Difference between revisions of "1983 AIME Problems/Problem 2"
(→Solution) |
(→Solution) |
||
| Line 13: | Line 13: | ||
<math> 0 < p < 15 </math> | <math> 0 < p < 15 </math> | ||
| − | <math> p\ | + | <math> p\leq x\leq 15 </math> |
== See Also == | == See Also == | ||
Revision as of 04:53, 8 February 2013
Problem
Let 
, where 
. Determine the minimum value taken by 
for 
in the interval 
.
Solution
It is best to get rid of the absolute value first.
Under the given circumstances, we notice that 
, 
, and 
.
Adding these together, we find that the sum is equal to 
, of which the minimum value is attained when 
.
Edit: 
can equal 
or 
(for example, if 
and 
, 
). Thus, our two "cases" are
(if 
) and 
(if 
). However, both of these cases give us 
as the minimum value for , which indeed is the answer posted above.
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 | ||
