JEE Advanced 2013 · Paper 1, Mathematics question 53

Critical points. $f'(x)=\sin\pi x+\pi x\cos\pi x=0$ gives $\tan\pi x=-\pi x$.

In $(n,n+1)$. The right side is negative, and $\tan\pi x<0$ only on $(n+\frac12,n+1)$, where it increases from $-\infty$ to 0 while $-\pi x$ decreases: exactly one crossing.

So there is a unique zero in $(n+\frac12,n+1)$ and hence a unique one in $(n,n+1)$.

(B), (C)