If $$f\left( x \right) = {x^\alpha }\log \,x$$    and $$f\left( 0 \right) = 0,$$   then the value of $$\alpha $$ for which Rolle's theorem can be applied in $$\left[ {0,\,1} \right]$$  is :

Solution :
For Rolle's theorem in $$\left[ {a,\,b} \right],\,f\left( a \right) = f\left( b \right),\,{\text{ln}}\left[ {0,\,1} \right] \Rightarrow f\left( 0 \right) = f\left( 1 \right) = 0$$
$$\because $$  the function has to be continuous in $$\left[ {0,\,1} \right]$$
$$\eqalign{ & \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 0 \cr & \Rightarrow \mathop {\lim }\limits_{x \to 0} \,{x^\alpha }\log \,x = 0 \cr & \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\log \,x}}{{{x^{ - \alpha }}}} = 0 \cr & {\text{Applying L}}{\text{.H}}{\text{. Rule, }}\mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{x}}}{{ - \alpha {x^{ - \alpha - 1}}}} = 0 \cr & \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{ - {x^\alpha }}}{\alpha } = 0 \cr & \Rightarrow \alpha > 0 \cr} $$