{"id":30160,"date":"2022-03-29T12:00:39","date_gmt":"2022-03-29T06:30:39","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=30160"},"modified":"2022-03-29T12:37:56","modified_gmt":"2022-03-29T07:07:56","slug":"ncert-solutions-for-class-12-maths-chapter-5-miscellaneous-exercise","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-5-miscellaneous-exercise\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Miscellaneous Exercise"},"content":{"rendered":"

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

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

\"NCERT<\/p>\n

Question 1.
\n(3x\u00b2 -9x + 5)9<\/sup>
\nSolution:
\nLet y = (3x\u00b2 -9x + 5)9<\/sup>
\n\u2234 \\(\\frac { dy }{ dx }\\) = (3x\u00b2 -9x + 5)8<\/sup>. \\(\\frac { d }{ dx }\\)(3x\u00b2 -9x + 5)
\n= 9(3x\u00b2 -9x + 5)8<\/sup> (6x – 9)
\n= 27 (3x\u00b2 -9x + 5)8<\/sup> (2x – 3)<\/p>\n

Question 2.
\nsin\u00b3x + cos6<\/sup> x
\nSoL
\nLet y = sin\u00b3x + cos6<\/sup> x
\n\u2234 \\(\\frac { dy }{ dx }\\) = 3 sin\u00b2x . cosx + 6cos5x (- sinx)
\n= 3 sinx cosx (sinx – 2 cos4<\/sup>x)<\/p>\n

\"NCERT<\/p>\n

Question 3.
\n(5x)3cos2x<\/sup>
\nSolution:
\nLet y = (5x)3cos2x<\/sup>
\nTaking logarithmon both sides,
\n\u2234 log y = 3 cos 2x log 5x
\nDifferentiating both sides, w.r.t. x,
\n\"NCERT<\/p>\n

Question 4.
\nsin-1<\/sup> (x\\(\\sqrt{x}\\)), 0 \u2264 x \u2264 1
\nSolution:
\n\"NCERT<\/p>\n

Question 5.
\n\\(\\frac{\\cos ^{-1} \\frac{x}{2}}{\\sqrt{2 x+7}},-2<x<2\\)
\nSolution:
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

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\"NCERT<\/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

\"NCERT<\/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

\"NCERT<\/p>\n

Question 10.
\nxx<\/sup> + xa<\/sup> + ax<\/sup> + aa<\/sup> for some fixed a > 0 and x > 0.
\nSolution:
\n\"NCERT<\/p>\n

Question 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\"NCERT
\nTaking logarithm on bath sides,
\n\u2234 log v = x\u00b2log(x – 3)
\nDifferentiating both sides w.r.t. x,
\n\"NCERT<\/p>\n

Question 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\"NCERT<\/p>\n

\"NCERT<\/p>\n

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\"NCERT<\/p>\n

Question 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

\"NCERT<\/p>\n

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\"NCERT<\/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\"NCERT<\/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\"NCERT<\/p>\n

\"NCERT<\/p>\n

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\"NCERT
\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\"NCERT
\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

\"NCERT<\/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\"NCERT<\/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

\"NCERT
\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

\"NCERT<\/p>\n

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\"NCERT<\/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\"NCERT<\/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

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