Quadratic Equations
Positive definite quadratics
Grade 11
Question:
<p>Suppose that \(f(x)\) is a quadratic expression positive for all real \(x\). If \(g(x) = f(x) + f'(x) + f''(x)\), then for any real \(x\) (where \(f'(x)\) and \(f''(x)\) represent 1st and 2nd derivative, respectively)</p>
<p>(1) \(g(x) < 0\)</p>
<p>(2) \(g(x) > 0\)</p>
<p>(3) \(g(x) = 0\)</p>
<p>(4) \(g(x) \geq 0\)</p>
Step-by-Step Solution
Key Concept: Since f(x) is a quadratic positive for all x, we can write f(x) = a(x-h)² + k where a > 0. The key insight is computing f'(x) and f''(x), then analyzing g(x) = f(x) + f'(x) + f''(x) by recognizing that the constant term f''(x) = 2a dominates the behavior.
<p><strong>Step 1:</strong> Write f(x) = ax² + bx + c where a > 0 (since f(x) > 0 for all real x).</p><p><strong>Step 2:</strong> Compute derivatives: f'(x) = 2ax + b and f''(x) = 2a.</p><p><strong>Step 3:</strong> Form g(x) = f(x) + f'(x) + f''(x) = ax² + bx + c + 2ax + b + 2a = ax² + (b+2a)x + (c+b+2a).</p><p><strong>Step 4:</strong> Since f(x) > 0 for all x, the discriminant of f(x) satisfies: b² - 4ac < 0, so b² < 4ac.</p><p><strong>Step 5:</strong> For g(x), check its discriminant: Δ_g = (b+2a)² - 4a(c+b+2a) = b² + 4ab + 4a² - 4ac - 4ab - 8a² = b² - 4ac - 4a² < 0 (since b² - 4ac < 0 and -4a² < 0).</p><p><strong>Step 6:</strong> Since Δ_g < 0 and a > 0, g(x) > 0 for all real x.</p><p>∴ Answer: B</p>
Correct Answer: B