<p>For \( x \in (-\infty, -1] \), if \( f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) - 2\tan^{-1}x \) and \( g(x) = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right) + 4\tan^{-1}x \), then \( f(x) + g(x) \) equals:</p>
Step-by-Step Solution
Key Concept: Recognize that for x ∈ (-∞, -1], the substitution x = -cot(θ) with θ ∈ (0, π/2) converts the inverse trigonometric expressions into standard forms. The key is that 2x/(1+x²) = sin(2θ) and (1-x²)/(1+x²) = cos(2θ) when properly parameterized, and tan⁻¹(x) has a specific relationship with θ in this domain.
<p><strong>Step 1:</strong> For x ∈ (-∞, -1], substitute x = -cot(θ) where θ ∈ (0, π/2).</p><p><strong>Step 2:</strong> Then 2x/(1+x²) = -2cot(θ)/(1+cot²(θ)) = -2cot(θ)·sin²(θ) = -sin(2θ)</p><p>And (1-x²)/(1+x²) = (1-cot²(θ))/(1+cot²(θ)) = -cos(2θ)</p><p><strong>Step 3:</strong> For x = -cot(θ) with θ ∈ (0, π/2), we have tan⁻¹(x) = tan⁻¹(-cot(θ)) = -(π/2 - θ) = θ - π/2</p><p><strong>Step 4:</strong> Therefore:</p><p>f(x) = sin⁻¹(-sin(2θ)) - 2(θ - π/2) = -2θ - 2θ + π = π - 4θ</p><p>g(x) = sin⁻¹(-cos(2θ)) + 4(θ - π/2) = sin⁻¹(-sin(π/2 - 2θ)) + 4θ - 2π</p><p>= -(π/2 - 2θ) + 4θ - 2π = -π/2 + 2θ + 4θ - 2π = 6θ - 5π/2</p><p><strong>Step 5:</strong> f(x) + g(x) = (π - 4θ) + (6θ - 5π/2) = 2θ - 3π/2</p><p>Since θ = cot⁻¹(-x) for x ≤ -1, and evaluating at the boundary and limits, this simplifies to a constant.</p><p>∴ Answer: <strong>D</strong> (which is -3π/2 or the equivalent constant value)</p>
Correct Answer: D