Applications of Derivatives
Rolle's Theorem
Grade 12
Question:
<p>Given quadratic equation is <br/> \(ax^2 + bx + c = 0\) <br/> Let \(f'(x) = ax^2 + bx + c\), so \(f(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx\). Clearly, \(f(0) = 0\) and \(f(1) = \frac{1}{6}(2a + 3b + 6c) = 0\). Given that \(f(0) = 0 = f(1)\), by Rolle's theorem, \(f'(x)\) has at least one root in \((0, 1)\). The root of \(ax^2 + bx + c = 0\) lies in:</p>
<p>\((0, 1)\)</p>
<p>\((1, 2)\)</p>
<p>\((-1, 0)\)</p>
<p>\((2, 3)\)</p>
Step-by-Step Solution
Key Concept: Rolle's theorem guarantees that if f(0) = f(1) = 0, then f'(x) = ax² + bx + c must have at least one root in the open interval (0,1). This directly translates the condition on f into a condition on its derivative.
<p><strong>Step 1:</strong> Given f'(x) = ax² + bx + c, integrate to get f(x) = (ax³/3) + (bx²/2) + cx + k.</p><p><strong>Step 2:</strong> Apply initial condition: f(0) = 0, which gives k = 0.</p><p><strong>Step 3:</strong> Apply second condition: f(1) = a/3 + b/2 + c = 0, which simplifies to (2a + 3b + 6c)/6 = 0.</p><p><strong>Step 4:</strong> Since f is continuous on [0,1] and differentiable on (0,1), with f(0) = f(1) = 0, Rolle's theorem guarantees that f'(ξ) = 0 for at least one ξ ∈ (0,1).</p><p><strong>Step 5:</strong> Since f'(x) = ax² + bx + c, this means the quadratic equation ax² + bx + c = 0 has at least one real root in the open interval (0,1).</p><p>∴ <strong>Answer:</strong> The root of ax² + bx + c = 0 lies in <strong>(0, 1)</strong></p>
Correct Answer: A