Difference between revisions of "2018 AMC 12A Problems/Problem 9"
(→Solution 2) |
(→Solution 3) |
||
Line 12: | Line 12: | ||
==Solution 3== | ==Solution 3== | ||
+ | If we plug in <math>\pi</math>, we can see that <math>\sin(x+\pi) \le \sin(x)</math>. Note that since <math>\sin(x)</math> is always nonnegative, <math>\sin(x+\pi)</math> is always nonpositive. So, the inequality holds true when <math>y=\pi</math>. The only interval that contains <math>\pi</math> in the answer choices is <math>\boxed{\textbf{(E) } 0\le y\le \pi}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2018|ab=A|num-b=8|num-a=10}} | {{AMC12 box|year=2018|ab=A|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:31, 18 April 2018
Problem
Which of the following describes the largest subset of values of within the closed interval for which for every between and , inclusive?
Solution 1
On the interval sine is nonnegative; thus for all . The answer is . (CantonMathGuy)
Solution 2
Expanding, Let , . We have that Comparing coefficients of and gives a clear solution: both and are less than or equal to one, so the coefficients of and on the left are less than on the right. Since , that means that this equality is always satisfied over this interval, or .
Solution 3
If we plug in , we can see that . Note that since is always nonnegative, is always nonpositive. So, the inequality holds true when . The only interval that contains in the answer choices is .
See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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.