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JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Application of Derivatives

Let the maximum value of

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

for

x \in [- \frac{3}{2}, \frac{1}{2}]

be

\frac{m}{n} π^{2}

, where

gcd (m, n) = 1

.

Then

m + n

is equal to ____.

Answer: 65

Solution

Let

α = sin^{- 1} x

. Then

cos^{- 1} x = \frac{π}{2} - α

.

(sin^{- 1} x)^{2} + (cos^{- 1} x)^{2} = α^{2} + (\frac{π}{2} - α)^{2}

= 2 α^{2} - π α + \frac{π ^{2}}{4} = 2 (α - \frac{π}{4})^{2} + \frac{π ^{2}}{8}

For

x \in [- \frac{3}{2}, \frac{1}{2}]

, we have

α \in [- \frac{π}{3}, \frac{π}{4}]

.

The function is maximized at the endpoint farthest from

\frac{π}{4}

, which is

α = - \frac{π}{3}

:

2 (- \frac{π}{3} - \frac{π}{4})^{2} + \frac{π ^{2}}{8} = 2 (\frac{7 π}{12})^{2} + \frac{π ^{2}}{8} = \frac{29 π ^{2}}{36}

So

m = 29, n = 36

, and

m + n = 65

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Integer answer

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