<p>The range of the function \( f(x) = \sin^{-1}\!\left(\log_2 \dfrac{x^2}{2}\right) \) is:</p>
<p>\( \left[\dfrac{-\pi}{2}, \dfrac{\pi}{2}\right] \)</p>
<p>\( [0, \pi] \)</p>
<p>\( [-\pi, \pi] \)</p>
<p>\( \left[0, \dfrac{\pi}{2}\right] \)</p>
Step-by-Step Solution
Key Concept: The range of f(x) = sin⁻¹(g(x)) depends on the range of g(x) within [-1, 1]. Find where log₂(x²/2) lies in [-1, 1], then apply sin⁻¹ to get the final range.
<p><strong>Step 1: Find domain constraint</strong></p><p>For f(x) to be defined, we need: -1 ≤ log₂(x²/2) ≤ 1</p><p><strong>Step 2: Solve the inequality</strong></p><p>From -1 ≤ log₂(x²/2):</p><p>2⁻¹ ≤ x²/2 → x² ≥ 1 → |x| ≥ 1</p><p>From log₂(x²/2) ≤ 1:</p><p>x²/2 ≤ 2 → x² ≤ 4 → |x| ≤ 2</p><p>Combined domain: x ∈ [-2, -1] ∪ [1, 2]</p><p><strong>Step 3: Find range of the inner function</strong></p><p>When x ∈ [-2, -1] ∪ [1, 2], we have x² ∈ [1, 4]</p><p>So log₂(x²/2) ∈ [log₂(1/2), log₂(2)] = [-1, 1]</p><p><strong>Step 4: Apply sin⁻¹</strong></p><p>As log₂(x²/2) varies from -1 to 1:</p><p>f(x) = sin⁻¹(t) where t ∈ [-1, 1]</p><p>Therefore, range of f(x) = [-π/6, π/6]</p><p>∴ Answer: A</p>
Correct Answer: A