Let f (x) = |cos x|. Then,
(a) $f(x)$ is everywhere differentiable
(b) $f(x)$ is everywhere continuous but not differentiable at $x=n \pi, n \in Z$
(c) $f(x)$ is everywhere continuous but not differentiable at $x=(2 n+1) \frac{\pi}{2}, n \in Z$.
(d) none of these
(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 $(2 n+1) \frac{\pi}{2}$.
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