Question

If $f\left(x\right)=|x|+|\sin x|$ for $x\in (-\frac{\pi }{2},\frac{\pi }{2}),$ then which of the following options is/are CORRECT?

• A. $f$ is discontinuous at three points in $(-\frac{\pi }{2},\frac{\pi }{2})$
• B. $f$ is continuous for all $x\in (-\frac{\pi }{2},\frac{\pi }{2})$
• C. $f$ is differentiable for all $x\in (-\frac{\pi }{2},\frac{\pi }{2})$
• D. $f$ is differentiable everywhere except at $x=0$

Answer 1

Answer: (B), (D)

$f$ is differentiable everywhere except at $x=0$

$f\left(x\right)=|x|+|\sin x|,x\in (-\frac{\pi }{2},\frac{\pi }{2})$
$f\left(x\right)=\{\begin{array}{cc}-x-\sin x,& -\frac{\pi }{2}<x<0\\ x+\sin x,& 0\le x<\frac{\pi }{2}\end{array}$

For continuity at $x=0:$
$L.H.L.=\underset{x\to 0^{-}}{lim}-(x+\sin x)$
$=\underset{h\to 0}{lim}-(-h-\sin h)$
$=0$
$R.H.L.=\underset{x\to 0^{+}}{lim}(x+\sin x)$
$=\underset{h\to 0}{lim}(h+\sin h)$
$=0$
$\Rightarrow L.H.L=R.H.L.=f\left(0\right)=0$
$\therefore f$ is continuous at $x=0$

Now, $L.H.D.=f^{\prime }\left(0^{-}\right)$
$=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$
$=\underset{h\to 0}{lim}\frac{h+\sin h}{-h}$
$=\underset{h\to 0}{lim}-(1+\frac{\sin h}{h})=-2$

$R.H.D.=f^{\prime }\left(0^{+}\right)$
$=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$
$=\underset{h\to 0}{lim}(1+\frac{\sin h}{h})=2$

$\Rightarrow L.H.D.\ne R.H.D.$
So, $f$ is not differentiable at $x=0.$