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

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== Problem ==
 
== Problem ==
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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?
  
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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?
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<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
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\textbf{(D) }\ 2 < x\le 4\qquad\textbf{(E) }\ 4 < x\le 6 </math>
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<math> \textbf{(A) }-4\le x\le -2\qquad\textbf{(B) }-2 < x\le 0\qquad\textbf{(C) }0 < x\le 2
 
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\textbf{(D) }2 < x\le 4\qquad\textbf{(E) }4 < x\le 6 </math>
 
  
 
== Solution ==
 
== Solution ==
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We can write our [[equation]] as
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<math> x= \left(\frac{1}{x} \right) \cdot (-x) +2 = -1+2 = 1</math>.
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Therefore, <math> \boxed{\textbf{(C) }0 < x\le 2} </math>.
  
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We can write our equation as
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==Video Solution by Daily Dose of Math==
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<math> x=(\frac{1}{x}) \times -x +2 </math>.
 
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<math> \frac{1}{x} \times -x = -1 </math>.
 
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We can substitute that product for <math> -1 </math>. Therefore,
 
  
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<math> x=-1+2 \implies x=1 </math>, which is in the interval
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https://youtu.be/6wBIYhmCfo8?si=tRO5zDw7syrXCPsS
  
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<math> \boxed{\textbf{(C) }0 < x\le 2} </math>
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~Thesmartgreekmathdude
  
 
== See Also ==
 
== See Also ==
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{{AMC10 box|year=2001|num-b=1|num-a=3}}
  
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{{AMC10 box|year=2001|num-b=1|num-a=3}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

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 (Problems • Answer Key • Resources)
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Problem 1 		Followed by
Problem 3
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All AMC 10 Problems and Solutions

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