2004 AMC 10A Problems/Problem 15
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.
Solution 3
As , we know that the denominator of our given fraction is negative. So to achieve the greatest value possible, our numerator, or 
must also be as small as possible.
So we pick our smallest value for , which is 
.
Now if we if set our value of to its lowest, our expression becomes 
Solution 3
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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