JEE MAIN - Mathematics (2025 - 22nd January Morning Shift - No. 18)
Explanation
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 $
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