<p>If \( \sin^{-1} x = 2\sin^{-1} a \) has a solution, then \( |a| \) satisfies:</p>
<p>(A) \( |a| \leq \dfrac{1}{\sqrt{2}} \) and \( |a| \leq 1 \)</p>
<p>(B) \( 0 \leq |a| \leq \dfrac{1}{\sqrt{2}} \) and \( |a| \leq 1 \)</p>
<p>(C) \( |a| \geq \dfrac{1}{\sqrt{2}} \)</p>
<p>(D) \( |a| = 1 \)</p>
Step-by-Step Solution
Key Concept: For sin⁻¹(x) = 2sin⁻¹(a) to have a solution, we need x = sin(2sin⁻¹(a)) to lie in [-1,1], and we must use the double angle formula: sin(2θ) = 2sin(θ)cos(θ) with appropriate domain constraints.
<p><strong>Step 1:</strong> Let sin⁻¹(a) = θ, so a = sin(θ) where θ ∈ [-π/2, π/2].</p><p><strong>Step 2:</strong> Then sin⁻¹(x) = 2θ requires 2θ ∈ [-π/2, π/2], which means θ ∈ [-π/4, π/4].</p><p><strong>Step 3:</strong> Since a = sin(θ) and θ ∈ [-π/4, π/4], we have a ∈ [sin(-π/4), sin(π/4)] = [-1/√2, 1/√2].</p><p><strong>Step 4:</strong> This gives |a| ≤ 1/√2 or equivalently |a| ≤ √2/2.</p><p><strong>Step 5:</strong> We can verify: x = sin(2sin⁻¹(a)) = 2sin(sin⁻¹(a))cos(sin⁻¹(a)) = 2a√(1-a²). For |a| ≤ 1/√2, we have |x| ≤ 2·(1/√2)·√(1-1/2) = 2·(1/√2)·(1/√2) = 1 ✓</p><p>∴ Answer: <strong>|a| ≤ 1/√2</strong> (or |a| ≤ √2/2)</p>
Correct Answer: B