JEE Mathematics 2025 Question with Solution

Given: We have $16\left[(\sec^{-1}x)^2 + (\csc^{-1}x)^2\right]$.

Find: The sum of its maximum and minimum values.

Let $a = \sec^{-1}x$. Then $\csc^{-1}x = \frac{\pi}{2} - a$.

Substitute into the expression:

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

To find the minimum value, differentiate the quadratic inside the bracket:

$$\frac{d}{da}\left(2a^2 - \pi a + \frac{\pi^2}{4}\right) = 4a - \pi$$

Set it equal to zero:

$$4a - \pi = 0$$ $$a = \frac{\pi}{4}$$

Now,

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

For the maximum value, check the endpoints.

At $a = \pi$,

$$16\left[2\pi^2 - \pi(\pi) + \frac{\pi^2}{4}\right] = 16\left[\frac{5\pi^2}{4}\right] = 20\pi^2$$

At $a = 0$,

$$16\left[2(0)^2 - \pi(0) + \frac{\pi^2}{4}\right] = 16\left[\frac{\pi^2}{4}\right] = 4\pi^2$$

So the maximum value is $20\pi^2$.

Therefore,

$$\text{Sum} = 2\pi^2 + 20\pi^2 = 22\pi^2$$

The correct option is D.

Given: $16\left[(\sec^{-1}x)^2 + (\csc^{-1}x)^2\right]$

Find: The sum of the maximum and minimum values.

Let

$$\theta = \sec^{-1}x \Rightarrow x = \sec\theta$$

Then

$$\csc^{-1}x = \sin^{-1}(\cos\theta)$$

For $\theta \in [0, \pi/2]$,

$$\sin^{-1}(\cos\theta) = \frac{\pi}{2} - \theta$$

So,

$$f(\theta) = \theta^2 + \left(\frac{\pi}{2} - \theta\right)^2 = 2\theta^2 - \pi\theta + \frac{\pi^2}{4}$$

Differentiate:

$$f'(\theta) = 4\theta - \pi = 0$$

Hence,

$$\theta = \frac{\pi}{4}$$

Minimum value:

$$f_{\min} = 2\left(\frac{\pi}{4}\right)^2 - \pi\left(\frac{\pi}{4}\right) + \frac{\pi^2}{4} = \frac{\pi^2}{8}$$

Maximum value at $\theta = \pi$:

$$f_{\max} = 2\pi^2 - \pi^2 + \frac{\pi^2}{4} = \frac{5\pi^2}{4}$$

Therefore,

$$16(f_{\max} + f_{\min}) = 16\left(\frac{5\pi^2}{4} + \frac{\pi^2}{8}\right) = 22\pi^2$$

Hence the required sum is $22\pi^2$, so the correct option is D.