{"id":17387,"date":"2020-01-03T11:32:10","date_gmt":"2020-01-03T06:02:10","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=17387"},"modified":"2021-07-02T15:31:28","modified_gmt":"2021-07-02T10:01:28","slug":"2nd-puc-maths-question-bank-chapter-1-ex-1-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-question-bank-chapter-1-ex-1-2\/","title":{"rendered":"2nd PUC Maths Question Bank Chapter 1 Relations and Functions Ex 1.2"},"content":{"rendered":"

Students can Download Maths Chapter 1 Relations and Functions Ex 1.2 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Maths Question Bank Chapter 1 Relations and Functions Ex 1.2<\/h2>\n

2nd PUC Maths Relations and Functions NCERT Text Book Questions and Answers Ex 1.2<\/h3>\n

Question 1.
\nShow that the function f : R0<\/sub> \u2192 R0<\/sub> defined by \\(f(x)=\\frac{1}{x}\\)\u00a0 is one-one and onto, where R0<\/sub> is the set of all non-zero real numbers. Is the result true, if the domain R0<\/sub> is replaced by N with co-domain being same as R0<\/sub>?
\nAnswer:
\n\"2nd
\nf : N \u2192 R0,\u00a0 <\/sub>then the function will remains one – one however mot onto.
\nHere the range of \\(\\mathrm{f}=\\left\\{1, \\frac{1}{2}, \\frac{1}{3}, \\ldots \\ldots .\\right\\} \\neq \\mathrm{R}_{\\mathrm{e}}\\)<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nCheck the injectivity and surjectivity of the following functions:
\n(i) f : N \u2192 N given by f (x) = x2
\n<\/sup>Answer:
\n\"2nd
\nhence f is injective, for some elements like, 2, 3 etc has no preimage in N such that
\nf (x) = 2 hence not surjective.<\/p>\n

(ii) f : Z \u2192 Z given by f(x) = x2
\n<\/sup>Answer:
\nf is not one-one or injective as f (1) = f (-1) \u21d2 1 = -1
\nf is not on to or surjective as \\(f(x)=2 \\Rightarrow x=\\sqrt{2} \\notin Z\\)<\/p>\n

(iii) f : R \u2192 R given by f(x) = x2
\n<\/sup>Answer:
\nf is not injective
\nas f (x1<\/sub>) = f (x2<\/sub>) \u21d2 x1<\/sub>= \u00b1 x2
\n<\/sub>Range off does not contain any negative real. Hence not surjective.<\/p>\n

(iv) f : N \u2192 N given by f(x) = x2
\n<\/sup>Answer:
\nLet x1<\/sub>, x2<\/sub> \u2208 N
\nf(x1<\/sub>) = f(x2<\/sub>) \u21d2 \\(x_{1}^{3}=x_{2}^{3}\\)
\nhence one-one
\nRange of f= {1,8,27,…… } \u2260 N
\nf is not surjective.<\/p>\n

\"KSEEB<\/p>\n

(v) f : Z \u2192 Z given by f (x) = x3
\n<\/sup>Answer:
\nLet x1<\/sub>, x2<\/sub> \u2208 Z
\n\\(\\begin{aligned} f\\left(x_{1}\\right)=f\\left(x_{2}\\right) & \\Rightarrow x_{1}^{3}=x_{2}^{3} \\\\& \\Rightarrow x_{1}=x_{2} \\end{aligned}\\)
\nf is one-one or injective
\nRange of f = {0, \u00b11, \u00b18, \u00b127,…………….. } \u2260Z
\nf is not surjective<\/p>\n

(vi) f : R \u2192 R given by f (x) = x3
\n<\/sup>Answer:
\n\"2nd<\/p>\n

Question 3.
\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.
\nAnswer:
\n\u2200 x \u2208 [0, 1)
\nf (x) = [x] = 0
\nf is not one-one, or injective
\nex: \\(0, \\frac{1}{2} \\in R \\Rightarrow f(0)=[0]=0\\)
\n\\(f\\left(\\frac{1}{2}\\right)=\\left[\\frac{1}{2}\\right]=0\\)
\nRange of f is the set of integers \u2260 R hence f is not onto.<\/p>\n

Question 4.
\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.
\nAnswer:
\nLet x1<\/sub>, x2<\/sub> \u2208 R f (x.) = f (x1<\/sub>)
\n\\(\\begin{aligned}f\\left(x_{1}\\right)=f\\left(x_{2}\\right)&\\Rightarrow\\left|x_{1}\\right|=\\left|x_{2}\\right| \\\\& \\Rightarrow x_{1}=\\pm x_{2} \\end{aligned}\\)
\n\\(|-2|=|+2|=2\\)hence f is not one-one.
\nRange of functions is only non negative real numbers
\nRange of f = {0, \u221e) \u2260 R
\n\u2234 f is not onto.<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nShow that the Signum Function f : R \u2192 R, given by
\n\\(f(x)=\\left\\{\\begin{array}{ll}{1} & {\\text { if } x>0} \\\\{0} & {\\text { if } x=0} \\\\{-1} & {\\text { if } x<0}\\end{array}\\right.\\)
\nis neither one-one nor onto.
\nAnswer:
\nf (x) = 1 \u2200 x \u2208 [0, \u221e )
\nhence f is not one-one Range of f = {-1,0, 1} hence f is not onto.<\/p>\n

Question 6.
\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.
\nAnswer:
\nx1<\/sub>, x2<\/sub> \u2208 A, f (x1<\/sub>) = f (x2<\/sub>)
\n\u21d2 x1<\/sub> = x2
\n<\/sub>\u2234 hence one-one
\nf (1) = 4, f (2) = 5, f (3) \u2260 R, hence f is one-one.<\/p>\n

Question 7.
\nIn each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
\n(i) f : R \u2192 R defined by f(x) = 3 – 4x
\nAnswer:
\nLet x1<\/sub> , x2<\/sub> \u2208 R
\nf (x1<\/sub>) = f (x2<\/sub>) \u21d2 3- 4x1<\/sub>, = 3- 4x2
\n<\/sub>\u21d2 x1<\/sub>\u00a0= x2<\/sub> hence one-one function
\n\"2nd
\nhence f is not onto<\/p>\n

(ii) f : R\u2192 R defined by f(x) = 1 + x2<\/sup>
\nAnswer:
\nLet x1<\/sub>, x2<\/sub> \u2208 R
\nf (x1<\/sub>) = f (x2<\/sub>)
\n\u21d2 1 + x1<\/sub>2<\/sup> = 1 + x2<\/sub>2
\n<\/sup>\u21d2 x1<\/sub> = \u00b1x2
\n<\/sub>\u2234\u00a0 f is not one-one
\nAlso range of f contains only real numbers which is greater than or equal to 1\u2260 R
\n\u2234 f is not onto hence f is not bijective.<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nLet A and B be sets. Show that
\nf : A x B \u2192 B x A such that
\nf (a, b) = (b, a) is bijective function.
\nAnswer:
\nLet (p, q), (r, s) g A x B f (p,q) = f (r, s)
\n\u21d2 (q, p) = (s, r) \u21d2 q = s and p = r
\n\u21d2 (p, q) = (r, s), hence one-one
\n\u2200 (b, a) \u2208 B x A, 3 (a, b) \u2208 A x B
\nsuch that f (a, b) = f (b, a), hence on to hence f is bijective function<\/p>\n

Question 9.
\nLet f : N \u2192 N be defined by
\n\\(\\mathbf{f}(\\mathbf{n})=\\left\\{\\begin{array}{c} {\\frac{\\mathbf{n}+\\mathbf{1}}{2}, \\text { if } \\mathbf{n} \\text { is odd }} \\\\{\\mathbf{All}\u00a0 \\mathbf{n} \\in \\mathbf{N}} \\\\{\\frac{\\mathbf{n}}{2}, \\text { if } \\mathbf{n} \\text { is even }}\\end{array}\\right.\\)
\nState whether the function \/ is bijective. Justify your answer.
\nAnswer:
\n\\(f(2 n-1)=\\frac{2 n-1+1}{2}=n\\)
\n\\(f(2 n)=\\frac{2 n}{2}=n\\)
\nf (2n – 1) = f (2n) = n, but 2n -1 \u2260 2n
\nen, f (3) = 2, f (4) = 2, but 3 \u2260 4
\nhence f is not one-one
\nHere range of f = N
\nas for any N \u2208 N, \u220b 2n \u2208 N such that
\nf(2n) = n
\nhence f is on to. But it is not bijective<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\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 onto? justify your answer.
\nAnswer:
\n\"2nd<\/p>\n

Question 11.
\nLet f : R \u2192 R be defined as f(x) = x4<\/sup>. Choose the correct answer.
\n(A) f is one-one onto
\n(B) f is many-one onto
\n(c) f is one-one but not onto
\n(D) f is neither one-one nor onto.
\nAnswer:
\nf (x,) = f (x2<\/sub>) \u21d2 x1<\/sub>4<\/sup> = x2<\/sub>4
\n<\/sup>= x1<\/sub>4<\/sup> = x2<\/sub>4 <\/sup>= 0
\n= (x1<\/sub> – x2<\/sub>) (x1<\/sub> + x2<\/sub>) (x1<\/sub>2<\/sup> = x2<\/sub>2<\/sup>)
\nx1<\/sub> = x2<\/sub> or x1<\/sub> = – x2<\/sub>
\n\u2234 f is not one-one
\nf (x) = x4<\/sup> \u2265 o \u2200 x \u2208 R
\nRange of f = [0, \u221e) \u2208 R
\nhence not onto
\n\u200d\u2234 correct Answer is \u201cD\u201d<\/p>\n

Question 12.
\nLet f : R \u2192 R be defined as f (x) = x5<\/sup>. Choose the correct answer.
\n(A) f is one-one onto
\n(B) f is many-one onto
\n(C) f is one-one but not onto
\n(D) f is either one-one nor onto.
\nAnswer:
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 1 Relations and Functions Ex 1.2 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Maths Question Bank Chapter 1 Relations and Functions Ex 1.2 2nd PUC Maths …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/17387"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=17387"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/17387\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=17387"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=17387"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=17387"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}