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PrintBelarusian Mathematical Olympiad
Belarus algebra
Problem
Find all real numbers for which there exists a function defined on the set of all real numbers which takes as its values all real numbers exactly once and satisfies the equality for all real .
Solution
Answer: .
Substituting such that , we get .
Substituting , we get .
Finally, substituting , we get .
Since takes all real values exactly once, which is equivalent to , i.e. .
Clearly, for the function satisfies the conditions of the problem.
Substituting such that , we get .
Substituting , we get .
Finally, substituting , we get .
Since takes all real values exactly once, which is equivalent to , i.e. .
Clearly, for the function satisfies the conditions of the problem.
Final answer
a = 0
Techniques
Functional EquationsInjectivity / surjectivityExistential quantifiers