Difference between revisions of "2000 AMC 8 Problems/Problem 2"
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− | (A) The number 0 has no reciprocal, and 1 and -1 are their own reciprocals. This leaves only 2 and -2. The reciprocal of 2 is 1/2, but 2 is not less than 1/2. The reciprocal of -2 is -1/2, and -2 is less than -1/2. | + | ==Problem== |
+ | |||
+ | Which of these numbers is less than its reciprocal? | ||
+ | |||
+ | <math>\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2</math> | ||
+ | |||
+ | ==Solution== | ||
+ | ===Solution 1=== | ||
+ | |||
+ | The number <math>0</math> has no reciprocal, and <math>1</math> and <math>-1</math> are their own reciprocals. This leaves only <math>2</math> and <math>-2</math>. The reciprocal of <math>2</math> is <math>1/2</math>, but <math>2</math> is not less than <math>1/2</math>. The reciprocal of <math>-2</math> is <math>-1/2</math>, and <math>-2</math> is less than<math> -1/2</math>, so it is <math>\boxed{A}</math>. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | |||
+ | The statement "a number is less than its reciprocal" can be translated as <math>x < \frac{1}{x}</math>. | ||
+ | |||
+ | Multiplication by <math>x</math> can be done if you do it in three parts: <math>x>0</math>, <math>x=0</math>, and <math>x<0</math>. You have to be careful about the direction of the inequality, as you do not know the sign of <math>x</math>. | ||
+ | |||
+ | If <math>x>0</math>, the sign of the inequality remains the same. Thus, we have <math>x^2 < 1</math> when <math>x > 0</math>. This leads to <math>0 < x < 1</math>. | ||
+ | |||
+ | If <math>x=0</math>, the inequality <math>x < \frac{1}{x}</math> is undefined. | ||
+ | |||
+ | If <math>x<0</math>, the sign of the inequality must be switched. Thus, we have <math>x^2 > 1</math> when <math>x < 0</math>. This leads to <math>x < -1</math>. | ||
+ | |||
+ | Putting the solutions together, we have <math>x<-1</math> or <math>0 < x < 1</math>, or in interval notation, <math>(-\infty, -1) \cup(0, 1)</math>. The only answer in that range is <math>\boxed {\text{(A)}\ -2}</math> | ||
+ | |||
+ | ===Solution 3=== | ||
+ | |||
+ | Starting again with <math>x < \frac{1}{x}</math>, we avoid multiplication by <math>x</math>. Instead, move everything to the left, and find a common denominator: | ||
+ | |||
+ | <math>x < \frac{1}{x}</math> | ||
+ | |||
+ | <math>x - \frac{1}{x} < 0</math> | ||
+ | |||
+ | <math>\frac{x^2 - 1}{x} < 0</math> | ||
+ | |||
+ | <math>\frac{(x+1)(x-1)}{x} < 0</math> | ||
+ | |||
+ | Divide this expression at <math>x=-1</math>, <math>x=0</math>, and <math>x=1</math>, as those are the three points where the expression on the left will "change sign". | ||
+ | |||
+ | If <math>x<-1</math>, all three of those terms will be negative, and the inequality is true. Therefore, <math>(-\infty, -1)</math> is part of our solution set. | ||
+ | |||
+ | If <math>-1 < x < 0</math>, the <math>(x+1)</math> term will become positive, but the other two terms remain negative. Thus, there are no solutions in this region. | ||
+ | |||
+ | If <math>0 < x < 1</math>, then both <math>(x+1)</math> and <math>x</math> are positive, while <math>(x-1)</math> remains negative. Thus, the entire region <math>(0, -1)</math> is part of the solution set. | ||
+ | |||
+ | 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>. | ||
+ | ==Solution 4== | ||
+ | We can find out all of their reciprocals. Now we compare and see that the answer is <math>\boxed{A}</math> | ||
+ | ==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== | ||
+ | |||
+ | {{AMC8 box|year=2000|num-b=1|num-a=3}} | ||
+ | {{MAA Notice}} |
Latest revision as of 12:36, 10 July 2019
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 | |
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 |
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