JEE Main 2026 Jan 21 Shift 2, Mathematics Q23: Maxima & Minima

Let $u={\sin }^{-1}x$, so ${\cos }^{-1}x=\frac{\pi }{2}-u$. Then $({\sin }^{-1}x{)}^2+({\cos }^{-1}x{)}^2=u^2+{\left(\frac{\pi }{2}-u\right)}^2=2u^2-\pi u+\frac{{\pi }^2}{4}$.

This is a parabola in $u$ opening upward, with vertex at $u=\pi /4$. So the maximum on a closed interval is reached at an endpoint of the $u$-range.

For $x\in [-\sqrt{3}/2,1/\sqrt{2}]$: $u={\sin }^{-1}x\in [-\pi /3,\pi /4]$. The endpoint farther from $\pi /4$ is $u=-\pi /3$.

Evaluate at $u=-\pi /3$: $2{\left(\frac{\pi }{3}\right)}^2+\pi \cdot \frac{\pi }{3}+\frac{{\pi }^2}{4}=\frac{2{\pi }^2}{9}+\frac{{\pi }^2}{3}+\frac{{\pi }^2}{4}$.

Common denominator 36: $\frac{8{\pi }^2+12{\pi }^2+9{\pi }^2}{36}=\frac{29{\pi }^2}{36}$.

So $m=29$, $n=36$, $\gcd (29,36)=1$, giving $m+n=65$.