2009 AMC 12B Problems/Problem 15
Contents
Problem
Assume . Below are five equations for . Which equation has the largest solution ?
Solution 1
(B) Intuitively, will be largest for that option for which the value in the parentheses is smallest.
Formally, first note that each of the values in parentheses is larger than . Now, each of the options is of the form . This can be rewritten as . As , we have . Thus is the largest for the option for which is smallest. And as is an increasing function, this is the option for which is smallest.
We now get the following easier problem: Given that , find the smallest value in the set .
Clearly is smaller than the first and the third option.
We have , dividing both sides by we get .
And finally, , therefore , and as both sides are positive, we can take the square root and get .
Thus the answer is .
Solution 2
As stated in the first solution, will be largest when the value in the parenthesis is smallest.
We can plug in a value for (for instance, ) and see which parenthesis has the smallest value. Doing so, we see that B is the answer.
-Made_in_2004
See Also
2009 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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