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For the function f(x) = α log|x| + βx² + x, if x = −1 and x = 2 are stationary points, then which pair of values of α and β is correct?
- α = 2, β = −1/2
- α = 2, β = 1/2
- α = −6, β = 1/2
- α = −6, β = −1/2
Correct answer: α = 2, β = −1/2
Solution
f'(x) = alpha/x + 2*beta*x + 1. At x = -1: -alpha - 2*beta + 1 = 0; at x = 2: alpha/2 + 4*beta + 1 = 0. Solving gives alpha = 2, beta = -1/2.
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