Let $f(x) = x \sin \pi x$, $x > 0$. Then for all natural numbers $n$, $f'(x)$ vanishes at

$f(x) = x\sin \pi x,\quad x > 0$ $\text{Study the zeros of } f'(x)$ $f'(x) = 0 \text{ at which interval?}$ $\text{for every } n \in \mathbb{N}$ $\text{Differentiate } f, \text{ set } f'(x)=0$ $\text{use signs and monotonicity to locate roots}$ $f'(x) = \sin \pi x + \pi x \cos \pi x$ $\sin \pi x + \pi x \cos \pi x = 0$ $\tan \pi x + \pi x = 0$ $\tan \pi x = -\pi x$ $\cos \pi x \neq 0 \Rightarrow x \neq n + \dfrac{1}{2}$ $f'\!\left(n+\dfrac{1}{2}\right) = \sin\!\left(n\pi+\dfrac{\pi}{2}\right) = \pm 1 \neq 0$ $g(x) = \tan \pi x + \pi x$ $g'(x) = \pi \sec^{2}\pi x + \pi > 0$ $y = \tan \pi x \text{ meets } y = -\pi x$ $\tan \text{ rises } -\infty \to +\infty,\quad -\pi x \text{ falls}$ $g(n) = \tan n\pi + n\pi = n\pi > 0$ $g(0) = 0 \text{ but } x > 0 \text{ excludes it}$ $\text{On } \left(n, n+\dfrac{1}{2}\right):\ g \nearrow \text{ and } g > 0 \Rightarrow \text{no root}$ $x \to \left(n+\dfrac{1}{2}\right)^{+} \Rightarrow g \to -\infty$ $g(n+1) = (n+1)\pi > 0$ $g \nearrow\ -\infty \to +\infty \Rightarrow \text{exactly one root}$ $f'(x) = 0 \text{ at a unique point in } \left(n+\dfrac{1}{2}, n+1\right)$ $\text{Answer: option (2)}$