All the values of x satisfying the equation 2 tan -1 2 x= sin -1((4 x/1+4 x2)) lie in the interval

Question

All the values of x satisfying the equation 2 tan -1 2 x= sin -1((4 x/1+4 x2)) lie in the interval (A) [-(1/2), (1/2)] (B) [-1,1] (C) [(1/2,) ∞) (D) [-∞,-(1/2))

Tardigrade Admin · Accepted Answer

We have, sin -1((4 x/1+4 x2)) ⇒ sin -1((2(2 x)/1+(2 x)2)) =2 tan -1 2 x, when -1 ≤ 2 x ≤ 1 ∴ -1 ≤ 2 x ≤ 1 ⇒ (-1/2) ≤ x ≤ (1/2) ∴ x ∈ [(-1/2), (1/2)]

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

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

$[- 1, 1]$

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

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

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

Q. All the values of x satisfying the equation 2tan−12x=sin−1(1+4x24x​) lie in the interval

[−21​,21​]

[−1,1]

[2,1​∞)

[−∞,−21​)