Question:
If Rolle's theorem holds for the function \(f(x) = 2x^3 + bx^2 + cx\), \(x \in [-1, 1]\) at the point \(x = \frac{1}{2}\), then \(2b + c\) is equal to
If Rolle's theorem holds for the function \(f(x) = 2x^3 + bx^2 + cx\), \(x \in [-1, 1]\) at the point \(x = \frac{1}{2}\), then \(2b + c\) is equal to
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Rolle's theorem needs \(f(-1) = f(1)\) and \(f'(1/2) = 0\).
Updated On: Nov 11, 2026
- \(2\)
- \(1\)
- \(-1\)
- \(-3\)
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The Correct Option is C
Solution and Explanation
\(f(-1) = f(1)\): \(-2 + b - c = 2 + b + c\), so \(c = -2\).
\(f'(x) = 6x^2 + 2bx + c\); \(f'(1/2) = 0\): \(\frac{6}{4} + b + c = 0\), so \(b = -\frac{3}{2} + 2 = \frac{1}{2}\).
\(2b + c = 1 - 2 = -1\).
Answer: (3) \(-1\)
\(f'(x) = 6x^2 + 2bx + c\); \(f'(1/2) = 0\): \(\frac{6}{4} + b + c = 0\), so \(b = -\frac{3}{2} + 2 = \frac{1}{2}\).
\(2b + c = 1 - 2 = -1\).
Answer: (3) \(-1\)
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