Question

Difference between the maximum and the minimum value of the function $f\left(x\right)=({\sin }^{-1}x{)}^2+({\cos }^{-1}x{)}^2$ is :

• A. $\frac{9{\pi }^2}{8}$
• B. $\frac{10{\pi }^2}{8}$
• C. $\frac{{\pi }^2}{8}$
• D. $2$

Answer 1

The correct option is A $\frac{9{\pi }^2}{8}$

$f\left(x\right)=({\sin }^{-1}x{)}^2+({\cos }^{-1}x{)}^2$
Put ${\sin }^{-1}x=t$ where $t\in [\frac{-\pi }{2},\frac{\pi }{2}]$$=(t{)}^2+{(\frac{\pi }{2}-t)}^2(\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2})=2{[t-\frac{\pi }{4}]}^2+\frac{{\pi }^2}{8}$
$\Rightarrow$ min. value is $\frac{{\pi }^2}{8}$
and max. value will be $\frac{10{\pi }^2}{8}$