{"id":29967,"date":"2022-03-29T12:00:54","date_gmt":"2022-03-29T06:30:54","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=29967"},"modified":"2022-03-29T12:35:24","modified_gmt":"2022-03-29T07:05:24","slug":"ncert-solutions-for-class-12-maths-chapter-5-ex-5-8","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-5-ex-5-8\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 5 Continuity and Differentiability Ex 5.8 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-5-ex-5-8\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.8<\/h2>\n

\"NCERT<\/p>\n

Ex 5.7 Class 12 Maths Ncert Solutions Question 1.<\/strong>
\nVerify Rolle\u2019s theorem for the function
\nf(x) = x\u00b2 + 2x – 8, x \u2208 [- 4, 2]
\nSolution:
\nNow f(x) = x\u00b2 + 2x – 8 is a polynomial
\n\u2234 It is continuous and derivable in its domain x \u2208 R.
\nHence it is continuous in the interval [- 4, 2] and derivable in the interval (- 4, 2)
\nf(-4) = (- 4)\u00b2 + 2(- 4) – 8 = 16 – 8 – 8 = 0,
\nf(2) = 2\u00b2 + 4 – 8 = 8 – 8 = 0
\nConditions of Rolle\u2019s theorem are satisfied.
\nf'(x) = 2x + 2
\n\u2234 f’ (c) = 2c + 2 = 0
\nor c = – 1, c = – 1 \u2208 [- 4, 2]
\nThus f’ (c) = 0 at c = – 1.<\/p>\n

Ex 5.7 Class 12 NCERT Solutions Question 2.<\/strong>
\nExamine if Rolle\u2019s theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle\u2019s theorem from these example?
\n(i) f(x) = [x] for x \u2208 [5, 9]
\n(ii) f (x) = [x] for x \u2208 [-2, 2]
\n(iii) f (x) = x\u00b2 – 1 for x \u2208 [1, 2]
\nSolution:
\n(i) In the interval [5, 9], f (x) = [x] is neither continuous nor derivable at x = 6, 7, 8 Hence Rolle\u2019s theorem is not applicable
\n(ii) f (x) = [x] is not continuous and derivable at – 1, 0, 1. Hence Rolle\u2019s theorem is not applicable.
\n(iii) f(x) = (x\u00b2 – 1), f(1) = 1 – 1 = 0,
\nf(2) = 22 – 1 = 3
\nf(a) \u2260 f(b)
\nThough it is continous and derivable in the interval [1,2].
\nRolle\u2019s theorem is not applicable.
\nIn case of converse if f (c) = 0, c \u2208 [a, b] then conditions of rolle\u2019s theorem are not true.
\n(i) f (x) = [x] is the greatest integer less than or equal to x.
\n\u2234 f(x) = 0, But fis neither continuous nor differentiable in the interval [5, 9].<\/p>\n

(ii) Here also, theough f (x) = 0, but f is neither continuous nor differentiable in the interval [- 2, 2].<\/p>\n

(iii) f (x) = x\u00b2 – 1, f'(x) = 2x. Here f'(x) is not zero in the [1, 2], So f (2) \u2260 f’ (2).<\/p>\n

Question 3.
\nIf f: [- 5, 5] \u2192 R is a differentiable function and if f (x) does not vanish anywhere then prove that f (- 5) \u2260 f (5).
\nSolution:
\nFor Rolle\u2019s theorem
\nIf (i) f is continuous in [a, b]
\n(ii) f is derivable in [a, b]
\n(iii) f (a) = f (b)
\nthen f’ (c) = 0, c \u2208 (a, b)
\n\u2234 f is continuous and derivable
\ni.e., f'(c) \u2260 0. Hence \\(\\frac{f(5)-f(-5)}{10}\\)
\nbut f (c) \u2260 0 \u21d2 f(a) \u2260 f(b)
\n\u21d2 f(-5) \u2260 f(5)<\/p>\n

Question 4.
\nVerify Mean Value Theorem, if
\nf (x) = x\u00b2 – 4x – 3 in the interval [a, b], where a = 1 and b = 4.
\nSolution:
\nf (x) = x\u00b2 – 4x – 3. It being a polynomial it is continuous in the interval [1, 4] and derivable in (1,4), So all the condition of mean value theorem hold.
\nthen f’ (x) = 2x – 4,
\nf'(c) = 2c – 4
\nf(4) = 16 – 16 – 3 = – 3,
\nf(1) = 1 – 4 – 3 = – 6
\n\u2234 f'(c) = 0 \\(\\frac{f(b)-f(a)}{b-a}\\) = \\(\\frac{f(4)-f(1)}{4-1}\\)
\n\u21d2 2c – 4 = \\(\\frac{-3-6}{4-1}\\)
\n\u21d2 2c – 4 = 1 \u21d2 c = \\(\\frac{5}{2}\\) \u2208 (1, 4)
\n\u2234 Mean Value Theorem is verified for f(x) on (1, 4)<\/p>\n

Question 5.
\nVerify Mean Value Theorem, if f (x) = x3<\/sup>\u00a0– 5x2<\/sup> – 3x in the interval [a, b], where a = 1 and b = 3. Find all c \u2208 (1, 3) for which f’ (c) = 0.
\nSolution:
\nf (x) = x3<\/sup>\u00a0– 5x2<\/sup> – 3x
\nf'(x) = 3x\u00b2 – 10x – 3
\nSince f'(x’) exists, f(x) is continous on [1, 3]
\nf(x) is differentiable on (1, 3)
\nf'(c) = 3c\u00b2 – 10c – 3
\nf(b) = f(3) = – 27
\nf(a) = f(1) = – 7
\n\u2234 f'(c) = 0 \\(\\frac{f(b)-f(a)}{b-a}\\)
\n\u21d2 3c\u00b2 – 10c – 3 = \\(\\frac{-27-7}{3-1}\\)
\n\u21d2 3c\u00b2 – 10c – 3 = – 10
\n\u21d2 3c\u00b2 – 10c + 7 = 0
\n\u21d2 (c – 1)(3c – 7) = 0 \u21d2 c = 1 or c = \\(\\frac{7}{3}\\)
\n\\(\\frac{7}{3}\\) \u2208 (1, 3)
\n\u2234 Mean Value Theorem is verified for f(x) on (1, 3)<\/p>\n

Question 6.
\nExamine the applicability of Mean Value theroem for all three functions given in the above Question 2.
\nSolution:
\n(i) F (x)= [x] for x \u2208 [5, 9], f (x) = [x] in the interval [5, 9] is neither continuous, nor differentiable.
\n(ii) f (x) = [x], for x \u2208 [-2, 2],
\nAgain f (x) = [x] in the interval [-2, 2] is neither continous, nor differentiable.
\n(iii) f(x) = x\u00b2 – 1 for x \u2208 [1,2], It is a polynomial.
\nTherefore it is continuous in the interval [1,2] and differentiable in the interval (1,2)
\nf (x) = 2x, f(1) = 1 – 1 = 0 ,
\nf(2) = 4 – 1 = 3, f'(c) = 2c
\n\u2234 f'(c) = 0 \\(\\frac{f(b)-f(a)}{b-a}\\)
\n2c = \\(\\frac{3-0}{2-1}\\) = \\(\\frac{3}{1}\\)
\n\u2234 c = \\(\\frac{3}{2}\\) which belong to (1, 2)<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-5-ex-5-8\/ NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.8 Ex 5.7 Class 12 Maths Ncert Solutions Question 1. Verify Rolle\u2019s theorem for the function f(x) …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8<\/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 5 Continuity and Differentiability Ex 5.8 - 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-5-ex-5-8\/\" \/>\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 5 Continuity and Differentiability Ex 5.8 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-5-ex-5-8\/ NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.8 Ex 5.7 Class 12 Maths Ncert Solutions Question 1. 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