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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
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Verified Answer
The correct answer is:
$\left[-\frac{1}{2}, \frac{1}{2}\right]$
We have, $\sin ^{-1}\left(\frac{4 x}{1+4 x^2}\right) \Rightarrow \sin ^{-1}\left(\frac{2(2 x)}{1+(2 x)^2}\right)$
$$
\begin{aligned}
& =2 \tan ^{-1} 2 x \text {, when }-1 \leq 2 x \leq 1 \\
& \therefore \quad-1 \leq 2 x \leq 1 \Rightarrow \frac{-1}{2} \leq x \leq \frac{1}{2} \\
& \therefore \quad x \in\left[\frac{-1}{2}, \frac{1}{2}\right] \\
&
\end{aligned}
$$
$$
\begin{aligned}
& =2 \tan ^{-1} 2 x \text {, when }-1 \leq 2 x \leq 1 \\
& \therefore \quad-1 \leq 2 x \leq 1 \Rightarrow \frac{-1}{2} \leq x \leq \frac{1}{2} \\
& \therefore \quad x \in\left[\frac{-1}{2}, \frac{1}{2}\right] \\
&
\end{aligned}
$$
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