Question

If $f:R\to R$ is a continuous function and satisfies $f\left(x\right)=e^x+\underset{0}{\overset{1}{\int }}{e}^{x}f\left(t\right)dt$, then

• A. $f\left(0\right)<0$
• B. $f\left(x\right)$ is a decreasing function $\forall x\in R$
• C. $f\left(x\right)$ is a increasing function $\forall x\in R$
• D. $\underset{0}{\overset{1}{\int }}f\left(x\right)dx>0$

Answer 1

Answer: (A), (B)

$f\left(x\right)$ is a decreasing function $\forall x\in R$

We have
$f\left(x\right)=e^x+\underset{0}{\overset{1}{\int }}{e}^{x}f\left(t\right)dt\Rightarrow f\left(x\right)=e^x+e^x\underset{0}{\overset{1}{\int }}f\left(t\right)dt\Rightarrow f\left(x\right)=a{e}^x$
where $a=1+\underset{0}{\overset{1}{\int }}f\left(t\right)dt$
$\Rightarrow a=1+\underset{0}{\overset{1}{\int }}a{e}^{x}dt\Rightarrow a=1+a(e-1)\Rightarrow a=\frac{1}{2-e}$
Therefore, $f\left(x\right)=\frac{e^x}{2-e}$
$f\left(0\right)=\frac{1}{2-e}<0(\because e>2)$

Now, $f^{\prime }\left(x\right)=\frac{e^x}{2-e}<0(\because e^x>0)$
Therefore, $f\left(x\right)$ is a decreasing function $\forall x\in R$
$\underset{0}{\overset{1}{\int }}f\left(x\right)dx=\underset{0}{\overset{1}{\int }}\frac{e^x}{2-e}dx\Rightarrow \underset{0}{\overset{1}{\int }}f\left(x\right)dx=\frac{e-1}{2-e}<0$