Std. XI Commerce Mathematics & Statistics - I

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2 Written as per the revised syllabus prescribed by the Maharashtra State oard of Secondary and Higher Secondary Education, Pune. Std. I Commerce Mathematics & Statistics - I Salient Features Exhaustive coverage of entire syllabus. Topic-wise distribution of all textual questions and practice problems at the beginning of every chapter. Covers answers to all textual and miscellaneous exercises. Precise theory for every topic. Neat, labelled and authentic diagrams. Relevant and important formulae wherever required. Practice problems and Multiple Choice Questions for effective preparation. Printed at: Repro India Ltd. Mumbai Target Publications Pvt. Ltd. No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/udio Video Cassettes or electronic, mechanical including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher. 065_08_JUP P.O. No. 98

3 Preface Mathematics is not just a subject that is restricted to the four walls of a classroom. Its philosophy and applications are to be looked for in the daily course of our life. The knowledge of mathematics is essential for us, to explore and practice in a variety of fields like business administration, banking, stock exchange and in science and engineering. With the same thought in mind, we present to you Std. I Commerce: Mathematics and Statistics-I a complete and thorough book with a revolutionary fresh approach towards content and thus laying a platform for an in depth understanding of the subject. This book has been written according to the revised syllabus. t the beginning of every chapter, topic wise distribution of all textual questions including practice problems have been provided for simpler understanding of different types of questions. Neatly labelled diagrams have been provided wherever required. We have provided answer keys for all the textual questions and miscellaneous exercises. In addition to this, we have included practice problems based upon solved exercises which not only aid students in self evaluation but also provide them with plenty of practice. We ve also ensured that each chapter ends with a set of Multiple Choice Questions so as to prepare students for competitive examinations. We are sure this study material will turn out to be a powerful resource for students and facilitate them in understanding the concepts of Mathematics in the most simple way. The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we ve nearly missed something or want to applaud us for our triumphs, we d love to hear from you. Please write to us on: mail@targetpublications.org ours faithfully Publisher est of luck to all the aspirants! No. Topic Name Page No. Sets, Relations and Functions Complex Numbers 0 Sequence and Series 78 ngle and its Measurement 0 5 Trigonometric Functions 0 6 Plane Co-ordinate Geometry 5 7 Circle and Conics 89 8 Equations 9 9 Determinants 67 0 Limits Differentiation 7

4 0 Chapter 0: Sets, Relations and Functions Sets, Relations and Functions Type of Problems Exercise Q. Nos. To describe sets in Roster form To describe sets in Set-uilder form Operations on Sets Ordered Pairs Cartesian product of two Sets To find domain and range of a given relation. Q. (i., ii., iii.) Practice Problems (ased on Exercise.) Q. (i., ii., iii.). Q. (i., ii., iii.), Q.(i. to iv.) Practice Problems Q. (i., ii., iii.) (ased on Exercise.) Q. (i. to iv.) Miscellaneous Q. (i., ii., iii.) Practice Problems (ased on Miscellaneous) Q.(i., ii.). Q. to Q. Q. (i. to iv.), Q. (i., ii.) Practice Problems Q. to Q.0 (ased on Exercise.) Q. (i., ii.), Q. (i., ii.) Miscellaneous Q.,, Practice Problems (ased on Miscellaneous) Q.,,. Q.,, 6, Practice Problems (ased on Exercise.) Q.,, 5, 9. Q.,, 5 Practice Problems (ased on Exercise.) Q., Miscellaneous Q.5 Practice Problems (ased on Miscellaneous) Q.5. Q.7, 8, 9, 0 Practice Problems (ased on Exercise.) Q.6, 7, 8 Miscellaneous Q.6, 7 Practice Problems (ased on Miscellaneous) Q.6, 7

5 Std. I : Commerce (Maths I). Q.0, Types of Functions Practice Problems (ased on Exercise.) Q.8, 9 Miscellaneous Q.9, 0 Practice Problems (ased on Miscellaneous) Q.9, 0. Q., 6, 7, 8 To find values of the given function Practice Problems (ased on Exercise.) Q.,, 5, 6 Miscellaneous Q. to Q.7 Practice Problems (ased on Miscellaneous) Q. to Q.6 Operations on functions. Q. Practice Problems (ased on Exercise.) Q. (i. to iii.). Q. to Q.5 Composite function Practice Problems (ased on Exercise.) Q.,,, 5 Miscellaneous Q.8 to Q. Practice Problems (ased on Miscellaneous) Q.7 to Q.0. Q.6 Inverse function Practice Problems Q.6 (ased on Exercise.) Miscellaneous Q., Practice Problems (ased on Miscellaneous) Q.. Q.,, 5, 9 To find domain and range of a given function Practice Problems (ased on Exercise.) Q.0,,, 7 Miscellaneous Q.8,, Practice Problems (ased on Miscellaneous) Q.8

6 Chapter 0: Sets, Relations and Functions Syllabus:. Sets. Types of sets. lgebra of sets. Intervals.5 Cartesian product of sets.6 Relations.7 Functions.8 Particular types of functions and their graphs.9 Composite function.0 Inverse function. Functions in Economics. Some more functions and their graphs Introduction ll basic concepts of modern mathematics are based on set theory. The concepts involving logic can be explained more easily with the help of set theory. It plays a crucial role in the study of relations, functions, probability and is used extensively in various other branches of mathematics. We shall briefly revise and study some more concepts about sets.. Sets set is a well-defined collection of objects. These objects may be actually listed or may be specified by a rule. set is usually denoted by the capital letters,, C, N, R, etc. Each object in a set is called an element or a member of the set and is denoted by the small letters a, b, c, etc. If x is an element of set, then we write it as x and read it as x belongs to and if y is not an element of set, then we write it as y and read it as y does not belong to. If = {,, 6, 8}, then, 7, 8, 0 The set of natural numbers, whole numbers, integers, rational numbers and real numbers are denoted by N, W, I, Q and R respectively. Methods of Representation of Sets There are two methods of representing a set which are as follows: i. Roster method (Listing method): In this method all the elements are listed or tabulated. The elements are separated by commas and are enclosed within two braces(curly brackets). The set of all positive even integers less than 9 can be written as = {,, 6, 8}. ii. Set-uilder method: In this method, the set is described by the characteristic property of its elements. In general, if all the elements of set satisfy some property P, then write in set-builder notation as = {x/x has property P} and read it as is the set of all x such that x has the property P. Let = {,, 5, 6, 7, 8} Using the set-builder method, can be written as = {x/x N, x 8} Since = {,, 5, 6, 7, 8} can also be stated as the set of natural numbers from to 8 including and 8. Some standard sets are as follows: N = set of all natural numbers = {,,,..} Z or I = set of all integers = {.,,, 0,,, } Q = set of all rational numbers p = /p,q Z,q 0 q. Types of sets. Empty set: set which does not contain any element is called an empty set and it is denoted by or { }. It is also called null set. = {x/x N, < x < } = {x/x is a positive integer < } Note: The set {0} and {} are not empty sets as they contain one element, namely 0 and respectively.

7 Std. I : Commerce (Maths I). Singleton set: set which contains only one element is called a singleton set. = {5}, = {}, = {x/x N, < x < } The set = set of all integers which are neither positive nor negative is a singleton set since = {0}. Finite set: set which contains countable number of elements is called a finite set. = {a, b, c} = {,,,, 5} C = {a, e, i, o, u}. Infinite set: set which contains uncountable number of elements is called an infinite set. N = {,,, } Z = {,,, 0,,,,..} 5. Subset: Set is called a subset of set, if every element of set is also an element of set i.e., if x, then x. We denote this relation as and read it as is a subset for. It s clear that i. Every set is a subset of itself i.e.,. ii. n empty set is a subset of every set. If = {,, 6, 8} and = {,, 6, 8, 0, }, then. If, then is called a superset of, denoted by. 6. Proper subset: set is said to be a proper subset of a set if every element of set is also an element of the set and contains atleast one element which is not in. We denote it by.. = {x/x is a natural number less than 5} = {,,, }. = {x/x is a divisor of } = {,,,, 6, } Note: i. Every set is a subset of itself. ii. Empty set is a subset of every set. and are improper subsets of. 7. Universal set: non-empty set of which all the sets under consideration are subsets, is called a universal set. It is usually denoted by or U. If = {,,, }, = {, 8,, 5} and C = {,,,, 50} are sets under consideration, then the set N of all natural numbers can be taken as the universal set. Venn diagram: set is represented by any closed figure such as circle, rectangle, triangle, etc. The diagrams representing sets are called venn diagrams. i. = {, 6, 9} ii. = {a, b, c, d, e, f} = {b, e, f} 8. Equal sets: Two sets and are said to be equal if they have the same elements and we denote this as =. From this definition it follows that two sets and are equal if and only if and If = {,,, }, = {,,, }, then =. 9. Complement of a set: Let be a subset of a universal set then the set of all those elements of which do not belong to is called the complement of set and it is denoted by or c. Thus, = {x/x, x }..6.9 The shaded region in the above figure represents..b.e.f.a.d.c

8 Chapter 0: Sets, Relations and Functions Let = {,,,, 5, 6, 7, 8, 9} be an universal set and = {,, 5, 6, 8}. Then = {,, 7, 9} Properties: If is the universal set and, U, then i. () = ii. = iii. = iv. = v. =. lgebra of sets. Union of sets: If and are two sets, then the set of those elements which belong to or to or to both and is called the union of the sets and and is denoted by. i.e., = {x/x or x } The shaded portion in the below venn diagram represents. i. If = {,,, }, = {,, 6, 8}, then = {,,,, 6, 8} ii. If is the set of all odd integers and is the set of all even integers, then is the set of all integers. Properties: If,, C are any three sets, then i. = ii. = iii. = (Commutative law) iv. ( ) C = ( C) (ssociative law) v. = (Idempotent law) vi. If, then = vii. ( ), ( ). Intersection of sets: If and are two sets, then the set of those elements which belong to both and i.e., which are common to both and is called the intersection of the sets and and is denoted by. Thus, = {x/x and x } The shaded portion in the below venn diagram represents. If = {,,,, 5}, = {,, 5, 7, 9}, then = {,, 5} Properties: If,, C are any three sets, then i. = ii. = iii. = (Commutative law) iv. ( ) C = ( C) (ssociative law) v. = (Idempotent law) vi. If, then = vii. ( ), ( ) Distributive Properties of union and intersection If a, b, c R, then a (b + c) = (a b) + (a c) This is known as distributive property of multiplication over addition. In set theory, the operation of union and intersection of sets are both distributive over each other i.e., If,, C are any three sets, then i. ( C) = ( ) ( C) ii. ( C) = ( ) ( C) We verify these distributive laws using Venn diagrams shown below. The shaded portion in each figure shows the set obtained by performing the operation given below the figure. i. ii. C ( C) C = = ( ) ( C) ( C) ( ) ( C) C C 5

9 Std. I : Commerce (Maths I) De Morgan s laws If and are two subsets of a universal set, then i. ( ) = ii. ( ) = We verify these laws using Venn diagrams shown below. The shaded portion in each figure shows the set obtained by performing the operation below the figure: i. = ii. Disjoint sets: Two sets and are said to be disjoint, if they have no element in common i.e., =. If = {,, 6} and = {, 5, 7}, then = and are disjoint sets. The venn diagram of the disjoint sets and is shown below:. Difference of sets: If and are two sets then the set of all the elements of which are not in is called difference of sets and and is denoted by. Thus, = {x/x and x } Similarly, = {x/x and x } In the below venn diagrams, shaded region represents and. ( ) ( ) = = If = {,,,, 5, 6}, = {,, 6, 8}, then = {,, 5} and = {8} Note: i. The sets,, are mutually disjoint sets i.e. the intersection of any of these two sets is the null (empty) set. ii. = ( ) ( ) ( ) Number of elements in a set: Let be a set. Then the total number of elements in it is denoted by n(). Let = {8, 9, 0,, } n () = 5 The number of elements in the empty set is zero. i.e., n () = 0 Results: For given sets,. n( ) = n() + n() n( ). When and are disjoint sets then n( ) = n() + n(). n( ) + n( ) = n(). n( ) + n( ) = n() 5. n( ) + n( ) + n( ) = n( ) 6. For any sets,, C n( C) = n() +n() + n(c) n( ) n( C) n( C) + n( C) Power set: The set of all subsets of set is called the power set of and it is denoted by P(). If = {a, b, c}, then P() ={, {a}, {b}, {c}, {a, b}, {b, c}, {a, c}, {a, b, c}} Note: If contains n elements, then the power set of i.e., P() contains n elements.. Intervals Open interval: If p, q R and p < q, then the set {x/xr, p < x < q} is called open interval and is denoted by (p, q). Here all the numbers between p and q (p, q) except p and q. p x q R 6 (p, q) = {x/xr, p < x < q}

10 Chapter 0: Sets, Relations and Functions Closed interval: If p, q R and p < q, then the set {x/xr, p x q} is called closed interval and is denoted by [p,q]. Here all the numbers between p and q [p, q] including p, q. [p, q] = {x/xr, p x q} Semi-closed interval: If p, q R and p < q, then the set {x/xr, p x < q} is called semi-closed interval and is denoted by [p, q). [p, q) = {x/xr, p x < q} [p, q) includes p but excludes q. Semi-open interval: If p, q R and p < q, then the set {x/xr, p < x q} is called semi-open interval and is denoted by (p, q]. (p, q] = {x/x R, p < x q} (p, q] excludes p but includes q. Remarks: i. Set of all real numbers > p i.e., (p, ) = [x/xr, x > p} ii. p p Set of all real numbers p i.e., [p, ) = {x/xr, x p} p p p Set of all real numbers < q i.e., (, q) {x/xr, x < q} Set of all real numbers q i.e., (, q] = {x/xr, x q} iii. Set of all real numbers R is (, ) x R = (, ) = {x/x R, < x < } x x q q q q q R R R R R R R R Exercise.. Describe the following sets in Roster form: i. {x/x is a letter of the word MRRIGE } ii. 9 x/ xisaninteger, x iii. {x/x = n, x N, n N} i. Let = {x/x is a letter of the word MRRIGE } = {M,, R, I, G, E} 9 ii. Let = x/ xisan integer and x = {0,,,, } iii. Let C = {x/x = n, n N} C = {,, 6, 8,.}. Describe the following sets in Set-uilder form: i. {0} ii. {0,,, } iii.,,,,,, i. Let = {0} 0 is a whole number but it is not a natural number = {x / x W, x N} ii. Let = {0,,, } is the set of elements which belongs to Z from to = {x / x Z, x } iii. Let C =,,,,,, In the given set C, numerators are natural numbers from to 7 and denominator = (numerator) + n C = x/x,nn,n 7 n 7

11 Std. I : Commerce (Maths I). If = {x / 6x + x 5 = 0} = {x / x 5x = 0} C = {x / x x = 0} Find i. ( C) ii. ( C) = x/6x x 50 6x + x 5 = 0 6x + 0x 9x 5 = 0 x(x + 5) (x + 5) = 0 (x + 5) (x ) = 0 5 x = or x = 5 =, = {x/x 5x = 0} x 5x = 0 x 6x + x = 0 x(x ) + (x ) = 0 (x )(x + ) = 0 x = or x = =, C = {x/x x = 0} x x = 0 x x + x = 0 x(x ) + (x ) = 0 (x ) (x + ) = 0 x = or x = C =, Thus, i. C = 5,,, 5 =,,,, ii. C = { }.,, C are the sets of the letters in the words college, marriage and luggage respectively, verify that [ ( C)] = [( ) ( C)]. = {c, o, l, g, e} = {m, a, r, i, g, e,} C = {l, u, g, a, e} C = {m, a, r, i, g, e, l, u} ( C) = {c, o} = {c, o, l} C = {c, o} [( ) ( C)] = {c, o} = ( C) [ ( C)] = [( ) ( C)] 5. If = {,,, }, = {,, 5, 6}, C = {, 5, 6, 7, 8} and universal set = {,,,, 5, 6, 7, 8, 9, 0} verify the following: i. ( C) = ( ) ( C) ii. ( C) = ( ) ( C) iii. ( ) = iv. ( ) = v. = ( ) ( ) vi. = ( ) ( ) vii. n( ) = n() + n() n( ) = {,,, }, = {,, 5, 6}, C = {, 5, 6, 7, 8} ={,,,, 5, 6, 7, 8, 9, 0} i. ( C) = {, 5, 6} ( C)= {,,,, 5, 6} ( ) = {,,,, 5, 6} ( C) = {,,,, 5, 6, 7, 8} ( ) ( C) = {,,,, 5, 6} ( C) = ( ) ( C) ii. ( C) = {,, 5, 6, 7, 8} ( C) = {, } = {, } C = {} ( ) ( C) = {, } ( C) = ( ) ( C) iii. = {,,,, 5, 6} ( ) ={7, 8, 9, 0} = {5, 6, 7, 8, 9, 0}, = {,, 7, 8, 9, 0} = {7, 8, 9, 0} ( ) = 8

12 Chapter 0: Sets, Relations and Functions iv. = {, } ( ) = {,, 5, 6, 7, 8, 9, 0} = {5, 6, 7, 8, 9, 0} = {,, 7, 8, 9, 0} = {,, 5, 6, 7, 8, 9, 0} ( ) = v. = {, } = {, } ( ) ( ) = {,,, } = ( ) ( ) vi. = {, } = {5, 6} ( ) ( ) = {,, 5, 6} = ( ) ( ) vii. = {,,, }, = {,, 5, 6}, = {, }, = {,,,, 5, 6} n() =, n() =, n( ) =, n( ) = 6 n() + n() n( ) = + = 6 n( ) = n() + n() n( ) 6. If and are subsets of the universal set and n() = 50, n() = 5, n() = 0, n( ) = 5, find i. n( ) ii. n( ) iii. n( ) iv. n( ). n() = 50, n() = 5, n() = 0, n( ) = 5 i. n( ) = n() [n( )] = n() n( ) = 50 5 = 5 ii. n( ) = n() + n() n( ) = = 0 iii. n( ) = n() n( ) = 0 0 = 0 iv. n( ) = n() n( ) = 5 0 = 5 7. In a class of 00 students who appeared in a certain examinations, 5 students failed in MHT-CET, 0 in IEEE, 0 in IIT, 0 failed in MHT-CET and IEEE, 7 in IEEE and IIT, 5 in MHT-CET and IIT and 5 failed in all three examinations. Find how many students i. did not fail in any examination. ii. failed in IEEE or IIT. Let = set of students who failed in MHT-CET = set of students who failed in IEEE C = set of students who failed in IIT = set of all students n() = 00, n() = 5, n() = 0, n(c) = 0, n( )= 0, n( C) = 7, n( C) = 5, n( C) = 5 i. n( C) = n() + n() + n(c) n( ) n( C) n( C) + n( C) = = 68 ii. No. of students who did not fail in any exam = n() n( C) = = No. of students who failed in IEEE or IIT = n( C) = n() + n(c) n( C) = = 6 8. From amongst 000 literate individuals of a town, 70% read Marathi newspapers, 50% read English newspapers and.5% read both Marathi and English newspapers. Find the number of individuals who read i. at least one of the newspapers. ii. neither Marathi nor English newspaper iii. only one of the newspapers. Let M = set of individuals who read Marathi newspapers E = set of individuals who read English newspapers = set of all literate individuals n() = 000, n(m) = = n(e) = = n(m E) = =

13 Std. I : Commerce (Maths I) n(m E) = n(m) + n(e) n(m E) = = 750 i. No. of individuals who read at least one of the newspapers = n(m E) = 750. ii. iii. 0 No. of individuals who read neither Marathi nor English newspaper = n(m E) = n(m E) = n() n(m E) = = 50 No. of individuals who read only one of the newspaper = n(m E) + n(m E) = n(m E) n(m E) = = In a hostel, 5 students take tea, 0 students take coffee, 5 students take milk, 0 students take both tea and coffee. 8 students take both milk and coffee. None of them take tea and milk both and everyone takes atleast one beverage, find the number of students in the hostel. Let T = set of students who take tea C = set of students who take coffee M = set of students who take milk n(t) = 5, n(c) = 0, n(m) = 5, n(t C) = 0, n(m C) = 8, n(t M) = 0, n(t M C) = 0 No. of students in the hostel = n(t C M) = n(t) + n(c) + n(m) n(t C) ( M C) (T M) + n(t M C) = = 0. There are 60 persons with a skin disorder. If 50 had been exposed to the chemical, 7 to the chemical, and 6 to both chemicals and, find the number of persons exposed to i. Chemical but not Chemical ii. Chemical but not Chemical iii. Chemical or Chemical. Let = set of persons exposed to chemical = set of persons exposed to chemical = set of all persons n()=60, n()=50, n()= 7, n( )= 6 i. No. of persons exposed to chemical but not to chemical = n( ) = n() n( ) = 50 6 = ii. iii. No. of persons exposed to chemical but not to chemical = n( ) = n() n( ) = 7 6 = 8 No. of persons exposed to chemical or chemical = n( ) = n() + n() n( ) = = 88. If = {,, }, write down all possible subsets of i.e., the power set of. = {,, } The power set of is given by P() = {,{},{},{}, {, }, {, },{, }, {,, }}. Write the following intervals in Set-uilder form: i. (, 0) ii. [6, ] iii. (6, ] iv. [, 5). i. Let = (, 0) = {x/x R and < x < 0} ii. Let = [6, ] = {x/x R and 6 x } iii. Let C = (6, ] C = {x/x R and 6 < x } iv. Let D = [, 5) D = {x/x R and x < 5}

14 Chapter 0: Sets, Relations and Functions. In the Venn diagram below shade i. ( ) ii. iii. iv.. In the Venn-diagram below, shade i. ( C) ii. ( ) ( C) i. i. C C C C ( ) ii. C ii. ( C) iii. C ( ) C C C iv. C C C ( ) ( C)

15 Std. I : Commerce (Maths I) Ordered Pair: If (a, b) is a pair of numbers then the order in which the numbers appear is important, is called an ordered pair. Ordered pairs (a, b) and (b, a) are different. Two ordered pairs (a, b) and (c, d) are equal, if and only if a = c and b = d lso, (a, b) = (b, a) if and only if a = b.5 Cartesian product of two sets Let and be any two non-empty sets. The set of all ordered pairs (a, b) such that a and b is called the cartesian product of and and is denoted by. Thus, = {(a, b)/a, b } where a is called the first element and b is called the second element of the ordered pair (a, b). If, then If = or = or both and are empty sets, then =. If = {,, }, = {a, b}, then = {(, a), (, b), (, a), (, b), (, a), (,b)}.6 Relations Consider the following statements: i. Ram is taller than Shyam. ii. Harshal and Ravi have shirts of same colour. iii. 5 is the square of 5. iv. and are even integers. Here we can say that Ram is related to Shyam by the relation is taller than. Harshal is related to Ravi by the relation have shirts of same colour. 5 is related to 5 by the relation is the square of and is related to by the relation are even integers. Definition: If and are two non-empty sets, then any subset of is called relation from to and is denoted by capital letters P, Q, R, etc. Consider the following illustration: a b c R l m n o p Let us consider = {a, b, c}, = {l, m, n, o, p} = {(a, l), (a, m), (a, n), (a, o), (a, p), (b, l), (b, m), (b, n), (b, o), (b, p), (c, l), (c, m), (c, n), (c, o), (c, p)} In the above figure, the arrow starting from the element a and pointing to the elements l and n indicates that a is related to l and n. Similarly, b is related to m and o and c is related to p. This relation is also represented by set of ordered pairs, R = {(a, l), (a, n), (b, m), (b, o), (c, p)} This relation R is a subset of. Thus, relation from set to is a subset of i.e., R. If R is a relation and (x, y) R, then it is denoted by xry. y is called image of x under R and x is called pre-image of y under R. Domain: The set of all first components of the ordered pairs in a relation R is called the domain of the relation R. i.e., domain (R) = {a/(a, b) R} Range: The set of all second components of the ordered pairs in a relation R is called the range of the relation R. i.e., range (R) = {b/(a, b) R} Co-domain: If R is a relation from to, then set is called the co-domain of the relation R. inary relation on a set: Let be non-empty set then every subset of is binary relation on. Types of Relation i. One-One relation: If every element of has at most one image in and distinct elements in have distinct images in, then a relation R from to is said to be one-one. Let = {,, 5, 6}, = {, 5, 6, 7, 9} and R = {(, 5), (, 6), (5, 7)} 5 6 R

16 ii. iii. iv. Then R is a one-one relation from to. Here, domain of R = {,, 5} and range of R = {5, 6, 7} Many-one relation: If two or more than two elements in have same image in, then a relation R from to is said to be many-one. Let = {,,, }, = {,,, 5, 6, 7} and R = {(, ), (, 7), (, )} R Then R is a many-one relation from to. Here, domain of R = {,, } and range of R = {, 7} Into relation: If there exists at least one element in which has no pre-image in, then a relation R from to is said to be into relation. Let = {,, 0,,, }, = {0,,,, } and R = {(, ), (, ), (0, 0), (, ), (, )} 0 R 0 Then R is into relation from to. Here, domain of R = {,, 0,, } and range of R = {0,, } Onto relation: If every element of is the image of some element of, then a relation R from to is said to be onto relation. Let = {,,,, }, = {,, 9} and R = {(, ), (, ), (, ), (, 9)} Chapter 0: Sets, Relations and Functions Then R is onto relation from to. Here, domain of R = {,,, } and range of R = {,, 9} Range = co-domain () Note: i. Here, is a relation on and is called the empty or void relation on. ii. Here, is a relation on called the universal relation on. i.e. R = = {,, } Then R = = {(, ), (, ), (, ), (, ), (, ), (, ), (, ), (, ), (, )} and R = is the universal relation on. iii. The total number of relations that can be defined from a set to set is the number of possible subsets of. If n() = m and n() = m, then n( ) = m m mm and the total number of relations is.7 Functions R Definition: function from set to the set is a relation which associates every element of a set to unique element of set and is denoted by f:. If f is a function from to and (x, y) f, then we write it as y = f(x) Let = {,,, }, = {,,, 5, 6, 7, 8} f 8 Let a relation from to be given as twice of then we observe that every element x of set is related to one and only one element of set. Hence this relation is a function from set to set. In this

17 Std. I : Commerce (Maths I) case f() =, f() =, f() = 6, f() = 8 are the values of function f(x) = x at x =,,, respectively. The set of all values of function f is {,, 6, 8}. This set is called range of the function f. Range: If f is a function from set to set, then the set of all values of the function f is called the range of the function f. Thus the range set of the function f: is {f(x) / x } Note that the range set is a subset of co-domain. This subset may be proper or improper. In the figure for f: = {,,, } is domain, = {,, 9, 6, 5, 6} is co-domain and set {,, 9, 6} is the range of the function f. Types of functions f One-one function: function f: is said to be one-one function, if different elements in have different images in. Consider the function f: such that each element of its range set is the value of the function at only one element of the domain set. In this case, f: is one-one function.. Onto function: If the function f: is such that each element in is the image of some element in, then f is said to be a onto function. In this case range of function f is same as its co-domain. f Consider the function f: represented by the following arrow diagram a b c In this case, range is equal to co-domain = {l, m, n} Hence f: is onto function.. Into function: If the function f: is such that there exists at least one element in which is not the image of any element in, then f is said to be into function. In this case, the range of a function f is a proper subset of its co-domain. Consider the function f: represented by the following arrow diagram. In this case range = {, 5, 7, 9} is a proper subset of co-domain {, 5, 7, 9, } Hence, f: is into function.. Many-one function: f f l m n If the function f: is such that two or more elements in a set have the same image in set i.e. there is at least one element in which has more than one preimage in then the function f is called many-one function. The function f: represented by the following arrow diagram is such that the co-domain contains, and 9 each of which is the value of the function f at two distinct elements of the domain set.

18 Chapter 0: Sets, Relations and Functions Representation of functions. rrow diagram: In this diagram, we use arrows. rrows start from the element of domain and point out it s value.. Starting the rule: (In terms of formula): This is the most usual way of exhibiting function. Let = {,,,, 5}, = {5, 7, 9,, } and f: is a function represented by arrow diagram. In this case we observe that, if we take any element x of the set, then the element of the set related to x is obtained by adding to twice of x. pplying this rule we get in general f(x) = x +, for all x. This is the formula which exhibits the function f. If we denote the value of f at x by y, then we get y = x +, for all x.. Function as a set of ordered pairs: x y z f Consider the function f(x) = x + 5, Where = {0,,, }, = {5, 8,,, 7} we can form set of ordered pairs, as {(0, 5), (, 8), (, ), (, )} In each of the ordered pairs, first component is an element of set and second component is an element of set. the function f can be given as: f = {(0, 0), (, 8), (, ), (, )} function f is subset of. Here, we observed that no two pairs of this set have the same first component. p q r s Tabular form: If the sets and are finite and contain very few elements, then a function f: can be exhibited by means of a table of corresponding elements. Let f = {(, 7), (, 9), (, ), (, )} We can represent the above function in tabular form as follows: x f(x) 7 9 Real valued function: function whose co-domain is a set of real numbers R, is called a real valued function. Henceforth, we will deal with only real valued functions. Exercise.. If (x, y + ) = (, ), find the values of x and y. y the definition of equality of ordered pairs, we have (x, y + ) = (, ) x = and y + = x = and y = y. x, =,, find x and y. y the definition of equality of ordered pairs, we have y x, =, x + = and y = x = y and + x = 6 and y = 5. If = {a, b, c}, = {x, y}, find,,,. = {a, b, c}, = {x, y} = {(a, x), (a, y), (b, x), (b, y), (c, x), (c, y)} = {(x, a), (x, b), (x, c), (y, a), (y, b), (y, c)} = {(a, a), (a, b), (a, c) (b, a) (b, b), (b, c), (c, a), (c, b) (c, c)} = {(x, x), (x, y), (y, x), (y, y)} 5

19 Std. I : Commerce (Maths I). If P = {,, } and Q = {}, find sets P Q and Q P. P = {,, }, Q = {} P Q = {(, ), (, ), (, )} and Q P = {(, ), (, ), (, )} 5. Let = {,,, }, = {, 5, 6}, C = {5, 6} Find i. ( C) ii. ( ) ( C) iii. ( C) iv. ( ) ( C) = {,,, }, = {, 5, 6}, C = {5, 6} i. C = {5, 6} ( C) = {(, 5), (, 6), (, 5), (, 6), (, 5), (, 6), (, 5), (, 6)} ii. = {(, ), (, 5), (, 6), (, ), (, 5), (, 6), (, ), (, 5), (, 6), (, ), (, 5), (, 6)} C = {(, 5), (, 6), (, 5), (, 6), (, 5), (, 6), (, 5), (, 6)} ( ) ( C) = {(, 5), (, 6), (, 5), (, 6), (, 5), (, 6), (, 5), (, 6)} iii. C = {, 5, 6} ( C) = {(, ), (, 5), (, 6), (, ), (, 5), (, 6), (, ), (, 5), (, 6), (, ), (, 5), (, 6)} iv. ( ) ( C) = {(, ), (, 5), (, 6), (, ), (, 5), (, 6), (, ), (, 5), (, 6), (, ), (, 5), (, 6)} 6. Express {(x, y)/ x + y = 00, where x, y W} as a set of ordered pairs. {(x, y) / x + y = 00, where x, y W} We have, x + y = 00 When x = 0, y = 0 x + y = = 00 When x = 6, y = 8 x + y = = 00 When x = 8, y = 6 x + y = = 00 When x = 0, y = 0 x + y = = 00 Set of ordered pairs = {(0, 0), (6, 8), (8, 6), (0, 0)} 6 7. Write the domain and range of the following relations: i. {(a, b)/a is a natural number less than 6 and b = } ii. {(a, b)/a and b are natural numbers and a + b = } iii. {(, ), (, 5), (, 6), (, 7)}. i. Let R = {(a, b)/ a N, a < 6 and b = } Set of values of a are domain and set of values of b are range a N and a < 6 a =,,,, 5 and b = Domain (R ) = {,,,, 5} Range (R ) = {} ii. Let R = {(a, b)/a, b N and a + b = } Now, a, b N and a + b = When a =, b = When a =, b = 0 When a =, b = 9 When a =, b = 8 When a = 5, b = 7 When a = 6, b = 6 When a = 7, b = 5 When a = 8, b = When a = 9, b = When a = 0, b= When a =, b = Domain (R ) = {,,,, 5, 6, 7, 8, 9, 0, } Range (R ) = {, 0, 9, 8, 7, 6, 5,,,, } iii. Let R = {(, ), (, 5), (, 6), (, 7)} Domain (R ) = {} Range (R ) = {, 5, 6, 7} 8. Let = {6, 8} and = {,, 5}. Let R = {(a, b)/a, b, a b is an even}. Show that R is an empty relation from to. = {6, 8}, = {,, 5} R = {(a, b)/ a, b, a b is an even number} a a = 6, 8 b b =,, 5 When a = 6 and b =, a b = 5 which is odd When a = 6 and b =, a b = which is odd When a = 6 and b = 5, a b = which is odd When a = 8 and b =, a b = 7 which is odd When a = 8 and b =, a b = 5 which is odd When a = 8 and b = 5, a b = which is odd Thus, no set of values of a and b gives a b even R is an empty relation from to.

20 Chapter 0: Sets, Relations and Functions 9. Determine the domain and range of the following: i. R = {(a, a ) / a is a prime number less than 5} ii. R = a, / 0 a 5,aN a i. R = {(a, a ) / a is a prime number less than 5} a =,, 5, 7,, a =, 9, 5, 9,, 69 R = {(, ), (, 9), (5, 5), (7, 9), (, ), (, 69)} Domain (R ) = {a/a is a prime number less than 5} = {,, 5, 7,, } Range (R ) = {a /a is a prime number less than 5} = {, 9, 5, 9,, 69} ii. R = a, 0 a 5,a N a a =,,,, 5 a =,,,, 5 R =,,,,,,,, 5, 5 Domain (R ) = {a/0 < a 5, a N} = {,,,, 5} Range (R ) = { /0 a5,an a =,,,, 5 0. The domain of the relation R = {(a, b) / b = a +, a I, 0 < a < 5}. Find the range of R. R = {(a, b) / b = a +, a I, 0 < a < 5} a =,,, b =,,, 5 Range (R) = {,,, 5}. Write the following relations as sets of ordered pairs and find which of them are functions. i. {(x, y) / y = x, x {,, }, y {, 6, 9, }} ii. {(x, y) / y > x +, x = {, } and y = {,, 6}} iii. {(x, y) / x + y =, x, y {0,,, }} i. {(x, y)/y = x, x {,, }, y {, 6, 9, }} Here y = x When x =, y = () = When x =, y = () = 6 When x =, y = () = 9 Ordered pairs are {(, ), (, 6), (, 9)} Every element of domain is associated with unique element of codomain. It is a function ii. {(x, y) / y > x +, x =, and y =,, 6} Here, y > x + When x = and y =, + When x = and y =, > + When x = and y = 6, 6 > + When x = and y =, + When x = and y =, > + When x = and y = 6, 6 > + Ordered pairs are {(, ), (, 6), (, ), (, 6)} Since, and are associated with two elements i.e., and 6 It is not a function iii. {(x, y) / x + y =, x, y (0,,, )} Here, x + y = When x = 0, y = When x =, y = When x =, y = When x =, y = 0 Ordered pairs are {(0, ), (, ), (, ), (, 0)} Every element of domain is associated with unique element of codomain It is a function. Find the domain and range of the following functions: i. f(x) = x ii. f(x) = x x iii. f(x) = x x iv. f(x) = 9 x i. f(x) = x Domain = set of all real numbers Range = {x / x R and x 0} 7

21 Std. I : Commerce (Maths I) ii. f(x) = x x For this to exist (x ).( x) 0 (x ) 0 and x 0 x and x x and x x x [, ] or x 0 and x 0 x and x x and x Which is not possible Domain is [, ] and For range 8 Let y = f(x) = x ( x ) y = (x ) ( x) y = x + x x x + ( + y ) = 0 Disc > 0.( x is real) ( ) ()( + y ) > 0 6 y 0 y < y < < y < f (x) Range is [, ] iii. f(x) = x x f(x) is not defined, when x = 0 i.e., when x = Domain of f = R {} Let y = f(x) = x x xy y = x xy + x = + y x(y + ) = + y x = y y which is not defined, when y + = 0 i.e., when y = Range of f = R { } iv. f(x) = 9 x f(x) is defined, when 9 x > 0 9 x x 9 x and x x Domain of f = [, ] Now, x 0 x 9 0 x x x 9 9 x f(x ) Range of f = [, ]. Find the range of each of the following functions: i. f(x) = x, for x ii. f(x) = 9 x, for 5 x iii. f(x) = x 6x +, for all x R. i. f(x) = x for x s x x 9 x 9 7 x 5 7 f(x) 5 Range of f is [7, 5]. ii. f(x) = 9 x, for 5 x s 5 x 0 x 5 0 x 50 0 x x x 9 f(x) Range of f is [, 9] iii. f(x) = x 6x +, for all x R = (x 6x + 9) + = (x ) + ut (x ) 0, for all x R (x ) f(x) Range = [, )

22 Chapter 0: Sets, Relations and Functions. Solve the following: i. x f(x) =, find f(), f() x ii. f(x) = (x )(x + ), find f(), f(), f() iii. f(x) = x x, find f(x + ). i. x f(x) = x f() = = 7 = 6 = f() = = = 0 = 0 ii. f(x) = (x )(x + ) f() = ( )[() + ] = 0 f() = ( )[() + ] = 5 f() = ( )[ ( ) + ] = 0 iii. f(x) = x x f(x + ) = (x + ) (x + ) = (x + x + ) x 6 = x + 8x + 8 x 7 f(x + ) = x + 5x + From fig. every element of set is associated with unique element of set It is a function Domain = {,, 6, 8, 0,, } Range = {,,,, 5, 6, 7} lso, each element of domain is half of the corresponding element of co-domain. x Function is y = ii. Let f = {(, ), (, ), (5, )} 5 From fig. every element of set is associated with unique element of set It is a function Domain = {,, 5} and Range = {, } The function cannot be expressed in formula. iii. Let f = {(, ), (, ), (, 5), (5, )} 5. Which of the following relations are functions? Give reason, if it is a function. Determine its domain and range. lso express the function by a formula. i. {(, ), (, ), (6, ), (8, ), (0, 5), (, 6), (, 7)} ii. {(, ), (, ), (5, )} iii. {(, ), (, ), (, 5), (5, )} iv. {(0, 0), (, ), (, ), (, ), (,), (9, ), (9, ), (6, ), (6, )} i. Let f = {(, ), (, ), (6, ), (8, ), (0, 5), (, 6), (, 7)} iv. 5 Since has two images i.e. and 5, therefore it is not a function Let f = {(0,0), (, ), (, ), (, ), (, ), (9, ), (9, ), (6, ), (6, )} ,, 9, 6 have two images It is not a function 9

23 Std. I : Commerce (Maths I) 6. Find a, if f(x) = ax + 5 and f() = 8. f(x) = ax + 5 and f() = 8 f() = a() = a + 5 a = 7. If f(x) = f(x ), for f(x) = x x +, find x. f(x) = x x + lso, f(x) = f(x ) x x + = (x ) (x ) + x x = 9x 6x + x + 8x x + 5 = 0 8x x 0x + 5 = 0 x(x ) 5(x ) = 0 (x )(x 5) = 0 x = or x = 5 8. If f(x) = x x +, then find the value of x satisfying f(x) = f(x + ). f(x) = x x + lso, f(x) = f(x + ) x x + = (x + ) (x + ) + x x + = x + x + 6x + x + x = 0 x + x x = 0 x(x + ) (x + ) = 0 (x + )(x ) = 0 x = or x = 9. Let = {,,, } and Z be the set of integers. Define f: Z by f(x) = x + 7. Show that f is a function from to Z. lso find the range of f. = {,,, } f(x) = x + 7 when x =, f() = () + 7 = 0 when x =, f() = () + 7 = when x =, f() = () + 7 = 6 when x =, f() = () + 7 = 9 f = {(, 0), (, ), (, 6), (, 9)} It is a function because each element in has one and only one image in Z Range of f = {0,, 6, 9} 0. Find whether following functions are one-one, onto or not: i. f: RR given by f(x) = x + 5 for all x R ii. f: ZZ given by f(x) = x + for all x Z i. Let f: R R given as f(x) = x + 5 for all x R. First we have to show that f is one-one function For this we have to show that if f(x ) = f(x ), then x = x Here, f(x) = x + 5 Let f(x ) = f(x ) x + 5 = x + 5 x = x x = x f is one-one function Now we have to show that f is onto. For that we have to prove that for any y co-domain R, there exist an element x domain R such that f(x) = y. Let y R be such that y = f(x) y = x + 5 x = y 5 x = y 5R for any y co-domain R, there exist an element x = y 5 domain R such f(x) = y. f is onto function. f is one-one onto function. ii. Let f : Z Z given by f(x) = x + for all x Z Let x, x R be such that f(x ) = f(x ) x + = x + x = x x = x f is not one-one function. Here 0 co-domain Z, but there does not exist x domain Z such that f(x) = 0 f is not onto function. f is not one - one onto function. 0

24 Chapter 0: Sets, Relations and Functions. Find which of the functions are one-one onto, many-one onto, one-one into, manyone into. Justify your answer. i. f:rr given as f(x) = x + 7 for all x R ii. f: RR given as f(x) = x for all x R iii. f = {(, ), (, 6), (, 9), (, )} defined from to where = {,,, }, = {, 6, 9,, 5}. i. f: RR given as f(x) = x + 7 for all x R First we have to show that f is oneone For this we have to show that if f(x ) = f(x ) then x = x Here, f(x) = x + 7 Let f(x ) = f(x ) x + 7 = x + 7 x = x f is one-one function. Now we have to show that f is onto function. For that we have to prove that for any y co-domain R, there exist an element x domain R such that f(x) = y. Let y R be such that y = f(x) y = x + 7 y 7 = x y 7 x = R for any y co-domain R, there exist an element y 7 x = R such that f(x) = y. f is onto function. f(x) is one-one onto function. ii. f: RR given as f(x) = x for all x R To find whether it is one-one or many-one Let f(x ) = f(x ) x = x x = x f is not one-one function. f is many one function Now we have to show that f is onto function. For that we have to prove that for any y co-domain R, there exist an element x domain R such that f(x) = y. Let y R be such that y = f(x) y = x x = y R if y < 0 for any y co-domain R, there does not exist an element x domain R such that f(x) = y. f is into function Hence f is many-one into function. iii. f = {(, ), (, 6), (, 9), (, )} = {,,, }, = {, 6, 9,, 5} f is defined from to f: Here each and every element of have their distinct images in f is one-one function. lso element 5 in the co-domain don t have any pre-image in the domain. f is into function. f is one-one into function..8 Particular types of functions and their graphs. Constant function: function f defined by f(x) = k, for all x R, where k is a constant, is called a constant function. The graph of a constant function is a line parallel to the -axis, intersecting -axis at (0, k). (0, k) O P(x, k) (x, 0) f(x) = k For example, f(x) = 5 is a constant function.. Identity function: The function f(x) = x, where x R is called an identity function. The graph of the identity function is the line which bisects the first and the third quadrants. Observe the following table of some values of f. x 0 f(x) 0

25 Std. I : Commerce (Maths I) Let f: (R(0)) R : f (x) = x for all values. Polynomial function: function of the form f(x) = a 0 + a x + a x + + a n x n, where n is a non-negative integer and a 0, a, a,, a n R is called a polynomial function. f(x) = x x for x R x 0 f(x) = x x (, 5) (, ) (0, 0) (, ) (, ) 5 (, 0) (, 0) O (0, ) (, ) (, ). Rational function: The function of the type f( x ), where f(x) and g( x ) g(x) are polynomial functions of x, defined in a domain, where g (x) 0 is called a rational function. (, ) (, ) O (, ) (, 5) of x R (0) We have x f(x) = /x O (0.5, ) (0.5, ) (, ) (, 0.5) 5. Modulus function: Let f: R R, the function f(x) = x such that x,forx0 x = x,forx 0 is called modulus or absolute value function. The graph of the absolute value function consist of two rays having common end point origin and bisecting the first and second quadrant. Consider table of same values of f(x) = x x 0 f(x) 0 f(x) = x (, ) (, ) (, ) 6. Even function: function f is said to be an even function, if f(x) = f(x) for all x R Let f: R R : f (x) = x for all x R Domain of f = R, range of f = {x/x R, x 0} (, ) (, ) f(x) = x (, )

26 Chapter 0: Sets, Relations and Functions We have x 0 f(x) = x 0 7. Odd function: function f is called an odd function, if f(x) = f(x) for all x R Let f: R R : f (x) = x for all x R Then, domain of f = R and range of f = R. We have (, ) (, ) (, ) x 0 f(x) = x (, ) (, 8) O (, ) 8 (, 8) (, ) O Exponential function: Let f: R R +. The function f is defined by f(x) = a x, where a > 0, a is called an exponential function. f(x) = b x (b < 0) O For example, f(x) = x function. f(x) = a x (a > 0) is an exponential 9. Logarithmic function: Let a be a positive real number with a, if a y = x, x R then y is called the logarithm of x with base a and we write it as y = log a x. i.e. function f : R + R defined by f(x) = log a x is called logarithmic function. f = {(x, log a x)/x R, a > 0, a } O O y = log a x (a > )

27 Std. I : Commerce (Maths I).9 Composite function If f: and g: C are two functions then the composite function of f and g is the function gof: C given by (gof)(x) = g[f(x)], for all x Let z = g(y) then z = g(y) = g[f(x)] C gof C This shows that every element x of the set is related to one and only one element z = g[f(x)] of C. This gives rise to a function from the set to the set C. This function is called the composite of f and g. Note that (fog) (x) (gof)(x).0 Inverse function x f y If a function f: is one-one and onto function defined by y = f(x), then the function g: defined by g(y) = x is called the inverse of f and is denoted by f. Thus f : is defined by x = f (y) We also write if y = f(x) then x = f (y) Note that if the function is not one-one nor onto, then its inverse does not exist. The geometrical representation of the function is called graph of the function. We know that a function can be expressed as a set of ordered pairs. Let f: be a function, where and are non empty subsets of R. Let (x, y) be an element of f, where x, y. Since x, y are real numbers, we can plot the point (x, y) in a plane by choosing a suitable co-ordinate system. On plotting all such ordered pairs from the set representing f we get a geometrical representation of the function. This is called graph of the function f. g z Exercise.. Let f and g be two real valued functions defined by f(x) = x + and g(x) = x. Find i. f + g ii. f g iii. f g. f(x) = x + and g(x) = x i. (f + g) x = f(x) + g(x) = x + + x = x ii. (f g) x = f(x) g(x) = x + (x ) = x iii. f x = f( x ) x g g( x) x. Find i. gof ii. fog, where i. f(x) = x, g(x) = x + x + ii. f(x) = x, x 0, g(x) = x x x. i. f(x) = x and g(x) = x + x + (gof)(x) = g[f(x)] = g(x ) = (x ) + (x ) + = x x + + x 6 + = x x (fog)x = f[g(x)] = f(x + x + ) = x + x + = x + x ii. f(x) = x and g(x) = x x (gof) = g[f(x)] = g x x x x x x x (fog)x = f[g(x)] = f = = x+ x x x. If f(x) = x x, x, prove that fof is identity function. f(x) = x x (fof)(x) = f[f(x)] x x = f = x x x x

28 Chapter 0: Sets, Relations and Functions = x 6 9 x 6 6x 96x+ = x (fof)(x) = x fof is an identity function. = x. If f(x) = x x, x and g(x) = x x, x prove that (gof) (x) = (fog) (x) = x. f(x) = x x, g(x) = x x x x (gof)(x) = g[f(x)] = g = x x x x x8x = = x = x x8 x x x x (fog)(x) = f[g(x)] = f = x x x = x 6 8 x 6 = x = x x8x (gof)(x) = (fog)(x) = x 5. If f = {(, ), (, 6), (, 8), (5, 0), (6, )}, g = {(, ), (6, 9), (8, 5), (0, ), (, 7)}, find (gof). f = {(, ), (, 6), (, 8), (5, 0), (6, )} g ={(, ), (6, 9), (8, 5), (0, ), (, 7)} Let = {,,, 5, 6}, = {, 6, 8, 0, } and C = {, 9, 5,, 7} f(x) = x and g(x) = x + (gof)(x) = g[f(x)] = 9(x) = (x) + = 6x + (gof)() = 6() + = (gof)() = 6() + = 9 (gof)() = 6() + = 5 (gof)(5) = 6(5) + = (gof)(6) = 6(6) + = 7 (gof) = {(, ), (, 9), (, 5), (5, ), (6, 7)} 6. Show that f:r R given by f(x) = x is one-one and onto. Find its inverse function. lso find f (9) and f (). f: R R given by f(x) = x Let x, x R be such that f(x ) = f(x ) x = x x = x x = x f is one-one function. Now we have to show that f is onto function. Let y R be such that y = f(x) y = x y x = R for any y co-domain R, there exist an element y x = R such that f(x) = y. f is onto function. f is a one-one onto function. f exists f y (y) = f x (x) = f (9) = 9 = f () =. Functions in Economics. Demand Function: ccording to demand law, quantity of commodity demanded is inversely related to the price of, where other things remain the same. i.e., for higher price, the demand is less and for lower price, the demand is more. Demand D is function of the price p i.e., D = f (p) ccording to marshall law, the price P is function of demand D i.e., p = g (D) p = 0 D 00, p = D 5 6, p = 50 D, etc. 5

29 Std. I : Commerce (Maths I) 6 The graph of Demand and price or quantity demanded is as follows: It shows the diagrammatic representation of the functional relationship between the price of quantity and demand of quantity. lso, it shows an inverse or negative relationship between price and demand of quantity.. Supply Function: Supply is also related to price like demand. If p is the price and S is supply for the good, then price p is the function of supply S i.e., p = g(s) p = 9 + S, p = S + S, etc. The graph of supply and price for a good is as follows: Price of O Price of O Quantity Demanded of Quantity Supplied of Price and supply increase or decrease together i.e., supply S increases (decreases) with the price p. Thus, the supply curves are sloping upwards from left to right as shown in graph above.. Total Revenue Function: The total revenue function is obtained from the demand and price. If p is price and D is the demand for the goods. Then the total revenue (R) is given by R = pd If Demand D is the function of price p i.e., D = f(p), then total revenue function can be expressed as a function of p. i.e., R = p. f(p) If price p is the function of Demand D i.e., p = g(d), then total revenue function can be expressed as a function of D i.e., R = D. g(d) If p is a linear function of D i.e. if p = D Then the total revenue function R is given by R = pd = ( D) D R = 500 D + 7D The graph of total revenue function is as shown below: Total Revenue (TR) O Demand D Total revenue curve is parabolic in nature.. Total Cost Function: The total cost function is the combination of fixed cost and variable cost, for a quantity x of a certain good. Fixed cost does not depend on the quantity x of that good. It may be due to rent of the premises, expenses on the research laboratory, etc. Variable cost depends on x Cost function may be of the following type C = ax + b, C = ax + bx + c, where a is positive. 5. Profit Function: Profit function is the difference between the total revenue function and total cost function. If R is the total revenue function and C is the total cost function, then the profit function () is given as = R C

30 Chapter 0: Sets, Relations and Functions. Some more functions and their graphs. Graph of Exponential Function: Let f: R R +. The function f is defined by f(x) = a x, where a > 0, a is called an exponential function. f(x) = a x (a > 0) f(x) = b x (b < 0) O O For example, f(x) = x function. is an exponential. Graph of y = log e x: Let a be a positive real number with a, if a y = x, x R then y is called the logarithm of x with base a and we write it as y = log a x. i.e. function f : R + R defined by f(x) = log a x is called logarithmic function. f = {(x, log a x)/x R, a > 0, a } y = log a x (a > ) x / / 5/6 y In the above table, we have assumed =. and = 0.87 sin = = =. = 0.7 Using the result sin (x) = sin x, the table for the values of x between and 0 is obtained as follows: x 5/6 / / / y x / / /6 0 y The graph corresponding to these points is as given below: / O / / y = sin x Extension of the graph of sin x : Since sin ( + x) = sin x, one can also extend the graph of y = sin x as shown below. O y = sin ( + x) 5 5 O. Graph of y = sin x: The values of x and y = sin x are given in the following table: x 0 /6 / / / y

31 Std. I : Commerce (Maths I). Graph of y = cos x: The values of x and y = cos x are given in the following table: x 0 /6 / / / y x / / 5/6 y Using the result cos (x) = cos x, the table for the values of x between and 0 is obtained as follows: x 5/6 / / y x / / / /6 0 y The graph corresponding to these points is as given below: Extension of the graph of cos x : / O / O 5 5 y = cos x 5. Graph of y = tan x: tan x does not exist for x = increases from 0 to. i. sin x increases from 0 to and ii. cos x decreases from to 0. tan x = sin x cos x but as x will increase indefinitely as x (starting from the value 0) approaches. Similarly as x (starting from the value 0) approaches, tan x decreases indefinitely. The corresponding value of x and y = tan x are given in the following table: x / / /6 0 /6 / / y The graph of y = tan x Extension of the graph of tan x : O / / y = tan x When the graph of tan x is extended to values beyond, the entire curve shown in below figure repeats completely for the intervals,, 5, etc. as well as for intervals,, 5, etc. 8