JEE Advanced 2014 · Paper 2, Mathematics question 54

Symmetry. Substitute $t\to1-t$:
$$g(a)=\int_0^1(1-t)^{-a}t^{a-1}dt=g(1-a).$$
So $g$ is symmetric about $a=\frac12$, and since it is differentiable, $g'(a)=-g'(1-a)$; at $a=\frac12$ this gives $g'\left(\frac12\right)=0$.

(Check: $g(a)=\pi/\sin\pi a$ has $g'(a)=-\pi^2\cos\pi a/\sin^2\pi a$, which is 0 at $\frac12$.)

(D)