-1<\/sup> (x\\(\\sqrt{x}\\)), 0 \u2264 x \u2264 1 \nSolution: \n <\/p>\nQuestion 5. \n\\(\\frac{\\cos ^{-1} \\frac{x}{2}}{\\sqrt{2 x+7}},-2<x<2\\) \nSolution: \n <\/p>\n
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Question 6. \n\\(\\cot ^{-1}\\left[\\frac{\\sqrt{1+\\sin x}+\\sqrt{1-\\sin x}}{\\sqrt{1+\\sin x}-\\sqrt{1-\\sin x}}\\right], 0<x<\\frac{\\pi}{2}\\) \nSolution: \n <\/p>\n
Question 7. \\((\\log x)^{\\log x}, x>1\\) \nSolution: \nLet y = \\((\\log x)^{\\log x}\\) \nTaking logarithmon both sides, \n\u2234 log y = log x(log log x) \nDifferentiating both sides, w.r.t. x, \n\\(\\frac { 1 }{ y }\\) \\(\\frac { dy }{ dx }\\) = logx.\\(\\frac { 1 }{ log x }\\).\\(\\frac { 1 }{ x }\\) + \\(\\frac { 1 }{ x }\\).log log x \n\u2234 \\(\\frac { dy }{ dx }\\) = (log x)log x<\/sup> [\\(\\frac { 1 }{ x }\\) + \\(\\frac { log log.x }{ x }\\)]<\/p>\n <\/p>\n
Question 8. \ncos (a cos x + b sin x), for sorne constant a and b. \nSolution: \nLet y = cos (a cosx + b sinx) \n\\(\\frac { dy }{ dx }\\) = sin (a cosx + h sinx). \n\\(\\frac { dy }{ dx }\\)(a cosx + b sinx) \n= – sin (a cosx + b sinx) [- a sinx + b cosx] \n= (a sinx – b cosx).sin (a cosx + b sinx)<\/p>\n
Question 9. \n(sin x – cos x)sin x-cos x<\/sup>, \\(\\frac { \u03c0 }{ 4 }\\)< x < \\(\\frac { 3\u03c0 }{ 4 }\\) \nSolution: \nWhen \\(\\frac { \u03c0 }{ 4 }\\)< x < \\(\\frac { 3\u03c0 }{ 4 }\\), then sin x > cos x \nso that sin x – cos x is positive. \nLet y = (sinx – cosx)sin x-cos x<\/sup> \nTaking logarithm on both sides, \n\u2234 logy = (sinx – cosx) log (sinx – cosx) \nDifferentiating both sides w.r.t. x, \n\\(\\frac { 1 }{ y }\\) \\(\\frac { dy }{ dx }\\) \n=( sinx – cosx)\\(\\frac { 1 }{ sin x-cosx }\\) .\\(\\frac { d }{ dx }\\)(sinx-cosx) + log (sinx – cosx). (cosx + sinx) \n= (cosx + sinx) + log (sinx – cosx) . (cosx + sinx) \n= (cosx + sinx) [1 + log (sinx – cosx)] \n\u2234 \\(\\frac { dy }{ dx }\\) = (sin x – cos x)sin x-cos x<\/sup>(cosx + sinx) [1 + log(sinx – cosx)]<\/p>\n <\/p>\n
Question 10. \nxx<\/sup> + xa<\/sup> + ax<\/sup> + aa<\/sup> for some fixed a > 0 and x > 0. \nSolution: \n <\/p>\nQuestion 11. \n\\(x^{x^{2}-3}+(x-3)^{x^{2}}\\), for x > 3 \nSolution: \nLet u = x\\(x^{x^{2}-3}\\) and v = (x – 3)x\u00b2<\/sup> \nDifferentiating w.r.t. x, \n\u2234 \\(\\frac { dy }{ dx }\\) = \\(\\frac { du }{ dx }\\) + \\(\\frac { dv }{ dx }\\) … (1) \nu = xx\u00b2-3<\/sup> \nTaking logarithm on bath sides, \n\u2234 log u = (x\u00b2 – 3)log \nDifferentiating both sides w.r.t. x, \n \nTaking logarithm on bath sides, \n\u2234 log v = x\u00b2log(x – 3) \nDifferentiating both sides w.r.t. x, \n <\/p>\nQuestion 12. \nFind \\(\\frac { dy }{ dx }\\), if y = 12(1 – cos t), \nx = 10(t – sin t), \\(\\frac { – \u03c0 }{ 2 }\\) < t < \\(\\frac { \u03c0 }{ 2 }\\) \nSolution: \n <\/p>\n
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Question 13. \nFind \\(\\frac { dy }{ dx }\\), if \ny = sin-1<\/sup> x + sin-1<\/sup> \\(\\sqrt{1-x^{2}},-1 \\leq x \\leq 1\\). \nSolution: \n <\/p>\nQuestion 14. \nIf \\(x \\sqrt{1+y}+y \\sqrt{1+x}\\) = 0, for – 1 < x < 1, prove that \\(\\frac { dy }{ dx }\\) = – \\(\\frac{1}{(1+x)^{2}}\\) \nSolution: \n\\(x \\sqrt{1+y}+y \\sqrt{1+x}\\) = 0 \n\\(x \\sqrt{1+y}\\) = – y\\(\\sqrt{1+y}\\) \nSquaring both sides, \nx\u00b2(1 + y) = y\u00b2(1 + x) \nx\u00b2 + x\u00b2y = y\u00b2 + y\u00b2x \nx\u00b2 – y\u00b2 = y\u00b2x – x\u00b2y \n(x – y)(x + y) = xy(x – y) \n(x + y) = – xy \ny + xy = – x \ny(1 + x) = – x \n\u2234 y = \\(\\frac { – x }{ 1+x }\\) \nDifferentiating w.r.t. x, \n\\(\\frac { dy }{ dx }\\) = \\(\\frac{(1+x)(-1)-(-x) 1}{(1+x)^{2}}\\) \n= \\(\\frac{-1-x+x}{(1+x)^{2}}\\) = \\(\\frac{-1}{(1+x)^{2}}\\), x \u2260 – 1<\/p>\n
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Question 15. \nIf (x – a)\u00b2 + (y – b)\u00b2 = c\u00b2, for some c > 0, prove that \n\\(\\frac{\\left[1+\\left(\\frac{d y}{d x}\\right)^{2}\\right]^{\\frac{3}{2}}}{\\frac{d^{2} y}{d x^{2}}}\\) \nis a constant, independent of a and b. \nSolution: \n <\/p>\n
Question 16. \nIf cos y = x cos (a + y), with \ncos a \u2260 \u00b1 1, prove that \\(\\frac { dy }{ dx }\\) = \\(\\frac{\\cos ^{2}(a+y)}{\\sin a}\\) \nSolution: \n <\/p>\n
Question 17. \nIf x = a(cos t + t sin t) and y = a(sin t – t cos t), find \\(\\frac{d^{2} y}{d x^{2}}\\) \nSolution: \n <\/p>\n
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Question 18. \nIf f(x) = |x|\u00b3, show that f”(x) exists for all real x and find it. \nSolution: \nf(x) can be redefined as \nf(x) = \\(\\left\\{\\begin{aligned} \nx^{3}, & x \\geq 0 \\\\ \n-x^{3}, & x<0 \\end{aligned}\\right.\\) For x > 0 and x < 0, fix) is a polynomial function. Hence f(x) is differentiable for x > 0 and x < 0. \u2234 For x > 0, f'(x) = 3x\u00b2 and f”(x) = 6x \nFor x < 0, f'(x) = – 3x\u00b2 and f”(x) = – 6x \n \nor f”(x) = 6|x| exists for all x \u2208 R<\/p>\n
Question 19. \nUsing mathematical induction, prove that \\(\\frac { d }{ dx }\\)(x\u207f) = \\(n x^{n^{-1}}\\) for all positive integers n. \nSolution: \n \nHence P(k + 1) is true. \ni.e., P(k + 1) is true whenever P(k) is true. \nHence by the principle of mathematical induction, \\(\\frac { d }{ dx }\\)(x\u207f) = nxn-1<\/sup> is true for positive integer n.<\/p>\n <\/p>\n
Question 20. \nUsing the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines. \nSolution: \nsin (A + B) = sin A cos B + cos A sin B \nDifferentiating both sides w.r.t. x, \n <\/p>\n
Question 21. \nDoes there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer. \nSolution: \nYes. \ni. Let f(x) = |x – 1| + |x – 2| \nLet g(x) = |x|, h(x) = x – 1, k(x) = x – 2. \nThen g(x), h(x) and k(x) are continuous functions since h(x) and k(x) are polynomial functions and g(x) \u00a1s the modulus function.<\/p>\n
\nat x = 1, since Lf\u2019(1) \u2260 Rf\u2019(1) \nSimilarly we can show that f(x) \u00a1s not differentiable at x = 2. \nThus f(x) = |x – 1| + |x – 2| is continuous everywhere and not differentiable at exactly two points, namely at x = 1 or x = 2.<\/p>\n
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Question 22. \nIf y = \\(\\left|\\begin{array}{ccc} \nf(x) & g(x) & h(x) \\\\ \nl & m & n \\\\ \na & b & c \n\\end{array}\\right|\\), prove that \\(\\frac { dy }{ dx }\\) = \\(\\left|\\begin{array}{ccc} \nf^{\\prime}(x) & g^{\\prime}(x) & h^{\\prime}(x) \\\\ \nl & m & n \\\\ \na & b & c \n\\end{array}\\right|\\) \nSolution: \n <\/p>\n
Question 23. \nIf y = \\(e^{a \\cos ^{-1} x}\\), – 1 \u2264 x \u2264 1, show that (1 – x\u00b2)\\(\\frac{d^{2} y}{d x^{2}}-x \\frac{d y}{d x}-a^{2} y\\) = 0 \nSolution: \nf(x) = x + \\(\\frac { 1 }{ x }\\) \nf(x) is a continuous function in [1, 3] \nAx) is differentiable in(1,3) \nf\u2019(x) = 1 + \\(\\\\frac{-1}{x^{2}}\\) exists for x \u2208 (1, 3) \n <\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-5-miscellaneous-exercise\/ NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Miscellaneous Exercise Question 1. (3x\u00b2 -9x + 5)9 Solution: Let y = (3x\u00b2 -9x + 5)9 \u2234 = …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Miscellaneous Exercise<\/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 Miscellaneous Exercise - 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