JEE Advanced 2014 · Paper 1, Mathematics question 57

Base. Put $h=x-1\to0$:
$$\frac{-ah+\sin h}{h+\sin h}\to\frac{1-a}{2}.$$

Exponent. $\dfrac{1-x}{1-\sqrt x}=1+\sqrt x\to2$.

Limit. $\left(\dfrac{1-a}2\right)^2=\dfrac14$ gives $a=0$ or $a=2$.

Which is allowed. For $a=2$ the base tends to $-\frac12$, and a negative base raised to the non-integer powers $1+\sqrt x$ is not real near $x=1$, so the limit does not exist. For $a=0$ the base tends to $\frac12$ and the limit is $\frac14$. The largest admissible value is 0.

Answer: 0