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>\nNCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.8<\/h2>\n <\/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>\nEx 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>\nQuestion 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 \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n