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 [0, 1] is

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