<p>Given \(f'(x) = x^2 - 2x\) and \(f(2) = 0\), the curve \(y = f(x)\) has a minimum at \(x = 2\). If the point of minimum ordinate is \((a, b)\), find \(a + 6b\).</p>
Step-by-Step Solution
Key Concept: A point is a minimum only if f'(x) = 0 AND f''(x) > 0 at that point. Use the condition that x = 2 is a minimum to verify consistency, then integrate f'(x) using f(2) = 0 to find the constant, and finally evaluate f(2) to get the minimum point coordinates.
<p><strong>Step 1: Verify x = 2 is a minimum point.</strong></p><p>Given f'(x) = x² - 2x. For a critical point: f'(2) = 4 - 4 = 0 ✓</p><p>Check f''(x) = 2x - 2, so f''(2) = 4 - 2 = 2 > 0. Thus x = 2 is indeed a minimum.</p><p><strong>Step 2: Find f(x) by integrating f'(x).</strong></p><p>f(x) = ∫(x² - 2x)dx = x³/3 - x² + C</p><p><strong>Step 3: Use initial condition f(2) = 0 to find C.</strong></p><p>f(2) = 8/3 - 4 + C = 0</p><p>8/3 - 12/3 + C = 0 → C = 4/3</p><p>Therefore: f(x) = x³/3 - x² + 4/3</p><p><strong>Step 4: Find the minimum point (a, b).</strong></p><p>Since the minimum occurs at x = 2: a = 2</p><p>b = f(2) = 0 (given)</p><p>So (a, b) = (2, 0)</p><p><strong>Step 5: Calculate a + 6b.</strong></p><p>a + 6b = 2 + 6(0) = 2</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2