<p>\(2\tan^{-1} x = \tan^{-1}\dfrac{2x}{1-x^2}\) is true if</p>
<p>(a) \(x > 0,\ x < 1\)</p>
<p>(b) \(x > 1\)</p>
<p>(c) \(x < 0\)</p>
<p>(d) \(x \in \mathbb{R}\)</p>
Step-by-Step Solution
Key Concept: The double angle formula for tangent states tan(2θ) = 2tan(θ)/(1-tan²(θ)). This identity holds only when 2θ lies in the range of tan⁻¹, which restricts the domain of x to |x| < 1 to ensure the equation is valid.
<p><strong>Step 1:</strong> Recognize the double angle formula. If we let θ = tan⁻¹(x), then tan(2θ) = 2tan(θ)/(1-tan²(θ)) = 2x/(1-x²).</p><p><strong>Step 2:</strong> This means 2θ = tan⁻¹(2x/(1-x²)), so 2tan⁻¹(x) = tan⁻¹(2x/(1-x²)).</p><p><strong>Step 3:</strong> However, tan⁻¹ has range (-π/2, π/2). For 2tan⁻¹(x) to equal tan⁻¹(2x/(1-x²)), we need 2tan⁻¹(x) ∈ (-π/2, π/2).</p><p><strong>Step 4:</strong> This requires tan⁻¹(x) ∈ (-π/4, π/4), which means <strong>|x| < 1</strong>.</p><p><strong>Step 5:</strong> Additionally, 1-x² ≠ 0, so <strong>x ≠ ±1</strong>.</p><p>∴ Answer: A (the condition is |x| < 1)</p>
Correct Answer: A