Difference between revisions of "2018 AMC 12A Problems/Problem 9"
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<cmath>\textbf{(A) } y=0 \qquad \textbf{(B) } 0\leq y\leq \frac{\pi}{4} \qquad \textbf{(C) } 0\leq y\leq \frac{\pi}{2} \qquad \textbf{(D) } 0\leq y\leq \frac{3\pi}{4} \qquad \textbf{(E) } 0\leq y\leq \pi </cmath> | <cmath>\textbf{(A) } y=0 \qquad \textbf{(B) } 0\leq y\leq \frac{\pi}{4} \qquad \textbf{(C) } 0\leq y\leq \frac{\pi}{2} \qquad \textbf{(D) } 0\leq y\leq \frac{3\pi}{4} \qquad \textbf{(E) } 0\leq y\leq \pi </cmath> | ||
− | == Solution == | + | == Solution 1 == |
On the interval <math>[0, \pi]</math> sine is nonnegative; thus <math>\sin(x + y) = \sin x \cos y + \sin y \cos x \le \sin x + \sin y</math> for all <math>x, y \in [0, \pi]</math>. The answer is <math>\boxed{\textbf{(E) } 0\le y\le \pi}</math>. (CantonMathGuy) | On the interval <math>[0, \pi]</math> sine is nonnegative; thus <math>\sin(x + y) = \sin x \cos y + \sin y \cos x \le \sin x + \sin y</math> for all <math>x, y \in [0, \pi]</math>. The answer is <math>\boxed{\textbf{(E) } 0\le y\le \pi}</math>. (CantonMathGuy) | ||
+ | |||
+ | ==Solution 2== | ||
+ | Expanding as before, <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>. | ||
==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 13:55, 9 February 2018
Contents
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 as before, 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 .
See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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All AMC 12 Problems and Solutions |
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