Difference between revisions of "2001 AMC 10 Problems/Problem 2"
(→Problem) |
|||
| Line 1: | Line 1: | ||
== 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? | + | 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 (TIP: After finding x, search in the intervals)? |
<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 | <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 | ||
Latest revision as of 11:29, 31 August 2025
Problem
A number 
is 
more than the product of its reciprocal and its additive inverse. In which interval does the number lie (TIP: After finding x, search in the intervals)?
Solution
We can write our equation as
.
Therefore, 
.
Video Solution by Daily Dose of Math
https://youtu.be/6wBIYhmCfo8?si=tRO5zDw7syrXCPsS
~Thesmartgreekmathdude
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
