Relations & Functions
Inequalities involving modulus
Grade 12
Question:
<p>Let \( |x - 2| = y \). The solution set of \( \dfrac{y-1}{y-2} \leq 0 \) is:</p>
<p>\( (0,1) \cup (3,4) \)</p>
<p>\( (0,1] \cup [3,4) \)</p>
<p>\( (0,1) \cup [3,4) \)</p>
<p>\( [0,1) \cup (3,4] \)</p>
Step-by-Step Solution
Key Concept: Solve the inequality for y first, then use the definition y = |x - 2| to find which x-values satisfy the condition. The critical insight is that |x - 2| is always non-negative, which restricts the valid range of y.
<p><strong>Step 1:</strong> Solve the inequality (y-1)/(y-2) ≤ 0 for y.</p><p>The critical points are y = 1 (numerator = 0) and y = 2 (denominator = 0).</p><p><strong>Step 2:</strong> Use sign analysis on intervals:</p><ul><li>For y < 1: (negative)/(negative) = positive ✗</li><li>For 1 ≤ y < 2: (non-negative)/(negative) = non-positive ✓</li><li>For y > 2: (positive)/(positive) = positive ✗</li></ul><p>Therefore: 1 ≤ y < 2</p><p><strong>Step 3:</strong> Apply the constraint y = |x - 2| ≥ 0. Since 1 ≤ y < 2 is already in the valid range, substitute back:</p><p>1 ≤ |x - 2| < 2</p><p><strong>Step 4:</strong> Solve |x - 2| ≥ 1: gives x ≤ 1 or x ≥ 3</p><p>Solve |x - 2| < 2: gives 0 < x < 4</p><p><strong>Step 5:</strong> Find intersection: (0 < x < 4) ∩ (x ≤ 1 or x ≥ 3) = (0, 1] ∪ [3, 4)</p><p>∴ Answer: A</p>
Correct Answer: A