Question

Let $f:[\frac{1}{2},1]\to R$ 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 $\underset{\frac{1}{2}}{\overset{1}{\int }}f\left(x\right)dx$ 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 $(0,\frac{e-1}{2})$

Given that

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

As $f\left(x\right)$ is positive function, so Integrating both sides, we get

$\Rightarrow \underset{1/2}{\overset{x}{\int }}\frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx<2\underset{1/2}{\overset{x}{\int }}dx\Rightarrow \ln \left|f\right(x\left)\right|-\ln \left|f\left(\frac{1}{2}\right)\right|<2x-1\Rightarrow \ln f\left(x\right)<2x-1\Rightarrow 0<f\left(x\right)<e^{2x-1}$
Now,
$0<\underset{\frac{1}{2}}{\overset{1}{\int }}f\left(x\right)dx<{\left[\frac{e^{2x-1}}{2}\right]}_{1/2}^1\Rightarrow 0<\underset{\frac{1}{2}}{\overset{1}{\int }}f\left(x\right)dx<\frac{(e-1)}{2}$