Trigonometry & Inverse Trigonometry
Domain of Inverse Trigonometric Functions
Grade None
Question:
<p>The largest interval lying in \(\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\) for which \( f(x) = 4^{-x^2} + \cos^{-1}\left(\dfrac{x}{2}-1\right) + \log(\cos x) \) is defined, is:</p>
<p>(A) \(\left[-\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)\)</p>
<p>(B) \(\left[0, \dfrac{\pi}{2}\right)\)</p>
<p>(C) \(\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\)</p>
<p>(D) \(\left[0, \pi\right]\)</p>
Step-by-Step Solution
Key Concept: For f(x) to be defined, we need THREE conditions simultaneously: (1) cos x > 0 for the logarithm, (2) -1 ≤ (x/2 - 1) ≤ 1 for inverse cosine, and (3) the exponential 4^(-x²) is always defined. The intersection of these constraints determines the domain.
<p><strong>Step 1: Constraint from cos⁻¹(x/2 - 1)</strong></p><p>For inverse cosine to be defined: -1 ≤ (x/2 - 1) ≤ 1</p><p>-1 ≤ x/2 - 1 ⟹ 0 ≤ x/2 ⟹ x ≥ 0</p><p>x/2 - 1 ≤ 1 ⟹ x/2 ≤ 2 ⟹ x ≤ 4</p><p>So: <strong>0 ≤ x ≤ 4</strong></p><p><strong>Step 2: Constraint from log(cos x)</strong></p><p>For logarithm to be defined: cos x > 0 (strictly positive)</p><p>In the interval (-π/2, π/2), cos x > 0 throughout since cos x = 0 only at the endpoints.</p><p>So: <strong>-π/2 < x < π/2</strong></p><p><strong>Step 3: Intersection of all constraints</strong></p><p>We need: [0 ≤ x ≤ 4] ∩ [-π/2 < x < π/2] ∩ (exponential always defined)</p><p>Since π/2 ≈ 1.57 and 4 > π/2, the intersection is: <strong>0 ≤ x < π/2</strong></p><p><strong>Step 4: Express as interval</strong></p><p>The largest interval lying in (-π/2, π/2) where f(x) is defined is: <strong>[0, π/2)</strong></p><p>∴ Answer: B</p>
Correct Answer: B