{"id":30116,"date":"2022-03-29T12:00:50","date_gmt":"2022-03-29T06:30:50","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=30116"},"modified":"2022-03-29T12:32:05","modified_gmt":"2022-03-29T07:02:05","slug":"ncert-solutions-for-class-12-maths-chapter-5-ex-5-7","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-5-ex-5-7\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 5 Continuity and Differentiability Ex 5.7 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-5-ex-5-7\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.7<\/h2>\n

\"NCERT<\/p>\n

Ex 5.7 Class 12 Maths Ncert Solutions Question 1.<\/strong>
\nx\u00b2 + 3x + 2
\nSolution:
\nLet y = x\u00b2 + 3x + 2
\nDifferentiating both sides w.r.t. x
\n\\(\\frac { dy }{ dx }\\) = 2x + 3 and \\(\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \\) = 2<\/p>\n

Ex 5.7 Class 12 NCERT Solutions Question 2.<\/strong>
\nx20<\/sup> = y
\nSolution:
\nLet y = x20<\/sup>
\nDifferentiating both sides w.r.t. x
\n\\(\\frac { dy }{ dx } ={ 20 }x^{ 19 }\\quad \u21d2\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =20\\times { 19x }^{ 18 }={ 380 }x^{ 18 }\\qquad \\)<\/p>\n

Question 3.
\nx.cos x = y(say)
\nSolution:
\nLet y = x.cos x
\nDifferentiating both sides w.r.t. x
\n\\(\\frac { dy }{ dx } \\) = x(- sinx) + cosx.1 = – xsinx + cosx
\n\\(\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \\) = – xcosx – sinx – sinx = – xcosx – 2sinx<\/p>\n

Question 4.
\nlog x = y (say)
\nSolution:
\nLet y = log x
\nDifferentiating both sides w.r.t. x
\n\\(\\frac { dy }{ dx } =\\frac { 1 }{ x } \u21d2 \\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-\\frac { 1 }{ { x }^{ 2 } } \\)<\/p>\n

Question 5.
\nx\u00b3 log x = y (say)
\nSolution:
\nLet y = x\u00b3 log x
\nDifferentiating both sides w.r.t. x
\n\\( \u21d2\\frac { dy }{ dx } ={ x }^{ 3 }.\\frac { 1 }{ x } +logx\\times { 3x }^{ 2 }={ x }^{ 2 }+{ 3x }^{ 2 }logx \\)
\n\\(\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =2x+{ 3x }^{ 2 }.\\frac { 1 }{ x } +logx.6x=x(5+6logx) \\)<\/p>\n

Question 6.
\nex<\/sup> sin 5x = y
\nSolution:
\n\"NCERT<\/p>\n

Question 7.
\ne6x<\/sup> cos3x
\nSolution:
\nLet y = e6x<\/sup> cos3x
\nDifferentiating both sides w.r.t. x
\n\"NCERT<\/p>\n

Question 8.
\ntan-1<\/sup> x
\nSolution:
\nLet y = tan-1<\/sup> x
\nDifferentiating both sides w.r.t. x
\n\\(\\frac { dy }{ dx } =\\frac { 1 }{ 1+{ x }^{ 2 } } \u21d2\\frac { { d }^{ 2y } }{ { dx }^{ 2 } } =\\frac { -2x }{ { ({ 1+x }^{ 2 }) }^{ 2 } } \\)<\/p>\n

Question 9.
\nlog(logx)
\nSolution:
\nLet y = log(logx)
\nDifferentiating both sides w.r.t. x
\n\"NCERT<\/p>\n

Question 10.
\nsin(log x)
\nSolution:
\nLet y = sin(log x)
\nDifferentiating both sides w.r.t. x
\n\"NCERT<\/p>\n

Question 11.
\nIf y = 5 cos x – 3 sin x, prove that \\(\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0\\)
\nSolution:
\nLet y = 5 cos x – 3 sin x
\nDifferentiating both sides w.r.t. x
\n\\(\\frac { dy }{ dx } \\) = – 5sinx – 3cosx
\n\\(\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \\) = – 5cosx +3sinx = – y
\n\\(\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \\) + y = 0
\nHence proved<\/p>\n

Question 12.
\nIf y = cos-1<\/sup> x, Find \\(\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \\) in terms of y alone.
\nSolution:
\nLet y = cos-1<\/sup> x
\nDifferentiating both sides w.r.t. x
\n\\(\\frac { dy }{ dx } = – { \\left( { 1-x }^{ 2 } \\right) }^{ -\\frac { 1 }{ 2 } }\\)
\n\\(\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\\frac { -cosy }{ { \\left( { sin }^{ 2 }y \\right) }^{ \\frac { 3 }{ 2 } } } = -coty\\quad { cosec }^{ 2 }y\\)<\/p>\n

Question 13.
\nIf y = 3 cos (log x) + 4 sin (log x), show that
\n\\({ x }^{ 2 }{ y }_{ 2 }+{ xy }_{ 1 } \\) + y = 0
\nSolution:
\nLet y = 3 cos (log x) + 4 sin (log x)
\nDifferentiating both sides w.r.t. x
\n\"NCERT<\/p>\n

Question 14.
\nIf A = Aemx<\/sup> + Benx<\/sup>, show that \\(\\frac{d^{2} y}{d x^{2}}-(m+n) \\frac{d y}{d x}+m n y\\) = 0
\nSolution:
\nLet y = Aemx<\/sup> + Benx<\/sup>
\nDifferentiating both sides w.r.t. x
\n\\(\\frac { d }{ dx }\\) = Aemx<\/sup> + Benx<\/sup>
\nDifferentiating both sides w.r.t. x
\n\\(\\frac{d^{2} y}{d x^{2}}\\) = Aemx<\/sup>(m) + Benx<\/sup>(n)
\n\\(\\frac{d^{2} y}{d x^{2}}-(m+n) \\frac{d y}{d x}+m n y\\)
\n= Aemx<\/sup> . (m\u00b2) + Benx<\/sup>(n\u00b2) – (m + n) [Aemx<\/sup>(m) + Benx<\/sup>(n)] + mny
\n= Am\u00b2emx<\/sup> + Bn\u00b2enx<\/sup> – [Am\u00b2emx<\/sup> – Bn\u00b2enx<\/sup> +mny
\n= – mn[Aemx<\/sup> + Benx<\/sup>] + mny
\n= – mny + mny = 0 = RHS<\/p>\n

Question 15.
\nIf y = 500e7x<\/sup> + 600e-7x<\/sup>, show that \\(\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \\) = 49y.
\nSolution:
\nLet y = 500e7x<\/sup> + 600e-7x<\/sup>
\nDifferentiating both sides w.r.t. x
\n\"NCERT<\/p>\n

Question 16.
\nIf ey<\/sup>(x+1) = 1, show that \\(\\frac{d^{2} y}{d x^{2}}=\\left(\\frac{d y}{d x}\\right)^{2}\\)
\nSolution:
\nLet y = ey<\/sup>(x+1)
\nDifferentiating both sides w.r.t. x
\n\"NCERT<\/p>\n

Question 17.
\nIf y = (tan-1<\/sup> x)\u00b2 show that (x\u00b2 + 1)\u00b2y2<\/sub> + 2x(x\u00b2 + 1)y1<\/sub> = 2
\nSolution:
\nLet y = (tan-1<\/sup> x)\u00b2
\nDifferentiating both sides w.r.t. x
\n\"<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-5-ex-5-7\/ NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.7 Ex 5.7 Class 12 Maths Ncert Solutions Question 1. x\u00b2 + 3x + 2 Solution: Let …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7<\/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 5.7 - 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-5-7\/\" \/>\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 5.7 - 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