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

$\frac{1}{2}$

Clearly, f(x) is continuous and differentiable on (0, 1) for $\alpha>0$

Also, f(0) = 0 = f(1)

For f(x) to be continuous at x = 0, we must have

$\lim\limits_{x \rightarrow 0^{+}} f(x)=f(0)$ i.e. $\lim\limits_{x \rightarrow 0^{+}} x^\alpha \log x=0$

Now,

$\lim\limits_{x \rightarrow 0^{+}} x^\alpha \log x$           $[0 \times \infty$ form for $\alpha>0]$

$=\lim\limits_{x \rightarrow 0^{+}} \frac{\log x}{\frac{1}{x^\alpha}}$          [$\frac{\infty}{\infty}$ form]

$=\lim\limits_{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{-\frac{\alpha}{x^{\alpha+1}}}=\lim\limits_{x \rightarrow 0^{+}}-\frac{x^\alpha}{\alpha}=0$         $[∵ \alpha>0]$

So, f(x) is continuous at x = 0 for $\alpha>0$

Hence, Rolle's theorem can be applied on $f(x)$ in $[0,1]$ for all $a>0$. So, option (d) is correct.