2014 AMC 12A Problems/Problem 18
Contents
Problem
The domain of the function is an interval of length , where and are relatively prime positive integers. What is ?
Solution 1 (Generalization)
For all real numbers and such that and note that:
- is defined if and only if
- For we conclude that:
- if and only if
- if and only if
For we conclude that:
- if and only if
- if and only if
Therefore, we have The domain of is an interval of length from which the answer is
Remark
This problem is quite similar to 2004 AMC 12A Problem 16.
~MRENTHUSIASM
Solution 2 (Substitution)
For simplicity, let , and .
The domain of is , so . Thus, . Since we have . Since , we have . Finally, since , .
The length of the interval is and the answer is .
Solution 3 (Calculus)
The domain of is the range of the inverse function . Now can be seen to be strictly decreasing, since is decreasing, so is decreasing, so is increasing, so is increasing, therefore is decreasing.
Therefore, the range of is the open interval . We find: Similarly, Hence the range of (which is then the domain of ) is and the answer is .
See Also
2014 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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All AMC 12 Problems and Solutions |
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