<p>Let \(f(x)\) be a polynomial of the second degree which is positive definite. If \(g(x) = f(x) + f'(x) + f''(x)\) then for any real \(x\)</p>
Step-by-Step Solution
Key Concept: Since f(x) is a positive definite quadratic, we can write f(x) = a(x-h)² + k where a > 0 and k > 0. Computing f'(x), f''(x), and analyzing g(x) = f(x) + f'(x) + f''(x) reveals the sign pattern through completing the square or examining the leading coefficient structure.
<p><strong>Step 1:</strong> Let f(x) = ax² + bx + c where a > 0 (since f is positive definite).</p><p><strong>Step 2:</strong> Compute derivatives: f'(x) = 2ax + b and f''(x) = 2a > 0.</p><p><strong>Step 3:</strong> Form g(x) = ax² + bx + c + 2ax + b + 2a = ax² + (b+2a)x + (c+b+2a).</p><p><strong>Step 4:</strong> For g(x) to be positive definite, check discriminant: Δ = (b+2a)² - 4a(c+b+2a) = b² + 4ab + 4a² - 4ac - 4ab - 8a² = b² - 4a(c+2a) = b² - 4ac - 8a².</p><p><strong>Step 5:</strong> Since f(x) is positive definite: b² - 4ac < 0, so b² < 4ac. Then Δ = (b² - 4ac) - 8a² < 0 - 8a² < 0.</p><p><strong>Step 6:</strong> Since a > 0 (coefficient of x²) and Δ < 0, g(x) > 0 for all real x.</p><p>∴ Answer: B</p>
Correct Answer: B