{"id":17508,"date":"2020-01-03T14:23:31","date_gmt":"2020-01-03T08:53:31","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=17508"},"modified":"2021-07-02T15:31:25","modified_gmt":"2021-07-02T10:01:25","slug":"2nd-puc-maths-question-bank-chapter-1-ex-1-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-question-bank-chapter-1-ex-1-3\/","title":{"rendered":"2nd PUC Maths Question Bank Chapter 1 Relations and Functions Ex 1.3"},"content":{"rendered":"

Students can Download Maths Chapter 1 Relations and Functions Ex 1.3 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 1.3<\/h2>\n

2nd PUC Maths Relations and Functions NCERT Text Book Questions and Answers Ex 1.3<\/h3>\n

Question 1.
\nLet f: {1, 3, 4} \u2192\u00a0 {1,2, 5} and g : {1, 2, 5} {1,3} be given by f= {(1,2), (3,5), (4,1)} and
\ng = {(1, 3), (2, 3), (5,1)}. Write down gof.
\nAnswer:
\ngof (1) = g (f (1)) = g (2) = 3 gof (3)
\n= g {f (3)} = g (5) = 1
\n= gof (4) = g (f (4)) = g (1) = 3
\n\u2234 gof = {(1,3), (3,1), (4,3)}<\/p>\n

Question 2.
\nLet f, g and h be functions from R to R. Show that
\n(f + g)oh =foh + goh
\n(f.g)oh = (foh) . (goh)
\nAnswer:
\n\u2200 x \u2208 R
\n{(f + g)oh} (x) = (f + g) h (x)
\n= f (h (x)) + g (h (x)) = foh (x) + goh (x)
\n= {foh + goh} (x)
\n\u2234 (f + g) oh = foh + goh
\n((fg )oh) (x) = (f.g) (h (x))
\n= fh (x). gh (x)
\n= {foh (x)} . {goh (x)}
\n{foh. goh} (x)
\n(fg) oh = (foh) (goh)<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nFind gof and fog, if
\n(i) f(x) = |x| and g(x) = |5x – 2|
\nAnswer:
\n\"2nd<\/p>\n

(ii) f\u00a0(x) = 8x3<\/sup> and g(x) = \\(x^{\\frac{1}{3}}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 4.
\nIf \\(f(x)=\\frac{(4 x+3)}{(6 x-4)}, x \\neq \\frac{2}{3}\\),show that fof(x)= x,for all
\n\\(x \\neq \\frac{2}{3}\\),what is the inverse of f?
\nAnswer:
\n\"2nd<\/p>\n

Question 5.
\nState with reason whether following functions have inverse
\n(i) f : {1,2, 3, 4} \u2192 {10} with
\nf = {(1, 10), (2, 10), (3, 10), (4, 10)}<\/p>\n

(ii) g : {5, 6, 7, 8} \u2192 {1, 2, 3, 4} with
\ng = {(5, 4), (6, 3), (7, 4), (8, 2)}<\/p>\n

(iii) h : {2, 3, 4, 5} \u2192 {7, 9,11,13} with
\nh = {(2, 7), (3, 9), (4, 11), (5, 13)}
\nAnswer:
\n(i) f (1) = f (2) = f (3) = f (4) = 10
\n\u21d2 f is not one-one
\n\u2234 f does not have an inverse<\/p>\n

(ii) g(5) = 4\u00a0 g(6) = 3 g(7) = 4 g(8) = 2 g(5) = g(7) = 4
\n\u2234 g is not one-one, hence g does not have inverse<\/p>\n

(iii) h(2) = 7, h(3) = 9, h(4) = 11, h(5)=13,
\n\u2234\u00a0 h is one-one<\/p>\n

Also range of h = {7,9,11,13} = co-domain
\n\u21d2 h is onto
\nHence h has an inverse
\nh-1<\/sup> = {(7,2), (9,3), (11,4), (13, 5)}.<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nShow that f: [-1,1] \u2192 R, given by f(x) =\\(\\frac{x}{(x+2)}\\) is one-one. Find the inverse of the function f : [-1,1] \u2192 Range f.
\n(Hint: For \\(y \\in \\text { Range } f, y=f(x)=\\frac{x}{x+2}\\) for some x in [-1,1]
\n\\(x=\\frac{2 y}{(1-y)}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 7.
\nConsider f : R \u2192 R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nConsider f : R+<\/sub> \u2192 [4, \u221e) given by f(x) = x2<\/sup> + 4. Show that f is invertible with the
\ninverse f-1<\/sup> of\/given by \\(f^{-1}(y)=\\sqrt{y-4}\\) . where R+<\/sub> is the set of ail non-negative real numbers.
\nAnswer:
\n\"2nd<\/p>\n

Question 9.
\nConsider f : R+<\/sub> \u2192 [- 5, \u221e) given by f (x) = 9x3 <\/sup>+ 6x – 5. Show that f is invertible with
\n\\(f^{-1}(y)=\\left(\\frac{(\\sqrt{y+6}) \\cdot 1}{3}\\right)\\)
\nAnswer:
\n\"2nd<\/p>\n

 <\/p>\n

Question 10.
\nLet f : X \u2192 Y be an invertible function. Show that\u00a0 f has unique inverse.\u00a0 (Hint: suppose g1<\/sub>, and g2<\/sub> are two inverses of f. Then for all y \u2208 Y, fog1<\/sub>(y) as 1Y<\/sub>(y) = fog2<\/sub>(y). Use one-one ness of A
\nAnswer:
\nLet g1<\/sub> and g2<\/sub> are two inverse of f, then for all y \u2208 y
\nfog1<\/sub> (y) = f [g (y)] = y = Iy<\/sub> (y)
\nfog2<\/sub> (y) = f [g2<\/sub> (y)] = y = Iy<\/sub> (y)
\n(fog1<\/sub>) (y) = (fog2<\/sub>) (y) \u2200 y \u2208 y
\nf(g1,<\/sub>(y)) = f (g2<\/sub> (y)) \u2200 y \u2208 y
\n\u2234 g, (y) = g2<\/sub> (y) (\u2234 f is one-one)
\ng1<\/sub> =g2
\n<\/sub>\u21d2 Inverse of f is unique.<\/p>\n

Question 11.
\nConsider f : {1, 2, 3} \u2192 {a, b, c} given by f(1) = a,f (2) = b and f(3) = Find f-1 <\/sup>and show that
\n(f -1<\/sup>)-1<\/sup> = f
\nAnswer:
\nGiven that
\nf = {(1. a), (2, b), (3, c)}
\nf is one-one and onto
\n\u2234 f is invertible f-1<\/sup> (a) = 1, f (b) = 2, f-1<\/sup> (c) = 3,
\n\u2234 f1<\/sup> (a, b, c} \u21d2 {1,2, 3} is also one and onto
\n\u2235 Inverse of f is (f)-1<\/sup> exist
\n(f1<\/sup>)-1<\/sup> = {(1. a), (2, b), (3, c)}=f hence (f-1<\/sup>)-1<\/sup> = f.<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nLet f : X \u2192 Y be an invertible function. Show that the inverse of f-1<\/sup> is f, i.e., that (f -1<\/sup>)-1<\/sup> = f
\nAnswer:
\nLet f : X \u2192 Y is invertible
\n\u21d2 f is one and onto and
\nf-1<\/sup> : Y \u21d2 X is defined as f-1<\/sup>(y) = x
\ny : f (x) \u2200 x \u2208 X and y \u2208 Y
\nlet y1<\/sub>,y2\u00a0<\/sub>\u2208 y
\nf-1<\/sup> (y1<\/sub>) = f-2<\/sup>(y2<\/sub>)
\nfof1<\/sup> (y1<\/sub>) = fof1<\/sup> (y2<\/sub>)
\nIy (y1<\/sub>) = Iy(y2<\/sub>)
\n\u21d2 y1<\/sub>\u00a0= y2 <\/sub>\u2234 f1<\/sup> is one-one
\n\u2200 x \u2208 X, \u220b y \u2208 Y such that
\nf1<\/sup> (y) = x, hence f1<\/sup> is onto
\nhence invertible.
\nlet g = (f1<\/sup>)-1
\n<\/sup>gof-1<\/sup> = Iy and f-1<\/sup>og = lx
\n\u2200 x \u2208 X, Ix <\/sub>(x) = x
\nfof-1<\/sup>(x) = f-1<\/sup> [g(x)] =\u00a0 x
\nfof-1<\/sup> [g (x)] = f (x)
\n(fof-1<\/sup>) (g (x)) = f (x)
\ng (x) = f (x)
\ng = f
\n<\/strong>(f-1<\/sup>)-1<\/sup> = f.
\n<\/strong><\/p>\n

Question 13.
\nIf f: R \u2192 R be given by \\(f(x)=\\left(3-x^{3}\\right)^{\\frac{1}{3}}\\),then fof (x) is
\n(A)\\(x^{\\frac{1}{3}}\\)
\n(B) x3<\/sup>
\n(C) x
\n(D) (3 – x3<\/sup>)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 14.
\nLet\u00a0 \\(f: \\mathbf{R}-\\left\\{-\\frac{4}{3}\\right\\} \\rightarrow \\mathbf{R}\\),be a function as \\(f(x)=+\\frac{4 x}{3 x+4}\\).The inverse off is the mag g : Range
\n\\(f \\rightarrow \\mathbf{R}-\\left\\{-\\frac{4}{3}\\right\\}\\) given by
\n(A) \\(g(y)=\\frac{3 y}{3-4 y}\\)
\n(B) \\(g(y)=\\frac{4 y}{4-3 y}\\)
\n(C) \\(g(y)=\\frac{4 y}{3-4 y}\\)
\n(D) \\(g(y)=\\frac{3 y}{4-3 y}\\)
\nAnswer:
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 1 Relations and Functions Ex 1.3 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 1.3 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\/17508"}],"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=17508"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/17508\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=17508"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=17508"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=17508"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}