<p>Find all real numbers \(a\) for which the equation \(x^2 + (a-2)x + 1 = 3|x|\) has exactly three distinct real solutions in x.</p>
Step-by-Step Solution
Key Concept: For exactly 3 distinct solutions, the parabola y = x² + (a-2)x + 1 must intersect the V-shaped curve y = 3|x| at exactly 3 points. By symmetry considerations, one intersection must occur at x = 0, and the other two must be symmetric about the origin.
<p><strong>Step 1: Split into cases using |x|.</strong></p><p>For x ≥ 0: x² + (a-2)x + 1 = 3x, giving x² + (a-5)x + 1 = 0</p><p>For x < 0: x² + (a-2)x + 1 = -3x, giving x² + (a+1)x + 1 = 0</p><p><strong>Step 2: Analyze conditions for exactly 3 solutions.</strong></p><p>For exactly 3 distinct real solutions total, one equation must have 2 distinct positive/negative roots and the other must have exactly 1 root (a repeated root or no real roots in that region).</p><p><strong>Step 3: Case - Right branch has 2 solutions, left branch has 1.</strong></p><p>For x ≥ 0: Need Δ₁ = (a-5)² - 4 > 0, so (a-5)² > 4, giving a < 3 or a > 7</p><p>For x < 0: Need Δ₂ = (a+1)² - 4 = 0 for tangency, so (a+1)² = 4, giving a = 1 or a = -3</p><p><strong>Step 4: Verify conditions.</strong></p><p>If a = 1: Right branch gives x² - 4x + 1 = 0 with solutions x = 2 ± √3 (both positive ✓). Left branch gives x² + 2x + 1 = (x+1)² = 0 with x = -1 (one negative solution ✓). Total: 3 solutions.</p><p>If a = -3: Right branch gives x² - 8x + 1 = 0 with solutions x = 4 ± √15 (both positive ✓). Left branch gives x² - 2x + 1 = (x-1)² = 0 with x = 1 (not in x < 0 region ✗).</p><p><strong>Step 5: Check the other case.</strong></p><p>If left branch has 2 solutions and right has 1: a = 7 gives x² + 2x + 1 = 0 (one solution), but (a+1)² = 64 > 4 (two solutions on left). This requires a = 7 checking: right equation x² + 2x + 1 = (x+1)² gives one repeated root at x = -1 (not in x ≥ 0). Doesn't work.</p><p>∴ Answer: <strong>a = 1</strong></p>
Correct Answer: A