2000
Problem - 2444
Let $f(x)$ be a function defined on $\mathbb{R}$. If for every real number $x$, the relationships $$f(x+3)\le f(x)+3\quad\text{and}\quad f(x+2)\ge f(x)+2$$ always hold.
1) Show $g(x) = f(x)-x$ is a periodic function.
2) If $f(998)=1002$, compute $f(2000)$
Let's start by trying a few numbers.
\begin{align}
g(x+2) &= f(x+2) - (x+2) \\
&\ge f(x) + 2 - (x+2)\\
&= f(x) - x \\
\\
g(x+3) &= f(x+3) - (x+3) \\
&\le f(x) + 3 - (x+3) \\
&= f(x) - x \\
\\
g(x+4) &= g((x+2)+2) \\
&\ge f(x+2) - (x+2) \\
&\ge f(x) + 2 - (x+2) \\
&= f(x) - x \\
\\
g(x+6) &= g((x+2)+4) \\
&\ge f(x+2) - (x+2) \\
&\ge f(x) +2 - (x+2)\\
&= f(x) - x \\
\\
g(x+6) &= g((x+3)+3) \\
&\le f(x+3) - (x+3) \\
&\le f(x) + 3 - (x+3) \\
&= f(x) - x
\end{align}
Therefore, we have $$f(x)-x \le g(x+6) \le f(x)-x \implies g(x+6) = f(x)-x$$
By definition, $f(x)-x=g(x)$. Hence, $g(x+6)=g(x)$ which means function $g(x)$ is periodic.
It follows that
\begin{align*}
f(2000) &= g(2000) + 2000 \\
&= g(998 + 167\times 6) + 2000\\
&= g(998) + 2000\\
&= f(998) - 998 + 2000\\
&= 1002 - 998 + 2000\\
&= \boxed{2004}
\end{align*}