Functional Inequality

Write the inequality in the form $f(x)≥f(x+19)-19$ .

Replacing $x$ by $x-19$ , we get $f(x-19)≥f(x)-19$ .

Similarly, $f(x-94)≤f(x)-94$ .

Write the inequality in the form $f(x)≥f(x+19)-19$ .

Replacing $x$ by $x-19$ , we get $f(x-19)≥f(x)-19$ .

Similarly, $f(x-94)≤f(x)-94$ .

Using induction, one can get

a. $f(x+19n)≤f(x)+19n$ and $f(x-19n)≥f(x)-19n$ .

b. $f(x+94n)≥f(x)+94n$ and $f(x-94n)≤f(x)-94n$ where $n$ is an integer.

Now, $1=5$ x $19-94$

So, $f(x+1)=f(x-94+5$ x $19)$ $≤f(x-94)+5$ x $19$ $≤f(x)-94+5$ x $19$ $=$ $f(x)+1$ .

Similarly,using $1=18$ x $94-89$ x $19$ ,show that $f(x+1)≥f(x)+1$

Hence, $f(x+1)=f(x)+1$ .

So, $f(n)=f(0)+n$ for any natural number $n$ .

Conflict in equations leads to conclude that eqation is f(x)=x ,now its easy to give ans