Question:

<p><strong>Step 1:</strong> Find the range of the innermost expression 16sin²x + 1.</p><p>Since sin²x ∈ [0, 1], we have:</p><p>16sin²x ∈ [0, 16]</p><p>Therefore, 16sin²x + 1 ∈ [1, 17]</p><p><strong>Step 2:</strong> Apply the first logarithm log₂(16sin²x + 1).</p><p>Since 16sin²x + 1 ∈ [1, 17]:</p><p>log₂(16sin²x + 1) ∈ [log₂(1), log₂(17)] = [0, log₂(17)]</p><p>Note: log₂(17) ≈ 4.09</p><p><strong>Step 3:</strong> Apply (2 - log₂(16sin²x + 1)).</p><p>Since log₂(16sin²x + 1) ∈ [0, log₂(17)]:</p><p>2 - log₂(16sin²x + 1) ∈ [2 - log₂(17), 2 - 0] = [2 - log₂(17), 2]</p><p>Since 2 - log₂(17) ≈ 2 - 4.09 = -2.09 &lt; 0, and we need the argument of log₂ to be positive, we check: when log₂(16sin²x + 1) = 0, we get 2 - 0 = 2 ✓</p><p>When log₂(16sin²x + 1) = 1 (which occurs when 16sin²x + 1 = 2, i.e., sin²x = 1/16), we get 2 - 1 = 1 ✓</p><p>The maximum of (2 - log₂(16sin²x + 1)) is 2 (when sin²x = 0).</p><p><strong>Step 4:</strong> Ensure the argument of outer log₂ is positive: 2 - log₂(16sin²x + 1) > 0.</p><p>This requires log₂(16sin²x + 1) < 2, so 16sin²x + 1 < 4, giving sin²x < 3/16.</p><p>When sin²x = 0: 2 - log₂(1) = 2 - 0 = 2, so log₂(2) = 1</p><p>When 16sin²x + 1 → 4: log₂(16sin²x + 1) → 2, so 2 - log₂(16sin²x + 1) → 0⁺, giving log₂(0⁺) → -∞</p><p>However, when sin²x ≤ 3/16, the expression is defined, and as sin²x increases from 0 to 3/16:</p><p>(2 - log₂(16sin²x + 1)) decreases from 2 to approaching 0</p><p>Therefore log₂(2 - log₂(16sin²x + 1)) ranges from log₂(2) = 1 down to log₂(0⁺) = -∞, but only where the expression is defined.</p><p><strong>Step 5:</strong> Reconsider for all valid sin²x ∈ [0, 1].</p><p>For sin²x ∈ [0, 3/16]: argument 2 - log₂(16sin²x + 1) ∈ (0, 2]</p><p>Thus f(x) = log₂(2 - log₂(16sin²x + 1)) ∈ (-∞, log₂(2)] = (-∞, 1]</p><p><strong>∴ Answer:</strong> a</p>