Q67 / 150 · MHT CET 2026 (11 April Shift 1) · Mathematics
The minimum value of $(\sin^{-1} x)^2 + (\cos^{-1} x)^2$ is............
- Correct answerA$\dfrac{\pi^2}{8}$
- B$\dfrac{3\pi^2}{8}$
- C$\dfrac{5\pi^2}{8}$
- D$\dfrac{7\pi^2}{8}$
Solution
Let $y = (\sin^{-1} x)^2 + (\cos^{-1} x)^2$
Using the identity $\sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2}$, we can write $\cos^{-1} x = \dfrac{\pi}{2} - \sin^{-1} x$
Substituting this into the expression for $y$:
$y = (\sin^{-1} x)^2 + \left(\dfrac{\pi}{2} - \sin^{-1} x\right)^2$
Let $t = \sin^{-1} x$, where $t \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$
$y = t^2 + \left(\dfrac{\pi}{2} - t\right)^2$
$y = 2t^2 - \pi t + \dfrac{\pi^2}{4}$
$y = 2\left(t^2 - \dfrac{\pi}{2}t\right) + \dfrac{\pi^2}{4}$
$y = 2\left(t - \dfrac{\pi}{4}\right)^2 - 2\left(\dfrac{\pi}{4}\right)^2 + \dfrac{\pi^2}{4}$
$y = 2\left(t - \dfrac{\pi}{4}\right)^2 + \dfrac{\pi^2}{8}$
The minimum value occurs when $t = \dfrac{\pi}{4}$, which lies in the interval $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$
Therefore, the minimum value is $\dfrac{\pi^2}{8}$
Answer: $\dfrac{\pi^2}{8}$
Using the identity $\sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2}$, we can write $\cos^{-1} x = \dfrac{\pi}{2} - \sin^{-1} x$
Substituting this into the expression for $y$:
$y = (\sin^{-1} x)^2 + \left(\dfrac{\pi}{2} - \sin^{-1} x\right)^2$
Let $t = \sin^{-1} x$, where $t \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$
$y = t^2 + \left(\dfrac{\pi}{2} - t\right)^2$
$y = 2t^2 - \pi t + \dfrac{\pi^2}{4}$
$y = 2\left(t^2 - \dfrac{\pi}{2}t\right) + \dfrac{\pi^2}{4}$
$y = 2\left(t - \dfrac{\pi}{4}\right)^2 - 2\left(\dfrac{\pi}{4}\right)^2 + \dfrac{\pi^2}{4}$
$y = 2\left(t - \dfrac{\pi}{4}\right)^2 + \dfrac{\pi^2}{8}$
The minimum value occurs when $t = \dfrac{\pi}{4}$, which lies in the interval $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$
Therefore, the minimum value is $\dfrac{\pi^2}{8}$
Answer: $\dfrac{\pi^2}{8}$
Hence, the correct answer is option: 'A'.