JEE Main 2025 — 22 January, Morning Shift — Question 5

$16\left(\sec ^{-1} \mathrm{x}\right)^{2}+\left(\operatorname{cosec}^{-1} \mathrm{x}\right)^{2}$

$\operatorname{Sec}^{-1} \mathrm{x}=\mathrm{a} \in[0, \pi]-\left\{\frac{\pi}{2}\right\}$

$\operatorname{cosec}^{-1} x=\frac{\pi}{2}-a$

$=16\left[a^{2}+\left(\frac{\pi}{2}-a\right)^{2}\right]=16\left[2 a^{2}-\pi a+\frac{\pi^{2}}{4}\right]$

$\max ]_{a=\pi}=16\left[2 \pi^{2}-\pi^{2}+\pi \frac{2}{4}\right]=20 \pi^{2}$

$\min ]_{\mathrm{a}=\frac{\pi}{4}}=16\left[\frac{2 \times \pi^{2}}{16}-\frac{\pi^{2}}{4}+\frac{\pi^{2}}{4}\right]=2 \pi^{2}$

Sum $=22 \pi^{2}$