Question

Let $K$ be the set of all real values of $x$ where the function $f\left(x\right)=\sin |x|-|x|+2(x-\pi )\cos |x|$ is not differentiable. Then the set $K$ is equal to :

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

Answer 1

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

$f\left(x\right)=\sin |x|-|x|+2(x-\pi )\cos |x|$

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

$f^{\prime }\left(x\right)=\{\begin{array}{cc}\cos x-1-2(x-\pi )\sin x+2\cos x,& x>0\\ -\cos x+1-2(x-\pi )\sin x+2\cos x,& x<0\end{array}$

$f^{\prime }\left(0^{+}\right)=1-1-2(0-\pi )\sin 0+2\cos 0=2f^{\prime }\left(0^{-}\right)=-\cos 0+1-2(0-\pi )\sin 0+2\cos 0=2\therefore f^{\prime }\left(0^{+}\right)=f^{\prime }\left(0^{-}\right)=2$
Hence, $f\left(x\right)$ is differentiable at $x=0$
So, it is differentiable everywhere.