Single Choice

The value of $$x$$ which satisfies equation $$\displaystyle 2\:\tan^{-1}2x=\sin^{-1}\frac{4x}{1+4x^{2}}$$ is valid in the interval

A$$\displaystyle \left[\frac{1}{2},\infty\right)$$
B$$\displaystyle \left(-\infty ,-\frac{1}{2}\right]$$
C$$\displaystyle \left[-1,1\right]$$
D$$\displaystyle \left[-\frac{1}{2},\frac{1}{2}\right]$$
Correct Answer

Solution

Given:

$$2tan^{-1}{2x=sin^-{1}\dfrac{4x}{1+4x^2}}$$

$$2x=tan \theta $$

$$x= \dfrac{tan \theta}{2} $$

$$2tan^{-1} (tan \theta)=sin^{-1}\dfrac{2tan \theta}{1+tan^{2} \theta}$$

$$2tan^{-1} (tan \theta)=sin^{-1}(sin2 \theta)$$

$$ \dfrac{-\pi}{2} \leq 2 \theta \leq \dfrac{\pi}{2} $$

$$ \dfrac{-\pi}{4} \leq \theta \leq \dfrac{\pi}{4} $$

$$-1 \leq tan \theta \leq +1$$

$$\dfrac{-1}{2} \leq \dfrac{tan \theta}{2} \leq \dfrac{+1}{2}$$

$$\dfrac{-1}{2} \leq x \leq \dfrac{+1}{2}$$




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