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Let f:[1/2, 1] -> R be a positive, non-constant, differentiable function with f'(x) < 2 f(x) and f(1/2) = 1. Then the integral of f(x) from x = 1/2 to 1 lies in which interval?

1. (2e - 1, 2e)
2. (e - 1, 2e - 1)
3. ((e - 1)/2, e - 1)
4. (0, (e - 1)/2)

Correct answer: (0, (e - 1)/2)

Solution

Since f(x) e^(-2x) decreases, f(x) < e^(2x-1), and integrating from 1/2 to 1 gives the bound (e-1)/2, so the integral lies in (0, (e-1)/2).

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