Question

$\text{Let}f\left(x\right)=\frac{\sin \pi x}{x^2},x>0.$
$\text{Let}{x}_1<x_2<x_3<...<x_n<...$ be all the points of local maximum of $f$ and $y_1<y_2<y_3<...<y_n<...$ be all the points of local minimum of $f.$
Then which of the following options is/are correct?

• A. $x_1<y_1$
• B. $x_{n+1}-x_n>2$ for every $n$
• C. $x_n\in (2n,2n+\frac{1}{2})$ for every $n$
• D. $|x_n-y_n|>1$ for every $n$

Answer 1

Answer: (B), (C), (D)

$|x_n-y_n|>1$ for every $n$

$f\left(x\right)=\frac{\sin \pi x}{x^2}$
$f^{\prime }\left(x\right)=\frac{\pi x^2\cos \pi x-2x\sin \pi x}{x^4}=\frac{2\cos \pi x(\frac{\pi x}{2}-\tan \pi x)}{x^3}$ where $x\ne 0$
For maxima and minima: $f^{\prime }\left(x\right)=0$
$\Rightarrow \cos \pi x(\frac{\pi x}{2}-\tan \pi x)=0$
Let $g\left(x\right)=\cos \pi x$ or, $h\left(x\right)=\tan \pi x-\frac{\pi x}{2}$
From the above graph we can observe that product of both $g\left(x\right)$ and $h\left(x\right)$ changes its sign at given interval
$\therefore \pi x=(2n+1)\frac{\pi }{2}$
$\Rightarrow x=\frac{2n+1}{2},n\in I$
From the graph, we can see that for all $x=\frac{2n+1}{2},$ doesn't change sign. Also we can see $x_1\text{is point of maxima and}{y}_1\text{is point of minima and}$ $x_1>y_1$
Same is the case for all $x\text{satisfying}\frac{\pi x}{2}=\tan \pi x$
$\text{where}{y}_n\in (2n-1,2n-\frac{1}{2})\forall n=1,2,\mathrm{3....}$
$\text{and}{x}_n\in (2n,2n+\frac{1}{2})\forall n=1,2,\mathrm{3....}$
$\therefore |x_n-y_n|>1$ for every $n$
$x_1>y_1$
$x_{n+1}-y_{n+1}>1$ and $y_{n+1}-x_n>1$
$\Rightarrow x_{n+1}-x_n>2$