Applications of Derivatives
Rolle's Theorem
Grade 12
Question:
<p>If the Rolle's theorem holds for the function \(f(x) = 2x^3 + ax^2 + bx\) in the interval \([-1, 1]\) for the point \(c = \dfrac{1}{2}\), then the value of \(2a + b\) is</p>
<p>1</p>
<p>\(-1\)</p>
<p>2</p>
<p>\(-2\)</p>
Step-by-Step Solution
Key Concept: If Rolle's theorem holds at c = 1/2, then f(-1) = f(1) and f'(1/2) = 0. These two conditions give us two equations to solve for a and b.
<p><strong>Step 1:</strong> Apply the condition f(-1) = f(1) from Rolle's theorem.</p><p>f(-1) = 2(-1)³ + a(-1)² + b(-1) = -2 + a - b</p><p>f(1) = 2(1)³ + a(1)² + b(1) = 2 + a + b</p><p>Setting them equal: -2 + a - b = 2 + a + b</p><p>This gives: -2 - b = 2 + b → -4 = 2b → <strong>b = -2</strong></p><p><strong>Step 2:</strong> Apply the condition f'(c) = 0 at c = 1/2.</p><p>f'(x) = 6x² + 2ax + b</p><p>f'(1/2) = 6(1/4) + 2a(1/2) + b = 0</p><p>3/2 + a + b = 0</p><p>3/2 + a - 2 = 0 → a - 1/2 = 0 → <strong>a = 1/2</strong></p><p><strong>Step 3:</strong> Calculate 2a + b.</p><p>2a + b = 2(1/2) + (-2) = 1 - 2 = <strong>-1</strong></p><p>∴ Answer: B</p>
Correct Answer: B