Difference between revisions of "2017 AMC 12B Problems/Problem 7"
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Start by noting that <math>\cos(-x)=\cos(x)</math>. Then realize that under this function the negative sine values yield the same as their positive value, so take the absolute value of the sine function to get the new period. This has period <math>\pi</math>, so the answer is, surprisingly, <math>\boxed{(B)}</math>. | Start by noting that <math>\cos(-x)=\cos(x)</math>. Then realize that under this function the negative sine values yield the same as their positive value, so take the absolute value of the sine function to get the new period. This has period <math>\pi</math>, so the answer is, surprisingly, <math>\boxed{(B)}</math>. | ||
Revision as of 21:38, 18 September 2019
Problem 7
The functions and are periodic with least period . What is the least period of the function ?
The function is not periodic.
Solution
Start by noting that . Then realize that under this function the negative sine values yield the same as their positive value, so take the absolute value of the sine function to get the new period. This has period , so the answer is, surprisingly, .
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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