Math Book Cambodia: If `g(x)=f(x)+1-x` , find the value of `g(2020)`

Thursday, July 8, 2021

It is given function $f(x)$ determine on $R$ , satisfied that:

$f(1)=1$

$f(x+5)\geqf(x)+5$

$f(x+1)\leqf(x)+1$

If $g(x)=f(x)+1-x$ , find the value of $g(2020)$ $\forallx,y\inR$

Solution

From the problem above, we had:

$f(x+5)\geqf(x)+5$

$f(x+1)\leqf(x)+1$

We will have: $f(x)+5\leqf(x+5)\leqf(x+4)+1$

Meanwhile $f(x+4)\leqf(x+3)+1$

Then, $f(x)+5\leqf(x+1)+4\leqf(x)+5$

Thus, $f(x+1)+4=f(x)+5$ or $f(x+1)-f(x)=1$

$x=1$ $f(2)-f(1)=1$

$x=2$ $f(3)-f(2)=1$

.....................................

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

Plus side to side: $f(2020)-f(1)=2019$

Then, $f(2020)=2019+f(1)$ or $f(2020)=2020$

Let's $g(x)=f(x)+1-x$

If $x=2020$ then, $g(2020)=f(2020)+1-2020=1$

Hence, $g(2020)=1$

Math Book Cambodia