2000 AMC 8 Problems/Problem 2
Contents
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 .
Solution 4
We can find out all of their reciprocals. Now we compare and see that the answer is
Solution 5
Look at each number. Notice that the number must be negative. The number cannot be -1, 0, 1, ... . -2 is all that is left
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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