Question

A function $f\left(x\right)$ satisfies the relation $f\left(x\right)=e^x+{\int }_0^1{e}^{x}f\left(t\right)dt$Then which of the following is correct?

• A. $f\left(0\right)<0$
• B. $f\left(x\right)$ is a decreasing function
• C. $f\left(x\right)$ is an increasing function
• D. ${\int }_0^{1}f\left(x\right)dx>0$

Answer 1

Answer: (A), (B)

The correct options are
A $f\left(0\right)<0$
B $f\left(x\right)$ is a decreasing function

$f\left(x\right)=e^x+{\int }_0^1{e}^{x}f\left(t\right)dt=e^x+e^x{\int }_0^{1}f\left(t\right)dt=e^x+e^{x}II={\int }_0^{1}f\left(t\right)dt={\int }_0^{1}f\left(x\right)dx={\int }_0^1(e^x+e^{x}I)dx={[e^x+e^{x}I]}_0^1=e+eI-1-I\Rightarrow I=\frac{e-1}{2-e}$

Now,

$f\left(x\right)=e^x(1+I)=e^x(1+\frac{e-1}{2-e})=\frac{e^x}{2-e}f\left(0\right)=\frac{1}{2-e}<0I={\int }_0^{1}f\left(x\right)dx={\int }_0^1\frac{e^x}{2-e}dx={\left[\frac{e^x}{2-e}\right]}_0^1=\frac{e}{2-e}-\frac{1}{2-e}=\frac{e-1}{2-e}<0$