IIT-JEE 2008 · Paper 2, Mathematics question 14

Derivative. Write $f=1-\dfrac{2ax}{D}$ with $D=x^2+ax+1$:
$$f'(x)=\frac{2a(x^2-1)}{D^2}.$$
So $f'(1)=f'(-1)=0$: (C) and (D) are false.

Second derivative at ±1. The factor $x^2-1$ vanishes there, so only its derivative survives:
$$f''(1)=\frac{4a}{(2+a)^2},\qquad f''(-1)=-\frac{4a}{(2-a)^2}.$$
Hence $(2+a)^2f''(1)+(2-a)^2f''(-1)=4a-4a=0$.

(A)