Explanation

(d)
To satisfy Rolle’s theorem, it should be continuous in [0, 1].
$\Rightarrow \lim_{n \rightarrow 0^{+}}{f(x) = f(0)}$
$\Rightarrow \lim_{n \rightarrow 0^{+}}{\frac{\log x}{x^{- a}} = 0}$
$\Rightarrow \lim_{n \rightarrow 0^{+}}{\frac{1/x}{- \alpha x^{- a - 1}} = 0\ \ \ \ }\lbrack using\ L^{'}Hospital^{'}s\ rule\rbrack$
$\Rightarrow \lim_{n \rightarrow 0^{+}}{- \frac{1}{\alpha x^{- \alpha}}}$=0
$\Rightarrow \lim_{n \rightarrow 0^{+}} - \frac{1}{\alpha}x^{\alpha}$=0
Which shows $\alpha > 0$otherwise, it would be discontinuous also when
$\alpha > 0,\ f(x)\ is\ differentiable\ in\ (0,1)and\ f(1) = f(0) = 0.$
$Clearly\ \alpha > 0,thus\ \alpha = \frac{1}{2}$is the possible answer.