Equation using $f(x)=\frac{\sin\pi x}{x^2} $ If $f(x)= \frac{\sin \pi x}{x^2}$, $x>0$
Let $x_1<x_2<x_3<\cdots<x_n<\cdots$ be all the points of local maximum of $f$.
Let $y_1<y_2<y_3<\cdots<y_n<\cdots$ be all the points of local minimum of $f$.
Then which of the options is(are) correct.
(A) $x_1<y_1$
(B) $x_{n+1}-x_n>2$ for all $n$
(C) $x_n \in (2n,2n+\frac{1}{2})$ for every $n$
(D) $|x_n-y_n|>1$ for every $n$
The official answer is $B,C,D$
My approach is as follow
$$y' = \frac{x^2 \times \pi  \times \cos \pi x - 2x\sin \pi x}{x^4}$$
$$y' = \frac{x^2 \times \pi  \times \cos \pi x - 2x\sin \pi x}{x^4} = \frac{x \times \pi  \times \cos \pi x - 2\sin \pi x}{x^3} = \frac{2\cos \pi x \left( \frac{\pi x}{2} - \tan \pi x \right)}{x^3}$$
We should consider $x\in\frac{1}{2},\frac{3}{2},\ldots.$ as inflection point even though $\tan\pi x$ is not defined because we need to consider the numerator part $\pi x\cos \pi x - 2\sin \pi x$
 A: Only the option $A)$ is correct.
$$x_1=\frac 12\;\;,\; y_1=\frac 32$$
Let $$f(x)=\frac{\sin(\pi x)}{x^2}$$
it is clear that
$$(\forall x>0)\;\; -\frac{1}{x^2}\le f(x)\le \frac{1}{x^2}$$
The local maximums are such that
$$f(x_i)=\frac{1}{x_i^2}$$
or
$$\sin(\pi x_i)=1$$
thus
$$x_i=\frac 12+2n$$
with $n\ge 0$.
So, $$x_i\in\left\{\frac 12,\frac 52,\frac 92,\ldots\right\}$$
by the same,
$$y_i\in\left\{\frac 32,\frac 72,\frac{11}{2},\ldots\right\}$$
A: $$f'(x)=\frac{\pi  x \cos (\pi  x)-2 \sin (\pi  x)}{x^3}$$
$$f'(x)=0\to \sin(\pi x)=\pi x\cos(\pi x)\tag{1}$$
Second derivative is
$$f''(x)=-\frac{\left(\pi ^2 x^2-6\right) \sin (\pi  x)+4 \pi  x \cos (\pi  x)}{x^4}$$
To understand which are max/min we use second derivative test using relation $(1)$ in the second derivative
$$f''(x^*)-\frac{\left(\pi ^2 x^2-6\right) \sin (\pi  x)+4 \sin(\pi x)}{x^4}=-\frac{\left(\pi ^2 x^2+4 \pi -6\right) \sin (\pi  x)}{x^4}$$
As the factor $\left(\pi ^2 x^2+4 \pi -6\right) $ is positive for any $x$ we have that second derivative is negative when $\sin(\pi x)>0$ that is $2k\pi <\pi x<(2k+1)\pi$ $\forall k\in\mathbb{N}$ and finally $2k<x<2k+1$.
In other words the solutions of $(1)$
are max when
$ 2k<x<2k+1$ and are minimum when $k+1<x<2k$
So we can see that statement (A) is false.
