Find all real numbers a for which ax² + (a-1)x + a + 1 ≥ 0 for all x.
Correct: D
1. Quadratic inequality holds for all x iff leading coefficient > 0 and discriminant ≤ 0. 2. Leading coefficient a ≠ 0. Case 1: a > 0. Discriminant: (a-1)² -4a(a + 1) ≤ 0 → -3a² -2a +1 ≤ 0. Solve quadratic inequality → a ∈ [–1, 1/3]. Intersect with a > 0 → a ∈ (0,1/3]. Case 2: a < 0 invalid (inequality not always ≥ 0). Thus, no solution. Correct answer requires rechecking.