Difference between revisions of "1984 AIME Problems/Problem 12"

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== Problem ==
 
== Problem ==
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A function <math>\displaystyle f</math> is defined for all real numbers and satisfies <math>\displaystyle f(2+x)=f(2-x)</math> and <math>\displaystyle f(7+x)=f(7-x)</math> for all <math>\displaystyle x</math>. If <math>\displaystyle x=0</math> is a root for <math>\displaystyle f(x)=0</math>, what is the least number of roots <math>\displaystyle f(x)=0</math> must have in the interval <math>\displaystyle -1000\leq x \leq 1000</math>?
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== Solution ==
 
== Solution ==
 
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Revision as of 01:30, 21 January 2007

Problem

A function $\displaystyle f$ is defined for all real numbers and satisfies $\displaystyle f(2+x)=f(2-x)$ and $\displaystyle f(7+x)=f(7-x)$ for all $\displaystyle x$. If $\displaystyle x=0$ is a root for $\displaystyle f(x)=0$, what is the least number of roots $\displaystyle f(x)=0$ must have in the interval $\displaystyle -1000\leq x \leq 1000$?

Solution

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See also

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