Mathematics & Statistics Commerce Part 1 Standard XI PDF
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Uploaded by PhenomenalForethought9808
Swami Vivekanand High School
2019
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This is a mathematics textbook for Standard XI Commerce students, published by the Maharashtra State Bureau of Textbook Production and Curriculum Research in 2019. It covers topics in algebra, coordinate geometry, and complex numbers.
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The Coordination Committee formed by GR No. Abhyas - 2116/(Pra.Kra.43/16) SD - 4 Dated 25.4.2016 has given approval to prescribe this textbook in its meeting held on 20.06.2019 and it has been decided to implement it from the educational year 2019-20. MATHEMATICS...
The Coordination Committee formed by GR No. Abhyas - 2116/(Pra.Kra.43/16) SD - 4 Dated 25.4.2016 has given approval to prescribe this textbook in its meeting held on 20.06.2019 and it has been decided to implement it from the educational year 2019-20. MATHEMATICS AND STATISTICS (COMMERCE) Part-I STANDARD XI 2019 Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune - 411 004 Download DIKSHA App on your smartphone. If you scan the Q.R.Code on this page of your textbook, you will be able to access full text. If you scan the Q.R.Code, you will be able to access audio-visual study material relevant to each lesson, provided as teaching and learning aids. First Edition : 2019 © Maharashtra State Bureau of Textbook Production First Reprint : 2020 and Curriculum Research, Pune- 411 004. Maharashtra State Bureau of Textbook Production and Curriculum Research reserves all rights relating to the book. No part of this book should be reproduced without the written permission of the Director, Maharashtra State Bureau of Textbook Production and curriculum Research, Pune. Commerce Mathematics Paper I Cover, Illustrations and Committee Members Computer Drawings Dr. Sharad Gore (Chairman) Shri. Sandip Koli, Artist, Mumbai Dr. Mangala Narlikar (Member) Smt. Prajakti Gokhale (Member) Co-ordinator Shri. Prahallad Chippalagatti (Member) Ujjwala Shrikant Godbole Shri. Prasad Kunte (Member) I/C Special Officer for Mathematics Shri. Sujit Shinde (Member) Shri. Milind Patil (Member) Typesetter Shri. Pradipkumar Bhavsar (Member) Shri. Ramakant Sarode (Member) Baladev Computers, Mumbai Smt. Ujjwala Godbole (Member-Secretary) Production Sachchitanand Aphale Commerce Mathematics Paper I Chief Production Officer Study Group Members Sanjay Kamble Production Officer Dr. Ishwar Patil Dr. Pradeep Mugale Dr. Pradnyankumar Bhojankar Shri. Devanand Bagul Prashant Harne Shri. Uday Mahindrakar Shri. Balkrishna Mapari Asst. Production Officer Shri. Prafullachandra Pawar Shri. Sachin Batwal Smt. Nileema Khaladkar Shri. Swapnil Shinde Smt. Swati Powar Smt. Deepti Kharpas Paper Smt. Mahadevi Mane Shri. Pramod Deshpande 70 GSM Cream wove Shri. Dilip Panchamia Shri. Vinayak Godbole Print Order No. Shri. Amit Totade Smt. Gauri Prachand Shri. Dhananjay Panhalkar Smt. Supriya Abhyankar Printer Shri. Sharadchandra Walagade Publisher Chief Co-ordinator Vivek Uttam Gosavi, Controller Smt. Prachi Ravindra Sathe Maharashtra State Textbook Bureau, Prabhadevi Mumbai- 400 025 NATIONAL ANTHEM PREFACE Dear Students, Welcome to the eleventh standard! You have successfully completed your secondary education and have entered the higher secondary level. You will now need to learn certain mathematical concepts and acquire some statistical skills to add accuracy and precision to your work. Maharashtra State Bureau of Text Book Production and Curriculum Research has modified and restructured the curriculum in Mathematics and Statistics for the Commerce stream in accordance with changing needs of the society. The curriculum of Mathematics and Statistics is divided in two parts. Part-1 covers topics in Algebra, Co-ordinate Geometry, Complex Numbers, Sets and Relations. Functions and Calculus. Part-2 covers Combinatorics and Statistics. There is a special emphasis on applications. Activities are added at the end of chapters for creative thinking. Some material will be made available on E-balbharati website (ebalbharati.in). It contains a list of specimen practical problems on each chapter. Students should complete the practical exercises under the guidance of their teachers. Journals are to be maintained by students and assessed by teachers. You are encouraged to use modern technology in your studies. Explore the Internet for more recent information on topics in the curriculum. Get more examples and practice- problems from the Internet. You will enjoy learning if you follow three simple principles: a thorough study of the textbook, learning based on activities, and continuous practice of solving problems. On the title page Q.R. code is given. It will help you to get more knowledge and clarity about the contents. This textbook is prepared by mathematics subject commitee and study group. This book has also been reviewed by senior teachers and eminent scholars. The Bureau would like to thank all of them for their contribution in the form of creative writing, constructive criticism, and valuable suggestions in making this book useful. Also the Bureau is grateful to members of the mathematics subject commitee, study group and reviewers for sparing their valuable time in preparing this book. The Bureau hopes that the textbook will be received well by all users in the right spirit. You are now ready to study. Best wishes for a happy learning experience. (Dr. Sunil Magar) Pune Director Date : 20 June 2019 Maharashtra State Bureau of Textbook Indian Solar Date : 30 Jyeshtha 1941 Production and Curriculum Research, Pune. XI Mathematics Commerce Part I Competency statements Sr. Area Topic Competency Statements No The student will be able to work with sets and set functions. Sets construct sets from given conditions/description/ rule. Sets and solve problems using set theory. 1 Relations identify the types of relations. Relations use relations to associate different sets. verify equality, equivalence or other relation- ships between given sets. work with function defined on different 2 Functions Functions domains. identify different types of functions. carry out complicated operations on functions. Complex Complex simplify algebraic expressions involving 3 Numbers Numbers complex numbers. identify the type of a given sequence. Sequence find the general term of given sequence. Sequence and 4 identify the type of a given series series Series find the nth term of a given series find the sum of the first n terms of a given series find the sum to infinite terms of a given series find equation of a straight line satisfying given Locus and Locus and 5 conditions Straight Line Straight Line identify properties of given set of straight lines find value of a determinant. 6 Determinats Determinants simplify determinant. solve linear equations in 2/3 variables, find area of triangle using determinants. find limit of a function 7 Limits Limits determine whether a given function has a limit determine whether a given function is continuous at a given point 8 Continuity Continuity determine whether a given function is continuous over a specified interval identify points of discontinuity of a given function 9 Differentiation Differentiation differentiate algebraic functions Sr. No. Topic Page No. 1 Sets and Relations 1 2 Functions 20 3 Complex Numbers 33 4 Sequences and Series 44 5 Locus and Straight Line 65 6 Determinants 81 7 Limits 97 8 Continuity 107 9 Differentiation 115 10 Answers 126 1. SETS AND RELATIONS ‘Happy people’, ‘Clever student’ are all relative terms. Here, the objects are not well-defined. Let’s study. In the last two collections. We can determine the objects clearly. Thus, we can say that objects are Representation of a Set well-defined. Intervals 1.1 SET: Types of sets Definition: Operations on sets A Collection of well-defined objects is called Ordered pair a set. Relations Object in a set is called its element or member. Let’s recall. We denote sets by capital letters A,B,C. etc. The elements of a set are represented by small The concept of a set was developed by German letters a, b, c, x, y, z etc. If x is an element of a mathematician George Cantor (1845-1918) set A we write x∈A, and read as ‘x belongs to A’. If x is not an element of a set A, we write x∉A, You have already learnt about sets and some and read as ‘x does not belong to A.’ basic operations involving them in the earlier standards. For example, zero is a whole number but not a natural number. In everyday life, we generally talk about group ∴ 0∈W and 0∉N or a collection of objects. Surely you must have used the words such as team, bouquet, bunch, Representation of a set: flock, family for collection of different objects. 1) Roster method: It is very important to determine whether a In the Roster method, we list all the given object belongs to a given collection or not. elements of the set within brackets, and separate Consider the following collections: the elements by commas. i) Successful persons in your city. Example : State the sets using Roster method. ii) Happy people in your town. i) B is the set of all days in a week. iii) Clever students in your class. B = {Monday, Tuesday, Wednesday, iv) Days in a week. Thursday, Friday, Saturday, Sunday} v) First five natural numbers. ii) C is the set of all vowels in English First three collections are not examples of sets, alphabets. but last two collections represent sets. This is C = {a, e, i, o, u} because in first three collections, we are not sure of the objects. The terms ‘successful persons,’ 1 2) Set-Builder method: In the set builder method, we describe the Let’s learn. elements of the set by specifying the property 1.2 INTERVALS: which determines the elements of the set uniquely. 1) Open Interval: Let a, b ∈R and a < b then Example : State the sets using set-builder the set {x / x ∈ R a < x < b} is called open method. interval and is denoted by (a,b). All the i) Y is the set of all months of a year. numbers between a and b belong to the Y = {x | x is month of a year} open interval (a,b) but a, b themselves do not belong to this interval. ii) B is the set of perfect squares of natural numbers. B = {x∈N/x is perfect square} Fig. 1.4 3) Venn Diagram: ∴ (a,b) = {x / x ∈ R, a < x < b} The pictorial representation of a set is called Venn diagram. Generally, the geometrical closed 2) Closed Interval: Let a, b∈R and aa} 1) Empty Set: A set containing no element is called an empty or a null set and is denoted by the Fig. 1.8 symbol f or { } or void set. ii) The set of all real numbers greater than e.g. A = {x / x ∈ N, 1< x < 2 }, n(A) = 0 or equal to a 2) Singleton set: i.e. [a,∞) = {x/x∈R, x ≥ a} A Set containing only one element is called a singleton set. Fig. 1.9 e.g. Let A be the set of all integers which are neither positive nor negative. 6) i) The set of all real numbers less than b. ∴A = {0}, n (A) = 1 ie. (-∞, b) 3) Finite set: x A set in which the process of counting of Fig. 1.10 elements comes to an end is called a finite set. ∴(-∞, b) = {x/x∈R, x < b} e.g. the set of letters in the word 'beautiful'. ii) The set of all real numbers less than or equal to b i.e. (-∞, b] A = {b,e,a,u,t,i,f,l}, n (A) = 8 A is a infinite set 4) Infinite set: Fig. 1.11 A set which is not finite, is called an infinite ∴(-∞, b] = {x/x∈R, x ≤ b} set. 7) The set of all real numbers i.e. (-∞, ∞) e.g. set of natural numbers. Let’s Note. Fig. 1.12 ∴ (-∞,∞) = {x/x∈R, -∞ 0) f(x) is represented in many different ways. Many mathematicians have used different notations for Solution : derivatives. Let f(x) = ax Most commonly used is the Leibniz’s notation. f(x + h) = ax+h dy By the method of first principle, The symbols dx, dy and were introduced by dx Gottfried Wilhelm Leibnitz. f ( x h) f ( x ) f’'(x) = lim h 0 h Note : If we consider the graph of a function in a( x h) a x XY-plane, then the derivative of the function f at = h 0 lim h a point ‘a’ , is the slope of the tangent line to the curve y = f(x) at x = a. (We are going to study this a x (a h 1) = h 0 lim in detail in next level) h x ah 1 = a lim Let’s learn. h 0 h Derivatives of some standard functions ax 1 f '( x ) a x log a.......... Q lim log a Ex. (1) Find the derivative of xn w. r. t. x. (n∈N) x 0 x Solution : Try the following Let f(x) = xn 1 –n (1) If f(x) = n , then prove that f ′(x) = x xn+1 f(x + h) = (x + h)n (2) If f(x) = ex , then prove that f ′(x) = ex By the method of first principle, 116 SOLVED EXAMPLES 4x h 4x f '(x) = h 0 lim h EX. 1. Find the derivatives of the following by 4 x (4h 1) using the method of first principle = lim h 0 h (i) x (ii) 4 x (iii) logx x 4h 1 = 4 lim Solution : h 0 h (i) Let f(x) = x ax 1 f '( x ) 4 x log 4.......... Q lim log a x+h x 0 x f(x + h) = By the method of first principle, (iii) Let f(x) = logx f ( x h) f ( x ) f(x+h) = log (x+h) f '(x) = lim h 0 h f ( x h) f ( x ) f '(x) = lim ( x h) x h 0 h = lim h 0 h log( x h) log( x) = lim ( x h) x ( x h) x h 0 h = lim h 0 h ( x h) x x h x h x = lim log x lim h 0 = h 0 h ( x h) x h h = lim h h 0 h ( x h) x lim log 1 x = h 0 1 h = lim .... (As h → 0, h ≠ 0) h 0 x h x h 1 log 1 1 = x+0 + x lim x = h 0 h x x 1 f '(x) = 2 x 1 log(1 x) = (1) .... Qlim 1 x x 0 x (ii) Let f (x) = 4x 1 f (x + h) = 4x + h = x By method of first principle, 1 ∴ f '(x) = f ( x h) f ( x ) x f '(x) = lim h 0 h 117 EX. 2. Find the derivative of x2 + x +2 at x = − 3 Theorem 3. Derivative of Product of functions. Let f(x) = x2 + x +2 If u and v are differentiable functions of x such For x = −3, f(− 3) = (− 3)2 – 3 + 2 = 9 – 3 + dy dv du that y = u.v then u v 2 = 8 dx dx dx f(−3+h) = (−3 + h)2 + (−3 + h) + 2 Corollary 2 : If u,v and w are differentiable = h2 − 6h + 9 − 3 + h + 2 = h2 − 5h + 8 functions of x such that y = u.v.w then By method of first principle, dy dw dv du uv uw vw dx dx dx dx f ( a h) f ( a ) f '(a) = lim Theorem 4. Derivative of Quotient of functions. h 0 h f ( 3 h) f ( 3) If u and v are differentiable functions of x such lim du dv ∴ f '(−3) = h 0 h v. u. u dy that y = where v ≠ 0 then dx 2 dx h 2 5h 8 8 v dx v = h 0 lim h Derivatives of Algebraic Functions h(h 5) = lim y dy/dx h 0 h Sr. No. f(x) f '(x) = lim (h − 5) = −5 (As h → 0, h ≠ 0) x→0 01 xn nxn-1 ∴ f ′(−3) = − 5 9.4 Rules of Differentiation (without proof) 02 1 n − Theorem 1. Derivative of Sum of functions. xn xn+1 If u and v are differentiable functions of x such 1 dy du dv 03 x that y = u + v then 2 x dx dx dx Theorem 2. Derivative of Difference of functions. 04 c 0 If u and v are differentiable functions of x such dy du dv SOLVED EXAMPLES that y = u − v then dx dx dx Ex. 1: Differentiate the following functions w.r.t.x. Corollary 1 : If u,v,w.... are differentiable 3 functions of x such that y = k1u ± k2v ± k3w i) y = x4 − 2x3 + x − −8 x2 ±......... where k1, k2, k3....... are constants then ii) y = (2x + 3) (x3 - 7x + 4) iii) x logx dy du dv dw k1 k2 k3........ iv) y = logx (ex + x) dx dx dx dx 118 Solution : d x d 1 3 = log x (e ) ( x) + (ex + x) 1) y = x4 − 2x3 + x – 2 −8 dx dx x x x x 1 ∴ y = x4 − 2x3 + x1/2 − 3x−2 − 8 = log x(e + 1) + (e + x) x Differentiate w.r.t.x. x 1 = e log x log x 1 dy d 4 d d x ( x ) 2 ( x 3 ) ( x1/ 2 ) dx dx dx dx 2) Differentiate the following functions w.r.t.x d d −3 ( x −2 ) − (8)....... (by rule 1) dx dx x2 7 x 9 i) y 1 x2 1 = 4x3 − 2(3x2) + ( x−1/2) − 3(−2x−3) − 0 2 x ii) y 1 x 3 = 4x3 − 6x2 + ( x−1/2) + 6x−3 2 l log x 16 iii) y = 4x3 − 6x2 + + 3 x 2 x x Solution: ii) y = (2x +3) (x3 − 7x + 4) x2 7 x 9 N Differentiating w.r.t.x i) y , say. x2 1 D dy d = (2x + 3) (x3 − 7x + 4) Differentiating w.r.t.x. dx dx d d d D. ( N ) N ( D) + (x − 7x + 4) 3 dy dx dx dx (2x + 3) dx D 2 = (2x + 3)(3x2 − 7) + (x3 − 7x + 4) (2) (by quotient rule) = 6x3 + 9x2 − 14x − 21 + 2x3 − 14x + 8 ( x 2 1)(2 x 7) ( x 2 7 x 9)(2 x) = = 8x3 + 9x2 − 28x − 13 ( x 2 1) 2 iii) y = x.log x 2 x3 2 x 7 x 2 7 2 x 3 14 x 2 18 x = ( x 2 1) 2 Differentiating w.r.t.x 7 x 2 16 x 7 dy d d = =x (logx) + logx (x) ( x 2 1) 2 dx dx dx 1 = x + logx.1 = 1 + logx x x ii) y x 3 iv) y = log x (ex + x) Differentiating w.r.t.x. Differentiating w.r.t.x d d ( x 3) x x ( x 3) dy d d dy dx dx = log x (ex + x) + (ex + x) (log x) dx dx dx dx ( x 3) 2 119 ( x 3) 1 x(1) (II) Differentiate the following w. r. t. x. = ( x 3) 2 (1) x5 + 3x4 3 (2) x x + logx − ex = ( x + 3) 2 2 7/2 5 2/5 (3) x5/2 + 5x7/5 (4) x + x l log x 7 2 iii) y x (5) x (x2 + 1)2 Differentiating w.r.t.x., we get (III) Differentiate the following w. r. t. x dy d d x (1 log x) (1 log x) ( x) 5 dx = dx dx (1) x3 log x (2) x 2 ex x 2 1 (3) ex log x (4) x3.3x x 0 (1 log x)(1) x = (IV) Find the derivatives of the following w. r. t. x. x2 x2 a2 3x 2 − 5 1 − 1 − log x (1) (2) = x2 a2 2 x3 − 4 x2 − log x log x 3e x 2 = (3) x3 − 5 (4) 3e x 2 x2 Derivatives of Logarithmic and Exponential xe x (5) functions x + ex y dy/dx (V) Find the derivatives of the following functions Sr.No. by the first principle. f(x) f '(x) i) 3x2 + 4 ii) x x 01 log x 1/x 1 x 1 02 ex ex iii) iv) 2x + 3 2x 7 03 ax(a>0) ax log a SOMEAPPLICATIONS OF DERIVATIVES: Supply and demand are perhaps one of the EXERCISE 9.1 most fundamental concepts of economics and it is the backbone of a market economy. Supply (I) Find the derivatives of the following functions represents how much the market can offer. The w. r. t. x. quantity supplied refers to the amount of a certain (1) x12 (2) x−9 (3) x3/2 goods producers are willing to supply when (4) 7x x (5) 35 receiving a certain price. Demand is the quantity of goods or commodity demanded. Similarly Cost, Revenue and Profit etc are most commonly used terminologies in business and economics. 120 DEMAND FUNCTION : Total Revenue (R): As Demand (D) is a function of Price (P), we can The Total Revenue is given by R = P.D where P is express it as D = f(P). price and D is quantity of goods demanded. Marginal Demand (MD): R P.D The rate of change of demand with respect to Average Revenue = = = P i.e. price it D D price is called the Marginal Demand (MD). self. dD So, MD = = f '(p) Total Profit (P) = Revenue − Cost dp P = R−C SUPPLY FUNCTION (S): Supply (S) is also a function of Price (P), we can SOLVED EXAMPLES express it as S = g(p). Marginal Supply (MS): Ex. (1) The demand D of a goods at price p is The rate of change of supply with respect to price 32 given by D = P2 +. is called the Marginal Supply (MS). P dS Thus, MS = = g'(p) Find the marginal demand when price is Rs.4. dp 32 Solution : Demand is given by D = P2 + P TOTAL COST FUNCTION (C): Diff. w. r. t. p. A Cost (C) function is a mathematical formula used to express the production expenses of dD 32 = 2P – 2 number of goods produced. It can be express as dP P C = f(x) where x is the number of goods produced. dD 32 Marginal Cost (MC): = 2(4) − =8−2=6 dP p 4 16 The rate of change of Cost with respect to number Ex. (2) The cost C of an output is given as of goods i.e. x is called Marginal Cost (MC). dC C= 2x3 + 20x2 − 30x +45. What is the rate of Therefore, MC = dx = f '(x) change of cost when the the output is 2 ? Solution : Average Cost (AC) is the cost of production of C C = 2x3 + 20x2 − 30x +45 each goods. So, AC = Diff. w. r. t. x. x dC Revenue and Profit Functions: = 6x2 + 40x − 30 dx If R(x) is the revenue received from the sale of x units of some goods (or commodity), then the dC = 6(2) + 40(2) − 30 = 74 2 derivative, R'(x) is called the Marginal Revenue. dx x 2 C Average Cost = AC = x 121 2 x3 20 x 2 30 x 45 ∴ Marginal Supply, when P = 4. = x 45 = 2x2 + 20x − 30 + x EXERCISE 9.2 Ex. (3) The demand D for a chocolate is inversely proportional to square of it’s price P. It is (I) Differentiate the following functions w.r.t.x observed that the demand is 4 when price is 4 per chocolate. Find the marginal demand x x2 + 1 1 (1) (2) (3) x when the price is 4. x +1 x e +1 Solution : Given that Demand is inversely ex x 2x (4) x (5) (6) proportional to square of the price e +1 log x log x i.e D a 12 (2e x 1) ( x 1)( x 1) P (7) (8) (2e x 1) (e x 1) D = k2 where k > 0...... (1) P II) Solve the following examples: Also , D = 4, when P = 4 , from (1) (1) The demand D for a price P is given as D = 27, find the rate of change of demand P Equation (1) becomes when price is 3. (2) If for a commodity; the price-demand P+5 Diff. w. r. t. P. relation is given as D =. Find the P-1 marginal demand when price is 2. ∴ when P = 4, (3) The demand function of a commodity is dD 128 2 given as P = 20 + D−D2. Find the rate at dP P 4 64 which price is changing when demand is 3. Ex (4) The relation between supply S and price (4) If the total cost function is given by; P of a commodity is given as S = 2P2 + P - 1. C = 5x3 + 2x2 + 7; find the average cost and Find the marginal supply at price 4. the marginal cost when x = 4. Solution : (5) The total cost function of producing n Given , S = 2P + P - 1 2 notebooks is given by C= 1500 − 75n + 2n2 Diff. w. r. t. P. n3 +. Find the marginal cost at n = 10. 5 ∴ 122 (6) The total cost of ‘t’ toy cars is given by l Standard Derivatives C=5(2t)+17. Find the marginal cost and f(x) f ′(x) average cost at t=3. xn nx n–1 for all n∈N x 1 (7) If for a commodity; the demand function 1/x – 1/ x2 is given by, D = 75−3P. find the marginal 1 demand function when P = 5 x 2 x (8) The total cost of producing x units is given k (constant) 0 by C=20ex, find its marginal cost and ex ex average cost when x = 4 ax ax log a log x 1/x (9) The demand function is given as P = 175 + 9D + 25D2. Find the revenue, average MISCELLANEOUS EXERCISE - 9 revenue and marginal revenue when demand is 10. I. Differentiate the following functions.w.r.t.x. (10) The supply S for a commodity at price P is given by S = P2 + 9P − 2. Find the marginal (1) x5 (2) x−2 (3) x supply when price is 7. 1 (4) x x (5) (6) 7x x (11) The cost of producing x articles is given by C = x2 + 15x + 81. Find the average cost and dy marginal cost functions. Find marginal cost II. Find if dx when x = 10. Find x for which the marginal cost equals the average cost. 1 2 (1) y = x2 + (2) y x 1 x2 Let’s remember! 2 1 (3) y x x l If u and v are differentiable functions (4) y x 3 2 x 2 x 1 of x and if x dy du dv (5) y x 2 1 2 (i) y = u ± v then = ± dx dx dx (6) y (1 x)(2 x) dy dv du (ii) y = u.v then =u +v dx dx dx 1 x (log x 1) (7) y (8) y 2 x x du dv v −u ex u dy dx dx (9) y = (iii) y = , v ≠ 0 then = log x v dx v2 (10) y x log x( x 2 1) 123 III. Solve the following. (10) The cost of producing x articles is given by (1) The relation between price (P) and demand C = x2 + 15x +81. Find the average cost and 12 marginal cost functions. Find the marginal (D) of a cup of Tea is given by D =. Find P cost when x = 10. Find x for which the the rate at which the demand changes when marginal cost equals the average cost. the price is Rs. 2/- Interpret the result. Gottfried Wilhelm Leibnitz (1646 – 1716) (2) The demand (D) of biscuits at price P is given Gottfried Wilhelm 64 Leibniz was a prominent by D = 3 , find the marginal demand when P German polymath and price is Rs. 4/-. philosopher in the history of mathematics and the history (3) The supply S of electric bulbs at price P of philosophy. His most is given by S = 2P3 + 5. Find the marginal notable accomplishment supply when the price is Rs. 5/- Interpret the was conceiving the ideas of differential and integral calculus, independently of result. Isaac Newton’s contemporaneous developments. (4) The marginal cost of producing x items is Calculus was discovered and developed independently by Sir Isaac Newton (1642 - 1727) given by C = x2 +4x + 4. Find the average cost in England and Gottfried Wilhelm Leibnitz (1646 and the marginal cost. What is the marginal - 1716) in Germany, towards the end of 17th cost when x = 7. century. (5) The Demand D for a price P is given as ACTIVITIES 27 D= , Find the rate of change of demand P when the price is Rs. 3/-. Activity 9.1: 1 (6) If for a commodity; the price demand relation Find the derivative of by first principle. x2 P 5 is given be D = . Find the marginal 1 1 P 1 f(x) = , f(x + h) = x2 ( x + h) 2 demand when price is Rs. 2/-. f '(x) = lim (7) The price function P of a commodity is given h 0 h as P = 20 + D − D2 where D is demand. Find the rate at which price (P) is changing when 1 1 demand D = 3. = lim ( x h) 2 x 2 h 0 h (8) If the total cost function is given by C = 5x3 + 2x2 + 1; Find the average cost and the marginal cost when x = 4. = lim h 0 h ( x h) 2 x 2 (9) The supply S for a commodity at price P is x 2 ( x 2 2 xh h 2 ) given by S= P2 + 9P − 2. Find the marginal = lim h 0 h( x h) 2. supply when price Rs. 7/-. 124 Activity 9.3: = lim h 0 h( x h) 2.x 2 The relation between price (P) and demand (D) at 32 = = a cup of Tea is given as D =. Find the rate of p which the demand changes when the price is Rs. Activity 9.2: 4/-. Interpret the result. The supply S of pens at price P is given by 32 S = 2P3 + 7. Find the marginal supply when price Given : D = p is Rs.5/-. Interpret the result. dD Given : S = 2p3 + 7 ∴ =............. =............. dP ds ∴ =............. =............. dp dD = = ∴ Marginal supply when P = 5/- dp P 4 ds Interpretation : ___________________ = = = dp p 5 Interpretation : ___________________ v v v 125 ANSWERS 1. SETS AND RELATIONS Exercise 1.2 1) x = 2, y = −2 Exercise 1.1 15 1) i) A = {M, A, R, I, G, E} 2) x = 0, y = 2 ii) B = {0, 1, 2, 3, 4} 3) i) A×B = {(a,x), (a,y), (b,x), (b,y), (c,x), iii) C = {2, 4, 6, 8,..............} (c,y)} 2) i) {x/x∈W, x ∉ N} ii) B×A = {(x,a) , (x,b), (x,c), (y,a), (y,b), ii) {x/−3 ≤ x ≤ 3, x ∈ Z} (y,c)} n iii) A×A = {(a,a) , (a,b), (a,c), (b,a), (b,b), iii) {x/x = , n ∈ N, n ≤ 7} n +1 2 (b,c), (c,a), (c,b), (c,c)} -5 1 3 iv) B×B = {(x,x) , (x,y), (y,x), (y,y)} 3) i) A∪B∪C = { 3 , −1, − 2 , 2 , 3} 4) i) P×Q = {(1,6), (2,6), (3,6) (1,4), (2,4), ii) A∩B∩C = { } (3,4)} 6) i) 45 ii) 10 iii) 10 iv) 25 ii) Q×P = {(6,1), (6,2), (6,3) (4,1), (4,2), (4,3)} 7) i) 132 ii) 63 5) i) A×(B∩C) = {(1,5), (1,6), (2,5), (2,6), 8) i) 1750 ii) 250 iii) 1100 (3,5), (3,6), (4,5), (4,6)} 9) 42 ii) {(1,5), (1,6), (2,5), (2,6), (3,5), (3,6), 10) i) 114 ii) 38 iii) 188 (4,5), (4,6)} iii) {(1,4), (1,5), (1,6), (2,4), (2,5), (2,6), 11) P(A) = {{1}, {2}, {3}, {1,2}, {2,3}, {1,3}, (3,4), (3,5), (3,6), (4,4), (4,5), (4,6)} {1,2,3}, {φ}} iv) {(1,4), (1,5), (1,6), (2,4), (2,5), (2,6), 12) i) {x / x ∈ R , −3 < x < 0} (3,4), (3,5), (3,6), (4,4), (4,5), (4,6)} ii) {x / x ∈ R , 6 ≤ x ≤ 12} 6) {(0,10), (6,8), (8,6), (10,0)} iii) {x / x ∈ R , 6 < x ≤ 12} iv) {x / x ∈ R , −23 ≤ x < 5} 7) i) Domain = {1,2,3,4,5}; Range = {4} ii) Domain = {1,2,3,4,5,6,7,8,9,10,11}; Range = {11,10,9,8,7,6,5,4,3,2,1)} iii) Domain = {2}; Range = {4,5,6,7} 126 9) i) R1 = {(2,4), (3,9), (5,25), (7,49),(11,121), (A×B) ∩ (B×A) = {(2,2)} (13,169)} ii) A×A×A = {(−1,−1,−1), (−1,−1,1), Domain = {2,3,5,7,11,13} (−1,1,−1), (1,−1,−1), Range = {4,9,25,49,121,169} (1,−1,1), (1,1,−1), (1,1,1), 1 1 1 1 (−1,1,1)} ii) R2 = {(1,1), (2, 2 ), (3, 3 ), (4, 4 ), (5, 5 )} Domain = {1,2,3,4,5} 6) i) R1 is a relation 1 1 1 1 ii) R2 is a relation Range = {1, 2 , 3 , 4 , 5 } iii) R3 is a relation 10) Range = {2,3,4,5} iv) R4 is not a relation 11) i) {(1,3), (2,6), (3,9)} 7) Domain = {1,2,3,4} Range = {4} ii) {(1,4), (1,6), (2,4), (2,6)} iii) {(0,3), (1,2), (2,1), (3,0)} 2. FUNCTION MISCELLANEOUS EXERCISE - 1 1) i) A = {x/x = 10n, n∈N, n≤5} Exercise 2.1 ii) B = {x/x is vowel of English alphabets} 1) a) It is a function iii) C = {x/x represents day of a week} b) It is not a function 2) i) A∪B = {1,2,4,6,7,10,11} c) It is not a function ii) B∩C = {} = φ 2) a) It is not a function iii) A-B = {1,10} b) It is a function iv) B–C = {2,4,6,7,11} c) It is not a function v) A∪B∪C = {1,2,3,4,5,6,7,8,9,10,11,12} d) It is a function vi) A∩ (B∪C) = {4,7} 1 3) a) 1 b) 19 c) − d) x2 − x −1 3) 230 4 e) x2 + 3x +1 4) 12 6 5) i) A×B = {(1,2), (1,4), (2,2), (2,4), 4) a) b) x = ±3 5 (3,2), (3,4)} 1 −2 B×A = {(2,1), (2,2), (2,3), (4,1), c) x = 2 or x = 3 (4,2), (4,3)} A×A = {(1,1), (1,2), (1,3), (2,1), 5) x = 0 or x = ±3 (2,2), (2,3), (3,1), (3,2), (3,3)} 6) a) f(3) = 22 b) f(2) = 7 c) f(0) = 3 B×B = {(2,2), (2,4), (4,2), (4,4) 127 7) a) f(−4) = −18 b) f(−3) = −14 −7 1 2) i) a = −4, b = −3 ii) a= 2 ,b= 2 c) f(1) = 5 d) f(5) = 25 3 −1 −8 8) a) 9x + 4 b) 0 iii) a = 10 , b = 10 iv) a = 29 , b = 0 3x 5 1 c) 238 d) domain = R −{ 6 } 6x 1 11 2 3 v) a = ,b= 19 19 9) a) 50x2 − 40x + 11 b) 10x2 + 13 c) 8x4 + 24x2 + 21 d) 25x − 12 vi) a = 13, b = 0 15 vii) a = 23 , b = MISCELLANEOUS EXERCISE - 2 13 13 1) i) Yes, Domain = {2,4,6,8,10,12,14} 4) i) −i ii) 1 iii) i Range = {1,2,3,4,5,6,7} iv) 1 v) −i vi) −1 ii) Not a function vii) 0 iii) Yes, Domain = {1,3,5}, 6) i) 2i ii) 0 Range = {1,2} 7) 1 5 (x − 2) 2) f−1(x) = 3 8) i) x = 1, y = 2 3) f(−1) = 1 ii) x = – 2, y = 2 f(−2) = −3 9) i) 7 f(0) = does not exist ii) 2 4) 2 Exercise 3.2 5) 3x2 − 11x + 15 6) a = 4, f(4) = 16 1) i) ± (1 − 3i) ii) ± (4 + 3i) 7) a = 3, b = – 2 iii) ± (2 + 3 i) iv) ±( 5 + 2 i) 3. COMPLEX NUMBERS v) ± ( 3 – i) 2) i) – 1 ± 7 i ii) 3 ± 5i Exercise 3.1 8 4 iii) 7 ± 11 i iv) 2 ± 3 i 1) i) 3−i ii) 3+i 6 iii) − 5 + 7i iv) 5 i 3) i) x = 2 i or x = – 5 i v) −5i vi) 5 +i 1 ii) x = i or x = – 2 i 2 vii) 2 − 3 i iii) x = – 2 i iv) x = – 2 i 128 4) i) x = 3 – i or x = – 1 + 2 i 4. SEQUENCES AND SERIES ii) x = 3 2 or x = 2 i Exercise 4.1 iii) x = 3 – 4 i or x = 2 + 3 i 4 2 iv) x = 1 – i or x = – i 1) i) tn = 2 (3n−1) ii) tn = (−5)n−1 5 5 iii) tn = (5)3/2 – n iv) it is not a G.P. Exercise 3.3 v) it is not a G.P. 1) i) 7 ii) 65 iii) w2 1 7 2) i) t7 = 81 ii) t3 = 2187 2) i) –1 ii) 0 iii) – 1 iv) 0 v) 1 iii) t6 = − 1701 iv) r = 3 3) t10 = 510 MISCELLANEOUS EXERCISE - 3 4 4) x=± 9 1) −1 4 5 5) G.P. with a = 25 , r = 2 2) −3 2 3) i) 3 + 8i ii) −4 + 0i 6) 3,6,12 and 12,6,3 15 1 1 1 1 iii) 14 − 5i iv) − 10i 7) , 27 3 , 3, 27 or 27, 3, 3 , 27 2 1 7 8) 1, 2, 4, 8, 16 or 1, -2, 4, -8, 16 v) −30 + 10i vi) 2 + 2 i −35 45 1 15 Exercise 4.2 vii) 26 − 26 i viii) 4 + i 4 8 56 1) i) Sn = 3 (2n −1) ix) −i + i p 2 n q n p n x) 5 25 ii) Sn = q p 4) i) x = 2, y = 1 ii) 3,2 28 3 266 iii) 17,19 iv) , 2) i) S6 = 243 61 61 v) 4,−2 ii) a = 3 5) i) 1 ii) −2 iii) −3 3) i) n=5 3 6) i) ± (3 + 5i) ii) ± (4 − i) ii) r = 5 iii) ± ( 3 + i) iv) ± (3 + 3i) 4) i) 635 v) ± (2 − i) vi) ± 2 (2 + i) ii) S10 = 2046 1 10 5 i) { (10n –1)–n} 3 9 129 8 10 15 15 15 ii) { (10n –1)–n} 5) 9 9 4 , 16 , 64 , ……………….. 4 1 6 i) {n − [1− (0.1)n } 9 9 Exercise 4.4 7 1 ii) {n − [1− (0.1)n } 1) i) Given series is a H.P.. 9 9 ii) Given series is a H.P. 5 iii) Given series is a H.P. 7 i) tn = [1−(0.1)n] 9 2) i) 1 , 1 2 3n − 1 23 ii) tn = {1−(0.1)n} 9 4 n ii) 1 ,1 8) tn = (3 ) 2n + 2 18 3 iii) 1 , 1. 5n 40 Exercise 4.3 3) A=5 24 1) i) Sum to infinity = 1 4) H= 5 ii) Sum to infinity = 6 5) G =60 −9 1 1 iii) 4 6) 9 and 11 iv) Sum to infinity does not exists. 7) −3 and 9 32 2) i) 0.32 = 8) 4 and 9 99 9) 14 and 56 32 ii) 3.5 = 9 Exercise 4.5 iii) 4.18 = 46 n(4n 2 9n 1) 1) 11 6 342 19 n(2n 2 + n + 1) iv) 0.345 = 2) = 2 990 55 3422 1711 n(n + 3) 3) v) 3.456 = = 4 990 495 3) a=4 n n 1 (n 2) 4) 6 12 4) r = 11 n(16n 2 + 48n + 41) 5) 3 130 2n(n + 1)(2n + 1) 17) r = ±15 6) 3 18) k=2 7) 2485 19) 1 8) n(6n + 8n + 3n −2) 3 2 5. STRAIGHT LINE 9) n = 48 Exercise 5.1 MISCELLANEOUS EXERCISE - 4 1. 2x − 4y + 5 = 0 1) t10 = 3072. 2. 9x − y + 6 = 0 3 3. 3x2 + 3y2 + 4x − 24y + 32 = 0 2) r= 4 4. x2 + y2 − 11x − 11y + 53 = 0 49 5 5. 3x + 4y − 41 = 0 3) a= 5,r = 7 6. x2 + y2 − 4x − 11y + 33 = 0 4) 5,10,20 or 20, 10, 5 7. (a) (−1,0) (b) (0,2) 1 1 1 1 8. (a) (6,7) (b) (4,6) 5) , 27 3 , 3,27 or 27, 3, 3 , 27 9. (−3, 11) 1 1 10. (a) 3X − Y + 6 = 0 6) , 1, 3, 9, 27, or 27, 9, 3, 1, 3 3 (b) X2 + Y2 + X + 4Y − 5 = 0 7) The sequence is a G.P. r = 7 (c) XY = 0 2 10 8) [ (10n–1) –n] Exercise 5.2 9 9 2 9) tn = [1–(0.1)n] 1. a) Slope of the line AB = 2 3 4 b) Slope of the line CD = n(10n 2 27 n 1) 7 10) c) Its slope is not defined. 6 11) n(n 1)(3n 17 n 26) 2 d) Slope of the line is 0. 12 3 2. − n(n + 1)(n + 2) 2 12) 18 1 3. n(n + 1)(2n + 1) 3 13) 24 4. 1 14) 2n(n+1)(n+2) 5. −1. 15) 2364 7. 1 16) 1275 8. k=1 131 9. 8x + 13y − 24 = 0 Exercise 5.3 10. 2x + y + 13 = 0, x − 9y + 73 = 0, 1. a) y = 5 b) x = −5 c) y = −1 and y = 7 11x − 4y − 52 = 0 , 1 , 10 2. a) y = 3 b) x = 4 19 19 3. a) x = 2 b) y = −3 MISCELLANEOUS EXERCISE - 5 4. 4x − y − 8 = 0 5. m = 1, c = −1 7 1 1. a) − b) − c) −1 d) 4 6. a) 2x + y − 4 = 0 2 4 b) 2x − 5y + 14 = 0 1 4 1 2. a) b) c) − c) 2x + 4y − 13 = 0. 3 3 2 7. a) X- intercept 3, Y-intercept 2 5 3. a) 22 b) c) 1 3 2 3 b) X- intercept , Y-intercept 3 2 8 4. y = −2x − , slope = −2. c) X- intercept −6, Y-intercept 4 3 6. 1 8. x+y−7=0 7. 1 9. a) 5x + y − 15 = 0 8. No, point does not satisfy the equation. b) 3x + 4y − 14 = 0 9. (d) 2x − y = 0. c) 2x − 3y − 1 = 0 10. a) y + 3 = 0 b) x + 2 = 0 c) y = 5 d) x = 3 Exercise 5.4 11. a) y = 3 b) y = 4 1. a) 2 slope − , X-intercept 3, Y-intercept 2 c) x=2 3 1 12. a) 5x − y + 7 = 0 b) 13x − y = 25 b) slope − , both the intercepts 0 c) x = 7 d) x = 0 2 e) 3x − 2y = 0 2. a) 2x − y − 4 = 0, b) 0x + 1y − 4 = 0 13. 4x − 3y + 12 = 0 c) 2x + y − 4 = 0 d) 2x − 3y + 0 = 0 14. a) 5x − y − 25 = 0 b) 3x y 4 0 15. a) BC : 3x + y = 9, CA : x = 1, 4. P = ±24 AB : x + y = 5 5. (1, −1) b) Median AD : x − y +3 = 0, 6. x + 3y = 3 Median BE : 2x + y = 7, 7. 4 units Median CF: 5x + y = 11 25 c) x − 3y + 12 = 0, y = 5, x − y + 2 = 0, 8. 117 units d) x − 3y + 11 = 0, y = 3, x − y + 5 = 0 132 6. DETERMINANTS Exercise 6.3 Exercise 6.1 1) i) x= 5 , y = 1, z = –4 3 3 1) i) 49 ii) -358 1499 520 332 ii) x = ,y= , z= iii) −27+9i iv) – 20 447 447 447 v) −10 vi) 46 iii) x = 4, y = 7, z = 6 vii) abc + 2fgh−af −bg2−ch2 2 3 3 1 viii) 0 iv) x , y z 5 5, 2 14 2) i) x = 2 ii) x = v) x=1, y = −2, z= 2 5 iii) x = 1 or x = 2 or x = 3 2) Rs. 1750 , Rs. 1500, Rs. 1750 3) i) x = 2 or x = −4 ii) x = −1 or x = 2 3) Consistent 4) x = −2 22 4) i) k = 16 ii) k = 5 5) x = 11 and y = 52 35 5) i) 13 sq.unit ii) sq. unit 2 Exercise 6.2 iii) 25 sq. unit –7 6) k=3;k= 1) i) 0 ii) 0 iii) 0 3 35 2) 4abc 7) sq. unit 2 7 8) A(DPQR) = 0 3) x=− 3 4) x = 0, or x = 12 9) 3,5,7 are the three required numbers 1 2 1 MISCELLANEOUS EXERCISE - 6 5) 10 3 1 7 3 2 6 1) i) –113 ii) −76 6) i) 0 ii) 0 –1 2 2) i) x= or x = 2 ii) x = 3 3 c1 a1 b1 7) (i) c2 a2 b2 3) 0 c3 a3 b3 4) i) 0 ii) LHS = RHS 1 x x2 iii) LHS = RHS iv) 0 (ii) 1 y y2 5) i) x = 1, y = 2, z = 1 1 z z2 3 3 1 ii) x , y z 5 5, 2 133 9 3 1 iii) x = , y=− , z= 2 2 2 2 II. 1. 2. – 8 3 3 14 6) i) k=5 ii) k = or k = 2 5 7 III. 1. 2. 1 3. 24 4. –24 7) i) 4 sq. unit 2 25 13 ii) sq. unit iii) sq. unit −1 2 2 IV. 1. 2 2. 3 8) i) k = 0 ; k = 8 ii) k = 34 ; k = 1 Exercise 7.4 7. LIMITS 1 9 15 I. 1. log 2. log 3. 1 log 4 5 2 Exercise 7.1 2 −2 II. 1. (log3)2 2. e 3 3. 2. −3 3 2 3 I. 1. 1 3. 4. ± 16 125 3 2 III. 1. 1 log a 3 (log 2) 2 2. II. 2 1. 3 3( 7 ) 2. 4 3. 4 4 b log 3 1 −1 3 3. (log3) (log5) 4. 6 III. 1. 2. 24 3. (a+2)1/2 6 2 15 4. IV. 1. (log5)2 2. (log7 – log5)2 2 MISCELLANEOUS EXERCISE - 7 Exercise 7.2 I. 1) n = 5 −1 −1 −1 −1 I. 1. 2. 3. 4. 4 2 2 2 5 II. 1) (a + 2)2/3 2) n 3) 1 3 4 II. 1. 2. 0 3. 0 3 1 3 4) 5) 1 6) − 7) log5 7 3 III. 1. 44 2. 3 3. –3 4. 8 1 5 8) e 5 9) 9 10) 3 11) log(abc) 12) 1 13) 1 Exercise 7.3 2 log a 1 14) 2 (loga)2 15) (log5)2 16) I. 1. 2. – 1 3. 2 log b 2 6 −1 17) 100 18) 19) 3 2 134 8. CONTINUITY 9. DIFFERENTIATION Exercise 8.1 Exercise 9.1 1) i) Continuous at x = −2 I) 1) 12x11 2) −9x−10 ii) Continuous on R except at x = 3 3 21 3) x 4) x 2 2 2) i) Discontinuous at x = 2 5) 0 ii) Continuous at x = 1 3 x 1 x i) Discontinuous at x = 2 II) 1) 5x4 + 12x3 2) e 3) 2 x ii) Continuous at x = 2 8 5 x 3/ 2 iii) Continuous at x = 3) + 7 x 2/5 4) x5/ 2 x 3/ 5 3 2 iv) Continuous at x = 3 9 7/2 1 5) x + 5 x 3/ 2 + 3 2 2 x 4) i) k = 2 ii) k = (log5)2 5 / 2 5 3/ 2 x III) 1) x2 + 3x2logx 2) x x e iii) a = 2, b = − 4 2 1 1 iv) a = 2 , b = 1 2 3) e x log x 4) 3x x 2 x log 3 3 x MISCELLANEOUS EXERCISE - 8 −4a 2 x IV) 1) ( x 2 − a 2 )2 I) 1) Continuous on its domain except at x = 5 6 x 4 30 x 2 24 x 2) 2) Continuous ( 2 x 3 4) 2 3) Continuous 1 ( x3 − 5) − log x.3 x 2 4) Discontinuous 3) x 5) Discontinuous ( x3 − 5) 2 3 12e x II) 1) k = e6, 2) k = 125 3) k = 2 4) (3e x + 2) 2 III) 1) a = 1 , b = −1 x ( x e )( x 1) x (1 e ) x x 5) e 2) a = −1 , b = −22 ( x e x )2 1 3 3 x 3) a = 3 , b= V) 1) 6x 2) 2 2
