<p>Solve <br>\(|x^2 - 2x| + |x - 4| > |x^2 - 3x + 4|\)</p>
<p>\(x \in (0, 2) \cup (4, \infty)\)</p>
<p>\(x \in (2, 4)\)</p>
<p>\(x \in (-\infty, 0) \cup (2, 4)\)</p>
<p>\(x \in (0, 4)\)</p>
Step-by-Step Solution
Key Concept: Analyze the inequality by identifying critical points where expressions inside absolute values change sign (x = 0, 2, 4), then test each interval separately to determine where the inequality holds.
<p><strong>Step 1: Identify critical points</strong></p><p>Critical points where expressions change sign: x = 0, 2, 4</p><p>These divide the real line into intervals: (-∞,0), (0,2), (2,4), (4,∞)</p><p><strong>Step 2: Analyze each interval</strong></p><p><strong>Interval 1: x < 0</strong><br>|x² - 2x| = x² - 2x, |x - 4| = 4 - x, |x² - 3x + 4| = x² - 3x + 4<br>x² - 2x + 4 - x > x² - 3x + 4<br>x² - 3x + 4 > x² - 3x + 4 ✗ (False)</p><p><strong>Interval 2: 0 ≤ x < 2</strong><br>|x² - 2x| = 2x - x², |x - 4| = 4 - x, |x² - 3x + 4| = x² - 3x + 4<br>2x - x² + 4 - x > x² - 3x + 4<br>x - x² + 4 > x² - 3x + 4<br>4x > 2x²<br>2x > x² ⟹ x(2 - x) > 0 ✓ (True for 0 < x < 2)</p><p><strong>Interval 3: 2 ≤ x < 4</strong><br>|x² - 2x| = x² - 2x, |x - 4| = 4 - x, |x² - 3x + 4| = x² - 3x + 4<br>x² - 2x + 4 - x > x² - 3x + 4<br>-3x + 4 > -3x + 4 ✗ (False, equality holds)</p><p><strong>Interval 4: x ≥ 4</strong><br>|x² - 2x| = x² - 2x, |x - 4| = x - 4, |x² - 3x + 4| = x² - 3x + 4<br>x² - 2x + x - 4 > x² - 3x + 4<br>-x - 4 > -3x + 4<br>2x > 8 ⟹ x > 4 ✓ (True for x > 4)</p><p><strong>Step 3: Combine solutions</strong></p><p>The inequality holds when: (0, 2) ∪ (4, ∞)</p><p>∴ Answer: <strong>x ∈ (0, 2) ∪ (4, ∞)</strong></p>
Correct Answer: A