Difference between revisions of "2018 AMC 12A Problems/Problem 9"
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Which of the following describes the largest subset of values of <math>y</math> within the closed interval <math>[0,\pi]</math> for which | Which of the following describes the largest subset of values of <math>y</math> within the closed interval <math>[0,\pi]</math> for which | ||
<cmath>\sin(x+y)\leq \sin(x)+\sin(y)</cmath>for every <math>x</math> between <math>0</math> and <math>\pi</math>, inclusive? | <cmath>\sin(x+y)\leq \sin(x)+\sin(y)</cmath>for every <math>x</math> between <math>0</math> and <math>\pi</math>, inclusive? | ||
− | < | + | |
+ | <math>\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 </math> | ||
== Solution 1 == | == Solution 1 == |
Revision as of 00:40, 1 June 2023
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
and equality only occurs when
, which is cosine's maximum value. 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
.
Solution 4 (Answer Choices)
Notice that for there is equality and nowhere else because the angle is not being rotated enough. We can just try all the subset of values from the answer choices to try out preferably the largest value for each subset that works as they've special trig values and are the upper bound for the set with values of the
interval.
From this we notice has the largest bound from the answer choices that works. Hence our answer is
.
~Batmanstark
Solution 5
Rewrite it as then move the terms such that it appears as this
We know that on the interval of
,
, and
, such that the whole thing on the right hand side is always greater than or equal to
. On the left hand side,
and we want
, which
satisfies.
~azure123456
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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