<p>Let <em>f</em> be a quadratic function such that: <em>f</em>(<em>x</em>) = 0 has 2 real solutions and <em>f</em>(<em>f</em>(<em>x</em>)) = 0 has 3 real solutions. What is the maximum number of solutions for <em>f</em>(<em>f</em>(<em>f</em>(<em>x</em>))) = 0?</p>
Step-by-Step Solution
Key Concept: If f(x) = 0 has roots α and β, then f(f(x)) = 0 means f(x) ∈ {α, β}. Each equation f(x) = α and f(x) = β can have 0, 1, or 2 real solutions. The constraint that f(f(x)) = 0 has exactly 3 real solutions means one root has 2 pre-images and the other has 1 pre-image. For f(f(f(x))) = 0, we need f(f(x)) ∈ {α, β}, so we must count pre-images of the 3 solutions of f(f(x)) = 0 under f.
<p><strong>Step 1:</strong> Let f(x) = ax² + bx + c with f(x) = 0 having roots α and β.</p><p><strong>Step 2:</strong> For f(f(x)) = 0: we need f(x) = α or f(x) = β. Each is a quadratic equation, giving 0, 1, or 2 solutions. Since f(f(x)) = 0 has exactly 3 real solutions total, one equation contributes 2 solutions and the other contributes 1 solution.</p><p><strong>Step 3:</strong> Without loss of generality, suppose f(x) = α has 2 solutions (call them x₁, x₂) and f(x) = β has 1 solution (call it x₃). So the 3 solutions of f(f(x)) = 0 are {x₁, x₂, x₃}.</p><p><strong>Step 4:</strong> For f(f(f(x))) = 0, we need f(f(x)) ∈ {α, β}. This means we solve f(x) = z for each z ∈ {x₁, x₂, x₃}.</p><p><strong>Step 5:</strong> Each of the three equations f(x) = xᵢ can have 0, 1, or 2 real solutions. To maximize, assume each contributes 2 solutions (achievable when each xᵢ is in the range of f and satisfies discriminant conditions).</p><p><strong>Step 6:</strong> Maximum = 2 + 2 + 2 = <strong>6 solutions</strong>.</p><p>∴ Answer: <strong>6</strong></p>
Correct Answer: 6