JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Application of Derivatives

We have $f^{\prime }(x)=\sin (\pi x)+\pi x\cos (\pi x)$. At $x=n$, $f^{\prime }(n)=0+\pi n\cos (n\pi )=\pi n(-1{)}^n$, which is non-zero. At $x=n+\frac{1}{2}$, $f^{\prime }(n+\frac{1}{2})=\sin (\pi (n+\frac{1}{2}))+0=\pm 1$, which is non-zero. At $x=n+1$, $f^{\prime }(n+1)=0+\pi (n+1)(-1{)}^{n+1}$, which is non-zero. Since $f^{\prime }$ is continuous and changes sign between $n+\frac{1}{2}$ and $n+1$, there is exactly one zero in $(n+\frac{1}{2},n+1)$ by the Intermediate Value Theorem. This is also the unique zero in $(n,n+1)$. Thus options B and C are correct.