Let f(x) = sin x ^ 2 , x > 0 . Let x _ 1 Then which of the...

$f'$\left(x\right) = \dfrac{2xcos \pix\left(\dfrac{\pix}{2} - tan \pix\right)}{{x}^{4}}$…(i)<br/>$⇒$for maxima/minima${f}^{'}\left(x\right) = 0$<br/>$\begin{cases}⇒ cos \pix = 0 & or\dfrac{\pix}{2} = tan \pix\end{cases}$<br/>$\begin{cases}∵ cos \pix \ne 0 & & \left\{∵ tan \pi xwillnotbedefined\right\}\end{cases}$<br/>$∴$maxima/minima will occur<br/>Where$tan $\left(\pix\right) = \dfrac{\pix}{2}$

Case $I:$
\left(i\right)$For example<br/>In$x \in $\left(2n, 2n + \dfrac{1}{2}\right)$ at point ${P}_{2}, x \in \left(2, \dfrac{5}{2}\right)$
$\cos \left(\pix\right) > 0$
and at ${P}_{2}^{+}, \tan \left(\pix\right) > \dfrac{\pix}{2}$
at ${P}_{2}^{-}, \tan \left(\pix\right) < \dfrac{\pix}{2}$
hence from equation (i)
in $x \in \left(2n, 2n + \dfrac{1}{2}\right), {f}^{'}\left(x\right)$ 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 $x \in \left(2n, 2n + \dfrac{1}{2}\right)$
Case $II$
In $x \in \left(2n + 1, 2n + \dfrac{3}{2}\right)$
\left(ii\right)$for example at${P}_{1}, x \in \left(1, \dfrac{3}{2}\right)$<br/>$\cos \left(\pix\right) < 0$<br/>at${P}_{1}^{+}, \tan \left(\pix\right) > \dfrac{\pix}{2}$<br/>and at${P}_{1}^{-}, \tan \left(\pix\right) < \dfrac{\pix}{2}$<br/>hence from equation (i)<br/>in$x \in \left(2n + 1, 2n + \dfrac{3}{2}\right), {f}^{'}\left(x\right)$goes from negative to positive<br/>$⇒$so, for the minima${y}_{1}: 1 < {y}_{1} < \dfrac{3}{2}$<br/>${y}_{2}: 3 < {y}_{2} < \dfrac{7}{2}$<br/>${y}_{3}: 5 < {y}_{3} < \dfrac{11}{2}$<br/>and for the maxima${x}_{1}: 2 < {x}_{1} < \dfrac{5}{2}$<br/>${x}_{2}: 4 < {x}_{2} < \dfrac{9}{2}$<br/>${x}_{3}: 6 < {x}_{3} < \dfrac{13}{2}$<br/>Hence${x}_{n} > {y}_{n}$… \left(a\right)$
$⇒$ 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$<br/>$⇒ {y}_{2} > {x}_{1}$<br/>$\tan \left(\pi{y}_{2}\right) > \tan \left(\pi{x}_{1}\right)$<br/>$\tan \left(\pi{y}_{2}\right) > \tan \left(\pi + \pi{x}_{1}\right) ⇒ {y}_{2} > {x}_{1} + 1$<br/>$⇒ {y}_{n + 1} > {x}_{n} + 1$<br/>Therefore${x}_{n} 1$…(c)<br/>Now consider${x}_{n + 1} - {x}_{n} = \left({x}_{n + 1} - {y}_{n + 1}\right) + \left({y}_{n + 1} - {x}_{n}\right)$<br/>From (b) and (c)<br/>${x}_{n + 1} - {x}_{n} > 2