Let $f:\left[\frac{1}{2}, 1\right] \in R$ (the set of all real numbers) be a positive, non-constant and differentiable function such that $f'(x)<2 f(x)$ and $f\left(\frac{1}{2}\right)=1$. Then the value of $\int\limits_{1 / 2}^1 f(x) d x$ lies in the interval
Answer & explanation
Correct answer: option 4
We have,
$f'(x)-2 f(x)<0$
$\Rightarrow f'(x) e^{-2 x}-2 e^{-2 x} f(x)<0 e^{-2 x}$
$\Rightarrow \frac{d}{d x}\left(f(x) e^{-2 x}\right)<0$
$\Rightarrow f(x) e^{-2 x}$ is decreasing on $[1 / 2,1]$
$\Rightarrow f(x) e^{-2 x}<f\left(\frac{1}{2}\right)$ for all $x \in[1 / 2,1]$
$\Rightarrow f(x) e^{-2 x}<\frac{1}{e}$ for all $x \in[1 / 2,1]$
$\Rightarrow f(x)<e^{2 x-1}$ for all $x \in[1 / 2,1]$
Thus, we have,
$0<f(x)<e^{2 x-1}$ for all $x \in[1 / 2,1]$
$\Rightarrow 0<\int\limits_{1 / 2}^1 f(x) d x<\int\limits_{1 / 2}^1 e^{2 x-1}$
$\Rightarrow 0<\int\limits_{1 / 2}^1 f(x) d x<\left[\frac{1}{2} e^{2 x-1}\right]_{1 / 2}^1$
$\Rightarrow 0<\int\limits_{1 / 2}^1 f(x) d x<\frac{e-1}{2}$
$\Rightarrow \int\limits_{1 / 2}^1 f(x) d x \in\left(0, \frac{e-1}{2}\right)$