Question

Let $f:[\frac{1}{2},1]\to 1$ (the set of all real numbers) 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 ${\int }_{1/2}^{1}f\left(x\right)dx$ lies in the interval

• A. $(0,\frac{e-1}{2})$
• B. $(\frac{e-1}{2},e-1)$
• C. $(e-1,2e-1)$
• D. $(2e-1,2e)$

Answer 1

The correct option is A $(0,\frac{e-1}{2})$

$\frac{dy}{dx}<2y$
$e^{-2x}\frac{dy}{dx}<2y{e}^{-2x}$
$\frac{d}{dx}\left(y{e}^{-2x}\right)<0$
$\Rightarrow y{e}^{-2x}$ is decreasing function
As, $\frac{1}{2}<x<1$
$\Rightarrow e^{-1}>y{e}^{-2x}>y\left(1\right)e^{-2}$
$\Rightarrow e^{2x-1}>y>y\left(1\right)e^{2x-2}$
$\Rightarrow {\int }_{1/2}^1{e}^{2x-1}dx>{\int }_{1/2}^{1}ydx>{\int }_{1/2}^{1}y\left(1\right)e^{2x-2}>0$
$\therefore 0<{\int }_{1/2}^{1}ydx<\frac{e-1}{2}$