20<\/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>\nQuestion 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 <\/p>\nQuestion 7. \ne6x<\/sup> cos3x \nSolution: \nLet y = e6x<\/sup> cos3x \nDifferentiating both sides w.r.t. x \n <\/p>\nQuestion 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>\nQuestion 9. \nlog(logx) \nSolution: \nLet y = log(logx) \nDifferentiating both sides w.r.t. x \n <\/p>\n
Question 10. \nsin(log x) \nSolution: \nLet y = sin(log x) \nDifferentiating both sides w.r.t. x \n <\/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>\nQuestion 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 <\/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>\nQuestion 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 <\/p>\nQuestion 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 <\/p>\nQuestion 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 \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n