Let the maximum value of ( ^-1x)^2+( ^-1x)^2 for x [-√32,1√2] be (m)/(n) ^2, where gcd(m,n…

Question

Let the maximum value of ${({\sin }^{-1}x)}^{2}+{({\cos }^{-1}x)}^{2}$ for $x\in [-\frac{\sqrt{3}}{2},\frac{1}{\sqrt{2}}]$ be $\frac{m}{n}{\pi }^{2}$, where $gcd(m,n)=1$. Then $m+n$ is equal to $____$

Answer: 65

Accepted Answer

$f(x)={({\sin }^{-1}x)}^{2}+{({\cos }^{-1}x)}^{2},\, x\in [-\frac{\sqrt{3}}{2},\frac{1}{\sqrt{2}}]$

For $x\in [-1,1]$,

${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}.$

Let $a={\sin }^{-1}x$. Then ${\cos }^{-1}x=\frac{\pi }{2}-a$, and

$f={a}^{2}+{(\frac{\pi }{2}-a)}^{2}=2{a}^{2}-\pi a+\frac{{\pi }^{2}}{4}.$

This is a quadratic in $a$ opening upward, so its maximum on the allowed interval occurs at an endpoint of $a$.

Now,

$x=-\frac{\sqrt{3}}{2}\Rightarrow a=-\frac{\pi }{3},\, x=\frac{1}{\sqrt{2}}\Rightarrow a=\frac{\pi }{4}.$

Evaluate $f$:

At $a=\frac{\pi }{4}$: this is the vertex, giving the minimum.

At $a=-\frac{\pi }{3}$:

$f={(-\frac{\pi }{3})}^{2}+{(\frac{5\pi }{6})}^{2}=\frac{{\pi }^{2}}{9}+\frac{25{\pi }^{2}}{36}=\frac{29{\pi }^{2}}{36}.$

Hence the maximum value is

$\frac{m}{n}{\pi }^{2}=\frac{29}{36}{\pi }^{2},$

with $gcd(29,36)=1$.

$m+n=65$

Solutions · 1