Question

Let $f(x)=\left(\sin ^{-1} x\right)^2-\left(\cos ^{-1} x\right)^2$. If range of $f(x)$ equals $\left[\frac{a \pi^2}{4}, \frac{b \pi^2}{4}\right]$ where $a, b \in I$, then find $(b-a)$.

Answer 1

$\text { Given, } f ( x )=\left(\sin ^{-1} x \right)^2-\left(\cos ^{-1} x \right)^2 $
$\left.=\left(\sin ^{-1} x +\cos ^{-1} x \right)\left(\sin ^{-1} x -\cos ^{-1} x \right)=\frac{\pi}{2}\left(\sin ^{-1} x -\left(\frac{\pi}{2}-\sin ^{-1} x \right)\right)\right)$
$=\left(\sin ^{-1} x+\cos ^{-1} x\right)\left(\sin ^{-1} x-\cos ^{-1} x\right)=\frac{\pi}{2}\left(\sin ^{-1} x-\left(\frac{\pi}{2}-\sin ^{-1} x\right)\right)$
$\therefore f ( x )=\frac{\pi}{2}\left(2 \sin ^{-1} x -\frac{\pi}{2}\right), \forall x \in[-1,1] $ [As, $f$ is increasing function]
So, $f _{\max }( x =1)=\frac{\pi}{2}\left(2\left(\frac{\pi}{2}\right)-\frac{\pi}{2}\right)=\frac{\pi^2}{4}$.
Also, $f_{\min }(x=-1)=\frac{\pi}{2}\left(-2\left(\frac{\pi}{2}\right)-\frac{\pi}{2}\right)=\frac{-3 \pi^2}{4}$
$\Rightarrow \text { range of } f ( x )=\left[\frac{-3 \pi^2}{4}, \frac{\pi^2}{4}\right]=\left[\frac{ a \pi^2}{4}, \frac{ b \pi^2}{4}\right] \text { (Given) } $
$\therefore a =-3 \text { and } b =1$
$\text { Hence, }( b - a )=1-(-3)=4 $