question_answer

Let \[f\,\,(x)\,\,\left\{ \begin{align}   & {{x}^{n}}\sin \frac{1}{x} \\  & \,\,\,\,\,\,\,\,\,\,\,0, \\ \end{align} \right.\,\,\,\,\]\[\begin{align}   & x\ne 0 \\  & x\,\,\,=0 \\ \end{align}\]Then, f is continuous but not differentiable at \[x=0\]f

A)  \[n\,\,\in (0,1]\]    

B)  \[n\,\,\in [1,\infty )\]   

C)  \[n\in \text{ }(-\infty ,\infty )\]    

D)  \[n=1\]