2001 AMC 10 Problems/Problem 2

Problem

A number $x$ is $2$ more than the product of its reciprocal and its additive inverse. In which interval does the number lie?

$\textbf{(A) }-4\le x\le -2\qquad\textbf{(B) }-2 < x\le 0\qquad\textbf{(C) }0 < x\le 2 \textbf{(D) }2 < x\le 4\qquad\textbf{(E) }4 < x\le 6$

We can write our equation as

$x=(\frac{1}{x}) \times -x +2$ .

$\frac{1}{x} \times -x = -1$ .

We can substitute that product for $-1$ . Therefore,

$x=-1+2 \implies x=1$ , which is in the interval

$\boxed{\textbf{(C) }0 < x\le 2}$

2001 AMC 10 (Problems • Answer Key • Resources)
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