Q. If 2$\sin^{- 1}{x = \sin^{- 1}{\left( 2x\sqrt{1 - x^{2}} \right),}}$ then $x$ is equal to
A) $\lbrack - 1,1\rbrack$
B) $\left\lbrack - \frac{1}{\sqrt{2}},1 \right\rbrack$
C) $\left\lbrack - \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right\rbrack$
D) None of these
Explanation
(c)
Since, $2\sin^{- 1}{= \sin^{- 1}{(2x\sqrt{1 - x^{2})}}}$
Range of right hand side is $\left\lbrack - \frac{\pi}{2},\frac{\pi}{2} \right\rbrack$
$\Rightarrow - \frac{\pi}{2} \leq 2\sin^{- 1}{x \leq \frac{\pi}{2}}$
$\Rightarrow \frac{\pi}{4} \leq \sin^{- 1}{x \leq}\frac{\pi}{4}$
$\Rightarrow x \in \left\lbrack - \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.