Question

$2{\cos }^{-1}x={\sin }^{-1}(2x\sqrt{1-x^2)}$ is valid for all values of $x$ satisfying

• A. $-1\le x\le 1$
• B. $0\le x\le 1$
• C. $\frac{1}{\sqrt{2}}\le x\le 1$
• D. $0\le x\le \frac{1}{\sqrt{2}}$

Answer 1

Answer: (C)

Put $co{s}^{-1}x=y,sothat$x = cos{\,}y$<br/><br/>Then,$0 {\,}\le$y\le \pi$ and $\left|x\right|\le 1\dots (i)$
and the RHS of given equation becomes
$si{n}^{-1}\left(2cosysiny\right)=si{n}^{-1}\left(sin2y\right)=2y$
Since, $si{n}^{-1}\left(2x\sqrt{1-x^2}\right)$ lies between $-\frac{\pi }{2}$ and $\frac{\pi }{2}.$
$\therefore$ 2y lies between $-\frac{\pi }{2}$ and $\frac{\pi }{2}$.
i.e., y lies between - $\frac{\pi }{4}$ and $\frac{\pi }{4}$.
$\therefore -\frac{\pi }{4}\le y\le \frac{\pi }{4}.\dots (ii)$
On combining Eqs. (i) and (ii), we get
$0\le y\le \frac{\pi }{4}$
$\Rightarrow 1\ge cosy\ge \frac{1}{\sqrt{2}}$
$\Rightarrow \frac{1}{\sqrt{2}}\le x\le 1$
$\Rightarrow x\in \left[\frac{1}{\sqrt{2}},1\right]$