<p><strong>100.</strong> If \(f(x) = x^3 - 3x + 1\), then minimum number of real roots of \(f(f(x)) = 0\) is:</p>
Step-by-Step Solution
Key Concept: Find roots of f(x) = 0 first, then for each root α, solve f(x) = α. The total number of real roots of f(f(x)) = 0 equals the sum of real roots across all these equations, determined by analyzing critical points and monotonicity of f.
<p><strong>Step 1:</strong> Find roots of f(x) = 0 where f(x) = x³ - 3x + 1</p><p>f'(x) = 3x² - 3 = 3(x² - 1), so critical points at x = ±1</p><p>f(-1) = -1 + 3 + 1 = 3 (local maximum)</p><p>f(1) = 1 - 3 + 1 = -1 (local minimum)</p><p>Since f(-1) = 3 > 0 and f(1) = -1 < 0, and f is continuous, f(x) = 0 has exactly 3 real roots: one in (-∞, -1), one in (-1, 1), and one in (1, ∞). Call these α₁, α₂, α₃.</p><p><strong>Step 2:</strong> f(f(x)) = 0 means f(x) ∈ {α₁, α₂, α₃}</p><p>For each equation f(x) = αᵢ, count real solutions using the shape of f:</p><p>• f(x) = α₁ where α₁ > 3: No real solutions (f(x) never exceeds 3 for x > -1 in the relevant region)</p><p>• f(x) = α₂ where -1 < α₂ < 3: 3 real solutions (horizontal line crosses cubic three times)</p><p>• f(x) = α₃ where -1 < α₃ < 3: 3 real solutions</p><p><strong>Step 3:</strong> More careful analysis: Since -1 < αᵢ < 3 for roots in the middle and right regions, and considering the local maximum/minimum, each equation f(x) = αᵢ has 3 real solutions when -1 < αᵢ < 3.</p><p>One root α₁ < -1 gives 1 solution; α₂, α₃ in (-1, 3) each give 3 solutions.</p><p>∴ Minimum number = 1 + 3 + 3 = <strong>7</strong> (Answer: B)</p>
Correct Answer: B