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

Answer: 0004

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