Question

If $\left(\sin ^{-1} x\right)^2-\left(\cos ^{-1} x\right)^2=a \pi^2$ then the range of $a$ is

• A. $\left[\frac{-3}{4}, \frac{1}{4}\right]$
• B. $\left[\frac{-3}{4}, \frac{3}{4}\right]$
• C. $[-1,1]$
• D. $\left[-1, \frac{3}{4}\right]$

Answer 1

Answer: (A)

$\frac{\pi}{2}\left(2 \sin ^{-1} x-\frac{\pi}{2}\right)=a \pi^2, \frac{\pi}{2}\left(2 \sin ^{-1} x-\frac{\pi}{2}\right)=a \pi^2 $
$\therefore \sin ^{-1} x=\frac{2 a \pi+\frac{\pi}{2}}{2}=(4 a+1) \frac{\pi}{4}$
$\frac{-\pi}{2} \angle(4 a+1) \frac{\pi}{4} \leq \frac{\pi}{2} $
$-2 \leq(4 a+1) \leq 2$
$-3 \leq 4 a \leq 1 $
$a \in\left[\frac{-3}{4}, \frac{1}{4}\right] $