<p><strong>329.</strong> 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 two distinct roots in (0,1), we need f(0) > 0, f(1) > 0, vertex in (0,1), and discriminant > 0. These conditions combined with a,b,c ∈ ℕ force a minimum value of abc, which bounds log₅(abc) from below.
<p><strong>Step 1: Set up conditions for roots in (0,1)</strong></p><p>For ax² - bx + c = 0 to have two distinct roots in (0,1):</p><ul><li>f(0) = c > 0 ✓ (always true since c ∈ ℕ)</li><li>f(1) = a - b + c > 0, so b < a + c</li><li>Discriminant: b² - 4ac > 0, so b² > 4ac</li><li>Vertex in (0,1): 0 < b/(2a) < 1, so 0 < b < 2a</li><li>Maximum value at vertex: negative, automatically satisfied by discriminant > 0</li></ul><p><strong>Step 2: Find minimum abc</strong></p><p>From b² > 4ac and b < 2a, we get: 4ac < b² < 4a²</p><p>So: c < a, meaning c ≤ a-1 (since both are natural numbers)</p><p>Try smallest values: a = 5, c = 1</p><ul><li>Need b² > 20, so b ≥ 5</li><li>Need b < 10</li><li>Need b < 5 + 1 = 6, so b = 5</li><li>Check: 25 > 20 ✓, discriminant exists</li><li>abc = 5·5·1 = 25</li></ul><p><strong>Step 3: Verify this is minimum</strong></p><p>For a = 4: c ≤ 3, need b² > 16c ≥ 48, so b ≥ 7. But b < 8, so b = 7. Check b < a+c: 7 < 7 fails.</p><p>For a = 5: a = 5, c = 1 gives abc = 25 (minimum)</p><p><strong>Step 4: Calculate bound on λ</strong></p><p>abc ≥ 25, so log₅(abc) ≥ log₅(25) = 2</p><p>Therefore λ ≤ 2</p><p>Non-negative integral values: <strong>λ ∈ {0, 1, 2}</strong></p><p>∴ Answer: A, B, C (corresponding to λ = 0, 1, 2)</p>
Correct Answer: A,B,C