JEE Advanced 2017 · Paper 2, Mathematics question 11

Left of 1. Here $|1-x|=1-x$. The numerator is $1-x(2-x)=(1-x)^2$, so
$$f(x)=(1-x)\cos\frac1{1-x}\to0$$
(bounded × 0). (A) is true.

Right of 1. Here $|1-x|=x-1$. The numerator is $1-x^2=-(x-1)(x+1)$, so
$$f(x)=-(x+1)\cos\frac1{1-x}.$$
This oscillates between about $-2$ and $2$, so the limit does not exist. (D) is true.

(A), (D)