Inverse Function Calculator<\/a>. Use this free tool to find the inverse of a function<\/p>\nQuestion 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 <\/p>\n
Another method: \nWithout using the result fof(x) = x, we can directly find the inverse off. \n <\/p>\n
<\/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 \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 <\/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>\nOnto \nLet y \u2208 range of f such that y = f(x) \n \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>\nQuestion 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 <\/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 \nHence gof = fog = Identity function. Hence f is invertible and \nf-1<\/sup>(y) = \\(\\left(\\frac{(\\sqrt{y+6})-1}{3}\\right)\\)<\/p>\nQuestion 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>\nQuestion 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 <\/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>\nQuestion 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 <\/p>\n <\/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. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-1-ex-1-3\/ 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} …<\/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":{"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 1.3 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n