The minimum value of ≤ft( ^{-1} x)²+≤ft( ^{-1} x)² is \_\_\_\_

Correct Answer is : π28{π²}{8}8π2

Correct Answer is : π28{π²}{8}8π2

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MHTCET 11 Apr 2026 Shift 1 Paper

Question : 115 of 150

Marks: +1, -0

\left(\sin ^{-1} x\right)^{2}+\left(\cos ^{-1} x\right)^{2}

\_\_\_\_

Solution:

Concept:

Use the identity

\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}

for

x \in [-1,1]

and convert the expression into a quadratic form.

Explanation:

Let

\sin^{-1} x = \theta

Then

\cos^{-1} x = \frac{\pi}{2} - \theta

So the expression becomes

\theta^2 + \left(\frac{\pi}{2} - \theta\right)^2

Simplify:

\theta^2 + \frac{\pi^2}{4} - \pi\theta + \theta^2 = 2\theta^2 - \pi\theta + \frac{\pi^2}{4}

This quadratic expression is minimum when

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

Substitute

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

\left(\frac{\pi}{4}\right)^2 + \left(\frac{\pi}{4}\right)^2 = \frac{\pi^2}{16} + \frac{\pi^2}{16} = \frac{\pi^2}{8}

Hence the minimum value is

\frac{\pi^2}{8}

Answer:

\frac{\pi^2}{8}

, which is Option A.

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