<p>Let \(1 \leq x \leq 256\) and <i>M</i> be the maximum value of \((\log_2 x)^4 + 16(\log_2 x)^2 \log_2\left(\frac{16}{x}\right)\). The sum of the digits of <i>M</i> is:</p>
Step-by-Step Solution
Key Concept: Substitute y = log₂(x) to convert the expression into a polynomial, then find its maximum on the restricted domain [0, 8] using calculus or algebraic manipulation.
<p><strong>Step 1: Set up substitution</strong></p><p>Let y = log₂(x). Since 1 ≤ x ≤ 256, we have 0 ≤ y ≤ 8 (since log₂(256) = 8).</p><p><strong>Step 2: Simplify the logarithmic term</strong></p><p>log₂(16/x) = log₂(16) - log₂(x) = 4 - y</p><p><strong>Step 3: Rewrite the expression</strong></p><p>f(y) = y⁴ + 16y²(4 - y) = y⁴ + 64y² - 16y³</p><p>Rearranging: f(y) = y⁴ - 16y³ + 64y²</p><p><strong>Step 4: Find critical points</strong></p><p>f'(y) = 4y³ - 48y² + 128y = 4y(y² - 12y + 32) = 4y(y - 4)(y - 8)</p><p>Critical points: y = 0, 4, 8</p><p><strong>Step 5: Evaluate at critical points and boundaries</strong></p><p>• At y = 0: f(0) = 0</p><p>• At y = 4: f(4) = 4⁴ - 16(4³) + 64(4²) = 256 - 1024 + 1024 = 256</p><p>• At y = 8: f(8) = 8⁴ - 16(8³) + 64(8²) = 4096 - 8192 + 4096 = 0</p><p><strong>Step 6: Identify maximum</strong></p><p>The maximum value is M = 256 at y = 4 (i.e., x = 2⁴ = 16)</p><p><strong>Step 7: Calculate sum of digits</strong></p><p>M = 256, sum of digits = 2 + 5 + 6 = 13</p><p>∴ Answer: d</p>
Correct Answer: d