Mathematics - Application of Derivative Question with Solution | TestHub

$f\text{'}\left(x\right)=\frac{2x\cos \pi x\left(\frac{\pi x}{2}-\tan \pi x\right)}{x^4}$…(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)
$x_{n+1}-x_n>2$