Let f (x) = |sin x|. Then,

(b) $f(x)$ is everywhere continuous but not differentiable at $x=n \pi, n \in Z$

We have,

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

$\Rightarrow f(x)= \begin{cases}0, & x=2 n \pi \\ \sin x, & 2 n \pi

When, $x$ is in first or second quadrant, i. e., $2 n \pi

$f(x)=\sin x$ which being a trigonometrical function is continuous and differentiable in $(2 n \pi,(2 n+1) \pi)$

When, $x$ is in third or fourth quadrant, i.e., $(2 n+1) \pi

$f(x)=-\sin x$ which being a trigonometrical function is continuous and differentiable in $((2 n+1) \pi,(2 n+2) \pi)$

Thus possible point of non-differentiability of $f(x)$ are $x=2 n \pi$ and $(2 n+1) \pi$

Now, LHD [at $x=2 n \pi]=\lim _{x \rightarrow 2 n \pi^{-}} \frac{f(x)-f(2 n \pi)}{x-2 n \pi}$

$=\lim _{x \rightarrow 2 n \pi^{-}} \frac{-\sin x-0}{x-2 n \pi}$

$=\lim _{x \rightarrow 2 n \pi^{-}} \frac{-\cos x}{1-0} \quad[$ By L' Hospital rule $]$

$=-1$

And $\operatorname{RHD}($ at $x=2 n \pi)=\lim _{x \rightarrow 2 n \pi^{+}} \frac{f(x)-f(2 n \pi)}{x-2 n \pi}$

$=\lim _{x \rightarrow 2 n \pi^{+}} \frac{\sin x-0}{x-2 n \pi}$

$=\lim _{x \rightarrow 2 n \pi^{+}} \frac{\cos x}{1-0} \quad$ [By L' Hospital rule]

$=1$

$\therefore \lim _{x \rightarrow 2 n_{\pi}^{-}} f(x) \neq \lim _{x \rightarrow 2 n_{\pi}^{+}} f(x)$

So $f(x)$ is not differentiable at $x=2 n \pi$

Now, LHD $[$ at $x=(2 n+1) \pi]=\lim _{x \rightarrow(2 n+1) \pi^{-}} \frac{f(x)-f((2 n+1) \pi)}{x-(2 n+1) \pi}$

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

$=\lim _{x \rightarrow(2 n+1) \pi^{-}} \frac{\cos x}{1-0} \quad$ [By L'Hospital rule]

$=-1$

And RHD (at $x=(2 n+1) \pi)=\lim _{x \rightarrow(2 n+1) \pi^{+}} \frac{f(x)-f((2 n+1) \pi)}{x-(2 n+1) \pi}$

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

$=\lim _{x \rightarrow(2 n+1) \pi^{+}} \frac{-\cos x}{1-0} \quad$ [By L'Hospital rule]

$=1$

$\therefore \lim _{x \rightarrow(2 n+1) \pi^{-}} f(x) \neq \lim _{x \rightarrow(2 n+1) \pi^{+}} f(x)$

So $f(x)$ is not differentiable at $x=(2 n+1) \pi$

Therefore, $f(x)$ is neither differentiable at $2 n \pi$ nor at $(2 n+1) \pi$

i. e. $f(x)$ is neither differentiable at even multiple of $\pi$ nor at odd multiple of $\pi$

i. e. $f(x)$ is not differentiable at $x=n \pi$

Therefore, $f(x)$ is everywhere continuous but not differentiable at $n \pi$.