1. [20] Two linear functions $f(x)$ and $g(x)$ satisfy the properties that for all $x$,
- $f(x)+g(x)=2$
- $f(f(x))=g(g(x))$
and $f(0)=2022$. Compute $f(1)$.
Answer: $\square$
Solution 1: Firstly, $f(x)$ and $g(x)$ must intersect - otherwise, $f(x)=g(x)=1$, which can't be true. Secondly, suppose they intersect at $a$, so that $f(a)=g(a)=c$. Then $f(c)=g(c) \Longrightarrow f(c)=g(c)=1$. But then, $a=c$, and $c=1$. So $f(1)=1$, and we're done.

Solution 2: We will solve the problem manually, setting $f(x)=a x+2022$ and $g(x)=-a x-2020$. Then $f(f(x))=a^{2} x+2022 a+2022$, while $g(g(x))=a^{2} x+2020 a-2020$.
Then, $a=-2021$, so then $f(1)=1$.