{"id":29227,"date":"2022-03-28T17:00:28","date_gmt":"2022-03-28T11:30:28","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=29227"},"modified":"2022-03-28T17:34:11","modified_gmt":"2022-03-28T12:04:11","slug":"ncert-solutions-for-class-12-maths-chapter-1-miscellaneous-exercise","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-1-miscellaneous-exercise\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Miscellaneous Exercise"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> 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\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Miscellaneous Exercise<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nLet f : R \u2192 R be defined as f(x) = 10x + 7. Find the function g : R \u2192 R such that g o f = fog = 1R<\/sub>.
\nSolution:
\nLet y \u2208R such that y = 10x + 7
\n\u21d2 10x = y – 7
\n\u21d2 x = \\(\\frac { y -7 }{ 10 }\\) \u2208 R
\nLet g(x) = \\(\\frac { y -7 }{ 10 }\\)
\n(fog)(x) = f(g(x)) = f(\\(\\frac { y -7 }{ 10 }\\))
\n= 10(\\(\\frac { y -7 }{ 10 }\\) ) + 7 = x
\nSimilarly (gof)(x) = x
\nThus (fog) = (gof) = IR<\/sub><\/p>\n

Question 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

\"NCERT<\/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>\n

Question 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\"NCERT
\nCase ii :
\n\"NCERT
\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>\n

Case iii :
\n\"NCERT
\nThus in any case, f(x1<\/sub>) \u2260 f(x2<\/sub>) when x1<\/sub> \u2260 x2<\/sub>
\nHence f is one-one<\/p>\n

Onto :
\nLet y \u2208 R
\n\"NCERT
\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\"NCERT
\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

\"NCERT<\/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>\n

Question 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>\n

Question 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

\"NCERT<\/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\"NCERT
\nF-1<\/sup> : T \u2192 S is defined as
\nF-1<\/sup> = {(3, a), (2, b),(1, c)}<\/p>\n

ii.
\n\"NCERT
\nAs 1 has 2 preimages, F is not bijective. F-1<\/sup> is not defined.<\/p>\n

\"NCERT<\/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\"NCERT
\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

\"NCERT<\/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\"NCERT
\nHence fa) = g(a) for all a \u2208 A.
\n\u2234 f and g are equal functions.<\/p>\n

\"NCERT<\/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

\"NCERT<\/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

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