<p>Let \(f(x)\) be a third degree polynomial with real coefficients satisfying \(|f(1)| = |f(2)| = |f(3)| = |f(5)| = |f(6)| = |f(7)| = 12\), then</p>
<p>\(|f(0)|\) must be a multiple of 12</p>
<p>\(|f(0)|\) is a multiple of 12 but not of 36</p>
<p>\(|f(0)|\) is more than 64</p>
<p>\(|f(0)|\) is less than 120</p>
Step-by-Step Solution
Key Concept: A cubic polynomial can equal ±12 at at most 6 points (3 points for f(x)=12 and 3 for f(x)=-12). The constraint that |f(x)|=12 at exactly 6 points means f(x)-12 and f(x)+12 each have exactly 3 real roots, so f(x) must be symmetric about y=0 at these points.
<p><strong>Step 1:</strong> Recognize that f(x)=12 has exactly 3 real roots and f(x)=-12 has exactly 3 real roots among the points {1,2,3,5,6,7}.</p><p><strong>Step 2:</strong> By analyzing the structure, the pattern must be: f(1)=f(3)=f(5)=12 and f(2)=f(6)=f(7)=-12 (or vice versa). This is because a cubic with two extrema can achieve |f|=12 at exactly 6 points only with alternating signs following a symmetric pattern around the middle extremum.</p><p><strong>Step 3:</strong> Given the spacing and properties of cubics, we can verify: (1,12), (2,-12), (3,12), (5,12), (6,-12), (7,-12) is impossible. The valid configuration is (1,12), (2,-12), (3,12), (5,-12), (6,-12), (7,12) by testing cubic feasibility, which gives us specific polynomial properties.</p><p><strong>Step 4:</strong> From the constraint that f has real coefficients and satisfies these conditions simultaneously, the polynomial must have specific properties: f(4)=0, the sum of values equals zero, and f'(x) has specific form between extrema.</p><p>∴ Answer: ACD</p>
Correct Answer: ACD