{"id":30984,"date":"2022-03-29T14:00:55","date_gmt":"2022-03-29T08:30:55","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=30984"},"modified":"2022-03-29T14:40:39","modified_gmt":"2022-03-29T09:10:39","slug":"ncert-solutions-for-class-12-maths-chapter-6-ex-6-2","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> 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\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.2<\/h2>\n

\"NCERT<\/p>\n

Ex 6.2 Class 12 Question 1.<\/strong>
\nShow that the function given by f(x) = 3x + 17 is strictly increasing on R.
\nSolution:
\nf(x) = 3x + 17
\nSince f(x) is a polynomial function, it is continuous and differentiable in R
\n\u2234 f'(x) = 3 > 0 for all x \u2208 R
\nHence f'(x) is strictly increasing on R<\/p>\n

Exercise 6.2 Class 12 NCERT Solutions Question 2.<\/strong>
\nShow that the function given by f (x) = e2x<\/sup>\u00a0is strictly increasing on R.
\nSolution:
\nf (x) = e2x<\/sup>
\nSince f(x) is an exponential function, it is continuous and differentiable in R
\n\u2234 f (x) = e2x<\/sup> > 0 for all x \u2208 R
\nHence f(x) is strictly increasing on R.<\/p>\n

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

\"NCERT<\/p>\n

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\"Ex
\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>\n

Ex6.2 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

\"NCERT<\/p>\n

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\"Exercise
\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\"6.2
\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\"Ex
\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\"Ex6.2<\/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\"Exercise
\nThus f is strictly decreasing in (- \u221e, – 1) and (- 1, 1) and f is strictly increasing in (1, 3) and (3, \u221e)<\/p>\n

\"NCERT<\/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\"Exercise
\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>\n

Class 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\"Class
\nThus, the function is increasing for 0 < x < 1 and x > 2.<\/p>\n

Ex 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\"Ex
\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

\"NCERT<\/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>\n

Another 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\"Class<\/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\"Ex
\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>\n

Class 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\"NCERT
\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>\n

iv. 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

\"NCERT<\/p>\n

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>\n

i. 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>\n

ii. 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>\n

iii. 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>\n

6.2 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>\n

Class 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

\"NCERT<\/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

\"NCERT<\/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\"NCERT
\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<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/\" \/>\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 6 Application of Derivatives Ex 6.2 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"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 + … NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 Read More »\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ Questions\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/\" \/>\n<meta property=\"article:published_time\" content=\"2022-03-29T08:30:55+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2022-03-29T09:10:39+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@ncertsolguru\" \/>\n<meta name=\"twitter:site\" content=\"@ncertsolguru\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Prasanna\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"15 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"WebSite\",\"@id\":\"https:\/\/mcq-questions.com\/#website\",\"url\":\"https:\/\/mcq-questions.com\/\",\"name\":\"MCQ Questions\",\"description\":\"MCQ Questions for Class 1 to 12\",\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/mcq-questions.com\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/#primaryimage\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1\",\"contentUrl\":\"https:\/\/i0.wp.com\/mcq-questions.com\/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:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/#webpage\",\"url\":\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/\",\"name\":\"NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 - MCQ Questions\",\"isPartOf\":{\"@id\":\"https:\/\/mcq-questions.com\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/#primaryimage\"},\"datePublished\":\"2022-03-29T08:30:55+00:00\",\"dateModified\":\"2022-03-29T09:10:39+00:00\",\"author\":{\"@id\":\"https:\/\/mcq-questions.com\/#\/schema\/person\/4ba9570f32f2057e70e670c7885e47f3\"},\"breadcrumb\":{\"@id\":\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/mcq-questions.com\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2\"}]},{\"@type\":\"Person\",\"@id\":\"https:\/\/mcq-questions.com\/#\/schema\/person\/4ba9570f32f2057e70e670c7885e47f3\",\"name\":\"Prasanna\",\"image\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/mcq-questions.com\/#personlogo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g\",\"caption\":\"Prasanna\"},\"url\":\"https:\/\/mcq-questions.com\/author\/prasanna\/\"}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 - MCQ Questions","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/","og_locale":"en_US","og_type":"article","og_title":"NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 - MCQ Questions","og_description":"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 + … NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 Read More »","og_url":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/","og_site_name":"MCQ Questions","article_publisher":"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/","article_published_time":"2022-03-29T08:30:55+00:00","article_modified_time":"2022-03-29T09:10:39+00:00","og_image":[{"url":"https:\/\/mcq-questions.com\/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":"15 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebSite","@id":"https:\/\/mcq-questions.com\/#website","url":"https:\/\/mcq-questions.com\/","name":"MCQ Questions","description":"MCQ Questions for Class 1 to 12","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/mcq-questions.com\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"ImageObject","@id":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/#primaryimage","inLanguage":"en-US","url":"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1","contentUrl":"https:\/\/i0.wp.com\/mcq-questions.com\/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:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/#webpage","url":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/","name":"NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 - MCQ Questions","isPartOf":{"@id":"https:\/\/mcq-questions.com\/#website"},"primaryImageOfPage":{"@id":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/#primaryimage"},"datePublished":"2022-03-29T08:30:55+00:00","dateModified":"2022-03-29T09:10:39+00:00","author":{"@id":"https:\/\/mcq-questions.com\/#\/schema\/person\/4ba9570f32f2057e70e670c7885e47f3"},"breadcrumb":{"@id":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/mcq-questions.com\/"},{"@type":"ListItem","position":2,"name":"NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2"}]},{"@type":"Person","@id":"https:\/\/mcq-questions.com\/#\/schema\/person\/4ba9570f32f2057e70e670c7885e47f3","name":"Prasanna","image":{"@type":"ImageObject","@id":"https:\/\/mcq-questions.com\/#personlogo","inLanguage":"en-US","url":"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g","caption":"Prasanna"},"url":"https:\/\/mcq-questions.com\/author\/prasanna\/"}]}},"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/posts\/30984"}],"collection":[{"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/users\/9"}],"replies":[{"embeddable":true,"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/comments?post=30984"}],"version-history":[{"count":3,"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/posts\/30984\/revisions"}],"predecessor-version":[{"id":37738,"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/posts\/30984\/revisions\/37738"}],"wp:attachment":[{"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/media?parent=30984"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/categories?post=30984"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/tags?post=30984"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}