The interval for $x$ which satisfies the equation $2 {tan}^{- 1} 2 x = {sin}^{- 1} \frac{4 x}{1 + 4 x^{2}}$ is

[\frac{1}{2}, \infty)

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(- \infty, - \frac{1}{2}]

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[- 1, 1]

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[- \frac{1}{2}, \frac{1}{2}]

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Solution

The correct option is D $[- \frac{1}{2}, \frac{1}{2}]$
Given : $2 {tan}^{- 1} 2 x = {sin}^{- 1} \frac{4 x}{1 + 4 x^{2}}$
Let $y = \frac{4 x}{1 + 4 x^{2}}$
$\Rightarrow 4 y x^{2} - 4 x + y = 0 ∵ x \in R \Rightarrow D \geq 0 \Rightarrow 16 - 16 y^{2} \geq 0 \Rightarrow y^{2} - 1 \leq 0 \Rightarrow y \in [- 1, 1]$

So,
${sin}^{- 1} \frac{4 x}{1 + 4 x^{2}} \in [- \frac{π}{2}, \frac{π}{2}]$
The above equality is possible only when,
$- \frac{π}{2} \leq 2 {tan}^{- 1} 2 x \leq \frac{π}{2} \Rightarrow - \frac{π}{4} \leq {tan}^{- 1} 2 x \leq \frac{π}{4} \Rightarrow - 1 \leq 2 x \leq 1 \Rightarrow - \frac{1}{2} \leq x \leq \frac{1}{2}$

Hence, the required values of $x$ is
$x \in [- \frac{1}{2}, \frac{1}{2}]$

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