If \(g(x)=f(x)+1-x\), find the value of \(g(2020)\).

It is given that \(f(x)\) is a function defined on \(\mathbb{R}\), satisfying \(f(1)=1\), \(f(x+5)\geq f(x)+5\), and \(f(x+1)\leq f(x)+1\). If \(g(x)=f(x)+1-x\), find the value of \(g(2020)\).

Solution

From the given conditions, we have \(f(x+5)\geq f(x)+5\) and \(f(x+1)\leq f(x)+1\).

Since \(f(x+5)\leq f(x+4)+1\) and \(f(x+4)\leq f(x+3)+1\leq f(x+2)+2\leq f(x+1)+3\), we get

$$ f(x+5)\leq f(x+1)+4. $$

Therefore,

$$ f(x)+5\leq f(x+5)\leq f(x+1)+4\leq f(x)+5. $$

Hence,

$$ f(x+1)+4=f(x)+5, $$

so

$$ f(x+1)-f(x)=1. $$

For \(x=1,2,\ldots,2019\), we have

$$ f(2)-f(1)=1,\quad f(3)-f(2)=1,\quad\cdots,\quad f(2020)-f(2019)=1. $$

Adding all equations,

$$ f(2020)-f(1)=2019. $$

Since \(f(1)=1\),

$$ f(2020)=2020. $$

Now,

$$ g(x)=f(x)+1-x. $$

For \(x=2020\),

$$ g(2020)=f(2020)+1-2020=2020+1-2020=1. $$

Therefore,

$$ \boxed{g(2020)=1}. $$

Solution by: Thin Sokkean