Let $f\left(x\right)=\left\{\begin{array}{cc}{x}^{\alpha }\log x,& \text{ if }x>0\\ 0,& \text{ if }x=0\end{array}\right.$.If Rolle’s theorem can be applied to $f$ on $[0, 1]$ then value of $\alpha$ can be

For the Rolle’s theorem to be applicable on $[0, 1],$ we must have that$f$ is continuous on $[0, 1]$ and in particular at $x = 0$ from the right.

So,      $\underset{x\to 0+}{lim} x^{\alpha }\log x=f\left(0\right)=0$

$\Rightarrow \underset{x\to 0+}{lim} \frac{\log x}{x^{-\alpha }}=0$

$\Rightarrow \underset{x\to 0+}{lim} \frac{(1/x)}{(-\alpha )x^{-\alpha -1}}=0\Rightarrow \underset{x\to 0+}{lim} \frac{x^{\alpha }}{-\alpha }=0$

This is possible only if $\alpha > 0.$