Difference between revisions of "2001 AMC 10 Problems/Problem 2"

Revision as of 12:15, 16 March 2011

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)
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 AMC 10 Problems and Solutions

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 <math> \boxed{\textbf{(C) }0 < x\le 2} </math>
+== See Also ==
+{{AMC10 box|year=2001|num-b=1|num-a=3}}