2000 AMC 8 Problems/Problem 2
Problem
Which of these numbers is less than its reciprocal?
Solution
Solution 1
The number has no reciprocal, and
and
are their own reciprocals. This leaves only
and
. The reciprocal of
is
, but
is not less than
. The reciprocal of
is
, and
is less than
, so it is
.
Solution 2
The statement "a number is less than its reciprocal" can be translated as .
Multiplication by can be done if you do it in three parts:
,
, and
. You have to be careful about the direction of the inequality, as you do not know the sign of
.
If , the sign of the inequality remains the same. Thus, we have
when
. This leads to
.
If , the inequality
is undefined.
If , the sign of the inequality must be switched. Thus, we have
when
. This leads to
.
Putting the solutions together, we have or
, or in interval notation,
. The only answer in that range is
Solution 3
Starting again with , we avoid multiplication by
. Instead, move everything to the left, and find a common denominator:
Divide this expression at ,
, and
, as those are the three points where the expression on the left will "change sign".
If , all three of those terms will be negative, and the inequality is true. Therefore,
is part of our solution set.
If , the
term will become positive, but the other two terms remain negative. Thus, there are no solutions in this region.
If , then both
and
are positive, while
remains negative. Thus, the entire region
is part of the solution set.
If , then all three terms are positive, and there are no solutions.
At all three "boundary points", the function is either or undefined. Therefore, the entire solution set is
, and the only option in that region is
, leading to
.
See Also
2000 AMC 8 (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 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.