Let f x = sin ⁡ π x x 2 ,   x > 0 . Let x 1 < x 2 < x 3 < … < x n…

1. $\left|x_n-y_n\right|>1$ for every $n$
2. $x_1<y_1$
3. $x_n\in \left(2n, 2n+\frac{1}{2}\right)$ for every $n$
4. $x_{n+1}-x_n>2$ for every $n$

Solution

$f\text{'}\left(x\right)=\frac{2x\cos \pi x\left(\frac{\pi x}{2}-\tan \pi x\right)}{x^4}$ …(i)
$\Rightarrow$ for maxima/minima $f^{\text{'}}\left(x\right)=0$
$\begin{array}{cc}\Rightarrow \cos \pi x=0& \mathrm{o}\mathrm{r}\frac{\pi x}{2}=\tan \pi x\end{array}$
$\begin{array}{ccc}\because \cos \pi x\ne 0& & \left\{\because \tan \pi x \mathrm{will}\mathrm{ }\mathrm{not}\mathrm{ }\mathrm{be}\mathrm{ }\mathrm{defined}\right\}\end{array}$
$\therefore$ maxima/minima will occur
Where $\tan \left(\pi x\right)=\frac{\pi x}{2}$

Case $\mathrm{I}:$
$\left(\mathrm{i}\right)$ For example
In $x\in \left(2n, 2n+\frac{1}{2}\right)$ at point $P_2, x\in \left(2,\frac{5}{2}\right)$
$\cos \left(\pi x\right)>0$
and at $P_2^{+}, \tan \left(\pi x\right)>\frac{\pi x}{2}$
at $P_2^{-}, \tan \left(\pi x\right)<\frac{\pi x}{2}$
hence from equation (i)
in $x\in \left(2n, 2n+\frac{1}{2}\right), f^{\text{'}}\left(x\right)$ goes from positive to negative
hence $P_2$ is maxima
Similarly $P_2, P_4, P_6$ … are point of maxima and all lies in $x\in \left(2n, 2n+\frac{1}{2}\right)$
Case $\mathrm{I}\mathrm{I}$
In $x\in \left(2n+1, 2n+\frac{3}{2}\right)$
$\left(\mathrm{i}\mathrm{i}\right)$ for example at $P_1, x\in \left(1,\frac{3}{2}\right)$
$\cos \left(\pi x\right)<0$
at $P_1^{+}, \tan \left(\pi x\right)>\frac{\pi x}{2}$
and at $P_1^{-}, \tan \left(\pi x\right)<\frac{\pi x}{2}$
hence from equation (i)
in $x\in \left(2n+1, 2n+\frac{3}{2}\right), f^{\text{'}}\left(x\right)$ goes from negative to positive
$\Rightarrow$ so, for the minima $y_1:1<y_1<\frac{3}{2}$
$y_2:3<y_2<\frac{7}{2}$
$y_3:5<y_3<\frac{11}{2}$
and for the maxima $x_1:2<x_1<\frac{5}{2}$
$x_2:4<x_2<\frac{9}{2}$
$x_3:6<x_3<\frac{13}{2}$
Hence $x_n>y_n$ … $\left(\mathrm{a}\right)$
$\Rightarrow$ now,
$x_1>y_1$
$\tan \left(\pi x_1\right)>\tan \left(\pi y_1\right)$
$\tan \left(\pi x_1\right)>\tan \left(\pi +\pi y_1\right)$
$\tan \left(\pi x_1\right)>\tan \left(\pi \left(1+y_1\right)\right)$
$\pi x_1>\pi \left(1+y_1\right)$
$x_1>1+y_1$
Similarly, $x_n>1+y_n$
$x_n-y_n>1$ …(b)
$\left|x_n-y_n\right|>1$
$\Rightarrow y_2>x_1$
$\tan \left(\pi y_2\right)>\tan \left(\pi x_1\right)$
$\tan \left(\pi y_2\right)>\tan \left(\pi +\pi x_1\right)\Rightarrow y_2>x_1+1$
$\Rightarrow y_{n+1}>x_n+1$
Therefore $x_n<y_{n+1}<x_{n+1}\Rightarrow y_{n+1}-x_n>1$ …(c)
Now consider $x_{n+1}-x_n=\left(x_{n+1}-y_{n+1}\right)+\left(y_{n+1}-x_n\right)$
From (b) and (c)
$x_{n+1}-x_n>2$

Asked in: JEE Advanced 2019 (Paper 2)

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