, 0) (-90°, -1), (0°, 0), (90°, 1), (180°, 0), (270°, -1) …., (360°, 0) coordinates to plot on the graphs paper. Join, these coordinates by free hand. We get the graphs as below., , 97
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At x = -90° and x = 180°, the graph has no break and no jump, so the function is, continuous at x = -90° and x = 180°, Example 2, Examine the continuity or discontinuity of the, graph defined from x = -3 to x = 8., , Solution, The function is defined from x = -3 to x = 8, in the graph. The graph has different steps (i.e. 5 steps.), The graphs is discontinuous at x = -1, x = 0, x = 1 and x = 4, But it is piece wise, continuous for -3 < x < -1, -1 < x < 0, o < x < 1, 1 < x < 4 and 4 < x < 8, Exercise 2.2, 1., , From the following graphs, find:, (a) domain of the function, , (b) Point of discontinuity, , (c) one point of continuity for the following graphs of function., (a), , (b), , 98
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(c), , (d), , (e), , 2., , 3., , (f), , Draw graph of the following functions and discuss about continuity at different, points at most 3 points., (a) y = x + 2(-4 < x < 5), , (b) y = x2(-6 < x < 6), , (c) y = x3 (-10 < x < 10), , (d) y = cosx(-180° < x < 360°), , Collect the different daily life examples of continuity and discontinuity and make a, short report., , 99
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2.3 Symbolic representation of continuity, Let us suppose a function 𝑓(𝑥) =, , 𝑥 2 −1, 𝑥−1, , defined for - < x < . Also take different, , situations at x = 2 and x = 1, At x = 2, f(2) =, , 22 −1, 2−1, , =3, , for x < 2, let, x = 1.99, f(1.99) =, , (1.99)2 −1, 1.99−1, , =, , 3.9601−1, =, 0.99, , 2.99 = 3 (nearly), , for x > 2, let x = 2.01(say), f(2.01) =, (2.01)2 −1, 2.01−1, , =, , 4.0401−1, 1.01, , = 3.01 = 3 (App.), , At x = 2, lim– 𝑓(𝑥) = 2.99 = 3 (App.), 𝑥→2, , lim 𝑓(𝑥) = 3.01 = 3 (App.), , 𝑥→2+, , f(2) = 3, So, the function has no jump, hole, break at, x = 2. the function is continuous at x = 2, At x = 1, f(1) =, , (1)2 −1, 1−1, , 0, 0, , = (does not exist), , There is gap at x = 1, But for x < 1 and x > 1 let us take x = 0.99 and x = 1.01, we get f(0.99) = 1.99 = 2 (nearly), and f(1.01) = 2.01 = 2(nearly), lim 𝑓(𝑥) = lim+ 𝑓(𝑥) = 2, , 𝑥→2–, , 𝑥→2, , The function is discontinuous at x = 1, Let y = f(x), be a function defined at x = a, f(x) is said to be continuous at x = a if lim −𝑓(𝑥), 𝑥→2, , = lim+ 𝑓(𝑥) = f(a). lim− 𝑓(𝑥) is read as when x is nearly a from left to the value of f(x) is, 𝑥→𝑎, , 𝑥→𝑎, , nearly approaches to f(a). lim+ +𝑓(𝑥) is read as when x is nearly approaches a from right,, 𝑥→𝑎, , f(x) is nearly approaches f(a)., , 100
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Example 1, 1., , A function f(x) is defined as follows:, f(x) = 3x + 5 for 0 < x < 1, , 9x – 1 for x > 1, , Examine the continuity at x = 1, Solution:, for x = 1, f(x) = 9x – 1, Now, f(1) = 9 x 1 – 1 = 9 – 1 = 8, for x > 1, f(x) = 9x – 1, let us take x = 1.01 (nearly x = 1), f(1.01) = 9 x (1.01) – 1 = 9.09 – 1 = 8.09 = 8 (nearly), since, lim— 𝑓(𝑥) = 𝑓(1) = lim+ 𝑓(𝑥), so the function is continuous at x = 1, 𝑥→1, , 𝑥→1, , Exercise 2.2, 1., , 2., , 3., , a., , Explain lim— 𝑓(𝑥), lim+ 𝑓(𝑥) and f(2) in words., , b., , If f(x) = x + 2, what is f(2)?, , c., , If f(x) = 2x – 1, what is lim— 𝑓(𝑥)? (take x = 0.99), , d., , If f(x) = 3x + 1, What is lim+ 𝑓(𝑥) (take x = 1.01), , 𝑥→2, , 𝑥→2, , 𝑥→2, , 𝑥→2, , If f(x) = 3x + 2,, a., , find f(2.001), f(2.0001), f(1.999), f(1.9999), , b., , find lim— 𝑓(𝑥), lim+ 𝑓(𝑥) 𝑎𝑛𝑑 𝑓(2), , c., , Is it continuous at x = 2?, , 𝑥→2, , If f(x) =, , 𝑥→2, , 𝑥 2 −9, 𝑥−3, , , find:, , a., , f(2.999) and f(3.001), , b., , f(2.999) and f(3.001) equal after approximation, , c., , Discuss about continuity at x = 3, , 101
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4., , Examine the continuity at points mentioned below:, a., , f(x) =, , 2 - x2 , for x < 2,, , at x = 2, , x – 4, for x > 2, x, for x < 0, b., , f(x) =, , 0, for x = 0,, , at x = 0, , x2 , for x > 0, c., , f(x) =, , 2, 5−𝑥, , for x < 3,, , at x = 3, , 5 – x for x > 3, 5., , Take a quadratic function and test its continuity at particular point., , 102
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Unit 3, , Matrix, , 3.0 Review, Observe the following table:, Articles/Shops, , A, , B, , Pen, , Rs. 40, , Rs. 50, , Copy, , Rs. 35, , Rs. 30, , Bag, , Rs. 400, , Rs. 450, , If we interchange the position of shops and arties, what will change in matrix?, Discuss., The above table shows the price of three articles in two shops. The profit from each, unit of pen, copy and bag are Rs 4. Rs. 6 and Rs. 50 respectively., a) Write the information in the above table as a 3x2 matrix A., b) Write a row matrix B that represents the profit, per units of each type of, product., c) Find the product of B and A., d) State what the elements of BA represent?, 40, 50, The product of matrix can be written as (4 6 50) ( 35, 30 ), 400 450, Answer the above questions., 7.1 Determinant of a Matrix., a11, A = [a, , 21, , a12, pen book, 1 2, a b, a22 ] , B = [copy eraser] , C = [c d] , D = [3 8], , i), , What kinds of matrix are these?, , ii), , What is the order of each matrix?, , iii), , Corresponding to each matrix, is there a number?, , Determinant is a function which associates each square matrix with a number. The, determinants of above matrices are denoted by det (A), det (B), det (C), det (D) or, |A|, |B|, |C|, |D| ‘’ is the single notation for determinant of any square matrix., 103
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i.e. If A be a square matrix, its determinant is denoted by det (A) or |A| and is a, number associated to that matrix A., A matrix whose determinant is Zero is said to be singular. Let A = [a11] be a square, matrix of order 1x1 then det (A) is a11, If a11 = -5 then det (A) = -5 ., If a11 = 5, det (A) = 5., [The determinant of 1x1 matrix is simply the element of the matrix.], a11 a12, if, A = [a, ], 21 a 22, a11 a12, The determinant of A is denoted by |A| = |a, | = a11a22 – a21a12, 21 a 22, Product of elements leading diagonal – product of elements of secondary diagonal., 2 1, 1 2, Let us take two matrices A = [, ] and B = [, ], 0 2, 2 4, 2 1, Now, determinant of A = |, | =2x2–1x0=4–0=4, 0 2, 1 2, and determinant of B = |, | =1x4–2x2=4–4=0, 2 4, A is non-singular matrix but B is singular matrix., Example 1, Evaluate, a) |, , 2 3, |, 4 5, , −2 −√5, b) |, |, −√5, 3, , c) |, , 𝑥, 𝑥, , 4, |, 𝑥2, , Solution, 2 3, a) |, | = 2 × 5 − 3 × 4 = 10 – 12 = − 2, 4 5, −2 −√5, | = (−2) × 3— (−√5)(−√5) = −6 − 5 = − 11, −√5, 3, 𝑥 4, c) |, | = 𝑥 2 . 𝑥 – 4𝑥 = 𝑥 3 – 4𝑥, 𝑥 𝑥2, b) |, , Example 2, If A = ( 2, −3, , 1, ) then find |A|, −4, , Solution, Here, A = ( 2, , 1, ), −3 −4, 1, Now, |A| = | 2, | = 2 × (−4) – 1 × (−3) = −8 + 3 = − 5, −3 −4, , 104
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Example 3, Solve for x, |, , 𝑥−1, 𝑥2 + 1, , 𝑥, |=0, 𝑥2 + 𝑥 + 1, , Solution, 𝑥−1, Here, | 2, 𝑥 +1, , 𝑥, |=0, 𝑥2 + 𝑥 + 1, , or, (x – 1) (x2 + x + 1) – x(x2 + 1) = 0, or, x3 – 1 – (x3 + x) = 0, or, - 1 – x = 0, or, -1 = x, x=-1, Example 4, If I is the identity matrix of order 2 x 2 and 𝐴 = (, , 2, 3, , 4, ), find the determinant of 4A + 3I, 4, , Solution: Here,, , 2 4, 1 0, A=(, ) and I = (, ), 3 4, 0 1, 2 4, 1 0, Now, 4A + 3I, = 4(, ) + 3(, ), 3 4, 0 1, 8 16, 3 0, =(, )+(, ), 12 16, 0 3, 8 + 3 16 + 0, 11 16, =(, )=(, ), 12 + 0 16 + 3, 12 19, |4A + 3I| = |11 16|, 12 19, = 11 × 19 − 16 × 12, = 209 − 192, = 17, Exercise 3.1, 1. a) Define determinant of a square matrix., b) What do you mean by singular matrix?, 2. a), , b), , Evaluate, 2 3, i) |, |, 4 5, , 5, 3, ii) |, |, −1 −4, , iii) |√5, √2, , 3 4, If 𝐴 = (, ), then, find |A|, 5 2, , 𝑦2, −√2, | iv) |, 𝑦, √5, −𝑎, c) If 𝐵 = (, 𝑏, , 105, , −2, |, −3, −𝑏, ), then find |B|, 𝑎
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d), , 𝑐𝑜𝑠𝜃, If 𝐴 = (, −𝑠𝑖𝑛𝜃, , 𝑠𝑖𝑛𝜃, ), then find |A|, 𝑐𝑜𝑠𝜃, , e), , 𝑐𝑜𝑠50° 𝑠𝑖𝑛10°, If 𝐴 = (, ), then find |A|, 𝑠𝑖𝑛50° 𝑐𝑜𝑠10°, , 3. Solve for x, 𝑥 5, a) |, | = 24, −4 𝑥, 𝑥+3 4, c) |, |=7, 𝑥−3 5, , 4𝑥 2, b) |, | = 12, 6 3, 𝑥−1 𝑥−2, d) |, |=0, 𝑥, 𝑥−3, , 4. If I is the identity Matrix of order 2 x 2, find the following:, 2 1, a) If A = (, ) , then |2A + 3I|, 1 2, 2 0, b) If A = (, ) , then|A2 + 2A + 4I|, −1 3, 4 2, c) B= (, ) , then |B 2 − 2B + 3I|, 1 −2, m + 2 −3, 5. If A = (, ) and |A| = 3m – 1 then find A2 + 3A, m + 5 −4, 6. Construct a matrix A of order 2 x 2 whose element aij = 3i + 2j and find |3A2+2A|, 2, 7. If 𝐴 = (, 4, , 3, −1 2, ) and 𝐵 = (, ) verify that |AB| = |A| |B|, 5, 2 1, , 3.2 Inverse of a matrix, 1., 2., , 1, , What is the product of a and 𝑎?, 4 −3, −2 3, If 𝐴 = (, ) 𝑎𝑛𝑑 𝐵 = (, ), Find AB and BA. What conclusion do you get, 3 −2, −3 4, from AB and BA? Discuss in the class., , [For an n x n matrix A, if there is a matrix B for which AB = I = BA, then B is the inverse of, A. B is written as A-1 and A is written as B–1, e f, b, ] and A−1 = [, ], g h, d, ae + bg af + bh, b e f, ][, ] =[, ], g, h, ce + dg cf + dh, d, 1 0, We know, AA−1 = [, ], 0 1, a, Let, A = [, c, a, AA−1 = [, c, , 106
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Let A and B are two non-singular matrices, then, i) A . A-1 = A-1 . A = I, Example 1, If A = (, , ii) (AB)-1 = B–1 . A-1, , iii) (A-1) = A, , iv), , (AT)–1, , 2 3, ) , does A−1 exist?, 4 5, , Solution: Here,, 2 3, ), 4 5, 2 3, |A|=|, | = 2 x 5 – 3 x 4 = 10 – 12 = - 2, 4 5, 𝐴=(, , Since |A|≠0, so, A-1 exists, Example 2, 𝑦−3, For what value of y, the matrix (, 𝑦, solution: Here,, , 𝑦−1, ) does not have its inverse?, 2(𝑦 − 1), , y−3, y−1, Let A = (, ), y, 2(y − 1), y−3, y−1, |A| = |, | = 2(y - 1) (y - 3) –y (y - 1), y, 2(y − 1), = 2 (y2 – y -3y + 3) – (y2 – y) = 2y2 – 8y + 6 – y2 + y = y2 + 7y + 6, -1, if A does not exist, then |A| = 0, or, y2 – 7y + 6 = 0, or, y2 – 6y – y +6 = 0, or, y(y – 6) -1 (y – 6) = 0, or, (y – 6) (y – 1) =0, either, y- 6 = 0 or y – 1 = 0, either, y = 6 or y = 1, 𝑦−3, 𝑦−1, for y = 6 or y = 1, the matrix (, ) does not have its inverse., 𝑦, 2(𝑦 − 1), , 108, , =, , (A–1)T
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1, 2, , by 3 × equation (iii) – equation (iv), 𝑎12 + 6𝑎22 = 0, 1, 𝑎12 + 3𝑎22 =, – –, – 2, 3𝑎22 = −, 𝑜𝑟, 𝑎22 = −, , 1, 6, , 1, 2, , by 3 × equation (iii) – equation (iv) we get,, 𝑎12 + 6𝑎22 = 0, 2𝑎12 + 6𝑎22 = 1, , –, , –a12, , –, , –, , = –1, , a12 = 1, 𝑎11, Hence, A−1 = (𝑎, 21, , −3 1, 𝑎12, 1, ), =, (, 𝑎22, 1 − ), 6, , Alternatively:, 1, |𝐴| = |3, 2, , 2| = 1 × 6 − 2 × 2 = −2, 3, 6, , A-1 exists, 1, 𝐴 = (3 2), 2 6, Interchanging the elements of leading diagonal and changing sign of secondary, 6, , diagonal, we get. adj. of A = (−2, , −2, 1, , ), , 3, , 1, , Now, 𝐴−1 = |𝐴| (adj. of A), 6 −2, 6 −2, 1, 1, 1 )=, 1), (, (, |𝐴| −2, (−2) −2, 3, 3, −1, −1, −3 1, ( )×6, ( ) × (−2), 2, 2, −1), =(, )=(, −1, −1, 1, 1, 6, ( ) × (−2), ( )×, 2, 2, 3, 𝐴−1 =, , 110
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Example 4, 2 3, 3 1, If𝐴 = (, ) and B = (, ) find A-1 and B-1. Verify that (AB)–1 = B–1.A–1, 4 5, 4 5, Solution: Here,, 2 3 |A|, 2, A=(, ),, =|, 4 5, 4, , 3, | = 10 − 12 = −2 0, 5, , So A-1 exists, 3 1, ), 4 5, 3 1, |B| = |, | = 15 − 4 = 11 0, 4 5, B=(, , So, B-1 exists, Now,, 1 5 −1, (, ), 11 −4 3, 1 5 −3, A−1 =, (, ), −2 −4 2, 1, 5 −1, 5 −3, B −1 A−1 =, (, )(, ), −4 2, −22 −4 3, 1 5 × 5 + (−1) × (−4) 5 × (−3) + (−1) × 2, =, (, ), −22 (−4) × 5 + 3 × (−4) (−4) × (−3) + 3 × 2, B −1 =, , 1, 25 + 4, −15 − 2, (, ), −22 −20 − 12 12 + 6, 1 29 −17, =− (, ), 22 −32 18, 2 3 3 1, AB = (, )(, ), 4 5 4 5, 2×3+3×4 2×1+3×5, =(, ), 4×3+5×4 4×1+5×5, 6 + 12 2 + 15, =(, ), 12 + 20 4 + 25, 18 17, =(, ), 32 29, 18 17, |AB| = |, |, 32 29, =, , = 18 × 29 − 17 × 32, = 522 − 544, , 111
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= −22, (AB)−1 =, , 1, 29, (, |AB| −32, , −17, ), 18, , 29, −32, , −17, ), 18, , =, , 1, −22, , (, , (AB)−1 = B −1 A−1 proved., , Exercise 3.2, 1. a) Define inverse of a matrix., , 𝑎11, b) Under which condition does A-1 exist for 𝐴 = (𝑎, , 21, , 2., , 𝑎12, 𝑎22 )?, , -1, , Does A exist? Give reason., a) A = (−8 6), , b) 𝐴 = (tanA, −4 3, SecA, −a −b, c) A = (, ) ; a ≠ 0, b ≠ 0, b −a, , SecA, ), tanA, , 3. Find the inverse of each of the following matrices, if exist., a) A = [−2 3], −3, , 4, , b) B = [ 4, , −3, , −7, ], 2, , c) C = [−2 −5], −3, , −8, , 2 3, 4. a) If P = (1 0) and 𝑄 = [, ] then find (PQ)-1, 0 1, 1 2, b) If M = [0 1] 𝑎𝑛𝑑 N = [2 3] find (NM)-1, 1, , 0, , 1, , 2, , 2 3, ] then find 𝐴2 + 𝐴𝐴−1 + 2𝐼, where I is the 2x2 identify matrix., 4 7, 2 6, 2 1, 6. (a) If A = (, ) and B = (, ) show that (AB)–1 = B–1A–1., 3 10, 5 3, 5. If 𝐴 = [, , (b) If A = (, , 2, 5, , 3, 7, ) and B = (, 6, 14, , 3, ) verify that (AB)–1 = B–1A–1., 6, , 3.3 Solutions of system of linear equations by using matrix method, Let us take a matrix equation:, 2 −3 𝑥, 1, (, ) (𝑦) = ( ), 1 1, 2, When we multiply,, 2x – 3y = 1 …………… (i), , 112
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x + y = 2 ………………. (ii), equation (i) and (ii) are simultaneous equation in x and y, The above equations can be written as AX = B, x, Where A = (2 −3) , B = (1) and X = (y), 1, , 1, , 2, , We have,, AX = B, A-1(AX) = A-1B [Multiplying by A-1 on both sides], or, (A-1A)X = A-1B, [By associative law], -1, or, IX = A B, [A-1A = I], or, X = A-1B, [IX = X], Since, matrix multiplication is not commutative (AB ≠ BA) in general, case must be taken, to multiply on the left by A-1, Example 1, Solve the given system of equations by matrix method: x = 2, x + y = 7, Solution, Here, x + 0.y = 2……. (i), x + y = 7 ………….. (ii), Writing equation (i) and (ii) in matrix form, we have,, , 1 0 𝑥, 2, )( ) = ( ), 1 1 𝑦, 7, AX=B, or, X = A–1B, 1, 1, Now, A−1 = |A| ( 1 0) = 1 ( 1 0) = ( 1 0), −1 1, −1 1, −1 1, So, X = 𝐴−1 B, 1×2+0×7, 2+0, 1 0 2, 2, =(, )( ) = (, )=(, )=( ), (−1) × 2 + 1 × 7, −2 + 7, −1 1 7, 5, 𝑥, 2, (𝑦 ) = ( ), 5, (, , Example 2, Solve the following equation by using matrix method 2x + 5 = 4 (y + 1) – 1; 3x + 4 =, 5 (y + 1) – 3, Solution: Here,, 2x + 5 = 4(y + 1) – 1, or, 2x + 5 = 4y + 4 – 1, or, 2x – 4y = -2 ………….. (i), Again, 3x + 4 = 5(y + 1) – 3, or, 3x + 4 = 5y +5 – 3, , 113
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or, 3x – 5y = -2 ………… (ii), Writing equation (i) and (ii) in matrix form, we have, , 2 −4 𝑥, −2, )( ) = ( ), 3 −5 𝑦, −2, or, Ax = B, 𝑥, Let, 𝐴 = (2 −4) , X = (𝑦) 𝐵 = (−2), 3 −5, −2, or, x = A-1B, 2 −4, |𝐴| = |, | = (-5) x 2 – (-4) x 3 = -10 + 12 = 2 (A-1 Exists), 3 −5, 1 −5 4, so, 𝐴−1 = (, ), 2 −3 2, Now, x = A–1B, 1 −5 4 −2, 1 (5) × (−2) + 4 × (−2), 𝑥= (, )( ) = (, ), 2 −3 2 −2, 2 (−3) × (−2) + 2 × (−2), 1 10 − 8, 1 2, 𝑥, (𝑦 ) = (, )= ( ), 2 6−4, 2 2, 1, ×2, 𝑥, 1, ( 𝑦 ) = (2, )=( ), 1, 1, ×2, 2, 𝑥, 1, (𝑦) = ( ), 1, (, , Comparing corresponding components of equal matrices, we get x = 1 and y = 1, Example 3, Solve:, , 2𝑥+4, 5, , =𝑦=, , 40−3𝑥, 4, , Solution: Here,, 𝑦=, , 2𝑥+4, 5, , 2, 4, or, 𝑦 = 𝑥 +, 5, 5, 2, 4, or, 𝑥 − 𝑦 = − …….. (i), 5, 5, 40 − 3𝑥, 𝑦=, 4, 40 3, or, 𝑦 =, − 𝑥, 4 4, 3, or, 𝑥 + 𝑦 = 10 ……….(ii), 4, , writing equation (i) and (ii) in matrix form we have,, 2, −4, −1 𝑥, (5, ) (𝑦 ) = ( 5 ), 3, 10, 1, 4, Let, AX = B, , 114
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X = A-1B ……………. (iii), for A-1, 2, |𝐴| = |5, 3, 4, , −1, |=, 1, , 𝐴−1 = 1 ÷, , 2, 3 2 3 8 + 15 23, × (1) + = + =, =, 5, 4 5 4, 20, 20, , 23 1, ( 3, 20 −, 4, , 1, 20 1, 2) =, ( 3, 23 −, 5, 4, , 1, 2), 5, , so, from equation (iii), X = A-1B, 20 1 1 − 4, ( 3 2) ( 5), 23 −, 4 5, 10, 4, 1 × (− ) + 1 × 10, 20, 5, = ( −3, ), −4, 2, 23, ( ) × ( ) + × 10, , 𝑜𝑟, X =, , 4, , =, , 5, , −4+50, 20, 5, ( 3+20, ), 23, 5, , 𝑥, 8, (𝑦) = ( ), 4, , 5, , =, , 46, 20 5, ( ), 23 23, 5, , =, , 20, ×, (23, 20, ×, 23, , 46, 5, 23), 5, , x = 8 and y = 4, Example 4, The sum of two numbers is 20 and their difference is 4., a) Write the equation in matrix form., b) Solve them by matrix method., Solution, Let the numbers be x and y (x>y), by question: x + y = 20 ………. (i), x – y = 4 ……… (ii), Writing equation (i) and (ii) in matrix form, 𝑥, 1 1, 20, (, ) (𝑦 ) = ( ), 1 −1, 4, Let, AX = B, or, X = A-1B ……… (iii), 1 1, For A-1 : |𝐴| = |, |, 1 −1, = 1 × (−1) − 1 × 1 = −1 − 1 = −2 0 (A-1 exists), Now,, 1 −1 −1, 𝐴−1 =, (, ), −2 −1 1, , 115, , 8, =( ), 4
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Now, X = A−1 B, 1 −1 −1 20, 1 (−1) × 20 + (−1) × 4, 𝑥, (𝑦 ) =, (, )( ) =, (, ), −1, 1, 4, −2, −2 (−1) × 20 + 1 × 4, 1 −20 − 4, 1 −24, =, (, )=, (, ), −2 −20 + 4, −2 −16, 1, (− ) × (−24), =( 2, ), 1, (− ) × (−16), 2, 𝑥, 12, (𝑦 ) = ( ), 8, The required numbers are 12 and 8, , Exercise 3.3, 1. a) If AA-1 = A-1A = I and AX = B, Write X in terms of A-1 and B., b) Write the matrix form of, 2x + y = 4, 3x + 2y = 7, 2. Solve the following system of linear equations using matrix method:, (a) x + y = 5, x – y = 1, (b) 3x + 5y = 3, 4x + 3y = 4, (c), , (e) (, (g), , =3, , (d) ( 2, , 2, 5 𝑥, )( ) = ( ), 1, 4 𝑦, , (f) +, , 3𝑥+5𝑦, 8, , 8, 7, , 5, 𝑦, , 1, , =, , 5𝑥−2𝑦, 3, , = 𝑥 − 1,, , 5, 𝑦, , −3, , 1, 𝑥, , 1, 2𝑦, , 1 𝑥, 1, )( ) = ( ), 1 𝑦, 2, , = 8,, , 1, 2𝑥, , 1, , −𝑦+1=0, , 2, , =𝑥−4, , h) 4(x – 1) + 5(y + 2) = 10, 5(x – 1) –3 (y + 2) +6 = 0, (i) 5x + 7y = 31xy, (j) 7x +5y = 29xy, , 3. Write equations and solve the following by matrix method:, a), , The cost of a pen and copy is 120. The cost of 2 copy is Rs. 100., , b), , The sum of present ages of a father and Son is 53 years. After 2 years, the age, of father will be 47 years., , 4. Write yourself two simultaneous equations in (x, y) related to your daily life and, solve them by matrix method., , 116
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The method of solving system of linear equations by using determinant is known a, Cramer's rule., Note:, (i) Grabriel Cramer is a Swiss mathematician who introduced technique to solve a, system of linear equations using determinant., (ii) If D = 0, Dx ≠ 0, Dy ≠ 0, the system of equations has no solutions, because 0x ≠ 0,, 0y ≠ 0., (iii) We always write the equation in the form 𝑎1 𝑥 + b1 y = c and 𝑎2 𝑥 + 𝑏2 𝑦 = 𝑐2, c1, i.e. for finding Dx(D1), substitute first column of D by (c ) and for finding Dy (D2),, 2, c1, substitute the second column by (c )., 2, Example 1, Solve 2(x -1) = y and 3 (x – 1) = 4y using Cramer’s rule, Solution, Here, 2(x – 1) = y, or, 2x – y = 2 …………. (i), 3(x – 1) = 4y, or, 3x – 4y = 3 ………… (ii), Now, arranging the coefficients and constant term, Coefficient of x, , Coefficient of y, , Constant term, , 2, , -1, , 2, , 3, , -4, , 3, , 2 −1, Now, 𝐷 = |, | = 2 x (-4) – 3 x (-1) = - 8 + 3 = - 5, 3 −4, 2 −1, 𝐷𝑥 = |, | = −5, 3 −4, 2 2, 𝐷𝑦 = |, |=2×3−2×3=0, 3 3, 𝐷, −5, 𝑥 = 𝐷𝑥 = −5 = 1, 𝑦 =, , 𝐷𝑦, 𝐷, , 0, , = −5 = 0, , (𝑥, 𝑦) = (1,0), , 118
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Example 2, Solve the following system of linear equations using Cramer’s rule., 2, 1, 𝑥+𝑦 =1 𝑥+𝑦 =1, 3, 2, Solution, Coefficient of x, Coefficient of y, 2, 1, 3, 1, 1, 2, 2, 1, 𝐷 = |3, |, 1, 1, 2, 2, 1 4−3 1, = ×1−1× =, = (≠ 0), 3, 2, 6, 6, 1 1, 𝐷𝑥 = |1, |, 1, 2, 1, 1 1, =1×1−1× =1− =, 2, 2 2, 2, 1, 𝐷𝑦 = |3, |, 1 1, 2 2, 2 1, 1 1 1 2 − 3 −1, = × −1× = − =, =, 3 2, 2 3 2, 6, 6, 1, 𝐷𝑋 2 6, 𝑁𝑜𝑤, 𝑥 =, = = =3, 1 2, 𝐷, 6, −1, 𝐷𝑦, 𝑥=, = 6 = −1, 1, 𝐷, 6, (x, y) = (3, 1), , 119, , Constant term, 1, 1, 2
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Unit 4, , Coordinate Geometry, , 4.0 Review, In a class four friends are seated at, the points A, B, C and D as shown in, graph. Sarita and Bimal walk into the, class and after observing for a few, minute Sarita asked following, questions to Bimal., , 11, , Discuss about these answers in, group and write conclusions., , 6, , 1. Find the equation of AB, BC, CD, and AD, , 4, , 10, 9, 8, , D, , 7, , 5, B, , 3, , C, , 2. Find the slope of each of the, 2, lines AB, BC, CD and AD., 1, , A, The slope of a line parallel to X-axis is, 0. The slope of Y-axis is not defined. The, 1 2 3 4 5 6 7 8 9 10 11, slope is independent of the sense of the line segment (i.e. slope of AB = slope of BA). The, equation of a straight line is the relation between x and y, which satisfies the co-ordinates, of each and every point on the line and not by those of any other point. The equation of, a line with slope m and passing through a point (x1, y1) is given by, , y – y1 = m(x – x1) and the equation of the line through the two given points (x1, y1) and, 𝑦 −𝑦, (x2, y2) is given by y - y1 = 2 1 (x-x1)., 𝑥2 −𝑥1, , 4.1 Angle between two straight lines, Let AB and CD be two straight lines with inclination θ2 and θ1 respectively. Answer the, following questions:, i., , What is the slope (m1) of CD in terms of θ1?, , ii., , What is the slope (m2) of AB in terms of θ2?, , iii., , What is the relation between θ, θ1, θ2?, , As we know θ1 and θ2 are angles made by the lines with, positive direction, so, , 121
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Exercise: 4.1, 1. If θ be the angle between two straight lines with slopes m1 and m2 then, a. Write the expression for tanθ., b. Write the relation between m1 and m2 when θ = 0° or 180°., c. Write the relation between m1 and m2 when θ = 90°., 2. If be the angle between the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0,, a. Express tan in terms of a1, a2, b1 and b2., b., , Identify the relation when both lines are parallel., , c. Identify the relation when both of the lines are perpendicular to each-other., 3. Find the acute angle between the lines given as below:, a. 2x – y + 7 = 0 and x – 3y + 6 = 0, b. 3x – 2y – 5 = 0 and 4x + y – 7 = 0, c. √3x – y + 6 = 0 and y + 3 = 0, 4. Find the obtuse angle between the lines given as below:, a. x –2y + 1 = 0 and x + 3y – 2 = 0, b. x – √3y – 5 = 0 and √3x – y – 6 = 0, c. x – 5y – 3 = 0 and x – 3y – 4 = 0, 5., , 6., , a., , Show that the straight lines (i) 5x + 12y + 13 = 0 and 12x – 5y – 18 = 0, (ii) 27x – 18y + 25 = 0 and 2x + 3y + 7 = 0 are perpendicular to each-other., , b., , Show that the straight lines (i) x – 2y + 3 = 0 and 2x – 4y +9 = 0 (ii) x + 2y – 9 = 0, and 2x + 4y + 5 = 0 are parallel to each other., , c., , If the lines 3y + kx – 8 = 0 and 2x – 3y – 11 = 0 are parallel to each-other, find the, value of k., , d., , If the lines ax – y – 7 = 0 and 3y + x – 9 = 0 are perpendicular to each-other, find, the value of a., , e., , If the lines ax – 5y – 2 = 0 and 4x – 2y + 3 = 0 are perpendicular, find the value of, a., , a., , Find the equation of the line passes through (1, 2) and parallel to 3x – 4y – 12=0., , 129
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b., , Find the equation of the line passes through (–3, –2) and parallel to 3x + 4y —, 11 = 0, , 7., , 8., , 9., , c., , Find the equation of the line passes through the point (2, 3) and perpendicular, to 5x – 4y + 3 = 0, , d., , Find the equation of the line passes through the point (-1, 3) and perpendicular, to the line 5x + 7y + 18 = 0., , a., , Find the equations of the lines passes through the point (2, 3) and make angle, 45° with x – 3y – 2 = 0, , b., , Find the equations of straight lines passing through the point (1,-4) and inclined, at 45° to the straight line 2x + 3y + 5 = 0, , c., , Find the equations of straight line passing through the point of intersection of, 2x – 3y + 1 = 0 and x + y – 2 = 0 and make 45° with x +2y – 5 = 0, , a., , Find the equation of perpendicular bisector of the line segment joining the, points (2, 3) and (4, -1)., , b., , Find the equation of perpendicular bisectors of the triangle having vertices, A(8, -10), B(7, -3) and C(0, -4)., , c., , Two opposite corners of a square are (3, 2) and (3, 6). Find the equations of, diagonals of square., , Draw a rectangle in graph paper. Find the equations of diagonals. Also find the angle, between each diagonal and sides of rectangle. Present your result in classroom., , 10. Find out a parallelogram shape in your surroundings, trace it and find the, co-ordinates of its vertices. Find the angle between their diagonals., , 4.2 Equation of pair of straight lines, y, , Consider the equation x – y = 0 and x + y = 0. Multiply both, of the equation. Find the power of x and y., Both of the equations x – y = 0 and x + y = 0 pass through the, origin. The combined equation of the lines x – y = 0 and, x + y = 0 is x2 – y2 = 0. The equation x2 – y2 = 0 represents a, pair of straight lines passing through the origin. The, equation x2 – y2 = 0 contains only second degree terms., , x-y=0, , x+y=0, , x’, , O, , x, , y’, , 130
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d., , Take examples of two straight lines passes through origin, multiply and get, single equation. Also find the angle between them. Share the solution with, friends., , e., , Discuss about the shape of homogenous pair of second degree equation in x, and y., , 4.3 Conic sections, What plane shapes may you get if you cut a carrot by a knife with different angles?, Right circular cone:, Let AB and CD be two fixed lines intersecting at the point, O at an angle (0° < θ < 90°). If the plane containing these, two lines AB and CD is rotated about the line CD, then the, surface generated by AB is called right circular cone. O is, vertex, CD is vertical axis, θ is vertical angle and AB is, generator of the cone., Take the solid cone and cut them by sharp knife as shown, below., (i) Cut the cone parallel to base. The section in between cone and plane is said to be, circle., , (ii) Cut the cone parallel to slant edge of the cone. The section in between cone and plane, is said to be parabola., , (iii) Cut the cone in such a way that the angle between plane and axis of cone is greater, than vertical angle of cone. But less than 90°. the section in between cone and the, plane is said to be ellipse., , 138
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ellipse, , (iv) Intersect double circular cone parallel to edge of the cone (slant edge). The pair of parabola, or the sections formed between double cone and plane is said to be hyperbola., , hyperbola, , Exercise: 4.3, 1. Name the different parts of cone indicated by a, b, c, d and e., , 2. Name the conic section from following figures., , a), , b), , 139
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d), , c), , 3. Take a solid circular double cone of either local made or from market. Cut it in, different positions and name the conic section obtained., 4.4 Circle, , A figure formed by the intersection of a plane and a, circular cone is conic section. As in figure when we, intersect circular cone parallel to the base we get a, conic section, which is known as circle., , Take a thumb pin and rubber band. Fix the thumb pin at the, position O and rotate the rubber band equal distance. The, positions A, B, C and D are obtained. ABCD is circle and O is, centre of the circle. Circle is the locus of a point which moves, C, so that its distance from a fixed point is always constant., , B, , o, , The fixed point is called the centre and the constant distance, is the radius of the circle., a. Equation of circle with centre O (0, 0) and radius ‘r’, units, Let any point P(x, y) on the circumference of circle., Now, OP = r, OP 2 = r2, or, (x − 0)2 + (y − 0)2 = r 2, or, x 2 + y 2 = r 2, Hence, the required equation of circle is 𝑥 2 + 𝑦 2 = 𝑟 2 ., , 140, , D, , A
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or, (h − 1)2 + (k − 1)2 = (h − 4)2 + (k − 4)2, or, h2 − 2h + 1 + k 2 − 2k + 1 = h2 − 8h + 16 + k 2 − 8k + 16, or, 6h + 6k = 30, or, h + k = 5, or, h = 5 − k … … … … . (i), Again, MA2 = MC 2, or, (h − 1)2 + (k − 1)2, = (h − 5)2 + (k − 1)2, or, h2 − 2h + 1 = h2 − 10h + 25, or, 8h = 24, or, h = 3, From equation (i) 3 = 5 − k, or, k = 2, ∴ Centre of the circle (h, k) = (3, 2), MA2 = (3 − 1)2 + (2 − 1)2, =4+1=5, Radius of circle = √5 units, Equation of circle is (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2, 𝑜𝑟,, 𝑜𝑟,, , (𝑥 − 3)2 + (𝑦 − 2)2 = 5, 𝑥 2 − 6𝑥 + 9 + 𝑦 2 − 4𝑦 + 4 = 5, , 𝑜𝑟, 𝑥 2 + 𝑦 2 − 6𝑥 − 4𝑦 + 8 = 0 is the required equation of circle., Note: If circle passes through four points, find the equation of the circle from any three, points and satisfy the equation by fourth point. Four points are said to be con-cyclic., Example 6, Find the equation of such a circle that touches both the axes in the first quadrant at, distance of 4 units from the origin., Y, Solution: Here,, Let (h, k) be the centre of the circle. Since it touches both the, axes, so (h, k) = (4, 4), Radius = 4 units, , 4, , Now, Equation of circle is, , O, , 144, , X, , 4
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c., 2., , 3., , (x1, y1) and (x2, y2) are the end points of a diameter of a circle. What is the, equation of circle in terms of (x1, y1) and (x2, y2)?, , Find the centre and radius of the circle from the following equations of the circle:, a., , 𝑥2 + 𝑦2 = 9, , b., , (𝑥 − 2)2 + 𝑦 2 = 25, , c., , (𝑥 + 1)2 − (𝑦 − 3)2 = 16, , d., , 𝑥 2 − 2𝑥 + 𝑦 2 − 5 = 0, , e., , 𝑥 2 + 𝑦 2 − 4𝑥 − 6𝑦 − 23 = 0, , f., , 𝑥 2 + 𝑦 2 − 8𝑥 + 2𝑦 − 24 = 0, , Find the equation of circle in each case:, a. Centre at origin, radius 5 units, 1, 2, , b. Centre at origin, radius 1 units, c. Centre at (3, 4), radius 3 units, d. Centre at (0, -4), radius 4 units, e. Centre at (1, 2), radius 5 units, f., 4., , Centre at (-1, -3), radius 6 units, , Find the centre, radius and equation of circle, if the end points of a diameter are given, as:, a. (2, 0) and (0, 6), 7 3, , b. (-1, 2) and (7, -4), , 1 1, , c. (3 , 2) 𝑎𝑛𝑑 (− 2 , 2), 5., , 6., , 7., , a., , Find the equation of circle passing through the points (1, 2), (3, 1) and (-3,-1)., , b., , Find the equation of circle passing through the points (2, -2), (6, 6) and (5, 7)., , c., , If (3, 3), (6, 4), (7, 1) and (4, 6) are the vertices of a cyclic quadrilateral, find the, equation of circle., , d., , Prove that the points (1, 0), (2, -7), (8, 1) and (9, -6) are con-cyclic., , Find the equation of circles in each of the following cases:, a., , Touches both axes, radius 5 units and centre in first quadrant., , b., , Touches x-axis, centre at (-3, 4), , c., , Touches y-axis, centre at (-4, -5), , Draw a circle of diameter PQ, where P (8, 1) and Q (8, 11). Take a point R (3, 6) on the, circle., a., , Calculate the distance PQ, PR and QR., 146
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8., , b., , Is (PQ)2 = (PR)2 + (QR)2 ?, , c., , What conclusion do you get?, , d., , Also find the equation of the circles., , The Vertices of a triangle are A (0, 1), B (-2, 0) and C (1, 0)., a. Find the equation of sides of the triangle., b. Find the equation of the circle drawn on that sides of the triangle lying in the first, quadrant as diameter., , 9., , Find the equation of the circle passing through the points (5,3) and (-2, 2) and the, centre lies on the line 𝑥 + 3𝑦 + 1 = 0, , 10. Make a circular piece of paper. Trace this circle in a graph paper. Find the centre,, radius and equation of the circle. Present your findings in classroom., , 147
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Unit 5, , Trigonometry, , 5.0 Review, If A and B are two different angles, what are their compound angles? How can we find, their Trigonometric ratios of Sine, Cosine, tangent and cotangent? Discuss in a group and, compare with the following results., sin(A + B) = sinA cosB + cosA sinB, , sin(A − B) = sinA cosB − cosA sinB, , cos(A + B) = cosA cosB − sinAsinB, , cos(A − B) = cosA cosB + sinA sinB, , tan(A + B) =, , tanA + tanB, 1 − tanA tanB, , tan(A − B) =, , tan A − tanB, 1 + tanA tanB, , cot(A + B) =, , cotA cotB − 1, cotB + cotA, , cot(A − B) =, , cotA cotB + 1, cotB − cotA, , The Trigonometric value of standard angles., 0°, sin, , 0, , cos, , 1, , tan, , 0, , cosec, , Undefined, , sec, , 1, , cot, , Undefined, , 30°, , 45°, , 60°, , 1, 2, √3, 2, 1, , 1, , √2, , √3, 2, 1, 2, , 1, , √3, , √2, 1, , √3, 2, , √2, , 2, , √2, , √3, , 1, , √3, , 2, √3, 2, 1, √3, , 90°, 1, 0, Undefined, 1, Undefined, 0, , 5.1.1 Trigonometric ratios of multiple angles, What are the multiple angles of an angle A? Discuss in groups., a. Trigonometric ratios of multiple angle 2A, With the help of compound angle, we can find the trigonometric ratios of Sine, Cosine,, tangent and cotangent of the angle 2A:, 148
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Example 1, If sinA =, , 5, then find the value of sin2A, cos2A and tan2A., 13, , Solution: Here,, sinA =, , 5, 13, , We know that,, cosA = √1 − sin2 A = √1 − (, , 5 2, 25, 144 12, ) = √1 −, =√, =, 13, 169, 169 13, , Now, by formula, i) sin2A = 2sinA cosA = 2 ×, , 5 12 120, ×, =, 13 13 169, , 5 2, 2 25 169 − 50 119, ii) cos2A = 1 − 2sin2 A = 1 − 2 ( ) = 1 −, =, =, 13, 169, 169, 169, 120, sin2A 169 120, iii) tan2A =, =, =, cos2A 119 119, 169, Example 2, If sinθ =, , 1, then find the value of sin3θ, cos3θ and tan3θ, 2, , Solution: Here, sinθ =, , 1, 2, , By formula,, sin3θ = 3sinθ − 4sin3 θ, 1, 1 3 3, 1 3 1, =3× −4×( ) = −4× = −, 2, 2, 2, 8 2 2, ∴ sin3θ = 1, 1 2, 1 √3, Cosθ = √1 − ( ) = √1 − =, 2, 4, 2, , 154
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c), , 𝜋𝑐 𝜃, − ), 4 4, 𝜋𝑐 𝜃, 2, 1+tan ( − ), 4 4, , 1−tan2(, , 𝜋𝑐, , 𝜃, , d) cos 2 (, , = sin 2, , 4, , 𝜋𝑐, , 𝜃, , 𝜃, , 𝜃, , − 4) − sin2 ( 4 − 4) = sin 2, , 9. Prove that:, a) (cos A − cos B)2 + (sin A − sin B)2 = 4 sin2, , A−B, 2, , b) (cos A + cos B)2 + (sin A + sin B)2 = 4 cos2, , A−B, 2, , 10. Prove that:, 𝐴, 2, , a) tan (45° − ) = √, , 1−sin 𝐴, 1+sin 𝐴, , =, , cos 𝐴, 1+sin 𝐴, , 𝐴, , b) tan (45° + 2 ) = sec 𝐴 + tan 𝐴, 𝜋𝑐, , 𝜋𝑐, , 𝜃, , 𝜃, , c) tan ( 4 + 2 ) + tan ( 4 − 2 ) = 2 sec 𝜃, 𝜃, , 𝜃, , 2 cos 𝜃, , d) cot (2 + 45°) − tan (2 − 45°) = 1+sin 𝜃, 11. Prove that:, a), , sin 2𝐴, 1+cos 2𝐴, , . 1+cos 𝐴 = tan 2, , b), , 1+cos 2𝜃, sin 2𝜃, , ., , c) (1 + sin, , cos 𝐴, , 1+cos 𝜃, cos 𝜃, , 𝐴, , 𝜃, , = cot 2, , 𝜋𝑐, 3𝜋𝑐, 5𝜋𝑐, )(1 + sin, )(1 − sin, )(1 −, 8, 8, 8, , 𝜃, , 𝜃, , 𝜃, , 1, , sin, , 7𝜋𝑐, ), 8, , =, , 1, 8, , 𝜃, , d) (cos6 4 − sin6 4 ) = cos 2 (1 − 4 sin 2 ), 𝐴, , 𝐵, , 13 cos 𝐵−5, , 12. If 2 tan 2 = 3 tan 2 then prove that cos 𝐴 = 12−5 cos 𝐵, , Transformation of trigonometric formula, The sum or difference form of Sine and Cosine ratios can be transformed into the product, form and the product form can be transformed into the sum or difference form. Let's, discuss about it., i., , Transformation of products into sum or difference, We have, the compound angles formulas of Trigonometric ratios as, sin A cos B + cos A sin B = sin(A + B)………………..(i), , 169
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sin A cos B − cos A sin B = sin(A − B)………………..(ii), cos A cos B − sin A sin B = cos(A + B)…………………(iii), cos A cos B + sinA sin B = cos(A − B)…………………(iv), Now, adding equation (i) and (ii) we get,, sin A cos B + cos A sin B + sin A cos B − cos A sin B = sin(A + B) + sin(A − B), 2sinAcosB = sin(A + B) + sin(A − B) ………………………..(v), Subtracting equation (ii) from equation (i), we get, sin A cos B + cos A sin B − (sin A cos B − cos A sin B) = sin(A + B) − sin(A − B), or, sin A cos B + cos A sin B − sin A cos B + cos A sin B = sin(A + B) − sin(A − B), or, 2 cos A sin B = sin(A + B) − sin(A − B)………………………..(vi), Similarly, adding equation (iii) and (iv), we get, cos A cos B − sin A sin B + cos A cos B + sin A sin B = cos(A + B) + cos(A − B), or, 2 cos A cos B = cos(A + B) + cos A − B)………………..(vii), Again, subtracting equation (iii) from (iv), we get, cos A cos B + sin A sin B − (cos A cos B − sin A sin B) = cos(A − B) − cos(A + B), or, cos A cos B + sin A sin B − cos A cos B + sin A sin B = cos(A − B) − cos(A + B), or, 2 sin A sin B = cos(A − B) − cos(A + B)…………………….(viii), ii., , Transformation of sum or difference into product, From above relation (v), (vi), (vii) and (viii), we have, sin(A + B) + sin(A − B) = 2 sin A cos B…………………………………….(a), sin(A + B) − sin(A − B) = 2 cos A sin B…………………………………….(b), cos(A + B) + cos(A − B) = 2 cos A cos B…………………………………..(c), cos(A − B) − cos(A + B) = 2 sin A sin B…………………………………….(d), Let, A + B = C and A − B = D then, adding them we get, 2A = C + D, ∴A=, , C+D, 2, , Again, subtracting A – B = D from A + B = C, we get, , 170
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2B = C − D, ∴B=, , C−D, 2, , Substituting the values of A, B, (A + B) and (A – B) in equation (a), (b), (c) & (d), we get, sin C + sin D = 2 sin, , C+D, C−D, cos 2 ………………………..(ix), 2, , sin C − sin D = 2 cos, , C+D, C−D, sin, …………………………(x), 2, 2, , cos C + cos D = 2 cos, , C+D, C−D, cos 2 ……………………….(xi), 2, , cos D − cos C = 2 sin, , C+D, C−D, sin 2 ………………………..(xii), 2, , Technique to remember the above formulae., Remember!, , Remember!, , S + S = 2SC, , S = sin, , S – S = 2CS, , C = cosine, , C + C = 2CC, C – C = 2SS, , Example 1, Express the following products into sum or difference:, a) sin25°cos15°, , b) cos4Acos5A, , Solution:, a), , Here,, = sin25°cos15°, 1, 2, , = (2sin25°cos15°), 1, = [sin(25° + 15°) + sin(25° − 15°)], 2, 1, , = 2 (𝑠𝑖𝑛40° + 𝑠𝑖𝑛10°), , 171, , c) 2sin42°sin22°
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ii. A + B + C = πc, Dividing both sides by 2, we get, A + B + C πc, =, 2, 2, A B, πc C, a. ( + ) = ( − ), 2 2, 2 2, B C, πc A, b. ( + ) = ( − ), 2 2, 2 2, C A, πc B, c. ( + ) = ( − ), 2 2, 2 2, iii. A + B + C = πc, Multiplying both sides by 2, we get, 2(A + B + C) = 2πc, a. 2A + 2B = 2πc − 2C, b. 2B + 2C = 2πc − 2A, c. 2C + 2A = 2πc − 2B, For the above relations formed on (i), (ii) and (iii) make a list of relations with the, Trigonometric Ratios Sine, Cosine and tangent and discuss in class., Example 1, A, B, C, A, B, C, If A + B + C = πc , Prove that; Cot + Cot + Cot = Cot Cot Cot, 2, 2, 2, 2, 2, 2, Solution: Here,, A + B + C = πc, A B πc C, [∵ Dividing both sides by 2], + = −, 2 2, 2 2, , or,, , Taking Trigonometric Ratio cotangent on both sides, we get, A B, πc C, cot ( + ) = cot ( − ), 2 2, 2 2, or,, , A, B, cot 2 cot 2 − 1, C, = tan, B, A, 2, cot 2 + cot 2, , 182
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Let us observe the following figure., , , What are the angles range in each, quadrant?, , , , Which trigonometric ratios are positive, in 1st quadrant?, , , , Which trigonometric ratios are positive, and which are negative in 2nd, quadrant?, , , , Similarly, discuss in 3rd quadrant and 4th, quadrant., , The standard angles 0° , 30° , 45° , 60° lies in 1st quadrant. What are their values with Sine,, cosine and tangent? Write them in tabular form., Steps to find the angles, 1. At first we determine the quadrant where the angle falls for this use the CAST rule., 2. Find the least positive angle of trigonometric function in the first quadrant for the, given relation., For example:, √3, 2, , the least positive angle in the first quadrant is 30° ., , i., , If cos 𝜃 =, , ii., , If tan 𝜃 = √3, the least positive angle in the first quadrant for tan 𝜃 = √3 is 60° ., , iii., , If sin 𝜃 = −1 the least positive angle in the first quadrant for sin 𝜃 is 90° ., , 3. When '𝜃' is the least positive angle in the first quadrant then, i), , The angle in the second quadrant =180° − 𝜃, , ii), , The angle in the 3rd quadrant = 180° + 𝜃, , iii), , The angle in the fourth quadrant = 360° − 𝜃, , 4. To find the angle more than 360° we add or subtract the least positive angles of 1st, quadrant in even multiple of 90° ., What are the smallest and greatest values of sin 𝜃 and cos 𝜃? Discuss., If 𝛼 ≤ sin 𝜃 ≤ 𝛽 then what are the values of 𝛼 and 𝛽. Are 𝛼 = −1 and 𝛽 = 1? Find, it., , 190
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Example 1, (0° < < 90°), , Solve: √3 tan 𝜃 = 3, Solution: Here,, √3 tan 𝜃 = 3, 3, √3, , or,, , tan 𝜃 =, , or,, , tan = √3, , Since tan is positive, so it lies in 1st and 3rd quadrants, Now, in 1st quadrant, tan = tan 60°, 𝜃 = 60°, But, in third quadrant it is out of range, Hence, 𝜃 = 60°, Example 2, 4𝑐𝑜𝑠 2 𝜃 − 1 = 0 (0° ≤ 𝜃 ≤ 90°), , Solve for 𝜽:, Solution: Here,, , 4𝑐𝑜𝑠 2 𝜃 − 1 = 0, or, 4𝑐𝑜𝑠 2 𝜃 = 1, 1, , or, 𝑐𝑜𝑠 2 𝜃 = 4, 1, 4, , or, cos 𝜃 = ±√, 1, , or, cos 𝜃 = ± 2, Taking positive sign, cos 𝜃 =, , 1, 2, , or, cos 𝜃 = cos 60°, 𝜃 = 60°, 1, , Taking negative sign, cos 𝜃 = − 2, cos 𝜃 = cos(180° − 60°), cos( 360° − 60°), 𝜃 = 120°, 300° but 0° ≤ 𝜃 ≤ 90°), , Hence, 𝜃 = 60°, , 191
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Example 3, Solve for x:, , (8 sin 𝑥 + 4)(2 cos 𝑥 + 1) = 0 (0° ≤ 𝑥 ≤ 180°), , Solution: Here,, (8 sin 𝑥 + 4)(2 cos 𝑥 + 1) = 0, Either, 8 sin 𝑥 + 4 = 0 ……… (i), , or, 2 cos 𝑥 + 1 = 0 ……… (ii), , From equation (i), 8 sin 𝑥 = −4, 4, , sin 𝑥 = − 8, sin 𝑥 = −, , or,, , 1, 2, , Least positive angle (P.A.) = 30°. Since the value sinx is (-ve), so it lies in 3rd and 4th, quadrant., ∴ sin 𝑥 = sin(180° + 30°), sin(360° − 30°), 𝑥 = 210°, 330°, But, (0° ≤ 𝑥 ≤ 180°), From equation (ii) 2 cos 𝑥 + 1 = 0, 2cos 𝑥 = −1, cos 𝑥 = −, , 1, 2, , Least positive angle = 60°, since, cos 𝑥 is (-ve) so, it lies in 2nd and 3rd quadrant., ∴ cosx = cos(180° – 60°), cos (180° + 60°), 𝑥 = 120°, 240°, But 𝑥 = 240°, , (out of range), , Hence, 𝑥 = 120°, Example 4, Solve:, , 𝑡𝑎𝑛2 𝑥 − 3 sec 𝑥 + 3 = 0, , (0° ≤ 𝜃 ≤ 360°), , Solution: Here,, 𝑡𝑎𝑛2 𝑥 − 3 sec 𝑥 + 3 = 0, or, 𝑠𝑒𝑐 2 𝑥 − 1 − 3 sec 𝑥 + 3 = 0, , 192
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5.6 Height and distance, Let's observe the following figures and discuss:, E, , E, , C, , Υ, , D, , B, , A Man, , E, Observer, , Figure (I), , B, B, , A, , Figure (II), F, , β, , E, , D, , A, , C, Figure (IV), , i), , What do 𝜃, 𝛾, 𝛼 𝑎𝑛𝑑 𝛽 stand for?, , ii), , What do AB and AC called in figure (iii)?, , iii), , Are there any similarity between , 𝛾, 𝛼 𝑎𝑛𝑑 𝛽 ?, , iv), , What is the relation between AB and CD in Figure (II)?, , v), , What is the relation between FBD and BDE in figure (IV)?, , From above figures, 𝜃 𝑎𝑛𝑑 𝛾 are called angle of elevation and 𝛼 𝑎𝑛𝑑 𝛽 are called, angle of depression., Let's define angle of elevation and angle of depression, Angle of elevation: - In the adjoining figure, C be the, position of observer and A be the position of an object., CA be the line of sight or line of observation. BC be the, horizontal line through the observation point C. Then, ∠𝐴𝐶𝐵 = 𝜃 is said to be angle of elevation., , A(Object), , e, Lin, , Sig, of, , ht, , When an observer observes an object lying above the, horizontal line (eye level), the angle formed by the line, θ, of sight with the horizontal line is called an angle of, elevation. Angle of elevation is also called altitude of A. C (Observer) Horizontal Line, , 198, , B
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Angle of depression, In the adjoining figure, 'B' be the position of an, eye of an observer and 'C' be the position of an, object. BC be the line of sight or line of, observation. BD be the horizontal line through D,, which is parallel to the horizontal line AC. C lies, below the position of an eye, then ∠DBC is said, to be the angle of depression., , When an observer observes an object lying below the horizontal line (eye level), the angle, formed by the line of sight with the horizontal line parallel to the ground is called an angle, of depression., When the actual measurement of the height of an object or distance between two points, (object) is not easy or even not possible, as an application of the trigonometry, a, technique is used to find them with the help of the angle/angles subtended at a point by, the object/objects whose distance or height is to be determined. The instruments like, theodolites or clinometer are used to measure the angle. This method is mostly used in, surveying, map making, aviation and astronomy etc., Example 1, Find the height of a building, when it is found that on walking towards it 40m in a, horizontal line through its base the angular elevation of its top changes from 30° to 45°., Solution: Here,, A, , Let AC = h be the height of building. Let, D and C be, the position of two points such that CD = 40m. The, angle of elevation from point C and D to the points A, are 45° and 30° respectively., , h, , ∴ ACB = 45°and ADB = 30°, , To find: Height of building (AB), , D, , Now, in right angled triangle ABC,, tan45° =, Or, 1 =, , AB, BC, , ℎ, 𝐵𝐶, , BC = h …………………………. (i), , 199, , 30·, 40m, , 45·, C, , B
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or, √3 =, , 200𝑚, 𝐴𝐶, , or, 𝐴𝐶 =, , 200𝑚, √3, , = 115.47𝑚, , Again, in rt. angled 𝐴𝐵𝐷,, 𝐴𝐵, 𝐴𝐷, 200𝑚, or, 1 =, 𝐴𝐷, or, AD = 200m, 𝑡𝑎𝑛45° =, , ∴ The distance between the cars (CD) = AD – AC = 200m – 115.47m = 84.52m, ii), , Let, AB = 200m be the height of house. Let, C, and D be the position of two rest cars on the, opposite side of house. Drawn, EF ∥, DC through B., , ∴ EBD = BDA = 45° and FBC = BCA = 60°, , Since, AC = 115.47m [calculation from (i)], AD = 200m, When the cars are on the opposite sides of house, the distance between the two, cars (CD) = AD + AC = 200m + 115.47m = 315.47m, Example 3, The length of the shadow of a tower standing on level place is found to be 30m longer, when the sun's altitude is 30°, then when it was 45°. Prove that the height of tower is, 15(√3 + 1)𝑚, Solution: Here,, , P, , Let, PQ = h be the height of tower. Let, R and S be, the two points such that, h, , RS = 30m, PSQ = 30° and PRQ = 45°., To Prove: Height of tower(PQ) = 15(√3 + 1)𝑚, In rt. angled 𝑃𝑄𝑅, 𝑡𝑎𝑛45° =, or, 1 =, , S, , PQ, QR, , h, QR, , 201, , 30o, 30m, , 45, R, , o, , Q
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or, ℎ =, , 7𝑚, √3−1, , = 9.56𝑚, , Hence, the height of tower (AB) = h = 9.56m., Exercise 5.6, 1. a) Define angle of elevation. Illustrate it with figure., b) Define angle of depression. Illustrate it with figure., 2. From the given figures, find the values of x, y and :, A, , a), , x, , x, , 45o, , 30o, D, , P, , b), , C, y, , 20m, , 45o, , 60o, , B, , Q, , y, , S, , R, , 60m, , d), , c), , A, θ, , 100m, 45, C, , e), , 204, , o, , 30, x, , B, y, , o, , D
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3. a), , The angle of elevation of a tower from a point was 60°. From a point on walking, 300 meter away from the point it was found 30°. Find the height of the, tower., , b), , The angle of elevation of the top of a house from a point on the ground was, observed to be 60° on walking 60 m away from that point it was found to be, 30°. If the house and these points are in the same line of the same plain, find, the height of house., , c), , The elevation of the top of a tower of height 60m from two places on the same, horizontal line due west of it are 60° and 45°. Find the distance between the, two places., , 4. a), , From the top of a tower 100 meters high the measures of the angles of, depression of two object due east of the tower are found to be 45° and 60°, Find the distance between the objects., , b), , From the top of 21m high cliff, the angles of depression of the top and the, bottom of a tower are observed to be 45°𝑎𝑛𝑑 60° respectively. Find the height, of tower., , c), , From the top of a tower 192 meter high the angle of depression of two vehicles, on a road at the same level as the base of tower and on the same side of it, 3, 1, are 𝑥° and 𝑦° when tan𝑥° = and tany° = . Calculate the distance between, 4, 3, them., , 5. a), , The shadow of a tower standing on a level ground is found to be 40m longer, when the sun's altitude is 30° then when it is 60°. Find the height of the tower., , b), , The shadow of a tower on the level ground increases in length by 'x' meter, when the sun's altitude is 30° then when it is 45°. If the height of the tower is, 25 meter, find the value of x., , 6. a), , The angles of elevation of the top of a tower as observed from the distance of, 4m and 16m from the foot of the tower are complementary. Find the height of, tower., , b), , Two lamp posts are 200m apart and height of one is double of the other. From, the midpoint of the line joining their feet an observer finds the angle of, elevation of their tops to be complementary. Find the height of the both post., , 7. a), , A flagstaff stands on the top of a post 20m high. From a point on the ground, the angles of elevation of the top and bottom of the flagstaff are found to be, 60° and 45° respectively. Find the length of the flagstaff., , b), , From a point P on the ground the angle of elevation of the top of a 10m tall, building is 30°. A flag is hoisted at the top of the building and the angle of, , 205
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elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff, and the distance of the building from the point P., 8. a), , A vertical pole is divided at a point in the ratio of 1:9 from the base. If both, parts of the pole subtend equal angles at a point 20m from the foot of the pole,, find the height of the pole., , b), , AB is a vertical tower with its top A. C is a point on AB such that AC:CB = 13:5., If the parts AC and CB subtend equal angles at a point on the ground which is, at a distance of 30 meters from the foot of the tower. Find the height of the, tower., , 9. a), , An aeroplane flying horizontally 900m above the ground is observed at an, elevation of 30°. After 10 seconds the elevation is observed to be 45°. Find the, speed of the aeroplane in km/hr., , b), , Two poles of equal heights are standing opposite each other on either side of, the road, which is 80m wide. From a point between them on the road, the, angles of elevation of the top of the poles are 60°𝑎𝑛𝑑 30° respectively. Find, the height of the poles and the distance of the point from the poles., , 10. a), , A ladder 10m long reaches a point 10m below the top of a vertical flagstaff., From the foot of the ladder, the elevation of the flagstaff is 60°. Find the height, of the flagstaff., , b), , From the top of a 7m high building, the angle of elevation of the top of a cable, tower is 60° and the angle of depression of its foot is 45°. Determine the height, of tower., , c), , The angle of depression and elevation of the top of a pole 25m high observed, from the top and bottom of the tower are 60°𝑎𝑛𝑑 30° respectively. Find the, height of tower., , 11. Make the clinometer by different group of students. From a point of the ground, measure the distance between the point and the base of building. What is the height, of your school building? Find it. Discuss in your calculation., , 206
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Unit 6, , Vector, , 6.0 Review, , -, , What is the direction of aeroplane when it is flying and landing?, , -, , If ⃗⃗⃗⃗⃗, OA = (2,3) and ⃗⃗⃗⃗⃗, OB = (5, 2), then find, , i), , ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗, OA, OB in row form, , ii), , ⃗⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗, OA, OB in column form, , the value of, , -, , What does 𝑖 and 𝑗 represent?, , -, , What are the types of vectors? Write one example of each., , -, , What are the important roles of vectors in our daily life? Let us discuss with some, examples., , In a vector there are two types of products. They are dot product and cross product. Dot, product is a scalar quantity but cross product is a vector quantity. In this grade we will, study only scalar product., When an aeroplane covers 20km distance in air but it doesn't cover that distance on the, ground, the perpendicular drawn from the positon of an aeroplane to the ground is called, the projection of an aeroplane to the ground., y, , 6.1 Scalar or dot product of two vectors, Study the given graph and answer the, following questions:, i), , What are the coordinates of A, and B?, , ii), , What are the position vectors of, A and B?, , iii), , Multiply X-coordinates of the, points A and B and Y coordinates, of the points A and B separately., , A, B, , x’, , 207, , x, , o, , y’
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iv), , What is the sum of the product of the x-coordinates and y-coordinates of the, points A and B?, , v), , Can we show the result obtained from (iv) in the same graph?, Again, Discuss on the following questions:, (a) What is the difference between zero (0) and origin (0, 0)?, (b) Is the product of two numbers zero?, (c) Can we find the coordinates of the origin (0, 0) by adding and subtracting, of two vectors?, (d) Is the product of two vectors is 0 (0, 0) or not?, , ∴ The dot product of two vectors gives the result in scalar form so it is known as, scalar product., If 𝑎 and 𝑏⃗ are two vectors and 𝜃 is the angle between them, then dot product is defined, as, 𝑎. 𝑏⃗ = |𝑎|. |𝑏⃗|𝑐𝑜𝑠𝜃, = 𝑎𝑏𝑐𝑜𝑠𝜃, , B, , (0° ≤ 𝜃 ≤ 𝜋 𝑐 ), b, , Where, 𝑎 = |𝑎| and 𝑏 = |𝑏⃗| are the magnitudes of, 𝑎 and 𝑏⃗ respectively., , θ, , Thus, 𝑎. 𝑏⃗ = |𝑎||𝑏⃗|𝑐𝑜𝑠𝜃, , A, , O, , a, , ∴The scalar product of two vectors a⃗ and ⃗b is defined as the product of the magnitudes of, two vectors multiplied by the cosine of the angle between them., 6.1.1 Geometrical interpretation of a scalar product., B, , ⃗⃗⃗⃗⃗ = 𝑎 and OB, ⃗⃗⃗⃗⃗ = 𝑏⃗. Let 𝜃 be the angle, Let 𝑂𝐴, between the two vectors 𝑎 𝑎𝑛𝑑 𝑏⃗., From A and B draw AD and BE perpendiculars, to OB and OA respectively., , b, , Now, 𝑎𝑏⃗ = |𝑎||𝑏⃗|𝑐𝑜𝑠𝜃 (where 0° ≤ 𝜃 ≤ 𝜋 𝑐 ), θ, , = 𝑎𝑏𝑐𝑜𝑠𝜃, , O, , = (𝑂𝐴)(𝑂𝐵 𝑐𝑜𝑠𝜃), = 𝑂𝐴 × 𝑂𝐸, , 208, , a, , E, , A
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4., , a., , The position vectors of M and N are 4𝑖 + 6𝑗 and −𝑖 + 3𝑗 respectively. Find, the position vector of the point Q which divides MN internally in the ratio of, K, 3:2., , b., , In the adjoining figure, the point N, divides MK in the ratio 2:3. If, ⃗⃗⃗⃗⃗, ON = −3i − 8j and ⃗⃗⃗⃗⃗, OK = 2i + 5j., ⃗⃗⃗⃗⃗⃗ ., Find OM, , 3, , 5., , 6., , N, , 2, M, , O, , a., , If the position vectors of the points A and B are 5𝑖 + 2𝑗 and 3𝑖 + 6𝑗, respectively, find the position vector of the point T which divides AB externally, in the ratio of 5:2., , b., , A and B are two points with coordinates (-4, 8) and (3, 7) respectively. Find, the position vector of Q which divides AB externally in the ratio of 4:3., , a., , A(-1, 1), B(5, -1) and C(2, 5) are the three vertices of ∆ABC. Find the position, vector of the centroid of ∆ABC., , b., , ⃗⃗⃗⃗⃗ = 3𝑖 − 5𝑗, 𝑂𝐵, ⃗⃗⃗⃗⃗ = −7𝑖 + 4𝑗 and the position vector of the, In a ∆ABC, 𝑂𝐴, centroid G is 2𝑖 + 𝑗. Find the position vector of C., O, , 7., , a., , In the given figure, ⃗⃗⃗⃗⃗, 𝑂𝐴 = 𝑎, ⃗⃗⃗⃗⃗, 𝑂𝐵 = 𝑏⃗,, ⃗⃗⃗⃗⃗⃗ = 𝑚, 𝑂𝑀, ⃗⃗ and M divides BA in the ratio, 1, of 3:2, then prove that 𝑚, ⃗⃗ = 5 (3𝑎 + 2𝑏⃗)., B, , b., , In the figure, UM is the median where, the position vectors of U and M are, 3𝑖 − 2𝑗 and −3𝑖 − 4𝑗 respectively. Find, the position vector of centroid G., , M, , 2, , A, , U, , G, , V, , 223, , 3, , M, , W
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Exercise 6.3, 1., , a), , B, , In ∆BOY, M and N are the midpoints of sides BO, and BY respectively. Write the relation between, MN and OY., M, , N, , Y, , O, , b), , In the given ∆AXE, AE = AX and EP = XP., Write the relation between AP and EX., , P, , E, , X, , E, , A, , c), , In the given ∆PEN, 𝑃𝑍 = 𝑁𝑍 = 𝐸𝑍., Write the relation between EP and NP., , Z, , N, , P, , A, , 2., , a), , In the given figure, P and Q are the middle, points of AB and AC respectively of the, ∆ABC. Prove by vector method that, i), , 𝐵𝐶 = 2𝑃𝑄, , ii), , 𝑃𝑄 ∥ 𝐵𝐶, , P, , Q, , C, , B, M, , b), , In the given ∆MAN, UV is a line segment joining, the midpoints of sides AN and MN then prove by, vector method that:, i), , 1, 𝐴𝑀, 2, , ii), , 𝐴𝑀 ∥ 𝑈𝑉, , V, , = 𝑈𝑉, , N, , 228, , U, , A
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3., , a), , In the given ∆GUN, GU = GN and, UT =TN then prove by vector method, that : 𝐺𝑇 ⊥ 𝑈𝑁., , G, , U, , T, , N, , T, =, , b), , In the given ∆BUT, BT = BU and, TM = UM, then prove by vector method, that ∠BMT = ∠BMU = 90°., , B, , M, , =, , U, , C, , 4., , a), , In the given triangle CAT, ∠𝐶𝐴𝑇 = 90°,, TW = CW, prove by vector method that:, , W, , CW = TW = AW, T, , A, , B, , b), , In a right angled ∆BAT, ∠𝐵𝐴𝑇 = 90°, S is the, midpoint of BT. Prove by vector method that, TS = BS., S, A, , T, , 229
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Both diagonals AC and BD have a common midpoint M. Hence, it is proved that, diagonals of parallelogram bisect to each other., Theorem 6, Statement: The diagonals of a rhombus bisect each other at right angle., Solution:, , D, , C, , Given: In a rhombus ABCD, AC and BD are the diagonals., To Prove: ABCD is a rhombus., We know that, rhombus is also a parallelogram and the, diagonals of the parallelogram bisect each other., , A, , B, , So, in rhombus also, diagonals bisect each other, Again, in ∆ABC, by ∆ law of vector addition, ⃗⃗⃗⃗⃗ + 𝐵𝐶, ⃗⃗⃗⃗⃗ = 𝐴𝐶, ⃗⃗⃗⃗⃗, 𝐴𝐵, , ……………………..(i), , Similarly, in ∆DAB, by ∆ law of vector addition, ⃗⃗⃗⃗⃗, 𝐷𝐴 + ⃗⃗⃗⃗⃗, 𝐴𝐵 = ⃗⃗⃗⃗⃗⃗, 𝐷𝐵, , ……………………….(ii), , Now, taking dot product of (i) and (ii), we get, ⃗⃗⃗⃗⃗ + BC, ⃗⃗⃗⃗⃗ )(DA, ⃗⃗⃗⃗⃗ + AB, ⃗⃗⃗⃗⃗ ) = AC, ⃗⃗⃗⃗⃗ DB, ⃗⃗⃗⃗⃗, (AB, or,, , ⃗⃗⃗⃗⃗ + BC, ⃗⃗⃗⃗⃗ ) (AB, ⃗⃗⃗⃗⃗ − AD, ⃗⃗⃗⃗⃗ ) = AC, ⃗⃗⃗⃗⃗ . DB, ⃗⃗⃗⃗⃗, (AB, , or,, , ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗, ⃗⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗, (AB, BC) (AB, BC) = ⃗⃗⃗⃗⃗, AC ⃗⃗⃗⃗⃗, DB, , or,, , ⃗⃗⃗⃗⃗ DB, ⃗⃗⃗⃗⃗, AB 2 − BC 2 = AC, , or,, , ⃗⃗⃗⃗⃗ DB, ⃗⃗⃗⃗⃗, AB 2 − BC 2 = AC, , or,, , AB 2 − AB 2 = ⃗⃗⃗⃗⃗, AC ⃗⃗⃗⃗⃗, DB, , or,, , ⃗⃗⃗⃗⃗, AC. ⃗⃗⃗⃗⃗, DB = 0, , [ ⃗⃗⃗⃗⃗, AD = ⃗⃗⃗⃗⃗, BC], , [ Being sides of rhombus AB = BC], , Since, the dot product of two vectors is zero, they are at 90°. So, 𝐴𝐶 ⊥ 𝐷𝐵., Hence, it is proved that the diagonals of a rhombus bisect each other at 90°., , 232
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Exercise 6.4, 1., , a), , In the given figure, L, A, M and P are the, midpoints of sides KI, IT, TE and EK, respectively of quadrilateral KITE. What type, of quadrilateral LAMP is it?, , M, , E, , A, , P, , K, , b), , I, , L, , U, , In the figure, O is the centre of semicircle PUT. PT is a diameter. Write the, value of ∠PUT., P, , 2., , T, , a), , In a rhombus ABCD, AC and BD are two, ⃗⃗⃗⃗⃗ . BD, ⃗⃗⃗⃗⃗ ?, diagonals. What is the value of AC, , b), , The given figure is a semi-circle with center, ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗, ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗, O. Prove that: (QO, OP). (QO, OR) = 0, , T, , O, , Q, , P, , O, , R, , S, , G, , 3., , 4., , a), , In the given figure, B, E, S and T are the, midpoints of RI, IN, NG and GR respectively., Prove by vector method that BEST is a, parallelogram., , E, , T, , R, , B, , b), , Prove by vector method that the diagonals of a rectangle ROSE are equal., , a), , Prove that the diagonals of a parallelogram TAPE bisect to each, other., , 235, , N, , I
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b), , Prove that the diagonals of a rhombus HOME bisect to each other at right, angle., E, , 5., , a), , In the given figure, A is the centre of, semi-circle DEN. DN is a diameter. Prove, that DE is perpendicular to EN., D, , A, , N, , E, , b), , In the given figure, BIKE is a parallelogram., BK and EI are two diagonals, then prove, that ET=TI and BT=TK., , T, , B, , 6., , K, , I, , What is the difference between theorems which are proved in vector geometry and, geometry? Make a short report with examples and present in the class., , 236
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Transformation, , Unit 7, 7.0 Review, , Write about reflection, rotation, translation, enlargement of a geometric figure. Discuss, in small group of students and list down formulae about transformation that you have, learned in previous classes., 7.1 Composition of transformation/combined transformation, From the adjoining graph, write the co-ordinate of, points A, A', A'' and B, B', B'' in your exercise book., Observe the single transformations that changes, position A to A” and B to B”. Discuss about the, situation that is found in the graph., T1 (02), , Here, A(2,1) →, T2 (−2, 0), , A′(2,3) →, , T3 (−2, 2), , A(2,1) →, , A′ (2,3), , A" (0,3), , A" (0,3), , T1 + T2 = (02) + (−2, ) = (−2, ) = T3, 0, 2, So, T1 + T2 = T3 = T1 OT2, In combination of transformations, one form of, transformation can be combined with another, form of transformation., 𝑆𝑜, 𝑇1 + 𝑇2 = 𝑇3 = 𝑇1 𝑂𝑇2, Similarly, B changes its position to B and B’, changes its position to B”. The corresponding, transformations are as:, 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛, , 𝐵(−3, 2) →, , 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛, , B′(−3, 2) →, , 𝐵′ (−3, −2), 𝐵" (3, −2), , But, B” (3, -2) is obtained after rotation of B through 180° about the origin., , 237
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Again, let us reflect the point A(3, 4) about, x = 1 and then about x = -3, , y, A(3,4), A”(-3,4), , In the graph when A(3, 4) is reflected about, x = 1, the image A'(-1, 4) is obtained. Again,, when we reflects A'(-1,4) about x = -2 we get, the new image point A''(-3, 4)., The distance between x = 1 and x = -2 is 3, units., , A’(-1,4), , x’, , x, , o, , When A(34) is translated by (−6, ) then we get, 0, 𝑇(−6, 0), , 𝐴(3,4) →, , 𝐴"(−3,4), , x=1, , x=-2, y’, , Hence, reflection about x = 1 followed by, reflection about x = -2 is equivalent to translation by (−6, ), we find x = 1 and x = -2 are, 0, parallel lines., So, if the axis of reflections are parallel, a reflection followed by another reflection is, equivalent to the translation., Example 1, Let the reflection in y-axis be r1 and reflection in the line x = 2 be r2, find the images, under following transformations., a) r1or2 (-2, 3), b) r2or1 (0, 5), Solution: We know,, r1(x, y), (-x, y), r2(x, y), (4 - x, y), So, r1 or2 (-2, 3) = r1(r2(-2,3)), = r1(4 + 2, 3), = r1 (6, 3), =(-6, 3), (−4, 0), , Note: It is equivalent to (−2,3) →, , (−6,3), , r2 or1 = T(−4, ), 3, b) r2 or1(0, 5) = r2(r1(0, 5)), = r2 (0, 5), = (4 – 0, 5), = (4, 5), , 238
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Example 2, Draw a ABC having vertices A(3, 1), B(6, -2) and C(0, 3) in a graph. Find the single, transformation or transformation equivalent to composition of reflection on X-axis, followed by reflection about Y-axis i.e. (ryrx). Transform the ABC by ryrx and represent, the image A'B'C' in same graph., Solution:, (ryrx)(a, b) = ry(rx(a, b)), , = ry (a – b), = (–a, –b), (a, b), (–a, –b) is equivalent to rotation, about (0, 0) through +180o, [+180°,(0,0)], , A(3, 1) →, , A′ (−3, −1), , [+180°,(0,0)], , B(6, −2) →, , [+180°,(0,0)], , C(0, 3) →, , B ′ (−6, −2), , C ′ (0, −3), , Example 3, Let E denotes the enlargement with centre at origin and scale factor 2 and R denotes the, reflection through y + x = 0. Find a) (ER) (2, 5) b)(RE) (-3,4), Solution:, 𝐸[(0,0),2], , We know, 𝑝(𝑥, 𝑦) →, , 𝑃′ (2𝑥, 2𝑦), , 𝑅[𝑥+𝑦] = 0, , 𝑝(𝑥, 𝑦) →, , 𝑃′ (−𝑦, −𝑥), , a) (ER) (2, 5), , = E(R(2,5)), = E(-5, -2), = (-10, -4), , b) (RE) (-3, 4), , = R(E(-3, 4)), = R(-6, 8), = (-8, 6), , Example 4, A(-2, 0), B(0, 4) and C(3, 0) are vertices of ABC. 𝑇(−3, ) and R[+90o, (0, 0)] are two, 1, transformations, find RoT(ABC). Represent the object and image in same graph., , 239
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Solution: we know,, (RT) (x, y), = R(T(x,y)), 𝑥−3, = R(𝑦+1, ), , [Since, 𝑇(−3, ) is given], 1, , = [–(y + 1), x – 3)], Now,, (RT) (–2, 0) = (-(0 + 1), -2 -3) = (-1, -5), (RT) (0, 4) = (-(4 + 1), 0-3) = (-5, -3), (RT) (3, 0) = (-(0 + 1), 3 – 3) = (-1, 0), A' (-1, -5), B'(-5, -3) and C' (-1, 0) are the images of A(2, 4) B(3, 3) and C(1, 5), under RoT, Representing the object and image figure in same graph, we get the following:, , Exercise 7.1, 1. a) Define composition of two transformations., b) If T1 (𝑎𝑏) and T2(𝑑𝑐 ) are two translations then find T1T2 (x, y)., c) If the axis of two reflections are parallel to each other, which transformation, is equivalent to the two reflections?, 2. Let, R1 represents the reflection on x-axis., R2 represents the reflection about x + y = 0., r1 represents the rotation through +90o about origin., , 240
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r2 represents the rotation through 180o about origin., E1: represents the enlargement about (0, 0) with scale factor 2., E2: [(0, 0), -3], Compute the following transformations:, a) (R1R2) (1, 2), b) (R2R1) (-2, 3), c) (r1R2) (-4, 3), d) (R2r2) (8, 9), e) E1R1) (1, 2), f) (E1E2) (2, 4), g) r1E1) (-3, 2), h) (E2r2) (-6, 2), 3. a) A(1, 1), B(3, 5) and C(5, -1) are the vertices of ABC. r1 is reflection about the line, x = 2 and r2 is reflection about the line x = -1. Find the single transformation, equivalent to r1r2. Transform ABC by r1r2. Represent the object and image, triangles in same graph., b) A(1, 2), B(4, -1) and C(2, 5) are vertices of ABC. r1 and r2 are the reflection about, the line x = 1 and y = -1 find the transformation equivalent to r1r2. Transform, ABC by r1r2 represent the object and image triangles in the same graph., c) P(2, 2), Q(1, -1) and R(3,0) are vertices of PQR. r1 is reflection about x-axis and r2, is rotation about origin through +90o. Find the rule that is equivalent to r2r1., Transform PQR by r2r1. Represent the object and image in same graph., d) If R1 is the rotation through + 90o about origin and R2 is the rotation through -270o, about origin then find the transformation equivalent to R1R2. Transform XYZ, with vertices X(1, 2), Y(-2, 3), Z(2, 5) by R1R2. Represent the object triangle and, image triangle in same graph., 3, , 4. a) A(2, 0), B(0, 2) and C(-3,0) are vertices of ABC. E1 [(0, 0), 2] and E2 [(0, 0), 2] are, two enlargements. Transform ABC by E1E2. Represent the object and image, triangles in same graph., b) A(1, 3), B(1, 6) C(3, 5) and D(4, 2) are the vertices of quadrilateral E[(0, 0), 2] and, T(−3, ) are two transformations. Find the rule related to ET. Transform ABCD by, 2, ET. Represent the objects and the image in same graph., c) AB is a line segment with end-points A(1, 2) and B(2, 1). It is enlarged by E2E1, 1, , where E2 [(0, 0), 2 2] and E1 [(0, 0), 2] are two enlargements. Transform AB by, E2E1, , 241
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5. If R1 is the reflection about x – 3 = 0 and R2 is the reflection about y + x = 0, show that, R2R1 and R1R2 gives rotation. Are R1R2 and R2R1, give same result, if not why? Give, reason., 6. Write the different situations of combined transformation (composition of, transformation) in brief. Verify these situations taking suitable example., 7. Investigate the different situations of combined transformations in our daily life and, make a report., , 7.2 Inversion transformation and inversion circle, , In the figure, O is centre of circle ABC. Q is any point on circle. P and P' are interior, and exterior points on circle. Measure, OP, OP'. P, Q and P' lie on same line. Examine, whether OP x OP' = OQ x OQ or not., Take yourself a suitable radius. (eg. OQ = 6cm, OP = 3cm,, OP' = 12cm), In the above figure, for any point P, different from centre, point (O), the inverse point P with respect to the circle, ABC is unique. It satisfies the condition, OP x OP’ =, (radius)2, i) The circle is said to be circle of inversion., ii) The point O is called the centre of inversion., iii) r or OQ is called the radius of inversion., iv) The point O, P and P' are collinear, v) If the radius of circle is ‘1’ unit, (OP) × (OP') = 1, 1, , or, OP = 𝑂𝑃′, 7.2.1 Characteristics or features of inversion, i), , To each point of the plane except centre, these corresponds on inverse point., , ii), , If any point lies on the circumference of the circle, its inversion point also lies, on the circumference, , iii), , If P is inside the circle, inversion point of P lies outside the circle., , iv), , The Point and its inversion point can be always interchanged., , v), , If P' is image of P then P is image of P'., , 242
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400 500, , (x', y') = (, , 41, , ,, , 41, , ), , Example 3, Find the inverse image of the point (3, 4) about the circle (x – 2)2 + (y – 2)2 = 36, Solution: Here,, Centre of circle (h, k) = (2, 2) and radius of the circle (r) = 6 units, object point (x, y) = (3, 4) and inversion point (x', y'), We know that, (𝑥 ′ , 𝑦 ′ ) = (, , 𝑟 2(𝑥−ℎ), , 𝑟 2 (𝑦−𝑘), , (𝑥−ℎ)2 +(𝑦−𝑘)2, , 𝑜𝑟, (𝑥 ′ , 𝑦 ′ ), , + ℎ, (𝑥−ℎ)2, , +(𝑦−𝑘)2, , 36(3−2), , 36×1, , 36, , = ( 5 + 2,, 46 82, ), 5, , 36(4−2), , = ((3−2)2 +(4−2)2 + 2, (3−2)2 +(4−2)2 + 2), = ( 1+4 + 2,, , Hence, ( 5 ,, , + 𝑘), , 36×2, + 2), 1+4, , 72, +, 5, , 46 82, ), 5, , 2) = ( 5 ,, , is the required image of the point (3, 4) about the given circle., , Exercise 7.2, 1. In the figure, O(0, 0) is the centre of the circle and r be, the radius of the circle, a) Write the relation between OP, OP' and r [P' is, inversion point of P], b) If P(x, y) is given, Find P’ (x',y') in terms of x, y and r., , P’, P, o, r, , 2. Complete the following table after calculation, S.N., , Point, , Equation of inversion circle, , (a), , (2, 3), , x2 + y 2 = 1, 2, , 2, , Inversion Point, ?, , (b), , (-1, 2), , x +y =4, , ?, , (c), , (0, 4), , x2 + y 2 = 9, , ?, , 2, , 2, , (d), , (2, -3), , x + y = 16, , ?, , (e), , (1, 5), , x2 + y2 = 25, , ?, , 2, , 2, , (f), , (4, 3), , x + y = 64, , ?, , (g), , (-1, -3), , (x-2)2 + (y + 1)2 = 16, , ?, , (h), , (2, -5), , 2, , 2, , x + y – 2x – 6y + 6 = 0, , 245, , ?
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3. a), , Find the inverse of point (6, 10) with respect to the circle x2 + y2 = 64., , b), , Find the inverse image of the point (4, 6) in respect to the circle x2 + 2x + y2 = 8., , c), , Find the inverse image of point K(4, -3) with respect to the circles:, i) x2 + y2 + 6x – 8y – 11 = 0, , ii) x2 + y2 – 4x – 6y – 23 = 0, , 4. Write two equations of circle having centre at origin and other than origin use these, circles to find the images of inversion point of A (4, 5) and B(-6, 7)., , 7.3 Transformation Using Matrices, , Y, , a) Transformation using 2 x 1 matrix, n, 5u, , C, , its, , In the right figure, A to C can be moved as 4 units horizontal, 3units, and then 3 units vertical. If the position of A is (x, y), write, A(x,y) 4units B, X, down the position of B and C in column matrix, i) The column matrix for point A is (𝑦𝑥 ), ii) The column matrix for point B is (𝑥+4, ) = (𝑦𝑥 ) + (40), 𝑦, 𝑥+4, iii) The column matrix for point C is (𝑦+3, ) = (𝑦𝑥 ) + (43), , So, the 2 x 1 matrix that transform point A to point B is (40) and the transformation, matrix that transform point A to C is (43), , [If T(𝑎𝑏) be the 2 x 1 translation vector, then P(x, y) will be translated into (x + a, y + b)], b) Transformation using 2 x 2 matrix, , y, , In the figure shown in a graph, write the coordinate, of A, B, C, A', B' and C' represent the ABC and, A'B'C' in matrix form., Which matrix can pre-multiply the matrix of ABC to, get image A'B'C'? Discuss., Here, the matrix for ABC, 1 4 3, 1 4 3, =(, ) and that for A'B'C' is (, ), 3 3 0, 6 6 0, 𝑎 𝑏, Let us suppose a matrix (, ) which pre-multiply, 𝑐 𝑑, , x’, , o, , x, , C = C', y’, , 246
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Now, a – 1 = 0; b = 0, 𝑎, c = 0; d + 1 = 0 gives (, 𝑐, , 1 0, 𝑏, )=(, ), 0 −1, 𝑑, , Similarly, we can perform the other translation and obtain the following result, Matrix, 1 0, [, ], 0 −1, −1 0, [, ], 0 1, 0 1, [, ], 1 0, 0 −1, [, ], −1 0, −1 0, [, ], 0 −1, , Result, , Geometric transformation, , (x, y) →(x, -y), , Reflection in X-axis, , (x, y)→(-x, y), , Reflection in Y-axis, , (x, y)→(y, x), , Reflection in the y = x, , (x, y) →(-y, –x), , Reflection in the line y = –x, , (x, y)→(-x, -y), , Rotation through 180o, about the origin, , −1, ], 0, , (x, y)→(-y, x), , Anti-clock wise rotation, through 90o about origin, , 0 1, [, ], −1 0, , (x, y)→(y, -x), , Clock wise rotation through, 90o about origin, , (x, y)→(mx, my), , Enlargement with scale, factor, and centre at origin, , 0, [, 1, , [, , 𝑚, 0, , 0, ], 𝑚, , Example 1, A(4, 5) and B(6, 7) are end-points of line segment AB. Translate A and B using matrix, (−4, ), 3, Solution: Here,, A and B has matrix (45) and (67) respectively., We know,, , 𝑇(𝑎, 𝑏), 𝑥, (𝑦) → (𝑥+𝑎, ), 𝑦+𝑏, , −4, , 4 𝑇( 3 ) −4 + 4, 0, So, ( ) →, (, )=( ), 3+5, 5, 8, −4, , 6 𝑇( 3 ) −4 + 6, 2, ( )→, (, )=( ), 3+7, 7, 10, Hence, the required image points are A'(0, 8) and B'(2, 10)., , 248
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Example 2, A(1, 1), B(3, 2), C(3, 5) and D(1, 4) are the vertices of a parallelogram ABCD. Find the coordinates of the vertices of the images of parallelogram ABCD under the transformation, 2 0, by the matrix (, ), 0 −1, Solution: Here,, object matrix = (1 3 3 1), I, 1, , 2, , 5 4, , 2, transformation matrix = (, 0, , 0, ), −1, , image matrix = ?, We know, Image matrix = (transformation matrix) × (object matrix), 2, or, Image matrix = (, 0, =(, =(, , 0, 1, )(, −1 1, , 2×1+0×1, 0 × 1 + (−1) × 1, , 2, 6, −1 −2, , 3, 2, , 3 1, ), 5 4, , 2×3+0×2, 0 × 3 + (−1) × 2, , 2×1+0×4, 2×3+0×5, ), 0 × 3 + (−1) × 5 0 × 1 + (−1) × 4, , 6 2, ), −5 −4 2×4, , Required image points are 𝐴′ (2, −1), 𝐵′ (6, −2), 𝐶 ′ (6, −5) and 𝐷′ (2 − 4), Example 3, 0, Find the transformation matrix in which a unit square (, 0, 0 3 4 1, into the parallelogram (, ), 0 1 3 2, , 1 1 0, ) is transformed, 0 1 1, , Solution: Here,, 𝑎, Let, the transformation matrix be (, 𝑐, , 𝑏, ), 𝑑, , We know that, transformation matrix × object matrix = Image matrix, 𝑎, By the question (, 𝑐, , 0 1, 𝑏, ), (, 𝑑 2×2 0 0, , 1 0, 0, )=(, 1 1, 0, , 249, , 3 4 1, ), 1 3 2
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d + 1 = 0; d = –1, 𝑎, Hence, (, 𝑐, , 1 1, 𝑏, )=(, ), 1 −1 2×2, 𝑑, , Example 5, The vertices of ∆ABC are A(2, 3), B(4, 5) and C(6, 2). If E1 = [(0, 0), 2] and E2 = [(0, 0), 2], then find the co-ordinates of images of ∆ABC under E1E2 using matrix method., Solution: Here,, E1 = [(0,0), 2] and E2 = [(0,0), 2 ], We know 𝐸 = 𝐸1 𝑜𝐸2 = 𝐸2 𝑜𝐸1 = [(0,0), 2 × 2] = [(0,0), 4], 4, The matrix represented by [(0, 0), 4] is [, 0, , 0, ], 4, , Now,, 4, [, 0, , 0 2 4 6, ][, ], 4 3 5 2, 4×2+0×3 4×4+0×5 4×6+0×2, =[, ], 0×2+4×3 0×4+4×5 0×6+4×2, 8 16 24, =[, ], 12 20 8, Hence, the co-ordinates of images are A′ (8, 12), B ′ (16, 20) and C ′ (24, 8)., Exercise 7.3, 1., , Write the 2 × 2 matrix related to following transformations:, a. Reflection in X-axis, b. Reflection in the line x + y = 0, c. Rotation through +90° about the origin, d. Rotation through 180° about the origin, e. Enlargement about (0,0) and scale factor 2, , 2., , Find the image of following points:, a., b., , −1, 6, A(2, 3) by T1T2 , where T1 = ( ) and T2 = ( ), 2, 2, 3, 0, P(-4, 5) by T1T2, where T1 = ( ) and T2 = ( ), 0, 4, , 251
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2 −1, 3.a. A(1, 0), B(2, 2) and C(0, 3) are vertices of ∆ABC . ∆ABC is transformed by (, )., 0 1, Find the images of ∆ABC., b., , A(2, 0), B(4, 4), C(2, 5) and D(0, 4) are vertices of kite. Find the co-ordinates of, 2 1, images of kite under the transformation by matrix (, )., 0 2, , c., , O(0, 0), P(6, 2), Q(8, 6) and R(2, 4) are vertices of parallelogram OPQR. Find the, 1 −2, images of OPQR when it is transformed by the matrix (, )., 2 1, 0 −1 −1 0, 4.a. Find the transformation matrix which transforms a square (, ) into, 0 0, 1 1, 0 3 4 1, the quadrilateral (, )., 0 3 4 1, b., , A quadrilateral ABCD with vertices A(0, 3), B(1, 1), C(3, 2) and D(2, 4) is mapped into, the quadrilateral A’(3, 0), B’(1, -1), C’(2, -3) and D’(4, -2). Find the 2 × 2 transformation, matrix., , c., , Find the 2 × 2 transformation matrix which transform ∆ABC with vertices A(1, 3),, B(4, 3), C(3, 0) into ∆A′B′C′ with vertices A’(1, 6), B’(4, 6) and C’(3, 0)., , 0, 5.a. Find the transformation matrix which transformed rectangle (, 0, unit square., b., , c., , 6., , 3 4, 0 1, , 1, ) into, 1, , 0 −1 −1 0, Find the transformation matrix which transform a quadrilateral (, ), 0 0, 1 1, 0 3 4 1, into the quadrilateral (, )., 0 3 4 1, 0 2 2 0, Find the transformation matrix which transform the quadrilateral (, ), 0 0 2 2, 0 0 2 2, into (, )., 0 2 2 0, Find the 2x2 matrix which transform the points are indicated below, a. A(𝑥, 𝑦) → A′ (−𝑥, −𝑦), , b. B(𝑥, 𝑦) → B ′ (3𝑥, 3𝑦), , c. C(𝑥, 𝑦) → C ′ (𝑥, −2𝑦), , 7.a. Show by matrix method that a reflection about the line y – x = 0 followed by the, rotation about origin through + 90° is a reflection in x = 0. Discuss about the order of, matrix multiplication., b., , Write any one difference between transformation using 2 x 1 matrix and 2 x 2 matrix., Show by using matrix method that reflection on the line about X-axis followed by the, Y-axis is the rotation about the origin through 180°., , 252
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Unit 8, , Statistics, , 8.0 Review, The marks obtained by two students of grade 9 in eight subjects are given below:, Subject, English, Nepali, C. Maths, Science, Social Studies, Population, Opt. Maths, Computer Science, , Students, A, 40, 50, 65, 60, 58, 62, 55, 46, , B, 50, 65, 80, 35, 55, 70, 85, 25, , Representing their marks in a graph,, , Study the above graph and answer the following questions:, i., ii., iii., iv., v., , What is the total scores of student A and student B? Also, find their average marks., Whose scores are more Scattered?, Whose achievement is better? And why?, What are the methods to measure the consistency and variability of the data? Which, one is the best among them?, How can we compare the obtained marks of these two students?, , 253
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In a data, measures of central tendency gives the idea about the concentration of the, items around the central value. Dispersion means scatterness, variability, deviation or, fluctuation etc. But the average (i.e. mean, median and mode) is not sufficient for clear, information about the distribution. We study measure of dispersion which shows the, scattering of data. It tells the variation of the data from one another and gives a clear idea, about the distribution of the data. The measure of dispersion shows the homogeneity or, the heterogeneity of the distribution of the observations., Can you get an idea about the distribution if we get to know about the dispersion of the, observations from one another, within and between the data? The main idea about the, measure of dispersion is to get to know how the data are spread. It shows how much the, data vary from their average value., Classification of measures of dispersions, The measure of dispersion is categorized as:, a. An absolute measure of dispersion, - The measures which expresses the scattering of observation in terms of distance, i.e. range, quartile deviation., - The measures which expresses the variations in terms of the average deviation, of the observations like mean deviation and standard deviation., b. A relative measure of dispersion, We use a relative measure of dispersion comparing distribution of two or more data, set and for unit free comparison. They are the coefficient of range, the coefficient of, quartile deviation, the coefficient of mean deviation, the coefficient of standard, deviation and the coefficient of variation. Here, we will study to find quartile, deviation, mean deviation and standard deviation of the continuous series only., 8.1 Quartile deviation (Semi-interquartile range), The marks obtained by grade 9 students in mathematics are given belowM, Marks, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80, 80-90, Obtained, No. of Students 4, 5, 10, 15, 8, 7, 1, What are the values of quartiles from the above data? What do these values, represents?, The values of quartiles means first quartiles (Q1), second quartiles (Q2) and third, quartile (Q3). To find the quartiles of continuous data, the data are tabulating below, in ascending order:, , 254
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Marks Obtained, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80, 80-90, , No. of Students (f), 4, 5, 10, 15, 8, 7, 1, ∑f = N = 50, , Cumulative frequency ( C f), 4, 9, 19, 34, 42, 49, 50, , Total number of students (N) =50, 𝑁 𝑡ℎ, , 50 𝑡ℎ, , The position of 1st quartile = ( 4 ) item = ( 4 ) item = 12.5𝑡ℎ item, In c.f column 19 is just greater than 12.5. So its corresponding class is 40-50. To find, Q1, the formula is, 𝑁, − 𝐶𝑓, 𝑄1 = 𝑙 + 4, × 𝑖 … … … … … … … … … (𝑖), 𝑓, Where, l = lower limit of the Q1 Class, C.f = cumulative frequency of just before Q1 class, f = frequency of the Q1 class, i = width of class interval, From the above table l = 40, f = 10, C.f =9, i = 10, From equation (i),, 𝑄1 = 40 +, , 12.5 − 9, × 10 = 40 + 3.5 = 43.5, 10, 𝑁 𝑡ℎ, , Again, the position of third quartile = 3 ( 4 ) item = 3 × 12.5𝑡ℎ item = 37.5𝑡ℎ item, In C.f Column, 42 is just greater than 37.5 so its corresponding class is 60-70. The, formula to find the Q3 value of Q3 is, 𝑄3 = 𝑙 +, , 𝑁, 3 ( 4 ) − 𝐶𝑓, 𝑓, , × 𝑖 … … … … … . . (𝑖𝑖𝑖), , 255
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Where. l = lower limit of the Q3 Class, Cf = cumulative frequency of just before Q3 class, f = frequency of the Q3 class, i = width of class interval, From the above table, l = 60, C.f = 34, f = 8, and i = 10, from the equation (iii),, 𝑄3 = 60 +, , 37.5 − 34, 3.5 × 10, × 10 = 60 +, = 64.375, 8, 8, , The quartiles of the given data are Q1 = 43.5 and Q3 = 64.375. Half of the difference, between upper quartile (Q3) and lover quartile (Q1) is called quartile deviation. It is, invented by Galton., ∴ Quartile deviation(QD) =, =, , Q 3 − Q1, 2, , 64.375 − 43.5 20.875, =, = 10.437, 2, 2, , Again, the relative measure based on lower and upper quartile known as coefficient, of quartile deviation. It is given by, Coefficient of Quartile deviation =, , 𝑄3 −𝑄1, 𝑄3 +𝑄1, , 64.375−43.5, , 20.875, , = 64.375+43.5 = 107.875 = 0.193, , Hence, Coefficient of QD = 0.193, Can we show the relationship of Q1, Q2, Q3 by graph? Discuss on it., , 256
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𝑁 𝑡ℎ, 4, , - The lower quartile (Q1) means, the value of ( ) item or the value of the 25%, of the total frequency of the distribution., 𝑁 𝑡ℎ, , - The middle quartile (2nd quartile) (Q2) means the value of 2 ( 4 ) item or the, value of 50% of the total frequency of the distribution., 𝑁 𝑡ℎ, , - The upper quartile (Q3) means the value of 3 ( 4 ) item or the value of 75% of, the total frequency of the distribution., Semi-interquartile range or Quartile deviation =, , 𝑄3 −𝑄1, 2, , 𝑄 −𝑄, , and coefficient of QD = 𝑄3 +𝑄1, 3, , 1, , Example 1, Find the quartile deviation and the coefficient of quartile deviation from the following, data:, Marks, No. of Students, , 20-30, 4, , 30-40, 5, , 40-50, 2, , 50-60, 4, , 60-70, 3, , 70-80, 2, , Solution: Here,, Tabulating the given data in ascending order for the calculation of QD., Marks, , No. of Students (f), , Cumulative frequency (Cf), , 20-30, , 4, , 4, , 30-40, , 5, , 9, , 40-50, , 2, , 11, , 50-60, , 4, , 15, , 60-70, , 3, , 18, , 70-80, , 2, , 20, , ∑f = N = 20, 𝑁 𝑡ℎ, , 20 𝑡ℎ, , The position of first quartile = ( 4 ) item = ( 4 ) item = 5𝑡ℎ item, In Cf column, 9 is just greater than 5 so its corresponding class is 30-40., l = 30, f = 5, Cf = 4, i = 10, , 257
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By formula,, 𝑁, − 𝐶𝑓, 5−4, 𝑄1 = 𝑙 + 4, × 𝑖 = 30 +, × 10 = 30 + 2 = 32, 𝑓, 5, 𝑁 𝑡ℎ, 4, , Again, the position of third quartile = 3 ( ) item = 3 × 5𝑡ℎ item = 15𝑡ℎ item, In Cf Column, 15th item corresponding class is 50-60., l = 50, f = 4, Cf = 11, i = 10, Now,, 3𝑁, − 𝐶𝑓, 15 − 11, 𝑄3 = 𝑙 + 4, × 𝑖 = 50 +, × 10 = 50 + 10 = 60, 𝑓, 4, Now, quartile deviation (QD) =, 𝑄 −𝑄, , 𝑄3 −𝑄1, 2, , =, , 60−32, 2, , 60−32, , =, , 28, 2, , = 14, , 28, , And Coefficient of QD = 𝑄3 +𝑄1 = 60+32 = 92 = 0.304, 3, , 1, , Exercise 8.1, 1., , 2., , 3., , a., , What is dispersion?, , b., , Write the various measures of dispersion., , a., , Define quartile deviation and write the formula to calculate quartile deviation., , b., , What do you mean by coefficient of quartile deviation?, , c., , Write the difference between quartile deviation and the coefficient of quartile, deviation., , a., , In a Continuous data, the first quartile and third quartile are 40 and 60, respectively, find the quartile deviation., , b., , In a continuous series, the lower quartile is 25 and its quartile deviation is 10,, find the upper quartile., , 258
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4., , a., , 1, , In a continuous series, the coefficient of quartile deviation is and its upper, 2, , quartile is 60, find its first quartile., b., , In a continuous series, the coefficient of quartile deviation is, , 3, 7, , and its first, , quartile is 20, find its upper quartile., 5., , Find the quartile deviation and its coefficient from the following data:, a., , Marks obtained, No. of Students, , 60-65, 7, , b. Weight (kg), No. of persons, c., , Class interval, Frequency, , d., , Size, Frequency, , e., , Height (in Inches), No. of students, , f, , 6., , 20-30, 5, , 4-8, 6, , Expenditure, No. of Workers, , 8-12, 10, , 65-70, 5, , 70-75, 8, , 75-80, 4, , 80-85, 3, , 30-40, 15, , 40-50, 10, , 50-60, 8, , 60-70, 6, , 20-30, 8, , 30-40, 16, , 40-50, 4, , 50-60, 4, , 12-16, 18, , 60-62, 4, , 16-20, 30, , 62-64, 6, , 20-24, 15, , 64-66, 8, , 24-28, 12, , 66-68, 12, , 85-90, 3, 70-80, 2, 60-70, 3, , 28-32, 10, , 68-70, 7, , 32-36 36-40, 6, 5, , 70-72, 2, , 0 ≤ x < 10, , 10 ≤ x < 20, , 20 ≤ x < 30, , 30 ≤ x < 40, , 40 ≤ x < 50, , 5, , 15, , 10, , 8, , 6, , The following are the marks obtained by class 9 students in their internal, examination. Taking class interval of (10-20) as first class, prepare a frequency, distribution table and find the quartile deviation. Also find its coefficient from the, following data:, a. 22, 25, 46, 34, 57, 69, 44, 36, 12, 27, 50, 36, 35, 62, 46, 52, 54, 61, 66, 55, 29,, 39, 40, 33, 14, 41, 25, 20, 16., b. 21, 45, 60, 57, 15, 41, 48, 50, 34, 29, 56, 40, 14, 62, 28, 70, 22, 30, 38, 74, 13,, 47, 20, 53, 64, 34, 75, 66., , 259
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8.2 Mean deviation, We know that range depends on the largest and smallest value of the distribution., Quartile deviation depends on first quartile and third quartile. They are not based on all, the observations and they do not measure the scatteredness of the items from the, average value. Thus, they are not considered as good measure of dispersion. But mean, deviation measures the variation of each observation of the total distribution from the, average., Mean deviation is defined as the average of the absolute values of the deviation of each, item from mean, median or mode. It is also known as average deviation. Mean deviation, calculated from mean is called mean deviation from mean. Similarly mean deviation from, median is known as mean deviation from median. Mean deviation from mean is, considered more reliable., Mean deviation of continuous series, Let, m1, m2, m3, ..., mn be the middle values of the corresponding classes x1, x2, x3, ..., xn, and f1 , f2 , f3 , ..., fn be their respective frequencies., i) Mean deviation from mean =, ̅,, Where, D = m − X, , ∑𝑓|𝐷|, 𝑁, , ̅ = mean of the class, m = mid value, X, , f = frequency of the corresponding term, N = ∑f = total of the frequency, ii) Mean deviation from median =, , ∑𝑓|𝐷|, 𝑁, , Where, D = m − Md , Md = median value of the class, , N = ∑f = total of the frequency, , Mean deviation is an absolute measure. So to compare two or more series having, different units, the relative measure corresponding to mean deviation is used, which is, called coefficient of mean deviation. Coefficient of mean deviation is calculated from, mean and median., M.D from mean, ̅), Mean (X, , i), , Coefficient of MD from mean =, , ii), , Coefficient of MD from median =, , M.D from median, Median (Md), , Steps to compute mean deviation, i., ii., iii., iv., , Calculate the value of mean or median., Take deviations from the mean or median of each term., Ignore the negative signs of the deviation., Apply the formula to get average deviation from mean or median., , 260
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Example 1, Find the mean deviation from mean and also calculate its coefficient:, Marks, No. of Students, , 0-10, 3, , 10-20, 5, , 20-30, 7, , 30-40, 3, , 40-50, 4, , Solution: Here,, Tabulating the given data in ascending order for the calculation of mean deviation and it, coefficient from mean, No. of, ̅, Mid value, 𝑚−X, Marks, Students, fm, |D|, F|D|, (m), =D, (f), 0-10, 3, 5, 15, -20, 20, 60, 10-20, 5, 15, 75, -10, 10, 50, 20-30, 7, 25, 175, 0, 0, 0, 30-40, 3, 35, 105, 10, 10, 30, 40-50, 4, 45, 180, 20, 20, 80, ∑𝑓 = 𝑁, ∑𝑓𝑚, ∑𝑓|𝐷|, = 22, = 550, = 220, ∑𝑓𝑚 550, =, = 25, 𝑁, 22, ∑f|D|, 220, Now, mean deviation from mean = N = 22 = 10, ̅) =, Mean (X, , Again, Coefficient of MD from mean =, , MD, ̅, X, , =, , 10, 25, , = 0.4, , Example 2, Calculate the mean deviation from median and its coefficient from the following data:, Weight (in kg), No. of People, , 10-20, 6, , 20-30, 8, , 30-40, 11, , 40-50, 14, , 50-60, 8, , 60-70, 3, , Solution: Here,, Tabulating the given data in ascending order for the calculation of mean deviation and its, coefficient from median., No. of, Mid-value m – md =, Weight, Cf, |D|, f|D|, People (f), (m), D, 10-20, 6, 6, 15, -25, 25, 150, 20-30, 8, 14, 25, -15, 15, 120, 30-40, 11, 25, 35, -5, 5, 55, , 261
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40-50, 50-60, 60-70, , 14, 8, 3, ∑𝑓 = 𝑁, = 50, , 39, 47, 50, , 45, 55, 65, , 5, 15, 25, , 𝑁 𝑡ℎ, , 5, 15, 25, , 70, 120, 75, ∑𝑓|𝐷|, = 590, , 50 𝑡ℎ, , Now, the position of median = ( 2 ) item = ( 2 ) item = 25𝑡ℎ item, In Cf column, the corresponding class of 25th item is 30-40, median class is 30-40, Now, l = 30,, f = 11, Cf = 14,, i = 10, N, − Cf, 25 − 14, 11, ∴ Median (Md) = 𝑙 + 2, × i = 30 +, × 10 = 30 +, × 10 = 40, f, 11, 11, Now, mean deviation from median =, Then coefficient of MD =, , MD, Md, , =, , 11.8, 40, , ∑𝑓|𝐷|, 𝑁, , =, , 590, 50, , = 11.8, , = 0.295, , Exercise 8.2, 1., , a., , Define mean deviation., , b., , What do you mean by coefficient of mean deviation?, , 2. a., , 3., , 4., , 5., , Write the formula to find the mean deviation (MD) from mean., , b., , Write the formula to calculate mean deviation and its coefficient from median., , c., , What are the methods to find the mean deviation and its coefficient? Which, one is the best? Write with reason., , In a continuous series, ∑𝑓m = 1000, N = 50 and ∑𝑓|𝑚 − ̅, X| = 308 then find, the mean deviation from mean and its coefficient., ̅| = 680, mean deviation (MD) =17 find ∑𝑓., b., In a continuous series,∑𝑓|𝑚 − X, a., In a continuous series, median (Md) = 40, ∑𝑓 = 50 and ∑𝑓|m − Md| = 530, then find the mean deviation from median and its coefficient., b., In a continuous series, coefficient of mean deviation is 0.5 and median = 40, then find the mean deviation (MD)., Find the mean deviation from mean and also calculate its coefficient from the, following data:, a., , a., , Marks, No. of Students, , 0-10, 5, , 10-20, 8, , 262, , 20-30, 15, , 30-40, 16, , 40-50, 6
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b. Ages (in yrs.), No. of Students, , 6., , c., , Class interval, No. of Students, , d., , Daily, Income (Rs.), No. of, Workers, , 4-8, 7, , 0-10, 7, , 8-12, 10, , 10-20, 12, , 12-16, 15, , 20-30, 18, , 30-40, 28, , 16-20, 7, 40-50, 16, , 20-24, 6, 50-60, 14, , b., , 100-200, , 200-300, , 300-400, , 400-500, , 500-600, , 600-700, , 700-800, , 4, , 6, , 10, , 20, , 10, , 6, , 4, , Weight (in kg), No. of Students, Marks, No. of Students, , 20-25, 7, 0-10, 5, , 25-30, 3, , 10-20, 3, , 20-30, 7, , 30-35, 6, 30-40, 5, , 35-40, 4, , 40-45, 8, , 45-50, 2, , 40-50, 10, , 50-60, 3, , 60-70, 2, , The following distribution gives the marks of 30 students in a certain examination:, Marks, 20-30, 30-40, 40-50, 50-60, 60-70, No. of Students, 7, 12, 18, 28, 16, Find the mean deviation from median and also find its coefficient., , 8., , 9., , 60-70, 8, , Find the mean deviation from median and also calculate its coefficient from the, following data:, a., , 7., , 0-4, 7, , 70-80, 14, , Find the mean deviation from mean and its coefficient from the following distribution:, Class interval 5 ≤ x < 10 10 ≤ x < 15 15 ≤ x < 20 20 ≤ x < 25 25 ≤ x < 30, Frequency, 7, 4, 5, 6, 3, In a survey of 28 students, the following are the marks obtained in a certain, examination:, 48, , 56, , 28, , 38, , 20, , 75, , 45, , 50, , 40, , 70, , 74, , 53, , 66, , 57, , 34, , 14, , 22, , 13, , 64, , 60, , 15, , 29, , 62, , 30, , 47, , 34, , 21, , 41, , 263
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Construct a frequency distribution table taking 10-20 as first class interval. Then, calculate i) Mean deviation from mean and its coefficient ii) Mean deviation from, median and its coefficient., 10. a., Find the mean deviation from the mean of the following frequency, distribution. Also, find its coefficient. The people having ages less than 70 years, and above 10 years are participating in a survey:, Ages, No. of People, , b., , Less than 20, , 3, , Less than 30, , 7, , Less than 40, , 15, , Less than 50, , 20, , Less than 60, , 26, , 30, Less than 70, The students who obtained marks more than 0 and less than 48 are, participating in a survey. The following frequency distribution is the marks, obtained by 50 students in a certain class examination:, Marks, No. of Students, , i., ii., , More than 0, , 50, , More than 8, , 42, , More than 16, , 35, , More than 24, , 30, , More than 32, , 18, , More than 40, , 6, , Find the mean deviation from mean and its coefficient., Find the mean deviation from median and its coefficient., , 264
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8.3 Standard deviation, We can find the dispersion of above data from various methods. But, Standard deviation, is the most popular and useful measure of dispersion used in practice. It removes the, drawback presented in other measure of dispersion. It gives the uniform, correct and, stable result., A Standard deviation (SD) is defined as the positive square root of mean of the square of, deviation from the arithmetic mean. It is also known as root mean square deviation. It is, denoted by Greek letter ‘σ ’ (read as sigma). It was first introduced by Karl Pearson in, 1823., Standard deviation is the best measure of dispersion because, i., , Its value is based on all the observation., , ii., , The deviation of each item is taken from mean., , iii., , All algebraic sign are also considered., , The smaller value of Standard deviation (SD) means a high degree of uniformity of the, observation as well as homogeneity of the series., Calculation of Standard deviation from continuous series, We can find the standard deviation for continuous series by following methods:, a., b., c., d., , Actual mean method, Assumed mean method (Short cut method), Step deviation method, Direct method, , a. Actual mean method, Let, 𝑚1 , 𝑚2 , 𝑚3 , … , 𝑚𝑛 be the mid values of the corresponding classes 𝑋1 , 𝑋2 , 𝑋3 , … , 𝑋𝑛, and 𝑓1 , 𝑓2 , 𝑓3 , … … … 𝑓𝑛 be their respective frequencies and 𝑋̅ be the actual mean then, ∑𝑓(𝑚 − 𝑋̅)2, ∑𝑓𝑑2, Standard deviation (𝜎) = √, =√, 𝑁, 𝑁, where d = 𝑚 − 𝑋̅ and N = ∑𝑓, , S, , 265
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Steps to be used in Actual mean method:, i., ii., , Find the midpoint of each class interval, ∑𝑓𝑚, Find actual mean (𝑋̅) by formula 𝑋̅ =, , iii., iv., v., vi., vii., , Find the difference between each mid value and mean (𝑋̅) i.e. 𝑑 = (𝑚 − 𝑋̅), Find the square of d i.e d2, Multiply each value of d2 by corresponding ‘f’ and denoted by fd2., Find ∑fd2, Use formula,, , 𝑁, , SD(𝜎) = √, , ∑𝑓𝑑2, 𝑁, , b. Assumed mean method (Short cut method), This method is also called short cut method. To calculate Standard deviation, actual mean, method is difficult and takes more time. In assumed mean method, by supposing a, number as mean we will calculate standard deviation., Let, 𝑚1 , 𝑚2 , 𝑚3 , … … … … 𝑚𝑛 be the mid value of the class intervals, 𝑓1 , 𝑓2 , 𝑓3 … … … 𝑓𝑛 be, their corresponding frequencies and ‘A’ be the assumed mean then., Standard deviation (SD) or (σ) = √, , ∑𝑓𝑑2, ∑𝑓𝑑 2, −(, ), 𝑁, 𝑁, , where d = m − A, , Steps to be used in assumed mean method:, i. Consider as assumed mean ‘A’ from mid value., ii. Take deviation of each item from ‘A’ and denoted it by d., iii. Multiply each deviation ‘d’ by the corresponding frequency f and denoted by them, by fd. Find ∑fd., iv. Find the square of each value of d and denote it by d2, v. Multiply each values of d2 by the corresponding frequency (f) and denote them by fd2., vi. Find ∑fd2 ., vii. Use formula, SD (σ) = √, , ∑𝑓𝑑2, ∑𝑓𝑑 2, −(, ), N, N, , 266
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Example 1, Find the standard deviation and its coefficient from the following data:, Marks obtained, No. of students, , 0-10, 3, , 10-20, 5, , 20-30, 4, , 30-40, 3, , 40-50, 2, , 50-60, 3, , By, , i. Actual mean method, , ii. Direct method, , iii. Assumed mean method, , iv. Step-deviation method, , Solution: Here,, Tabulating the given data in ascending order for the calculation of SD and its coefficient, i., , By actual mean method, Marks, , No. of, students (f), , 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, , 3, 5, 4, 3, 2, 3, ∑𝑓 = 𝑁, = 20, , Mid, value, (m), 5, 15, 25, 35, 45, 55, , fm, , 𝑑 = 𝑚 − 𝑋̅, , d2, , fd2, , 15, 75, 100, 105, 90, 165, , -22.5, -12.5, -2.5, 7.5, 17.5, 27.5, , 506.25, 156.25, 6.25, 56.25, 306.25, 756.25, , 1518.75, 781.25, 25, 168.75, 612.50, 2268.75, , ∑𝑓𝑚 = 550, , Now,, ̅) =, mean(X, , ∑𝑓𝑚 550, =, = 27.5, 𝑁, 20, , By formula,, ∑𝑓𝑑2, 5375, SD (𝜎) = √, =√, = √268.75 = 16.39, 𝑁, 20, , 268, , ∑𝑓𝑑 2 = 5375
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ii., , By direct method, No. of, students (f), 3, 5, 4, 3, 2, 3, , Marks, 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, , Mid, value (m), 5, 15, 25, 35, 45, 55, , ∑𝑓 = 𝑁 = 20, , fm, , m2, , fm2, , 15, 75, 100, 105, 90, 165, , 25, 225, 625, 1225, 2025, 3025, , 75, 1125, 2500, 3675, 4050, 9075, ∑𝑓𝑚2 = 20500, , ∑𝑓𝑚 = 550, , By formula,, ∑𝑓𝑚2, ∑𝑓𝑚 2, 20500, 550 2, SD (𝜎) = √, −(, ) =√, −(, ) = √1025 − 756.25, 𝑁, 𝑁, 20, 20, = √268.75, ∴ 𝜎 = 16.39, iii. By assumed mean method, Marks, 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, , No. of, students (f), 3, 5, 4, 3, 2, 3, ∑𝑓 = 𝑁, = 20, , Mid, value, (m), 5, 15, 25, 35, 45, 55, , d=m–A, , fd, , d2, , fd2, , -30, -20, -10, 0, 10, 20, , -90, -100, -40, 0, 20, 60, , 900, 400, 100, 0, 100, 400, , 2700, 2000, 400, 0, 200, 1200, , ∑𝑓𝑑 = −150, , ∑𝑓𝑑2 = 6500, , Now,, Let, assumed mean (A) = 35, By formula,, ∑𝑓𝑑2, ∑𝑓𝑑 2, 6500, 150 2, √, √, SD (𝜎) =, −(, ) =, − (−, ) = √325 − 56.25 = √268.75, 𝑁, 𝑁, 200, 20, ∴ 𝜎 = 16.69, , 269
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iv. By step deviation method, Marks, 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, , No. of, students (f), 3, 5, 4, 3, 2, 3, ∑𝑓 = 𝑁, = 20, , Mid value, (m), 5, 15, 25, 35, 45, 55, , d=m–, A, -30, -20, -10, 0, 10, 20, , 𝑑′ =, , 𝑑, 𝑖, , -3, -2, -1, 0, 1, 2, , fd', , d'2, , fd'2, , -9, -10, -4, 0, 2, 6, ∑𝑓𝑑′, = −15, , 9, 4, 1, 0, 1, 4, , 27, 20, 4, 0, 2, 12, ∑𝑓𝑑′2, = 65, , Let, assumed mean (A) = 35, class interval (i) = 10, By formula,, ∑𝑓𝑑′2, ∑𝑓𝑑′ 2, 65, 15 2, SD (𝜎) = √, −(, ) × 𝑖 = √ − (− ) × 10, 𝑁, 𝑁, 20, 20, = √3.25 − 0.5625 × 10 = √2.6975 × 10 = 1.6393 × 10 = 16.39, And for the coefficient of standard deviation,, ̅) = A +, Actual mean (X, the coefficient of SD =, , ∑fd′, 15, × i = 35 + (− ) × 10 = 35 − 7.5 = 27.5, N, 20, , σ 16.39, =, = 0.596, ̅, 27.5, X, , Note:, ̅) = A +, 1. Actual mean (X, , ∑fd, N, , ̅) = A +, 2. Actual mean (X, , ∑fd, ×, N, , i, , 270
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Coefficient of variation, It is purely a number and independent of unit. If the coefficient of standard deviation is, multiplied by 100, it is known as coefficient of variation. Thus coefficient of variation, denoted by C.V is defined by, standard deviation, × 100%, mean, σ, or, CV = × 100%, ̅, X, CV =, , Greater the coefficient of variation, greater will be variation and less will be the, consistency or uniformity. Less the coefficient of variation, greater will be the consistency, or uniformity or more homogenous or more stable. Hence, to compare the two given, distribution, we use the coefficient of variation., Example 1, The following table gives the marks obtained by 20 students in a certain examination:, Marks obtained, No. of students, , 30-40, 2, , 40-50, 3, , 50-60, 6, , 60-70, 5, , 70-80, 4, , Find the arithmetic mean, standard deviation, coefficient of standard deviation and, coefficient of variation., Solution: Here,, Tabulating the given data in ascending order for the calculation of standard deviation and, its coefficient., Marks, obtained, 30-40, 40-50, 50-60, 60-70, 70-80, , i., , No. of, students (f), 2, 3, 6, 5, 4, ∑𝑓 = 𝑁, = 20, , mean (𝑋̅) =, , ∑𝑓𝑚, 𝑁, , =, , Mid, value (m), 35, 45, 55, 65, 75, , 1160, 20, , fm, 70, 135, 330, 325, 300, ∑𝑓𝑚, = 1160, , = 58, , 271, , 𝑑, = 𝑚 − 𝑋̅, -23, -13, -3, 7, 17, , d2, , fd2, , 529, 169, 9, 49, 289, , 1058, 507, 54, 245, 1156, ∑𝑓𝑑 2, = 3020
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∑𝑓𝑑 2, 𝑁, , 3020, 20, , ii., , Standard deviation (𝜎) = √, , iii., , Coefficient of SD = 𝑋̅ =, , iv., , Coefficient of variation (CV) = 𝑋̅ × 100% = 0.21 × 100% = 21%, , 𝜎, , 12.29, 58, , =√, , = √151 = 12.29, , = 0.21, 𝜎, , Example 2, An analysis of monthly wages paid to the works in two firm A and B belonging to the same, industry given the following results:, Average monthly wage, Standard deviation of distribution of wage, , A, Rs. 586, Rs. 9, , B, Rs. 575, Rs. 10, , i., , Examine which firm A or B has greater variability in wage distribution, , ii., , Which firm has more homogeneity? Give Reason., , Solution: Here,, For Firm A, ̅, X = Rs. 586, , σ = Rs. 9, , C. V =?, , We have,, σ, Rs. 9, CV = × 100% =, × 100% = 1.53%, ̅, Rs. 586, X, For Firm B, ̅, X = Rs. 575,, σ = Rs. 10, C. V =?, σ, Rs. 10, CV = × 100% =, × 100% = 1.73%, ̅, Rs. 575, X, i. Since, the CV for the firm B is higher, so the variability of wage distribution is firm B, is greater., ii. The CV for the firm A is less then B, so the firm A has more homogeneity., Note:, more uniformity, Less cv, , more consistency, , more variability, , more stability, , more CV, , more homogenity, , more scatterness, , 272
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Exercise 8.3, 1., , a., , Define standard deviation., , b., , What is coefficient of standard deviation? Also, write its formula., , c., , Define coefficient of variation. Also, write its formula., , 2. a., , 3., , b., , Write the difference between coefficient of standard deviation and the, coefficient of variation., , a., , In a continuous series, ∑𝑓(𝑚 − 𝑋̅)2 = 484, N = 24 𝑎𝑛𝑑 ̅, X = 25 find the, standard deviation and its coefficient., , b., , In a continuous series ∑𝑓𝑑 = 0, ∑𝑓𝑑2 = 848, 𝑁 = 100, assumed mean, (A) = 12 then find the standard deviation and its coefficient., , 4. a., b., 5., , 6., , Write the difference between standard deviation and mean deviation., , If ∑𝑓𝑑′2 = 125, ∑𝑓𝑑′ = −4, 𝑁 = 20, 𝐶 = 10, find SD (σ), What is the meaning a low coefficient of variation?, , Find the standard deviation and its coefficient from the following data:, a., , Class interval, Frequency, , b., , Marks obtained, No. of students, , c., , Daily Wages (Rs.) 100-125, No. of workers, 75, , a., , 0-4, 7, , 4-8, 7, , 8-12, 10, , 12-16, 15, , 16-20, 7, , 20-24, 6, , 0-10, 8, , 10-20, 12, , 20-30, 20, , 30-40, 40, , 40-50, 12, , 50-60, 8, , 125-150, 57, , 150-175, 81, , 175-200 200-225, 19, 12, , The following table gives the marks obtained by 40 students of class 10 in a, unit test of Opt. Mathematics. Calculate the standard deviation and coefficient, of variation., Marks obtained, No. of Students, , 30-40, 4, , 40-50, 8, , 273, , 50-60, 10, , 60-70, 16, , 70-80, 6, , 80-90, 6
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b., , Calculate the coefficient of standard deviation and coefficient of variation from, the following data:, Marks obtained, No. of students, , 7., , a., , 0-10, 10, , 10-20, 15, , 20-30, 25, , 30-40, 25, , 50-60, 10, , 60-70, 5, , The following table gives per day salary of 50 high school teachers:, Salary, , 10001100, , 11001200, , 12001300, , No. of, 10, 15, 5, teachers, i., Find the average salary of teachers., , b., , 40-50, 10, , 13001400, , 14001500, , 15001600, , 8, , 10, , 2, , ii. Find the standard deviation of the above data., iii. Find the coefficient of variation of the data., The following distribution gives the marks of 30 students in a certain, examination:, Marks, 20-30, 30-40, 40-50, 50-60, 60-70, No. of students, 3, 5, 6, 8, 4, Find the standard deviation and coefficient of variation by, , 8., , a., , i., , Actual mean method, , ii., , Assumed mean method, , iii., , Step deviation method, , 70-80, 4, , A purchasing agent obtained samples of incandescent lamps from two, suppliers. He had the samples tested in his own laboratory for length of life, with the following results:, Lengths of life (hrs), Sample, A, of, B, company, , 700-900, , 900-1100, , 1100-1300, , 1300-1500, , 10, , 16, , 26, , 8, , 3, , 42, , 12, , 3, , i. Which company’s bulb gives a higher average life?, ii. Find the standard deviation of both companies., iii. Which company’s lamps are more uniform?, , 274
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b., , An analysis of marks obtained by boys students and girls students are given, below:, No. of students, , Boys, , Girls, , Average marks, , 63, , 54, , Standard deviation of marks, , 9, , 6, , i. Which group has greater variability in marks obtained?, ii. Which group has more uniform marks obtained?, 9., , The weight (in kg) of 20 people are given below. Construct a frequency distribution, table taking (20-30) as the first class interval. Also, find the standard deviation and, the coefficient of variations., , 59,, 71,, 45,, 44,, 35,, 21,, 29,, 49,, 42,, 37,, 58,, 69,, 55,, 39,, 79,, 50,, 65,, 52,, 60,, 64, 10. Construct a frequency distribution table taking (0-10) as the first class interval. Then,, find the standard deviation and its coefficient., 25,, 24,, 45,, 28,, 33,, 10,, 20,, 5,, 11,, 30,, 25,, 40,, 15,, 31,, 2,, 29,, 23,, 17,, 14,, 26,, 30,, 13,, 41,, 35,, 7, 11. Collect the data related to your locality and find coefficient of variation of the data., Present your finding in your class., , 275
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Answer, Note: Remaining answers: show to your subject teacher, Exercise 1.1.1(a), 2.(a) constant function (b) Identify function, (e) cubic function, , (f) quadratic function, , 4.(a) y = 4x – 130, , (b) 118 pounds, , (c) cubic function (d) quadratic function, , Exercise 1.1.1. (b), 1(a) +1, -1, , (b) +1, -1, , (c) No maximum, no minimum values, , 2(a) 3, , b() 2, , (c) , , Exercise 1.1.2, 2.(a) (i) {3, 5, 9} (ii) {4, 6, 10}, , (iii) does not exists, , (iv) {16, 36, 100}, , (b) (i) {1, 2, 3}, , (iii) does not exist, , (iv) {3, 4, 5}, , (ii) {2, 3 4}, , 3.(a) 2x2 – 4x + 7, 4x2 – 24x + 41 (b) 37, 9, , (c) 30, 21, , 4.(a) 2x and 2x (b) 2(x2 – 2x + 3)3 + 1, 4(x2 + 2)2 – 16(x2 + 2) + 18, 8(a) 3t (b) r2 (c) (Ar)(t) = 9r2, Exercise 1.1.3, 2(a) f—1 = {(2,1), (3,2), (5,4)}, , (b) g—1 = {(4, 1), (5, 2), (6, 3)}, , (c) h—1 = {(4, -2), (9, -3) (36, -6)}, 1, 3, , 3(a) (i) (x + 5) (ii), 1, 2, , 5, 3, , (iii) −, , 11, 2, , (iii), , (b) (i) (x + 5), , (ii), , (c) (i) 2x – 1, , (ii) 11, , (d (i), , 2(1+𝑥), (1−𝑥), 1, , 4.(a) 2(x +2), , (ii), , 14, 5, , (b) 2(x-1), , 1, 4, , (iv) -1, , 9, 8, , (iv), 1, , (iii) − 2, (iii), , 3, 2, , (iv) -5, , 10, 3, , (iv), , 5, , 5.(a) 2, , −2, 3, , (b) -11, 5, , 276, , (d) 55, 201
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1, , 6., , Volume: f(x) =, , 4, , x3, , 3, , ,f, , 3𝑥 3, (x) = ( ), 4𝜋, , —1, , 𝑥, , Surface area: f(x) = 4px2, f–1(x) = ±√4𝜋, Exercise 1.2.1, 2. (a) x – 7; 0, , (b) x2 + x – 6; 0, , 3.(a) x2 + x – 6; -1, (c) 2x3 + x2 +, , 11, 𝑥, 2, , 4(a) yes, , (c) x2 + 2x + 4; 0, , (d) x2 + 6x + 9; 0, , (b) x2 – 15x + 91; -429, 1 −57, 8, , (d) y4 + 3y3 + 10y2 + 30y + 89; 267, , - 4;, , (b) yes (c) No, , (d) No (e) No, , Exercise 1.2.2, 2.(a) quotient = x2 – 2x + 4, remainder = 0, , (b) quotient = 2x3 + x2 – 3x + 10, , (c) quotient = 4x2 – 11x + 23, remainder = -37, , (d) quotient = 2x2 + x – 3, remainder = 1, , (e) quotient = 8x2 + 8x + 10, remainder = -2, Exercise 1.2.3(A), 2(a) – 7, , (b) 4, , 3(a) – 12, , (b), , 11, 5, , (c) 3, (c), , −6, 7, , (d), , 645, 32, , (d), , 31, 2, , −283, 8, , (e) 456, , (f), , (d) No, , (e) No, , (g), , 680, 81, , Exercise 1.2.3 (B), 2(a) yes, , (b) No, , 3(a) 11 (b) -4, , (c), , (c) No, , −13, 2, , 4.(a) (x – 1) (x + 2) (2x + 1), , (b) (x – 1) (x + 1) (x + 2), , (c) (y + 1) (y – 2) (y – 5), , (d) (x + 1) (x + 10) (x +2), , (e) (x – 1) (x + 1) (2x + 1), , (f) (x – 1) (x – 10) (x – 12), , (g) (x + 2) ( 2x – 3) (x + 6), 5.(a) 1, 5, -2, , (b) -3, 2, 1, , (c) -1, 1,, , 1, , (d) -1, -1,5, , 3, , 277, , (e) -3, 2, 4 (f) 1, 1, 6
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Exercise 1.3.1, 3.a) -1, -2, 1 – 2n, -9, -11, , (b) 2, 4, 4x – 2, 18, 22, , (c) 7, 10, 10n – 3, 47, 57, , (d) ,, , 4.(a) tn = a + (n – 1) d, , (b) yes, common difference is constant, , 5.a) 13 (b) 112, , 5 −1 6−𝑛 1, 4, , 4, , 6.(a) 11, , ,, , 4, , , 4, 0, , (b) -5, , 7.(a) 6, , (c) first term, (b) – 2, , 8.(a) 6, , (b) 5, , 9.(a) 23, , (b) 70, , 10.(a) 312, , 11.(a) -10, , (b) 2, , 12(a) 34, , (b) 27, , (c) 11, , 13.(a) yes, , (b) No 14.(a) 2, , (b) b =, , 15, ,, 2, , (b) 4, , 60, 43, 22, , 15)(a) (i) -32, 7 (ii) 178 (iii) -32, -25, -18, ….., (b) (i) 8, -2, , (ii) 5, , 16. a) 46, , (iii) No, because, no of terms is zero, (b) 78, , 19) -13, -8, -3, , 18) a) 13, , 20.(a) 2030, , (b) 22, , (b) 9 weeks, , Exercise 1.3.2, 1.a) no of mean, 3.(a) 2, , (b) last term, (b) 5, , 4.(a) 8, , 2.(a) 0, b) t6, , 5.(a) 110, , 6.a) 15, 20, (b) -8, -3, 2, (b) 70, 40, , 7.(b) 40, , 9.(a) 20, 30, 40, 50, 60, , (b) 15, 19, 23, 27, 31, 35, , 10.(a) 5, 3:4, , (b) a, , (c) 16, 24, , (b) n = 5, 13, 17, 25, 29, , 11.(a) 17, , (b) 15, 8.(a) 10, 16, , (b) 3, 31, , Exercise 1.3.3, 𝑛, , 1.(a) Sn = 2 [2a + (n – 1)d], (c), , 2093, 2, , (d) -8930, −35, 3, , 5.(a) 380, , (b), , 7.(a) 1089, , (b) 945 (c) 625, , (b) 8, , 2(a) 25 (b) 30, , 3.(a) -180, , 4.(a) 120, , (b) 147, , 6.(a) d = 6, , (b) 7, , 8.(a) 6, , 278, , (b) 8, , (b) 670, , 9.(a) n = 10 or 11
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(b)(i) 10, 4, , (ii) 10 + 14 + 18 + …., , (iii) 960, , 10)(a) (i) 17, 3 (ii) 1325, (b) (i) 1, 2, , (ii) 1 + 3 + 5 + …, , (iii) 400, , 11.(a) 6, 10, 14 or 14, 10, 6, , (b) 2, 4, 6 or 6, 4 2, , 12.(i) Rs. 160, , (iii) 120, , (ii) 140, , (iv) 40, , Exercise 1.3.4, 1, 2 and 3 consult with your teacher, 4.(a) 3 (b) 64, 7.(a) (i), (b)(i), , −2, ,, 3, , −1, ,, 5, , (5) (a) 2, , -81, , -3750, , 8.(a) 6561, , (b) 4, , 3, , (c) 6, , 6.(a) -8 (b) 2, 2, −512, 243, , (ii) -81, 54, -36, ………, , (iii), , (ii) -3750, 750, -150, …., , (iii) 125, , 1, , (b) 6561, , (c) ±2, , 6, , 3, , 19683, , 9.(ii) r = 4, , (iii) 262144, , Exercise 1.3.5, 1.(b) √𝑝𝑞, , (c) 1, , 2.(a) 5 (b) 36 (c) 5, , 3.(a) 10, 8, 2, , 4.(a) 12, 24, , (c) 1 4, , (b) 10, 20, 40, , 1 1 1, , 15, , (b) 12, , 5.(a) 8, 4, 2, 1, 2, 4, 8, 18, , (b) 1, 2, 4, 8, 16 and 31, , 6.(a) q = 972, 12, 108, 324, , (b) n = 3, 10 , 20, , 7.(a) 80, 20 or 20, 80, , (b) 81, 9 or 9, 81, , 8.(a) 7, , (b) 6, , Exercise 1.3.6, 1.(a) 10, 3.(i), , 3069, 64, , 4.(a) ±3, , (b) 31, , 2.(a) 4679, , (ii) 129, , (iii), , (b) ±2 5.(a) 17, , 1, , (b) r = 2 (c) 81, 3, , 665, 9, , (iv), (b) 4, , 279, , 255√2, 128, , (c), , (v) 8745 (vi) 1533, 634, 3
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6.(a) 6, , (b) 6, , 7.(a) 510, , (b)(i) 3, 1, , (iii) 3280, , 8.(a)(i) 2, 6, , (ii) 6 + 12 + 24 + …., , (iii) 378 (b), , 9.(a) 4, 8, 16 or 16, 8, 4, 10.(a) 32, 16, 8, , (ii) 3 + 9 + 27 + …, 255, 4, , (b) 1, 3, 9 or 9, 3, 1, (b) 3, 5, 7 or 12, 5, -2, , Exercise 1.4.1, 1, 2, 3, 4, 5 and 6 show the subject teacher, 7.(a) Max. value = 59 at B(3, 10), Min. value = 9 at A(3, 0), (b), , Max. value = 26 at B(2,4), Min. value = 0 at 0(0, 0), , 8.(i) (2, 0), (3, 0), (2, 1), , (ii) 2, -1), (4, 0), (9/2 , 1), , (iii) (2, 0), (4, 0), (4/5, 12/5), , (iv) (0,0), (1, 0), (3, 1), (0, 4), , 9.(i) Max = 96 (ii) Max = 51 at (1, 5/2), , (iii) 36 at (6, 4), , (iv) 10 at (2, 2), , 10(i) 0 at (0, 0), , (ii) 9 at (3, 0), , (iv) 15 at (1, 1), , (ii) 2 at (2, 0), , 11. Max = 8, Y > 0, x + y = 2, 2x – 3y > –6, 12) Max = 16, Min = –12, x – 3y > -4, x + 2y > –4, x < 2, Exercise 1.5.1, Show 1 and 2 to the subject teacher., 3.(i) (0,0), x = 0, , (ii) (–1, 1), x = –1, , (ii) (–1, 6) x = –1, , 4.(a) Turning upwards from origin, (b) A curve line from 2nd quadrant to fourth quadrant through origin., 6.(a) (–1, 1), , (ii) (6, 0), , −3 −1, ), 4, , (iv) ( 2 ,, , (iii) (0, 0), , 8. (i) x = –3, 1, , (ii) x = 0, 2, , y = 9, 1, , y = 0, 2, , (v) (3, –4), , (iii) x = 1, 2, y = -1, 0, , (iv) x = 0.3, –2.3, , (v) x = –4, 2, , y = 0.3, –2.3, , y = –7, 5, , 280, , (vi) x = 10 , –1, y = –6, 5
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9.(i) –3, 1, , (ii) 2, 3, , 1, , (v) 3, 2, , (iii) 5, –3, , (iv) –1,, , −2, 3, , (vi) –5, –1, , Exercise 2.1, 1., , (a) N = {1, 2, 3, ….} (b) W = {0, 1, 2, 3, ……..}, , (c) I = {……, -2, -1, 0, 1, 2, ……}, , (d) Q = {x : x = 1/q, p and q are integers and q ≠ 0}, (e) Q = {x : x Q}, , (f) R = {x : -a < x <q}, , 2, 3, 4 and 5 show to your teacher, Exercise 2.2, 1(a) –4 < x < 4, at x = 2 discontinuous., (b) -4 < x < 3, at x = 1 discontinuous, at remaining points continuous., (c) -8 < x < 10, at x = 5 discontinuous at remaining points continuous., (d) 6 < x < 7, at x = 0, discontinuous, at remaining points continuous., (e) - < x < , at x = 0 discontinuous at remaining points continuous., (f) -6 < x < 6, at x = –2 discontinuous, at remaining points continuous., 2, 3 show to your teacher., Exercise 3.1, 2.(a) (i) -2, , (ii) -17, , (iii) 3, , (iv) -3y2 + 2y, , (b) -14, , (c) b2 – a2, , (d) 1, , (e) 2, , 3.(a) ±2 (b) 2, , (c) -20 (d) 1.5, , 4.(a) 45, , (b) 228, , 1, , 7, 5) [, 21, , (c) 185, , −9, ], −7, , 6) -168, , Exercise 3.2, 2.(a) |A| = 0, does not exist, 2 −3, 4.(a) (, ), −1 2, , (b) |A|≠ 0, exists, , 3 −2, (b) (, ), −2 3, , 19, 5) [, 36, , 281, , (c) |A| ≠ 0, exists, 27, ], 68
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Exercise 3.3, 2(a) (3, 2), , (b) (1, 0), , (f) (-1,2), , [correction in q. no. 2], , (g) (1/5, 3/5), , (f) (-1,2), , 1 1, , (g) (6 , 4), , 1 5, , (h) (3 , 2), , (c) (3, 3), , (d) 3, 7), , (e) (1/5, 3/5), , [correction in q. no. 2], 3(a) Pen: Rs 70, copy: Rs 50, , (b) 49 years, 4 years, , Exercise: 3.4, 2(a) (2,1), , 7 21, ), 2, , (b) (2 ,, , 23, , (c) (251 ,, , −23, ), 72, , (d) (1, 0), , 32 96, , (e) ( 7 , 13), , Exercise 4.1, 1. Show to your teacher, a b −a b, , 2.(a) tan = ± ( a2 a1 b 1b 2 ), , (b) a2b1 = a1b2, , 3.(a) 45°, , (b) 47.72°, , (c) 60° 4.(a) 135°, , 5.(c) -2, , (d) 3, , (e), , 1 2 1 2, , 6.(a) 3x – 4y + 5 = 0, , (c) a1a2 + b1b2 = 0, (b) 150°, , (c) 172.87°, , −5, 2, , (b) 3x + 4y + 17 = 0, , (c) 4x + 54y – 23 = 0, , (d) 7x – 5y + 22 = 0, 7(a) x – y + 1 = 0, x + y – 5 = 0, , (b) x – 5y – 21 = 0, 5x – y + 1 = 0, , (c) x – y + 1 = 0, x + y – 3 = 0, 8(a) x – 2y – 1 = 0, , (b) x – 7y – 53 = 0, 7x + y – 21 = 0, , (c) x – 3 = 0, y – 4 = 0, Exercise 4.2, 2.(a) x2 – 6xy + 5y2 = 0 (b) 6x2 – 7xy – 3y2 = 0, , (c) x2 – 4y2 = 0, , (d) x2cos - (1 + sincos)xy + y2sin = 0, 3.(a) x + 2y = 0, 3x + y = 0, , (b) 2x + y = 0, 3x + y = 0, , (c) x - √3𝑦 = 0, √3𝑥 - y = 0, (d) y = (cosec + tan)x; y = (cosec - cot)x, , 282, , (e) y = (sec + tan)x; y = (sec - tan)x
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(d) tan3 =, 3(a), , 3𝑡𝑎𝑛𝜃−𝑡𝑎𝑛3 𝜃, 1−3𝑡𝑎𝑛2 𝐴, , 7, 25, , 3, , 24, , (b) 5, , (g) 1, 0, , (h), , 120 119, , 24, , (d) 169 , 169, , (c) 25, , (e) 25, , (f) 1,, , 11, 2, , Exercise 5.2, 2, , 2.(a) 9 (b) 0, 3.(a), , √3 1, ,, 2 2, , (c) = 1, , , √3, , (b), , 4.(a) -1, 0, , 24 7 24, , ,, 25 25 7, , (b) 1, , (c), , 24 7 24, , ,, 25 25 7, , (c) –1, , Exercise 5.3, 1 and 2 show to your subject teacher, 3.(a) 1, (c), , 1, √2, , (b) 1, , 4.(a) –2sin55°.sin15°, , 1, , (d) 4, , 1, 2, , 1, √2, , (c), , −1, sin100°, 2, , 1, , (e) 1, , 5.(a) [cos18° – cos104°], , b), , (f) 4, , 1, 2, , 1, 2, , (b) [cos12° - ], 1, , (f) 2 [sin16 + sin2], , (d) sin80° + sin16°, , (e) sin7 + sin3, , (g) cos14 + cos8, , (h) cos4 – cos10, , 6.(a) √2cos20°, , (b) 2cos33°sin13°, , (c) 2cos50°.cos20°, , (d) 2cos5.sin2, , (e) 2sin7.cos4, , (f) 2cos10α.cos5α, , Exercise 5.5, 1 show to your subject teacher., 2.(a) 60°, , (b) 360° – , , (c) Max. value = 1, Min. value = 0, , 3.(a) = 60°, , (b) = 60°, , (c) = 60°, , (f) = 90°, , (g) = 45°, , (h) = 60°, , 4.(a) = 120°, , (b) = 135°, , (c) = 60°, 120°, , (d) = 45°, , 284, , (e) = 30°, , (d) = 60°, , −9√3, 16
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(e) = 60°, 120, , (f) = 150°, , 5.(a) α = 45°, 135°, , (b) α = 60°, 120°, , (c) α = 60°, 120°, , (d) = 60°, 120°, , (e) α = 60°, 120°, , (f) = 60°, 120°, , 6.(a) = 90°, 150°, , (b) = 30°, 150°, , (c) = 30°, 150°, , (e) = 30°, 90°, 150°, , (f) = 0°, 60°, 180°, , (g) = 10°, 50°, 90°, 130°, 170°, , (d) = 30°, , (h) = 15°, 45°, 75°, 105°, 135°, 165°, 1, , 7.(a) 0°, cos–1(3), 360°, (d) 0°, 180°, 360°, , (b) 120°, 240°, , (c) 45°, 225°, , (e) 45°, 135° (f) 45°, 135°, 225°, 315°, , (g) 30°, 135°, 210°, 315°, , (h) 120°, 150°, 300°, 330, , (i) 60°, 135°, 240°, 315°, , (j) 0°, 60°, 300°, 360°, , 8.(a) 0°, 120°, 360°, , (b) 90°, 330°, , (c) 45°, , (d) 0°, 60°, 360°, , (e) 105°, 345°, , (f) 60°, , (g) 300°, 360°, , (h) 60°, 180°, , (i) 0°, 120°, 360°, 9.(a) 0°, 60°, 90°, 120°, 180°, (d) 30°, 60°, 90°, 11.(a) 60°, , (b) 0°, 60°, 180°, , (e) 0°, 180°, , (f) 18°, 90°, , (b) 45°, 315°, , Exercise 5.6, 1 and 2 show to your teacher, 3.(a) 150√3m, , (b) 30√3m, , (c) 25.35 m, , 4.(a) 42.26m, , (b) 8.87m, , (c) 320m, , 5.(a) 38.97m, , (b) 18.30m, , 6.(a) 8m, , (b) 50√2m, 100√2m, , 7.(a) 14.64 m, , (b) 7.32m, 10√3m, , 8.(a) 178.9m, , (b) 288m, , 285, , (c) 45°, 60°, 135°, 300°, (g) 0°, 90°, , 10) 20°, 30°, 80°
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9.(a) 136.96km/hr, , (b) 29√3m, it is 20m from on pole and 60 m form another, , 10. (a) 15 m, , (b) 19.12m, , (c) 100m, , Exercise 6.1, 1, 2 and 3 show to your teacher, 4.(a) 7, , (b) 3-√3, , (c) 19, , 5.(i) ⃗⃗⃗⃗⃗, AB = i − 4j, ⃗⃗⃗⃗⃗, BC = 4𝑖 + j,⃗ ⃗⃗⃗⃗⃗, CD= -i + 4j, ⃗⃗⃗⃗⃗, DA = 4i + j, ⃗⃗⃗⃗⃗, BD = 3i + 5j, (ii) -17 (iii) 34, 17, 6.(a) -14, obtuse, 7.(a) (i) 60°, , (b) (i) 1, , (ii) 150°, , (c) (i) 60°, , (ii) -2, , (iii) 6, , (iv) 5, , (v) 13, , (b) (i) 90°, , (ii) 0°, , (iii) 74.74°, , (iv) 45°, , (ii) 41.56°, , 8.(b) (i) –3, , (ii) 6, , 10.(a)(i) 0°, , (iii) 3, (ii) 60°, , 9.(a) (i) 8, (b)(i) 90°, , (ii) 7, 1, 4, , (ii) cos-1 ( ), , Exercise 6.2, 1 and 2 show to your teacher., 1, , (b) 2 (𝑖 + 𝑗), , 3.(a) 4𝑖 – 2𝑗, 1, , 4.(a) 5 (11i + 21j), 1, , 5.(a) 3 (5𝑖 + 26𝑗), 7.(b) −𝑖, ⃗⃗⃗⃗ −, , (vi) 4, , (c) 2𝑖 + 4𝑗, , 1, , (b) 3 (−19𝑖 − 50𝑗), 5, , (b) 24𝑖 + 4𝑗, , 6.(a) 2𝑖 + 3 𝑗, , (b) 10𝑖 + 4𝑗, , (d) 9m 8), , (e) (2, –4), , 10, 𝑗, 3, , Exercise 7.1, 1. show to your teacher., 2.(a) (-2, 1), , (b) (3, 2), , (c) (-4, - 3), , (f) (-12, -24), , (g) (-4, -6), , (h) (6, 18), , 3, 4, 5, 6, 7 show to your teacher., , 286, , (iii) 48
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Exercise 7.2, 2, , 3, , −4 8, , 2.(a) (13 , 13), , 9, , (b) ( 5 , 5), , 25 125, ), 26, , (c) (0, 4), , 256 192, ), 25, , (e) (26 ,, , −22 45, , (f) ( 25 ,, , 48 80, , (g) ( 13 , 13), , −25 54, , 3.(a) (17 , 17), , 32 −48, ), 13, , (d) (13 ,, , 69 163, ), 65, , (h) (65 ,, , −75 142, ), 49, , (b) ( 61 , 61), , (c) ( 49 ,, , 4. Show to your teacher., Exercise 7.3, 1. Show to your teacher., 2.(a) A' (7, 7), , (b) p'(-1, 9), , 3.(a) A'(2, 0), B'(4, 2), C'(-3, 3), , (b) A'(2, 0), B'(12, 8), C'(9, 10), D'(4, 8), , (c) O(0,0), P'(2, 14), Q'(8, 16), R'(-6, 8), −3 1, 4.(a) (, ), −3 1, 1, , 5.(a) ( 3, 0, , 0, (b) (, 1, , 1, , −3, ), 0, , −1 0, 6.(a) (, ), 0 −1, , 1, ), 0, , 1 0, (c) (, ), 0 0, , −3 1, (b) (, ), −3 1, , 0 1, (c) (, ), 1 0, , 3, (b) (, 0, , 0, ), 3, , 1 0, (c) (, ), 0 −2, , 4.(a) 20, , (b) 50, , 7 and 8 show to your teacher., Exercise 8.1, 1 and 2 show to your teacher., 3.(a) 10, , (b) 45, , 5.(a) 6.31, 0.09, 0.273, , (b) 10.65kg, 0.23, , (c) 7.39, 0.19, , (e) 1.0225 inches, 0.015, , (f) 9.875, 0.41, , 6.(a) 13.21, 0.327, , (b) 16, 0.36, , (d) 5.5,, , Exercise 8.2, 3.(a) 6.16, 0.308, , (b) 40, , 4.(a) 10.6, 0.265, , 287, , (b) 20
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5.(a) 9.44, 0.35, , (b) 5.08, 0.423, , (c) 12.82, 0.36, , 9.(i) 15.81, 0.35, , (ii) 15.85, 0.36, , (d) 113.3, 0.25, , 6.(a) 727, 0.213, 7. 11.53, 0.21, 8. 10.08, 0.30, , 10. Show to your teacher., Exercise 8.3, 3.(a) 4.49, 0.17, , (b) 2.91, 0.242, , 4.(a) 24.91, , (b) greater will be the consistency, , 5.(a) 6.05, , (b) 12.96, 0.41, , 6.(a) 11.23, 23%, , (b) 0.51, 51.41%, , 7.(a)(i) 1248.12 (ii) 155.55, (b)(i) 14.98, 29.57%, 8.(a)(i) A, (b) (i) Boys, , (c) 28.35, , (iii) 12.46%, , (ii) 14.985, 29.57%, , (ii) 184.27, 124.49, , (iii) B, , (ii) Girls, , 9 and 10 show to your teacher., , 288, , (iii) 14.98, 29.57%
