2008 IMO Problems/Problem 4
Contents
Problem
Find all functions (so is a function from the positive real numbers) such that
for all positive real numbers satisfying
Solution
Considering and which satisfy the constraint we get the following equation:
At once considering we get and knowing that the only possible solution is since is impossible.
So we get the quadratic equation:
Solving for as a function of we get:
At once we see that for one value of , can only take one of 2 possible values:
.
Take into consideration that but verifies the quadratic equation and thus so far we can't say that or alternatively . This is indeed the case but we haven't proved it yet.
To prove the previous assertion consider 2 values such that while having
Consider now the original functional equation with which verifies the constraint. Substituting we have:
Now either or . (notice that by hypothesis)
If then we have and since the only solution is .
If then we have and since the only solution is .
So the only solutions are or in which case both alternatives imply . Thus we conclude that solutions to the functional equation are a subset of .
Finally, plug each of these 2 functions into the functional equation and verify that they indeed are solutions.
This is trivial since is an obvious solution and for we have:
provided that which verifies the original constraint.
So the functional equation has 2 solutions:
Video Solution
https://youtu.be/wb2gp8uoGfM [Video Solution by little fermat]
See Also
2008 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |