Comment on points of maxima and minima for $ f(x) = \frac{\sin(\pi x)}{x^2} $ Let $$f(x)=\frac{\sin(\pi x)}{x^2}$$ where $x>0$. Let  $x_{1}<x_{2}<x_{3}<...<x_{n}<...$  be the points of local maxima and let  $y_{1}<y_{2}<..<y_{n}<...$ 
be the points of local minima. Then which of the following options are correct:
(A)$|x_{n}-y_{n}|>1$ for every $n$
(B)$x_{1}<y_{1}$
(C)$x_{n}\in\left(2n,2n+\frac{1}{2}\right)$ for every $n$
(D)$|x_{n+1}-x_{n}|>2$ for every $n$.
My Attempt:
$$f'(x)=\frac{\pi x\cos(\pi x)-2\sin(\pi x)}{x^3}$$
So $x_{k}$ and $y_{k}$ are roots of the equation $$\frac{\pi x}{2}=\tan(\pi x)$$.
Beyond this not able to decipher
 A: Let $z_n$ be the zeros of $f$ which are clearly at $n\pi$. Between $z_n$ and $z_{n+1}$, the function is either negative or positive and this alternates. So on each interval $[z_{n},z_{n+1}]$ the function takes on a minimum or maximum, alternatingly.  
Now look at the equation $\pi x/2 = \tan(\pi x)$. It's easy to see that the derivative of the RHS is always bigger than the derivative of the LHS, so on each interval $(n-1/2,n+1/2); n\geq1$ the equation can have only one solution. Since $\tan(\pi x)$ is surjective on this interval, the solution indeed exists. Denote it by $s_n$. Let's analyse these.
Firstly, since $\pi x/2>0$, we have $s_n\in(n,n+1/2)$; since $x_n = s_{2n}$ (recall the alternating property and the fact that $s_1$ = $y_1$), certainly (C) is correct.  
Secondly, we'll show $s_{n+1}-s_n>1$. Indeed, we know that $\tan(\pi s_n)=\tan(\pi(s_n+1))$, so we have 
$$\tan(\pi(s_n+1))=\tan(\pi s_n)=\pi s_n/2 < \pi(s_n+1)/2$$
Reading just the far left and far right sides, we conclude that there must be a solution on the interval $(s_{n}+1,(n+1)+1/2)$ and by uniqueness, $s_{n+1}\in(s_n+1,n+3/2)$, so clearly $s_{n+1}-s_n>1$. (A) and (D) immediately follow.  
Thus the correct answers are (A), (C) and (D).
