Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.2, Exercise 5.3

Exercise 5.2

Differentiate the functions with respect to x in Questions 1 to 8.

Question 1.
sin(x² + 5)
Solution:
Let y = sin(x² + 5),
put x² + 5 = t
y = sint
t = x²+5
$\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx }$
$\frac { dy }{ dx } =cost.\frac { dt }{ dx } =cos({ x }^{ 2 }+5)\frac { d }{ dx } ({ x }^{ 2 }+5)$
= cos (x² + 5) × 2x
= 2x cos (x² + 5)

Question 2.
cos (sin x)
Solution:
let y = cos (sin x)
put sinx = t
∴ y = cost,
t = sinx
∴ $\frac { dy }{ dx } =-sin\quad t,\frac { dt }{ dx } =cos\quad x$
$\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } =(-sint)\times cosx$
Putting the value of t, $\frac { dy }{ dx } =-sin(sinx)\times cosx$
$\frac { dy }{ dx } =-[sin(sinx)]cosx$

Question 3.
sin(ax+b)
Solution:
let = sin(ax+b)
put ax+bx = t
∴ y = sint
t = ax+b
$\frac { dy }{ dt } =cost,\frac { dt }{ dx } =\frac { d }{ dx } (ax+b)=a$
$Now\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } =cost\times a=acos\quad t$
$\frac { dy }{ dx } =acos(ax+b)$

Question 4.
sec(tan(√x))
Solution:
let y = sec(tan(√x))
by chain rule
$\frac { dy }{ dx } =sec(tan\sqrt { x } )tan(tan\sqrt { x } )\frac { d }{ dx } (tan\sqrt { x } )$
$\frac { dy }{ dx } =sec(tan\sqrt { x } ).tan(tan\sqrt { x } ){ sec }^{ 2 }\sqrt { x } .\frac { 1 }{ 2\sqrt { x } }$

Question 5.
$\\ \frac { sin(ax+b) }{ cos(cx+d) }$
Solution:
y = $\\ \frac { sin(ax+b) }{ cos(cx+d) }$ = $\\ \frac { v }{ u }$
u = sin(ax+b)
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 5

Question 6.

cos x³ . sin²(x5) = y(say)
Solution:
Let u = cos x³ and v = sin² x5,
put x³ = t
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 6

Question 7.
$2\sqrt { cot({ x }^{ 2 }) } =y(say)$
Solution:
$2\sqrt { cot({ x }^{ 2 }) } =y(say)$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 7

Question 8.
cos(√x) = y(say)
Solution:
cos(√x) = y(say)
$\frac { dy }{ dx } =\frac { d }{ dx } cos\left( \sqrt { x } \right) =-sin\sqrt { x } .\frac { d\sqrt { x } }{ dx }$
$=-sin\sqrt { x } .\frac { 1 }{ 2 } { (x) }^{ -\frac { 1 }{ 2 } }=\frac { -sin\sqrt { x } }{ 2\sqrt { x } }$
vedantu class 12 maths Chapter 5 Continuity and Differentiability 8

Question 9.
Prove that the function f given by f (x) = |x – 1|,x ∈ R is not differential at x = 1.
Solution:
The given function may be written as
$f(x)=\begin{cases} x-1,\quad if\quad x\ge 1 \\ 1-x,\quad if\quad x<1 \end{cases}$
$R.H.D\quad at\quad x=1\quad =\underset { h\rightarrow 0 }{ lim } \frac { f(1+h)-f(1) }{ h }$

Question 10.
Prove that the greatest integer function defined by f (x)=[x], 0 < x < 3 is not differential at x = 1 and x = 2.
Solution:
(i) At x = 1
$R.H.D=\underset { h\rightarrow 0 }{ lim } \frac { f(1+h)-f(1) }{ h }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 10

Exercise 5.3

Excercise 5.3

Find $\\ \frac { dy }{ dx }$ in the following

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

Question 2.
2x + 3y = siny
Solution:
2x + 3y = siny
Differentiating w.r.t x,
$2+3.\frac { dy }{ dx } =cosy\frac { dy }{ dx }$
=> $\frac { dy }{ dx } =\frac { 2 }{ cosy-3 }$

Question 3.
ax + by² = cosy
Solution:
ax + by² = cosy
Differentiate w.r.t. x,
$a+2\quad by\quad \frac { dy }{ dx } =-siny\frac { dy }{ dx }$
=> $or\quad (2b+siny)\frac { dy }{ dx } =-a=>\frac { dy }{ dx } =-\frac { a }{ 2b+siny }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 3

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

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

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

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

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

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
$y={ sin }^{ -1 }\left( \frac { 2tan\theta }{ { 1+tan }^{ 2 }\theta } \right) ={ sin }^{ -1 }(sin2\theta )=2\theta$
$y={ 2sin }^{ -1 }x\quad \therefore \frac { dy }{ dx } =\frac { 2 }{ 1+{ x }^{ 2 } }$

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

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
$y={ cos }^{ -1 }\left( \frac { 1-tan^{ 2 }\quad \theta }{ 1+{ tan }^{ 2 }\quad \theta } \right) ={ cos }^{ -1 }(cos2\theta )=2\theta$
$y={ 2tan }^{ -1 }x\quad \therefore \frac { dy }{ dx } =\frac { 2 }{ 1+{ x }^{ 2 } }$

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

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

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
$y=sin^{ -1 }\left( 2sin\quad \theta \sqrt { 1-{ x }^{ 2 } } \right)$
$y=sin^{ -1 }\left( 2sin\theta \quad cos\theta \right) \quad ={ sin }^{ -1 }(sin2\theta )\quad =2\theta$
$y=2sin^{ -1 }x\quad \therefore \frac { dy }{ dx } =\frac { 2 }{ \sqrt { { 1-x }^{ 2 } } }$

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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