Question

The minimum value of the function $M=\frac{{\pi }^2}{16{\sin }^{-1}(-x)}+{\cos }^{-1}x$ is

• A. $0$
• B. $\pi$
• C. $\frac{\pi }{2}$
• D. $-\frac{\pi }{8}$

Answer 1

The correct option is D $-\frac{\pi }{8}$

$M=\frac{{\pi }^2}{16{\sin }^{-1}(-x)}+{\cos }^{-1}x$
Let
${\sin }^{-1}x=t,x\in [-1,1],\&t\in [-\frac{\pi }{2},\frac{\pi }{2}]$
$M=-\frac{{\pi }^2}{16t}+\frac{\pi }{2}-t(\because {\sin }^{-1}(-x)=-{\sin }^{-1}(x)\&{\cos }^{-1}x=\frac{\pi }{2}-{\sin }^{-1}x)$

$\frac{dM}{dt}=-\frac{{\pi }^2}{16}(-\frac{1}{t^2})-1=0$
$\Rightarrow t=\pm \frac{\pi }{4}$ are critical points.
and $t=\pm \frac{\pi }{2}$ are end points

at $t=\frac{\pi }{2}$
$M_{min}=-\frac{\pi }{8}$