JEE Main 2025 — Inverse Trigonometric Functions Question with Solution

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

We can express $f(x)$ as:

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

This simplifies to:

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

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:

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