Mathematics - Continuity - Differentiability Question with Solution | TestHub

$f\left(x\right)=\left|x-\pi \right|.\left(e^{\left|x\right|}-1\right).\sin \left|x\right|$

In the neighbourhood of $x=\pi$, $f\left(x\right)=\left|x-\pi \right|.\left(e^x-1\right).\sin x$

$R{f}^{\prime }\left(\pi \right)=\underset{h\to 0}{\lim }\frac{f\left(\pi +h\right)-f\left(\pi \right)}{h}=\underset{h\to 0}{\lim }\frac{\left|h\right|\sin \left(\pi +h\right)\left(e^{\pi +h}-1\right)}{h}=0$

$L{f}^{\prime }\left(\pi \right)=\underset{h\to 0}{\lim }\frac{f\left(\pi \right)-f\left(\pi -h\right)}{h}=-\underset{h\to 0}{\lim }\frac{\left|-h\right|\sin \left(\pi -h\right)\left(e^{\pi -h}-1\right)}{h}=0$

Differentiable at $x=\pi$

$R{f}^{\prime }\left(0\right)=\underset{h\to 0}{\lim }\frac{f\left(h\right)-f\left(0\right)}{h}=\underset{h\to 0}{\lim }\left|h-\pi \right|\times \frac{\sin \left|h\right|\left(e^{\left|h\right|}-1\right)}{h}=0$

$L{f}^{\prime }\left(0\right)=\underset{h\to 0}{\lim }\frac{f\left(0\right)-f\left(-h\right)}{h}=-\underset{h\to 0}{\lim }\left|-h-\pi \right|\times \frac{\sin \left|-h\right|\left(e^{\left|-h\right|}-1\right)}{h}=0$

Differentiable at $x=0$

Hence, $f\left(x\right)$ is differentiable everywhere.

$\Rightarrow S=\varphi$