<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 inverse trigonometric 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). The domains and ranges of inverse functions must be carefully matched to the double angle substitution.
<p><strong>Step 1: Substitute x = tan(θ)</strong> where θ ∈ (-π/2, π/2)</p><p>Then: 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: Analyze f(x) = sin⁻¹(sin(2θ))</strong></p><p>For x ∈ [0,1): 2θ ∈ [0, π/2), so f(x) = 2θ = 2tan⁻¹(x) ✓</p><p>For x ∈ (-1,0): 2θ ∈ (-π/2, 0), so f(x) = 2tan⁻¹(x) ✓</p><p><strong>Step 3: Analyze g(x) = cos⁻¹(cos(2θ))</strong></p><p>For x ∈ ℝ: 2θ ∈ (-π, π), and cos⁻¹ returns values in [0,π]</p><p>For x ∈ (-∞, -1) ∪ (1, ∞): g(x) = 2π - 2tan⁻¹(x) or adjustments apply</p><p>For x ∈ [-1,1]: g(x) = 2tan⁻¹(|x|) with proper range consideration</p><p><strong>Step 4: Analyze h(x) = tan⁻¹(tan(2θ))</strong></p><p>For x ∈ (-1,1): 2θ ∈ (-π/2, π/2), so h(x) = 2tan⁻¹(x) ✓</p><p>For |x| > 1: tan(2θ) undefined or h(x) requires domain restriction</p><p><strong>Conclusion:</strong> Statements A, B, D are correct based on domain-restricted relationships between f(x), g(x), h(x) and the 2tan⁻¹(x) form.</p>
Correct Answer: A,B,D