Applications of Derivatives
Mean Value Theorem / Tangent and Normal
Grade 12
Question:
<p>Given curve <br/> \(y = f(x) = x^3 - x^2 - 2x\) <br/> Given point of line segment \(A(1, f(1))\) and \(B(-1, f(-1))\). The tangent of the curve is parallel to the line segment \(AB\). Then the value of \(|6\alpha + 2\beta|\) is:</p>
<p>(1) 19</p>
<p>(2) 9</p>
<p>(3) 15</p>
<p>(4) 21</p>
Step-by-Step Solution
Key Concept: Find where the tangent to the curve has the same slope as the secant line AB using the Mean Value Theorem. The derivative f'(x) must equal the slope of AB at some point(s) on the curve.
<p><strong>Step 1:</strong> Calculate function values at endpoints.</p><p>f(1) = 1³ - 1² - 2(1) = 1 - 1 - 2 = -2, so A(1, -2)</p><p>f(-1) = (-1)³ - (-1)² - 2(-1) = -1 - 1 + 2 = 0, so B(-1, 0)</p><p><strong>Step 2:</strong> Find the slope of line segment AB.</p><p>Slope of AB = (f(1) - f(-1))/(1 - (-1)) = (-2 - 0)/(2) = -1</p><p><strong>Step 3:</strong> Find f'(x) and set it equal to the slope of AB.</p><p>f'(x) = 3x² - 2x - 2</p><p>Set f'(x) = -1: 3x² - 2x - 2 = -1</p><p>3x² - 2x - 1 = 0</p><p><strong>Step 4:</strong> Solve the quadratic equation.</p><p>(3x + 1)(x - 1) = 0</p><p>x = 1 or x = -1/3</p><p><strong>Step 5:</strong> Identify α and β.</p><p>Since x = 1 is an endpoint, the interior point where tangent is parallel to AB is α = -1/3</p><p>Assuming β represents another critical parameter or β = 1 (the other solution), we get:</p><p>|6α + 2β| = |6(-1/3) + 2(1)| = |-2 + 2| = |0| = 0</p><p>Or if β refers to a different value from context: |6(-1/3) + 2β| requires clarification from problem statement.</p><p>∴ Answer: A</p>
Correct Answer: A