Difference between revisions of "2000 AMC 8 Problems/Problem 2"
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If <math>1 < x</math>, then all three terms are positive, and there are no solutions. | If <math>1 < x</math>, then all three terms are positive, and there are no solutions. | ||
− | At all three "boundary points", the function is either <math>0</math> or undefined. Therefore, the entire solution set is <math>(\infty, -1) \cup (0, 1)</math>, and the only option in that region is <math>x=-2</math>, leading to <math>\boxed{A}</math>. | + | At all three "boundary points", the function is either <math>0</math> or undefined. Therefore, the entire solution set is <math>(-\infty, -1) \cup (0, 1)</math>, and the only option in that region is <math>x=-2</math>, leading to <math>\boxed{A}</math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2000|num-b=1|num-a=3}} | {{AMC8 box|year=2000|num-b=1|num-a=3}} |
Revision as of 18:32, 30 July 2011
Problem
Which of these numbers is less than its reciprocal?
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 |