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NCERT Solutions For Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2

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Here, Below you all know about NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2 Question Answer. I know many of you confuse about finding Chapter 5 Continuity and Differentiability Ex 5.2 Of Class 12 NCERT Solutions. So, Read the full post below and get your solutions.

TextbookNCERT
BoardCBSE
CategoryNCERT Solutions
ClassClass 12
SubjectMaths
ChapterChapter 5
ExerciseClass 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.2
Number of Questions Solved10
NCERT Solutions For Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2

Table of Contents

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NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2

NCERT TEXTBOOK EXERCISES

Ex 5.2 Class 12 Maths 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)

Ex 5.2 Class 12 Maths 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, $\frac { dy }{ dx } =-sin(sinx)\times cosx$
$\frac { dy }{ dx } =-[sin(sinx)]cosx$

Ex 5.2 Class 12 Maths Question 3.
sin(ax+b)

Solution:

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

Ex 5.2 Class 12 Maths Question 4.
sec(tan(√x))

Solution:

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

Ex 5.2 Class 12 Maths Question 5.
$\ \frac { sin(ax+b) }{ cos(cx+d) }$

Solution:

y = $\ \frac { sin(ax+b) }{ cos(cx+d) }$= v/u
u = sin(ax+b)

Ex 5.2 Class 12 Maths Question 6.
cos x³ . sin²(x5) = y(say)

Solution:

Let u = cos x³ and v = sin² x5,
put x³ = t

Ex 5.2 Class 12 Maths Question 7.
$2\sqrt { cot({ x }^{ 2 }) } =y(say)$

Solution:

$2\sqrt { cot({ x }^{ 2 }) } =y(say)$

Ex 5.2 Class 12 Maths Question 8.
cos(√x) = y(say)

Solution:

cos(√x) = y(say)
$\frac { dy }{ dx } =\frac { d }{ dx } cos\left( \sqrt { x } \right) =-sin\sqrt { x } .\frac { d\sqrt { x } }{ dx }$
=$=-sin\sqrt { x } .\frac { 1 }{ 2 } { (x) }^{ -\frac { 1 }{ 2 } }=\frac { -sin\sqrt { x } }{ 2\sqrt { x } }$

Ex 5.2 Class 12 Maths 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
$f(x)=\begin{cases} x-1,\quad if\quad x\ge 1 \ 1-x,\quad if\quad x<1 \end{cases}$
$R.H.D\quad at\quad x=1\quad =\underset { h\rightarrow 0 }{ lim } \frac { f(1+h)-f(1) }{ h }$

Ex 5.2 Class 12 Maths 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 Exercise 5.2 PDF

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