QJEEM25Mab498: JEE Main 2025 Mathematics - Class 12 JEEM25Mab498 Solution: Using the principal values of the inverse trigonometric…

Let $f(x) = 16\left[\left(\sec^{-1} x\right)^2 + \left(\operatorname{cosec}^{-1} x\right)^2\right]$ .

We can express $f(x)$ as:

$f(x) = 16\left[\left(\sec^{-1} x + \operatorname{cosec}^{-1} x\right)^2 - 2\left(\sec^{-1} x\right)\left(\frac{\pi}{2} - \sec^{-1} x\right)\right]$

This simplifies to:

$f(x) = 16\left[\frac{\pi^2}{4} - \pi \sec^{-1} x + 2 \left(\sec^{-1} x\right)^2\right], \quad \text{where } \sec^{-1} x \in [0, \pi] - \left\{\frac{\pi}{2}\right\}$

Further simplification gives:

$f(x) = 16\left[2\left(\sec^{-1} x - \frac{\pi}{4}\right)^2 + \frac{\pi^2}{4} - \frac{\pi^2}{8}\right]$

For the maximum value when $\sec^{-1} x = \pi$ :

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

For the minimum value when $\sec^{-1} x = \frac{\pi}{4}$ :

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

Therefore, the sum of the maximum and minimum values is:

$\text{Sum} = 22\pi^2$