<p>A function \(y = f(x)\) has a second order derivative \(f''(x) = 6(x - 1)\). If its graph passes through the point (2,1) and at the point the tangent to the graph is \(y = 3x - 5\), then the function is</p>
Step-by-Step Solution
Key Concept: Integrate the second derivative twice using initial conditions (point and tangent slope) to uniquely determine the function. The tangent line gives us f'(2) = 3, and the point (2,1) gives us f(2) = 1.
<p><strong>Step 1: Find f'(x) by integrating f''(x)</strong></p><p>Given: f''(x) = 6(x - 1)</p><p>Integrating: f'(x) = ∫6(x - 1)dx = 6 · (x - 1)²/2 + C₁ = 3(x - 1)² + C₁</p><p><strong>Step 2: Use the tangent condition to find C₁</strong></p><p>At point (2, 1), the tangent line is y = 3x - 5, so f'(2) = 3</p><p>Substituting x = 2: f'(2) = 3(2 - 1)² + C₁ = 3(1) + C₁ = 3 + C₁</p><p>Since f'(2) = 3: 3 + C₁ = 3 ⟹ C₁ = 0</p><p>Therefore: f'(x) = 3(x - 1)²</p><p><strong>Step 3: Find f(x) by integrating f'(x)</strong></p><p>f(x) = ∫3(x - 1)²dx = 3 · (x - 1)³/3 + C₂ = (x - 1)³ + C₂</p><p><strong>Step 4: Use the point condition to find C₂</strong></p><p>The graph passes through (2, 1), so f(2) = 1</p><p>Substituting: f(2) = (2 - 1)³ + C₂ = 1 + C₂ = 1</p><p>This gives: C₂ = 0</p><p><strong>Step 5: Write the final function</strong></p><p>f(x) = (x - 1)³</p><p><strong>Verification:</strong> f(2) = (2-1)³ = 1 ✓ and f'(2) = 3(2-1)² = 3 ✓</p><p><strong>∴ Answer:</strong> b</p>
Correct Answer: b