Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>If non-zero real numbers \(b\) and \(c\) are such that \(\min f(x) > \max g(x)\), where \(f(x) = x^2 + 2bx + 2c^2\) and \(g(x) = -x^2 - 2cx + b^2\) (\(x \in \mathbb{R}\)); then \(\left|\dfrac{c}{b}\right|\) lies in the interval:</p>
<p>\(\left(0, \dfrac{1}{2}\right)\)</p>
<p>\(\left[\dfrac{1}{2}, \dfrac{1}{\sqrt{2}}\right)\)</p>
<p>\(\left[\dfrac{1}{\sqrt{2}}, \sqrt{2}\right]\)</p>
<p>\(\left(\sqrt{2}, \infty\right)\)</p>
Step-by-Step Solution
Key Concept: The minimum of f(x) occurs at its vertex and the maximum of g(x) occurs at its vertex. Since f is upward-opening and g is downward-opening, we need min f(x) > max g(x), which creates a constraint relating b and c through their vertex values.
<p><strong>Step 1:</strong> Find the minimum of f(x) = x² + 2bx + 2c²</p><p>Vertex x-coordinate: x = -b</p><p>min f(x) = f(-b) = b² - 2b² + 2c² = 2c² - b²</p><p><strong>Step 2:</strong> Find the maximum of g(x) = -x² - 2cx + b²</p><p>Since coefficient of x² is negative, parabola opens downward</p><p>Vertex x-coordinate: x = -c</p><p>max g(x) = g(-c) = -c² + 2c² + b² = c² + b²</p><p><strong>Step 3:</strong> Apply the condition min f(x) > max g(x)</p><p>2c² - b² > c² + b²</p><p>c² > 2b²</p><p>c²/b² > 2</p><p>|c/b| > √2</p><p><strong>Step 4:</strong> Since b and c are non-zero real numbers, |c/b| ∈ (√2, ∞)</p><p>∴ Answer: D</p>
Correct Answer: D