please solve rd sharma class 12 chapter 9 Differentiability exercise Multiple choice question, question 7 maths textbook solution

: (b)

HINTS: Understand the  definition of continuity and differentiability

SOLUTION:

$f(x)=|\cos x|$

let

\begin{aligned} f(x) &=|x| \\ g(x) &=\cos x \\ h &=\operatorname{fog}(x) \\ &=f(g(x)) \\ &=f(\cos x) \\ &=|\cos x| \end{aligned}

As cosx  and \begin{aligned} &|x| \end{aligned} are continuous function.Hence \begin{aligned} &|\cos x| \end{aligned} is also continuous function For differentiability
$f(x)= \begin{cases}-\cos x & x<(2 n+1) \frac{\pi}{2} \\ 0 & x=(2 n+1) \frac{\pi}{2} \\ \cos x & x>(2 n+1) \frac{\pi}{2}\end{cases}$

At $(2 n+1) \frac{\pi}{2}$

\begin{aligned} & \lim _{x \rightarrow(2 n+1) \frac{\pi}{2}} \frac{f(x)-f(2 n+1) \frac{\pi}{2}}{x-(2 n+1) \frac{\pi}{2}} \\ \end{aligned}

\begin{aligned} =& \lim _{h \rightarrow 0} \frac{f(2 n+1) \frac{\pi}{2}-h-0}{(2 n+1) \frac{\pi}{2}-h-(2 n+1) \frac{\pi}{2}} \\ \end{aligned}

\begin{aligned} =& \lim _{h \rightarrow 0} \frac{-\cos (2 n+1) \frac{\pi}{2}-h}{-1} \\ \end{aligned}

\begin{aligned}=& \sin (2 n+1) \frac{\pi}{2} \\ & \lim _{x \rightarrow(2 n+1) \frac{\pi^{-1}}{2}^{-1}} \frac{f(x)-f(2 n+1) \frac{\pi}{2}}{x-(2 n+1) \frac{\pi}{2}} \\ \end{aligned}

\begin{aligned} =& \lim _{h \rightarrow 0} \frac{f\left((2 n+1) \frac{\pi}{2}-h\right)-0}{(2 n+1) \frac{\pi}{2}+h-(2 n+1) \frac{\pi}{2}} \\ \end{aligned}

\begin{aligned} =& \lim _{h \rightarrow 0} \frac{\cos (2 n+1) \frac{\pi}{2}+h}{-1} \\ =&-\sin (2 n+1) \frac{\pi}{2} \end{aligned}

As LHD $\neq$ RHD

f(x) is not differentiate at $x=(2 n+1) \frac{\pi}{2}$