<p><strong>174.</strong> Solution set of the equation \(\sqrt{4^x - 2^{x+1} + 1} + \sqrt{4^x - 2^{x+3} + 16} = 3\) is:</p>
Step-by-Step Solution
Key Concept: Recognize that the expressions under the square roots are perfect squares by substituting t = 2^x, then express them as (t-1)² and (t-4)² to convert the equation into |t-1| + |t-4| = 3.
<p><strong>Step 1:</strong> Let t = 2^x where t > 0. Then 4^x = t².</p><p><strong>Step 2:</strong> Rewrite the equation:<br/>√(t² - 2t + 1) + √(t² - 8t + 16) = 3<br/>√(t-1)² + √(t-4)² = 3<br/>|t - 1| + |t - 4| = 3</p><p><strong>Step 3:</strong> Analyze cases based on critical points t = 1 and t = 4:</p><p><strong>Case 1:</strong> If 0 < t < 1: (1-t) + (4-t) = 3 → 5 - 2t = 3 → t = 1 (boundary, discard)</p><p><strong>Case 2:</strong> If 1 ≤ t ≤ 4: (t-1) + (4-t) = 3 → 3 = 3 ✓ (true for all t ∈ [1,4])</p><p><strong>Case 3:</strong> If t > 4: (t-1) + (t-4) = 3 → 2t - 5 = 3 → t = 4 (boundary, discard)</p><p><strong>Step 4:</strong> From Case 2: 1 ≤ 2^x ≤ 4<br/>2^0 ≤ 2^x ≤ 2^2<br/>0 ≤ x ≤ 2</p><p>∴ Answer: x ∈ [0, 2]</p>
Correct Answer: D