Difference between revisions of "2004 AMC 10A Problems/Problem 15"
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We also know that <math>\frac{y}x<0</math> because <math>x</math> and <math>y</math> are of opposite sign. | We also know that <math>\frac{y}x<0</math> because <math>x</math> and <math>y</math> are of opposite sign. | ||
− | Therefore, <math>1+\frac{y}x</math> is maximized when <math>\frac{y}x</math> is minimized, which occurs when <math>|x|</math> is the largest and <math>|y|</math> is the smallest. | + | Therefore, <math>1+\frac{y}x</math> is maximized when <math>|\frac{y}x|</math> is minimized, which occurs when <math>|x|</math> is the largest and <math>|y|</math> is the smallest. |
− | This occurs at (-4,2), so <math>\frac{x+y}x=1-\frac12=\frac12\Rightarrow \mathrm{(D)}</math>. | + | This occurs at <math>(-4,2)</math>, so <math>\frac{x+y}x=1-\frac12=\frac12\Rightarrow\boxed{\mathrm{(D)}\ \frac{1}{2}}</math>. |
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+ | == Solution 2== | ||
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+ | If the answer choice is valid, then it must satisfy <math>\frac{(x+y)}x</math>. We use answer choices from greatest to least since the question asks for the greatest value. | ||
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+ | Answer choice <math>\text{(E)}</math>. We see that if <math>\frac{(x+y)}x = 1</math> then | ||
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+ | <math>x+y=x</math> and <math>y=0</math>. However, <math>0</math> is not in the domain of <math>y</math>, so <math>\text{(E)}</math> is incorrect. | ||
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+ | Answer choice <math>\text{(D)}</math>, however, we can find a value that satisfies <math>\frac{x+y}{x}=\frac{1}{2}</math> which simplifies to <math>x+2y=0</math>, such as <math>(-4,2)</math>. | ||
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+ | Therefore, <math>\boxed{\text{(D)}}</math> is the greatest. | ||
==See also== | ==See also== |
Revision as of 12:46, 14 December 2019
Contents
Problem
Given that and , what is the largest possible value of ?
Solution
Rewrite as .
We also know that because and are of opposite sign.
Therefore, is maximized when is minimized, which occurs when is the largest and is the smallest.
This occurs at , so .
Solution 2
If the answer choice is valid, then it must satisfy . We use answer choices from greatest to least since the question asks for the greatest value.
Answer choice . We see that if then
and . However, is not in the domain of , so is incorrect.
Answer choice , however, we can find a value that satisfies which simplifies to , such as .
Therefore, is the greatest.
See also
2004 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 10 Problems and Solutions |
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