NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3

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Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Maths
Chapter	Chapter 5
Chapter Name	Continuity and Differentiability
Exercise	Ex 5.3
Number of Questions Solved	15
Category	NCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exc 5.3

Find in the following

Ex 5.3 Class 12 Maths Question 1.
2x + 3y = sinx
Solution:
2x + 3y = sinx
Differentiating w.r.t x,
$2+3frac { dy }{ dx } =cosx$
=> $frac { dy }{ dx } =frac { 1 }{ 3 } (cosx-2)$

Ex 5.3 Class 12 Maths Question 2.
2x + 3y = siny
Solution:
2x + 3y = siny
Differentiating w.r.t x,
$2+3.frac { dy }{ dx } =cosyfrac { dy }{ dx }$
=> $frac { dy }{ dx } =frac { 2 }{ cosy-3 }$

Ex 5.3 Class 12 Maths Question 3.
ax + by² = cosy
Solution:
ax + by² = cosy
Differentiate w.r.t. x,
$a+2quad byquad frac { dy }{ dx } =-sinyfrac { dy }{ dx }$
=> $orquad (2b+siny)frac { dy }{ dx } =-a=>frac { dy }{ dx } =-frac { a }{ 2b+siny } “><br><img loading=$

Ex 5.3 Class 12 Maths Question 4.
xy + y² = tan x + y
Solution:
xy + y² = tanx + y
Differentiating w.r.t. x,
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 4

Ex 5.3 Class 12 Maths Question 5.
x² + xy + y² = 100
Solution:
x² + xy + xy = 100
Differentiating w.r.t. x,
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 5

Ex 5.3 Class 12 Maths Question 6.
x³ + x²y + xy² + y³ = 81
Solution:
Given that
x³ + x²y + xy² + y³ = 81
Differentiating both sides we get
byjus class 12 maths Chapter 5 Continuity and Differentiability 6

Ex 5.3 Class 12 Maths Question 7.
sin² y + cos xy = π
Solution:
Given that
sin² y + cos xy = π
Differentiating both sides we get
$2quad sinquad yfrac { dquad siny }{ dx } +(-sinxy)frac { d(xy) }{ dx } =0$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 7

Ex 5.3 Class 12 Maths Question 8.
sin²x + cos²y = 1
Solution:
Given that
sin²x + cos²y = 1
Differentiating both sides, we get
byjus class 12 maths Chapter 5 Continuity and Differentiability 8

Ex 5.3 Class 12 Maths Question 9.
$y={ sin }^{ -1 }left( frac { 2x }{ { 1+x }^{ 2 } } right)$
Solution:
$y={ sin }^{ -1 }left( frac { 2x }{ { 1+x }^{ 2 } } right)$
put x = tanθ
$y={ sin }^{ -1 }left( frac { 2tantheta }{ { 1+tan }^{ 2 }theta } right) ={ sin }^{ -1 }(sin2theta )=2theta$
$y={ 2sin }^{ -1 }xquad therefore frac { dy }{ dx } =frac { 2 }{ 1+{ x }^{ 2 } }$

Ex 5.3 Class 12 Maths Question 10.
$y={ tan }^{ -1 }left( frac { { 3x-x }^{ 3 } }{ { 1-3x }^{ 2 } } right) ,-frac { 1 }{ sqrt { 3 } } <x<frac { 1 }{ sqrt { 3 } }$
Solution:
$y={ tan }^{ -1 }left( frac { { 3x-x }^{ 3 } }{ { 1-3x }^{ 2 } } right)$
put x = tanθ
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 10

Ex 5.3 Class 12 Maths Question 11.
$y={ cos }^{ -1 }left( frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } right) ,0<x<1$
Solution:
$y={ cos }^{ -1 }left( frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } right) ,0<x<1$
put x = tanθ
$y={ cos }^{ -1 }left( frac { 1-tan^{ 2 }quad theta }{ 1+{ tan }^{ 2 }quad theta } right) ={ cos }^{ -1 }(cos2theta )=2theta$
$y={ 2tan }^{ -1 }xquad therefore frac { dy }{ dx } =frac { 2 }{ 1+{ x }^{ 2 } }$

Ex 5.3 Class 12 Maths Question 12.
$y={ sin }^{ -1 }left( frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } right) ,0<x<1$
Solution:
$y={ sin }^{ -1 }left( frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } right) ,0<x<1$
put x = tanθ
we get
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 12

Ex 5.3 Class 12 Maths Question 13.
$y={ cos }^{ -1 }left( frac { 2x }{ 1+{ x }^{ 2 } } right) ,-1<x<1$
Solution:
$y={ cos }^{ -1 }left( frac { 2x }{ 1+{ x }^{ 2 } } right) ,-1<x<1$
put x = tanθ
we get
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 13

Ex 5.3 Class 12 Maths Question 14.
$y=sin^{ -1 }left( 2xsqrt { 1-{ x }^{ 2 } } right) ,-frac { 1 }{ sqrt { 2 } } <x<frac { 1 }{ sqrt { 2 } }$
Solution:
$y=sin^{ -1 }left( 2xsqrt { 1-{ x }^{ 2 } } right) ,-frac { 1 }{ sqrt { 2 } } <x<frac { 1 }{ sqrt { 2 } }$
put x = tanθ
we get
$y=sin^{ -1 }left( 2sinquad theta sqrt { 1-{ x }^{ 2 } } right)$
$y=sin^{ -1 }left( 2sintheta quad costheta right) quad ={ sin }^{ -1 }(sin2theta )quad =2theta$
$y=2sin^{ -1 }xquad therefore frac { dy }{ dx } =frac { 2 }{ sqrt { { 1-x }^{ 2 } } }$

Ex 5.3 Class 12 Maths Question 15.
$y=sin^{ -1 }left( frac { 1 }{ { 2x }^{ 2 }-1 } right) ,0<x<frac { 1 }{ sqrt { 2 } }$
Solution:
$y=sin^{ -1 }left( frac { 1 }{ { 2x }^{ 2 }-1 } right) ,0<x<frac { 1 }{ sqrt { 2 } }$
put x = tanθ
we get
$y=sec^{ -1 }left( frac { 1 }{ { 2cos }^{ 2 }theta -1 } right) ={ sec }^{ -1 }left( frac { 1 }{ cos2theta } right)$
$y=sec^{ -1 }(sec2theta )=2theta ,quad y=2{ cos }^{ -1 }x$
$therefore frac { dy }{ dx } =frac { -2 }{ sqrt { { 1-x }^{ 2 } } }$

All Chapter NCERT Solutions For Class12 Maths

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NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exc 5.3

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