<p><b>For Problems 1–3:</b> If roots of the equation \(f(x) = x^6 - 12x^5 + bx^4 + cx^3 + dx^2 + ex + 64 = 0\) are positive, then</p><p><b>Question 1:</b> Which has the greatest absolute value?</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas for a degree-6 polynomial: the product of all roots equals the constant term (64), and the sum equals the coefficient of x⁵ (12). Since all roots are positive and their product is 64 while their sum is 12, at least one root must be significantly smaller than others.
<p><strong>Step 1:</strong> Apply Vieta's formulas. For f(x) = x⁶ - 12x⁵ + bx⁴ + ..., if roots are r₁, r₂, r₃, r₄, r₅, r₆:</p><p>• Product: r₁·r₂·r₃·r₄·r₅·r₆ = 64</p><p>• Sum: r₁ + r₂ + r₃ + r₄ + r₅ + r₆ = 12</p><p><strong>Step 2:</strong> By AM-GM inequality: (r₁ + r₂ + ... + r₆)/6 ≥ ⁶√64 = 2</p><p>This gives: Sum ≥ 12, and equality holds when all roots equal 2.</p><p><strong>Step 3:</strong> However, since the sum equals exactly 12 (the minimum), we must have r₁ = r₂ = r₃ = r₄ = r₅ = r₆ = 2.</p><p><strong>Step 4:</strong> Each root has absolute value |2| = 2, so all roots have equal absolute value.</p><p>∴ Answer: A (All roots have equal absolute value, which is 2)</p>
Correct Answer: A