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

 
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<math> x= \left(\frac{1}{x} \right) \cdot (-x) +2 = -1+2 = 1</math>.
 
<math> x= \left(\frac{1}{x} \right) \cdot (-x) +2 = -1+2 = 1</math>.
 
Therefore, <math> \boxed{\textbf{(C) }0 < x\le 2} </math>.
 
Therefore, <math> \boxed{\textbf{(C) }0 < x\le 2} </math>.
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==Video Solution by Daily Dose of Math==
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https://youtu.be/6wBIYhmCfo8?si=tRO5zDw7syrXCPsS
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~Thesmartgreekmathdude
  
 
== See Also ==
 
== See Also ==

Latest revision as of 15:09, 15 July 2024

Problem

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

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

Solution

We can write our equation as $x= \left(\frac{1}{x} \right) \cdot (-x) +2 = -1+2 = 1$. Therefore, $\boxed{\textbf{(C) }0 < x\le 2}$.

Video Solution by Daily Dose of Math

https://youtu.be/6wBIYhmCfo8?si=tRO5zDw7syrXCPsS

~Thesmartgreekmathdude

See Also

2001 AMC 10 (ProblemsAnswer KeyResources)
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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