Question

Find the range of the given
$f\left(x\right)=\left({\sin }^{-1}x\right)\left({\cos }^{-1}x\right)$

Answer 1

Given :

$f\left(x\right)=\left(si{n}^{-1}x\right)\left(co{s}^{-1}x\right)$

we know that $si{n}^{-1}+co{s}^{-1}x=\frac{\pi }{2}$

so, $co{s}^{-1}x=(\frac{\pi }{2}-si{n}^{-1}x)$

$f\left(x\right)=si{n}^{-1}x[\frac{\pi }{2}-si{n}^{-1}x]$

$f\left(x\right)=\frac{\pi }{2}si{n}^{-1}x-(si{n}^{-1}x{)}^2$

$f\left(x\right)=2\left(\frac{\pi }{4}\right)si{n}^{-1}x-(si{n}^{-1}x{)}^2$

$=-\left[\right(si{n}^{-1}x{)}^2-2\left(\frac{\pi }{4}\right)si{n}^{-1}x+\frac{{\pi }^2}{16}-\frac{{\pi }^2}{16}]$

$=-\left[\right(si{n}^{-1}x-\frac{\pi }{4}{)}^2-\frac{{\pi }^2}{16}]$

$f\left(x\right)=\frac{{\pi }^2}{16}-(si{n}^{-1}x-\frac{\pi }{4}{)}^2$

maximum, when $si{n}^{-1}x-\frac{\pi }{4}=0$

so, $f(x{)}_{max}=\frac{{\pi }^2}{16}$

$f(x{)}_{minimum}$, when $si{n}^{-1}x=(-\frac{\pi }{2})$

so, $f(x{)}_{min}=\frac{{\pi }^2}{16}-(-\frac{\pi }{2}-\frac{\pi }{4}{)}^2=\frac{{\pi }^2}{16}-(-\frac{3\pi }{4}{)}^2$

$f(x{)}_{min}=\frac{{\pi }^2}{16}-\frac{9{\pi }^2}{16}=\frac{-8{\pi }^2}{16}=\frac{-{\pi }^2}{2}$

so, Range of $f\left(x\right)=[-\frac{{\pi }^2}{2},\frac{{\pi }^2}{16}]$