Let the maximum value of $(\sin^{-1}x)^{2}+(\cos^{-1}x)^{2}$ for $x\in\left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right]$ be $\frac{m}{n}\pi^{2}$, where $\gcd(m,n)=1$. Then $m+n$ is equal to:
Solution:
Let $f(x) = (\sin^{-1}x)^{2}+(\cos^{-1}x)^{2}$
We know that $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$
$\therefore \cos^{-1}x = \frac{\pi}{2} – \sin^{-1}x$
$f(x) = (\sin^{-1}x)^{2} + \left(\frac{\pi}{2} – \sin^{-1}x\right)^{2}$
$f(x) = (\sin^{-1}x)^{2} + \frac{\pi^{2}}{4} + (\sin^{-1}x)^{2} – \pi \sin^{-1}x$
$f(x) = 2(\sin^{-1}x)^{2} – \pi \sin^{-1}x + \frac{\pi^{2}}{4}$
$f(x) = 2\left[ (\sin^{-1}x)^{2} – \frac{\pi}{2} \sin^{-1}x + \frac{\pi^{2}}{16} \right] – \frac{\pi^{2}}{8} + \frac{\pi^{2}}{4}$
$f(x) = 2\left( \sin^{-1}x – \frac{\pi}{4} \right)^{2} + \frac{\pi^{2}}{8}$
Given $x \in \left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right]$
$\Rightarrow \sin^{-1}x \in \left[-\frac{\pi}{3}, \frac{\pi}{4}\right]$
For maximum value of $f(x)$, $\left( \sin^{-1}x – \frac{\pi}{4} \right)^{2}$ should be maximum.
This occurs at $\sin^{-1}x = -\frac{\pi}{3}$.
Max value $= 2\left( -\frac{\pi}{3} – \frac{\pi}{4} \right)^{2} + \frac{\pi^{2}}{8}$
$= 2\left( -\frac{7\pi}{12} \right)^{2} + \frac{\pi^{2}}{8}$
$= 2\left( \frac{49\pi^{2}}{144} \right) + \frac{\pi^{2}}{8}$
$= \frac{49\pi^{2}}{72} + \frac{9\pi^{2}}{72}$
$= \frac{58\pi^{2}}{72} = \frac{29\pi^{2}}{36}$
Comparing with $\frac{m}{n}\pi^{2}$, we get $m=29, n=36$.
$\gcd(29, 36) = 1$
$\therefore m+n = 29 + 36 = 65$
Ans. 65