{"id":93552,"date":"2019-09-18T10:59:42","date_gmt":"2019-09-18T05:29:42","guid":{"rendered":"https:\/\/www.learncbse.in\/?p=93552"},"modified":"2019-09-18T10:59:42","modified_gmt":"2019-09-18T05:29:42","slug":"ncert-solutions-for-class-12-maths-chapter-5-ex-5-3","status":"publish","type":"post","link":"https:\/\/www.learncbse.in\/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":"
Get Free NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Continuity and Differentiability Exercise 5.3 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 5 Class 12 Continuity and Differentiability Ex 5.3 provided in NCERT Textbook.<\/p>\n
Free download NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.<\/p>\n
The topics and sub-topics included in the Continuity and Differentiability chapter are the following:<\/strong><\/p>\n There are total eight exercises and one misc exercise(144 Questions fully solved<\/b>) in the class 12th maths chapter 5 Continuity and Differentiability.<\/p>\n Find \\(\\\\ \\frac { dy }{ dx } \\) in the following<\/strong><\/p>\n Ex 5.3 Class 12 Maths Question 1.<\/strong><\/span> Ex 5.3 Class 12 Maths\u00a0Question 2.<\/strong><\/span> Ex 5.3 Class 12 Maths\u00a0Question 3.<\/strong><\/span> Ex 5.3 Class 12 Maths\u00a0Question 4.<\/strong><\/span> Ex 5.3 Class 12 Maths\u00a0Question 5.<\/strong><\/span> Ex 5.3 Class 12 Maths\u00a0Question 6.<\/strong><\/span> Ex 5.3 Class 12 Maths\u00a0Question 7.<\/strong><\/span> Ex 5.3 Class 12 Maths\u00a0Question 8.<\/strong><\/span> Ex 5.3 Class 12 Maths\u00a0Question 9.<\/strong><\/span> Ex 5.3 Class 12 Maths\u00a0Question 10.<\/strong><\/span> Ex 5.3 Class 12 Maths\u00a0Question 11.<\/strong><\/span> Ex 5.3 Class 12 Maths\u00a0Question 12.<\/strong><\/span> Ex 5.3 Class 12 Maths\u00a0Question 13.<\/strong><\/span> Ex 5.3 Class 12 Maths\u00a0Question 14.<\/strong><\/span> Ex 5.3 Class 12 Maths\u00a0Question 15.<\/strong><\/span> NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Get Free NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Continuity and Differentiability Exercise 5.3 Class 12 Maths […]<\/p>\n","protected":false},"author":30,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":""},"categories":[2],"tags":[],"yoast_head":"\n\n
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3<\/h3>\n
\n2x + 3y = sinx
\nSolution:<\/strong><\/span>
\n2x + 3y = sinx
\nDifferentiating w.r.t x,
\n\\(2+3\\frac { dy }{ dx } =cosx \\)
\n=>\\(\\frac { dy }{ dx } =\\frac { 1 }{ 3 } (cosx-2)\\)<\/p>\n
\n2x + 3y = siny
\nSolution:<\/strong><\/span>
\n2x + 3y = siny
\nDifferentiating w.r.t x,
\n\\(2+3.\\frac { dy }{ dx } =cosy\\frac { dy }{ dx } \\)
\n=>\\(\\frac { dy }{ dx } =\\frac { 2 }{ cosy-3 } \\)<\/p>\n
\nax + by\u00b2 = cosy
\nSolution:<\/strong><\/span>
\nax + by\u00b2 = cosy
\nDifferentiate w.r.t. x,
\n\\(a+2\\quad by\\quad \\frac { dy }{ dx } =-siny\\frac { dy }{ dx } \\)
\n=>\\(or\\quad (2b+siny)\\frac { dy }{ dx } =-a=>\\frac { dy }{ dx } =-\\frac { a }{ 2b+siny } \\)
\n<\/p>\n
\nxy + y\u00b2 = tan x + y
\nSolution:<\/strong><\/span>
\nxy + y\u00b2 = tanx + y
\nDifferentiating w.r.t. x,
\n<\/p>\n
\nx\u00b2 + xy + y\u00b2 = 100
\nSolution:<\/strong><\/span>
\nx\u00b2 + xy + xy = 100
\nDifferentiating w.r.t. x,
\n<\/p>\n
\nx\u00b3 + x\u00b2y + xy\u00b2 + y\u00b3 = 81
\nSolution:<\/strong><\/span>
\nGiven that
\nx\u00b3 + x\u00b2y + xy\u00b2 + y\u00b3 = 81
\nDifferentiating both sides we get
\n<\/p>\n
\nsin\u00b2 y + cos xy = \u03c0
\nSolution:<\/strong><\/span>
\nGiven that
\nsin\u00b2 y + cos xy = \u03c0
\nDifferentiating both sides we get
\n\\(2\\quad sin\\quad y\\frac { d\\quad siny }{ dx } +(-sinxy)\\frac { d(xy) }{ dx } =0\\)
\n<\/p>\n
\nsin\u00b2x + cos\u00b2y = 1
\nSolution:<\/strong><\/span>
\nGiven that
\nsin\u00b2x + cos\u00b2y = 1
\nDifferentiating both sides, we get
\n<\/p>\n
\n\\(y={ sin }^{ -1 }\\left( \\frac { 2x }{ { 1+x }^{ 2 } } \\right) \\)
\nSolution:<\/strong><\/span>
\n\\(y={ sin }^{ -1 }\\left( \\frac { 2x }{ { 1+x }^{ 2 } } \\right) \\)
\nput x = tan\u03b8
\n\\(y={ sin }^{ -1 }\\left( \\frac { 2tan\\theta }{ { 1+tan }^{ 2 }\\theta } \\right) ={ sin }^{ -1 }(sin2\\theta )=2\\theta \\)
\n\\(y={ 2sin }^{ -1 }x\\quad \\therefore \\frac { dy }{ dx } =\\frac { 2 }{ 1+{ x }^{ 2 } } \\)<\/p>\n
\n\\(y={ tan }^{ -1 }\\left( \\frac { { 3x-x }^{ 3 } }{ { 1-3x }^{ 2 } } \\right) ,-\\frac { 1 }{ \\sqrt { 3 } } <x<\\frac { 1 }{ \\sqrt { 3 } } \\)
\nSolution:<\/strong><\/span>
\n\\(y={ tan }^{ -1 }\\left( \\frac { { 3x-x }^{ 3 } }{ { 1-3x }^{ 2 } } \\right) \\)
\nput x = tan\u03b8
\n<\/p>\n
\n\\(y={ cos }^{ -1 }\\left( \\frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \\right) ,0<x<1 \\)
\nSolution:<\/strong><\/span>
\n\\(y={ cos }^{ -1 }\\left( \\frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \\right) ,0<x<1 \\)
\nput x = tan\u03b8
\n\\(y={ cos }^{ -1 }\\left( \\frac { 1-tan^{ 2 }\\quad \\theta }{ 1+{ tan }^{ 2 }\\quad \\theta } \\right) ={ cos }^{ -1 }(cos2\\theta )=2\\theta \\)
\n\\(y={ 2tan }^{ -1 }x\\quad \\therefore \\frac { dy }{ dx } =\\frac { 2 }{ 1+{ x }^{ 2 } } \\)<\/p>\n
\n\\(y={ sin }^{ -1 }\\left( \\frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \\right) ,0<x<1\\)
\nSolution:<\/strong><\/span>
\n\\(y={ sin }^{ -1 }\\left( \\frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \\right) ,0<x<1\\)
\nput x = tan\u03b8
\nwe get
\n<\/p>\n
\n\\(y={ cos }^{ -1 }\\left( \\frac { 2x }{ 1+{ x }^{ 2 } } \\right) ,-1<x<1 \\)
\nSolution:<\/strong><\/span>
\n\\(y={ cos }^{ -1 }\\left( \\frac { 2x }{ 1+{ x }^{ 2 } } \\right) ,-1<x<1 \\)
\nput x = tan\u03b8
\nwe get
\n
\n<\/p>\n
\n\\(y=sin^{ -1 }\\left( 2x\\sqrt { 1-{ x }^{ 2 } } \\right) ,-\\frac { 1 }{ \\sqrt { 2 } } <x<\\frac { 1 }{ \\sqrt { 2 } } \\)
\nSolution:<\/strong><\/span>
\n\\(y=sin^{ -1 }\\left( 2x\\sqrt { 1-{ x }^{ 2 } } \\right) ,-\\frac { 1 }{ \\sqrt { 2 } } <x<\\frac { 1 }{ \\sqrt { 2 } } \\)
\nput x = tan\u03b8
\nwe get
\n\\(y=sin^{ -1 }\\left( 2sin\\quad \\theta \\sqrt { 1-{ x }^{ 2 } } \\right) \\)
\n\\(y=sin^{ -1 }\\left( 2sin\\theta \\quad cos\\theta \\right) \\quad ={ sin }^{ -1 }(sin2\\theta )\\quad =2\\theta \\)
\n\\(y=2sin^{ -1 }x\\quad \\therefore \\frac { dy }{ dx } =\\frac { 2 }{ \\sqrt { { 1-x }^{ 2 } } } \\)<\/p>\n
\n\\(y=sin^{ -1 }\\left( \\frac { 1 }{ { 2x }^{ 2 }-1 } \\right) ,0<x<\\frac { 1 }{ \\sqrt { 2 } } \\)
\nSolution:<\/strong><\/span>
\n\\(y=sin^{ -1 }\\left( \\frac { 1 }{ { 2x }^{ 2 }-1 } \\right) ,0<x<\\frac { 1 }{ \\sqrt { 2 } } \\)
\nput x = tan\u03b8
\nwe get
\n\\(y=sec^{ -1 }\\left( \\frac { 1 }{ { 2cos }^{ 2 }\\theta -1 } \\right) ={ sec }^{ -1 }\\left( \\frac { 1 }{ cos2\\theta } \\right) \\)
\n\\(y=sec^{ -1 }(sec2\\theta )=2\\theta ,\\quad y=2{ cos }^{ -1 }x \\)
\n\\(\\therefore \\frac { dy }{ dx } =\\frac { -2 }{ \\sqrt { { 1-x }^{ 2 } } } \\)<\/p>\nNCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Hindi Medium Ex 5.3<\/h3>\n
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\n<\/p>\nNCERT Class 12 Maths Solutions<\/h3>\n
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