<p>Given \(P(x) = x^4 + ax^3 + bx^2 + cx + d\) such that \(x = 0\) is the only real root of \(P'(x) = 0\). If \(P(-1) < P(1)\), then in the interval \([-1, 1]\)</p>
<p>\(P(-1)\) is the minimum and \(P(1)\) is the maximum of \(P\).</p>
<p>\(P(-1)\) is not minimum but \(P(1)\) is the maximum of \(P\).</p>
<p>\(P(-1)\) is the minimum and \(P(1)\) is not the maximum of \(P\).</p>
<p>Neither \(P(-1)\) is the minimum nor \(P(1)\) is the maximum of \(P\).</p>
Step-by-Step Solution
Key Concept: If x = 0 is the only real root of P'(x) = 0, then P'(x) = 4x³ + 3ax² + 2bx + c must have x = 0 as a root with multiplicity ≥ 1, and all other critical points must be complex. This severely constrains the coefficients and the behavior of P(x).
<p><strong>Step 1:</strong> Since x = 0 is the only real root of P'(x) = 4x³ + 3ax² + 2bx + c, we have P'(0) = 0, giving c = 0.</p><p><strong>Step 2:</strong> So P'(x) = 4x³ + 3ax² + 2bx = x(4x² + 3ax + 2b). For x = 0 to be the only real root, the quadratic 4x² + 3ax + 2b must have no real roots. Thus Δ = 9a² - 32b < 0, or b > (9a²)/32.</p><p><strong>Step 3:</strong> Since 4x² + 3ax + 2b > 0 for all x ≠ 0, the sign of P'(x) depends only on x: P'(x) < 0 for x < 0 and P'(x) > 0 for x > 0. Therefore P is strictly decreasing on (-∞, 0) and strictly increasing on (0, ∞).</p><p><strong>Step 4:</strong> This means P(-1) > P(0) and P(1) > P(0). Given P(-1) < P(1), combined with P being decreasing on (-1, 0) and increasing on (0, 1), we have P(0) < P(-1) < P(1), confirming the ordering. The answer depends on the specific constraint—typically comparing P(-1), P(0), P(1) or evaluating specific properties.</p><p>∴ Answer: B</p>
Correct Answer: B