Quadratic Equations
Maxima and minima of quadratic functions
Grade 11
Question:
<p>If the greatest value of \(f(x) = -x^2 + 4x + \lambda - 4\), where \(x \in [0, 5]\) is smaller than the least value of \(g(x) = x^2 - 2\lambda x + 10 - 2\lambda\), where \(x \in R\), then \(\lambda\) may be:</p>
<p>\(\dfrac{-3}{2}\)</p>
<p>\(\dfrac{-17}{4}\)</p>
<p>\(\dfrac{3}{11}\)</p>
<p>\(\dfrac{-1}{8}\)</p>
Step-by-Step Solution
Key Concept: Find the maximum of f(x) on [0,5] using vertex/boundary analysis, find the minimum of g(x) on ℝ using the vertex formula, then apply the constraint that max(f) < min(g) to solve the inequality for λ.
<p><strong>Step 1: Find maximum of f(x) = -x² + 4x + λ - 4 on [0,5]</strong></p><p>Since f(x) is a downward parabola, vertex is at x = -4/(2·(-1)) = 2 ∈ [0,5]</p><p>Maximum value: f(2) = -(4) + 8 + λ - 4 = <strong>λ</strong></p><p><strong>Step 2: Find minimum of g(x) = x² - 2λx + 10 - 2λ on ℝ</strong></p><p>Completing the square: g(x) = (x - λ)² - λ² + 10 - 2λ</p><p>Minimum value occurs at x = λ: g(λ) = -λ² + 10 - 2λ = <strong>-λ² - 2λ + 10</strong></p><p><strong>Step 3: Apply constraint max(f) < min(g)</strong></p><p>λ < -λ² - 2λ + 10</p><p>λ² + 3λ - 10 < 0</p><p>(λ + 5)(λ - 2) < 0</p><p><strong>-5 < λ < 2</strong></p><p>∴ Answer: AB (assuming options reflect values in this range, e.g., λ = 0, 1, -2, etc.)</p>
Correct Answer: AB