Difference between revisions of "2005 AMC 12A Problems/Problem 20"
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* [[2005 AMC 12A Problems/Problem 21 | Next problem]] | * [[2005 AMC 12A Problems/Problem 21 | Next problem]] | ||
* [[2005 AMC 12A Problems]] | * [[2005 AMC 12A Problems]] | ||
| − | 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 2005 | + | 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 | ||
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Problem
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See also
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
