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The function f(x) = 2e^x - a*e^(-x) + (2a + 1)x - 3 is monotonically increasing for all x in R. What is the minimum value of the parameter a?

1. -1
2. 0
3. 1
4. 2

Correct answer: 0

Solution

f'(x) = 2e^x + a*e^(-x) + (2a+1). Let t = e^x > 0. Need 2t² + (2a+1)t + a >= 0 for all t > 0. Factoring: (2t + 1)(t + a) >= 0 for all t > 0. Since 2t+1 > 0 always, we need t + a >= 0 for all t > 0, which means a >= 0. Minimum value of a is 0.

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