Let $f:\left[\frac{1}{2},1\right]\to \mathbb{R}$ (the set of all real number) be a positive, non-constant and differentiable function such that $f'(x)<2f(x)$ and $f\left(\frac{1}{2}\right)=1$. Then the value of $\int_{1/2}^{1} f(x)\,dx$ lies in the interval

$f:\left[\dfrac12,1\right]\to\mathbb{R}, \quad f>0, \text{ non-constant, differentiable}$ $f'(x)<2f(x), \qquad f\!\left(\dfrac12\right)=1$ $\text{Find the interval containing } \displaystyle\int_{1/2}^{1} f(x)\,dx$ $\text{Options: } (2e-1,2e),\ (e-1,2e-1),\ \left(\tfrac{e-1}{2},e-1\right),\ \left(0,\tfrac{e-1}{2}\right)$ $F(x)=e^{-2x}f(x)$ $F'(x)=e^{-2x}(f'-2f)<0 \Rightarrow F \text{ decreasing}$ $\Rightarrow f(x)<e^{2x-1}, \text{ then integrate}$ $F(x)=e^{-2x}f(x)$ $F'(x)=e^{-2x}\big(f'(x)-2f(x)\big)<0$ $x>\dfrac12 \Rightarrow F(x)<F\!\left(\dfrac12\right)=e^{-1}$ $e^{-2x}f(x)<e^{-1} \Rightarrow f(x)<e^{2x-1}$ $\displaystyle\int_{1/2}^{1} f(x)\,dx<\int_{1/2}^{1} e^{2x-1}\,dx$ $=\left[\dfrac12 e^{2x-1}\right]_{1/2}^{1}=\dfrac{e-1}{2}$ $f(x)>0 \Rightarrow \displaystyle\int_{1/2}^{1} f(x)\,dx>0$ $0<\displaystyle\int_{1/2}^{1} f(x)\,dx<\dfrac{e-1}{2}$ $\dfrac{e-1}{2}\approx 0.859, \quad \displaystyle\int_{1/2}^{1} f\,dx\in\left(0,\dfrac{e-1}{2}\right)$ $\Rightarrow \text{option (4)}$ $\boxed{\text{Answer: option (4)}}$