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)

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