{"id":29221,"date":"2023-10-31T16:33:55","date_gmt":"2023-10-31T11:03:55","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=29221"},"modified":"2023-11-01T11:42:54","modified_gmt":"2023-11-01T06:12:54","slug":"ncert-solutions-for-class-12-maths-chapter-1-ex-1-3","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-1-ex-1-3\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 1 Relations and Functions Ex 1.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions 1.3<\/h2>\n

\"NCERT<\/p>\n

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

Question 2.
\nLet f, g and h be functions from R to R. Show that
\ni. (f + g)oh = foh + goh
\nii. (f. g)oh = (foh) . (goh)
\nSolution:
\ni. (f + g)oh(x) = (f + g)h(x)) = f(h(x)) + g(h(x)) = (foh)(x) + (goh)(x)
\n\u2234 (f + g)oh = foh + goh<\/p>\n

ii. (f. g)oh(x) = (f.g)(h(x))
\n= f(h(x)).g(h(x))
\n= (foh)(x). (goh)(x)
\n= [(foh).(goh)](x)
\n\u2234 (f. g)oh = (foh) . (goh)<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nFind gof and fog, if
\ni. f(x) = |x| and g(x) = |5x – 2|
\nii. f(x) = 8x and g(x) = x\\(x^{\\frac{1}{3}}\\)
\n(March 2014, SAY 2015, March 2016)
\nSolution:
\ni. (gof)(x) = g(f(x)) = g(|x|) = |5|x| – 2|
\n(fog)(x) = f(g(x)) = f(|5x – 2|)
\n= ||5x – 2|| = |5x – 2|<\/p>\n

ii. (gof)(x) = g(f(x)) = g(8x\u00b3) = (8x\u00b3 )A1\/3<\/sup> = 2x
\n(fog)(x) = f(g(x)) = f(x1\/3<\/sup>) = 8(x1\/3)\u00b3 = 8x<\/sup><\/sup><\/p>\n

Inverse Function Calculator<\/a>. Use this free tool to find the inverse of a function<\/p>\n

Question 4.
\nIf f(x) = \\(\\frac{(4 x+3)}{(6 x-4)}\\), x \u2260 3, show that fof (x) = x, for all x \u2260 \\(\\frac{2}{3}\\). What is the inverse of f?
\nSolution:
\n\"NCERT<\/p>\n

Another method:
\nWithout using the result fof(x) = x, we can directly find the inverse off.
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nState with reason whether following functions have inverse
\ni. f : {1,2, 3,4} \u2192 {10} with
\nf = {(1, 10), (2, 10), (3, 10), (4, 10)}
\nii. g: {5, 6, 7, 8} \u2192 {1, 2, 3, 4} with g ={(5, 4), (6, 3), (7, 4), (8, 2)}
\niii. h : {2, 3,4, 5} \u2192 {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}
\nSolution:
\ni. The elements 1,2,3 and 4 have the same image.
\n\u2234 g is not one-one.
\nHence g has no inverse.<\/p>\n

ii. The image of the elements 5 and 7 is the same
\n\u2234 g is not one-one
\nHence g has no inverse.<\/p>\n

iii.
\n\"NCERT
\nFrom the arrow diagram, it is clear that h is one-one and onto.
\nHence h has inverse.<\/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.
\nSolution:
\nLet x1<\/sub>, x2<\/sub> \u2208 [-1, 1]
\nf(x1<\/sub>) = f(x2<\/sub>) \u21d2 \\(\\frac{x_{1}}{2+x_{1}}\\) = \\(\\frac{x_{2}}{2+x_{2}}\\)
\n\u21d2 x1<\/sub>(2 + x2<\/sub>) = x2<\/sub>(2 + x1<\/sub>)
\n\u21d2 2x1<\/sub> + x1<\/sub>x2<\/sub> = 2x2<\/sub> + x1<\/sub>x2<\/sub>
\n\u21d2 2x1<\/sub> = 2x2<\/sub>
\n\u21d2 x1<\/sub> = x2<\/sub>
\n\u2234 f is one-one
\nLet y = \\(\\frac{x}{x+2}\\) \u21d2 y(x + 2)= x
\n\u21d2 xy – x = – 2y
\n\u21d2 x (1 – y) = 2y
\n\u21d2 x = \\(\\frac{2y}{1-y}\\), y \u2260 1
\n\u2234 f-1<\/sup>(x) = \\(\\frac{2x}{1-x}\\)<\/p>\n

\"NCERT<\/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.
\nSolution:
\nf(x) = 4x + 3, x \u2208 R
\nDomain of f = R and range of f = R
\nOne-one
\nLet x1<\/sub>, x2<\/sub> \u2208 R
\nf(x1<\/sub>) = f(x2<\/sub>) \u21d2 4x1<\/sub> + 3 = 4x2<\/sub> + 3
\n\u21d2 4x1<\/sub> = 4x2<\/sub>
\n\u21d2 x1<\/sub> = x2<\/sub>
\n\u2234 f is one-one.<\/p>\n

Onto
\nLet y \u2208 range of f such that y = f(x)
\n\"NCERT
\ni.e., corresponding to every y \u2208 R, there exists a real number \\(\\frac{y-3}{4}\\) such that f(\\(\\frac{y-3}{4}\\)) = y
\n\u2234 f is onto
\nHence f is a bijection and f-1<\/sup> exists
\nTo find f-1<\/sup>
\nLet y = f(x)
\nThen x = \\(\\frac{y-3}{4}\\) from (1)
\n\u2234 Inverse of\/is given by f-1<\/sup>(x) = \\(\\frac{x-3}{4}\\)<\/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 inverse f-1<\/sup> of f given by f-1<\/sup>(y) = \\(\\sqrt{y-4}\\), where R+<\/sub> is the set of all non-negative real numbers.
\nSolution:
\nLet x1<\/sub>, x2<\/sub> \u2208 R+<\/sub> such that f(x1<\/sub>) = f(x2<\/sub>)
\n\u21d2 x\u00b21<\/sub> + 4 = x\u00b22<\/sub> + 4
\n\u21d2 x\u00b21<\/sub> = x\u00b22<\/sub> \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> \u2208 R
\n\u21d2 x1<\/sub> = x2<\/sub>
\n\u2234 f is one-one.
\nLet y \u2208 [4, \u221e) such that y = f(x)
\n\u21d2 y = x2<\/sup> + 4 \u21d2 x2<\/sup> = y – 4
\n\u21d2 x = \\(\\sqrt{y – 4}\\) since x \u2208 R+<\/sub>
\ni.e., x = \\(\\sqrt{y – 4}\\) \u2208 R+<\/sub>+ …. (1)
\nf(x) = f(\\(\\sqrt{y – 4}\\)) = (\\(\\sqrt{y – 4}\\))\u00b2 + 4
\n= y – 4 + 4 = y
\n\u2234 f is onto
\nSince f is one-one and onto, f is invertible.
\nTo find f-1<\/sup>
\nLet y = x2<\/sup> + 4
\n\u21d2 x = \\(\\sqrt{y – 4}\\) from(1)
\n\u2234 f-1<\/sup>(y) = \\(\\sqrt{y – 4}\\)<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nConsider f : R+<\/sub> \u2192 [- 5, \u221e) given by f(x) = 9x2<\/sup> + 6x – 5. Show that f is invertible with f-1<\/sup>(y) = \\(\\left(\\frac{(\\sqrt{y+6})-1}{3}\\right)\\)
\nSolution:
\nLet y = 9x2<\/sup> + 6x – 5 = (3x + 1)\u00b2
\n\u21d2 (3x + 1)\u00b2 = y + 6
\n\u21d2 3x + 1 = \\(\\sqrt{y + 6}\\) \u21d2 3x = \\(\\sqrt{y + 6}\\) – 1
\n\u21d2 x = \\(\\frac{\\sqrt{y+6}-1}{3}\\)
\n\"NCERT
\nHence gof = fog = Identity function. Hence f is invertible and
\nf-1<\/sup>(y) = \\(\\left(\\frac{(\\sqrt{y+6})-1}{3}\\right)\\)<\/p>\n

Question 10.
\nLet f : X \u2192 Y be an invertible function. Show that f has unique inverse.
\nSolution:
\nLet g1<\/sub> and g2<\/sub> be two inverses of f.
\nThen for all y \u2208 Y
\n(fog1<\/sub>)y = y and (fog2<\/sub>)y = y
\n\u21d2 (fog1<\/sub>)y = (fog2<\/sub>)(y)
\n\u21d2 f(g1<\/sub>(y)) = f(g2<\/sub>(y))
\n\u21d2 g1<\/sub>(y) = g2<\/sub>(y) since f is one-one
\n\u21d2 g1<\/sub> = g2<\/sub>
\nHence the 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) = c. Find f-1<\/sup> and show that (f-1<\/sup>)-1<\/sup> = f.
\nSolution:
\nf = {(1, a), (2, b), (3, c)}
\nf-1<\/sup> = {(a, 1), (b, 2), (c, 3))
\nClearly f is bijective
\n(f-1<\/sup>)-1<\/sup> = {(1, a), (2, b), (3, c)} = f<\/p>\n

\"NCERT<\/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., (f-1<\/sup>)-1<\/sup> = f.
\nSolution:
\nf : X \u2192 Y is invertible. Then f-1<\/sup> : Y \u2192 X
\nLet x \u2208 X , then y = f(x) \u2208 Y such that f-1<\/sup> (y) = x
\nNow (f-1<\/sup>of)(x) = f-1<\/sup>(f(x)) = f-1<\/sup>(y) = x
\nand (fof-1<\/sup>)(y) = f(f-1<\/sup>(y)) = f(x) = y
\ni.e., f-1<\/sup> of and fof-1<\/sup> are identity functions.
\nHence they are inverse of each other.
\ni.e., the inverse of f-1<\/sup> is f.
\ni.e., (f-1<\/sup>)-1<\/sup> = f.<\/p>\n

Question 13.
\nIf f : R \u2192 R be given by f(x) = (3 – x3<\/sup>)\\(\\frac { 1 }{ 3 }\\)<\/sup>, then fof (x) is
\na. x\\(\\frac { 1 }{ 3 }\\)<\/sup>
\nb. x\u00b3
\nc. x
\nd. (3 – x\u00b3)
\nSolution:
\nc. x
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 14.
\nLet f : R – { – \\(\\frac { 4 }{ 3 }\\) } \u2192 be a function defined as f(x) = \\(\\frac { 4x }{ 3x+4 }\\). The inverse of f is the map g : Range f \u2192 R – \\(\\frac { 4 }{ 3 }\\) given by
\na. g(y) = \\(\\frac { 3y }{ 3-4y }\\)
\nb. g(y) = \\(\\frac { 4y }{ 4-3y }\\)
\na. g(y) = \\(\\frac { 4y }{ 3-4y }\\)
\na. g(y) = \\(\\frac { 3y }{ 4-3y }\\)
\nSolution:
\nb. g(y) = \\(\\frac { 4y }{ 4-3y }\\)
\nLet \\(\\frac { 4x }{ 3x+4 }\\) = y \u21d2 4x = 3xy + 4y
\n\u21d2 4x – 3xy = 4y \u21d2 x = \\(\\frac { 4y }{ 4-3y }\\)
\n\u2234 g(y) = \\(\\frac { 4y }{ 4-3y }\\).<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions 1.3 Question 1. Let f : {1,3,4} \u2192 (1,2, 5} and g: (1, 2, 5} \u2192 {1, 3} be …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3<\/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":{"spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3 - NCERT Solutions<\/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:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-1-ex-1-3\/\" \/>\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 1.3 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3 Questions and Answers are prepared by our highly skilled subject experts. 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NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions 1.3 Question 1. Let f : {1,3,4} \u2192 (1,2, 5} and g: (1, 2, 5} \u2192 {1, 3} be … NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3 Read More »","og_url":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-1-ex-1-3\/","og_site_name":"NCERT Solutions","article_publisher":"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/","article_published_time":"2023-10-31T11:03:55+00:00","article_modified_time":"2023-11-01T06:12:54+00:00","og_image":[{"url":"https:\/\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png"}],"twitter_card":"summary_large_image","twitter_creator":"@ncertsolguru","twitter_site":"@ncertsolguru","twitter_misc":{"Written by":"Prasanna","Est. reading time":"9 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Organization","@id":"https:\/\/ncertsolutions.guru\/#organization","name":"NCERT Solutions","url":"https:\/\/ncertsolutions.guru\/","sameAs":["https:\/\/www.facebook.com\/NCERTSolutionsGuru\/","https:\/\/www.instagram.com\/ncertsolutionsguru\/","https:\/\/www.linkedin.com\/in\/ncertsolutionsguru\/","https:\/\/in.pinterest.com\/ncertsolutionsguru\/","https:\/\/en.wikipedia.org\/wiki\/National_Council_of_Educational_Research_and_Training","https:\/\/twitter.com\/ncertsolguru"],"logo":{"@type":"ImageObject","@id":"https:\/\/ncertsolutions.guru\/#logo","inLanguage":"en-US","url":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1","contentUrl":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1","width":460,"height":45,"caption":"NCERT Solutions"},"image":{"@id":"https:\/\/ncertsolutions.guru\/#logo"}},{"@type":"WebSite","@id":"https:\/\/ncertsolutions.guru\/#website","url":"https:\/\/ncertsolutions.guru\/","name":"NCERT Solutions","description":"NCERT Solutions for Class 1 to 12","publisher":{"@id":"https:\/\/ncertsolutions.guru\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/ncertsolutions.guru\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"ImageObject","@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-1-ex-1-3\/#primaryimage","inLanguage":"en-US","url":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1","contentUrl":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1","width":170,"height":17,"caption":"NCERT Solutions"},{"@type":"WebPage","@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-1-ex-1-3\/#webpage","url":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-1-ex-1-3\/","name":"NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3 - 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