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

Solution:

(c) $f(x)$ is everywhere continuous but not differentiable at $x=(2 n+1) \frac{\pi}{2}, n \in Z$.

We have,

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

$\Rightarrow f(x)=\left\{\begin{array}{rc}\cos x, & 2 n \pi \leq x<(4 n+1) \frac{\pi}{2} \\ 0, & x=(4 n+1) \frac{\pi}{2} \\ -\cos x, & (4 n+1) \frac{\pi}{2}

When, $x$ is in first quadrant, i.e. $2 n \pi \leq x<(4 n+1) \frac{\pi}{2}$, we have

$f(x)=\cos x$ which being a trigonometrical function is continuous and differentiable in $\left(2 n \pi,(4 n+1) \frac{\pi}{2}\right)$

When, $x$ is in second quadrant or in third quadrant, i. e., $(4 n+1) \frac{\pi}{2}

$f(x)=-\cos x$ which being a trigonometrical function is continuous and differentiable in $\left((4 n+1) \frac{\pi}{2},(4 n+3) \frac{\pi}{2}\right)$

When, $x$ is in fourth quadrant, i.e., $(4 n+3) \frac{\pi}{2}

$f(x)=\cos x$ which being a trigonometrical function is continuous and differentiable in $\left((4 n+3) \frac{\pi}{2},(2 n+2) \pi\right)$

Thus possible point of non-differentiability of $f(x)$ are $x=(4 n+1) \frac{\pi}{2},(4 n+3) \frac{\pi}{2}$

Now, LHD $\left[\right.$ at $\left.x=(4 n+1) \frac{\pi}{2}\right]=\lim _{x \rightarrow(4 n+1) \frac{\pi}{2}^{-}} \frac{f(x)-f\left((4 n+1) \frac{\pi}{2}\right)}{x-(4 n+1) \frac{\pi}{2}}$

$=\lim _{x \rightarrow(4 n+1) \frac{\pi}{2}^{-}} \frac{\cos x-0}{x-(4 n+1) \frac{\pi}{2}}$

$=\lim _{x \rightarrow(4 n+1) \frac{\pi}{2}^{-}} \frac{-\sin x}{1-0} \quad$ [By L'Hospital rule]

$=-1$

And $\operatorname{RHD}\left(\right.$ at $\left.x=(4 n+1) \frac{\pi}{2}\right)=\lim _{x \rightarrow(4 n+1) \frac{\pi}{2}^{+}} \frac{f(x)-f\left((4 n+1) \frac{\pi}{2}\right)}{x-(4 n+1) \frac{\pi}{2}}$

$=\lim _{x \rightarrow(4 n+1) \frac{\pi}{2}^{+}} \frac{-\cos x-0}{x-(4 n+1) \frac{\pi}{2}}$

$=\lim _{x \rightarrow(4 n+1) \frac{\pi}{2}^{+}} \frac{\sin x}{1-0} \quad$ [By L'Hospital rule]

$=1$

$\therefore \lim _{x \rightarrow(4 n+1) \frac{\pi}{2}^{-}} f(x) \neq \lim _{x \rightarrow(4 n+1) \frac{\pi}{2}^{+}} f(x)$

So $f(x)$ is not differentiable at $x=(4 n+1) \frac{\pi}{2}$

Now, LHD $\left[\right.$ at $\left.x=(4 n+3) \frac{\pi}{2}\right]=\lim _{x \rightarrow(4 n+1) \frac{\pi}{2}^{-}} \frac{f(x)-f\left((4 n+3) \frac{\pi}{2}\right)}{x-(4 n+3) \frac{\pi}{2}}$

$=\lim _{x \rightarrow(4 n+3) \frac{\pi}{2}^{-}} \frac{-\cos x-0}{x-(4 n+3) \frac{\pi}{2}}$

$=\lim _{x \rightarrow(4 n+3) \frac{\pi}{2}^{-}} \frac{\sin x}{1-0} \quad$ [By L'Hospital rule]

$=1$

And $\operatorname{RHD}\left(\right.$ at $\left.x=(4 n+3) \frac{\pi}{2}\right)=\lim _{x \rightarrow(4 n+3) \frac{\pi}{2}^{+}} \frac{f(x)-f\left((4 n+3) \frac{\pi}{2}\right)}{x-(4 n+3) \frac{\pi}{2}}$

$=\lim _{x \rightarrow(4 n+3) \frac{\pi}{2}^{+}} \frac{\cos x-0}{x-(4 n+3) \frac{\pi}{2}}$

$=\lim _{x \rightarrow(4 n+3) \frac{\pi}{2}^{+}} \frac{-\sin x}{1-0} \quad$ [By L'Hospital rule]

$=-1$

$\therefore \lim _{x \rightarrow(4 n+3) \frac{\pi}{2}^{-}} f(x) \neq \lim _{x \rightarrow(4 n+3) \frac{\pi}{2}^{+}} f(x)$

So $f(x)$ is not differentiable at $x=(4 n+3) \frac{\pi}{2}$

Therefore, $f(x)$ is neither differentiable at $(4 n+1) \frac{\pi}{2}$ nor at $(4 n+3) \frac{\pi}{2}$

i.e. $f(x)$ is not differentiable at odd multiples of $\frac{\pi}{2}$

i. e. $f(x)$ is not differentiable at $x=(2 n+1) \frac{\pi}{2}$

Therefore, $f(x)$ is everywhere continuous but not differentiable at $(2 n+1) \frac{\pi}{2}$.