Class 12 ncert solution math exercise 5.2

Exercise 5.2(Continuity and differentiability)

Differentiate the functions with respect to x in Exercises 1 to 8.(Class 12 ncert solution math exercise 5.2)

Question 1:- $\sin(x^2+5)$

Solution: $\sin(x^2+5)$

Differentiate w.r.t. x

$\frac{d}{dx} \sin(x^2+5)$

$= \cos(x^2+5)\frac{d}{dx}(x^2+5)$

$= \cos(x^2+5).2x$

$= 2x.\cos(x^2+5)$

Question 2:- $\cos(\sin x)$

Solution:- $\cos(\sin x)$

Differentiate w.r.t. x

$\frac{d}{dx}\cos(\sin x) = -\sin(\sin x)\frac{d}{dx}\sin x$

$= -\sin(\sin x) .\cos x$

$= -\cos x.\sin(\sin x)$

Question 3:- $\sin(ax+b)$

Solution:- $\sin(ax+b)$

Differentiate w.r.t. x

$\frac{d}{dx}\sin(ax+b) = \cos(ax+b).\frac{d}{dx}(ax+b)$

$= \cos(ax+b).(a.1+0)$

$= a\cos(ax+b)$

Question 4:- $\sec(\tan(\sqrt{x}))$

Solution:- $\sec(\tan(\sqrt{x}))$

Differentiate with respect to x

$\frac{d}{dx}\sec(\tan(\sqrt{x}))$

$= \sec(\tan(\sqrt{x})\tan(\tan(\sqrt{x}))\frac{d}{dx}\tan(\sqrt{x})$

$= \sec(\tan(\sqrt{x})\tan(\tan(\sqrt{x}))\sec^2(\sqrt{x})\frac{d}{dx}x^{1/2}$

$= \sec(\tan(\sqrt{x})\tan(\tan(\sqrt{x}))\sec^2(\sqrt{x})\frac{1}{2\sqrt{x}}$

$= \frac{1}{2\sqrt{x}}\sec^2(\sqrt{x}) \sec(\tan(\sqrt{x})\tan(\tan(\sqrt{x}))$

Question 5:- $\frac{\sin(ax+b)}{\cos(cx+d)}$

Solution: $\frac{\sin(ax+b)}{\cos(cx+d)}$

Differentiate with respect to x

$\frac{d}{dx}\left(\frac{\sin(ax+b)}{\cos(cx+d)}\right)$

$= \frac{\cos(cx+d).\frac{d}{dx}\sin(ax+b) - \sin(ax+b)\cos(cx+d)}{\cos^2(cx+d)}$

$= \frac{\cos(cx+d)\cos(ax+b)\frac{d}{dx}(ax+b)-\sin(ax+b)[-\sin(cx+d)].\frac{d}{dx}(cx+d)}{\cos^2(cx+d)}$

$= \frac{\cos(cx+d)\cos(ax+b).a-\sin(ax+b)[-\sin(cx+d)].c}{\cos^2(cx+d)}$

$= \frac{a.\cos(cx+d).\cos(ax+b)}{\cos^2(cx+d)}+\frac{c.\sin(ax+b).\sin(cx+d)}{\cos^2(cx+d)}$

$= a.\cos(ax+b).\sec(cx+d)+ c.\cos(ax+b).\tan(cx+d).\sec(cx+d)$

Question 6:- $\cos x^3.\sin^2(x^5)$

Solution: $\cos x^3.\sin^2(x^5)$

Differentiate with respect to x

$\frac{d}{dx}\cos x^3.\sin^2(x^5)$

$= \cos x^3\frac{d}{dx}\sin^2(x^5)+ \sin^2(x^5)\frac{d}{dx}\cos x^3$

$= \cos x^3.2\sin x^5\frac{d}{dx}\sin x^5+\sin^2(x^5).[-\sin x^3].\frac{d}{dx}x^3$

$= \cos x^3.2\sin x^5\cos x^5\frac{d}{dx}x^5 - \sin^2(x^5)\sin x^3.3x^2$

$= = \cos x^3.2\sin x^5\cos x^5.5x^4 - \sin^2(x^5)\sin x^3.3x^2$

$= 10x^4\sin x^5.\cos x^5.\cos x^3-3x^2\sin x^3.\sin^2(x^5)$

Question 7:- $2\sqrt{\cot(x^2) }$

Solution: $2\sqrt{\cot(x^2) }$

Differentiate with respect to x

$\frac{d}{dx}2\sqrt{\cot(x^2) }$

$= 2\frac{d}{dx}[\cot(x^2)]^{1/2}$

$= 2.\frac{1}{2[\cot(x^2)]^{1/2}}\frac{d}{dx}\cot(x^2)$

$= \frac{-\operatorname{cosec^2(x^2)}}{\sqrt{\cot(x^2)}}\frac{d}{dx}x^2$

$= \frac{-\operatorname{cosec^2(x^2)}}{\sqrt{\cot(x^2)}}.2x$

$= \frac{-2x}{\sin^2x^2.\sqrt{\cot(x^2)}}$

$= \frac{-2x}{\sin x^2.\sin x^2\sqrt{\frac{\cos x^2}{\sin x^2}}}$

$= \frac{-2x}{\sin x^2\sqrt{\sin x^2.\cos x^2}}$

$= \frac{-2\sqrt{2}x}{\sin x^2\sqrt{2\sin x^2\cos x^2}}$

$= \frac{-2\sqrt{2}x}{\sin x^2\sqrt{\sin 2x^2}}$

Question 8:- $\cos(\sqrt{x})$

Solution: $\cos(\sqrt{x})$

Differentiate with respect to x

$\frac{d}{dx}\cos(\sqrt{x})$

$= -\sin(\sqrt{x})\frac{d}{dx}x^{1/2}$

$= -\sin(\sqrt{x})\frac{1}{2\sqrt{x}}$

$= -\frac{\sin{x}}{2\sqrt{x}}$

Question 9: Prove that the function $f$ given by $f(x)=|x-1|, x \in \mathbf{R}$ is not differentiable at $x=1$ .

Solution: Given, $f(x)=|x-1|, x \in \mathbf{R}$

It is known that a function $f$ is differentiable at a point $x=c$ in

its domain if both
$\displaystyle \lim _{h \rightarrow 0^{-}} \frac{f(c)-f(c-h)}{h}$ and $\displaystyle\lim _{h \rightarrow 0^{+}} \frac{f(c+h)-f(c)}{h}$ are finite and equal.

To check the differentiability of the given function at $x=1$ ,

$LHD =\displaystyle \lim _{h \rightarrow 0^{-}} \frac{f(1)-f(1-h)}{h} =\lim _{h \rightarrow 0^{-}} \frac{f|1-1|-|1-h-1|}{h}$

$=\displaystyle \lim _{h \rightarrow 0^{-}} \frac{0-|h|}{h}$

$=\displaystyle \lim _{h \rightarrow 0^{-}} \frac{-h}{h} \quad(h<0 \Rightarrow|h|=-h)$

=-1

$RHD = \displaystyle \lim _{h \rightarrow 0^{+}} \frac{f(1+h)-f(1)}{h} =\lim _{h \rightarrow 0^{+}} \frac{f|1+h-1|-|1-1|}{h}$

$=\displaystyle \lim _{h \rightarrow 0^{+}} \frac{|h|-0}{h}$

$=\displaystyle \lim _{h \rightarrow 0^{+}} \frac{h}{h} \quad(h>0 \Rightarrow|h|=h)$

Since LHD and RHD at x=1 are not equal,

Therefore, f is not differentiable at $x=1$ .

Question 10: Prove that the greatest integer function defined by $f(x)=[x], 0<x<3$ is not differentiable at $x=1$ and $x=2$

Solution: Given, $f(x)=[x], 0<x<3$

It is known that a function $f$ is differentiable at a point $x=c$ in its domain if both $\lim _{h \rightarrow 0^{-}} \frac{f(c)-f(c-h)}{h}$ and $\lim _{h \rightarrow 0^{+}} \frac{f(c+h)-f(c)}{h}$ are finite and equal.
At $x=1$ ,