Difference between revisions of "2001 AMC 10 Problems/Problem 2"
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== Problem == | == Problem == | ||
| + | A number <math> x </math> is <math> 2 </math> more than the product of its [[reciprocal]] and its additive [[inverse]]. In which [[interval]] does the number lie? | ||
| − | + | <math> \textbf{(A) }\ -4\le x\le -2\qquad\textbf{(B) }\ -2 < x\le 0\qquad\textbf{(C) }0</math> <math>< x \le 2 \qquad | |
| − | + | \textbf{(D) }\ 2 < x\le 4\qquad\textbf{(E) }\ 4 < x\le 6 </math> | |
| − | <math> \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 </math> | ||
== Solution == | == Solution == | ||
| − | + | We can write our [[equation]] as | |
| − | We can write our equation as | + | <math> x= \left(\frac{1}{x} \right) \cdot (-x) +2 = -1+2 = 1</math>. |
| − | + | Therefore, <math> \boxed{\textbf{(C) }0 < x\le 2} </math>. | |
| − | <math> x=(\frac{1}{x}) \ | ||
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| − | <math> \boxed{\textbf{(C) }0 < x\le 2} </math> | ||
== See Also == | == See Also == | ||
| + | {{AMC10 box|year=2001|num-b=1|num-a=3}} | ||
| − | + | [[Category:Introductory Algebra Problems]] | |
Revision as of 21:52, 25 August 2011
Problem
A number 
is 
more than the product of its reciprocal and its additive inverse. In which interval does the number lie?
Solution
We can write our equation as
.
Therefore, 
.
See Also
| 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 | ||
