Difference between revisions of "2017 AMC 12B Problems/Problem 7"
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Solution by TheUltimate123 (Eric Shen) | Solution by TheUltimate123 (Eric Shen) | ||
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| + | ==Solution II== | ||
| + | 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>!!!! | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=B|num-b=6|num-a=8}} | {{AMC12 box|year=2017|ab=B|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:05, 3 February 2018
Contents
Problem 7
The functions 
and 
are periodic with least period 
. What is the least period of the function 
?
The function is not periodic.
Solution
has values 
at its peaks and x-intercepts. Increase them to 
. Then we plug them into . 
and 
. So, is 
Solution by TheUltimate123 (Eric Shen)
Solution II
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
