Question:

<p><strong>Step 1:</strong> Let u = |x-2| where u ≥ 0. The inequality becomes:</p><p>$$\frac{u-1}{u-2} \leq 0$$</p><p><strong>Step 2:</strong> Find critical points: u = 1 and u = 2 (denominator ≠ 0)</p><p><strong>Step 3:</strong> Analyze sign of rational expression:</p><ul><li>For 0 ≤ u < 1: numerator negative, denominator negative → positive ✗</li><li>For 1 ≤ u < 2: numerator non-negative, denominator negative → non-positive ✓</li><li>For u > 2: numerator positive, denominator positive → positive ✗</li></ul><p>So: 1 ≤ u < 2</p><p><strong>Step 4:</strong> Convert back to x using |x-2| = u:</p><p>$$1 \leq |x-2| < 2$$</p><p><strong>Step 5:</strong> Split absolute value:</p><ul><li>|x-2| ≥ 1 gives: x ≤ 1 or x ≥ 3</li><li>|x-2| < 2 gives: -2 < x-2 < 2, so 0 < x < 4</li></ul><p><strong>Step 6:</strong> Find intersection of both conditions:</p><p>(x ≤ 1 or x ≥ 3) AND (0 < x < 4)</p><p>= (0 < x ≤ 1) ∪ [3, 4)</p><p>∴ Answer: <strong>C</strong> — x ∈ (0, 1] ∪ [3, 4)</p>