Let $f\left(x\right)=\{x^p\sin \left(\frac{1}{x}\right),x\ne 0\}$  if $f\left(x\right)$ is continuous but not differentiable at $x=0$ then the number of integral values of $p$ is

For continuity  $\underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right)=0$
$\underset{x\to 0}{\lim }{x}^p\sin \frac{1}{x}=0$
This is possible only when $p>0$
$f^1\left(x\right)=\underset{x\to 0}{\lim }\frac{{(o+h)}^p\sin \left(\frac{1}{0+h}\right)-0}{h}=\underset{x\to 0}{\lim }{h}^{p-1}\sin \left(\frac{1}{h}\right)$
$\text{For\hspace{0.17em}}{f}^1\left(x\right)\text{\hspace{0.17em}to\hspace{0.17em}exist\hspace{0.17em}}p-1>0p>1$
f(x) will not be differentiable if $p\le 1$
$\therefore 0<p\le 1$