Question

Let $S$ be the set of all points where the function $f\left(x\right)=|x-\pi |(e^{|x|}-1)\sin |x|$ is not differentiable, then $S$ is

• A. $\left\{0\right\}$
• B. $\left\{\pi \right\}$
• C. $\{0,\pi \}$
• D. $\varphi$

Answer 1

The correct option is D $\varphi$

$f\left(x\right)=|x-\pi |(e^{|x|}-1)\sin |x|$
critical points $x=0,\pi$

At $x=0$
for LHD
$f\left(x\right)=-(x-\pi )(e^{-x}-1)\sin (-x)=(x-\pi )(e^{-x}-1)\sin \left(x\right)$
$f^{\prime }\left(x\right)=(x-\pi )(e^{-x}-1)\cos \left(x\right)+(x-\pi )(-e^{-x})\sin \left(x\right)+\left(1\right)(e^{-x}-1)\sin \left(x\right)$
$f^{\prime }\left(0\right)=0$
for RHD
$f\left(x\right)=(\pi -x)(e^x-1)\sin \left(x\right)$
$f^{\prime }\left(x\right)=(\pi -x)(e^x-1)\cos \left(x\right)+(\pi -x)\left(e^x\right)\sin \left(x\right)+(-1)(e^x-1)\sin \left(x\right)$
$f^{\prime }\left(0\right)=0$

At $x=\pi$
for LHD
$f\left(x\right)=(\pi -x)(e^x-1)\sin \left(x\right)$
$f^{\prime }\left(x\right)=(\pi -x)(e^x-1)\cos \left(x\right)+(\pi -x)\left(e^x\right)\sin \left(x\right)+(-1)(e^x-1)\sin \left(x\right)$
$f^{\prime }\left(\pi \right)=0$
for RHD
$f\left(x\right)=(x-\pi )(e^x-1)\sin \left(x\right)$
$f^{\prime }\left(x\right)=(x-\pi )(e^x-1)\cos \left(x\right)+(x-\pi )\left(e^x\right)\sin \left(x\right)+\left(1\right)(e^x-1)\sin \left(x\right)$
$f^{\prime }\left(\pi \right)=0$
By using the first principle of derivative we can say that $f\left(x\right)$ is differentiable at $x=0,\pi$
$\Rightarrow f\left(x\right)$ is differentiable for all $xϵR$