Tomi's Challenges - #15

$xF(x)-2021F(x(k+1))=0$

Let $x=2021$ .

$2021F(2021)-2021F(2021(k+1))=0$

$F(2021)-F(2021k+2021)=0$

$F(2021)=F(2021k+2021)$ .

By Rolle's theorem there exists an extreme value for some $k\neq 0$ . In order to guarantee the image to be a bounded subset of $\mathbb{R}$ we need to make it have both a supremum and an infimum, which will be guaranteed if we had infinitely many local extreme points. This wouldn't be the case if the derivative could be anything, but our absolute derivative is less than 2021, so we have a limit of how fast the function can go up or down. If $k$ can only have a finite amount of values, then the function will have that amount of extreme values, so we cannot guarantee anything. Other bounded sets only guarantee the existence of extreme values on their interval, so they are not an option. This eliminates A, C, D and F. If $k$ is a real number, $2021k$ is a real number aswell, and we have $\lim_{2021k\to0} \frac{F(2021k+2021)-F(2021)}{2021k}=0$ , which means that $F(x)$ is a constant and the cardinality of its image is $1$ . This leaves only B, which works because it's unbounded, has an infinite amount of extreme values, and it doesn't make the function a constant.