6.2 Class 12 NCERT Solutions Question 3.<\/strong> \nShow that the function given by f (x) = sin x is \n(a) strictly increasing in \\(\\left( 0,\\frac { \\pi }{ 2 } \\right) \\) \n(b) strictly decreasing in \\(\\left( \\frac { \\pi }{ 2 } ,\\pi \\right) \\) \n(c) neither increasing nor decreasing in (0, \u03c0) \nSolution: \nWe have f(x) = sinx \n\u2234 f’ (x) = cosx \n(a) f’ (x) = cos x is + ve in the interval \\(\\left( 0,\\frac { \\pi }{ 2 } \\right) \\) \n\u21d2 f(x) is strictly increasing on \\(\\left( 0,\\frac { \\pi }{ 2 } \\right) \\)<\/p>\n(b) f’ (x) = cos x is a -ve in the interval \\(\\left( \\frac { \\pi }{ 2 } ,\\pi \\right) \\) \n\u21d2 f (x) is strictly decreasing in \\(\\left( \\frac { \\pi }{ 2 } ,\\pi \\right) \\)<\/p>\n
(c) f’ (x) = cos x is +ve in the interval \\(\\left( 0,\\frac { \\pi }{ 2 } \\right) \\) \nwhile f’ (x) is -ve in the interval \\(\\left( \\frac { \\pi }{ 2 } ,\\pi \\right) \\) \n\u2234 f(x) is neither increasing nor decreasing in (0, \u03c0)<\/p>\n
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Ex 6.2 Class 12 Ncert Solutions Question 4.<\/strong> \nFind the intervals in which the function f given by f(x) = 2x\u00b2 – 3x is \n(a) strictly increasing \n(b) strictly decreasing \nSolution: \nf(x) = 2x\u00b2 – 3x \n\u21d2 f’ (x) = 4x – 3 \n\u21d2 f’ (x) = 0 at x = \\(\\frac { 3 }{ 4 }\\) \nThe point \\(x=\\frac { 3 }{ 4 }\\) divides the real \n \n(i) f(x) is strictly increasing in (\\(\\frac { 3 }{ 4 }\\), \u221e) \n(ii) f(x) is strictly increasing in (- \u221e, \\(\\frac { 3 }{ 4 }\\))<\/p>\nEx6.2 Class 12 Question 5.<\/strong> \nFind the intervals in which the function f given by f (x) = 2x\u00b3 – 3x\u00b2 – 36x + 7 is \n(a) strictly increasing \n(b) strictly decreasing \nSolution: \nf (x) = 2x\u00b3 – 3x\u00b2 – 36x + 7 \nDifferentiating w.r.t. x, \nf (x) = 6 (x – 3) (x + 2) \n\u21d2 f’ (x) = 0 at x = 3 and x = – 2 \nThe points x = 3, x = – 2, divide the real line into three disjoint intervals viz. (-\u221e, -2), (-2, 3), (3, \u221e) \nNow f’ (x) is +ve in the intervals (-\u221e, -2) and (3, \u221e). Since in the interval (-\u221e, -2) each factor x – 3, x + 2 is -ve. \n\u21d2 f’ (x) = + ve. \n(a) f is strictly increasing in (-\u221e, -2)\u222a(3, \u221e)<\/p>\n(b) In the interval (-2, 3), x+2 is +ve and x-3 is -ve. \nf (x) = 6(x – 3)(x + 2) = + x – = -ve \n\u2234 f is strictly decreasing in the interval (-2, 3).<\/p>\n
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Exercise 6.2 Class 12 Maths Ncert Solutions Question 6.<\/strong> \nFind the intervals in which the following functions are strictly increasing or decreasing: \n(a) x\u00b2 + 2x – 5 \n(b) 10 – 6x – 2x\u00b2 \n(c) – 2x3<\/sup>\u00a0– 9x\u00b2 – 12x + 1 \n(d) 6 – 9x – x\u00b2 \n(e) (x + 1)3<\/sup>(x – 3)3<\/sup> \nSolution: \n(a) Let f(x) = x\u00b2 + 2x – 5 \nDifferentiating w.r.t. x, \nf'(x) = 2x + 2 = 2(x + 1) \nf'(x) = 0 \u21d2 2(x + 1) = 0 = – 1 \nx = – 1 divides R into disjoint open intervals as (-\u221e, -1) and (-1, \u221e) \n \n\u2234 f is strictly decreasing in (-\u221e, -1) \nf is strictly increasing in (-1, \u221e)<\/p>\n(b) Let f(x) = 10 – 6x – 2x\u00b2 \nDifferentiating w.r.t. x, \nf'(x) = – 6 – 4x = – 2(2x + 3) \nf'(x) = 0 \u21d2 – 2(2x + 3) = 0 \n \n\u2234 f is strictly decreasing in (-\u221e, \\(\\frac { -3 }{ 2 }\\)) and \nf is strictly increasing in (\\(\\frac { 3 }{ 2 }\\), \u221e)<\/p>\n
(c) Let f(x) = – 2x3<\/sup> – 9x2<\/sup> – 12x + 1 \n\u2234 f’ (x) = – 6x2<\/sup> – 18x – 12 \n= – 6(x2<\/sup> + 3x + 2) \nf'(x) = – 6(x + 1)(x + 2), f’ (x) = 0 gives x = -1 or x = -2 \nThe points x = – 2 and x = – 1 divide the real line into three disjoint intervals namely ( – \u221e, – 2) ( – 2, – 1) and( – 1, \u221e). \n \nf (x) is increasing in (-2, -1) \nIn the interval (-1, \u221e) i.e.,-1 < x < \u221e,(x + 1) and (x + 2) are both positive. f’ (x) = (-) (+) (+) = -ve. \n\u21d2 f (x) is decreasing in (-1, \u221e) \nHence, f (x) is increasing for – 2 < x < – 1 and decreasing for x < – 2 and x > – 1.<\/p>\n(d) Let f(x) = 6 – 9x – x\u00b2 \nDifferentiating w.r.t. x, \n <\/p>\n
(e) Let f(x) = (x + 1)3<\/sup>(x – 3)3<\/sup> \nDifferentiating w.r.t. x, \nf'(x) = 3 (x\u00b2 – 2x – 3)\u00b2 (2x – 2) \n= 6 [(x + 1) (x – 3)]\u00b2 (x – 1) \n= 6 (x + 1)\u00b2 (x – 1 ) (x – 3)\u00b2 \nf'(x) = 0 \u21d2 x = – 1, 1 or 3 \nx = – 1, x = 1 or x = 3 divide R into disjoint open intervals (-\u221e, -1), (-1, 1), (1, 3) and (-3, – \u221e) \n \nThus f is strictly decreasing in (- \u221e, – 1) and (- 1, 1) and f is strictly increasing in (1, 3) and (3, \u221e)<\/p>\n <\/p>\n
Exercise 6.2 Class 12 Maths Question 7.<\/strong> \nShow that y = log(1 + x) – \\(\\frac { 2x }{ 2 + x }\\), x > – 1 is an increasing function of x throughout its domain. \nSolution: \n \nWhen x > – 1, we get y’ = (+)(+) = (+) i.e., \ny’ is positive for x > – 1 \n\u2234 When x > – 1, y = log(1 + x) – \\(\\frac { 2x }{ 2 + x }\\) an increasing function.<\/p>\nClass 12 Maths Ex 6.2 NCERT Solutions Question 8.<\/strong> \nFind the values of x for which y = [x (x – 2)]\u00b2 is an increasing function. \nSolution: \ny = x4<\/sup> – 4x3<\/sup> + 4x2<\/sup> \n\u2234 \\(\\frac { dy }{ dx }\\) = 4x3<\/sup> – 12x2<\/sup> + 8x \nFor the function to be increasing \\(\\frac { dy }{ dx }\\) >0 \n4x3<\/sup> – 12x2<\/sup> + 8x>0 \n\u21d2 4x(x – 1)(x – 2)>0 \n \nThus, the function is increasing for 0 < x < 1 and x > 2.<\/p>\nEx 6.2 Class 12 Maths Ncert Solutions Question 9.<\/strong> \nProve that y = \\(\\frac { 4sin\\theta }{ (2+cos\\theta ) } -\\theta \\) is an increasing function of \u03b8 in \\(\\left[ 0,\\frac { \\pi }{ 2 } \\right]\\) \nSolution: \n \n\u2234 \\(\\frac { dy }{ d\u03b8 }\\) > 0 since (4 – cos \u03b8) is positive. \nHence y is strictly increasing on (0, \\(\\frac { \u03c0 }{ 2 }\\)). \nAlso y is continuous at \u03b8 = 0 and \u03b8 = \\(\\frac { \u03c0 }{ 2 }\\). \nHence y is increasing on [0, \\(\\frac { \u03c0 }{ 2 }\\)]<\/p>\n <\/p>\n
Class 12 Maths Chapter 6 Exercise 6.2 Question 10.<\/strong> \nProve that the logarithmic function is strictly increasing on (0, \u221e). \nSolution: \nLet f (x) = log x. \nThe domain of f is (0, \u221e). \nNow, f’ (x) = \\(\\frac { 1 }{ x }\\) \nWhen takes the values x > 0 \n\\(\\frac { 1 }{ x }\\) > 0, when x > 0, \n\u2235 f’ (x) > 0 \nHence, f (x) is an increasing function for x > 0 i.e<\/p>\nAnother method: \nFrom the graph, we can observe that as x increases f(x) also increases. Hence logx is an increasing function on (0, \u221e). \n <\/p>\n
Ex 6.2 Class 12 Maths\u00a0 Question 11.<\/strong> \nProve that the function f given by f (x) = x\u00b2 – x + 1 is neither strictly increasing nor strictly decreasing on (- 1, 1). \nSolution: \nlet f (x) = x\u00b2 – x + 1 \nDifferentiating both sides w.r.t. x, \nf'(x) = 2x – 1 \nf'(x) = 2(x – \\(\\frac { 1 }{ 2 }\\)) … (1) \nf'(x) = 0 \u21d2 2(x – \\(\\frac { 1 }{ 2 }\\)) = 0 \u21d2 x = \\(\\frac { 1 }{ 2 }\\) \n\u2234 The domain of f(x) is (-1, 1) \n \n\u2234 f(x) is strictly increasing in (\\(\\frac { 1 }{ 2 }\\), 1) and strictly decreasing in (- 1, \\(\\frac { 1 }{ 2 }\\)) \nIn the interval (- 1, 1) the given function is neither strictly increasing nor strictly decreasing.<\/p>\nClass 12 Ex 6.2 NCERT Solutions Question 12.<\/strong> \nWhich of the following functions are strictly decreasing on \\(\\left[ 0,\\frac { \\pi }{ 2 } \\right] \\) \ni. cos x \nii. cos 2x \niii. cos 3x \niv. tan x \nSolution: \n \nsin 3x can be positive or negative. \n\u2234 f'(x) can be positive or negative. \nHence cos 3x is neither increasing nor decreasing on (0, (\\(\\frac { \u03c0 }{ 2 }\\)).<\/p>\niv. Let f(x) = tan x, x \u2208 (0, \\(\\frac { \u03c0 }{ 2 }\\)) \nDifferentiating w.r.t. x, \nf'(x) = sec\u00b2x is positive \ntan x is strictly increasing on (0, \\(\\frac { \u03c0 }{ 2 }\\)) \nThus cos x and cos 2x are strictly decreasing on (0, \\(\\frac { \u03c0 }{ 2 }\\))<\/p>\n
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6.2 Class 12 Maths Question 13.<\/strong> \nOn which of the following intervals is the function f given by f (x) = x100<\/sup> + sin x – 1 strictly decreasing ? \ni. (0, 1) \nii. \\(\\left[ \\frac { \\pi }{ 2 } ,\\pi \\right] \\) \niii. \\(\\left[ 0,\\frac { \\pi }{ 2 } \\right] \\) \niv. none of these \nSolution: \nf(x) = x100<\/sup> + sin x – 1 \nDifferentiating w.r.t x, \n\u2234 f’ (x)= 100x99 <\/sup>+ cos x<\/p>\ni. When x \u2208 (0, 1), x99<\/sup> > 0 and cosx > 0 \n\u2234 f'(x) is positive \ni.e.,\/(x) is strictly increasing on (0, 1)<\/p>\nii. When x \u2208 (\\(\\frac { \u03c0 }{ 2 }\\), \u03c0), 100x99<\/sup> > 100 and -1 < cosx < 0 \nf'(x)= 100x99<\/sup> + cosx > 100 + a negative number greater than – 1. \nf'(x) is positive \n\u2234 Hence f(x) is strictly increasing on (\\(\\frac { \u03c0 }{ 2 }\\), \u03c0)<\/p>\niii. When x \u2208 (0, \\(\\frac { \u03c0 }{ 2 }\\)), 100x99<\/sup> > 0 and cosx > 0 \nHence f'(x) is positive \nf(x) is strictly increasing on (0, \\(\\frac { \u03c0 }{ 2 }\\)) \n\u2234 f(x) is not strictly decreasing in any of the intervals (0, 1), (\\(\\frac { \u03c0 }{ 2 }\\), \u03c0) and (0, \\(\\frac { \u03c0 }{ 2 }\\)).<\/p>\n6.2 Maths Class 12 NCERT Solutions Question 14.<\/strong> \nFind the least value of a such that the function f given by f (x) = x\u00b2 + ax + 1 is strictly increasing on (1, 2). \nSolution: \nWe have f (x) = x\u00b2 + ax + 1 \n\u2234 f’ (x) = 2x + a. \nSince f (x) is an increasing function on (1, 2) \nf’ (x) > 0 for all 1 < x < 2 Now, f” (x) = 2 for all x \u2208 (1, 2) \u21d2 f” (x) > 0 for all x \u2208 (1, 2) \n\u21d2 f’ (x) is an increasing function on (1, 2) \n\u21d2 f’ (x) is the least value of f’ (x) on (1, 2) \nBut f’ (x)>0 \u2200 x\u2208 (1, 2) \n\u2234 f’ (1)>0 \u21d2 2 + a > 0 \u21d2 a > – 2 \n\u2234 Thus, the least value of ‘a’ is – 2.<\/p>\nClass 12 Maths Exercise 6.2 Question 15.<\/strong> \nLet I be any interval disjoint from (- 1, 1). Prove that the function f given by f(x) = x + \\(\\frac { 1 }{ x } \\) is strictly increasing on I. \nSolution: \nf(x) = x + \\(\\frac { 1 }{ x } \\) and I = R – (-1, 1) \nDifferentiating w.r.t. x, \nf'(x) = 1 – \\(\\frac{1}{x^{2}}\\) = \\(\\frac{x^{2}-1}{x^{2}}\\) \nWhen x\u00b2 > 1, f'(x) > 0 \nWhen x \u2208 R – (- 1, 1), x\u00b2 > 1 \nHence f'(x) is positive. \ni.e., when x \u2208 1, f(x) is strictly increasing<\/p>\n <\/p>\n
Question 16. \nProve that the function f given by f (x) = log sin x is strictly increasing on (0, \\(\\frac { \u03c0 }{ 2 }\\)) and strictly decreasing on (\\(\\frac { \u03c0 }{ 2 }\\), \u03c0) \nSolution: \nf'(x) = \\(\\frac { 1 }{ sin\\quad x }\\)cos x cot x \nwhen 0 < x < \\(\\frac { \\pi }{ 2 }\\), f’ (x) is +ve; i.e., increasing \nWhen \\(\\frac { \\pi }{ 2 }\\) < x < \u03c0, f’ (x) is – ve; i.e., decreasing, \n\u2234 f (x) is decreasing. Hence, f is increasing on (0, \u03c0\/2) and strictly decreasing on (\u03c0\/2, \u03c0).<\/p>\n
Question 17. \nProve that the function f given by f(x) = log cos x is strictly decreasing on (0, \\(\\frac { \u03c0 }{ 2 }\\))and strictly increasing on (\\(\\frac { \u03c0 }{ 2 }\\), \u03c0) \nSolution: \nGiven f(x) = log cos x \nf'(x) = \\(\\frac { 1 }{ cosx }\\) (-sinx) = – tanx \nIn the interval \\(\\left( 0,\\frac { \\pi }{ 2 } \\right)\\), f’ (x) = -ve \n\u2234 f is strictly decreasing. \nIn the interval \\(\\left( \\frac { \\pi }{ 2 } ,\\pi \\right)\\), f’ (x) is + ve. \n\u2234 f is strictly increasing in the interval.<\/p>\n
Question 18. \nProve that the function given by \nf (x) = x3<\/sup> – 3x2<\/sup> + 3x -100 is increasing in R. \nSolution: \nf’ (x) = 3x2<\/sup> – 6x + 3 \n= 3 (x2<\/sup> – 2x + 1) \n= 3 (x -1 )2<\/sup> \nNow x \u2208 R, f'(x) = (x – 1)2<\/sup> \u2265 0 \ni.e. f'(x) \u2265 0 \u2200 x \u2208 R; hence, f(x) is increasing on R.<\/p>\n <\/p>\n
Question 19. \nThe interval in which y = x2<\/sup>\u00a0e-x<\/sup> is increasing is \n(a) (-\u221e, \u221e) \n(b) (-2 0) \n(c) (2, \u221e) \n(d) (0, 2) \nSolution: \ny = x2<\/sup>\u00a0e-x<\/sup> \nDifferentiating w.r.t. x, we get \n \nx = 0 and x = 2 divide the domain R into disjoint open intervals as (-\u221e, 0), (0, 2) and (2, \u221e). \n\u2234 f is strictly increasing on (0, 2)<\/p>\n","protected":false},"excerpt":{"rendered":"These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/ NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.2 Ex 6.2 Class 12 Question 1. Show that the function given by f(x) = 3x + …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2<\/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 6 Application of Derivatives Ex 6.2 - 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