Question

If $f\left(x\right)=3x^2+{\int }_0^x{e}^{-t}f(x-t)dt,$ then

• A. $f\left(x\right)=0$ has $3$ solutions
• B. $f\left(x\right)$ is monotonically increasing
• C. $f\left(x\right)=4$ has $2$ solutions
• D. ${\int }_{-1}^{1}f\left(x\right)dx=2$

Answer 1

Answer: (C), (D)

The correct options are
C $f\left(x\right)=4$ has $2$ solutions
D ${\int }_{-1}^{1}f\left(x\right)dx=2$

$f\left(x\right)=3x^2+e^{-x}{\int }_0^{x}f\left(t\right).e^{t}dt\Rightarrow f\left(x\right)=3x^2+e^{-x}I\text{where}I\left(x\right)={\int }_0^{x}f\left(t\right).e^{t}dt{f}^{\prime }\left(x\right)=6x-e^{-x}I\left(x\right)+e^{-x}.I^{\prime }\left(x\right)f^{\prime }\left(x\right)=6x-e^{-x}\left(\frac{f\left(x\right)-3x^2}{e^{-x}}\right)+e^{-x}.e^{x}f\left(x\right)f^{\prime }\left(x\right)=3x^2+6xf\left(x\right)=x^3+3x^2+Cf\left(0\right)=0\Rightarrow C=0f\left(x\right)=x^3+3x^2$

$f\left(x\right)=0\Rightarrow x^3+3x^2=0$ has two solutions ($x=0\text{and}x=-3$)
$f\left(x\right)=4\Rightarrow x^3+3x^2=4$ has two solutions ($x=1\text{and}x=-2$)
$f^{\prime }\left(x\right)=3x^2+6x$, $f$ is not monotonically increasing.
${\int }_{-1}^{1}f\left(x\right)dx=2$