Question

If $f\left(x\right)={\int }_0^{1}f\left(xt\right)dt,$ where $f\left(x\right)$ is a continuos function such that $f\left(1\right)=2$, then

• A. $f\left(x\right)$ is a periodic function
• B. $f\left(4\right)=2$
• C. $f\left(x\right)$ is an even function
• D. $f\left(4\right)=5$

Answer 1

Answer: (A), (B), (C)

$f\left(x\right)$ is an even function

Put $xt=z\Rightarrow \frac{dz}{dt}=x$
$f\left(x\right)={\int }_0^x\frac{f\left(z\right)dz}{x}\Rightarrow xf\left(x\right)={\int }_0^{x}f\left(z\right)dz$
$\Rightarrow x{f}^{\prime }\left(x\right)+f\left(x\right)=f\left(x\right)$
$\Rightarrow x{f}^{\prime }\left(x\right)=0$
$\Rightarrow f^{\prime }\left(x\right)=0\Rightarrow f\left(x\right)=c\Rightarrow f\left(x\right)=2$