Q36 / 75 · JEE Main 2025 (22 Jan Shift 1) · Mathematics

Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16\left(\left(\sec ^{-1} x\right)^2+\left(\operatorname{cosec}^{-1} x\right)^2\right)$ is :

  • A
    $24 \pi^2$
  • Correct answer
    B
    $22 \pi^2$
  • C
    $31 \pi^2$
  • D
    $18 \pi^2$

Solution

$\begin{aligned} & 16\left(\sec ^{-1} x\right)^2+\left(\operatorname{cosec}^{-1} x\right)^2 \\ & \operatorname{Sec}^{-1} x=a \in[0, \pi]-\left\{\frac{\pi}{2}\right\} \\ & \operatorname{cosec}^{-1} x=\frac{\pi}{2}-a \\ & =16\left[a^2+\left(\frac{\pi}{2}-a\right)^2\right]=16\left[2 a^2-\pi a+\frac{\pi^2}{4}\right] \\ & \max ]_{a=\pi}=16\left[2 \pi^2-\pi^2+\pi \frac{2}{4}\right]=20 \pi^2 \\ & \min ]_{a=\frac{\pi}{4}}=16\left[\frac{2 \times \pi^2}{16}-\frac{\pi^2}{4}+\frac{\pi^2}{4}\right]=2 \pi^2 \\ & \operatorname{Sum}=22 \pi^2\end{aligned}$

Hence, the correct answer is option: 'B'.

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