{"id":29891,"date":"2022-03-29T12:00:26","date_gmt":"2022-03-29T06:30:26","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=29891"},"modified":"2022-03-29T12:18:21","modified_gmt":"2022-03-29T06:48:21","slug":"ncert-solutions-for-class-12-maths-chapter-5-ex-5-3","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-5-ex-5-3\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 5 Continuity and Differentiability Ex 5.3 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-5-ex-5-3\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.3<\/h2>\n

\"NCERT<\/p>\n

Ex 5.3 Class 12 NCERT Solutions Question 1.<\/strong>
\n2x + 3y = sinx
\nSolution:
\n2x + 3y = sinx
\nDifferentiating w.r.t x,
\n\\(2+3\\frac { dy }{ dx } =cosx \\)
\n\u21d2 \\(\\frac { dy }{ dx } =\\frac { 1 }{ 3 } (cosx-2)\\)<\/p>\n

Exercise 5.3 Class 12 Maths Ncert Solutions In Hindi Question 2.<\/strong>
\n2x + 3y = siny
\nSolution:
\n2x + 3y = siny
\nDifferentiating w.r.t x,
\n\\(2+3.\\frac { dy }{ dx } =cosy\\frac { dy }{ dx } \\)
\n\u21d2 \\(\\frac { dy }{ dx } =\\frac { 2 }{ cosy-3 } \\)<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nax + by\u00b2 = cosy
\nSolution:
\nax + by\u00b2 = cosy
\nDifferentiate both sides w.r.t. x,
\n\"NCERT<\/p>\n

Question 4.
\nxy + y\u00b2 = tan x + y
\nSolution:
\nxy + y\u00b2 = tan x + y
\nDifferentiate both sides w.r.t. x,
\n\"NCERT<\/p>\n

Question 5.
\nx\u00b2 + xy + y\u00b2 = 100
\nSolution:
\nx\u00b2 + xy + y\u00b2 = 100
\nDifferentiate both sides w.r.t. x,
\n\"<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nx\u00b3 + x\u00b2y + xy\u00b2 + y\u00b3 = 81
\nSolution:
\nx\u00b3 + x\u00b2y + xy\u00b2 + y\u00b3 = 81
\nDifferentiate both sides w.r.t. x,
\n\"NCERT<\/p>\n

Question 7.
\nsin\u00b2 y + cos xy = \u03c0
\nSolution:
\nsin\u00b2 y + cos xy = \u03c0
\nDifferentiate both sides w.r.t. x,
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nsin\u00b2x + cos\u00b2y = 1
\nSolution:
\nGiven that
\nsin\u00b2x + cos\u00b2y = 1
\nDifferentiating both sides, we get
\n\"NCERT<\/p>\n

Question 9.
\ny = \\({ sin }^{ -1 }\\left( \\frac { 2x }{ { 1+x }^{ 2 } } \\right)\\)
\nSolution:
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 10.
\ny = \\({ tan }^{ -1 }\\left( \\frac { { 3x-x }^{ 3 } }{ { 1-3x }^{ 2 } } \\right) ,-\\frac { 1 }{ \\sqrt { 3 } } <x<\\frac { 1 }{ \\sqrt { 3 } } \\)
\nSolution:
\n\"NCERT<\/p>\n

Question 11.
\ny = \\({ cos }^{ -1 }\\left( \\frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \\right) ,0<x<1\\)
\nSolution:
\ny = \\({ cos }^{ -1 }\\left( \\frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \\right) ,0<x<1\\)
\nput x = tan\u03b8
\ny = \\({ cos }^{ -1 }\\left( \\frac { 1-tan^{ 2 }\\quad \\theta }{ 1+{ tan }^{ 2 }\\quad \\theta } \\right) ={ cos }^{ -1 }(cos2\\theta )=2\\theta\\)
\n= 2\u03b8 = 2 tan-1 x
\n<\/sup>i.e., y = 2 tan-1x
\n<\/sup>Differentiating w.r.t x, \\(\\frac{d y}{d x}=\\frac{2}{1+x^{2}}\\)<\/p>\n

\"NCERT<\/p>\n

Question 12.
\ny = \\({ sin }^{ -1 }\\left( \\frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \\right) ,0<x<1\\)
\nSolution:
\n\"NCERT<\/p>\n

Question 13.
\ny = \\({ cos }^{ -1 }\\left( \\frac { 2x }{ 1+{ x }^{ 2 } } \\right)\\), – 1 < x < 1
\nSolution:
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 14.
\ny = \\(sin^{ -1 }\\left( 2x\\sqrt { 1-{ x }^{ 2 } } \\right) ,-\\frac { 1 }{ \\sqrt { 2 } } <x<\\frac { 1 }{ \\sqrt { 2 } } \\)
\nSolution:
\n\"NCERT<\/p>\n

Question 15.
\ny = \\(sin^{ -1 }\\left( \\frac { 1 }{ { 2x }^{ 2 }-1 } \\right) ,0<x<\\frac { 1 }{ \\sqrt { 2 } } \\)
\nSolution:
\ny = \\(sin^{ -1 }\\left( \\frac { 1 }{ { 2x }^{ 2 }-1 } \\right) ,0<x<\\frac { 1 }{ \\sqrt { 2 } } \\)
\nput x = tan\u03b8
\nwe get
\ny = \\(sec^{ -1 }\\left( \\frac { 1 }{ { 2cos }^{ 2 }\\theta -1 } \\right) ={ sec }^{ -1 }\\left( \\frac { 1 }{ cos2\\theta } \\right) \\)
\ny = \\(sec^{ -1 }(sec2\\theta )=2\\theta ,\\quad 2{ cos }^{ -1 }x \\)
\ni.e., y = 2 cos-1\u00a0<\/sup>x
\nDifferentiating w.r.t.x,
\n\u2234 \\(\\frac{d y}{d x}=\\frac{-2}{\\sqrt{1-x^{2}}}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-5-ex-5-3\/ NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.3 Ex 5.3 Class 12 NCERT Solutions Question 1. 2x + 3y = sinx Solution: 2x + …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3<\/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 5.3 - 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-5-3\/\" \/>\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 5.3 - 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