Question

The set of values of ${}^{\prime }{x}^{\prime }$ for which the formula $2{\sin }^{-1}x={\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$ is true, is

• A. $(-1,0)$
• B. $[0,1]$
• C. $[-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2}]$
• D. $[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$

Answer 1

The correct option is A $(-1,0)$

Consider the given equation,

$2{\sin }^{-1}x={\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$

Put, $x=\sin \theta \Rightarrow \theta ={\sin }^{-1}x$

$2{\sin }^{-1}\sin \theta ={\sin }^{-1}\left(2\sin \theta \sqrt{1-{\sin }^2\theta }\right)$

$2\theta ={\sin }^{-1}\left(2\sin \theta \sqrt{{\cos }^2\theta }\right)(\because {\sin }^{2}A+{\cos }^2)$

$2\theta ={\sin }^{-1}\left(2\sin \theta \cos \theta \right)$

$2\theta ={\sin }^{-1}\sin 2\theta$

$2\theta =2\theta$

$\sin x=\sin x$

Now,

$\because -1\le \sin x\ge 1$

$\therefore -\frac{\pi }{2}\le x\ge \frac{\pi }{2}$

$x\in [-\frac{\pi }{2},\frac{\pi }{2}]$

Hence, this is the answer.