Difference between revisions of "1983 AIME Problems/Problem 2"
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It is best to get rid of the [[absolute value]]s first. | It is best to get rid of the [[absolute value]]s first. |
Latest revision as of 21:19, 20 October 2024
Contents
[hide]Problem
Let , where . Determine the minimum value taken by for in the interval .
Solution
Solution 1
It is best to get rid of the absolute values first.
Under the given circumstances, we notice that , , and .
Adding these together, we find that the sum is equal to , which attains its minimum value (on the given interval ) when , giving a minimum of .
Solution 2
Let be equal to , where is an almost neglectable value. Because of the small value , the domain of is basically the set . plugging in gives , or , so the answer is
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |