Question

$f(x)=f(x)+\displaystyle\int_{0}^{1}f(x)dx$ and given $f(0) = 1$ , then $f(x) $ equals

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

Answer 1

Answer: (B)

Given $f'\left(x\right) = f\left( x \right) +\int_{0}^{1}f \left(x\right)dx \ldots\left(1\right)$
$\therefore f'' \left(x\right)=f'\left(x\right) \Rightarrow \frac{f'\left(x\right)}{f'\left(x\right)}=1$
$\therefore log f' \left(x\right)=x+log\,C$
$\therefore f'\left(x\right)=Ce^{x} \ldots\left(2\right)$
$=Ce^{x}+C_{1} \ldots\left(3\right)$
Since $f \left(0\right)=1$
$\therefore 1=Ce^{0}+C_{1}=C+C_{1}$
$\therefore C_{1}=1-C$
$\therefore f\left(x\right)=Ce^{x}+1-C \ldots\left(4\right)$
By $\left(1\right), \left(2\right), \left(4\right),$ we get
$Ce^{x}=Ce^{x}+1-C+\int\limits_{0}^{1} \left(Ce^{x}+1-C\right)dx$
$\Rightarrow 0 = 1-C + \left|ce^{x}+\left(1-C\right)x\right|_{0}^{1}$
$\Rightarrow 0 = 1 - C + Ce + \left(1-C\right)-C=2-3C + Ce$
$\therefore C\left(e-3\right)=-2$
$\Rightarrow C=\frac{2}{3-e}$
$C_{1}=1-C=1-\frac{2}{3-e}=\frac{3-e-2}{3-e}=\frac{1-e}{3-e}$
$\left(By \left(3\right)\right)$
Hence $ f \left(x\right)=C e^{x}+C_{1}=\frac{2e^{x}}{3-e}+\frac{1-e}{3-e}$