Question

Let $f\left(x\right)=x|\sin x|,x\in R$. Then

• A. $f$ is differentiable for all $x$, except at $x=n\pi ,n=1,2,3,\dots .$
• B. $f$ is differentiable for all $x$, except at $x=n\pi ,n=\pm 1,\pm 2,\pm 3,\dots .$
• C. $f$ is differentiable for all $x$, except at $x=n\pi ,n=0,1,2,3,\dots .$
• D. $f$ is differentiable for all $x$, except at $x=n\pi ,n=0,\pm 1,\pm 2,\pm 3,\dots .$

Answer 1

The correct option is B $f$ is differentiable for all $x$, except at $x=n\pi ,n=\pm 1,\pm 2,\pm 3,\dots .$

Given:
$f\left(x\right)=x|\sin x|$
$\therefore f\left(x\right)=\{\begin{array}{cc}x\sin x& x\in [2n\pi ,(2n+1\left)\pi \right],n\in Z\\ -x\sin x& x\in \left(\right(2n+1)\pi ,(2n+2\left)\pi \right),n\in Z\end{array}$

$f^{\prime }\left(x\right)=\{\begin{array}{cc}\sin x+x\cos x& x\in [2n\pi ,(2n+1\left)\pi \right]\\ -\sin x-x\cos x& x\in \left(\right(2n+1)\pi ,(2n+2\left)\pi \right)\end{array}$

For the function to be differentiable at $x=n\pi$,
$\sin n\pi +n\pi \cos n\pi =-\sin n\pi -n\pi \cos n\pi \Rightarrow \cos n\pi =-\cos n\pi \Rightarrow n=0$

Hence, $f$ is differentiable for all $x$, except at $x=n\pi ,n=\pm 1,\pm 2,\pm 3,\cdots$