11th Maths 2 Science PDF

Prepared according to the new syllabus prescribed by Maharashtra State Secondary and Higher Secondary Education Board, Pune effective from 2013 - 14 onwards. Revise...

Prepared according to the new syllabus prescribed by Maharashtra State Secondary and Higher Secondary Education Board, Pune effective from 2013 - 14 onwards. Revise d CHETANA Edition as per New P aper Pattern MATHEMATICS & STATISTICS PART - II STANDARD XI Powered by MT EDUCARE Salient Features : w According to the latest textbook and New Board Paper Pattern. w ‘Points to Remember’ - highlights theories and definitions of all the topics in the chapter. w Solutions to all textual exercises. w Additional Problems for Practice for thorough revision. w List of all important formulae given. w All chapters broken down into topics and sub-topics for easy understanding. w Includes marks for all questions in all sections. w Multiple Choice Questions (MCQs) given for every chapter. w Assessment Paper at the end of every chapter. w Termwise Model Question papers. CHETANA PUBLICATIONS (INDIA) LLP EDUCATIONAL PUBLISHERS 4th Floor, B Wing, Building E - Trade Link Kamala City, Senapati Bapat Marg, Lower Parel, Mumbai - 400 013. Tel.: +91-22-6121 6000 E-mail: [email protected] Website: www.chetanapublications.com MKE1180318  PREFACE Chetana’s Master Key Series is designed for F.Y.J.C (Std. XI) Science. The series follows the guidelines set by Maharasthra State Board of Secondary and Higher Secondary Education, Pune, Maharashtra. The main purpose of the Master Key is to enable every student to prepare himself / herself thoroughly for the Board Exams. In Mathematics Part I & II, a method of unit wise bifurcation of each chapter has been adopted. The sub-units of each chapter and the topics are illustrated and covered through adequate textual and additional exercises, Points to Remember, Solutions to Textual Exercise and Miscellaneous Exercise. This method enables the students to divide their study time into short durations and study the whole chapter, bifurcating it and thus making our Master Key Series a new generation student friendly way of learning. Understanding the needs and study environment of the new generation students, we have researched and accomplished this task by the co-ordination of highly experienced teachers, educationists and experts and made this series a PERFECT Key to every subject. Chetana assures to build confidence and trust in every student to face the challenges to unlock the doors of doubts and fears in educational field. We are grateful to educationist & teachers who have nourished us with ideas and advice to improve and helped us to grow in the educational sector. At the same time we heartily welcome all your suggestions in future. We wish you Good Luck and grace that you may succeed in all your future endeavours. - Publisher Disclaimer This reference book is transformative work based on textual contents published by Bureau of Textbook. We the publishers are making this reference book which constitutes as fair use of textual contents which are transformed by adding and elaborating, with a view to simplify the same to enable the students to understand, memorize and reproduce the same in examinations. This work is purely inspired upon the course work as prescribed by the Maharashtra State Board of Secondary and Higher Secondary Education, Pune. Every care has been taken in the publication of this reference book by the Authors while creating the contents, the Authors and the Publishers shall not be responsible for any loss or damages caused to any person on account of errors or omissions which might have crept in or disagreement of any third party on the point of view expressed in the reference book. ©️ reserved for all the contents created by our Authors. No copyright is claimed in the textual contents which are presented as part of fair dealing with a view to provide best supplementary study material for the benefit of students. The text of this publication, or any part thereof, may not be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, storing in retrieval system, or otherwise, without the prior written permission of the publishers. © Publishers Published by : CHETANA PUBLICATIONS (INDIA) LLP Educational Publishers 4th Floor, B Wing, Building E - Trade Link Kamala City, Senapati Bapat Marg, Lower Parel, Mumbai - 400 013. Tel.: +91-22-6121 6000 E-mail: [email protected] Website: www.chetanapublications.com Printed at : Graphic Printers, Sewri, Mumbai. (ii)  Paper Pattern General Instructions: (1) All Questions are compulsory. (2) The question paper consist of 30 questions divided in 4 sections A, B, C, D. (3) Section A contains 6 multiple choice questions (MCQ) of 1 mark each. Section B contains 8 questions of 2 marks each. (One of them will have internal option) Section C contains 6 questions of 3 marks each. (Two of them will have internal options) Section D contains 10 questions of 4 marks each. (Three of them will have internal options) (4) Use of logarithmic tables is allowed. (5) Use of calculator is not allowed. Section A : Question No. 1 to 6 Section B : Question No. 7 to 14 Section C : Question No. 15 to 20 Section D : Question No. 21 to 30 Weightage to types of questions Type Marks to each question Number of questions Marks Objectives (MCQ) 1 6 6 Very small Answer 2 8 16 Short Answer 3 6 18 Long Answer 4 10 40 Total 80 Weightage to assessment of Objectives Objectives Marks Knowledge 24 Understanding 32 Application 16 Skill 8 Total 80 Weightage to Difficulty Level of Questions Estimated Level Marks Difficult 16 Average 40 Easy 24 Total 80 (iii)  1 Sets, Relations And Functions a and b belong to the open interval (a, b) but Points to Remember : a, b themselves do not belong to this interval. R Set: a x b A well-defined collection of objects is ∴ (a, b) = {x/x ∈ R, a < x < b}. called a set. (2) Closed Interval: The object in a set is called its element or member. Let a, b ∈ R and a < b then the set {x/x ∈ R, a ≤ x ≤ b} is called closed interval We denote sets by capital letters A, B, C, and is denoted by [a, b]. All the numbers etc. The elements of a set are represented by small letters x, y, z etc. If x is an element between a and b belong to the closed interval of set A, we write x ∈ A, and read as ‘x [a, b]. Also a and b belong to this interval. belongs to A’. If x is not an element of set R A, we write x ∉ A, and read as ‘x does not a x b belong to A’. There are two methods of [a, b]= {x/x ∈R, a ≤ x ≤ b} representing a set, (3) Semi-closed interval: (1) Roster or Tabular Method or list method [a, b)={x/x ∈R, a ≤ x < b} (2) Set-Builder or Ruler Method R Subset: a x b A set A is said to be a subset of set B if every Note that a ∈ [a, b) and b ∉ [a, b). element of A is also an element of B and i.e., [a, b) exclude b but include a. we write A ⊆ B. Superset: (4) Semi-open interval: If A ⊆ B, then B is called a superset of A (a, b] = {x/x ∈R, a < x ≤ b} and we write, B ⊇ A. R a x b Proper subset: Note that a ∉ (a, b] but b ∈(a, b] i.e., (a, b] A non-empty set A is said to be a proper exclude a but include b. subset of the set B, if and only if all elements of set A are in set B and at least one element (5) (i) The set of all real numbers greater than of B is not in A. i.e. If A ⊆ B and A ≠ B then a i.e. (a, ∞)={x/x ∈R, x > a} A is called a proper subset of B and we R write A⊂ B. a (ii) The set of all real numbers greater than Remarks: or equal to a If there exists even a single element in A [a, ∞ )={x/x ∈R,x ≥ a} which is not in B, then A is not a subset of R B and we write, A ⊄ B. a Intervals : (6) (i) The set of all real numbers less than b (1) Open Interval: i.e. (– ∞, b) Let a, b ∈ R and a < b then the set {x/x ∈ R R, a < x < b} is called open interval and is b denoted by (a, b). All the numbers between (– ∞, b) ={x/x ∈R, x < b} (5) 6 Master Key Mathematics & Statistics - Part II (Std. XI) (ii) The set of all real numbers less than or consideration are subsets of a set say U, equal to b i.e. (– ∞ , b] then U is called the universal set. R Venn diagram: b The pictorial representation of a set is called (–∞ , b] ={x/x ∈ R, x ≤ b] Venn diagram. (7) The set of all real numbers R is (– ∞ , ∞ ) R= (– ∞ , ∞ ) ={x/x ∈R, –∞ < x < ∞ }. Types of Sets: (1) Empty Set : A set containing no element is called the empty set or the null set and is denoted by the symbol φ or { } or void set. Complement of a set: e.g. A = {x ∈ N/1 < x < 2}. If A is a subset of the universal set U, then (2) Singleton Set: A set containing only one the set of all elements in U which are not in element is called a Singleton Set. A is called the complement of the set. e.g. (i) Let A be a set of all integers which are neither positive nor negative. ∴ A = {0} (ii) Let B be a set of Capital of India. Properties of Complement of a set: ∴ B = {Delhi} (1) (A’)’ = A (3) Finite and Infinite Sets: A set in which the (2) φ’ = U (U is the universal set) process of counting of elements comes to (3) U’ = φ an end is called a finite set. Union of sets: A set which is not finite is called an infinite set. If A and B are two sets, then the set of those elements which belongs to A or to B or to e.g. both A and B is called union of set. Set of letters in the word ‘beautiful’ A = {a, b, e, l, f, t, i, u} Check whether you agree with following: (i) An empty set is a finite set. (ii) N, Z and set of all points on the circumference of a circle are infinite sets. Properties of Union of sets: (4) Equal Sets: Two sets are said to be equal if they contain the same elements i.e. if A ⊆ B (1) A ∪ B = B ∪ A Commutativity and B ⊆ A. (2) (A ∪ B) ∪ C =A ∪ (B ∪ C) Equivalent Sets: Associativity property Two finite sets A and B are said to be (3) A ∪ φ = A Identity for Union equivalent if n(A) = n(B). Equal sets are (4) A ∪ A = A Idempotent law always equivalent but equivalent sets need (5) A ∪ A’= U Complement law not be equal. (6) If A ⊂ B then A ∪ B = B Universal Set: (7) U ∪ A = U If in a particular discussion, all sets under (8) A ⊂ (A ∪ B ), B ⊂ (A ∪ B) Sets, Relations and Functions 7 Intersection of sets: A B If A and B are two sets, then the set of A–B B–A those elements which belong to both A and B, i.e. which are common to both A and B, is called Intersection of set. Similarly, B – A = {y/y ∈ B, y ∉A} Remarks: (1) A – B is a subset of A and B – A is a subset of B. (2) The sets A – B, A ∩ B, B – A are mutually Properties of Intersection of sets: disjoint sets, i.e. the intersection of any of (1) A∩B=B∩A Commutativity these two sets is the null (empty) set (2) (A ∩ B) ∩ C = A ∩ (B ∩ C) (3) A–B = A ∩ B’ Associativity B–A = A’ ∩ B (3) φ ∩A= φ (4) A ∪ B = (A – B) ∪ (A ∩ B) ∪ (B – A) (4) A∩A=A Idempotent law Number of elements in a set: (5) A ∩ A’ = φ Complement law The number of distinct elements contained (6) If A ⊂ B then A ∩ B = A in a finite set A is denoted by n(A) (7) U∩A=A Identity for Some important results: intersection For given sets A, B (8) A ∩ B ⊂ A, A ∩ B ⊂ B (1) n(A ∪ B) = n(A) + n(B) – n(A ∩ B) (9) (i) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) (2) When A and B are disjoint sets then n(A ∪ B ) = n(A) + n(B) (ii)A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) Distributive properties (3) n(A ∩ B’) + n (A ∩ B) = n(A) (4) n(A’ ∩ B ) + n(A ∩ B) = n(B ) De Morgan’’ s Laws: (5) n(A ∩ B’) + n(A ∩ B)+ n(A’ ∩ B) = n(A ∪ B) For any two sets A and B For any sets A, B, C (1) (A ∪ B)’= A’ ∩ B’ (6) n(A ∪ B ∪ C ) = n(A) + n(B ) + n(C) – (2) (A ∩ B)’ = A’ ∪ B’ n(A ∩ B ) – n(B ∩ C ) – n(A ∩ C)+ Disjoint sets: n (A ∩ B ∩ C) Power set: Two sets A and B are said to be disjoint, if Definition: they have no common element. The set of all subsets of a given set A is called i.e., A ∩ B = φ the power set A and is denoted by P(A). Thus, every element of power set A is a set. e.g. consider the set A = {a, b}, let us write all subsets of the set A. We know that φ is a subset of every set, so φ is a subset of A. Also {a}, {b} and {a, b} are also subsets of A. Difference of sets: Thus, the set A has in all four subsets The difference of set A and set B is the set viz. φ , {a}, {b}, {a, b} of elements which are in A but not in B and ∴ P(A) ={φ , {a}, {b}, {a, b} } is denoted by A – B. ∴ n[P(A)] = 4 = 22 in general, The shaded portion represents A – B If n(A) = m then we have, n[P(A)] = 2m ∴ A – B = {x/x ∈A, x ∉ B} 8 Master Key Mathematics & Statistics - Part II (Std. XI) MASTER KEY QUESTION SET - 1 ∴ C= EXERCISE - 1.1 (Textbook Page No. 10) (3) If A = {x / 6x2+ x – 15 = 0}, B = {x / 2x2 – 5x – 3 = 0}, (1) Describe the following sets in Roster form : C = {x / 2x2 – x – 3 = 0}, (i) {x / x is a letter of the word ‘MARRIAGE’} Find : (i) A ∪ B ∪ C (ii) A ∩ B ∩ C (2 marks each) (ii) {x/x is an integer and – x + 1, x = 1, 2 and 2, 3, 5, 7, 11, 13. y = 2, 4, 6} ∴ R1 = {(2, 4), (3, 9), (5, 25), (7, 49), (11, 121), Here y > x + 1, where x = 1, 2 and y = 2, 4, 6 (13, 169)} When x = 1, 4 > 1 + 1, 6 >1 + 1 Domain of R1 ={a/a is a prime number less ∴ y = 4, y =6 than 15} When x = 2, 4 > 2 + 1, 6 >2 + 1 = {2, 3, 5, 7, 11, 13} ∴ y = 4, y = 6 Range of R1 ={ a2/a is a prime number less ∴ R2 = {(1, 4), (1, 6), (2, 4), (2, 6)} than 15} (iii) Let R 3 = {(x, y)/x + y = 3, x, y ∈{0, 1, 2, 3}} = {4, 9, 25, 49, 121, 169} Here, x + y = 3, where x, y ∈ {0, 1, 2, 3} when x = 0, y = 3 (ii) R2 = when x = 1, y = 2 Domain of R2 = {a/0 < a ≤ 5, a ∈N} when x = 2, y = 1 = {1, 2, 3, 4, 5} when x = 3, y = 0 ∴ R3 = {(0, 3), (1, 2), (2, 1), (3, 0)} Range of R2 = Problems for Practice - 1.2 = (10) R = {(a, b)/b = a + 1, a ∈ Z, 0 < a < 5}. Find (1) Find x and y if (2 marks) the range of R. (2 marks) Solution: (2) Find A × B, B × A for the sets A = {1, 2, 3}, R = {(a, b)/b = a + 1, a ∈ Z, 0 < a < 5} B = {x, y} (3 marks) Range of R = {b/b = a + 1, a ∈ Z, 0< a x + 1, x = 1, 2 and y = 2, 4, 6} (v) {(3, 6), (4, 6), (5, 6), (6, 6)} (iii) {(x, y)/ x + y = 3, x, y ∈ {0, 1, 2, 3} (2 marks each) Sets, Relations and Functions 17 (5) Write the relation as a set of ordered pairs: relation which associates every element of (2 marks) a set A to unique element of set B and is {(x, y)/y = x + 2, x ∈{1, 2, 3} y ∈{3, 4, 5}} denoted by f : A → B. If f is a function from A to B and ANSWERS (x, y) ∈ f then we write it as y = f(x) Here y is called image of x under f and x is called, the preimage of y under f. The (1) x= , y= function f from A to B is denoted by (2) A × B = {(1, x), (1, y), (2, x), (2, y), (3, x), (3, y)} f : A → B. Let f be a function from the set A to the setB B × A = {(x, 1), (x, 2), (x, 3), (y, 1), (y, 2), (y, 3)} i.e. f :A → B then set A is the domain of the (3) A × A = {(2, 2), (2, 4), (2, 6), (2, 8), (4, 2), (4, 4), function and set B is co-domain of the (4, 6), (4, 8), (6, 2), (6, 4), (6, 6), (6, 8), function. (8, 2), (8, 4), (8, 6), (8, 8), } Range : Set of all images in B is called the (4) (i) Domain = {1, 2, 3, 4, 5}, range of f. ∴ Range = { f (x)/ x ∈A} Range = {2, 3, 4, 5, 6} Every function is a relation but every (ii) Domain = {5}, Range = {1, 2, 3, 4} relation need not be a function. If the (iii) Domain = {0, 1, 4}, domain and co-domain are not specified, they are assumed to be the set of real Range = {1, 2, 3, 4, 5} numbers. (iv) Domain = {1, 2, 3, 4, 5}, Equal function: Range = {4, 7, 10, 14, 19} Definition: Two functions f and g are said (v) Domain = { 3, 4, 5, 6 }, to be equal if (1) domain of f = domain of g Range = {6} (2) co-domain of f = co-domain of g and (5) (1,3), (2, 4), (3,5) (3) f (x) = g (x ) for every x belonging to their common domain. Points to Remember : Two functions f and g are equal, then we write f = g. Functions: Representation of function: Function as a special kind of relation: (1) Arrow diagram: Here the corresponding Let us first consider an example where A is elements are connected by arrows. the set of men. B is the set of positive real (2) Stating the rule (In terms of formula) numbers. f be a relation from A to B given by In this method, we state domain, f = {(x, y )/y is the weight of man x } codomain and the rule under which correspondence is to be established. Hence f relates from A a man to his weight (3) Function as a set of ordered pairs: We can i.e. consider the function f as a set of ordered (1) Every man has some weight associated pairs i.e. a subset of A × B. with him. We can observe, that no two pairs of this set (2) that weight is unique. have same first component, for otherwise Definition: it will mean that one element of A is related A function from set A to the set B is a to two different elements of set B. 18 Master Key Mathematics & Statistics - Part II (Std. XI) (4) Tabular form: If the sets A and B are finite (4) Onto Function: and contain very few elements, then If the function f : A → B is such that each function f: A → B can be represented by a element in B is the image of some elements table in which domain and range are written in A, then f is said to be a function of ‘A in two rows. onto B’ and expressed symbolically as Types of Functions: A B. In this case, the range of f is the (1) Constant Function: same as its co-domain B. A function f defined by f(x) = c, x ∈R, Note : An ‘onto’ mapping is also called where c is a constant, is called a constant subjective mapping. function. In a constant function, domain = R and For example, the function f3 and f4 are onto range = { c }. The graph of a constant functions. function is a line parallel to axis, intersecting the y-axis in (0, c). A constant function is many-to-one. (5) Into Function: If the function f : A → B is such that there exists atleast one element in B which is not (2) Identity Function: the image of any element in A, then f is said A function f : A → A defined by f (x) = x, to be a function of “A into B” and express x∈A, is called the identity function on A. it symbolically as It is denoted by I. f:A B For the identity function, domain = R. The In this case, the range of f is the proper graph of the identity function is the line subset of its co-domain B. y = x which passes through the origin and makes an angle of 45° with the +ve x-axis. For example, the function f1 and f2 are into functions. (3) One - One Function: (6) Odd and Even Functions: If the function f : A → B is such that each A function f is called an odd function if element in A has one and only one image f (–x) = –f (x ) for all x in the domain of f. in B and every image in B does not have A function is called an even function if more than one pre-image in A, then f is f (–x) = f(x), for all x in the domain of f. called a one-one function. For example, Note: Let f (x)= x3 +2x. A one-one function is also called injective mapping. We can have one-one and onto Then f (–x) = (–x)3 + 2(–x) function or one-one and into function. A = –x3 – 2x = – (x3 + 2x) = –f (x) ‘one-one and onto’ mapping is called ∴ f is an odd function. bijective mapping. Let f (x)= x4 + 5x2 +3 Sets, Relations and Functions 19 Then f (–x) = (–x)4 + 5(–x)2 +3 (b) The Greatest integer Function 4 2 = x + 5x + 3 = f (x) (Step Function): ∴ f is an even function. A function f defined by f(x) = [x], x ∈ R, where [ ] denotes the greatest integer less than Note: Every function need not be even or odd. or equal to x i.e. ‘the greatest integer not greater (7) Polynomial Function : than x’ is called the step function. A function f defined by Thus, f (x )= [x] = n, if n ≤ x < n +1, where 2 n f(x)= a0+ a1x + a2x +.......+anx , x ∈R, n is an integer. where a0, a1, a2,..., an are constants and n Thus, f [x] = –3, if – 3 ≤ x < –2, is a +ve integer is called a polynomial = –2, if – 2 ≤ x < –1, function. For example, = –1, if –1 ≤ x < 0, 3 2 f (x) = 2x + 5x – x + 3 = 0, if 0 ≤ x < 1, (8) Rational Function: = 1, if 1 ≤ x < 2, A function f defined by f (x) = , = 2, if 2 ≤ x < 3 etc. x∈R, where g(x) and h(x) are polynomial functions is called a rational function. For example, f (x) = (9) Absolute value Function (Modulus function): A function f defined by f (x) = |x| = x, if x ≥ 0, = – x , if x < 0 Here, domain = R and range = I. is called an absolute value function. The graph consists of segments parallel to X-axis, corresponding to different intervals. The right hand end points of the line segments do not lie on the graph. There are breaks in the graph at all values of x = n, where n is an integer. Since the graph resembles steps, it is known as the step function. Here domain = R and (10) Exponential Function: range = {x/x ≥ 0} = (0, ∞) A function f : R→R + (set of +ve real The graph consists of two lines y = x and numbers) given by f (x) = a x , a > 0 is called y = –x above the x-axis. i.e., in first and exponential function and ‘a’ is called base second quadrants. i.e. f ≡ {(x, ax)/ x ∈ R, a > 0} (10)(a) The Signum Function: The function f : R → R e.g. f (x) = 4x, f (x) = , for all x ∈ R f (x)= 1, if x >0 Graph of ex : = 0, if x = 0 x 0 0.5 1 1.5 2 2.5 = –1, if x < 0 ex 1 1.6487 2.7182 4.4817 7.3891 12.1825 is called the signum function. Clearly, dom (f ) = R and range (f ) = {–1, 0, 1} x 3 3.5 4 ex 20.0855 33.1155 52.5981 20 Master Key Mathematics & Statistics - Part II (Std. XI) Graph of y = log ex x 0.5 1.5 2 2.5 3 logex –0.6931 0.4055 0.6931 0.9163 1.0986 : Graph of e–x: x 0 0.5 1 1.5 2 2.5 3 –x e 1 0.6065 0.3679 0.2231 0.3153 0.0921 0.0498 Operation on Functions: Let X ⊂ R then, various operations on functions are defined below : (1) Sum of functions: Let f : X → R and g : X → R then, f + g : X → R is defined by (f + g) (x)= f(x) + g(x), for all x ∈X. (2) Difference of functions: Let f : X → R and g : X → R then, f – g : X → R is defined by (f – g) (x) = f (x) – g(x), for all x ∈X. (3) Product of functions: Let f : X → R and g : X → R then, (11) Logarithmic Function: f. g : X → R is defined by Let a be a positive real number with (f. g)(x)=f (x). g (x) for all x ∈X. y a ≠ 1, if a = x, x ∈R then y is called the (4) Quotient of functions: logarithm of x with base ‘a’ and we write it Let f : X → R and g : X → R as y = loga x. i.e. A function f : R+→ R defined by f (x) = loga x is called logarithmic Note that is not defined when function. g (x) = 0 ∴ f = {(x, loga x)/ x ∈ R, a > 0, a ≠1} Let X 0 = {x ∈ X/g (x) = 0}. Note : Whenever, we say that f (x ) = log x without Then : X – X0 → R is defined by specifying the base, it is understood that the base of this number is ‘e’. (x) = , for all x ∈ X such that g (x) ≠ 0 (i.e. for all x ∈ X – X0) Sets, Relations and Functions 21 (5) Scalar multiplication of a Function: Let g(a, b)= a * b such that (a, b)A × A and Let f : X → R and k be a scalar then a * b ∈A i.e. g(a, b)= a * b , where a, b, a * b ∈A (kf ) : X → R is given by (kf )(x) = kf (x), Here * is called the binary operation in A. For for all x ∈X. A = R (real numbers) * can be +, –, × i.e +, –, × Composite function: are examples of binary operations, ÷ is a Let f : A → B and g : B → C be two functions. binary operation on R – {0}. If * is a binary operation in Set A, then we say that the set A Then, the composition of f and g, denoted is closed with respect to binary operation *. by gof, is defined as the function. gof : A → C given by gof (x) = g (f (x)), x ∈ A Real Valued Functions of the real variable: Clearly, dom (gof ) = dom. (f ) Definition: Also, gof is defined only when A function f : A→ B is called a real valued range (f ) ⊆ dom (g) function of real variable if A and B both are subsets of R. i.e.,domain and co-domain both are subset of the set R of all real numbers. Domain of real functions: The domain of the real function, f (x ) is the set of all those real numbers for which the expression for f(x) or the formula for f(x) Inverse Function: assumes real values only. Let f : A→ B is one-one, onto and g : B → A Range of real functions: which is also one-one, onto such that The range of a real function of a real gof: A→ A and fog : B→ B are both identity variable is the set of all real values taken functions by f(x) at points in its domain. (i.e. gof (x)= x, fog (y)=y ) then f and g are Method to find range of a real function called as inverse functions of each other. f(x) : Function g is denoted by f –1 and is read as Put y = f (x) f - inverse. So we define a function f –1 as Solve the equation y = f (x) for x in terms of y. f –1 : B → A such that if f(x)= y then f –1 (y)= x. Let x= φ (y), find the values of y for which Note: the values of x obtained from x= φ (y), are (1) Domain of f = Range of f –1 and Range of f = real and in the domain of f. Domain of f –1 (2) When f and g are inverses of each other, EXERCISE - 1.3 (Textbook Page No. 29) both f and g are one-one and onto. (1) Find the domain and range of the following Binary Operation functions: (1) Unary Operation: (i) f (x) = x2 (ii) f (x) = A function f : A → A is called a unary operation in A. (2 marks) (3 marks) i.e. f = {(a, f(a))/a, f(a) ∈A)} (iii) f (x) = (iv) f (x) = (2) Binary Operation: A function g : A × A → A is called a binary (2 marks) (2 marks) operation in A i.e. g = {(a, b), g(a, b)/(a, b) ∈A × A and (v) f (x) = (vi) f (x) = g(a, b) ∈A} (3 marks) (4 marks) 22 Master Key Mathematics & Statistics - Part II (Std. XI) Solution: ∴ Domain = {x/x ∈ R, x >1 or x < –1} 2 2 (i) f (x) = x since x can be uniquely calculated = (– ∞, – 1) ∪ (1, ∞) for every x ∈ R, Let y =f (x), i.e., y = Domain = {x/x ∈ R} = R Since x2 ≥ 0 for every x ∈ R, ∴ y2 = ∴ x2 – 1 = Range ={x/x ∈ R, x ≥ 0} = R + ∪ {0}. ∴ x2 = +1 ∴x= (ii) f (x) = f (x) is defined, if (x – 1)(3 – x) ≥ 0 Clearly, x is not defined, if y = 0 i.e., if – x2 + 4x – 3 ≥ 0 Range = {y/y ∈ R, y ≠ 0} i.e., if – x2 + 4x ≥ 3 = R – {0} i.e., if x2 + 4x ≤ – 3 (iv) f (x) is not defined at x =3 i.e., if x2 + 4x + 4 ≤ – 3 + 4 ∴ Domain = {x/x ∈ R, x ≠ 3} i.e., if(x – 2)2 ≤ 1 = R – {3} i.e., if –1 ≤ x– 2 ≤ 1 Let y = i.e., if 1 ≤ x ≤ 3 ∴Domain = {x/x ∈ R, 1≤ x ≤ 3} ∴ xy – 3y = x +3 = [1, 3] ∴ xy – x = 3y +3 ∴ x (y – 1) = 3y +3 Now, f (x) = = ∴ x = = Clearly, x is not defined, if y =1 ∴ Range = {y/y ∈ R, y ≠ 1} = 1≤x≤3 = R – {1} ∴ 1–2≤x–2≤3–2 (v) f (x)= ∴ –1≤x– 2≤1 f (x) is defined, if 9 – x2 ≥ 0 ∴ 0 ≤ (x – 2) 2 ≤ 1 i.e., if 9 ≥ x2, i.e., if x2 ≤ 9 ∴ 0 ≥ –(x – 2) 2 ≥ –1 ∴ 0 + 1 ≥ 1 – (x – 2)2 ≥ 1 – 1 i.e., if – 3 ≤ x ≤ 3 ∴ 1 ≥ 1 – (x – 2)2 ≥ 0 ∴ Domain = {x/x ∈ R, – 3 ≤ x ≤ 3} ∴ 0 ≤ 1 – (x – 2)2 ≤ 1 = [–3, 3] ∴ –1 ≤ ≤1 When – 3 ≤ x ≤ 3, 0 ≤ x2 ≤ 9 ∴ 0 ≥ – x2 ≥ –9 ∴ –1 ≤ f (x) ≤1 ∴ 0 + 9 ≥ 9 – x2 ≥ 9 –9 ∴ –1 ≤ y ≤ 1,where y = f (x) ∴ 0 ≤ 9 – x2 ≤ 9 Range = {y/y ∈ R, –1 ≤ y ≤ 1} = [–1, 1] ∴ 0≤ ≤3 ∴ 0 ≤ f (x) ≤ 3 (iii) f (x) = ∴ 0 ≤ y ≤ 3, where y = f (x) f (x) is defined, if x2 – 1 > 0, ∴ Range = {y/y ∈ R, –0 ≤ y ≤3} = [0, 3]. i.e., if x2 > 1, i.e., if x >1 or x < –1 Sets, Relations and Functions 23 ∴ 0 ≤ 2x2 ≤ 50 (vi) f (x) = ∴ 0 ≥ –2x2 ≥ –50 f (x) is defined, if ≥ 0 and x ≠ 3 ∴ 0 + 9 ≥ 9 – 2x2 ≥ 9 – 50 ∴ 9 ≥ f (x) ≥ – 41 ≥ 0, if x – 2 ≥ 0 and 3 – x > 0 ∴ –41 ≤ y ≤ 9, where y = f (x) or x – 2 ≤ 0 and 3 – x < 0 ∴ Range of f = {y/y ∈R, – 41 ≤ y ≤ 9} If x – 2 ≤ 0 and 3 – x < 0, = [–41, 9]. then x ≤ 2 and 3 ≤ x [Note : Answer in the textbook is incorrect.] (iii) For all x ∈ R, (x – 3) ∈R ∴ > 0, if x – 2 ≥ 0 and 3 – x > 0 ∴ (x – 3)2 ≥ 0 i.e., if x ≥ 2 and 3 > x ∴ (x – 3)2 +2 ≥ 0 + 2 ∴ f (x) ≥ 2 i.e., if 2 ≤ x < 3 ∴ y ≥ 2, where y = f (x) ∴ Domain = {x/x ∈ R, 2 ≤ x < 3} ∴ Range of f = {y/y ∈R, y ≥ 2} = [2, 3) = [2, ∞). Let y = (3) (i) If f (x) = , then find f (–3), f (–1) ∴ y2 = (ii) If f (x) = (x – 1) (2x + 1), then find f(1), f(2), f(–3) ∴ 3y2 – xy2 = x –2 (iii) If f (x) = 2x2 – 3x– 1, then find f (x +2), 2 2 ∴ xy + x = 3y + 2 (2 marks each) ∴ x (y2 + 1) = 3y2 + 2 Solution: (i) f (x) = ∴ x = Clearly, x is defined for all real numbers. ∴ f (– 3) = = ∴ Range = {y/y ∈ R} = R [Note : Answer for domain is incorrect in = the textbook.] ∴ f (– 3) = –2.6 (2) Find the range of each of the following functions: Also, f (–1) = = = (i) f (x)= 3x – 4, for – 1 ≤ x ≤ 3 ∴ f (– 1) = 0 (ii) f (x) = 9 – 2x2 , for –5 ≤ x ≤ 3 (ii) If f (x) = (x – 1) (2x + 1) (iii) f (x)= x2 – 6x + 11, for all x ∈R. (2 marks each) ∴ f (1) = (1 – 1) (2 × 1 + 1) = 0(3) Solution: ∴ f (1) = 0 (i) –1≤ x ≤ 3 f (2) = (2 – 1) (2 × 2 + 1) = 1(5) ∴ –3 ≤ 3x ≤ 9 ∴ f (2) = 5 ∴ –3 – 4 ≤ 3x – 4 ≤ 9 –4 f (–3) = (–3 – 1) [2 (–3) + 1] = (–4) (–5) ∴ –7 ≤ f (x) ≤ 5 ∴ f (–3) = 20. ∴ –7 ≤ y ≤ 5, where y = f (x) (iii) f (x) = 2x2 – 3x –1 ∴ Range of f = {y/y ∈R, –7 ≤ y ≤ 5} ∴ f (x +2) = 2 (x + 2) 2 – 3 (x + 2) – 1 = [–7, 5] = 2(x2 + 4x + 4) – (3x + 6) – 1 (ii) –5 ≤ x ≤ 3 = 2x2 + 8x + 8 – 3x – 6 – 1 ∴ 0 ≤ x2 ≤ 25 ∴ f (x +2) = 2x2 +5x + 1. 24 Master Key Mathematics & Statistics - Part II (Std. XI) (4) Which of the following relations are function? (5) Find a, if f (x)= ax + 5 and f (1) = 8 (2 marks) Justify your answer. If it is a function, Solution: determine its domain and range. Also find the f (x) = ax + 5 function by formula (if possible). ∴ f (1) = a(1) + 5 = a + 5 (i) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)} ∴ f (1) = 8 gives a +5 = 8 ∴a=3 (ii) {(2, 1), (3, 1), (5, 2)} (6) If f (x) = f (3x – 1) for f (x) = x2 – 4x + 11, find x. (iii) {(2, 3), (3, 2), (2, 5), (5, 2)} (2 marks) (iv) {(0, 0), (1, 1), (1, –1), (4, 2), (4, –2), (9, 3), Solution: f (x) = x2 – 4x + 11 (9, –3), (16, 4),(16, –4)} (2 marks each) ∴ f (3x –1) = (3x – 1) 2 – 4(3x – 1) + 11 Solution: = (9x2 – 6x + 1) – (12x – 4) + 11 (i) Let f = {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, = 9x2 – 6x + 1 – 12x + 4 + 11 6), (14, 7)} = 9x2 – 18x + 16 f is a function because every element of the But f (x) = f (3x – 1) domain of f is related with one and only one ∴ x2 –4x + 11 = 9x2 – 18x + 16 element of the range of f. ∴ 8x2 – 14x + 5 = 0 Domain of f = {2, 4, 6, 8, 10, 12, 14} ∴ 8x2 – 10x – 4x + 5 = 0 Range of f = {1, 2, 3, 4, 5, 6, 7} ∴ 2x(4x – 5) –1(4x – 5)= 0 The second component of each ordered pair ∴ (4x – 5)(2x – 1) = 0 in f is half of the first component in the same ∴ 4x – 5 = 0 or 2x –1 = 0 ordered pair. If we denote any element of the ∴ x= or x = domain set by x, then it is related to in the range set. (7) If f (x) = x2 – 3x + 4, then find the value of x Hence, the formula to exhibit the satisfying f(x) = f(2x + 1). (2 marks) Solution: function f is f (x) = , for all x ∈ domain of f. f (x) = x2 – 3x + 4 (ii) Let f = {(2,1), (3,1), (5,2)} ∴ f (2x + 1) = (2x + 1)2 – 3(2x + 1) + 4 f is a function because every element of the = 4x2 + 4x + 1 – 6x – 3 + 4 domain of f is related with one and only one = 4x2 – 2x + 2 element of the range of f. Domain of f = {2, 3, 5} But f (x) = f (2x + 1) Range of f = {1, 2} ∴ x2 – 3x + 4 = 4x2 – 2x + 2 This function cannot be exhibited by formula. ∴ 3x2 + x – 2 = 0 ∴ 3x2 + 3x – 2x – 2 = 0 (iii) Let f = {(2, 3), (3, 2), (2, 5), (5, 2)} ∴ 3x (x + 1 ) – 2(x + 1 ) = 0 f is not a function because the element 2 of ∴ (3x – 2 ) (x + 1 ) = 0 the domain of f is related to two distinct elements 3 and 5 of the range. ∴ x= or – 1 (iv) Let f = {(0, 0), (1, 1), (1, –1), (4, 2), (4, –2), (9, 3) (8) Let A= {1, 2, 3, 4} and Z be the set of integers. (9, –3),(16, – 4)} Define f : A → Z by f (x) = 3x + 7. Show that f is f is not a function because the elements 1, 4, 9 a function from A to Z. Also find the range of f. (2 marks) and 16 of the domain of f are related to two Solution: distinct elements of the range. f (x) = 3x + 7, x ∈A where A = {1, 2, 3, 4} When x = 1, f (x) = f (1) = 3(1) + 7 = 10 Sets, Relations and Functions 25 When x = 2, f (x) = f (2) = 3(2) + 7 = 13 ∴ f is not one - one function. When x = 3, f (x) = f (3) = 3(3) + 7 = 16 Hence, f is not one - one and not onto When x = 4, f (x) = f (4) = 3(4) + 7 = 19 function. ∴ f = {(1, 10), (2, 13), (3, 16), (4, 19)} (10) Find which of the functions are one-one onto, Since each and every element of domain A many-one onto, one -one into, many-one into. is related with one and only one element of B, Justify your answer. f is a function from A to Z. Range of f = {10, 13, 16, 19} (i) f : R → R given as f (x) = 3x + 7 for all x ∈R. (9) Find whether following functions are one- (ii) f : R → R given as f (x) = x2 for all x ∈R. one, onto or not : (iii) f = {(1, 3), (2, 6), (3, 9), (4, 12)} defined from 3 (i) f : R → R given by f (x) = x + 5 for all x ∈R A to B where A = {1, 2, 3, 4} B = {3, 6, 9, 12, 15} (ii) f : Z → Z given by f (x) = x2 + 4 for all x ∈Z (3 marks each) (3 marks each) Solution: Solution: (i) Let x1, x2 ∈R be such that f (x1) = f (x2) (i) To show that the function is one-one, we have to show that if f (x1) = f (x2), ∴ 3x1 +7 = 3x2 + 7...[ f(x) =3 x + 7] then x1 = x2 ∴ 3x1 = 3x2 3 Given: f (x) = x + 5 ∴ x1 = x2 Let f (x1) = f (x2), then x1, x2 ∈ R ∴ f is one-one function. ∴ +5 = +5 Let y ∈R, we want to get x ∈R such that ∴ = ∴ x1 = x2 y = f (x), i.e y = 3x +7 ∴ f is one-one function. ∴ x= ∈R To show that the function is onto, let y be an arbitrary element of the codomain R and ∴ for any y ∈codomain R, there exist let y = f (x) an element x = ∈ domain R y = x3 +5 such that f (x) = y ∴ x= and ∈R ∴ f is one-one function. ∴ for any value of y ∈ codomain R, Hence, f is one-one onto function. there exist an element y = ∈R such that f (x) = y (ii) Let x1 , x2 ∈R be such that f (x1) = f (x2) ∴ f is onto function. ∴ =...[ f (x) = x2] Hence, f is one-one onto function. ∴ x1 = x2 (ii) To show that the function is one-one, we ∴ f is many-one function. have to show that if f (x1) = f (x2), Let y ∈R, we want to get x ∈R such that then x1 = x2 y = f (x), i.e y = x2 Given: f (x) = x2 +4 ∴ x= Let f (x1) = f (x2), x 1, x 2∈Z But ∉R if y < 0 ∴ +4= +4 ∴ for any y ∈ codomain R, there does not exist ∴ = ∴ x1 = x2 x ∈ domain R such that f is not one-one function. f (x) = y Also, 0 ∈Z (codomain) but there does not ∴ f is into function. exist x ∈ Z (domain) such that f (x) = 0 Hence, f is many-one into function. 26 Master Key Mathematics & Statistics - Part II (Std. XI) (iii) f = {(1, 3), (2, 6), (3, 9), (4, 12)} defined from = x2 + 3x +1 – 2 A to B = x2 + 3x –1 where A = {1, 2, 3, 4} B = {3, 6, 9, 12, 15} Since each and every element of A have their (ii) Given: f (x) = and g (x)= distinct images in B, f is one-one function. ∴ (gof) (x) = g[f (x)] Since 15 ∈ B but it does not have any pre-image in the domain A. ∴ f is into function. =g Hence, f is one-one into function. (11) Let f and g be two real valued functions = defined by f (x) = x + 1 and g(x) = 2x – 3. ∴ (fog) (x) = f [g (x)] Find (i) f + g (ii) f – g (iii) (3 marks) Solution: =f = Given: f (x) = x + 1 and g (x) = 2x –3 (i) (f + g) (x) = f (x) + g(x) = = (x + 1) + (2x – 3) = x +1+ 2x –3 = 3x –2 (ii) (f – g) (x) = f (x) – g(x) (13) If f (x) = , prove that fof is an identity = (x + 1) – (2x – 3) function. (2 marks) Solution: = x +1– 2x +3 = 4 – x f (x) = (iii) (x) = ∴ (fof) (x) = f [f (x)] = , x≠ (12) Find gof and fog, where =f = (i) f (x) = x – 2, g (x)= x + 3x +1 2 (ii) f (x) = , g (x) = (4 marks) = Solution: (i) Given: f (x) = x – 2 and g (x) = x2 +3x +1 = ∴ (gof) (x) = g [ f (x)] = g(x – 2) = =x = (x – 2)2 + 3(x – 2) +1 = x2 – 4x + 4 + 3x – 6 + 3 ∴ (fof) (x) = x = x2 – x +1 Hence, fof is an identity function. and (fog) (x) = f [g (x)] = f (x2 + 3x + 1) Sets, Relations and Functions 27 Solution: (14) If f (x) = and g(x) = , prove that Let A = {2, 3, 4, 5, 6}, B = {4, 6, 8, 10,12} (gof) (x) = (fog) (x) = x (3 marks) and C = {13, 19, 25, 31, 37} Solution: Now, f = {(2, 4), (3, 6),(4, 8), (5, 10), (6, 12)} Given : f (x) = and g(x)= Here, f (2) = 4, f(3) = 6, f(4) = 8, f(5) = 10, f(6) = 12 ∴ (gof) (x) = g [f(x)] Also, g = {(4,13), (6,19), (8,25), (10,31), (12,37)} Here, g(4) = 13, g (6) = 19, g (8) = 25, =g g(10) = 31, g(12) = 37. ∴ (gof)(2) = g[f(2)] = g(4) = 13 (gof)(3) = g[f(3)] = g(6) = 19 = (gof)(4) = g[f(4)] = g(8) = 25 (gof)(5) = g[f(5)] = g(10) = 31 (gof)(6) = g[f(6)] = g(12) = 37 ∴ (gof) = {(2, 13), (3, 19), (4, 25), (5, 31), (6, 37)} = (16) Show that f : R→ → R given by f (x) = 3x – 4 is one-one. Find its inverse function. Also find = = =x f–1(9) and f –1(–2). (3 marks) ∴ (fog) (x)= x... (i) Solution: and (fog) (x) = f [g (x)] Let x1, x2 ∈R be such that f (x1) = f (x2) ∴ 3x1 – 4 = 3x2 – 4...[ f(x) = 3x – 4] =f ∴ 3x1 = 3x2 ∴ x1 = x2 ∴ f is one-one function. Let y ∈R, we want to get x ∈R such that = y = f (x), i.e y = 3x – 4 ∴ x= ∈R ∴ for any y ∈ codomain R, there exist an = element x = ∈ R such that f(x) = y = = =x ∴ f is onto function. ∴ (fog) (x) = x... (ii) Hence, f is one-one onto function. From (i) and (ii), we get, Let y = f (x) = 3x – 4 (gof) (x) = (fog) (x)= x. ∴ x= (15) If f = {(2, 4), (3, 6), (4, 8), (5, 10), (6, 12)} g = {(4, 13), (6, 19), (8, 25), (10, 31), (12, 37)}, ∴ f–1 (y) = find (gof). (3 marks) 28 Master Key Mathematics & Statistics - Part II (Std. XI) (10) Show that f : R → R defined by f (x) = 5x + ∴ f–1 (x)= 7 is a function. Find domain the range of f. (2 marks) ∴ f–1 (9) = = (11) In each of the following case a relation f is given in terms of ordered pairs from A × B. and f–1 (– 2) = = State whether f is a function, if function, state whether it is 1 - 1, many-one, into or Problems for Practice - 1.3 onto. Describe function as formula. (i) A = {0, 1, 2} B = {0, 2, 3, 4} (1) If f(x) = find f = {(0, 0), (0, 3), (1, 4), (2, 2)} (ii) A = B = {0, 1, 2, 3} (i) f (2) (ii) f (3) (iii) f (3 marks) f = {(0, 0), (1, 1), (2, 2), (3, 3)} (2) A function f is defined as : (iii) A = {1, 3, 5, 7, 9}, B = {x/x is odd, 0 ≤ x ≤ 9} f (x) = –3 ≤ x < 3 f = {(1, 1), (3, 3), (5, 5), (7, 7), (9, 9)} =0 x=0 State the range of function (2 marks) (iv) A = {0, 1, 2, 3, 4, 5}, B = {0, 2, 4, 6, 8, 10} (3) If f (x) = (x – 1)(x – 3)(x – 5), –1 ≤ x < 5 f = {(0, 0), (1, 2), (2, 4), (3, 6), (4, 8), (5, 10)} find f(–2), f(4), f(5), if they exist. Also find x, (2 marks each) if f(x) = 0 (3 marks) (12) Functions f : R → R and g : R → R are given 2 (4) If f (x) = x + 5x + 6. Find the value of x for by f (x) = 3x + 5 and g(x) = 2x2 – 4x + 3. Find when f(x) = f(x + 2) (2 marks) (i) fof (ii) fog (iii) gof (3 marks) 2 (5) If f (x) = x – 2x + 4, solve the equation (13) Find gof and fog where f and g are given by f (x) = f (x – 2) (2 marks) (i) f (x)= x2+ 5, g(x) = x – 1 (6) Find the domain and range of each of the functions (ii) f (x) = , g(x) = (i) f(x) = 5x – 2 , – 5 ≤ x ≤ 2 (iii) f (x)= x2, g(x) = cos x (2 marks each) 2 (ii) f(x) = 3x – 2 , – 1 ≤ x ≤ 4 (2 marks) (14) If f (x) = 5x + k and g(x) = 4x + k and if (7) If f (x) = ax + 5, f (1) = 10, find a and f (5): fof = gof. find k. (2 marks) (2 marks) (15) If f(x) = ax2 + 3x + c, f(0) = – 2, f(–1) = – 5 (8) (i) If f (x) = 1 – x+x2 and f (x – 1) = f (x + 2), find a and c. (3 marks) find x (16) If f (x) = 3x + 1 and g(x) = show that fog(x) – gof (x) = (3 marks) (ii) If f (x) = x2 +4x + 3, f (x) = f (x + 1), find x (2 marks) (17) If f (x) = , show that (9) Show that f : R → R given by f (x) = 7x + 2 is one-one, into function. (2 marks) fof(x) = (3 marks) Sets, Relations and Functions 29 (18) Determine whether following functions are (19) k = 0 (20) yes odd or even : (21) yes, f–1(x) = (x – 5) (i) f (x) = cos (x3+ 1) (ii) sin2x (iii) f (x) = x2 + 3x (2 marks) (19) If f (x) = x3 – kx2+ 2x + 8, n ∈ R is an odd Points to Remember : function, find k. (3 marks) * A set is a collection of well defined objects. (20) A function f : R → R is defined by f (x) = * A set which does not contain any element 3x – 8, x ∈ R, Does the inverse of f exist? is called an empty set denoted by φ. Justify your answer. (2 marks) * A power set of a set A is collection of all (21) A function f : R → R is defined as f (x) = subsets of A denoted by P (A). 2x + 5. Does the inverse of f exist? Justify * A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) your answer, define f–1also (2 marks) * A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) * For any sets A and B ANSWERS (A∪ B)’ = A’ ∩ B’ [De’ Morgan’s laws] (A ∩ B)’ = A’∪ B’ (1) (i) –5 (ii) not defined (iii) * If A and B are finite sets such that A ∩ B = φ. then (2) (0, 3) n (A ∪ B) = n (A) + n (B) (3) f (– 2) = – 105, f(4) = 3, f(6) does not exist If n(A ∩ B) ≠ 0 then x = 1, 3 when f(x) = 0 n (A∪ B) = n(A) + n (B) – n(A ∩ B) * A × B = {(a, b) / a ∈A, b ∈ B} (4) – (5) x=2 In particular R × R = {(x, y) / x, y ∈ R} (6) (i) Domain [–5, 2], Range = [–27, 3] * If (a, b) = (x, y) then a = x and b = y * If n (A) = p and n(B) = q then (ii) Domain (–1, 4), Range = (1, 46) n (A × B) = pq. (7) a = 5, f (x) = 35 * A×φ=φ. (8) (i) x = 0 (ii) x = – * The image of an element x under a relation R is given by y, if (x, y) ∈ R. (11) (i) not a function * The domain of R is the set of all 1 st (ii) f is one-one function, f(x) = x. components of the ordered pairs in R (relation). (iii) f is one-one into function, f(x) = x. * The range of R (relation) is the set of all second (iv) f is one-one into function, f(x) = 2x. components of the ordered pairs in R. (12) (i) 9x + 20 (ii) 6x2 – 12x +14 * f (x) = f (x) if f is even and x ∈ R. (iii) 18x2 + 48x +33 * f (– x) = – f (x) if f is odd and x ∈ R. (14) k = (15) a = 0, c = – 2 (18) (i) even (ii) even (iii) odd 30 Master Key Mathematics & Statistics - Part II (Std. XI) (vi) (B ∪ C) = {2, 4, 6, 7, 11} ∪ {3, 5, 8, 9, 12} Miscellaneous Exercise - 1 (Textbook Page No. 30) = {2, 3, 4, 5, 6, 7, 8, 9, 11, 12} ∴ A ∩ (B ∪ C) = {1, 4, 7, 10} ∩ (1) Write down the following sets in set- {2, 3, 4, 5, 6, 7, 8, 9, 11, 12} builder form: = {4, 7} (i) {10, 20, 30, 40, 50} (ii) {a, e, i, o, u} (iii) {Sunday, Monday, Tuesday, (3) In a survey of 425 students in a school it was Wednesday, Thursday, found that 115 drink apple juice, 160 drink Friday, Saturday} (2 marks each) orange juice and 80 drink both apple as well as orange juice. How many drink neither Solution: apple juice nor orange juice? (4 marks) (i) Let A = {10, 20, 30, 40, 50} Solution: A is set of natural numbers which are 10 Let X ≡ set of all students times n where n ∈N and n ≤ 5 A ≡ set of students who drink apple juice ∴ A={x / x = 10n, n ∈N and n ≤ 5} B ≡ set of students who drink orange juice (ii) Let A = {a, e, i, o, u} We are given the following : B is the set of all vowels of English alphabet ∴ B = {x/ x is the vowels of English alphabet} n(X) = 425, n(A) = 115, n(B) = 160 and n (A ∩ B) = 80 (iii) Let C = {Sunday, Monday, Tuesday, The venn diagram is shown below: Wednesday, Thursday, Friday, Saturday} C is the set of all days of the week. ∴ C = {x / x represents the day of the week} (2) If U= {x /x ∈N, 1 ≤ x ≤ 12} A = {1, 4, 7, 10}, B = {2, 4, 6, 7, 11}, C = {3, 5, 8, 9, 12}, write down the following The number of students who drink neither sets: apple juice nor orange juice (i) A ∪ B (ii) B ∩ C (iii) A – B (iv) B ∩ C’ = n(A ∪ B)’ (v) A ∪ B ∪ C (vi) A ∩ (B ∪ C) = n(X) – n(A ∪ B) (1 mark each) = 425 – (35 + 80 + 80) Solution: U = {x / x ∈N, 1 ≤ x ≤ 12} = 425 – 195 = 230 ∴ U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} (4) In a school there are 20 teachers who teach Mathematics or Physics, of these, 12 teach (i) A ∪ B = {1, 4, 7, 10} ∪ {2, 4, 6, 7, 11} Mathematics and 4 teach both Physics and = {1, 2, 4, 6, 7, 10, 11} Mathematics. How many teach Physics? (ii) B ∩ C = {2, 4, 6, 7, 11} ∩ {3, 5, 8, 9, 12} (2 marks) =φ Solution: (iii) A – B = {1, 4, 7,10} – {2, 4, 6, 7, 11} Let M ≡ set of teachers who teach = {1,10} Mathematics P ≡ set of teachers who teach Physics. (iv) C’= {1, 2, 4, 6, 7, 10, 11} We are given following : ∴ B ∩ C’ = {2, 4, 6, 7, 11} ∩ {1, 2, 4, 6, 7, 10, 11} n(M ∪ P) = 20, n(M) = 12, n(M ∩ P) = 4 = {2, 4, 6, 7, 11} We know that (v) A ∪ B ∪ C = {1, 4, 7, 10} ∪ {2, 4, 6, 7, 11} n (M ∪ P) = n (M) + n(P) – n (M ∩ P) ∪ {3, 5, 8, 9, 12} ∴ n(P) = n(M ∪ P) + n(M ∩ P) – n(M) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} Sets, Relations and Functions 31 = 20 + 4 – 12 =12 Since all the elements of R3 are in A × B, Hence, the number of teachers who teach R3 ⊆ A × B physics = 12. ∴ R 3 is the relation from A to B (5) (i) If A = {1, 2, 3} and B = {2, 4} write down (iv) R4 = {(4, 2), (2, 6), (5, 1), (2, 4)} the following : A × A, A × B, B × A, B × B, Since (4, 2) ∉ A × B and (5, 1) ∉ A × B (A × B) ∩ (B × A) R4 ∉ A × B (ii) If A = {–1, 1}, find A × A × A ∴ R 4 is not the relation from A to B. (2 marks each) (7) Determine the domain and range of the Solution: following relations : (i) A = {1, 2, 3}, B = {2, 4} (i) R = {(a, b) /a ∈N, a < 5, b = 4} ∴ A × A = {(1, 1), (1, 2),(1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2),(3, 3) (ii) S = {(a, b) /b =|a – 1|, a ∈ Z, |a|≤ ≤ 3} (2 marks each) A × B = {(1, 2), (1, 4),(2, 2), (2, 4), (3, 2), (3, 4)} Solution: B × A = {(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)} (i) R= {(a, b)/a ∈ N, a < 5, b = 4} B × B = {(2, 2), (2, 4), (4, 2), (4, 4)} Domain of R = {a/a ∈ N, a < 5} (A × B) ∩ (B × A) = {(1, 2), (1, 4), (2, 2), (2, 4), = {1, 2, 3, 4} (3, 2), (3, 4)} ∩{(2, 1), (2, 2), (2, 3),(4, 1),(4, 2), (4, 3)} Range of R = {b/b = 4} = {4} = {(2, 2)} (ii) S = {(a, b)/b =|a – 1|, a ∈ Z, |a|≤ 3} (ii) A = {–1, 1} Domain of S = {a/a ∈Z,|a|≤ 3} ∴ A × A × A = {(–1, –1, –1), (–1, –1, 1), (–1, 1,–1), = {–3, –2, –1, 0, 1, 2, 3} (–1, 1, 1), (1, 1, 1), (1, 1, –1), Range of S = {b/b =|a - 1|, a ∈ Z, |a|≤ 3} (1, –1, 1),(1, –1, –1)} = {|–3 –1|,|–2, –1|,|–1 –1|, (6) If A = {1, 2, 3} and B = {4, 5, 6} which of the |0 – 1|, |1 – 1|, |2 – 1|, following are relations from A to B? |3 – 1|} (i) R 1 = {(1, 4), (1, 5),(1, 6)} = {4, 3, 2, 1, 0, 1, 2} (ii) R 2 = {(1, 5), (2, 4),(3, 6)} = {0, 1, 2, 3, 4} (iii)R 3 = {(1, 4), (1, 5),(3, 6),(2, 6),(3, 4)} (8) Which of the following relations are functions? If it is a function, determine its (iv) R4 = {(4, 2), (2, 6), (5, 1),(2, 4)} (2 marks each) domain and range: Solution: (i) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), A = {1, 2, 3}, B = {4, 5, 6} (14, 7)} ∴ A × B = {(1, 4), (1, 5), (1, 6), (2, 4),(2, 5), (ii) {(0, 0), (1, 1), (1, –1), (4, 2), (4, –2),(9, 3), (2, 6), (3, 4), (3, 5), (3, 6)} (9, –3), (16, 4), (16, –4) (i) R1 = {(1, 4), (1, 5), (1, 6)} Since all the elements (iii) {(2, 1), (3, 1), (5, 2)} (2 marks each) of R1 are in A × B, Solution: R1 ⊆ A × B (i) Refer to the solution of Q.4 of Exercise 1.3 ∴ R1 is the relation from A to B Function, Domain = {2, 4, 6, 8, 10, 12, 14}, (ii) R2 = {(1, 5), (2, 4), (3, 6)} Since all the elements Range ={1, 2, 3, 4, 5, 6, 7} of R2 are in A × B, (ii) Not a function R2 ⊆ A × B (iii) Function, Domain = {2, 3, 5}, ∴ R2 is the relation from A to B Range = {1, 2} (iii) R3 = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)} 32 Master Key Mathematics & Statistics - Part II (Std. XI) (9) Find whether following functions are one- But ∉R, if y < 3 one or not : ∴ for any y ∈ codomain R, there does not exist → R defined by f (x) = x2 + 5 (i) f : R→ x ∈domain R such that f (x) = y (ii) f : R – {3} → R defined by f (x) ∴ f is not onto function. = for x ∈R – {3} (2 marks each) (11) Let f : R→ → R be a function defined by f (x) = 5x3 – 8 for all x ∈R, show that f is one- Solution: one and onto. Hence find f –1. (4 marks) (i) Let x1, x2 ∈R be such that Solution: f (x1) = f (x2) Let x1 , x2 ∈ R be such that f (x1) = f (x2) ∴ +5= +5... [ f (x)= x2 + 5] ∴5 –8= 5 –8... [ f (x) = 5x3 – 8] ∴ = ∴ 5 = 5 ∴ x1 = x2 ∴ = ∴ f is not one-one function. ∴ x1 = x2 (ii) Let x1, x2 ∈ R – {3} be such that f is one-one function. f (x1) = f (x2) Let y ∈ R, we want to get x ∈ R such that y = f (x), i.e, y = 5x3 – 8 ∴ =... ∴ x= ∈R ∴ (5x1 + 7) (x2 – 3) = (5x2 + 7) (x1 – 3) ∴5x1x2 – 15x1 + 7x2 – 21 = 5x1x2 – 15x2 + 7x1 – 21 ∴ for any y ∈ codomain R, there exist an ∴ –15x1 + 7x2 = –15x2 +7x1 element x = ∈R such that ∴ – 22x1 = –22x2 ∴ x 1 = x2 f (x) = y ∴ f is one-one function. ∴ f is onto function. Hence, f is one-one, onto function. (10) Find whether the following functions are onto or not : Let y = f (x) = 5x3 – 8 (i) f : Z → Z defined by f (x) = 6x – 7 for all x ∈Z ∴ x= (ii)f : R → R defined by f (x) = x2 + 3 for all x ∈R (2 marks each) Solution: ∴ f -1(y) = (i) Let y ∈Z, we want to get x ∈Z such that y = f (x), i.e, y = 6x – 7 ∴ f -1(x) = x= (12) A function f : R→ → R defined by f (x) = +2 But ∉Z for all y ∈Z x ∈R. Show that f is one-one and onto. Hence ∴ for any y ∈ codomain Z, there does not exist find f –1. (4 marks) an element x ∈ domain Z, such that f(x)= y. Solution: ∴ f is not onto function. Let x1 , x2 ∈ R be such that f (x1) = f (x2) (ii) Let y ∈R, we want to get x ∈R such that ∴ +2= +2... y = f (x), i.e, y = x2 +3 ∴ x= ∴ = Sets, Relations and Functions 33 ∴ x 1 = x2 = 3(x4 – 4x3 + 6x2 – 4x +1) – 5(x2 – 2x +1) + 7 ∴ f is one-one function. = 3x4 – 12x3 + 18x2 – 12x +3 – 5x2 + 10x – 5 + 7 Let y ∈ R, we want to get x ∈R such that = 3x4 – 12x3 + 13x2 – 2x +5 y = f (x). (16) If f (x) = 3x + a and f (1) = 7 , find a and f (4). i.e y = +2 (2 marks) Solution: ∴ =y–2 f (x) = 3x + a