Difference between revisions of "2018 AMC 12A Problems/Problem 9"
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==Solution 2== | ==Solution 2== | ||
Expanding, <cmath>\cos y \sin x + \cos x \sin y \le \sin x + \sin y</cmath> Let <math>\sin x =a \ge 0</math>, <math>\sin y = b \ge 0</math>. We have that <cmath>(\cos y)a+(\cos x)b \le a+b</cmath> Comparing coefficients of <math>a</math> and <math>b</math> gives a clear solution: both <math>\cos y</math> and <math>\cos x</math> are less than or equal to one, so the coefficients of <math>a</math> and <math>b</math> on the left are less than on the right. Since <math>a, b \ge 0</math>, that means that this equality is always satisfied over this interval, or <math>\boxed{\textbf{(E) } 0\le y\le \pi}</math>. | Expanding, <cmath>\cos y \sin x + \cos x \sin y \le \sin x + \sin y</cmath> Let <math>\sin x =a \ge 0</math>, <math>\sin y = b \ge 0</math>. We have that <cmath>(\cos y)a+(\cos x)b \le a+b</cmath> Comparing coefficients of <math>a</math> and <math>b</math> gives a clear solution: both <math>\cos y</math> and <math>\cos x</math> are less than or equal to one, so the coefficients of <math>a</math> and <math>b</math> on the left are less than on the right. Since <math>a, b \ge 0</math>, that means that this equality is always satisfied over this interval, or <math>\boxed{\textbf{(E) } 0\le y\le \pi}</math>. | ||
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+ | ==Solution 3== | ||
==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 02:24, 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
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 |
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