<p>\(3\tan^{-1} x = \tan^{-1}\dfrac{3x - x^3}{1-3x^2}\) \((x > 0)\) if \(x < \underline{\quad}\).</p>
Step-by-Step Solution
Key Concept: Recognize that the RHS is the triple angle formula for tangent: tan(3θ) = (3tanθ - tan³θ)/(1 - 3tan²θ). The equation holds when 3tan⁻¹(x) lies in the principal range of tan⁻¹, requiring 3·tan⁻¹(x) ∈ (-π/2, π/2).
<p><strong>Step 1:</strong> Recognize the RHS as the triple angle tangent formula: if tan(θ) = x, then tan(3θ) = (3x - x³)/(1 - 3x²).</p><p><strong>Step 2:</strong> Set θ = tan⁻¹(x). Then 3θ = 3tan⁻¹(x) and the equation becomes: 3tan⁻¹(x) = tan⁻¹(tan(3θ)) = tan⁻¹[tan(3tan⁻¹(x))].</p><p><strong>Step 3:</strong> For the formula 3tan⁻¹(x) = tan⁻¹[(3x - x³)/(1 - 3x²)] to hold, we need 3tan⁻¹(x) to remain in the principal range of tan⁻¹, i.e., 3tan⁻¹(x) ∈ (-π/2, π/2).</p><p><strong>Step 4:</strong> The boundary condition is when 3tan⁻¹(x) = π/2 (for x > 0). This gives tan⁻¹(x) = π/6, so x = tan(π/6) = 1/√3.</p><p><strong>Step 5:</strong> Therefore, the formula holds when 0 < x < 1/√3.</p><p>∴ Answer: x < 1/√3</p>
Correct Answer: 1/√3