Question

All the values of $x$ satisfying the equation $2 \tan ^{-1} 2 x=\sin ^{-1}\left(\frac{4 x}{1+4 x^{2}}\right)$ lie in the interval

• A. $\left[-\frac{1}{2}, \frac{1}{2}\right]$
• B. $[-1,1]$
• C. $\left[\frac{1}{2,} \infty\right)$
• D. $\left[-\infty,-\frac{1}{2}\right)$

Answer 1

Answer: (A)

We have,
$\sin ^{-1}\left(\frac{4 x}{1+4 x^{2}}\right)$
$\Rightarrow \sin ^{-1}\left(\frac{2(2 x)}{1+(2 x)^{2}}\right)$
$=2 \tan ^{-1} 2 x,$ when $-1 \leq 2 x \leq 1$
$\therefore -1 \leq 2 x \leq 1$
$\Rightarrow \frac{-1}{2} \leq x \leq \frac{1}{2}$
$\therefore x \in \left[\frac{-1}{2}, \frac{1}{2}\right]$