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(b)

HINTS: understand the definition of continuity and differentiability

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

SOLUTION:

As $\cos x$ & $\left | x \right |$ are continuous function .Hence $\cos x$ is also  continuous function.

For differentiability ,

\begin{aligned} &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} \\ \end{aligned}

at \begin{aligned}&(2 n+1) \frac{\pi}{2} & \end{aligned}

\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}} \\ &=\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} \\ &=\sin (2 n+1) \frac{\pi}{2} \end{aligned}

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

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