R<\/sub><\/p>\nQuestion 2. \nLet f : W \u2192 W be defined as f(n) = n – 1, if n is odd and f(n) = n + 1, if n is even. Show that\/is invertible. Find the inverse off. Here, W is the set of all whole numbers. \nSolution: \nOne-one \nLet m and n be two distinct elements of W. \nCase i : \nWhen m and n are even \n\u2234 f(m) = m + 1 and f(n) = n + 1 \nSince m \u2260 n, then m + 1 \u2260 n + 1 \n\u21d2 f(m) \u2260 f(n)<\/p>\n
Case ii : \nWhen m and n are odd \nf(m) = m – 1 and f(n) = n – 1 \nSince m \u2260 n, m – 1 \u2260 n – 1 \n\u21d2 f(m) \u2260 f(n)<\/p>\n
Case iii : \nWhen m is odd and n is even \nf(m) = m – 1 and f(n) = n + 1 \nSince m \u2260 n, m – 1 \u2260 n + 1 \n\u21d2 f(m) \u2260 f(n)<\/p>\n
Case iv : \nWhen m is even and n is odd \nf(m) = m – 1 and f(n) = n – 1 \nSince m \u2260 n, m – 1 \u2260 n – 1 \n\u21d2 f(m) \u2260 f(n) \ni.e., in any case if m \u2260 n, then f(m) \u2260 f(n) \nHence f is one-one.<\/p>\n
Onto \nLet y \u2208 W, the co-domain of f \nIf y is odd, then y – 1 is even \nf(y – 1) = y – 1 + 1 = y \nIf y is even, then y + 1 is odd \nf(y + 1) = y + 1 – 1 = y \nHence every element in the co-domain has a preimage in the domain. \nHence f is onto. \ni.e., f is one-one and onto, hence f is invertible. \nWe have proved that f(n – 1) = n, if n is odd and f(n + 1) = n, if n is even \n\u2234 f-1<\/sup> = \\(\\left\\{\\begin{array}{l} \nn-1, \\text { if } n \\text { is odd } \\\\ \nn+1, \\text { if } n \\text { is even } \n\\end{array}\\right.\\) \ni.e., f-1<\/sup>(n) = f(n) \nHence inverse of f is f itself.<\/p>\n <\/p>\n
Question 3. \nIf f: R \u2192 R is defined by f(x) = x\u00b2 – 3x + 2, find f(f (x)). \nSolution: \nf(f(x)) = f(x\u00b2 – 3x + 2) \n= (x\u00b2 – 3x + 2)\u00b2 – 3(x\u00b2 – 3x + 2) + 2 \n= x4<\/sup> + 9x\u00b2 + 4 – 6x\u00b3 – 12x + 4x\u00b2 – 3x\u00b2 + 9x – 6 + 2 \n= x4<\/sup> – 6x\u00b3 + 10x\u00b2 – 3x<\/p>\nQuestion 4. \nShow that the function \nf : R \u2192 {x \u2208 R : – 1 < x < 1} defined by f(x) = \\(\\frac{x}{1+|x|}\\), x \u2208 R is one one and onto function. \nSolution: \nOne-one \nThe domain of f is R \nLet x1<\/sub> x2<\/sub> \u2208 R \nCase i : \nx1<\/sub> \u2265 0, x2<\/sub> \u2265 0 \nThen 1 + x1<\/sub> \u2260 1 + x2<\/sub> \n \nCase ii : \n \nsince |x1<\/sub>| = – x1<\/sub> and |x2<\/sub>| = – x2<\/sub> as x1<\/sub> \u2264 0 and x2<\/sub> \u2264 0 \n\u21d2 f(x1<\/sub>) \u2260 f(x2<\/sub>)<\/p>\nCase iii : \n \nThus in any case, f(x1<\/sub>) \u2260 f(x2<\/sub>) when x1<\/sub> \u2260 x2<\/sub> \nHence f is one-one<\/p>\nOnto : \nLet y \u2208 R \n \ni.e, when y \u2265 0, there is \\(\\frac{y}{1-y}\\) \u2208 R such that \\(f\\left(\\frac{y}{1-y}\\right)\\)<\/p>\n
case iii. \n– 1 < y \u2264 0 \n \ni.e, when y < 0, there is \\(\\frac{y}{1+y}\\) \u2208 R such that \\(f\\left(\\frac{y}{1-y}\\right)\\) = y<\/p>\n
<\/p>\n
Question 5. \nShow that the function f : R \u2192 R given by f(x) = x\u00b3 is injective. \nSolution: \nLet f(x1<\/sub>) = f(x2<\/sub>) for x1<\/sub>, x2<\/sub> \u2208 R \n\u21d2 x\u00b31<\/sub> = x\u00b32<\/sub> \u21d2 x\u00b31<\/sub> – x\u00b32<\/sub> = 0 \n\u21d2 (x1<\/sub> – x2<\/sub>) (x\u00b21<\/sub> + x1<\/sub>x2<\/sub> + x\u00b22<\/sub>) = 0 \n\u21d2 x1<\/sub> = x2<\/sub> \n\u2234 f is one-one (injective)<\/p>\nQuestion 6. \nGive examples of two functions f : N \u2192 Z and g : Z \u2192 Z such that gof is injective but g is not injective. \nSolution: \nLet f : N \u2192 Z defined by f(x) = x and \ng : Z \u2192 Z defined by g(x) = |x| \nWe have 1 \u2260 – 1, but |1| = |- 1| \n\u2234 g(1) = g(- 1) \nHence g is not injective \ngof: N \u2192 Z \nLet x1<\/sub>, x2<\/sub> \u2208 N \nsuch that (gof)(x1<\/sub>) = (gof)(x2<\/sub>) \n\u21d2 g(f(x1<\/sub>)) = g(f(x2<\/sub>)) \n\u21d2 g(x1<\/sub>) = g(x2<\/sub>) \n\u21d2 |x1<\/sub>| = |x2<\/sub>| \n\u21d2 x1<\/sub> = x2<\/sub> \nsince x1<\/sub>, x2<\/sub> \u2208 N \n\u2234 gof is injective<\/p>\nQuestion 7. \nGive examples of two functions f : N \u2192 N and g : N \u2192 N such that gof is onto but f is not onto. \nSolution: \nLet f, g : N \u2192 N defined by f(x) = x + 1 \ng(x) = \\(\\left\\{\\begin{array}{l} \nx-1 \\text { if } x>1 \\\\ \n1 \\quad \\text { if } x=1 \n\\end{array}\\right.\\) \nLet x \u2208 N, x \u2265 1 \u21d2 x + 1 \u2265 2 \n\u21d2 f(x) \u2265 2 for all x \u2208 N \nRange of f \u2260 N \n\u2234f is not onto \ngof : N \u2192 N \n(gof)(x) = g(f(x)) = g(x + 1) \n= (x + 1) – 1 since x + 1 > 1 = x \ni.e., (gof)x = x for all x \u2208 N \ni.e., gof is an identity function and hence onto.<\/p>\n
<\/p>\n
Question 8. \nGiven a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A \u2282 B. Is R an equivalence relation on P(X)? Justify your answer. \nSolution: \nARB if A \u2282 B \ni. ARA since A \u2282 A, for all A \u2208 P(X) Hence R is reflexive. \nii. ARB need not imply BRA since A \u2282 B \n\u21d2 B \u2284 A . Hence R is not symmetric. \nHence R is not an equivalence relation.<\/p>\n
Question 9. \nGiven a non-empty set X, consider the binary operation * : P(X) x P(X) \u2192 P(X) given by A * B = A \u2229B for all A, B in P(X), where P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation *. \nSolution: \nX \u2208 P(X) \nFor any A e P(X), A * X = A \u2229 X = A and \nX * A= X \u2229 A = A \n\u2234 X is the identity element in P(X). \nLet S be an invertible element in P(X). \nThen S * X = X * S = X \nS \u2229 X = X \u2229 S = X \nS = X \n\u2234 X is the only invertible element in P(X).<\/p>\n
Question 10. \nFind the number of all onto functions from the set {1, 2, 3,…, n} to itself. \nSolution: \nSince the functions are onto, every element of {1, 2,3 , n} has a pre-image. The element I can have pre-images in n ways. The element 2 can have pre-images in (n – 1) ways and so on. The last element n can have pre-iniage in only one way. Therefore number of ways we can have pre images is n(n – 1)(n – 2) …… 2.1 n! The total number of all onto functions from the set {1, 2, 3, …….. n} to itself is n!.<\/p>\n
Question 11. \nLet S = {a, b, c} and T = {1, 2, 3}. Find F-1<\/sup> of the following functions F from S to T, if it exists. \ni. F = {(a, 3), (b, 2), (c, 1)} \nii. F = {(a, 2), (b, 1), (c, 1)} \nSolution: \ni. \n \nF-1<\/sup> : T \u2192 S is defined as \nF-1<\/sup> = {(3, a), (2, b),(1, c)}<\/p>\nii. \n \nAs 1 has 2 preimages, F is not bijective. F-1<\/sup> is not defined.<\/p>\n <\/p>\n
Question 12. \nConsider the binary operations *: R x R \u2192 R and \u00b0 : R x R \u2192 R defined as a * b = |a – b| and a \u00b0 b = a, for all a, b \u2208 R. Show that * is commutative but not associative, \u00b0 is associative but not commutative. Further, I show that for all a, b, c \u2208 R, a * (b \u00b0 c) = (a * b) o (a * b). [If it is so, we say that the j operation * distributes over the operation \u00b0]. Does o distribute over *? Justify your answer. \nSolution: \nLet a, b \u2208 R, then \na * b = |a – b| = |b – a| = b * a \ni.e., a * b = b * a for all a, b \u2208 R \nHence * is commutative. \nLet 1, 2, 3 \u2208 R \n(1 * 2) * 3 = |1 – 2| * 3 \n= 1 * 3 \n= |1 – 3| = 2 \n1 * (2 * 3) = 1 * |2 – 3| = 1 * 1 \n= |1 – 1| = 0 \ni.e., (1 * 2) * 3 \u2260 1 * (2 * 3) \n\u2234 * is not associative \nThus * is commutative but not associative. \nLet a, b \u2208 R \na \u00b0 b – a and boa = b \n\u2234 a \u00b0 b \u2260 b \u00b0 a \nHence \u00b0 is not commutative. \nNow (a \u00b0 b) \u00b0 c = a \u00b0 c = a \na \u00b0 (b \u00b0 c) = a \u00b0 b = a \ni.e., a \u00b0 (b \u00b0 c) = (a \u00b0 b) \u00b0 c \ni.e., o is associative \nHence \u00b0 is associative but not commutative. \nLet a, b, c \u2208 R \na * (b\u00b0c) = a * b =|a – b| \n(a * b) o (a * c) = |a – b| \u00b0 |a – c| = |a – b| \ni.e., a * (b\u00b0c) = (a * b) \u00b0 (a * c) \ni.e., * is distributive over \u00b0 \nNow a\u00b0 (b * c) = a \u00b0 |b-c| = a \n(a\u00b0b) * (a\u00b0c) = a * a = |a – a| = 0 \ni.e., in general a\u00b0 (b * c) * (a \u00b0 b) * (a \u00b0 c) \ni.e., \u00b0 is not distributive over *.<\/p>\n
Question 13. \nGiven a non-empty set X, \nlet *: P(X) x P(X) \u2192 P(X) be defined as A * B = (A – B) \u222a (B – A), for all A, B \u2208 P(X). Show that the empty set (\u03a6) is the identity for the operation * and all the elements A of P(X) are invertible with A-1<\/sup> = A. \nSolution: \n \nHence (\u03a6) is the identity element in P(X) \nNow A * A = (A – A) \u222a (A – A) \n= \u03a6\u222a\u03a6 = \u03a6 \n\u2234 A is the inverse of itself, i.e., A-1<\/sup> = A. Hence all the elements of P(X) are invertible with A-1<\/sup> = A.<\/p>\n <\/p>\n
Question 14. \nDefine a binary operation * on the set {0,1,2,3,4, 5} as a * b = \\(\\left\\{\\begin{array}{l} \na+b, \\quad \\text { if } a+b<6 \\\\ \na+b-6, \\text { if } a+b \\geq 6 \n\\end{array}\\right.\\) Show that zero is the identity for this operation and each element a of the set is invertible with 6 – a being the inverse of a. \nSolution: \nThe operation table for * is given below \n \nFrom the table we observe that a * 0 = 0 * a = a for all a \u2208 {0,1,2,3.4,5}. Hence 0 is the identity element for the operation *. \nLet a \u2260 0 be any element of {0, 1, 2, 3, 4, 5}. \nThen 6 – a \u2208 {0, 1, 2, 3, 4, 5} \nNow a * (6 – a) = a + 6 – a – 6 = 0 \nand (6 – a) * a = 6 – a + a – 6 = 0 \ni.e., a * (6 – a) = (6 – a) * a = 0, identity element of * \n\u2234 6 – a is the inverse of a \u2260 0. \nHence a \u2260 0 is invertible. \nAlso the inverse of 0 is 0.<\/p>\n
Question 15. \nLet A = {- 1, 0, 1, 2}, B = {- 4, – 2, 0, 2} and f, g : A \u2192 B be functions defined by f(x) = x\u00b2 – x, x \u2208 A and g(x) = 2|x – \\(\\frac { 1 }{ 2 }\\)| – 1, x \u2208 A. Are f and g equal? Justify your answer. \nSolution: \n \nHence fa) = g(a) for all a \u2208 A. \n\u2234 f and g are equal functions.<\/p>\n
<\/p>\n
Question 16. \nLet A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is \na. 1 \nb. 2 \nc. 3 \nd. 4 \nSolution: \na. 1 \nLet R be the relation containing (1, 2) and 0,3) \n\u2234 (1, 2), (1, 3) \u2208 R \nSince R is reflexive, R contains (1, 1), (2, 2), (3, 3) \nSince R is symmetric, R contains (2, 1) and (3, 1) \n\u2234 R = {(1, 1), (1, 2), (1, 3), (2, 2), (2, 1), (3, 1), (3, 3)} \nSince R is not transitive (2, 1) and (1, 3) \u2208 R but (2, 3) \u2209 R. \nIf (3, 2) \u2208 R, then (2, 3) \u2208 R since R is symmetric. \n\u2234(3, 2) \u2209 R \n\u2234 There is only one relation which is reflexive, symmetric and not transitive.<\/p>\n
Question 17. \nLet A = {1, 2, 3}. Then number of equivalence relations containing (1,2) is \na. 1 \nb. 2 \nc. 3 \nd. 4 \nSolution: \nb. 2 \nGiven (1, 2) i.e., 1 is related to 2. \nThen there are two possibilities. \nCase i : \n1 is related to 3 \nA x A= {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3,1), (3, 2), (3, 3)} \nThen A x A is reflexive, symmetric and transitive. \n\u2234 A x A is an equivalence relation Containing (1, 2)<\/p>\n
Case ii : \n1 is not related to 3 \nLet R= {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3)} \nThen R is reflexive, symmetric and transitive, i.e., R is an equivalence relation containing (1 2). \nThus there are two equivalence relations containing (1, 2).<\/p>\n
Question 18. \nLet f : R \u2192 R be the Signum Function defined as \nf(x) = \\(\\left\\{\\begin{array}{c} \n1, x>0 \\\\ \n0, x=0 \\\\ \n-1, x<0 \n\\end{array}\\right.\\) \nand g : R \u2192 R be the Greatest Integer Function given by g (x) = [x], where [x] is greatest integer less than or equal fox. Then, does fog and gof coincide in (0, 1]? \nSolution: \n(fog)(x) = f(g(x)) = f([x]) x \u2208 (0, 1] \n= f(0) = 0 \n(gof)(x) = g(Ax)) = g(1) x \u2208 (0,1] \n= 1 \n\u2234 fog and gof do not coincide in (0, 1]<\/p>\n
<\/p>\n
Question 19. \nNumber of binary operations on the set {a, b} are \na. 10 \nb. 16 \nc. 20 \nd. 8 \nSolution: \nb. 16 \n{a, b) x {a, b) contains 4 elements. \nNumber of relations from {a, b} x {a, b) to {a, b) is 24<\/sup> = 16<\/p>\n","protected":false},"excerpt":{"rendered":"These NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-1-miscellaneous-exercise\/ NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Miscellaneous Exercise Question 1. Let f : R \u2192 R be defined as f(x) = 10x + 7. …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Miscellaneous Exercise<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Miscellaneous Exercise - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n