If $f(x)=x^{\alpha} \log x$ and $f(0)=0$, then the value of $\alpha$ for which Rolle's theorem can b

Question

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

Professor PC Thomas Classes · Answer

(d) : For function to satisfy the condition of Rolle's theorem, it should be continuous in $[0,1]$\ $\Rightarrow \quad \lim _{x ightarrow 0^{+}} f(x)=f(0)$\ $\Rightarrow \quad \lim _{x ightarrow 0^{+}} \frac{\log x}{x^{-\alpha}}=0$\ $\Rightarrow \quad \lim _{x ightarrow 0^{+}} \frac{1 / x}{-\alpha x^{-\alpha-1}}=0=\lim _{x ightarrow 0^{+}} \frac{-x^{\alpha}}{\alpha}=0$\ $\Rightarrow \quad \alpha>0$\ Also $\forall \alpha>0, f(x)$ is differentiable in $(0,1)$ and $f(1)=0=f(0)$.\