Difference between revisions of "1983 AIME Problems/Problem 2"
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<math>30-x</math> (if <math>x-p\leq15</math>) and <math>x-2p</math> (if <math>x-p\geq15</math>). However, both of these cases give us <math>\boxed{15}</math> as the minimum value for <math>f(x)</math>, which indeed is the answer posted above. | <math>30-x</math> (if <math>x-p\leq15</math>) and <math>x-2p</math> (if <math>x-p\geq15</math>). However, both of these cases give us <math>\boxed{15}</math> as the minimum value for <math>f(x)</math>, which indeed is the answer posted above. | ||
− | == See | + | == See Also == |
{{AIME box|year=1983|num-b=1|num-a=3}} | {{AIME box|year=1983|num-b=1|num-a=3}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] |
Revision as of 06:59, 16 April 2012
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 |