Question

Let $S=\{t\in R:f(x)=|x-\pi |\cdot (e^{|x|}-1\left)\sin |x|\text{is not differentiable at t}\right\}.$ Then the set $S$ is equal to

• A. $\{0,\pi \}$
• B. $\varphi$ (an empty set)
• C. $\left\{0\right\}$
• D. $\left\{\pi \right\}$

Answer 1

The correct option is B $\varphi$ (an empty set)

Given: $f\left(x\right)=|x-\pi |\cdot (e^{|x|}-1)\cdot \sin |x|$
Doubtful points $=0,\pi$
Now, at $x=0$
$f^{\prime }\left(0^{+}\right)=\underset{h\to 0^{+}}{lim}\left(\frac{|h-\pi |(e^{|h|}-1)\sin |h|}{h}\right)$
$\Rightarrow f^{\prime }\left(0^{+}\right)=\underset{h\to 0^{+}}{lim}\left(\frac{|\pi -h|(e^h-1)\sin h}{h}\right)$
$\Rightarrow f^{\prime }\left(0^{+}\right)=0$
$\Rightarrow f^{\prime }\left(0^{-}\right)=\underset{h\to 0^{+}}{lim}\left(\frac{|-h-\pi |(e^{|-h|}-1)\sin |-h|}{-h}\right)$
$\Rightarrow f^{\prime }\left(0^{-}\right)=0$

And, at $x=\pi$
$f^{\prime }\left({\pi }^{+}\right)=\underset{h\to 0^{+}}{lim}\left(\frac{|h|.(e^{|\pi +h|}-1)\sin |\pi +h|}{h}\right)$
$\Rightarrow f^{\prime }\left({\pi }^{+}\right)=\underset{h\to 0^{+}}{lim}\left(\frac{h(e^{\pi +h}-1)\cdot (-\sin h)}{h}\right)$
$\Rightarrow f^{\prime }\left({\pi }^{+}\right)=0$
$f^{\prime }\left({\pi }^{-}\right)=\underset{h\to 0^{+}}{lim}\left(\frac{h(e^{\pi -h}-1)\sin h}{-h}\right)$
$\Rightarrow f^{\prime }\left({\pi }^{-}\right)=0$

Hence, $f\left(x\right)$ is diff. for all $x\in R.$