Let the maximum value of ( ^ -1 x)^ 2 +( ^ -1 x)^ 2 for x ∈[- 3 2 , 1 2 ] be m n π^ 2 , where gcd(m, n)=1 . Then m+n is equal to \_\_\_\_ .

Correct answer: 0 Let α = ^ -1 x and β = ^ -1 x . Using α + β = (π)/(2) : f(x) = α^2 + β^2 = α^2 + ((π)/(2) - α)^2 = 2α^2 - πα + (π^2)/(4) This equals 2(α - (π)/(4))^2 + (π^2)/(8) , which is minimized at α = (π)/(4) . For x ∈ [- 3 2 , 1 2 ] , we have α ∈ [-(π)/(3), (π)/(4)] . Distance from -(π)/(3) to vertex: (π)/(4) + (π)/(3) = (7π)/(12) . Distance from (π)/(4) to vertex: 0. Maximum occurs at x = - 3 2 where α = -(π)/(3) : f(- 3 2 ) = 2((π^2)/(9)) + (π^2)/(3) + (π^2)/(4) = π^2((8 + 12 + 9)/(36)) = (29)/(36)π^2 Since (29, 36) = 1 , we have m = 29, n = 36 , so m + n = 65

JEE Main 202621 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 21 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

JEE Main 2026 · 21 January · Evening Shift · Mathematics · Inverse Trigonometric Functions

Let the maximum value of ( ^ -1 x)^ 2 +( ^ -1 x)^ 2 for x ∈[- 3 2 ,...

Let the maximum value of

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

for

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

\frac{\mathrm{m}}{\mathrm{n}} \pi^{2}

, where

\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1

. Then

\mathrm{m}+\mathrm{n}

is equal to

\_\_\_\_

Correct answer

Step-by-step solution

Let

\alpha = \sin^{-1}x

and

\beta = \cos^{-1}x

. Using

\alpha + \beta = \frac{\pi}{2}

f(x) = \alpha^2 + \beta^2 = \alpha^2 + (\frac{\pi}{2} - \alpha)^2 = 2\alpha^2 - \pi\alpha + \frac{\pi^2}{4}

This equals

2(\alpha - \frac{\pi}{4})^2 + \frac{\pi^2}{8}

, which is minimized at

\alpha = \frac{\pi}{4}

.
For

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

, we have

\alpha \in [-\frac{\pi}{3}, \frac{\pi}{4}]

.
Distance from

-\frac{\pi}{3}

to vertex:

\frac{\pi}{4} + \frac{\pi}{3} = \frac{7\pi}{12}

. Distance from

\frac{\pi}{4}

to vertex: 0.
Maximum occurs at

x = -\frac{\sqrt{3}}{2}

where

\alpha = -\frac{\pi}{3}

f(-\frac{\sqrt{3}}{2}) = 2(\frac{\pi^2}{9}) + \frac{\pi^2}{3} + \frac{\pi^2}{4} = \pi^2(\frac{8 + 12 + 9}{36}) = \frac{29}{36}\pi^2

Since

\gcd(29, 36) = 1

, we have

m = 29, n = 36

, so

m + n = 65

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