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(D) For Rolle's theorem to be applicable on $[0, 1]$,the function $f(x)$ must satisfy three conditions: $1$. $f(x)$ is continuous on $[0, 1]$. $2$. $f

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$1/2$

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(D) For Rolle's theorem to be applicable on $[0, 1]$,the function $f(x)$ must satisfy three conditions: $1$. $f(x)$ is continuous on $[0, 1]$. $2$. $f(x)$ is differentiable on $(0, 1)$. $3$. $f(0) = f(1)$. Given $f(x) = x^\alpha \log x$ and $f(0) = 0$. First,check the condition $f(0) = f(1)$: $f(1) = 1^\alpha \log(1) = 1 	imes 0 = 0$. Since $f(0) = 0$,the condition $f(0) = f(1)$ is satisfied for any $\alpha$. Next,check continuity at $x = 0$: $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} x^\alpha \log x = 0$. This limit exists and equals $0$ if $\alpha > 0$. Using $L$'$H$ôpital's rule: $\lim_{x 	o 0^+} \frac{\log x}{x^{-\alpha}} = \lim_{x 	o 0^+} \frac{1/x}{-\alpha x^{-\alpha-1}} = \lim_{x 	o 0^+} \frac{x^\alpha}{-\alpha} = 0$ (for $\alpha > 0$). Checking the options,$\alpha = 1/2$ is the only value greater than $0$.

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

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