<p>\(\tan^{-1} x + \tan^{-1}\dfrac{2x}{1-3x^2} = \pi + \tan^{-1}\dfrac{3x - x^3}{1-3x^2}\) \((x > 0)\) is true if</p>
<p>(a) \(x < \dfrac{1}{\sqrt{3}}\)</p>
<p>(b) \(\dfrac{1}{\sqrt{3}} < x < 1\)</p>
<p>(c) \(x > \dfrac{1}{\sqrt{3}}\)</p>
<p>(d) \(\dfrac{1}{\sqrt{3}} < x < \sqrt{3}\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a composition of the addition formula for inverse tangent: tan⁻¹A + tan⁻¹B = tan⁻¹((A+B)/(1-AB)) + nπ, where the π term appears when the denominator 1-AB is negative (indicating we're in the second quadrant of the arctangent composition).
<p><strong>Step 1:</strong> Use the addition formula: tan⁻¹A + tan⁻¹B = tan⁻¹((A+B)/(1-AB)) when 1-AB > 0, or tan⁻¹((A+B)/(1-AB)) + π when 1-AB < 0.</p><p><strong>Step 2:</strong> Here A = x and B = 2x/(1-3x²). Calculate AB = x · 2x/(1-3x²) = 2x²/(1-3x²).</p><p><strong>Step 3:</strong> Find 1 - AB = 1 - 2x²/(1-3x²) = (1-3x² - 2x²)/(1-3x²) = (1-5x²)/(1-3x²).</p><p><strong>Step 4:</strong> Calculate (A+B)/(1-AB) = [x + 2x/(1-3x²)] / [(1-5x²)/(1-3x²)] = [x(1-3x²) + 2x]/(1-5x²) = x(1-3x² + 2)/(1-5x²) = x(3-3x²)/(1-5x²) = (3x - 3x³)/(1-5x²).</p><p><strong>Step 5:</strong> The right side has (3x - x³)/(1-3x²). For the equation with the π term to hold, we need 1 - AB < 0, meaning (1-5x²)/(1-3x²) < 0.</p><p><strong>Step 6:</strong> Since x > 0 and 1-3x² must be negative (otherwise the B term doesn't match), we need 1-3x² < 0, so x > 1/√3. Also need 1-5x² > 0 for the ratio to be negative, so x < 1/√5 is impossible simultaneously. Actually, we need 1-5x² < 0 AND 1-3x² > 0, giving 1/√3 < x < 1/√5 is impossible. Or 1-5x² > 0 AND 1-3x² < 0, giving x > 1/√3.</p><p><strong>Step 7:</strong> The equality holds when <strong>x > 1/√3</strong> (or equivalently x > √3/3, or 3x² > 1).</p><p>∴ Answer: C</p>
Correct Answer: C