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Revision as of 23:35, 4 July 2013

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

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 $\boxed {\text{(A)}\ -2}$

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 $\boxed{A}$.

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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