All the values of x satisfying the equation 2 | vedclass

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Question

(A) The given equation is $2 	an^{-1} 2x = \sin^{-1} \left( \frac{4x}{1+4x^2} ight)$.
We know the identity $\sin^{-1} \left( \frac{2	heta}{1+	he

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given equation is $2 	an^{-1} 2x = \sin^{-1} \left( \frac{4x}{1+4x^2} ight)$.
We know the identity $\sin^{-1} \left( \frac{2	heta}{1+	heta^2} ight) = 2 	an^{-1} 	heta$, which holds true when $-1 \leq 	heta \leq 1$.
Here, let $	heta = 2x$. The equation becomes $2 	an^{-1} 2x = \sin^{-1} \left( \frac{2(2x)}{1+(2x)^2} ight)$.
This identity is valid if and only if $-1 \leq 2x \leq 1$.
Dividing by $2$, we get $-\frac{1}{2} \leq x \leq \frac{1}{2}$.
Thus, the values of $x$ lie in the interval $[-\frac{1}{2}, \frac{1}{2}]$.

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

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