Question:
If the Rolle's theorem holds for the function \(f(x) = 2x^3 + ax^2 + bx\) in the interval \([-1, 1]\) for the point \(c = \frac12\), then the value of \(2a + b\) is:
If the Rolle's theorem holds for the function \(f(x) = 2x^3 + ax^2 + bx\) in the interval \([-1, 1]\) for the point \(c = \frac12\), then the value of \(2a + b\) is:
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Use both Rolle conditions: \(f(-1) = f(1)\) and \(f'(\tfrac12) = 0\).
Updated On: Nov 11, 2026
- -1
- 2
- 1
- -2
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The Correct Option is A
Solution and Explanation
Rolle's theorem needs \(f(-1) = f(1)\): \(-2 + a - b = 2 + a + b\), so \(b = -2\).
And \(f'(c) = 0\) at \(c = \frac12\): \(f'(x) = 6x^2 + 2ax + b\), so \(\frac32 + a + b = 0\), giving \(a = -\frac32 + 2 = \frac12\).
Then \(2a + b = 1 - 2 = -1\).
Answer: (1) -1
And \(f'(c) = 0\) at \(c = \frac12\): \(f'(x) = 6x^2 + 2ax + b\), so \(\frac32 + a + b = 0\), giving \(a = -\frac32 + 2 = \frac12\).
Then \(2a + b = 1 - 2 = -1\).
Answer: (1) -1
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