Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with '== 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> \tex…')
 
 
(3 intermediate revisions by 3 users not shown)
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?
+
<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
 +
<math> x= \left(\frac{1}{x} \right) \cdot (-x) +2 = -1+2 = 1</math>.
 +
Therefore, <math> \boxed{\textbf{(C) }0 < x\le 2} </math>.
  
−
We can write our equation as
+
==Video Solution by Daily Dose of Math==
−
 
 
−
<math> x=(\frac{1}{x}) \times -x +2 </math>.
 
  
−
<math> \frac{1}{x} \times -x = -1 </math>.
+
https://youtu.be/6wBIYhmCfo8?si=tRO5zDw7syrXCPsS
  
−
We can substitute that product for <math> -1 </math>. Therefore,
+
~Thesmartgreekmathdude
  
−
<math> x=-1+2 \implies x=1 </math>, which is in the interval
+
== See Also ==
 +
{{AMC10 box|year=2001|num-b=1|num-a=3}}
  
−
<math> \boxed{\textbf{(C) }0 < x\le 2} </math>
+
[[Category:Introductory Algebra Problems]]
 +
{{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)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_2&oldid=222622"