Question

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

• A. $[-1,1]$
• B. $[-\frac{1}{\sqrt{2}},1]$
• C. $[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$
• D. $noneofthese$

Answer 1

The correct option is A $[-1,1]$

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

Given expression is an identity. Only we need to check is domain.
(i) Let $2{\sin }^{-1}x=\theta ⟹x=\sin \frac{\theta }{2}$

As $|\sin \theta |\le 1⟹|x|\le 1⟹x\in [-1,1]$
(ii)Let ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)=\theta ⟹2x\sqrt{1-x^2}=\sin \theta$

$|2x\sqrt{1-x^2}|\le 1⟹x^2(1-x^2)\le 1/4$
$⟹4x^4-4x^2+1\ge 0⟹(2x^2-1{)}^2\ge 0$

Hence using (i) and (ii) the domain is $x\in [-1,1]$