Let f: [1/2, 1] → R (the set of all real numbers) be a positive, non-constant and differentiable function such that f'(x) < 2f(x) and f(1/2) = 1. Then the value of ∫(1/2)^1 f(x)dx lies in the interval (2e⁻¹, 2e), (e⁻²/2, e⁻¹), (e⁻¹, 2e⁻¹), (0, e⁻²/2)

Given that f'(x) < 2f(x)
We have f'(x) - 2f(x) < 0
Multiplying by e⁻²ˣ, we get
e⁻²ˣf'(x) - 2e⁻²ˣf(x) < 0
This can be written as
d/dx(e⁻²ˣf(x)) < 0
Integrating from 1/2 to 1, we have
∫(1/2)^1 d/dx(e⁻²ˣf(x)) dx < ∫(1/2)^1 0 dx
e⁻²ˣf(x)^1 < 0
e⁻²f(1) - e⁻¹f(1/2) < 0
e⁻²f(1) - e⁻¹ < 0
e⁻²f(1) < e⁻¹
f(1) < e
Also, since f(x) is positive and f'(x) < 2f(x), we have
∫(1/2)^1 f(x) dx > 0
Let I = ∫(1/2)^1 f(x) dx
Since f(1/2) = 1 and f(x) is positive, we can say that f(x) is approximately 1 in the interval [1/2, 1].
Therefore, I ≈ ∫(1/2)^1 1 dx = 1/2
Also, we have f(1) < e
Then, I = ∫(1/2)^1 f(x) dx < ∫(1/2)^1 e dx = e/2 ≈ 1.359
Since f(x) is a positive function and I > 0, and I < e/2, we can conclude that I lies in the interval (0, e/2).
However, since f(1/2) = 1 and f'(x) < 2f(x), the value of f(x) will be less than e in the interval [1/2,1].
Therefore, the value of ∫(1/2)^1 f(x) dx will lie in the interval (e⁻¹, 2e⁻¹).