If 2 sin-1 x = sin-1(2x√1-x2), then x belongs to

Question

If 2 sin-1 x = sin-1(2x√1-x2), then x belongs to (A) [-1, 1] (B) [-(1/√2), 1] (C) [-(1/√2), (1/√2)] (D) None of these

Tardigrade Admin · Accepted Answer

Since, 2 sin-1 x = sin-1(2x√1-x2) Range of right hand side is [-(π/2),(π /2)]. ⇒ -(π /2) le2 sin-1 x le(π /2) ⇒ -(π /4) le sin-1 x le (π /4) ⇒ x ∈ [-(1/√2), (1/√2)]

Q. If $2 s i n^{- 1} x = s i n^{- 1} (2 x 1 - x^{2})$ , then $x$ belongs to

$[- 1, 1]$

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

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

None of these

Since, $2 s i n^{- 1} x = s i n^{- 1} (2 x 1 - x^{2})$
Range of right hand side is $[- \frac{π}{2}, \frac{π}{2}]$ .
$\Rightarrow - \frac{π}{2} \leq 2 s i n^{- 1} x \leq \frac{π}{2}$
$\Rightarrow - \frac{π}{4} \leq s i n^{- 1} x \leq \frac{π}{4}$
$\Rightarrow x \in [- \frac{1}{2}, \frac{1}{2}]$

Q. If 2sin−1x=sin−1(2x1−x2​), then x belongs to

[−1,1]

[−2​1​,1]

[−2​1​,2​1​]

None of these

Since, 2sin−1x=sin−1(2x1−x2​) Range of right hand side is [−2π​,2π​]. ⇒−2π​≤2sin−1x≤2π​ ⇒−4π​≤sin−1x≤4π​ ⇒x∈[−2​1​,2​1​]

Q. If $2 s i n^{- 1} x = s i n^{- 1} (2 x 1 - x^{2})$ , then $x$ belongs to

$[- 1, 1]$

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

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