2000 AMC 8 Problems/Problem 2
Which of these numbers is less than its reciprocal?
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 .
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
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 .
We can find out all of their reciprocals. Now we compare and see that the answer is
Look at each number. Notice that the number must be negative. The number cannot be -1, 0, 1, ... . -2 is all that is left
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