Let the maximum value of (sin^{-1}x)^{2} + (c | vedclass

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(B) Let $f(x) = (sin^{-1}x)^{2} + (cos^{-1}x)^{2}$.Since $cos^{-1}x = \frac{\pi}{2} - sin^{-1}x$,we have:$f(x) = (sin^{-1}x)^{2} + (\frac{\pi}{2} - si

Vedclass · Accepted Answer

$65$

Vedclass · Answer

(B) Let $f(x) = (sin^{-1}x)^{2} + (cos^{-1}x)^{2}$.Since $cos^{-1}x = \frac{\pi}{2} - sin^{-1}x$,we have:$f(x) = (sin^{-1}x)^{2} + (\frac{\pi}{2} - sin^{-1}x)^{2}$$f(x) = (sin^{-1}x)^{2} + \frac{\pi^{2}}{4} - \pi sin^{-1}x + (sin^{-1}x)^{2}$$f(x) = 2(sin^{-1}x)^{2} - \pi sin^{-1}x + \frac{\pi^{2}}{4}$$f(x) = 2[(sin^{-1}x)^{2} - \frac{\pi}{2} sin^{-1}x] + \frac{\pi^{2}}{4}$$f(x) = 2[(sin^{-1}x - \frac{\pi}{4})^{2} - \frac{\pi^{2}}{16}] + \frac{\pi^{2}}{4}$$f(x) = 2(sin^{-1}x - \frac{\pi}{4})^{2} + \frac{\pi^{2}}{8}$.Given $x \in [-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}]$,the range of $sin^{-1}x$ is $[-\frac{\pi}{3}, \frac{\pi}{4}]$.To maximize $f(x)$,we choose the value of $sin^{-1}x$ furthest from $\frac{\pi}{4}$,which is $-\frac{\pi}{3}$.Max value $= 2(-\frac{\pi}{3} - \frac{\pi}{4})^{2} + \frac{\pi^{2}}{8} = 2(-\frac{7\pi}{12})^{2} + \frac{\pi^{2}}{8} = 2(\frac{49\pi^{2}}{144}) + \frac{\pi^{2}}{8} = \frac{49\pi^{2}}{72} + \frac{9\pi^{2}}{72} = \frac{58\pi^{2}}{72} = \frac{29\pi^{2}}{36}$.Thus,$m = 29$ and $n = 36$. Since $\gcd(29, 36) = 1$,$m+n = 29 + 36 = 65$.

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 ........... .

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