Difference between revisions of "2000 AMC 8 Problems/Problem 2"

Revision as of 18:32, 30 July 2011

Problem

Which of these numbers is less than its reciprocal?

$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

Solution 1

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 $-1/2$ , so it is $\boxed{A}$ .

Solution 2

The statement "a number is less than its reciprocal" can be translated as $x < \frac{1}{x}$ .

Multiplication by $x$ can be done if you do it in three parts: $x>0$ , $x=0$ , and $x<0$ . You have to be careful about the direction of the inequality, as you do not know the sign of $x$ .

If $x>0$ , the sign of the inequality remains the same. Thus, we have $x^2 < 1$ when $x > 0$ . This leads to $0 < x < 1$ .

If $x=0$ , the inequality $x < \frac{1}{x}$ is undefined.

If $x<0$ , the sign of the inequality must be switched. Thus, we have $x^2 > 1$ when $x < 0$ . This leads to $x < -1$ .

Putting the solutions together, we have $x<-1$ or $0 < x < 1$ , or in interval notation, $(-\infty, -1) \cup(0, 1)$ . The only answer in that range is $\boxed {-2, A}$

Solution 3

Starting again with $x < \frac{1}{x}$ , we avoid multiplication by $x$ . Instead, move everything to the left, and find a common denominator:

$x < \frac{1}{x}$

$x - \frac{1}{x} < 0$

$\frac{x^2 - 1}{x} < 0$

$\frac{(x+1)(x-1)}{x} < 0$

Divide this expression at $x=-1$ , $x=0$ , and $x=1$ , as those are the three points where the expression on the left will "change sign".

If $x<-1$ , all three of those terms will be negative, and the inequality is true. Therefore, $(-\infty, -1)$ is part of our solution set.

If $-1 < x < 0$ , the $(x+1)$ term will become positive, but the other two terms remain negative. Thus, there are no solutions in this region.

If $0 < x < 1$ , then both $(x+1)$ and $x$ are positive, while $(x-1)$ remains negative. Thus, the entire region $(0, -1)$ is part of the solution set.

If $1 < x$ , then all three terms are positive, and there are no solutions.

At all three "boundary points", the function is either $0$ or undefined. Therefore, the entire solution set is $(\infty, -1) \cup (0, 1)$ , and the only option in that region is $x=-2$ , leading to $\boxed{A}$ .

@@ Line 37: / Line 37: @@
 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>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.

2000 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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