Question

Let $g\left(x\right)=xf\left(x\right)$, where $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}x\sin \frac{1}{x},& x\ne 0\end{array}\\ \begin{array}{cc}0,& x=0\end{array}\end{array}$. At $x=0$,

• A. $g$ is differentiable but $g^{\prime }$ is not continuous
• B. $g$ is differentiable while $f$ is not
• C. Both $f$ and $g$ are differentiable
• D. $g$ is differentiable and $g^{\prime }$ is continuous

Answer 1

Answer: (A), (B)

The correct options are
A $g$ is differentiable but $g^{\prime }$ is not continuous
B $g$ is differentiable while $f$ is not

We have, $g\left(x\right)=\{\begin{array}{c}\begin{array}{cc}{x}^2\sin \frac{1}{x},& x\ne 0\end{array}\\ \begin{array}{cc}0,& x=0\end{array}\end{array}$

For $x\ne 0,$

$g^{\prime }\left(x\right)=x^2\cos \frac{1}{x}(-\frac{1}{x^2})+2x\sin \frac{1}{x}=-\cos \frac{1}{x}+2x\sin \frac{1}{x}$

For $x=0$

$g^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{g\left(x\right)-g\left(0\right)}{x-0}=\underset{h\to 0}{lim}\frac{x^2\sin \frac{1}{x}-0}{x}=\underset{h\to 0}{lim}x\sin \frac{1}{x}=0$

$\therefore g^{\prime }\left(x\right)=\{\begin{array}{c}\begin{array}{cc}2x\sin \frac{1}{x}-\cos \frac{1}{x},& x\ne 0\end{array}\\ \begin{array}{cc}0,& x=0\end{array}\end{array}$

$g^{\prime }$ is not continuous at $x=0$ as $\cos \frac{1}{x}$ is not continuous at $x=0.$

Also, $f$ is not differentiable at $x=0.$