<p>Let the equation \(ax^2 - bx + c = 0\) has 2 distinct roots in the interval \((0, 1)\) where \(a, b, c \in \mathbb{N}\). If \(\lambda \leq \log_5(abc)\) for all choices of natural numbers \(a, b, c\), then non-negative integral values of \(\lambda\) can be:</p>
Step-by-Step Solution
Key Concept: For a quadratic ax² - bx + c = 0 to have both roots in (0,1), we need f(0) > 0, f(1) > 0, and the vertex inside (0,1) with discriminant positive. This forces strict inequalities on a, b, c that yield a minimum value of abc, giving an upper bound on λ.
<p><strong>Step 1:</strong> For both roots α, β ∈ (0,1) of ax² - bx + c = 0, we need:</p><ul><li>f(0) = c > 0 ✓ (c ∈ ℕ)</li><li>f(1) = a - b + c > 0, so b < a + c</li><li>Vertex at x = b/(2a) ∈ (0,1), so 0 < b < 2a</li><li>Discriminant Δ = b² - 4ac > 0, so b² > 4ac</li></ul><p><strong>Step 2:</strong> From b² > 4ac and b < 2a, we get: (2a)² > b² > 4ac, which gives 4a² > 4ac, hence a > c.</p><p><strong>Step 3:</strong> Since a, b, c ∈ ℕ, try smallest values. If c = 1, a ≥ 2. For a = 5, b² > 20, so b ≥ 5. With b = 5: check 25 > 20 ✓ and f(1) = 5 - 5 + 1 = 1 > 0 ✓</p><p><strong>Step 4:</strong> Minimum abc = 5 × 5 × 1 = 25. Therefore λ ≤ log₅(25) = log₅(5²) = 2.</p><p>Non-negative integral values of λ: <strong>λ ∈ {0, 1, 2}</strong></p><p>∴ Answer: A</p>
Correct Answer: A