NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3.
- Continuity and Differentiability Class 12 Ex 5.1
- Continuity and Differentiability Class 12 Ex 5.2
- Continuity and Differentiability Class 12 Ex 5.4
- Continuity and Differentiability Class 12 Ex 5.5
- Continuity and Differentiability Class 12 Ex 5.6
- Continuity and Differentiability Class 12 Ex 5.7
- Continuity and Differentiability Class 12 Ex 5.8
| 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 \(\\ \frac { dy }{ dx } \) in the following
Ex 5.3 Class 12 Maths Question 1.
2x + 3y = sinx
Solution:
2x + 3y = sinx
Differentiating w.r.t x,
\(2+3\frac { dy }{ dx } =cosx \)
=>\(\frac { dy }{ dx } =\frac { 1 }{ 3 } (cosx-2)\)
Ex 5.3 Class 12 Maths Question 2.
2x + 3y = siny
Solution:
2x + 3y = siny
Differentiating w.r.t x,
\(2+3.\frac { dy }{ dx } =cosy\frac { dy }{ dx } \)
=>\(\frac { dy }{ dx } =\frac { 2 }{ cosy-3 } \)
Ex 5.3 Class 12 Maths Question 3.
ax + by² = cosy
Solution:
ax + by² = cosy
Differentiate w.r.t. x,
\(a+2\quad by\quad \frac { dy }{ dx } =-siny\frac { dy }{ dx } \)
=>\(or\quad (2b+siny)\frac { dy }{ dx } =-a=>\frac { dy }{ dx } =-\frac { a }{ 2b+siny } \)
Ex 5.3 Class 12 Maths Question 4.
xy + y² = tan x + y
Solution:
xy + y² = tanx + y
Differentiating w.r.t. x,
Ex 5.3 Class 12 Maths Question 5.
x² + xy + y² = 100
Solution:
x² + xy + xy = 100
Differentiating w.r.t. x,
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
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
\(2\quad sin\quad y\frac { d\quad siny }{ dx } +(-sinxy)\frac { d(xy) }{ dx } =0\)
Ex 5.3 Class 12 Maths Question 8.
sin²x + cos²y = 1
Solution:
Given that
sin²x + cos²y = 1
Differentiating both sides, we get
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
\(y={ sin }^{ -1 }\left( \frac { 2tan\theta }{ { 1+tan }^{ 2 }\theta } \right) ={ sin }^{ -1 }(sin2\theta )=2\theta \)
\(y={ 2sin }^{ -1 }x\quad \therefore \frac { dy }{ dx } =\frac { 2 }{ 1+{ x }^{ 2 } } \)
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
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
\(y={ cos }^{ -1 }\left( \frac { 1-tan^{ 2 }\quad \theta }{ 1+{ tan }^{ 2 }\quad \theta } \right) ={ cos }^{ -1 }(cos2\theta )=2\theta \)
\(y={ 2tan }^{ -1 }x\quad \therefore \frac { dy }{ dx } =\frac { 2 }{ 1+{ x }^{ 2 } } \)
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
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
Ex 5.3 Class 12 Maths Question 14.
\(y=sin^{ -1 }\left( 2x\sqrt { 1-{ x }^{ 2 } } \right) ,-\frac { 1 }{ \sqrt { 2 } } <x<\frac { 1 }{ \sqrt { 2 } } \)
Solution:
\(y=sin^{ -1 }\left( 2x\sqrt { 1-{ x }^{ 2 } } \right) ,-\frac { 1 }{ \sqrt { 2 } } <x<\frac { 1 }{ \sqrt { 2 } } \)
put x = tanθ
we get
\(y=sin^{ -1 }\left( 2sin\quad \theta \sqrt { 1-{ x }^{ 2 } } \right) \)
\(y=sin^{ -1 }\left( 2sin\theta \quad cos\theta \right) \quad ={ sin }^{ -1 }(sin2\theta )\quad =2\theta \)
\(y=2sin^{ -1 }x\quad \therefore \frac { dy }{ dx } =\frac { 2 }{ \sqrt { { 1-x }^{ 2 } } } \)
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
\(y=sec^{ -1 }\left( \frac { 1 }{ { 2cos }^{ 2 }\theta -1 } \right) ={ sec }^{ -1 }\left( \frac { 1 }{ cos2\theta } \right) \)
\(y=sec^{ -1 }(sec2\theta )=2\theta ,\quad y=2{ cos }^{ -1 }x \)
\(\therefore \frac { dy }{ dx } =\frac { -2 }{ \sqrt { { 1-x }^{ 2 } } } \)
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