Question:

<p><strong>Step 1:</strong> Let f(x) = 0 have roots α and β (both real). Then f(f(x)) = 0 requires f(x) = α or f(x) = β.</p><p><strong>Step 2:</strong> Since f(f(x)) = 0 has exactly 3 real solutions, by pigeonhole principle: one equation (say f(x) = α) has 2 real solutions and f(x) = β has 1 real solution. This means one root has 2 preimages and one root has 1 preimage.</p><p><strong>Step 3:</strong> For f(f(f(x))) = 0, we need f(f(x)) ∈ {α, β}. Let S = {x : f(f(x)) = 0} = {x₁, x₂, x₃} (the 3 solutions from Step 2). We need to find how many solutions to f(y) ∈ {α, β} exist in the set S.</p><p><strong>Step 4:</strong> Among the 3 values in S, each can map to either α or β under f. To maximize solutions to f(f(f(x))) = 0, we need each element of S to be a root of either f(x) = α or f(x) = β. The maximum occurs when f maps elements of S such that all preimages are real. With optimal distribution: the element with 1 preimage under previous stage can contribute 2 preimages, and elements with 2 preimages can each contribute 2, giving at most 2 + 2 + 2 = 6 preimages, but we need to verify feasibility.</p><p><strong>Step 5:</strong> The actual maximum is bounded by the constraint structure. If f(x) = α has 2 solutions and f(x) = β has 1 solution in the original layer, then for the next composition: f(f(x)) = α and f(f(x)) = β each can have at most 2 solutions. From the 2 values with 2 preimages and 1 value with 1 preimage, we get at most 2 + 2 + 2 = <strong>6 solutions</strong>, but the tighter constraint from composition depth limits this to <strong>4</strong> maximum.</p><p>∴ Answer: <strong>4</strong></p>