Let the maximum value of ( ^-1 x)^2+( ^-1 x)^2 for x [-32, 12] be mn ^2, where gcd(m, n)=1. Then m+n is equal to \_\_\_\_ .

( ^-1 x)^2+( ^-1 x)^2\\ =( ^-1 x+ ^-1 x)^2-2 ^-1 x ^-1 x\\ =^24-2( ^-1 x)(2- ^-1 x)\\ =2( ^-1 x-4)^2+^28 where ^-1 x [-3, 4]\\ Then max value occurs at ^-1 x=-3\\ Which is 2(3+4)^2+^28=29 ^236\\ m=29 and n=36\\ m+n=65

Let the maximum value of sin 1 x 2 cos 1 x 2 for x in frac : null (Mathematics)

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

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

=\frac{\pi^{2}}{4}-2\left(\sin ^{-1} \mathrm{x}\right)\left(\frac{\pi}{2}-\sin ^{-1} \mathrm{x}\right)\\

=2\left(\sin ^{-1} \mathrm{x}-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8} \quad

where

\sin ^{-1} \mathrm{x} \in\left[\frac{-\pi}{3}, \frac{\pi}{4}\right]\\

Then max value occurs at

\sin ^{-1} x=\frac{-\pi}{3}\\

Which is

2\left(\frac{\pi}{3}+\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}=\frac{29 \pi^{2}}{36}\\

\Rightarrow \mathrm{m}=29

and

\mathrm{n}=36\\

\therefore \mathrm{m}+\mathrm{n}=65

Explanation

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