The function f(x)=x3 −62x2+ax+9 has a local maximum at x=1. Then the value of a is:

Given the function f(x)=x³ −62x²+ax+9. If f(x) has a local maximum at x=1, then the first derivative f′(x) must be zero at x=1, and the second derivative f′′(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 f′(1)=0: f′(1) = 3(1)² −124(1) + a = 0 3 − 124 + a = 0 −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)−124 f’’(1) = 6−124 f’’(1) = −118 Since f’’(1) = −118 < 0, there is indeed a local maximum at x=1. Therefore the final answer is option (B), a = 121.