Question

If $f^{\prime }\left(x\right)=f\left(x\right)+\underset{0}{\overset{1}{\int }}f\left(x\right)dx$ and $f\left(0\right)=1,$ then the value of $f\left({\log }_{e}2\right)$ is

• A. $\frac{1}{3+e}$
• B. $\frac{5-e}{3-e}$
• C. $\frac{2+e}{e-2}$
• D. $1$

Answer 1

The correct option is B $\frac{5-e}{3-e}$

Differentiating the given expression, we get $f^{\prime \prime }\left(x\right)=f^{\prime }\left(x\right)$
$\Rightarrow \int \frac{d{f}^{\prime }\left(x\right)}{f^{\prime }\left(x\right)}=\int dx\Rightarrow \ln |f^{\prime }\left(x\right)|=x+c\Rightarrow f^{\prime }\left(x\right)=\pm A{e}^x\cdots \left(i\right)$
(where $A=e^c$)
$\Rightarrow \underset{}{\overset{}{\int }}{f}^{\prime }\left(x\right)dx={\int }^{}(\pm A{e}^x)dx\Rightarrow f\left(x\right)=\pm A{e}^x+B\cdots \left(ii\right)$
Now, $f\left(0\right)=1\Rightarrow \pm A+B=1$
Case $1:$ taking the positive sign,
$\therefore f^{\prime }\left(x\right)=f\left(x\right)+\underset{0}{\overset{1}{\int }}(A{e}^x+B)dx$
$A{e}^x=(A{e}^x+1-A)+[A{e}^x+(1-A)x{]}_0^1$
$\Rightarrow 1-A+(Ae+1-A-A)=0$
$\Rightarrow A(e-3)=-2$
$\Rightarrow A=\frac{2}{3-e}$ and $B=1-\frac{2}{3-e}=\frac{1-e}{3-e}$
$\therefore f\left({\log }_{e}2\right)=2A+B=\frac{4}{3-e}+\frac{1-e}{3-e}=\frac{5-e}{3-e}$
Case $2:$ taking negative sign in $f\left(x\right)=\pm A{e}^x+B$
$\therefore f^{\prime }\left(x\right)=f\left(x\right)+\underset{0}{\overset{1}{\int }}(-A{e}^x+B)dx$
$-A{e}^x=(-A{e}^x+1+A)+[-A{e}^x+(1+A)x{]}_0^1$
$\Rightarrow 1+A+(-Ae+1+A+A)=0$
$\Rightarrow A(3-e)=-2$
$\Rightarrow A=\frac{-2}{3-e}$ and $B=1+\frac{-2}{3-e}=\frac{1-e}{3-e}$
$\therefore f\left({\log }_{e}2\right)=-2A+B=\frac{4}{3-e}+\frac{1-e}{3-e}=\frac{5-e}{3-e}$
Thus from Case $1$ and $2$,
$f\left({\log }_{e}2\right)=\frac{5-e}{3-e}$