<p>Let \(f(x) = \sin^{-1}\left(\dfrac{2x}{1+x^2}\right)\), \(g(x) = \cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)\) and \(h(x) = \tan^{-1}\left(\dfrac{2x}{1-x^2}\right)\), then:</p>
<p>\(f(x) + 2\tan^{-1}x = \pi \; \forall \; x \geq 1\)</p>
<p>\(\dfrac{f(x)}{g(x)} = 1 \; \forall \; x \in [0,1]\)</p>
<p>\(g(x) + h(x) = \forall \; x \in (-1,0)\)</p>
<p>\(\dfrac{\lim_{x \to 1^+}(f(x)+g(x)+h(x))}{\lim_{x \to 1^-}(f(x)+g(x)+h(x))} = 3\)</p>
Step-by-Step Solution
Key Concept: Recognize that f(x), g(x), and h(x) are disguised forms of double angle formulas: sin(2θ) = 2x/(1+x²), cos(2θ) = (1-x²)/(1+x²), and tan(2θ) = 2x/(1-x²) when x = tan(θ). Use the substitution x = tan(θ) to simplify each inverse function.
<p><strong>Step 1: Apply substitution x = tan(θ) where θ ∈ (-π/2, π/2)</strong></p><p>With this substitution:</p><p>• sin(2θ) = 2tan(θ)/(1+tan²(θ)) = 2x/(1+x²)</p><p>• cos(2θ) = (1-tan²(θ))/(1+tan²(θ)) = (1-x²)/(1+x²)</p><p>• tan(2θ) = 2tan(θ)/(1-tan²(θ)) = 2x/(1-x²)</p><p><strong>Step 2: Simplify f(x) = sin⁻¹(sin(2θ))</strong></p><p>For x ∈ (-1, 1), we have 2θ ∈ (-π/2, π/2), so f(x) = 2θ = 2tan⁻¹(x)</p><p><strong>Step 3: Simplify g(x) = cos⁻¹(cos(2θ))</strong></p><p>For x ∈ (-1, 1), we have 2θ ∈ (-π/2, π/2), so 2θ ∈ [0, π) after adjustment. Thus g(x) = 2θ = 2tan⁻¹(x) when properly accounting for range of arccos</p><p><strong>Step 4: Simplify h(x) = tan⁻¹(tan(2θ))</strong></p><p>For x ∈ (-1, 1), we have 2θ ∈ (-π/2, π/2), so h(x) = 2θ = 2tan⁻¹(x)</p><p><strong>Step 5: Verify relationships</strong></p><p>• f(x) = 2tan⁻¹(x) for |x| < 1 ✓</p><p>• g(x) = 2tan⁻¹(x) for |x| < 1 ✓</p><p>• h(x) = 2tan⁻¹(x) for |x| < 1 ✓</p><p>∴ Answer: A, B, D (Statements affirming these relationships are correct)</p>
Correct Answer: A,B,D