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 ${\int }_{\frac{1}{2}}^{1}f\left(x\right)dx$ 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 $(0,\frac{e-1}{2})$

Given that

$f^{\prime }\left(x\right)<2f\left(x\right)$

$\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)}<2$

(becasue $f\left(x\right)$ is positive function.)

Integrating both sides, we get

$\Rightarrow \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx<2\int dx$

$\Rightarrow \ln f\left(x\right)<2x+c$

$\Rightarrow f\left(\frac{1}{2}\right)=1$

$\Rightarrow c=-1$

Hence,

${\int }_{\frac{1}{2}}^{1}f\left(x\right)dx\Rightarrow {\int }_{\frac{1}{2}}^{1}f\left(x\right)dx$

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