Quadratic Equations
Conditions for Positive Quadratic
Grade 11
Question:
<p>Let <i>f</i>(<i>x</i>) be a quadratic expression positive for all real <i>x</i>. If <i>g</i>(<i>x</i>) = <i>f</i>(<i>x</i>) − <i>f</i>'(<i>x</i>) + <i>f</i>''(<i>x</i>), then for any real <i>x</i></p>
<p>(a) <i>g</i>(<i>x</i>) ≤ 0</p>
<p>(b) <i>g</i>(<i>x</i>) > 0</p>
<p>(c) <i>g</i>(<i>x</i>) ≥ 0</p>
<p>(d) <i>g</i>(<i>x</i>) ≤ 0</p>
Step-by-Step Solution
Key Concept: Since f(x) is quadratic and positive for all real x, we can write f(x) = ax² + bx + c where a > 0 and discriminant < 0. We then compute f'(x) and f''(x), substitute into g(x), and analyze the resulting expression.
<p><strong>Step 1:</strong> Express f(x) as a quadratic. Let f(x) = ax² + bx + c where a > 0 (since f(x) > 0 for all real x).</p><p><strong>Step 2:</strong> Compute the derivatives. f'(x) = 2ax + b and f''(x) = 2a (constant).</p><p><strong>Step 3:</strong> Find g(x) by substituting into g(x) = f(x) − f'(x) + f''(x):<br/>g(x) = (ax² + bx + c) − (2ax + b) + 2a<br/>g(x) = ax² + bx + c − 2ax − b + 2a<br/>g(x) = ax² + (b − 2a)x + (c − b + 2a)</p><p><strong>Step 4:</strong> Analyze the sign of g(x). Since a > 0, g(x) is a parabola opening upward. We need to check the discriminant of g(x):<br/>Δ = (b − 2a)² − 4a(c − b + 2a)<br/>Δ = b² − 4ab + 4a² − 4ac + 4ab − 8a²<br/>Δ = b² + 4a² − 4ac − 8a²<br/>Δ = b² − 4ac − 4a²</p><p><strong>Step 5:</strong> Use the condition that f(x) > 0 for all real x. This means the discriminant of f(x) is negative: b² − 4ac < 0, so b² < 4ac.</p><p><strong>Step 6:</strong> Substitute this inequality into the discriminant of g(x):<br/>Δ = b² − 4ac − 4a² < 0 − 4a² = −4a² < 0<br/>(since a > 0, we have −4a² < 0)</p><p><strong>Step 7:</strong> Since Δ < 0 and the coefficient of x² in g(x) is a > 0, the parabola g(x) has no real roots and opens upward. Therefore, g(x) > 0 for all real x.</p><p><strong>∴ Answer:</strong> B</p>
Correct Answer: B