{"id":29220,"date":"2022-03-28T17:00:01","date_gmt":"2022-03-28T11:30:01","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=29220"},"modified":"2022-03-28T17:18:11","modified_gmt":"2022-03-28T11:48:11","slug":"ncert-solutions-for-class-12-maths-chapter-1-ex-1-2","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-1-ex-1-2\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.2"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 1 Relations and Functions Ex 1.2 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-1-ex-1-2\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions 1.2<\/h2>\n

\"NCERT<\/p>\n

Ex 1.2 Class 12 NCERT Solutions Question 1.<\/strong>
\nShow that the function f : R*<\/sub> \u2192 R*<\/sub>, defined by f(x) = \\(\\frac { 1 }{ x }\\) is one-one and onto, where R*<\/sub> is the set of all non-zero real numbers. Is the result true, if the domain R*<\/sub> is replaced by N with co-domain being same as R*<\/sub>?
\nSolution:
\na. One-one
\nLet x1<\/sub>, x1<\/sub> \u2208 R, such that f(x1<\/sub>) = f(x2<\/sub>)
\n\"Ex
\nFor each y \u2208 R*<\/sub> there exists x = \\(\\frac { 1 }{ y }\\) \u2208 R,
\nsuch that f(x) = y.
\n\u2234 f is onto<\/p>\n

b. One-one
\nLet x1<\/sub>, x2<\/sub> \u2208 N such that f(x1<\/sub>) = f(x2<\/sub>)
\n\u21d2 \\(\\frac{1}{x_{1}}=\\frac{1}{x_{2}}\\) \u21d2 x1<\/sub> = x2<\/sub>
\n\u2234 f is one-one
\nOnto
\nThe domain of f is N = {1, 2, 3, ……. }
\nThe range of f = {1, \\(\\frac { 1 }{ 2 }\\), \\(\\frac { 1 }{ 3 }\\), \\(\\frac { 1 }{ 4 }\\), ……. } \u2260 R
\n\u2234 f is not onto
\nAnother method
\n2 \u2208 R*<\/sub>.
\nThe pre-image of 2 is \\(\\frac { 1 }{ 2 }\\) \u2209 N .
\nHence f is not onto.<\/p>\n

Exercise 1.2 Class 12 NCERT Solutions Question 2.<\/strong>
\nCheck the injectivity and surjectivity of the following functions
\ni. f : N \u2192 N given by f (x) = x\u00b2
\nii. f : Z \u2192 Z given by f (x) = x\u00b2
\niii. f : R \u2192 R given by f (x) = x\u00b2
\niv. f : N \u2192 N given by f (x) = x\u00b3
\nv. f : Z \u2192 Z given by f (x) = x\u00b3
\nSolution:
\ni. Injectivity
\nLet x1<\/sub>, x1<\/sub> \u2208 N such that f(x1<\/sub>) = f(x2<\/sub>)
\n\u21d2 x\u00b21<\/sub> = x\u00b22<\/sub>
\n\u21d2 x\u00b21<\/sub> – x\u00b22<\/sub> = 0
\n\u21d2 (x1<\/sub> – x2<\/sub>)(x1<\/sub> + x2<\/sub>) = 0
\n\u21d2 x1<\/sub> – x2<\/sub> = 0 since x1<\/sub> + x2<\/sub> \u2260 0 as x1<\/sub>, x1<\/sub> \u2208 N
\n\u21d2 x1<\/sub> = x2<\/sub>
\n\u2234 f is injective
\nSurjectivity
\nLet y = 3 \u2208 N, the co-domain of f
\nf(x) = 3 \u21d2 x\u00b2 = 3
\n\u21d2 x = \u00b1\\(\\sqrt{3}\\) \u2209 N , the domain of f
\n\u2234f is not surjective<\/p>\n

ii. Injectivity
\nLet x1<\/sub> = 2, x2<\/sub> = – 2
\nf(x1<\/sub>) = 2\u00b2 = 4, f(x2<\/sub>) = (- 2)\u00b2 = 4
\ni.e., f(x1<\/sub>) = f(x2<\/sub>) for x1<\/sub> \u2260 x2<\/sub>
\n\u2234 f is not injective<\/p>\n

Surjectivity
\nLet y = 2 \u2208 Z, the co-domain of f
\n\u2234 f(x) = 2 \u21d2 x\u00b2 = 2 \u21d2 x = \u00b1 \\(\\sqrt{2}\\) \u2209 Z, the domain of f
\n\u2234 f is not surjective.<\/p>\n

iii. Injectivity
\nLet x1<\/sub> = 2, x2<\/sub> = – 2
\nf(x1<\/sub>) = 2\u00b2 = 4, f(x2<\/sub>) = (- 2)\u00b2 = 4
\ni.e., f(x1<\/sub>) = f(x2<\/sub>) for x1<\/sub> \u2260 x2<\/sub>
\n\u2234 f is not injective (one-one)<\/p>\n

Surjectivity
\nLet y = – 1 \u2208 R, the co-domain of f
\nf(x) = – 1 \u21d2 x\u00b2 = -1
\n\u21d2 x = \u00b1 \\(\\sqrt{1}\\) \u2209 R, the domain of f
\n\u2234 f is not surjective (onto)<\/p>\n

iv. Injectivity
\nLet x1<\/sub>, x2<\/sub> \u2208 N such that f(x1<\/sub>) = f(x1<\/sub>) = f(x2<\/sub>)
\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>2) = 0
\n\u21d2 x1<\/sub> – x2<\/sub> = 0 since x\u00b21<\/sub> + x1<\/sub>x2<\/sub> + x\u00b22<\/sub> \u2260 0
\n\u21d2 x1<\/sub> = x2<\/sub>
\n\u2234f is injective (one-one)<\/p>\n

Surjectivity
\nLet y = 4 \u2208 N, the co-domain of f
\n\u21d2 f(x) = 4 \u21d2 x3<\/sub> = 4
\n\u21d2 x = 41\/3<\/sup> \u2209 N, the domain of f
\n\u2234f is not surjective (onto)<\/p>\n

v. Injectivity
\nLet x1<\/sub>, x2<\/sub> \u2208 Z such that f(x1<\/sub>) = f(x2<\/sub>)
\n\u21d2 x\u00b31<\/sub> = x\u00b32<\/sub> \u21d2 x1<\/sub> = x2<\/sub>
\n\u2234 f is injective (one-one)<\/p>\n

Surjectivity
\nLet y = 4 \u2208 Z, the co-domain of f such that
\nf(x) = 4
\n\u21d2 x3<\/sub> = 4 \u21d2 x = 41\/3<\/sup> \u2209 Z, the domain of f
\n\u2234 f is not surjective (onto)<\/p>\n

\"NCERT<\/p>\n

Exercise 1.2 Class 12 Maths NCERT Solutions Question 3.<\/strong>
\nProve that the Greatest Integer Function f : R \u2192 R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
\nSolution:
\nRefer Example 26
\nOne-one
\nLet x1<\/sub> = 2.1, x2<\/sub> = 2.5
\nf(x1<\/sub>) = f(2.1) = [2.1] = 2
\nf(x2<\/sub>) = f(2.5) = [2.5] = 2
\ni.e. f(x1<\/sub>) = f(x2<\/sub>) for x1<\/sub> \u2260 x2<\/sub>
\n\u2234 f is not one-one.<\/p>\n

Onto
\nLet y = 2.5 \u2208 R, the co-domain of f
\nf(x) = 2.5 \u21d2 [x] = 2.5, which is not possible.
\n\u2234 f is not onto<\/p>\n

Class 12 Maths Ncert Solutions Chapter 1 Ex 1.2 Question 4.<\/strong>
\nShow that the Modulus Function f : R \u2192 R, given by f(x) = |x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is – x, if x is negative.
\nSolution:
\nOne-one
\nLet x1<\/sub> = 1 and x2<\/sub> = – 1 \u2208 R
\nf(x1<\/sub>) = f(1) = |1| = 1
\nf(x2<\/sub>) = f(- 1) = |- 1|= 1
\ni.e., f(x1<\/sub>) = f(x2<\/sub>) for x1<\/sub> \u2260 x2<\/sub>
\n\u2234 f is not one-one Onto
\nLet y = – 1 \u2208 R, the co-domain of f
\nf(x) = – 1 \u21d2 |x|= – 1 which is not possible
\n\u2234 f is not onto.<\/p>\n

Class 12 Maths 1.2 NCERT Solutions\u00a0 Question 5.<\/strong>
\nShow that the Signum Function f : R \u2192 R, given by
\n\\(f(x)=\\left\\{\\begin{array}{c}
\n1, \\text { if } x>0 \\\\
\n0, \\text { if } x=0 \\\\
\n-1, \\text { if } x<0
\n\\end{array}\\right.\\)
\nis neither one-one nor onto.
\nSolution:
\nOne-one
\nLet x1<\/sub> = 1, x2<\/sub> = 2
\nf(x1<\/sub>) = f(1) = 1
\nf(x2<\/sub>) = f(2) = 1
\ni.e. f(x1<\/sub>) = f(x2<\/sub>) for x1<\/sub> \u2260 x2<\/sub>
\n\u2234 f is not one-one<\/p>\n

Onto
\nThe co-domain of f is R
\nThe range of f is {-1, 0, 1} \u2260 R
\n\u2234 f is not onto.<\/p>\n

\"NCERT<\/p>\n

Relation And Function Class 12 Exercise 1.2 Question 6.<\/strong>
\nLet A = {1, 2, 3}, B = (4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.
\nSolution:
\n\"Exercise
\nDifferent elements in A have different images in B. Hence f is one – one.<\/p>\n

Ex 1.2 Class 12 Maths NCERT Solutions Question 7.<\/strong>
\nIn each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
\ni. f : R \u2192 R defined by f(x) = 3 – 4x
\nii. f : R \u2192 R defined by f(x) = 1 + x2<\/sup>
\nSolution:
\ni. One-one
\nLet x1<\/sub>, x2<\/sub> 6 R such that f(x1<\/sub>) = f(x2<\/sub>)
\n\u21d2 3 – 4x1<\/sub> = 3 – 4x2<\/sub>
\n\u21d2 – 4x1<\/sub> = – 4x2<\/sub>
\n\u21d2 x1<\/sub> = x2<\/sub>
\n\u2234f is one-one
\nOnto
\nLet y \u2208 R, such that y = f(x)
\n\u21d2 y = 3 – 4x
\n\u21d2 4x = 3 – y
\n\u21d2 x = \\(\\frac { 3-y }{ 4 }\\) \u2208 R
\nf(x) = f\\(\\frac { 3-y }{ 4 }\\) = 3 – 4(\\(\\frac { 3-y }{ 4 }\\))
\n= 3 – 3 + y = y
\nFor each y \u2208 R there exists x = \\(\\frac { 3-y }{ 4 }\\) \u2208 R
\nsuch that f(x) = y
\n\u2234f is onto
\nSince f is one-one and onto, f is bijective.
\nAnother Method
\nf(x) = 3 – 4x is a linear function from R to R.
\n\u2234 f is one-one and onto.
\nHence f is bijective.<\/p>\n

ii. One-one
\nLet x1<\/sub> = 1, x2<\/sub> = – 1 \u2208 R
\nf(x1<\/sub>) = f(1) = 1 + 1\u00b2 = 2
\nf(x2<\/sub>)= f(-1) = 1 + (- 1)\u00b2 = 2
\ni.e., f(x1<\/sub>) = f(x2<\/sub>) for x1<\/sub> \u2260 x2<\/sub>
\n\u2234f is not one-one<\/p>\n

Onto
\nLet y = – 1 \u2208 R, the co-domain of f
\nf(x) = y \u21d2 1 + x2<\/sup> = – 1
\n\u21d2 x2<\/sup> = – 2
\n\u21d2 x = \u00b1 \\(\\sqrt{-2}\\) \u2209 R
\n\u2234f is not onto.
\nThus f is neither one-one nor onto.<\/p>\n

\"NCERT<\/p>\n

Class 12 Ex 1.2 NCERT Solutions Question 8.<\/strong>
\nLet A and B be sets. Show that f : A x B \u2192 B x A such that f(a, b) = (b, a) is bijective function.
\nSolution:
\nLet (a1<\/sub>, b1<\/sub>) and (a2<\/sub>, b2<\/sub>) \u2208 A x B
\nsuch that f(a1<\/sub>, b1<\/sub>) = f(a2<\/sub>, b2<\/sub>)
\n\u21d2 (b1<\/sub>, a1<\/sub>) = (b2<\/sub>, a2<\/sub>)
\n\u21d2 b1<\/sub> = b2<\/sub> and a1<\/sub> = a2<\/sub>
\n\u21d2 (a1<\/sub>, b1<\/sub>) = (a2<\/sub>, b2<\/sub>)
\n\u2234f is one-one.
\nLet (b, a) \u2208 B x A \u21d2 b \u2208 B and a \u2208 A
\n\u21d2 (a, b) \u2208 A x B
\nf(a, b) = (b, a)
\ni.e., corresponding to each (b, a) \u2208 B x A, there exists (a, b) \u2208 A x B such that f(a, b) = (b, a)
\n\u2234 f is onto
\nHence f is a bijection.<\/p>\n

Maths Class 12 Exercise 1.2 NCERT Solutions Question 9.<\/strong>
\nLet f : N \u2192 N be defined by
\nf(n) = \\(\\left\\{\\begin{array}{l}
\n\\frac{n+1}{2}, \\text { if } n \\text { is odd } \\\\
\n\\frac{n}{2}, \\text { if } n \\text { is even }
\n\\end{array} \\text { for all } n \\in \\mathbf{N}\\right.\\)
\nState whether the function \/ is bijective. Justify your answer.
\nSolution:
\nLet x1<\/sub> = 1 and x2<\/sub> = 2 \u2208 N
\nf(x1<\/sub>) = f(1) = \\(\\frac { 1+1 }{ 2 }\\) = 1 and f(x2<\/sub>) = f(2) = \\(\\frac { 2 }{ 2 }\\) = 1
\n\u2234 f(x1<\/sub>) = f(x2<\/sub>) for x1<\/sub> \u2260 x2<\/sub>
\nHence f is not one-one and hence not bijective.<\/p>\n

Class 12 Math Ex 1.2 NCERT Solutions Question 10.<\/strong>
\nLet A = R – {3} and B = R – {1}. Consider the function f : A \u2192 B defined by f(x) = \\(\\left(\\frac{x-2}{x-3}\\right)\\). Is f one-one and onto? Justify your answer.
\nSolution:
\nOne-one
\n\"Exercise<\/p>\n

One-one
\nLet y \u2208 B such that f(x) = y
\n\"Class
\nFor each y \u2208 B, there exists x = \\(\\left(\\frac{2-2y}{1-y}\\right)\\) \u2208 A
\nsuch that f(x) = y
\n\u2234 f is onto.<\/p>\n

\"NCERT<\/p>\n

Relations And Functions Class 12 Exercise 1.2 Question 11.<\/strong>
\nLet f : R \u2192 R be defined as f(x) = x4<\/sup>. Choose the correct answer.
\na. f is one-one onto
\nb. f is many-one onto
\nc. f is one-one but not onto
\nd. f is neither one-one nor onto.
\nSolution:
\nd. f is neither one-one nor onto.<\/p>\n

One-one
\nLet x1<\/sub> = 1, x2<\/sub> = – 1 \u2208 R
\nf(x1<\/sub>) = f(1) = (1)4<\/sup> = 1
\nf(x2<\/sub>) = f(-1) = (-1)4<\/sup>= 1
\nf(x1<\/sub>) = f(x2<\/sub>) for x1<\/sub> \u2260 x2<\/sub>
\n\u2234 f is not one-one<\/p>\n

Onto
\nThe co-domain of f is R.
\nThe range of f is [0, \u221e), the non-negative real numbers.
\nSince range of f \u2260 co-domain of f is not onto.
\nHence f is neither one-one nor onto.<\/p>\n

1.2 Class 12 NCERT Solutions Question 12.<\/strong>
\nLet f : R \u2192 R be defined as f (x) = 3x. Choose the correct answer.
\na. f is one-one onto
\nb. f is many-one onto
\nc. f is one-one but not onto
\nd. f is neither one-one nor onto.
\nSolution:
\na. f is one-one onto
\nf(x1<\/sub>) = f(x2<\/sub>) \u21d2 3x1<\/sub> = 3x2<\/sub> \u21d2 x1<\/sub> = x2<\/sub>
\n\u2234 f is one-one
\nFor y \u2208 co-domain off there exists \\(\\frac { y }{ 3 }\\) in the domain of f such that f(\\(\\frac { y }{ 3 }\\)) – 3(\\(\\frac { y }{ 3 }\\)) = y
\nHence f is onto.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.2 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-1-ex-1-2\/ NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions 1.2 Ex 1.2 Class 12 NCERT Solutions Question 1. Show that the function f : R* \u2192 R*, …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.2<\/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 Ex 1.2 - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-1-ex-1-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.2 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.2 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-1-ex-1-2\/ NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions 1.2 Ex 1.2 Class 12 NCERT Solutions Question 1. 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