Question

Let $g(x)=xf(x),$ where $f(x)=\{\begin{array}{ll}{x}^2\sin \frac{1}{x}& :x\ne 0\\ 0& :x=0\end{array}.$ At $x=0$,

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

Answer 1

Answer: (D)

We have, $g(x)=\{\begin{array}{ll}{x}^3\sin \frac{1}{x}& :x\ne 0\\ 0& :x=0\end{array}$
For $x\ne 0$ $g^{\prime }(x)=x^3\cos \frac{1}{x}\left(-\frac{1}{x^2}\right)+3x^2\sin \frac{1}{x}$
$=-x\cos \frac{1}{x}++3x^2\sin \frac{1}{x}$
For $x=0$ $g^{\prime }(0)={\lim }_{x\to 0}\frac{g(x)-g(0)}{x-0}={\lim }_{x\to 0}\frac{x^3\sin \frac{1}{x}-0}{x}$
$={\lim }_{x\to 0}{x}^2\sin \frac{1}{x}=0$
$\therefore g^{\prime }(x)=\{\begin{array}{c}3x^2\sin \frac{1}{x}-x\cos \frac{1}{x},x\ne 0\\ 0,x=0\end{array}$
$g^{\prime }$ is continuous at $x=0$
Also, $g$ is differentiable at $x=0$.