Question

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

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

Answer 1

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

$f^{\prime }\left(x\right)<2f\left(x\right)$
$\Rightarrow e^{-2x}{f}^{\prime }\left(x\right)-2e^{-2x}f\left(x\right)<0$
$\Rightarrow \frac{d}{dx}\left(e^{-2x}f\right(x\left)\right)<0$
$\Rightarrow e^{-2x}f\left(x\right)$ is a strictly decreasing function.

$\Rightarrow f\left(1\right)e^{-2}<e^{-2x}f\left(x\right)<f\left(\frac{1}{2}\right)e^{-1}$
$\Rightarrow f\left(1\right)e^{-2}<e^{-2x}f\left(x\right)<e^{-1}$
$\Rightarrow 0<f\left(1\right)e^{2x-2}<f\left(x\right)<e^{2x-1}$
$\Rightarrow \underset{1/2}{\overset{1}{\int }}0dx<\underset{1/2}{\overset{1}{\int }}f\left(x\right)dx<\underset{1/2}{\overset{1}{\int }}{e}^{2x-1}dx$
$\Rightarrow 0<\underset{1/2}{\overset{1}{\int }}f\left(x\right)dx<\frac{e-1}{2}$