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Approved by the Government of Nepal, Ministry of Education, Science and Technology, Curriculum Development Centre, Sanothimi, Bhaktapur as an Additional Learning Material EvMexdcAaenlTtaiHn OEMptAioTnICaSl 10Book Author Piyush Raj Gosain Hukum Pd. Dahal Editors P. L. Shah Tara Bdr. Magar vedanta Vedanta Publication (P) Ltd. jb] fGt klAns;] g kf| = ln= Vanasthali, Kathmandu, Nepal +977-01-4982404, 01-4962082 [email protected] www.vedantapublication.com.np

vEMexdcAaenlTtaiHn OEMptAioTnICaSl 10Book Author Piyush Raj Gosain All rights reserved. No part of this publication may be reproduced, copied or transmitted in any way, without the prior written permission of the publisher. ¤ Vedanta Publication (P) Ltd. First Edition: B.S. 2077 (2020 A. D.) Second Edition: B.S. 2078 (2021 A. D.) Layout and Design Pradeep Kandel Printed in Nepal Published by: Vedanta Publication (P) Ltd. j]bfGt klAns];g k|f= ln= Vanasthali, Kathmandu, Nepal +977-01-4982404, 01-4962082 [email protected] www.vedantapublication.com.np

Preface The series of Vedanta Excel in Opt. Mathematics for class 9 and 10 is completely based on the contemporary pedagogical teaching learning activities and methodologies. It is an innovative and unique series in the sense that the contents of each textbooks of the series are written and designed to ful ill the need of integrated teaching learning approaches. Vedanta Excel in Opt. Mathematics has incorporated applied constructivism the latest trend of learner centered teaching pedagogy. Every lesson of the series is written and designed in such a manner that makes the classes automatically constructive and the learner actively participate in the learning process to construct knowledge themselves, rather than just receiving ready made information from their instructor. Even teachers will be able to get enough opportunities to play the role of facilitators and guides shifting themselves from the traditional methods imposing instructions. The idea of the presentation of every mathematical item is directly or indirectly re lected from the writer's long experience, more than two decades, of teaching optional mathematics. Each unit of Vedanta Excel in Opt. Mathematics series is provided with many more worked out examples, arranged in the hierarchy of the learning objectives and they are re lective to the corresponding exercises. Vedanta Excel in Opt. Mathematics class 10 covers the latest syllabus of CDC, the government of Nepal, on the subject. My honest efforts have been to provide all the essential matter and practice materials to the users. I believe that the book serves as a staircase for the students of class 10. The book contains practice exercises in the form of simple to complex including the varieties of problems. I have tried to establish relationship between the examples and the problems set for practice to the maximum extent. In the book, every chapter starts with review concepts of the same topic that the students have studied in previous classes. Discussion questions in each topic are given to warm up the students for the topic. Questions in each exercise are categorized into three groups - Very Short Questions, Short Questions and Long Questions. The project works are also given at the end of exercise as required. In my experience, the students of class 10 require more practices on Trigonometry and Vector Geometry, the examples and the exercise questions are given to ful ill it in the corresponding topics. The latest syllabus of the subject speci ication grid and a model question issued by CDC are given at the end of the book. My hearty thanks goes to Mr. Hukum Pd. Dahal, Tara Bahadur Magar and P.L. Shah, the series editors, for their invaluable efforts in giving proper shape to the series. I am also thankful to my colleage Mr. Gyanendra Shrestha who helped me a lot during the preparation of the book. I am also thankful to my respected parents and my family members for their valuable support to bring the book out in this form. I would also like to express my hearty gratitude to all my friends, colleagues and beloved students who always encouraged me to express my knowledge, skill and experience in the form of books. I am highly obliged to all my known and unknown teachers who have laid the foundation of knowledge upon me to be such a person. Last but not the least, I am hearty thankful to Mr. Pradeep Kandel, the computer and designing senior of icer of the publication for his skill in designing the series in such an attractive form. Efforts have been made to clear the subject matter included in the book. I do hope that teachers and students will best utilize the series. Valuable suggestions and comments for its further improvement from the concerned will be highly appreciated. Piyush Raj Gosain

CONTENT Unit 1 Functions 5 25 Unit 2 Polynomials 44 84 Unit 3 Sequence and Series 97 113 Unit 4 Linear Programming 128 154 Unit 5 Quadratic Equations and Graphs 199 269 Unit 6 Continuity 299 347 Unit 7 Matrices Unit 8 Co-ordinate Geometry Unit 9 Trigonometry Unit 10 Vectors Unit 11 Transformations Unit 12 Statistics Syllabus Specification Grid Model Questions

vedanta Excel In Opt. Mathematics - Book 10 1 Functions 1.0 Review Study the following relations and state which are functions and which are not functions by giving your reasons: R1 = {(1,2), (2,3), (3,4)} R2 = {1,2}, (1,4), (1,6)} R3 = {(2,4), (3,6), (4,8)} Discuss the following types of functions with examples related to our daily life: (a) Onto Functions (b) Into Functions (c) One to one Functions (d) Many to one Functions 1.1 Algebraic Functions The functions which can be generated from a real variable x by a finite number of algebraic operations (addition, subtraction, multiplication, division, and extraction of roots) are called algebraic functions. Quite often algebraic functions are algebraic equations. Some examples of algebraic functions are f(x) = 4x + 5, y = f(x) = x2 + 3x + 5 f(x) = 5 etc. Note: function f(x) is also denoted by y. i.e., y = f(x) Simple Algebraic Functions The following are some simple algebraic functions: Y (a) Linear Function (b) Constant Function 5 y=x+3 4 (c) Identity Function (d) Quadratic Function (e) Cubic Function 3 (0, 3) 2 Let us define each of the above functions with their graphs. 1 X (a) Linear Function X' (-3, 0) O 1 2 3 4 5 An algebraic function of the first degree expressed Y' in the form of y = f(x) = mx + c is called linear function. Here, m and c are constants. The linear function always gives a straight line when plotted in a graph. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 5

vedanta Excel In Opt. Mathematics - Book 10 In the above function, put m = 1 and c = 3, we get y = x + 3. To draw graph of y = x + 3, we take points as (1, 4) (-3, 0), (0, 3). The graph is shown in above figure. Y (b) Constant Function 4 y=4 An algebraic function that is expressed in the form of y = f(x) = c i.e. y = c, (where c is a constant) is 3 called a constant function. It represents a straight 2 line parallel to X-axis. In constant function, the X' 1 X value of y is constant for different values of x. O 1234 For example, y = f(x) = 4, we get f(1) = 4, f(2) = 4, f(3) = 4. The graph of y = 4 is shown in the figure, which Y' is parallel to x-axis. (c) Identity Function Y An algebraic function expressed in the form of y=x is X' 2 y=x X called identity function in which every x is associated 1 with itself. In this function, the function has same 45° domain and range. -3 -2 -1 O 12 3 -1 The graph of identity function is straight line passing -2 through the origin and bisecting the angle between the axes. Note: An identity function is one to one onto function. Y' (d) Quadratic Function An algebraic function expressed in the form of y = f(x) = ax2 + bx + c, where a ≠ 0, a, b, and c are constants, is called a quadratic function. The graph of a quadratic function is a curve called parabola. The curve is symmetric b b 4ac - b2 about the line x = - 2a with vertex = – 2a , 4a The standard form of equation of parabola is y = a(x – h)2 + k, For example : y = x2 – 6x + 8 Comparing it to y = ax2 + bx + c, Y we get, a = 1, b = –6, c = 8   where vertex (h, k) = – b , 4ac - b2  y = x2 - 6x + 8 2a 4a   = – -6 , 4.1.8 - 36 = (3, -1)  2.1 4.1  from y = x2 - 6x + 8  x0123456 X' O  X y 8 3 0 -1 0 3 8 6 Y'Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 (e) Cubic Function Y A function expressed in the form of  y = ax3 + bx3 + cx + d, where a ≠ 0, a, b, c and d are  constants, is called a cubic function.   Examples of cubic functions are  y = x3, y = x3 + 3x, y = 4x3 + 3x2 + 4x + 5 etc.   Let's as draw a graph of cubic function. y = x3 .  X x –2 –1 0 1 2 X'  y –8 –1 0 1 8     Its graph is given as shown in the figure.     1.2 Trigonometric Functions  Y' A function involving trigonometric ratios like sine, cosine, tangent etc. is called trigonometric function. Trigonometric functions are related to the angles and their measurements. Trigonometric functions of a right angled triangle are defined in terms of ratios of sides of right angled triangle. 1. Discuss the values of trigonometric ratios of angles from -2Sc to 2Sc. 2. Note the maximum and minimum values of the trigonometric ratios of sine, cosine and tangent. 3. Do the values of the trigonometric function repeat for different values of angles ? Discuss it. Trigonometric functions do not satisfy the algebraic operations like addition, subtraction, multiplication, and division. They are different from algebraic functions. The functions which are not algebraic are called transcendental functions. Trigonometric functions are also transcendental functions. Trigonometric functions are periodic functions. A function f which satisfies f(x+k) = f(x) for all x belonging to its domain and smallest positive value of k. Examples : sin(x + 2Sc) = sinx cos(x +2Sc) = cosx tan(x+Sc) = tanx We can say that sinx, cosx, and tanx are the periodic function of periods 2Sc, 2Sc, and Sc respectively. Graphs of Trigonometric Functions (a) Graph of y = sinx, (-2Sc ≤ x ≤ 2Sc) For the sine function, y = sinx, take the values of x from -2Sc to 2Sc, we get the corresponding values of y as shown in the table. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 7

vedanta Excel In Opt. Mathematics - Book 10 x -2Sc -3Sc -Sc -Sc 0 Sc Sc 3Sc 2Sc 2 2 2 2 y 0 1 0 -1 0 1 0 -1 0 What are maximum and minimum values of sinx ? Plot the points (-2Sc,0), – 3Sc , 1 , Y 2 Sc Sc (Sc, 0), – 2 , - 1 , (0, 0) 2 , 1 , (Sc, 0), 3Sc , –1 , and (-2Sc, 0) in a -1 2 X' X graph paper and join the points. We -2cS 3Sc O Sc Sc - 2 -Sc Sc -1 2 3Sc 2Sc - 2 2 get a curve as shown in the figure alongside. (b) Grph of y = cosx, (-2Sc ≤ x≤ 2Sc) Y' For y = cosx, take the values of x from -2Sc to 2Sc, we get corresponding values of y as shown in the table. x -2Sc -3Sc -Sc -Sc 0 Sc Sc 3Sc 2Sc 2 2 1 2 2 y 1 0 -1 0 0 -1 0 -1 Note the maximum and minimum values of cosx. Plot the above points in the graph paper and join the points and we get the curve in the figure below : Y -1 X' -2Sc 3Sc 0 Sc X 2 -1 2 - -Sc - 3 Sc 3Sc 2Sc Sc 2 Y' (c) Graph of y = tanx, (-2Sc ≤ x ≤ 2Sc) For y = tanx, the value of y is not defined for the values x = - 32Sc, - Sc , Sc , 3Sc . 2 2 2 Image of the function for these values of x are not defined. x -2Sc -7Sc -3Sc -5Sc -Sc -3Sc -Sc -Sc 0 Sc Sc 3Sc Sc 5Sc 3Sc 7Sc 2Sc 4 2 4 4 2 4 4 2 4 4 2 4 y 0 1 ∞ -1 0 1 ∞ -1 0 1 ∞ -1 0 -1 ∞ -1 0 8 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Y (1.249, 3) 4 3 2 1 X' 3Sc O X - 2 -Sc - Sc Sc Sc 3Sc 2Sc 2 2 2 -1 -2 -3 Y' The graph of tanx is shown in the above graph. Worked out Examples Example 1. Name the following algebraic functions. Solution: (a) y = 5 (b) y = ax + b, Example 2. (c) y = x3 + 2x2 (d) y = 2x2 + x + 5 (a) Constant Function (b) Linear Function (c) Cubic Function (d) Quadratic Function Name the algebraic functions of the following graphs. Y Y X' O X X' O X Y' Y' (a) The given graph represents linear function. Solution : (b) The given graph represents quadratic function. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 9

vedanta Excel In Opt. Mathematics - Book 10 Exercise 1.1 Very Short Questions 1. Classify the following functions: (a) y = x (b) y = x2 (c) y = x3 (d) y = ax2 + bx + c (e) y = x3 - x2 (f) y = 7 (g) y = tanx 2. Define a function with an example. 3. Define an algebraic function. 4. What is a trigonometric function ? 5. What is a transcendental function ? 6. What is the minimum value of the cosine function y = cosx ? 7. Write the periods of the following functions: (a) y = sinx (b) y = cosx (c) y = tanx 8. For what values of x, the image of tangent function y = tanx is not defined ? 9. Define each of following functions with an example: (a) Linear Function (b) Constant Function (c) Identity Function (d) Quadratic Function (e) Cubic Function (f) Periodic Function 10. Name the functions represented by the following graphs: Y (a) Y (b) 4 2 3 1 2 1 45° X' X X' X -3 -2 -1 O 12 3 -3 -2 -1 O 12 3 Y -1 -1 -2 -2 -3 Y' -4 (c) Y (d) Y' -1 -1 X' 0 X X' -2Sc 3Sc 0 Sc Sc X 2 -1 2 -2Sc - 3Sc -Sc - 3 Sc Sc 3Sc - -Sc - Sc 3Sc 2Sc 2 Sc -1 2 2 2 2 10 Y' Y' Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Y (e) Y (f)        X' 0 X  X' O X   y = -3  Y'     Short Questions   11. Draw graph for each of the following table: Y' (a) x 2 3 4 5 6 (b) x 0 ±1 ±2 ±3 y 4 6 8 10 12 y0149 (c) x -3 -2 0 1 (d) x 1234 y -1 -2 -3 -4 y 6 4 0 -2 12. Draw graph of each of the following functions: (a) y = 2x - 1 (b) y = x + 1 (c) y = x2 (d) y = -x (e) y = x3 (f) y = -x3 13. Draw the graphs of the following trigonometric functions: (a) f(x) = sinx, (-Sc ≤ x ≤ Sc) (b) f(x) = cosx, (-Sc ≤ x ≤ Sc) (c) f(S) = tanx, (-Sc ≤ x ≤ Sc) Long Questions 14. Draw the graphs of the following trigonometric functions: (a) f(x) = sinx, (-2Sc≤ x ≤ 2Sc) (b) f(x) = cosx, (-2Sc≤ x ≤ 2Sc) (c) f(x) = tanx, (-2Sc ≤ x ≤ 2Sc) Project Work 15. Record your body weight continuously for a week at the same time in the morning. Taking days in X-axis and weight in Y-axis, draw the graph of it and discuss the type of function represented by the graph drawn. 16. List the price of petrol per litre for a week. Taking days in X-axis and the price in Y-axis, represent the information obtained in graph. Discuss the type of function represented by the graph drawn. 17. Where are the functions f(x) = sinx and f(x) = cosx used in our daily life ? Investigate and prepare a report and present in your class. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 11

vedanta Excel In Opt. Mathematics - Book 10 1. (a) Identity function (b) quadratic (c) cubic (d) quadratic (g) trigonometric (e) cubic (f) constant 6. –1 (c) Sc 7. (a) 2Sc (b) 2Sc (c) y = cosx 8. Sc 10. (a) identity (b) quadratic 2 (f) cubic (e) constant (d) y = sinx 1.3 Composite Function Let A = {1, 2, 3}, B = {2, 3, 4}, and C = {4, 6, 8} be three sets. Let us define two functions f and g such that f : A → B and g : B → C such that f = {(1, 2), (2, 3), (3, 4)} and g = {(2, 4), (3, 6), (4, 8)} By representing above function in mapping diagram, we write it as below. fg A BC 1 24 2 36 3 48 gof Here, f is a function defined from A to B. So write. 1  A o f(1) = 2  B 2  A o f(2) = 3  B and 3  A o f(3) = 4  B range of f = {2, 3, 4} which is domain for g. Again, g is a function from B to C. 2  B o g(2) = 4  C 3  B o g (3) = 6  C 4  B o g (4) = 8  C Now, let us define a new function from A to C. Such that, 1  A o 4 = g(2) = g(f(1))  C 2  A o 6 = g(3) = g (f(2))  C 4  B o 8 = g(4) = g (f(3))  C 12 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 The new function defined from A to C is called composite A gof function of f and g, and it is denoted by gof. we can draw for gof as below. 1 C 2 Writing the above composite function as a set of ordered pair, 3 4 6 we get, gof = {(1, 4), (2, 6), (3, 8)} 8 Definition Let f : A → B and g: B → C be two functions. Then a new function defined from set A to set C is called composite function of f and g. It is A B C denoted by gof and defined by gof : A → C. Then every element of A is associated with a x f(x) g(f(x)) unique element of set C. Symbolically, we write. If f : A → B and g: B → C, then the composite gof function from A to C is defined by gof : A → C, for every x  A, (gof) (x) = g (f(x))  C Note : (i) The composite function gof is read as \"f followed by g\". (ii) In general gof ≠ fog (iii) If (fog) (x) = x and (gof) (x) = x, then, the composite function is called identity function. Worked Out Examples Example 1. If f = {(1, 1), (2, 4), (3, 9)} and g = {(1, 2), (4, 8), and (9, 18)}, then, show the function gof in arrow diagram and find it in the set of ordered pairs. Solution: Here, f = {(1, 1), (2, 4), (3, 9)} g = {(1,2), (4,8), (9,18)} Arrow diagram of composite function gof is given below, f g BC A 11 2 24 8 3 9 18 gof 13 Writing above composite function in the ordered pairs, we get gof = {(1,2), (2, 8), (3, 18)} Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Example 2. Let f : R → R, f(x) = 4x + 3 and g : R → R, g(x) = x2 + 4, Find the following: Solution: (a) (fof) (x) (b) (gog) (x) (c) (fog) (x) (d) (got) (x) (e) (gof) (1) (f) (fog)(2) Also, check is (gof) (x) = (fog)(x) ? Here, f(x) = 4x + 3, g(x) = x2 + 4 (a) (fof)(x) = f(f(x)) = f(4x + 3) = 4(4x+3) + 3 = 16x + 12 + 3 = 16x + 15 (b) (gog)(x) = g(g(x)) = g(x2 + 4) = (x2 + 4)2 + 4 = x4 + 8x2 + 16 + 4 = x4 + 8x2 + 20 (c) (fog)(x) = f(g(x)) = f(x2 + 4) = 4(x2 + 4) + 3 = 4x2 + 16 + 3 = 4x2 + 19 (d) (gof) (x) = g(f(x))= g(4x + 3) =(4x + 3)2 + 4 = 16x2 + 24x + 9 + 4 = 16x2 + 24x + 13 (e) (gof) (1) = g(f(1)) = g(4.1 + 3) Alternative Method = g(7) we have, = 72 + 4 (gof)(x) = 16x2 + 24x + 23 = 49 + 4 (gof)(1) = 16.12 + 24.1 + 13 = 53 = 53 (f) (fog)(2) = f(g(2)) Alternative Method = f(22 + 4) we have, =f(8) (fog)(x) = 4x2 + 19 =4.8 + 3 ? (fog)(2) = 4.22 + 19 = 32 + 3 = 16 + 19 = 35 = 35 From above, we get, (fog) (x) = 4x2 + 19 (gof) (x) = 16x2 + 24x + 13 ? (gof)(x) ≠ (fog)(x)  14 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Example 3. If f: R → R, and g: R → R are defined by f(x) = 2x and g(x) = 3x + 4 and Solution: (gof)(x) = 22, find the value of x. Example 4. Here, f(x) = 2x and g(x) = 3x + 4 Solution: Now, (gof)(x) = g(f(x)) = g(2x) =3(2x) + 4 = 6x + 4 Example 5. Solution: By question, (gof)(x) = 22 Example 6. ie. 6x + 4 = 22 Solution: or, 6x = 22 - 4 or, x= 18 6 ? x = 3 If f(x) = px + 4 and g(x) = 5x and (gof)(2) =4, find the value of p. 2 px + 4 Here, f(x) = 2 and g(x) = 5x Now, (gof)(2) = 4 or, g(f(2)) = 4 or, g p.2+4 =4 2 or, g(p + 2) = 4 or, 5(p + 2) = 4 or, p+2= 4 5 4 or, p = 5 -2 or, p = 4 - 10 =- 6 5 5 6 ? p =- 5 If (fog)(x) = 7x - 1 and f(x) = 3x + 5, find g(x) where g(x) is a linear function. 3 We have, f(x) = 3x + 5 Now, (fog) (x) = 7x - 1 3 7x - 1 or, f[g(x)] = 3 or, 3g(x) + 5 = 7x- 1 3 7x - 1 or, 3g(x) = 3 – 5 g(x) = 7x - 16 9 If f: R → R ; f(x) = 8 - x, Show that (fof)(x) = x. Here, f(x) = 8 - x (fof)(x) = f(8 - x) = 8 - (8 - x) = 8 - 8 + x = x ? (fof)(x) = x, proved Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 15

vedanta Excel In Opt. Mathematics - Book 10 Exercise 1.2 Very Short Questions 1. Define a composite function of two functions. 2. Write the composite function of given functions f and g as shown in the mapping diagram. f g C A B x yz 3. From the adjoining figure, write the composite function gof of function f and g in the ordered pair form. f g C B A ad x be y cf z 4. Write the composite function gof of the following functions in the ordered pair form: (a) f = {(1, 2), (3, 4), (4, 5)} and g = {(2,6), (4,12), (5,15)} (b) f = {(1,2), (3,4), (5,6)} and g = {(2,3), (4,6), (6, 8)} (c) f = {(3,4), (4,5), (5,6)} and g = {(4,5), (5,6), (6.7)} 5. Write the composite function fog of the following functions in the ordered pair form: (a) f = {(3,4), (4,5), (5,6)} and g = {(2,3), (3,4), (4,5)} (b) f = {(4,2), (8,4), (16, 8)} and g = {(2,4), (4, 8), (8, 16)} (c) f = {(2, 8), (4, 64), (8, 256)} and g = {(1, 2), (2, 4), (4, 8)} Short Questions 6. (a) f = {(1,3), (2,1), (3,2)} and g = {(1,2), (2,3), and (3,1)}, find the domain and range of the composite function gof. (b) If f = {(1,2), (3,5), (4,1)} and g = {(2,3), (5,1), and (1,3)}, find the domain and range of the composite function gof. 7. (a) If f = (1,5), (2,1), (3,3), (5,2)} and g= {(1,3), (2,1), (3,2), (5,5)}, find gof and fog. (b) If h = {(5,2), (6,3), k = {(2,5), (3,6)} find hok and koh. 8. (a) From the function f = {(4,1), (5,3), (6,2)} and g = {(1,2), (3,6), (2,4)}, show gof in arrow diagram and write it in ordered pair form. (b) From the functions f = {(a,b), (b,c), (c,d)} and g = {(b,e), (c,f), (d,h)}, find gof in ordered pair form and also show in mapping diagram. 16 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 9, If (fog) = {(2,1), (4,2), (6,3), and (8,4)} and g = {(2,4), (4,8), (6,12), and (8,16)}, find f in ordered pair form. Also show fog in arrow diagram. 10. (a) If f(x+2) = 4x + 5, find f(x) and (fof) (x) (b) If g(x+5) = x + 20, find g(x) and (gog) (x). Long Questions 11. (a) Let f: R → R and g : R → R be defined by f(x) = x + 1 and g(x) = x3. Then (i) find gof and fog. (ii) Is (gof) (x) = (fog)(x)? (b) Let f: R → R and g: R→R be defined by f(x) = x3 - 1 and g(x) = x2 (i) Find (gof) (x) and (fog) (x) (ii) Is (gof)(x) = (fog)(x) 12. (a) If f: R → R and g: R → R are defined by f(x) = 2x + 3 and g(x) = 2x - 1, find (gof) (3) and (gof)(-2). (b) If f: R → R and g: R → R be defined by f(x) = x + 5 and g(x) = 5x - 5. Find (gof)(-2) and (gog)(2). 13. (a) Find the value of x if f(x) = 4x + 5 and g(x) = 6x + 3, (gof)(x) = 75 (b) Find the value of x if f(x) = 3x + 4 and g(x) = 2x + 5, (fog)(x) = 25. 14. (a) If f(x) = px + 8 and g(x) = 3x + 5, (gof)(3) = 60, find the value of p. (b) If f(x) = 7x + q and g(x) = x + 8, (fog) (4) = 24, find the value of q. (c) If g(x) = 62 and h(x)= ax2 - 1 and (goh)(3)= 17, find the value of a. x-2 (d) If f(x)= 3x + a and g(x) = 4x + 5 and (gof)(5) = 20, find the value of a. (e) If f(x) = 3x , g(x)= Ex - 4, and (gof)(1) = 10, find the value of E. x+2 15. (a) If f(x) = 4x - 5 and g(x) is a linear function and (gof) (x) = 5x + 1, find g(x). (b) If g(x) = 5x + 3 and (gof)(x) = 2x + 5 and f(x) is a linear function, find the value of f(x). (c) If g(x) = 2x and (fog)(x) = 6x-2, find f(x), where f(x) is a linear function. (d) If f(x) = 2x + 3 and (fog) (x) = 10x + 1, find the value of g(x). (e) If f(x) = 2x + 8, (fog)(x) = 3x+4, find the function g(x). (f) If f(x)= 2x -1, g(x) is a quadratic function and (fog)(x) = 6x2+4x + 1, find g(x). 16. (a) If g(x) = 10 - x, show that (gog)(x) = x. (b) If h(x) = 20 -x, show that (hoh)(x) = x 17. (a) If p(x) = 4x+5 and q(x)= 9x-5 prove that (poq)(x) is an identity function. 9 4 3x+2 (b) If p(x) = 3 and q(x) = 3x-2 , prove that (poq) (x) is an identity function. (a) If g(x3) = {(3,1), (5,1), (7,1)} prove that (gof) is constant 18. f = {(1,3), (4,5), (6,7)} and function. (b) If f = {(2,4), (4,5), (7,10)} and g = {(4,2), (5,2), (10,2)}. Show that gof is a constant function. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 17

vedanta Excel In Opt. Mathematics - Book 10 Project Work 19. Number of bacteria in food kept in a refrigerator is expressed by N(T)=20T2 - 80T + 500, (2 ≤ T ≤ 14), where T denotes temperature and T(t) = 4t + 2, (0 ≤ t ≤ 3), where t represents time in hour. (a) Find (NoT) (t) (b) What is the number of bacteria in 2 hours? (c) In how many hours does the number of bacteria in the food reach 3300 ? 2. gof 3. gof = {(a, x), (b, y), (c, z)} 4. (a) {(1, 6), (3, 12), (4, 15)} (b) {(1, 3), (3, 6), (5, 8)} (c) {(3, 5), (4, 6), (5, 7)} 5. (a) {(2, 4), (3, 5), (4, 6)} (b) {(2, 2), (4, 4), (8, 8) (c) {(1, 8), (2, 64), (4, 256)} 6. (a) D = {1, 2, 3), R = {1, 2, 3} (b) D = {1, 3, 4}, R = {3, 1} 7. (a) gof = {(1, 5), (2, 3), (3, 2), (5, 1)} fog = {(1, 3), (2, 5), (3, 1), (5, 2)} (b) hok = {(2, 2), (3, 3)} koh = {(5, 5), (6, 6)} 8. (a) gof = {(4, 2), (5, 6), (6, 4)} (b) fog = {(a, e), (b, f), (c, h)} 9. F = {(4, 1), (8, 2), (12, 3), (16, 4)} 10.(a) 4x – 3, 16x – 15 (b) x + 15, x + 30 11. (a) gof = x3 + 3x2 + 3x + 1, fog = x3 + 1, No. (b) gof = x6 – 2x3 + 1, fog = x6 – 1, No. 12. (a) 17, –3 (b) 10, 20 13. (a) 7 (b) 1 4 31 19 45 14. (a) 9 (b) –36 (c) 51 (d) – 4 (e) 14 15. (a) 5x + 29 (b) 2(x + 1) (c) 3x – 2 (d) 5x – 1 4 5 3x (e) 2 – 2 (f) 3x2 + 2x + 1 19. (a) 320t2 + 420 (b)1700 (c) 3 hours 1.4 Inverse Function Let A = {1, 2, 3} and B = {1, 8, 27} be two sets and the function f is defined from A to B by f = {(1,1), (2,8), (3,27)} A fB 11 28 3 27 Here, domain of f = {1,2,3} range of f = {1, 8, 27} 18 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 co-domain of f = {1, 8, 27} since range of f = co-domain of f, f is one to one onto function. (or bijective function) Now, let us define a new function g from B to A so that the domain and range of g are obtained by interchanging the domain and range of f respectively. Then g = {(1,1), (8,2), (27,3)} g 11 82 27 3 domain of g = {1, 8, 27} range of g = {1, 2, 3} Here, g is also one onto function. g is called inverse of function of f. This new function defined from B to A is called inverse of f. It is denoted by f -1. Therefore the function must be one to one onto, to have its inverse function. Let us take another function g as shown in the mapping diagram. g 1x 2y 3z Here, g is not a onto function. It is many to one into function. Now, let us interchange the domain and range of g. Is it again a function ? g -1 x1 y2 z3 This g-1 does not represent a function. Hence, to have inverse of a function, the function must be one to one onto, i.e., bijective. A function f is called bijective if it is one to and onto. Only the bijective functions have their inverse functions. Remember, f-1 ≠ 1 and f-1 (x) ≠ 1 f f(x) Definition: Let f: A → B be a one to one onto function. Then, the function f-1 : B → A is called inverse of f for every x  B there corresponds f-1 (x) = y  A Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 19

vedanta Excel In Opt. Mathematics - Book 10 In functions f and f-1, the roles of x and y are interchanged. Steps for finding inverse of an algebraic functions: (i) Put y in place of f(x) i.e. y = f(x) (ii) Interchange the roles of x and y. (iii) Get new y in terms of x solving equation obtained in (ii). Note : (i) If the two functions f and g are inverse to each other, f-1 = g or g-1 = f and (gof)(x) = x, (fog)(x) = x. (ii) If f is a bijective function, (f-1)-1 = f. (iii) If f–1 exists, then, domain of f is range of f–1 and range of f is domain of f–1. Worked Out Examples Example 1. Let f: A → B be a bijective function defined by f = {(1,2), (2,3), (3,4),(4,5)}. Solution: Find f-1 in ordered pair form. Example 2. Solution : Here, f = {(1,2), (2,3), (3,4), (4,5)} Interchanging the elements of ordered pairs, we get f-1 = {(2,1), (3,2), (4,3), (5,4)} Let f: A → B be one to one and onto function, defined by f = {(2,4), (3,6), (4,8), (5,10)}. Find f-1 and show f and f-1 in mapping diagrams. Here, f = {(2,4), (3,6), (4,8), (5,10)}. Interchanging the components in each ordered pair, we get f-1 = {(4,2), (6,3), (8,4), (10,5)} In mapping diagrams, we show that Af B B f A 2 44 2 3 66 3 4 88 4 5 10 10 5 Example 3. (a) If f(x) = 7x - 2, find f-1 (x) Solution: (b) If g : 5x - 3 , x ≠ - 1 , find g-1 (x) and g-1 (2) 2x + 1 2 (a) Here, f(x) = 7x - 2 To find f-1, y = 7x - 2 Interchanging the roles of x and y, we get x = 7y - 2 or, x + 2 = 7y y= x+2 7 x+2 ? f-1 (x) = 7 20 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 (b) Here, g (x) = 5x - 3 2x+ 1 To find g-1 (x), interchanging the roles of x and y, we get x = 5y-3 2y+1 or, 2xy + x = 5y - 3 or, 2xy - 5y = - x - 3 or, y(2x - 5) = - (x + 3) or, y = - (x+3) 2x - 5 x+3 5 y = 5 - 2x , x ≠ 2 g-1 (x) = x+3 5 - 2x 2+3 5 Also, g-1 (2) = 5-2×2 = 5 - 4 =5 ? g–1 (2) = 5 Example 5. If f = {(x,y) : y = 7x - 4} find f-1 (x) and f-1 (4). Solution: Here, y = 7x - 4, to find f-1(x), interchanging the roles of x and y, we get Example 6. Solution: x = 7y - 4 or, 7y = x + 4 y = x+4 7 x+4 ? f-1 (x) = 7 Now, f-1 (4) = 4 + 4 = 8 7 7 8 ? f–1(4) = 7 If f = {(x, y) : y = 3x - k} and f-1(5) = 4, find the value of k. Here, f = {(x,y) : y = 3x - k} y = 3x - k i.e. f(x) = 3x - k To find f-1(x), interchanging the roles of x and y, we get, x = 3y - k or, y = x + k 3 x+k f-1(x) = 3 and f-1 (5) = 5+k 3 But, f-1(5) = 4 or, 5+k =4 3 or, 5 + k = 12 ? k=7 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 21

vedanta Excel In Opt. Mathematics - Book 10 Example 7. Given a function f(x+3) = 3x + 5, find f-1(x) Solution: Here, f(x+3) = 3x + 5 Example 8. Solution: replacing x by x - 3, we get, f(x - 3 + 3) = 3(x - 3) + 5 or, f(x) = 3x - 4 or, y = 3x - 4 To find f-1 (x) , Interchanging the roles of x and y, we get, x = 3y - 4 or, y = x+4 3 x+4 ? f-1(x) = 3 Alternate Method. Here, f(x + 3) = 3x + 5 or, f(x + 3) = 3 (x + 3) - 4 f(D) = 3D - 4 where, D = x + 3 Since D is a dummy variable, it can be replaced by any other variable. f(x) = 3x - 4 or, y = 3x - 4 To find f-1 (x), interchanging the role of x and y, we get, x = 3y - 4 or, y = x+4 3 x + 4 ? f-1(x) = 3 If f(2x + 3) = 6x + 7, show that (fof-1) (x) is an identity function. Here, f(2x + 3) = 6x + 7 x is replaced by x , we get, 2 ( )f2. x =6× x 7 2 +3 2 + or, f(x + 3) = 3x + 7 Again, x is replaced by x - 3, we get, f(x) = 3(x - 3) + 7 or, f(x) = 3x - 2 or, y = 3x - 2 To find the f-1 (x), interchanging the role of x and y, we get, x = 3y – 2 22 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 y= x+2 3 ( )Now, (fof-1) (x) = f (f-1 (x)) = f x x+2 = 3. + 2 –2 3 3 = x + 2 - 2 = x. Since (fof-1) (x) = x Therefore, (fof-1)(x) is an identity function. Exercise 1.3 Very Short Questions 1. Define one-to-one function with an example. 2. What is the required condition for existence of inverse of a function ? 3. Do all the functions have their inverses ? 4. Find the inverse functions of the following functions. (a) f = {(a,x), (b,y), (c,z)} (b) g = {(2,6), (3,9), (4, 12), (5, 15)} (c) h = {(1,2), (2,3), (3,4), (4,5), (5,6)} 5. If f = {(2, 1), (4, 1), (5, 1)} Does f-1 exist ? 6. Let f : R →R be defined by f(x) = 2x - 3, find the formula that defines the inverse function f-1. 7. Let f : R → R, then, find the inverse function in each of the following cases: (a) f = {(x,y): y = 4x + 5} { }(b) g = 3x-1 3 , (x,y) : y = 3-2x , x≠ 2 { }(c) h = (x,y) : y = 4x+3 , x ≠ 1 , x-1 8. If f = {(a,x), (b,y), (c,z), (d,w)}, then, show f and f-1 in a mapping diagram. Short Questions 9. (a) If g(x) = 3x - 5, the find g-1 (5), where g is a one to one onto function. (b) If h(x) = 2x-3 , find h-1 (4), where h(x) is one to one. 4 2x+8 10. (a) If g(x) = 5 , find g-1(x) and g-1(10). (b) If k(x) = 5x+7 , x ≠ 2, find k-1 (x) and k-1(4). x-2 11. (a) If f(x) = 2x + k, f-1(4) = 20, find the value of k. (b) If f(x) = 7x + k, f-1(3) = 12, find the value of k. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 23

vedanta Excel In Opt. Mathematics - Book 10 12. (a) If f(x+2) = 4x + 5, find f-1(x) (b) If f(2x + 5) = 10x + 7, find f-1(x) (c) If f(4x + 5) = 12x + 20, find f-1 (x) 13. (a) If f(x + 2) = 7x + 13, find f-1(2) (b) If f(x + 5) = 10x + 11, find f-1(4) (c) If f(4x + 5) = 20x + 24, find f-1(x). Long Questions 14. (a) If f and g are two one-to-one and onto functions defined by f(x) = 2x – 3 and g(x) = 4x + 5, (fof)(x) = g-1 (x), find the value of x. (b) If f(x) = 4x - 7 and g(x) = 3x - 5 are two one to one onto functions, and (fog-1) (x) = 15, find the value of x. (c) If f(x) = x -2 and g(x) = 1 and f-1(x) = (gof)(x), find the value of x. 2x +1 x 2x + 8 (d) If f(x) 4x-17 and g(x)= 5 and (fof)(x)= g-1(x), find the value of x. 15. (a) If f(x) = 2x - 4, then prove that, (fof-1)(x) is an identify function. (b) If f(x) = 2x + 5 , x ≠ - 2, then, prove that (fof-1)(x) is an identity function. x+2 2x + 1, (c) If f(x) = 4 prove that (f-1of)(x) is an identify function. 16. If f(x) = x + 1 and g(x) = 3 - x , x ≠ 0, are two one-to-one onto function, then prove that (f-1og-1)(2) = 0 x 17. If f = {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10)}, g = {(4, 9), (5, 11), (6, 13), (7, 15), (8, 17)} and 3 (fog)–1(x) = x + 2, find the positive value of x. Project Work 18. State the formula of conversion of temperature from degree Celsius to degree Fahrenheit as a function. Write the inverse of the function. Convert 37°c into °F and vice versa. 2. One one and onto function 3. No. 4. (a) {(x, a), (y, b), (z, c)} (b) {(6, 2), (9, 3), (12, 4), (15, 5)} (c) {(2, 1), (3, 2), (4, 3), (5, 4), (6, 5)} 5. does not exist 6. x+3 7.(a) x–5 (b) 3x + 1 , x ≠ – 3 2 4 2x + 3 2 x+ 3 10 19 (c) x– 4 , x ≠ 4 9.(a) 3 (b) 2 10. (a) 21 (b) –15 11.(a) –36 (b) –81 (d) 6 12. (a) x+3 (b) x + 18 (c) x–5 4 5 3 3 43 x+1 13. (a) 7 (b) 10 (c) 5 14. (a) 31 (b) 23 (c) ± 1 17. 4 15 2 24 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 2 Polynomials 2.0 Review Discuss the following : (a) What is the difference between a polynomial and an algebraic expression ? (b) Find the sum and difference of f(x) and g(x), when f(x) = 4x2 + 3x - 5 and g(x) = 2x2 + 5x + 3. (c) Are the sum f (x) + g(x) and difference f(x) - g(x) again a polynomials ? 2.1 Division of polynomials Let f(x) = 7x3 + 8x2 - 5x + 10 be a polynomial. It is divided by x - 1. We have the process of divisions as follows. ) (x-1 7x3 + 8x2 - 5x + 11 7x2 + 15x + 10 7x3 - 7x2 -+ 15x2 - 5x 15x2 - 15x -+ 10x + 11 10x - 10 -+ 21 Now, by division algorithm, Dividend = Divisor × Quotient + Remainder i.e. f(x) = (x - 1). Q(x) + R Where, Q(x) = Quotient R = Remainder f(x) = Dividend. If R = 0, f(x) = (x - 1) × Q(x) Then, (x - 1) is called factor of f(x). Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 25

vedanta Excel In Opt. Mathematics - Book 10 Worked out Examples Example 1. If f(x) = x4 + 7x3 + x2 +2x + 7 is divided by d(x) = x2 + 4, find the quotient Solution: Q(x), and Remainder R(x). Example 2. Here, f(x) = x4 + 7x3 + 6x2 + 3x + 5 Solution: d(x) (x2 + 4) ) (Now, x2 + 4 x4 + 7x3 + 6x2 + 3x + 5 x2 + 7x + 2 2.1 x4 + 4x2 -- 7x3 + 2x2 + 3x 7x3 + 28x -- 2x2 - 25x + 5 2x2 + 8 -- - 25x - 3 quotient = x2 + 7x + 2 remainder R (x) = - 25x - 3 If quotient Q(x) = x3 + 4x2 + 5 and remainder R(x) = x - 4 and divisor d(x) = x2 + 3, find the polynomial. Here, quotient, Q(x) = x3 + 4x2 + 5 divisor, d(x) = x2 + 3 remainder, R(x) = x - 4 Then, by division algorithm, we have, dividend = divisor × quotient + remainder i.e. f(x) = (x2 + 3) × (x3 + 4x2 + 5) + (x - 4) = x5 + 4x4 + 5x2 + 3x3 + 12x2 + 15 + x - 4 = x5 + 4x4 + 3x3 + 17x2 + x + 11 Synthetic Division Synthetic division is the process of division which is used to find the quotient and remainder when a polynomial f(x) is divided by a linear polynomial x - a. For an example, let f(x) = 4x3 + 3x2 - 7x + 5 be a polynomial and it is to be divided by x - 2 by division method, we have, ) (x - 2 4x3 + 3x2 - 7x + 5 4x2 + 11x + 15 -4x3 +- 8x2 11x2 - 7x -1+11x52x+- 22x +5 - 15x +- 30 35 ? Divisor = d(x) = x - 2 26 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Quotient = Q(x) = 4x2 + 11x + 15 Remainder = R = 35 Again, polynomial f(x) can be divided by (x - 2) by writing the coefficients only. We write the coefficients in descending order of variable x. 24 3 -7 5 30 8 22 4 11 15 35 coefficient of x2 remainder coefficient of x coefficient of x°   ? Quotient = Q(x) = 4x2 + 11x + 15 Remainder R = 35 This process of division of a polynomial by a linear polynomial is called synthetic division. How do we operate synthetic division? Let us note the followings: Following Steps are performed for synthetic division. (i) Change the sign of the constant term in the linear divisor. For above example, x - 2 is compared with x - a, we get a = 2. (in divisor), (ii) Write down the coefficients of dividend f(x) with their signs. In above example. We write 4, 3, -7, 5 as coefficients of x3, x2, x constant term respectively. (iii) Bring down the leading coefficient (i.e., 4 in above example) (iv) Multiply it (4) by the constant term of the divisor with sign changed in above example 4 is multiplied by 2, we get 8. (v) Write the result under next coefficient (i.e., 3 in above example) and add them (we get 11 in above example) (vi) Again, multiply the result obtained (i.e., 11 is multiplied by 2 to get 22). (vii) Repeat the process until we get the last term. The last number so obtained is the remainder. The number just before the last term is constant term for quotient. Note : That the degree of quotient is 1 degree less than the dividend polynomial f(x), when f(x) is divided by a linear factor. In above example, the degree of f(x) is 3 and that of divisor is 1. (a) If the divisor is x+2, we write x+2 = x - (-2). Comparing it to x - a, we get a = -3, i.e., the constant term of the divisor is -3 with sign changed. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 27

vedanta Excel In Opt. Mathematics - Book 10 { ( )}(b) 2 If the divisor is 3x + 2, we write, 3x + 2 = 3 x – 3 . Then divide the polynomial ( )first by x =–2 , the quotient obtained is divided by 3 to get actual quotient. 3 Worked out Examples Example 1. What is degree of quotient when the polynomial of degree 4 is divided by a Solution: linear divisor? Example 2. Degree of polynomial = 4 Solution: Degree of linear polynomial = 1 Degree of quotient is less than degree of polynomial by 1. Degree of quotient = 4 - 1 = 3. Divide f(x) = 4x4 + 7x3 + 5x2 + 3x + 5 by linear polynomial d(x) = x - 2 Comparing x - 2 with x - a, we get a = 2, Now, by using synthetic division method, we get, 2 47 53 5 8 30 70 146 4 15 35 73 151 quotient = Q(x) = 4x3 + 15x2 + 35x + 73 remainder = R = 151 Example 3. Find the quotient and remainder when f(x) = 2x3 + 7x2 + x - 15 is divided by Solution: x + 2, by using synthetic division method. Here, f(x) = 2x3 + 7x2 + x - 15 divisor x + 2 = x -(-2), comparing it with x - a, we get a = -2 Writing the coefficients in order, -2 2 7 1 -15 -4 -6 10 23 -5 -5 Quotient = Q(x) = 2x2 + 3x - 5, Remainder = R = -5 Example 4. Find the quotient and remainder when 2x3 - 9x2 + 5x - 5 is divided by 2x - 3. Solution: Here, dividend, f(x) = 2x3 - 9x2 + 5x - 5 28 ( )divisor, d(x)= 2x-3=2 x - 3 x - a, we get a = 2 Comparing it with 3 . 2 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Writing the coefficients in order, we get 3 2 -9 5 -5 2 3 -9 -6 2 -6 -4 -11 quotient = Q(x) = 1 (2x2 - 6x - 4) = x2 - 3x - 2 2 remainder = R = -11 Example 5. Find the quotient and remainder when 4x3 + 2x2 - 4x + 3 is divided by 2x + 3. Solution: dividend = 4x3 + 2x2 - 4x + 3 Here, divisor = 2x + 3 = 2 {x – (– 3 )}, 3 2 2 The constant term in the divisor with sign changed is - . Writing the coefficients in order, we get -4 3 - 3 4 2 6 -3 2 -6 4 -4 20 Quotient = Q(x) = 1 (4x2 – 4x + 2) = 2x2 – 2x + 1 2 Remainder R = 0 Exercise 2.1 Very Short Questions 1. (a) If dividend = f(x), quotient = Q(x), divisior = d(x), and remainder = R, write their relation. (b) If x3 - 8 is divided by x - 2, find the quotient. (c) Is division of a polynomial by a polynomial again a polynomial ? Divide x3 + 27 by x2 and justify your answer. (d) What is the degree of quotient when polynomial of degree n is divided by a linear factor? 2. (a) What should be the degree of divisor in synthetic division method? (b) What should be the difference of degree of dividend and quotient in synthetic division ? 3. (a) When f(x) = x4 + 3x3 + 7x2 + 4x + 7 is divided by d(x) = x - 2, write the degree of quotient. (b) If f(x) = (x-2) × (x2 + 4x + 2) + 3, write the degree of f(x). Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 29

vedanta Excel In Opt. Mathematics - Book 10 Short Questions 4. Divide f(x) by d(x) in each of the following cases. Also find remainder if exists. (Use actual division method) (a) f(x) = x3 + 3x2 + 5x + 6, d(x) = x + 3 (b) f(x) = x2 + 5x + 6, and f(x) = x + 3 (c) f(x) = 3x4 - 9x2 - 4 and d(x) = x - 3 (d) f(x) = 2x3 + 3x2 - 5x + 6, and d(x) = 2x - 3. 5. (a) If 2x3 - 11x2 + 20x - 15 = (x - 6). Q(x) + R(x). Find Q(x) and R(x). (b) If 4x3 - 10x2 + 25x - 20 = (4x - 3). Q(x) + R(x). Find the value of Q(x) and R(x). 6. By using synthetic division, find the quotient and remainder when f(x) is divided by d(x) in each of the following by synthetic division: (a) f(x) = 3x3 - 2x2 + 13x + 10, d(x) = x - 2 (b) f(x) = 4x3 - x2 + 10x + 2, d(x) = x - 4 7. Find the quotient and remainder when f(x) is divided by d(x) in each of the following by synthetic division: (a) f(x) = x4 + 3x3 + 7x2 + 3x + 8, d(x) = x + 3 (b) f(x) = 2x3 + 4x2 + 3x + 7, d(x) = x + 2 8. Find the quotient and remainder in each of the following cases when f(x) divided by d(x) (use synthetic division method): (a) f(x) = 2x3 - 9x2 + 5x - 5, d(x) = 2x - 3. (b) f(x) = 8x3 + 4x2 + 6x - 7, d(x) = 2x - 1. (c) f(x)=3x3 - 3x2 + 2x + 5, d(x) = 2x – 1 (d) f(x)=4x3 -7x2 + 6x - 2, d(x) = 3x + 2 (e) f(x) = 4x3 - 3x2 + 7x + 8, d(x) = 2x + 3 (f) f(x) = 4x3 + 2x2 - 4x + 3, d(x) = 2x + 3 Long Questions 9. Let p(x) = x4 + x2 + 1, q(x) = x2 - x + 1, and r(x) = x2 + 2x + 5, find the value of {p(x) ÷ q(x)} + r(x). 10. If p(x) = x8 + x4 + 1, q(x) = x4 + x2 + 1, and r(x) = x4 + x2 + 6, then, find the value of {p(x) ÷ q(x)} + r (x). 11. (a) When the polynomial 2x3 + 9x2 - 7x + k, is exactly divisible by 2x + 3, find the value of k by using synthetic division. (b) When the polynomial 8x3 + 4x2 + 6x + p is exactly divisible by 2x - 1, find the value of p by using synthetic division. 12. Divide f(x) = x3 + 2x2 – 5x – 6 by (x + 1) and (x + 3) by using synthetic division. On the basis of remainder on the division process. Draw your conclusion. 30 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 1. (a) f(x) = d(x) . Q(x) + R (b) x2 + 2x + 4 (c) No (d) n – 1 2. (a) 1 (b) 1 3.(a) 3 (b) 3 4. (a) x2 + 5, – 9 (b) x + 2, 0 (c) 3x3 + 9x2 + 18x + 54, 158 (d) x2 + 3x + 2, 12 5. (a) 2x2 + x + 26, 141 (b) x2 – 7x + 79 , – 83 4 16 16 6. (a) 3x2 + 4x + 21, 52 (b) 4x2 + 15x + 70, 282 7. (a) x2 + 7x – 18, 62 (b) 2x2 + 3, 1 8. (a) x2 – 3x – 2, –11 (b) 4x2 + 4x + 5, – 2 (c) 3x2 – 3x + 5 , 45 (d) 4x2 – 29x + 112 , – 278 2 4 8 8 3 9 27 27 9x 41 91 (e) 2x2 – 2 + 4 , – 4 (f) 2x2 – 2x + 1, 0 9. 2x2 + 3x + 6 10. 2x4 + 7 11. (a) – 24 (b) –5 12. Both are factors of polynomial f(x). 2.3 Remainder Theorem Let f(x) = x2 + 4x + 5 be divided by x-2. Find the remainder and the value of f(x) at x = 2. Are they equal? Obviously, the answer is yes. Hence, a remainder can be calculated when a polynomial f(x) is divided by linear polynomial without actual division or synthetic division. Remainder Theorem Statement : If a polynomial f(x) is divided by x-a, then the remainder is R = f(a). Proof : Let Q(x) be the quotient when the polynomial f(x) is divided by (x-a), R be the remainder independent of x. Then by division algorithm, we get, f(x) = (x-a) × Q(x) + R This is an identity and is true for all values of x. So, put x = a, we get f(a) = (a -a) × Q(a) + R f(a) = R Hence, the remainder is R = f(a). Proved. Note : If a polynomial f(x) is divided by (x+a), then, the remainder is R = f(-a). Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 31

vedanta Excel In Opt. Mathematics - Book 10 Extension of the Remainder Theorem. ( )Statement : If a polynomial f(x) is divided by (ax-b), then, the remainder is R = f ba . ( )Proof : We have ax - b = a x - b . a Let Q(x) be the quotient when f(x) is divided by (ax-b) and the remainder R. Then, by division algorithm, f(x) = (ax - b) Q(x) + R ( )or, x- b f(x) = a a Q(x) + R Putting, x = b , we get, a f( b ) = a ( b - b ). ( )b +R a a a a ( )R = f b . a ( )Hence, the remainder is R = f b = value of f(x) at x = b . a a Note : ( )If a polynomial f(x) is divided by (ax+b), the remainder is R = f - b . a Remainder Theorem Table Dividend Divisor Remainder f(x) x-a f(a) f(x) x+a f(-a) f(x) ax-b f( b ) a f(x) ax+b f(- b ) a Worked out Examples Example 1. What is the remainder when polynomial f(x) is divided by x-5? Use the Solution: remainder theorem. Dividend = f(x) Divisor = x - 5 Comparing x - 5 with x - a, we get a = 5 by the remainder theorem, R = f(5). 32 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Example 2. By using the remainder theorem, find the remainder when 4x3 + 6x2 - 5x + 2 Solution: is divided by x - 2. Example 3. Here, f(x) = 4x3 + 6x2 - 5x + 2 = dividend, (x - 2) = divisor, Comparing Solution: it with x-a, Example 4. Solution: we get a = 2  By using the remainder theorem, Example 5. Solution: R = f(a) = 4.23 + 6.22 - 5.2 + 2 = 32 + 24 - 10 + 2 = 58 - 10 = 48 By using the remainder theorem, find the remainder when 4x4 + x3 + 2x - 50 is divided by 2x + 1. Let, f(x) = dividend = 4x4 + x3 + 2x - 50 divisor = 2x + 1 = 2{x – (– 1 )} 2 By the remainder theorem, ( )Remainder – 1 =R=f 2 = 4(– 1 )4 + (– 1 )3 +2 (– )1 - 50 2 2 1 1 2 4 8 = – – 51 = 2- 1 - 408 =- 407 8 8 Find the value of k when x3 + 3x2 - kx + 4 is divided by x + 2 if remainder is k(use the remainder theorem). Let, f(x) = x3 + 3x2 - kx + 4, dividend d(x) = x + 2 = x - (-2), divisor. By remainder theorem, R = f(-2) = (-2)3 + 3(-2)2 - k(-2) + 4 = -8 + 12 + 2k + 4 = 8 + 2k By question, R = 8 + 2k or, k = 8 + 2k ? k = -8 If f(x) = 4x4 - 8x2 + 3x - 20 is divided by g(x) = x + 2, prove that the remainder is 6. Here, f(x) = 4x4 - 8x2 + 3x - 20, dividend g(x) = x + 2, divisor = x - (-2), comparing it with x - a, we get a = -2 Now, by using remainder theorem, R = f(-2) = 4(-2)4 - 8(-2)2 + 3(-2) - 20 = 64 - 32 - 6 - 20 = 64 - 58 = 6 R = f(-2) = 6 Proved. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 33

vedanta Excel In Opt. Mathematics - Book 10 Exercise 2.2 Very Short Questions 1. (a) State the remainder theorem. (b) What is the remainder when f(x) is divided by x - b (use remainder theorem)? (c) What is remainder when polynomial f(x) is divided by ax + b? (d) What is the remainder when the polynomial p(x) is divided by mx + n2? 2. Find the remainder in each of the following when f(x) is divided by d(x), by using remainder theorem: (a) f(x) = x3 + 3x2 + 9x - 7 d(x) = x - 2 (b) f(x) = x3 + 4x2 - 7x - 20 d(x) = x - 4 (c) f(x) = 2x3 + 3x2 - 8x + 50 d(x) = x + 3 (d) f(x) = x3 + 4x2 – 20x + 20 d(x) = x + 2 (e) f(x) = 2x3 + 4x2 - 10x + 20 d(x) = 2x - 1 (f) f(x) = 16x3 + 12x2 + 24x + 5 d(x) = 2x + 3 3. Find the value of p in each of the following when f(x) is divided by d(x), with the help of remainder theorem: (a) f(x) = px3 + 4x - 10 d(x) = x + 3 R = 5 (b) f(x) = x3 + 3x2 - px + 4 d(x) = x - 2 R = 8 (c) f(x) =2x3 - 4x2 + 6x - p d(x) = x-2 R = 18 (d) f(x) = 4x3 - 3x2 + 3x - p d(x) = x - 2 R = 12 (e) f(x) = 2x3 - 3x2 + px - 8 R = f(3) = 10 (f) f(x) = x3 + 5x2 - px + 6 R = 2p d(x) = x - 1 (g) f(x) = x4 + x3 + =px2 + x + 20 f(2) = 20 Long Questions 4. (a) If 4x2 - 6x + p and x3 - 2x2 + px + 7 are polynomials, they are divided by x + 2 and remainders are equal. What is the value of p ? (b) If x3 - px2 + 10x + 11 and 2x3 - 2px2 + 7px + 23 are divided by x - 1, and each gives equal remainder, find the value of p. (c) If x3p2 + x2p - 8 is divided by x - 1, then, remainder is -2, find the values of p. (d) Find the value of p in polynomial 2x2p2 - 5px + 3 when it is divided by x-1 and remainder is zero. 34 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 1. (b) f(b) (c) f – b (d) f – n2 a m 2. (a) 31 (b) 80 (c) 47 (d) 68 (e) 65 (f) –58 3.(a) –1 (b) 8 4 13 (c) –6 (d) 14 (e) –3 (f) 4 (g) – 2 4. (a) – 37 (b) – 1 (c) 2, –3 (d) 1, 3 3 2 2 2.4 Factor Theorem (a) If f(x) = x2 - 25, then find the values of f(x) for x = 5 and x = -5 i.e. f(5) and f(-5). Discuss the following: (b) Is each of f(5) and f(-5) zero? (c) What are the factors of f(x)? We note that if f(5) = 0, then, (x-5) is a factor of f(x), and f(-5) = 0, (x+5) is also a factor of f(x). We say that (x+5) and (x-5) are the factors of f(x). This is called factor theorem. Factor Theorem Statement : If a polynomial f(x) in x is divided by (x-a) and the remainder Proof : R = f(a) = 0, then, (x-a) is a factor of f(x). Let Q(x) be the quotient when f(x) is divided by (x-a) and R is the Corollary (I) : remainder. Then, division algorithm, we get Corollary (II) : f(x) = (x-a) Q(x) + R If remainder R = f(a) = 0, then, (x-a) is a factor of f(x) If a polynomial f(x) in x is divided by (x+a) and the remainder R = f(-a) = 0, then, (x + a) is a factor of f(x) If a polynomial f(x) in x is divided by ax + b and the remainder –( )R = f b = 0, then, ax + b is a factor of f(x). a Converse of Factor Theorem If (x-a) is a factor of f(x), then, R = f(a) = 0 Proof : Let Q(x) be the quotient when a polynomial f(x) is divided by (x-a). Since (x-a) is a factor of f(x), we write by division algorithm, f(x) = (x-a).Q(x) or, f(x) = (x-a). Q(x) + 0 Also, f(x) = (x - a). Q(x) + R Comparing these two, we get R = 0. Proved. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 35

vedanta Excel In Opt. Mathematics - Book 10 Worked out Examples Example 1. Examine that (x-3) is a factor of f(x) = x3 - 27. Solution: Here, f(x) = x3 - 27 Example 2. comparing (x-3) with (x-a), Solution: Example 3. we get a = 3, Solution: Now, f(3) = 33 - 27 Example 4. R = f(x) = 0 Solution: By factor theorem, (x-3) is a factor of f(x). 36 Show that (x-3) is a factor of x3 - 6x2 + 11x - 6. Let f(x) = x3 - 6x2 + 11x - 6 comparing x - 3 with x-a, we get a = 3 f(3) = 33 - 6.32 + 11.3 - 6 = 27 - 54 + 33 - 6 = 60 - 60 = 0 Since R = f(3) = 0, (x-3) is 6 factor of f(x) What must be added to f(x) = x3 - 6x2 + 10x + 8, so that (x-3) is a factor of f(x)? Here, f(x) = x3 - 6x2 + 10x + 8. Let k be added, f(x) = x3 - 6x2 + 10x + 8 + k. comparing x - 3 with x - a, we get x = 3 f(3) = 33 - 6.32 + 10.3 + 8 + k = 27 - 54 + 30 + 8 = 65 - 54 + k = 11 + k By factor theorem, f(3) = 0 or, 11 + k = 0 k = –11 Therefore, -11 must be added. What must be subtracted from 3x3 + 5x2 - 6x + 7 so that x + 2 is a factor of it? Let f(x) = 3x3 + 5x2 - 6x + 7 Let k be subtracted from f(x). Then f(x) = 3x3 + 5x2 - 6x + 7 - k comparing (x+2) with x - a, we get a = -2 Now, f(-2) = 3(-2)3 + 5(-2)2 - 6(-2) + 7 - k = -24 + 20 + 12 + 7 - k = 15 - k Since, (x + 2) is a factor of f(x), by factor theorem, we get, R = f(-2) = 0 or, 15 - k = 0 ? k = 15 ? 15 should be subtracted. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Application of Factor Theorem and Remainder Theorem Factor theorem is used to factorize polynomials. Remainder theorem is used to find remainder when a polynomial is divided by linear polynomial without actual division. If remainder R = f(a) = 0, for real value of a, then, (x-a) is a factor of f(x). Example 5. Factorize f(x) = x3 - 3x2 - 6x + 8. Solution: Here, f(x) = x3 - 3x2 - 6x + 8 The possible factors of constant (8) are ± 1, ±2, ±4, ±8 For x = 1, f(1) = 13 - 3.12 - 6.1 + 8 =9-9=0 Hence, (x-1) is a factor of f(x). Now, f(x) = x3 - 3x2 - 6x + 8 = x3 - x2 - 2x2 + 2x – 8x + 8 = x2(x-1) - 2x(x-1) - 8(x-1) = (x – 1) (x2 – 2x – 8) = (x-1) (x2 - 4x + 2x - 8) = (x-1) {x(x-4) + 2(x-4)} ? f(x) = (x-1) (x-4) (x+2) Alternate Method : Here, f(x) = x3 - 3x2 - 6x + 8 f(1) = 13 - 3.12 - 6.1 + 8 =9-9=0 Hence, (x-1) is a factor of (x) Now, by using synthetic division. Writing the coefficients of x in order. 1 1 -3 -6 8 1 -2 -8 1 -2 -8 0 Example 6. Quotient, Q(x) = x2 - 2x - 8 = x2 - 4x + 2x - 8 Solution: = x(x-4) + 2(x-4) =(x-4) (x+2) ? f(x) = (x - 1) (x + 2) (x - 4) Factorize : 2x3 + 3x2 - 11x - 6 Let, f(x) = 2x3 + 3x2 - 11x - 6 The possible factors of constant term (-6), are ± 1, ±2, ±3, ±6 For x = 2, f(2) = 2.23 + 3.22 - 11.2 - 6 = 16 + 12 - 22 - 6 = 28 - 28 = 0 Hence (x - 2) is a factor of f(x) Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 37

vedanta Excel In Opt. Mathematics - Book 10 Now, by using the synthetic division, we get 22 3 -11 -6 4 14 6 27 30 Quotient, Q(x) = 2x2 + 7x + 3 = 2x2 + 6x + x + 3 = 2x(x+3) + 1(x + 3) = (x + 3) (2x + 1) ? f(x) = (x-2) (x+3) (2x + 1). Example 7. If f(x) = x3 + px2 - 6x + 8 has a factor (x-1), find the value of p. Then, Solution: factorize f(x) completely. Example 8. Since (x-1) is a factor of f(x), Solution: f(1) = 0 i.e. 12 + p.12 - 6.1 + 8 = 0 or, 1 + p - 6 + 8 = 0 or, p + 3 = 0 ? p = -3 Now, f(x) = x3 - 3x2 - 6x + 8 = x3 - x2 - 2x2 + 2x - 8x + 8 = x2 (x - 1) - 2x(x-1) -8(x-1) = (x-1) (x2 - 2x - 8) = (x-1){x(x-4) + 2(x-4)} = (x-1) (x+2) (x-4) Factorize (x + 1) (x + 3) (x + 5) (x + 7) + 16 Here, (x + 1) (x + 3) (x + 5)(x + 7) + 16 = {(x + 1)(x + 7)} {(x +3)(x + 5)} + 16 = (x2 + 8x + 7)(x2 + 8x + 15) + 16 = (D + 7) (D + 15) + 16, where D = x2 + 8x = D2 + 22x + 105 + 16 = D2 + 2.D.11 + (11)2 = (D + 11)2 = (x2 + 8x + 11)2 (replacing D = x2 + 8) = (x2 + 8x + 11) (x2 + 8x + 11) 2.5 Polynomial Equation Let f(x) = anxn + an-1xn-1 + .... + a0, an ≠ 0 be a polynomial of degree n of x. Then, f(x) = 0 is called the polynomial equation of degree n in x. In above polynomial an, an-1, .... a0  R. For examples, ax + b = 0, linear equation ax2 + bx + c = 0, quadratic equation. ax3 + bx2 + cx + d = 0, cubic equation. 38 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 If k be a real number such that f(k) = 0, k is called a root of the polynomial equation f(x) = 0, k is also called zero of polynomial. Remember that the degree of a polynomial, the highest exponent, dictates the maximum number of roots it can have. Example: f(x) = 2x3 + 13x2 - 36 = 0 Here, f(x) is of degree 3. Hence, it may have maximum three roots. It has roots -2, -6, 3 2 Example 9. Solve x3 - 13x - 12 = 0 Solution: Let, f(x) = x3 - 13x - 12 = 0 for x = - 1, f(-1) = (-1)3 - 13(-1) -12) = -1 + 13 - 12 = 13 - 13 = 0 Since f(-1) = 0, by factor theorem, (x + 1) is a factor of f(x). Now, f(x) = x3 + x2 - x2 - x - 12x - 12 = x2 (x + 1) - x(x + 1) - 12(x + 1) = (x +1)(x2 - x - 12) = (x + 1)(x2 - 4x + 3x - 12) =(x +1) {x(x-4) + 3(x-4)} = (x + 1) (x - 4) (x + 3) To solve, f(x) = 0 Alternative method or, (x + 1) (x - 4) (x + 3) = 0 Here, f(–1) = 0, x + 1 is a factor of f(x) by factor theorem. Either x + 1= 0....... (i) By using synthetic division method. or, x - 4 = 0..... (ii) -1 1 0 -13 -12 or, x + 3 = 0 .......(iii) From equation (i), x = -1 -1 1 12 From equation (ii), x = 4 1 -1 -12 0 From equation (iii), x =-3 Now, f(x) = 0 ? x = -3, -1, 4 or, (x + 1) (x2 – x – 12) = 0 Example 10. Solve: x3 - 9x2 + 24x - 20 = 0 or, (x + 1) (x2 – 4x + 3x – 12) = 0 Solution: or, (x + 1) (x – 4) (x + 3) = 0 Let f(x) = x3 - 9x2 + 24x - 20 =0 ? x = –3, –1, 4 For x =2, f(2) = 23 - 9.22 + 24.2 - 20 = 8 - 36 + 48 - 20 = 56 - 56 = 0 Hence, by factor theorem, (x-2) is a factor of f(x). Now, f(x) = x3 - 9x2 + 24x - 20 = x3 - 2x2 - 7x2 + 14x + 10x - 20 = x2(x -2) - 7x (x -2)+ 10(x-2) = (x-2) (x2 - 7x + 10) = (x-2) (x2 -5x - 2x + 10) = (x-2) {x(x-5) -2(x-5)} = (x-2)(x-5)(x-2) To solve, f(x) = (x-2)2 (x-5) = 0 Either (x - 2)2 = 0 .......(i) x-5 = 0 ........ (ii) From equation (i), x=2 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 39

vedanta Excel In Opt. Mathematics - Book 10 From equation (ii), x=5 ? x = 2, 5 Example 11. If (x-1) is a factor of x3 + ax2 - x + b and leaves remainder 15 when divided by x - 2. Find the values of a and b. Solution: Let f(x) = x3 + ax2 - x + b Since (x-1) is a factor of f(x), f(1) = 0 i.e., 13 + a.12 - 1 + b = 0 or, a + b = 0 ...............(i) Again, when f(x) is divided by x - 2, remainder (R) = 2 i.e., f(2) = 15 or, 23 + a.22 - 2 + b = 15 or, 4a + b = 9 .......... (ii) Solving equation (i) and (ii), we get, a+b=0 Subtracting -4a +- b =-9 - 3a = -9 ? a=3 Putting the value of a in equation (i), 3+b=0 b = -3 ? a = 3, b = -3 Exercise 2.3 Very Short Questions 1. (a) State factor theorem. (b) Is there any relation between the factor theorem and remainder theorem when a polynomial is factorized ? (c) If f(x) = (x - a) × Q(x) + R and x - a is a factor of f(x), what is the value of R? (d) If (x-a) is a factor of f(x), what is the value of R, f(a)? ( )(e) If f – b = 0 and f(x) is a polynomial, find one of its factor. a ( )(f) – 2 = 0. What is one of factor of f(x) ? If f(x) is a polynomial and f 3 2. (a) If (x-k) is a factor of f(x) = x2 - 9 What is the value of f(k) ? (b) If (x - a ) is a factor of f(x) = x2 - 121, what is the value of a? 40 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 (c) If (x - a) is a factor of f(x) = x3 - 27, find the value of a. 3. (a) If (x + 2) is a factor of f(x) = x3 - 8, find f(-2). (b) If (x - 3) is a factor of f(x) = x3 - 27, find the value of f(3). (c) If f(x) = 2x3 + 3x2 + - 11x - 6 has a factor (x-2), find f(2). 4. (a) By using factor theorem check, that (x - 5) is a factor of 2x2 - 11x + 5 or not. (b) By using factor theorem check, that (x-3) is a factor of x3 - 4x2 + x + 6 or not. (c) Show that (x + 1) is a factor of x3 - 4x2 + x + 6. 5. (a) What is meaning of roots of a polynomial equation f(x) = 0? (b) What is zero of polynomial ? (c) What is a polynomial equation ? (d) What is the difference between a zero polynomial and the zero of a polynomial ? (e) Give an example of each of zero polynomial and zero of polynomial. (f) What are the maximum number of roots of polynomial equation f(x) = 0, whose degree is 3 ? Short Questions 6. (a) What must be added to f(x) = 2x3 + 6x2 + 4x + 8 so that (x + 3) is a factor of it ? (b) What must be added to f(x) = x3 - 6x2 + 11x + 8 so that (x - 3) is a factor of it ? (c) What must be subtracted from f(x) = x3 + 8x2 + 4x + 10 so that (x+2) is a factor of it ? 7. (a) If 4x2 + px + 8 has a factor x - 2, find the value of p. (b) If 2x3 + 4x2 + px + 7 has a factor of (x + 2), find the value of p. (c) If f(x) = x3 + 4x2 + kx + 7 has a factor (2x - 1), find the value of k. 8 (d) If x3 - bx2 - 2x + b + 5 has a factor (x - b), find the value of b. (e) If f(x) = 3x3 + 3x + k and f(-1) = 0, find the value of k. Long Questions 8. (a) By using the factor theorem, show that (x + 1), (x-2), and (x-3) are the factors of f(x) = x3 - 4x2 + x + 6. (b) By using the factor theorem, show that (x + 2), (x + 6), and (2x - 3) are the factors of f(x) = 2x3 + 13x2 - 36. (c) If f(x) = 6x3 + 11x2 -26x - 15, determine which are the factors of f(x) given below: (i) x + 1 (ii) (x + 3) (iii) 2x + 1 (iv) 3x-5 (v) 2x - 1 9. Factorize : (a) x2 + 3x + 2 (b) 3x2 + 8x + 4 (c) x2 - 5x + 6 (d) x2 - 4x + 3 (e) x2 - 6x + 9 (f) x2 + 4x - 12 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 41

vedanta Excel In Opt. Mathematics - Book 10 10. Factorize by using the factor theorem : (a) x3 - 6x2 + 11x - 6 (b) x3 - 4x2 + x + 6 (c) x3 - 11x2 + 31x - 21 (d) x3 - 3x2 + 4x - 4 (e) x3 + 2x2 - 5x - 6 (f) 3x3 - 6x2 + 4x - 8 (g) 6x3 + 7x2 - x - 2 (h) 2x3 + 3x2 - 11x - 6 (i) x3 + 5x2 - 2x - 24 (j) x3 - 9x2 + 26x - 24 (k) x3 - 9x2 + 23x - 15 (l) x3 - 19x - 30 (m) x3 - 5x + 4 (n) 2x3 + 3x2 - 11x - 6 11. Factorize : (a) (x + 3) (x + 5) (x + 7) (x + 9) + 16 (b) (x + 2)(x - 3) (x - 1) (x - 6) + 56 (c) (x - 1) (2x2 + 5x + 5) + 4 (d) (x - 1) (2x2 + 15x + 15) - 21 (e) (x - 3) (x2 - 5x + 8) - 4x + 12 12. Solve the following polynomial equations : (a) x3 - 3x2 - 4x + 12 = 0 (b) x3 - 9x2 + 24x - 20 = 0 (c) x3 - 7x2 + 7x + 15 = 0 (d) x3 - 4x2 + x + 6 = 0 (e) 3x3 - 19x2 + 32x - 16 = 0 (f) 6x3 + 7x2 - x - 2 = 0 (g) x3 - 6x2 + 11x - 6 = 0 (h) 5x3 + 2x2 - 20x - 8 = 0 (i) y3 - 13y - 12 = 0 (j) y3 - 3y - 2 = 0 (k) 4y3 - 3y - 1 = 0 (l) y3 - 19y - 30 = 0 (m) x4 - x3 - 19x2 + 49x - 30 = 0 (n) 2x4 - 13x3 + 28x2 - 23x + 6 = 0. 13. Find the values of a and b in each of the following cases : (a) (x - 1) and (x - 2) are the factors of x3 - ax2 + bx - 8. (b) (x + 3) and (2x - 7) are factors of ax2 - bx - 21. (c) 2x3 + ax2 + bx - 2 has a factor (x + 2) and leaves remainder 7 when divided by 2x - 3. (d) 2x3 + ax2 - 11x + b leaves a remainder 0 and 42 when divided by (x-2) and (x-3) respectively. Project Work 14. Write two functions f(x) and g(x) of degree 4 both of which have two common factors. How many factors are there in each of the functions f(x) and g(x) ? 42 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 1. (c) 0 (d) 0 (e) ax + b (f) 3x + 2 2. (a) ± 3 (b) ± 11 (c) 3 3.(a) 0 (b) 0 (c) 0 4.(a) yes (b) yes 5. (f) 3 6.(a) 4 (b) –14 (c) 26 7. (a) –12 (b) 7 (c) –4 (d) 5 (e) 6 8. (c) (ii), (iii), (iv) 2 9. (a) (x + 1) (x + 2) (b) (x + 2) (3x + 2) (c) (x - 2) (x - 3) (d) (x - 3) (x - 1) (e) (x - 3) (x - 3) (f) (x + 6) (x - 2) 10. (a) (x - 1) (x - 2) (x - 3) (b) (x - 2) (x - 3) (x + 1) (c) (x - 1) (x - 3) (x - 7) (d) (x - 2) (x2 - x + 2) (e) (x + 1) (x - 2) (x + 3) (f) (x - 2) (3x2 + 4) (g) (x + 1) (2x - 1) (3x + 2) (h) (x - 2) (x + 3) (2x + 1) (i) (x - 2) (x + 3) (x + 4) (j) (x - 2) (x - 3) (x - 4) (k) (x - 1) (x - 3) (x - 5) (l) (x + 2) (x + 3) (x - 5) (m) (x - 1) (x2 + x - 4) (n) (x - 2) (x + 3) (2x + 1) 11. (a) (x2 + 12x + 31)2 (b) (x - 5) (x + 1) (x2 - 4x - 4) (c) (x + 1)2 (2x - 1) (d) (x + 2) (x + 6) (2x - 3) (e) (x - 1) (x - 3) (x - 4) 12. (a) ±2, 3 (b) 2, 5 (c) -1, 3, 5 (d) -1, 2, 3 (e) 1, 4, 4 (f) -1, - 23, 1 (g) 1, 2, 3 (h) ±2, - 2 3 2 5 1 (i) -1, 4, -3 (j) -1, 2 (k) 1, - 2 (l) -2, -3, 5 (m) 1, 2, 3, -5 (n) 1, 2, 3, 1 13. (a) 7, 14 2 (b) 2, 1 (c) 3, -3 (d) 3, -6 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 43

vedanta Excel In Opt. Mathematics - Book 10 3 Sequence and Series 3.0 Review Study the following pattern of numbers. (i) 1, 2, 3,4, 5, ................. (ii) 5, 10, 15, 20, 25 ............................, 100. (iii) 2, 4, 8, 16, 32,......................... (iv) 1, 2,4, 7, 11, 16 Now, answer the following questions: (a) Does the each of above pattern of number follow a certain rule? (b) What are the rules followed by each pattern of numbers? (c) Can we find next three terms in each pattern of numbers? (d) Can we find the nth term in each of these cases? (e) Among them, which are finite and which are infinite? (f) What is the series of each of above pattern of numbers that follows a certain rule? Discuss answers of above questions and arrive at the conclusions. Definition: A set of numbers presented in a definite order and formed according to a certain rule is known a sequence. Each number of a sequence is called term of the sequence. The terms of a sequence are denoted by t1, t2, t3, ......, tn in case of finite sequence. The nth term of the sequence is called the general term. The series of a sequence t1, t2, t ..... . tn is written as t1 + t2 +..... + tn. 3 Progression : A sequence (or series) is called a progression if the functional relations exist between its two successive terms is constant. Examples : (i) 3 + 6 + 9 + 12,........ series (ii) 7, 9, 11, 13, ...... sequence (iii) 2, 4, 8, 16, …, sequence. All of these three are examples of progression, why? 3.1 Arithmetic Sequence and Series. Let us consider the following sequence of number. (i) 1, 5, 9, 13, 17, ...... 44 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 (ii) 100, 95, 90, 85, 80,..... Each of above sequence of numbers are in a certain rule. In sequence (i), each term is obtained by adding 4 on the preceding term. Difference between a term and its preceding term is constant throughout the whole sequence. i.e., 5 - 1 = 4, 9 - 5 = 4, 13 - 9 = 4, 17 - 13 = 4,........, etc. The difference is called common difference and it is denoted by d. This type of sequence is called arithmetic sequence and the corresponding series is called arithmetic series(A.S). i.e., 1 + 5 + 9 +13 + 17 ...... is an arithmetic series. In sequence (ii) each term is obtained by subtracting 5 from the preceding term. Common difference is -5. Arithmetic series of above sequence is given by 100 + 95 + 90 + 85 + 80 + ................... Definition : A sequence of numbers with a constant difference between two consecutive terms is called an arithmetic sequence. The constant difference is called common difference and it is denoted by d. If ‘a’ is the first term and common difference d, a, a + d, a + 2d, a + 3d ...... are the terms of an arithmetic sequence. Common difference (d) = tk+1 - tk , k ≥ 1, k  N, for k = 1, d = t2 - t1 General Term of Arithmetic Sequence Let us consider an arithmetic sequence 5, 8, 11, 14, 17,..... In this sequence, Common difference (d) = 8-5 = 3 The first term (t1) = a = 5 Now, the first term (t1) = 5 + (1-1) 3 = a + (1 - 1)d the second term (t2) = 8 = 5 + (2-1)3 = 5 + (2-1)d the third term (t3) = 11 = 5 + (3-1)3 = a + (3-1)d the nth term (tn) = a + (n-1)d = 5 + (n - 1)3 = 3n + 2 The general term (tn) = a + (n - 1)d, where n is the number of terms. If ‘l’ is the last term of an arithmetic sequence, we write tn = l = a + (n - 1)d. Worked Out Examples Example 1. Show that each of the sequence is in arithmetic: Solution: (a) 2, 4, 6, 8, 10, 12..... (b) 120, 100, 80, 60,..... (a) In 2, 4, 6, 8, 10,12 Here, t2- t1= 4-2=2,t3-t2=6-4= 2, t4-t3=8-6=2t5, - t4=10-8 = 2, t6 - t5=12 – 10 =2 Therefore, the difference between any two successive terms is 2 throughout the whole sequence. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 45

vedanta Excel In Opt. Mathematics - Book 10 Given sequence is arithmetic, Here, 120, 100, 80, 60,...... (b) t2 - t1 = 100 - 120 = -20 t3 - t2 = 80 - 100 = -20 t4 - t3 = 60 - 80 = -20 Therefore, the difference between any two successive terms is -20 throughout the whole sequence. Given sequence is arithmetic. Example 2. Find the general term of the given arithmetic sequences: Solution: (a) 7, 14, 21, 28,....... (b) 10, 25, 40, ...... Example 3. Solution: (i) Here, the first term (a) = 7 Example 4. The common difference (d) = t2 - t1 = 14 - 7 = 7 Solution: The general term (tn) = a + (n - 1)7 = 7 + 7n - 7 = 7n (ii) Here, the first term (a) = 10 The common difference (d) = t2 - t1 = 25 - 10 = 15 Now, the general term (tn) = a + (n - 1) d = 10 + (n - 1) 15 = 10 + 15n - 15 = 15n - 5 ? tn = 5(3n - 1) Find the next two terms of given arithmetic sequence. 1, 4, 7, 10, 13,........ In given arithmetic sequence, the first term (a) = 1 the common difference (d) = 4 - 1 =3 By using formula, tn = a + (n - 1) d or, t6 = 1 + (6 - 1).3 = 1 + 15 = 16 and t7 = 1 + (7 - 1) 3 = 19 Find the common difference of an arithmetic sequence whose first term is 9 and fifth term is 17. Here, the first term (t1) = 9 the fifth term (t5) = 17 the common difference (d) = ? By using formula, tn = a + (n - 1) d or, t5 = 9 +(5 - 1) .d or, 17 = 9 + 4d or, 4d = 8 ? d=2 Therefore, the common difference (d) is 2 46 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Example 5. Find the number of terms in the series 13 + 16 + 19 + ........... + 79 Solution: Here, the first term (a) = 13 Example 6. Solution: the common difference (d) = 16 - 13 = 3 Example 7. the last term (tn) = 79 Solution: By using formula, tn = a + (n-1)d or, 79 = 13 + (n - 1) 3 or, 66 + 3 = 3n or, 3n = 69 or, n= 69 =23. 3 Therefore, number of terms (n) is 23. If the 3rd and 8th terms of an A.P. are 14 and 34 respectively, find the first term, the common difference, and the 12th term of the sequence. In given A.P., ( tn = a + (n – 1)d) the third term (t3) = 14 or, a + 2d = 14 .......... (i) and the 8th term (t8) = 34 or, a + 7d = 34.......... (ii) Subtracting (i) from (ii) we get, a + 7d = 34 –a +– 2d =–14 5d = 20 ?d = 2v50alu=e 4 d in equation (i), we get, Putting the of a + 2 × 4 = 14 or, a + 8 = 14 or, a = 14 - 8 ⸫ a=6 Again, t12 = a + (12 - 1)d = 6 + 11 (4) = 6 + 44 = 50 If 5th term of an AP is 28 and the 9th term is 48. Are 63 and 95 the terms of the AP? In given AP, the fifth term (t5) = 28 and the ninth term (t9) = 48 Then, by using formula, tn = a + (n - 1)d or, t5 = a + 4d. 28 = a + 4d ...........(i) and t9 = a + 8d or, 48 = a + 8d ....... (ii) Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 47

vedanta Excel In Opt. Mathematics - Book 10 Subtracting (i) from (ii), we get, 48 = a + 8d –28 =–a +– 4d 20 = 4d ? d=5 Putting the value of ‘d’ in (i), we get, 28 = a + 4 × 5 or, 28 = a + 20 or, a =28 - 20 = 8 Let tn = 63 if possible, then find n. or, a + (n - 1) d = 63 or, 8 + (n - 1).5 = 63 or, (n - 1) . 5 = 55 or. n -1 = 11 ? n = 12 Therefore, 63 is the 12th term of the A.P. Again, let tm = 95. If possible, then find the value of m. tm = a + (m - 1) d. or, 95 = 8 + (m-1).5 or, 95 - 8 = (m - 1) 5 or, m857==85m7 -1 = 87 + 5 = 92 = 18 2 . or. +1 5 5 5 The number of terms cannot be in fraction. Therefore, 95 is not the term of the given A.P. Example 8. The nth term of the series 9 + 7 + 5 + ....... is equal to the nth term of the series Solution: 15 + 12 + 9 + ........, find the value of n. For the first series, 9 + 7 + 5 + ..... the first term (a) = 9 the common difference (d) = 7 - 9 = - 2 tn = a + (n - 1) d = 9 + (n - 1) (-2) = 9 - 2n + 2 = 11 - 2n for the second series 15 + 12 + 9 + ....... the first term (a) = 15 the common difference (d) = 12 - 15 = - 3 tn = a +(n - 1) d = 15 + (n - 1) (-3) = 15 - 3n + 3 = 18 - 3n By question, 11 - 2n = 18 - 3n or, -2n + 3n = 18 - 11 ? n=7 Therefore, the 7th term of both series is equal. 48 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Example 9. The starting salary of a man is Rs. 7200 per month. If he gets an increment of Rs. 200 every year, what is his salary in the 10th year ? Solution: Starting salary = the first term (a) = Rs. 7200 yearly salary increment = d = Rs. 200. Salary in the 10th year is the 10th term of the arithmetic sequence. By using formula, tn = a + (n - 1) d or, t10 = 7200 + (10 - 1). 200 = 7200 + 1800 = 9000. Therefore, his salary in the 10th year is Rs. 9000. Exercise 3.1 Very Short Questions 1. (a) Define sequence and series. (b) Write a difference between a sequence and series with an example of each. (c) Define a progression with an example. (d) What is a general term of a sequence ? 2. (a) Define an arithmetic sequence. (b) Define common difference of an A.P. (c) If a = the first term, d = the common difference n = the number of terms of an A.P. Write formula for common difference in an A.P. 3. Find common difference in each of given A.P. (a) 2, 4, 6, 8,...... (b) 10, 15, 20, 25,....... (c) 20, 18, 16, 14, .......... 4. Which of the following sequences are in A.P. ? (a) 2, 7, 12, 17, ........... (b) 12, 8, 6, 4, .............. (c) 10, 20, 30, 40, ........ (d) 200, 180, 160, 140, ............. 5. (a) If the first term of an A.P. is 10 and the common difference 5. Find the 4th term of the progression. (b) If the 10th term of an A.P. is 100 and common difference is 10, find the first term. 6. A man deposits sequence of payments Rs. 200, Rs. 400, Rs. 600, Rs. 800, and Rs. 1000 for six days. What type of sequence is it ? Short Questions 7. Find the general terms for each of the following arithmetic sequence. (a) 3,4,5,6,7,....... (b) 7, 10, 13, 16, ..... (c) 6, 9, 12, 15,..... (d) 51, 45, 39, ..... 8. Compute the first term and the 10th term in each of the following sequence. (a) d = 3, t4 = 12 (b) d = -4, t6 = 24 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 49

vedanta Excel In Opt. Mathematics - Book 10 (c) d = 7, t8 = 63 (d) d = - 8, t5 = 80 9. Determine the number of terms in each of the following A.P. (a) the first term = 2, the common difference = 4, l = 102 (b) the first term = 3, the common difference 6, l = 57. 10. Find the number of terms in each of the following A.P.: (a) 2 + 4 + 6 + ....... + 100 (b) 3 + 6 + 9 + ......... + 99 11. (a) Which term of the series 2 + 4 + 6 + ..... + 512 is 256 ? (b) Which term of the series is 5 + 10 + 15 + ..... + 200 ? 12. (a) Is 75 a term of the series 9 + 11 + 13 + .....? (b) Is -100 a term of series - 10 - 20 -30 - ...... ? Long Questions 13. Find the common difference and the first term of arithmetic progression in each of the following cases: (a) t4 = 13, t6 = 7 (b) t5 = 15, t7 = 9 (c) t4 = 75, t10 = 117 (d) t3 = -40, t13 = 0 (e) t5 = 15, t7 =9 Also, find the 10th term in each of above progression. 14. (a) In an A.P. the 3rd and the 13th terms are 40 and 0 respectively, which term will be 28 ? (b) The 6th and the 9th terms of an A.P. 23 and 35. Which term is 67 ? 15. (a) If the nth term of the A.S. 7, 12, 17, 22,...... is equal to the nth term of another A.S. 27, 30, 33, 36,...... find the value of n. (b) If the nth term of an A.P. 23, 26, 29, 32, ...... is equal to the nth term of another A.P. 59, 58, 57, 56,....... , find the number of terms. 16. (a) In an A.P., the 7th term is four times the second terms and the 10th term is 29, find the progression. (b) In an A.P. the 10 times of 10th term is fifteen times the fifteen term. If the first term is 48, find the progression. 17. (a) If x + 6, 3x, and 2x + 9 are in A.P., find the value of x and the next three terms the progression. (b) If 4k + 10, 6k + 2, and 10k are in A.P., find the value of k and terms t7, t8, and t9. 18. (a) If the pth term of an AS is q and qth term is p. Show that mth term is p+q-m. (b) If m times the mth term of an A.P. equal to n times its nth term, show that (m+n)th term of the A.P. is zero. 19. If the pth, qth, and rth terms of an A.P. are a,b, and c respectively. prove that : p(b-c) + q (c-a) + r(a-b) = 0 50 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur