Explanation

(d)
To satisfy Rolle’s theorem it should be continuous in [0, 1]
$ie,\ \ \$RHL$= f(0)$
$\Rightarrow \ \lim_{x \rightarrow 0^{+}}\frac{\log x}{x^{- \alpha}} = 0$
$\Rightarrow \ \lim_{x \rightarrow 0^{+}}\frac{\frac{1}{x}}{- \alpha x^{- \alpha - 1}} = 0$ (using L’Hospital’s rule)
$\Rightarrow \ \lim_{x \rightarrow 0^{+}}\frac{1}{\alpha x^{- \alpha}} = 0$
$\Rightarrow \ \lim_{x \rightarrow 0^{+}}\frac{x^{\alpha}}{\alpha} = 0$
Which shows $\alpha > 0$ otherwise it would be continuous also
When $\alpha > 0,\ \ f(x)$ is differentiating in (0, 1) and $f(1) = f(0) = 0$
Clearly, $\alpha > 0$, thus $\alpha = \frac{1}{2}$ is the possible answer