Question

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

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

Answer 1

The correct option is B $x\in [-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$

If $x$ is substituted as $sin\theta$, we get $si{n}^{-1}sin2\theta =2si{n}^{-1}sin\theta$
This would be true only if $2\theta \in [\frac{-\pi }{2},\frac{\pi }{2}]$
Or, $\theta \in [\frac{-\pi }{4},\frac{\pi }{4}]$
$\Rightarrow x\in [\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$