Difference between revisions of "1968 IMO Problems/Problem 5"

Latest revision as of 13:33, 29 January 2021

Problem 5

Let $f$ be a real-valued function defined for all real numbers $x$ such that, for some positive constant $a$ , the equation $f(x + a) = \frac{1}{2} + \sqrt{f(x) - (f(x))^2}$ holds for all $x$ .

(a) Prove that the function $f$ is periodic (i.e., there exists a positive number $b$ such that $f(x + b) = f(x)$ for all $x$ ).

(b) For $a = 1$ , give an example of a non-constant function with the required properties.

Solution

(a) Since $f(x+a) \ge \frac{1}{2}$ is true for any $x$ , and $f(x+a)(1-f(x+a)) = \frac{1}{4} - (f(x)-(f(x))^2) = (\frac{1}{2}-f(x))^2$

We have: $f(x+2a) = \frac{1}{2} + \sqrt{(\frac{1}{2}-f(x))^2} = \frac{1}{2} + (f(x) - \frac{1}{2}) = f(x)$ Therefore $f$ is periodic, with $2a>0$ as a period.

(b) $f(x) = 1$ when $2n\le x < 2n+1$ for some integer $n$ , and $f(x)=\frac{1}{2}$ when $2n+1\le x < 2n+2$ for some integer $n$ .

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 (b) <math>f(x) = 1</math> when <math>2n\le x < 2n+1</math> for some integer <math>n</math>, and <math>f(x)=\frac{1}{2}</math> when <math>2n+1\le x < 2n+2</math> for some integer <math>n</math>.
+== See Also == {{IMO box|year=1968|num-b=4|num-a=6}}
 [[Category:Olympiad Algebra Problems]]
 [[Category:Functional Equation Problems]]

1968 IMO (Problems) • Resources
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Difference between revisions of "1968 IMO Problems/Problem 5"

Latest revision as of 13:33, 29 January 2021

Problem 5

Solution

See Also