Trigonometry & Inverse Trigonometry
Inverse trigonometric functions
Grade 12
Question:
<p>If \(f(x) = \sin^{-1}\!\left(\dfrac{2x}{1+x^2}\right) - 2\tan^{-1}x\) and \(g(x) = \sin^{-1}\!\left(\dfrac{1-x^2}{1+x^2}\right) + 4\tan^{-1}x\), then range of \((f(x) - g(x))\) for \(x \in (-\infty, -1]\) is:</p>
<p>\(\left[0, \dfrac{3\pi}{2}\right)\)</p>
<p>\(\left[\dfrac{-3\pi}{2}, \pi\right)\)</p>
<p>\(\left[-\pi, \dfrac{-\pi}{2}\right)\)</p>
<p>\(\left[\pi, \dfrac{7\pi}{2}\right)\)</p>
Step-by-Step Solution
Key Concept: Recognize that sin⁻¹(2x/(1+x²)) = 2tan⁻¹(x) for x ∈ [-1,1] and sin⁻¹((1-x²)/(1+x²)) = π - 2tan⁻¹(x) for x ≤ -1. These identities immediately simplify f(x) and g(x) to linear functions of tan⁻¹(x).
<p><strong>Step 1: Simplify f(x)</strong></p><p>For x ∈ ℝ: sin⁻¹(2x/(1+x²)) = 2tan⁻¹(x)</p><p>Therefore: f(x) = 2tan⁻¹(x) - 2tan⁻¹(x) = 0</p><p><strong>Step 2: Simplify g(x) for x ≤ -1</strong></p><p>For x ≤ -1: sin⁻¹((1-x²)/(1+x²)) = π - 2tan⁻¹(x) (since tan⁻¹(x) ∈ (-π/2, -π/4] when x ≤ -1)</p><p>Therefore: g(x) = π - 2tan⁻¹(x) + 4tan⁻¹(x) = π + 2tan⁻¹(x)</p><p><strong>Step 3: Find f(x) - g(x)</strong></p><p>f(x) - g(x) = 0 - (π + 2tan⁻¹(x)) = -π - 2tan⁻¹(x)</p><p><strong>Step 4: Find the range for x ∈ (-∞, -1]</strong></p><p>When x ∈ (-∞, -1]: tan⁻¹(x) ∈ (-π/2, -π/4]</p><p>So 2tan⁻¹(x) ∈ (-π, -π/2]</p><p>Therefore -2tan⁻¹(x) ∈ [π/2, π)</p><p>Thus -π - 2tan⁻¹(x) ∈ [-π/2, 0)</p><p><strong>∴ Answer: B (Range is [-π/2, 0))</strong></p>
Correct Answer: B