Question

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

• A. $\text{g}$ is differentiable but $\text{g}^{'}$ 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 $\text{g}^{'}$ is continuous

Answer 1

Answer: (D)

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