#### Need Solution for R.D.Sharma Maths Class 12 Chapter 9 Differentiability Exercise 9.2 Question 12 Maths Textbook Solution.

Answer: $|\sin x|$ is not differentiable at $x=n\pi$ and $|\cos x|$ is differentiable everywhere.

Hint:

Check LHD at $x=n\pi$ is equal to RHD at $x=n\pi$ or not for $n$ is even and $n$ is odd respect for the function$|\sin x|$

Given:

Given functions are $|\sin x|$ and $|\cos x|$.

Solution:

Let $f\left ( x \right )=|\sin x|$

$= \begin{cases}-\sin x & , x

Case  1

For $x=n\pi$ (Where $n$ is even)

LHD at $x=n\pi$

\begin{aligned} &=\lim _{x \rightarrow n-1} \frac{f(x)-f(n \pi)}{x-n \pi} \\ &=\lim _{h \rightarrow 0} \frac{-\sin (n \pi-h)-\sin n \pi}{n \pi-h-n \pi} \\ &=\lim _{h \rightarrow 0} \frac{\sinh -0}{-h} \end{aligned}

LHD at $x=n\pi =-1$                                                                                                 … (i)

RHD at $x=n\pi,$

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\sin (n \pi+h)-\sin n \pi}{h} \\ &=\lim _{h \rightarrow 0} \frac{\sinh }{h} \end{aligned}

RHD at $x=n\pi =1$                                                                                                ....(iii)

From(i) and (ii)

LHD at $x=n\pi \neq RHD \: at\: x=n\pi$

Case II

For $x=n\pi$ (Where $n$ is odd)

LHD at $x=n\pi$

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-\sin (n \pi-h)-\sin n \pi}{-h} \\ &=\lim _{h \rightarrow 0} \frac{-\sinh }{-h} \end{aligned}

RHD at $x=n\pi =-1$                                                                        .....(iv)

From (iii) and (iv)

LHD at $x=n\pi \neq RHD \: at\: x=n\pi$

Thus, $f\left ( x \right )=|\sin x|$ is not differentiable at $x=n\pi$

Now we need to check $\cos |x|$ is differentiable or not

Let,

$g\left ( x \right )=\cos |x|$

$\cos \left ( -x \right )=\cos x$

$g\left ( x \right )=\cos \: x$

$g\left ( x \right )=\cos |x|$  is differentiable everywhere.