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

Question:

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

Solution:

(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\$.