Quadratic Equations
Maxima/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 properties, find the minimum of g(x) over all reals, then impose the strict inequality max(f) < min(g) to constrain λ.
<p><strong>Step 1: Find max(f) on [0,5]</strong></p><p>f(x) = -x² + 4x + λ - 4 is a downward parabola with vertex at x = 2.</p><p>Since 2 ∈ [0,5], the maximum occurs at x = 2:</p><p>f(2) = -4 + 8 + λ - 4 = λ</p><p><strong>Step 2: Find min(g) over ℝ</strong></p><p>g(x) = x² - 2λx + 10 - 2λ is an upward parabola with vertex at x = λ.</p><p>The minimum value is:</p><p>g(λ) = λ² - 2λ·λ + 10 - 2λ = -λ² + 10 - 2λ</p><p><strong>Step 3: Apply the constraint</strong></p><p>We need: max(f) < min(g)</p><p>λ < -λ² + 10 - 2λ</p><p>λ² + 3λ + λ - 10 < 0</p><p>λ² + 3λ - 10 < 0</p><p>(λ + 5)(λ - 2) < 0</p><p>∴ -5 < λ < 2</p><p>Checking the options labeled AB would represent the interval (-5, 2).</p>
Correct Answer: AB