Let $f\left(x\right)=\frac{\sin \pi x}{x^2},x>0,$ Let $x_1<x_2<x_3<\mathrm{.....}<x_n<\mathrm{......}$ be all the points of local maximum of $f$ and $y_1<y_2<y_3<\mathrm{........}{y}_n<\mathrm{....}$ be all the points of local minimum of $f.$Then which of the following options is/are correct

$f\left(x\right)=\frac{\sin \pi x}{x^2},x>0$
For points of local max/min, put $f\text{'}\left(x\right)=0$
$\Rightarrow \frac{\pi x^2\cos \pi x-2x\sin \pi x}{x^4}=0$
$\Rightarrow \frac{\cos \pi x(\pi x-2\tan \pi x)}{x^3}=0$
$\cos \pi x=0\Rightarrow x=\frac{1}{2},\frac{3}{2},\frac{5}{2},\frac{7}{2},\mathrm{....}$and $\pi x-2\tan \pi x=0$ which can be solved by drawing the which can be solved by drawing the $y=\pi x$ and $y=2\tan \pi x,$as follows
 
Plotting the stationary points on number line and finding the sign of $f\text{'}\left(x\right)$ in different intervals we observe

i.e $x_{n+1}-x_n>2$ for every $n$
$x_n\in (2n,2n+\frac{1}{2})$ for every
$|x_n-y_n|>1$ for every $n$
$x_1>y_1$