Trigonometry & Inverse Trigonometry
Range of Inverse Trigonometric Functions
Grade 12
Question:
<p>The range of the function \( f(x) = \sin^{-1}\!\left(\log_2 \dfrac{x^2}{2}\right) \) is:</p>
<p>(a) \( \left[\dfrac{-\pi}{2}, \dfrac{\pi}{2}\right] \)</p>
<p>(b) \( [0, \pi] \)</p>
<p>(c) \( [-\pi, \pi] \)</p>
<p>(d) \( \left[0, \dfrac{\pi}{2}\right] \)</p>
Step-by-Step Solution
Key Concept: The domain of sin⁻¹ is [-1, 1], so we need -1 ≤ log₂(x²/2) ≤ 1. Solving this inequality gives the valid x-values, and then we evaluate the function on this restricted domain to find the range.
<p><strong>Step 1:</strong> For f(x) to be defined, the argument of sin⁻¹ must be in [-1, 1]:</p><p>-1 ≤ log₂(x²/2) ≤ 1</p><p><strong>Step 2:</strong> Convert to exponential form:</p><p>2⁻¹ ≤ x²/2 ≤ 2¹</p><p>1/2 ≤ x²/2 ≤ 2</p><p>1 ≤ x² ≤ 4</p><p>So x ∈ [-2, -1] ∪ [1, 2]</p><p><strong>Step 3:</strong> Find the range of log₂(x²/2) on this domain:</p><p>When |x| = 1: log₂(1/2) = -1</p><p>When |x| = 2: log₂(4/2) = log₂(2) = 1</p><p>As x² increases from 1 to 4, log₂(x²/2) increases from -1 to 1</p><p><strong>Step 4:</strong> The argument of sin⁻¹ takes all values in [-1, 1]</p><p>Therefore, the range of f(x) = sin⁻¹(t) where t ∈ [-1, 1] is:</p><p>∴ Answer: [-π/2, π/2]</p>
Correct Answer: A