JEE Main 2026 — Mathematics Trigonometry

Let $\alpha = \sin^{-1}x$α=sin−1x and $\beta = \cos^{-1}x$β=cos−1x. Using $\alpha + \beta = \frac{\pi}{2}$α+β=2π​:

$$f(x) = \alpha^2 + \beta^2 = \alpha^2 + (\frac{\pi}{2} - \alpha)^2 = 2\alpha^2 - \pi\alpha + \frac{\pi^2}{4}$$f(x)=α2+β2=α2+(2π​−α)2=2α2−πα+4π2​

This equals $2(\alpha - \frac{\pi}{4})^2 + \frac{\pi^2}{8}$2(α−4π​)2+8π2​, which is minimized at $\alpha = \frac{\pi}{4}$α=4π​.

For $x \in [-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}]$x∈[−23​​,2​1​], we have $\alpha \in [-\frac{\pi}{3}, \frac{\pi}{4}]$α∈[−3π​,4π​].

Distance from $-\frac{\pi}{3}$−3π​ to vertex: $\frac{\pi}{4} + \frac{\pi}{3} = \frac{7\pi}{12}$4π​+3π​=127π​. Distance from $\frac{\pi}{4}$4π​ to vertex: 0.

Maximum occurs at $x = -\frac{\sqrt{3}}{2}$x=−23​​ where $\alpha = -\frac{\pi}{3}$α=−3π​:

$$f(-\frac{\sqrt{3}}{2}) = 2(\frac{\pi^2}{9}) + \frac{\pi^2}{3} + \frac{\pi^2}{4} = \pi^2(\frac{8 + 12 + 9}{36}) = \frac{29}{36}\pi^2$$f(−23​​)=2(9π2​)+3π2​+4π2​=π2(368+12+9​)=3629​π2

Since $\gcd(29, 36) = 1$gcd(29,36)=1, we have $m = 29, n = 36$m=29,n=36, so $m + n = 65$m+n=65