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    Question

    The function f(x)=x3 βˆ’62x2+ax+9

    has a local maximum at x=1. Then the value of a is:
    A 100 Correct Answer Incorrect Answer
    B 121 Correct Answer Incorrect Answer
    C 169 Correct Answer Incorrect Answer
    D 144 Correct Answer Incorrect Answer

    Solution

    Given the function f(x)=x3 βˆ’62x2+ax+9. If f(x) has a local maximum at x=1, then the first derivative fβ€²(x) must be zero at x=1, and the second derivative fβ€²β€²(x) must be negative at x=1. First, find the first derivative of f(x): Since there is a local maximum at x=1, we must have fβ€²(1)=0: fβ€²(1) = 3(1)2 βˆ’124(1) + a = 0 3 βˆ’ 124 + a = 0 βˆ’121 + a = 0 a = 121 Now, we need to check the second derivative to ensure it's a local maximum. Find the second derivative of f(x): Evaluate the second derivative at x=1: f’’(1) = 6(1)βˆ’124 f’’(1) = 6βˆ’124 f’’(1) = βˆ’118 Since f’’(1) = βˆ’118 < 0, there is indeed a local maximum at x=1. Therefore the final answer is option (B), a = 121.

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