Difference between revisions of "2005 AMC 12A Problems/Problem 20"
| Line 1: | Line 1: | ||
| − | |||
| − | |||
== Solution == | == Solution == | ||
{{solution}} | {{solution}} | ||
| − | + | ||
| − | |||
| − | |||
For the two functions f(x)=2x (0<=x<=1/2) and f(x)=2-2x (1/2<x<=1),we can see that as long as f(x) is between 0 and 1, x will be in the right domain. Therefore, we don't need to worry about the domain of x. Also, every time we change f(x), the final equation will be in a different form and thus we will get a different value of x. Every time we have two choices for f(x) and altogether we have to choose 2005 times. | For the two functions f(x)=2x (0<=x<=1/2) and f(x)=2-2x (1/2<x<=1),we can see that as long as f(x) is between 0 and 1, x will be in the right domain. Therefore, we don't need to worry about the domain of x. Also, every time we change f(x), the final equation will be in a different form and thus we will get a different value of x. Every time we have two choices for f(x) and altogether we have to choose 2005 times. | ||
2*2*2...2=2^2005 | 2*2*2...2=2^2005 | ||
Revision as of 23:34, 24 December 2007
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
For the two functions f(x)=2x (0<=x<=1/2) and f(x)=2-2x (1/2<x<=1),we can see that as long as f(x) is between 0 and 1, x will be in the right domain. Therefore, we don't need to worry about the domain of x. Also, every time we change f(x), the final equation will be in a different form and thus we will get a different value of x. Every time we have two choices for f(x) and altogether we have to choose 2005 times. 2*2*2...2=2^2005
