Let $f(x)=\frac{\sin \pi x}{x^{2}}, x>0$ Let $x_{1}<x_{2}<x_{3}<\ldots<x_{n}<\ldots$ be all the points of local maximum of $f$ and $y_{1}<y_{2}<y_{3}<\ldots<y_{n}<\ldots$ be all the points of local minimum of $f$. Then which of the following options is/are correct?

Let f(x)=(sin π x)/x², x>0 Let x₁<x₂<x₃<…<xₙ<… be all the points of local maximum of f and y₁<y₂<y₃<…<yₙ<… be all the points of local minimum of f. Then which of the following options is/are correct? $f(x) = \dfrac{\sin \pi x}{x^2},\qquad x>0$ $x_n: \text{local maxima (increasing)},\qquad y_n: \text{local minima (increasing)}$ $\text{To find: which of } (1),\ (2),\ (3),\ (4) \text{ are correct}$ $\text{Plan: } f'=0 \Rightarrow \tan \pi x = \dfrac{\pi x}{2};\ \text{locate roots, then test each option}$ $f'(x) = \dfrac{\pi x\cos \pi x - 2\sin \pi x}{x^3}$ $f'(x)=0 \Rightarrow \tan \pi x = \dfrac{\pi x}{2}$ $\text{Graph of } f \text{ on } (0,9)$ $u = \pi x:\quad \tan u = \dfrac{u}{2}$ $\text{roots } u_n \in (n\pi,\ n\pi+\tfrac{\pi}{2}) \Rightarrow x \in (n,\ n+\tfrac12)$ $u_1,u_2,u_3,\ldots = \text{min, max, min, max, }\ldots$ $x_n = \dfrac{u_{2n}}{\pi};\qquad u_{2n}\in\left(2n\pi,\ 2n\pi+\dfrac{\pi}{2}\right)$ $\Rightarrow x_n\in\left(2n,\ 2n+\dfrac{1}{2}\right) \Rightarrow (2)\ \text{TRUE}$ $u_{2n+2} > u_{2n}+2\pi \Rightarrow x_{n+1}-x_n > 2$ $\Rightarrow (1)\ \text{TRUE}$ $y_n=\dfrac{u_{2n-1}}{\pi};\qquad u_{2n}-u_{2n-1}>\pi$ $\Rightarrow |x_n-y_n|>1 \Rightarrow (3)\ \text{TRUE}$ $y_1=\dfrac{u_1}{\pi}\approx 1.36,\quad x_1=\dfrac{u_2}{\pi}\approx 2.42$ $x_1 > y_1 \Rightarrow (4)\ \text{FALSE}$ $(1),\ (2),\ (3)\ \text{are correct}$ $\boxed{(1),\ (2),\ (3)}$