Let `f:[1/2,1]->R` (the set of all real numbers) be a positive, non-constant, and differentiable function such that `f^(prime)(x)<2f(x)` and `f(1/2)=1` . Then the value of `int f(x)dx` lies in the interval for x:[1/2,1] (a)`(2e-1,2e)` (b) `(3-1,2e-1)` (c)`((e-1)/2,e-1)` (d) `(0,(e-1)/2)`

Options

A. `(2e-1,2e)`
B. `(e-1,2e-1)`
C. `((e-1)/2,e-1)`
D. `(0,(e-1)/2)`

Explanation

Given `f'(x)-2f(x)lt0`
or `f'(x)e^(-2x)-2e^(-2x)f(x)lt0`
or `d/(dx)(f(x)e^(-2x))lt0`
Thus `g(x)=f(x)e^(-2x)` is decreasing function.
Also `f(1//2)=1`
Now `g(x)ltg(1//2)` or `f(x)e^(-2x)ltf(1//2)e^(-1)` or `f(x)lt e^(2x-1)`
or `0lt int_(1//2)^(1)f(x)dxlt int_(1//2)^(1)e^(2x-1)dx` or `0lt int_(1//2)^(1)f(x)dxlt (e-1)/2`