Question

$f(x)=f(x)+{\int }_0^{1}f(x)dx$ 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 $f^{\prime }\left(x\right)=f\left(x\right)+{\int }_0^{1}f\left(x\right)dx\dots \left(1\right)$
$\therefore f^{\prime \prime }\left(x\right)=f^{\prime }\left(x\right)\Rightarrow \frac{f^{\prime }\left(x\right)}{f^{\prime }\left(x\right)}=1$
$\therefore log{f}^{\prime }\left(x\right)=x+logC$
$\therefore f^{\prime }\left(x\right)=C{e}^x\dots \left(2\right)$
$=C{e}^x+C_1\dots \left(3\right)$
Since $f\left(0\right)=1$
$\therefore 1=C{e}^0+C_1=C+C_1$
$\therefore C_1=1-C$
$\therefore f\left(x\right)=C{e}^x+1-C\dots \left(4\right)$
By $\left(1\right),\left(2\right),\left(4\right),$ we get
$C{e}^x=C{e}^x+1-C+\underset{0}{\overset{1}{\int }}\left(C{e}^x+1-C\right)dx$
$\Rightarrow 0=1-C+{\left|c{e}^x+\left(1-C\right)x\right|}_0^1$
$\Rightarrow 0=1-C+Ce+\left(1-C\right)-C=2-3C+Ce$
$\therefore C\left(e-3\right)=-2$
$\Rightarrow C=\frac{2}{3-e}$
$C_1=1-C=1-\frac{2}{3-e}=\frac{3-e-2}{3-e}=\frac{1-e}{3-e}$
$\left(By\left(3\right)\right)$
Hence $f\left(x\right)=C{e}^x+C_1=\frac{2e^x}{3-e}+\frac{1-e}{3-e}$