Applications of Derivatives
Tangent and Normal
Grade 12
Question:
<p>Let \(S\) be the set of all values of \(x\) for which the tangent to the curve \(y = f(x) = x^3 - x^2 - 2x\) at \((x, y)\) is parallel to the line segment joining the points \((1, f(1))\) and \((-1, f(-1))\), then \(S\) is equal to:</p>
<p>\(\left\{\dfrac{1}{3}, 1\right\}\)</p>
<p>\(\left\{-\dfrac{1}{3}, -1\right\}\)</p>
<p>\(\left\{\dfrac{1}{3}, -1\right\}\)</p>
<p>\(\left\{-\dfrac{1}{3}, 1\right\}\)</p>
Step-by-Step Solution
Key Concept: Apply Rolle's Theorem by finding the slope of the chord joining two points, then solve f'(x) = slope to find where tangent is parallel to the chord.
<p><strong>Step 1:</strong> Calculate f(1) and f(-1).</p><p>f(x) = x³ - x² - 2x</p><p>f(1) = 1 - 1 - 2 = -2</p><p>f(-1) = -1 - 1 + 2 = 0</p><p><strong>Step 2:</strong> Find the slope of chord joining (1, -2) and (-1, 0).</p><p>m_chord = (0 - (-2))/(-1 - 1) = 2/(-2) = -1</p><p><strong>Step 3:</strong> Find f'(x) and set f'(x) = m_chord.</p><p>f'(x) = 3x² - 2x - 2</p><p>We need: 3x² - 2x - 2 = -1</p><p>3x² - 2x - 1 = 0</p><p><strong>Step 4:</strong> Solve the quadratic equation.</p><p>3x² - 2x - 1 = 0</p><p>(3x + 1)(x - 1) = 0</p><p>x = -1/3 or x = 1</p><p><strong>Step 5:</strong> Verify both solutions are valid (they are interior and boundary points where tangent exists).</p><p>∴ S = {-1/3, 1} or answer D</p>
Correct Answer: D