Trigonometry & Inverse Trigonometry
Inverse trigonometric functions
Grade 12
Question:
<p>Consider \(f(x) = \tan^{-1}\!\left(\dfrac{2x}{\sqrt{9-4x^2}}\right) - \cos^{-1}\!\left(\dfrac{x}{3}\right)\). Identify which of the following statement(s) is(are) <strong>correct</strong>?</p>
<p>(a) Number of solutions of the equation \(f(x) = \ln(-x)\) is 2.</p>
<p>(b) Number of solutions of the equation \(f(x) = \ln(-x)\) is 1.</p>
<p>(c) If \(f(x) - k = 0\) has a solution then number of integral values of \(k\) is 4.</p>
<p>(d) If \(f(x) - k = 0\) has a solution then number of integral values of \(k\) is 3.</p>
Step-by-Step Solution
Key Concept: Recognize that tan⁻¹(2x/√(9-4x²)) can be converted to sin⁻¹(2x/3) using the substitution 2x = 3sin(θ), and then use the complementary angle relationship sin⁻¹(y) + cos⁻¹(y) = π/2 to simplify f(x).
<p><strong>Step 1: Simplify tan⁻¹(2x/√(9-4x²))</strong></p><p>Let 2x = 3sin(θ), where θ ∈ [-π/2, π/2]. Then √(9-4x²) = √(9-9sin²(θ)) = 3cos(θ).</p><p>So tan⁻¹(2x/√(9-4x²)) = tan⁻¹(sin(θ)/cos(θ)) = tan⁻¹(tan(θ)) = θ = sin⁻¹(2x/3)</p><p><strong>Step 2: Apply the complementary angle identity</strong></p><p>f(x) = sin⁻¹(2x/3) - cos⁻¹(x/3)</p><p>Since sin⁻¹(y) + cos⁻¹(y) = π/2, we have sin⁻¹(2x/3) = π/2 - cos⁻¹(2x/3)</p><p><strong>Step 3: Use the relationship cos⁻¹(a) - cos⁻¹(b) formula or direct substitution</strong></p><p>f(x) = π/2 - cos⁻¹(2x/3) - cos⁻¹(x/3)</p><p>For x ∈ [-3/2, 3/2] (domain constraint), when cos⁻¹(2x/3) + cos⁻¹(x/3) = π/2, then f(x) = 0.</p><p>Using the identity: cos⁻¹(a) + cos⁻¹(b) = cos⁻¹(ab - √((1-a²)(1-b²))) when this equals π/2, we get ab = √((1-a²)(1-b²))</p><p><strong>Step 4: Verify specific behavior</strong></p><p>At x = 0: f(0) = sin⁻¹(0) - cos⁻¹(0) = 0 - π/2 = -π/2</p><p>At x = 3/2: f(3/2) = sin⁻¹(1) - cos⁻¹(1/2) = π/2 - π/3 = π/6</p><p>f(x) is continuous and strictly increasing on its domain.</p><p>∴ Statements about f being strictly increasing and its range being [-π/2, π/6] are correct.</p>
Correct Answer: B,D