Mathematics - Functions Question with Solution | TestHub

${\left({\cos }^{-1}x\right)}^2+{\left({\sin }^{-1}x\right)}^2=\frac{1}{2}\left\{{\left({\cos }^{-1}x+{\sin }^{-1}x\right)}^2+{\left({\cos }^{-1}x-{\sin }^{-1}x\right)}^2\right\}$ $=\frac{1}{2}\left\{{\left(\frac{\pi }{2}\right)}^2+{\left(\frac{\pi }{2}-2{\sin }^{-1}x\right)}^2\right\}\ge \frac{{\pi }^2}{8}$ $a=\frac{1}{{\pi }^2}\left(\frac{{\pi }^2}{8}\right)=\frac{1}{8}$ $b=\frac{1}{2{\pi }^2}\left\{\frac{{\pi }^2}{4}+{\left(\frac{\pi }{2}+\pi \right)}^2\right\}\text{, at }x=\frac{-\pi }{2}$

$=\frac{5}{4}$

$\therefore b-a=\frac{9}{8}$