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 "1958 AHSME Problems/Problem 46"

(→‎Solution)
(→‎Solution)
Line 15: Line 15:
 
From there, we get that <math>x</math> is either <math>2</math> or <math>0</math>. Substituting both of them in, you get that if <math>x=2</math>, then the value is <math>1</math>. If you plug in the value of <math>x=0</math>, you get the value of <math>-1</math>. So the answer is <math>\textbf{(D)}</math>
 
From there, we get that <math>x</math> is either <math>2</math> or <math>0</math>. Substituting both of them in, you get that if <math>x=2</math>, then the value is <math>1</math>. If you plug in the value of <math>x=0</math>, you get the value of <math>-1</math>. So the answer is <math>\textbf{(D)}</math>
  
−
Solution by: the_referee
+
Solution by: [b]the_referee[b]
  
 
== See Also ==
 
== See Also ==

Revision as of 11:23, 23 October 2018

Problem

For values of $x$ less than $1$ but greater than $-4$, the expression $\frac{x^2 - 2x + 2}{2x - 2}$ has:

$\textbf{(A)}\ \text{no maximum or minimum value}\qquad \\ \textbf{(B)}\ \text{a minimum value of }{+1}\qquad \\ \textbf{(C)}\ \text{a maximum value of }{+1}\qquad \\ \textbf{(D)}\ \text{a minimum value of }{-1}\qquad \\ \textbf{(E)}\ \text{a maximum value of }{-1}$

Solution

From $\frac{x^2 - 2x + 2}{2x - 2}$, we can further factor $\frac{x^2 - 2x + 2}{2(x - 1)}$ and then $\frac{(x-1)^{2}+1}{2(x - 1)}$ and finally $\frac{x-1}{2}+\frac{1}{2x-2}$. Using $AM-GM$, we can see that $\frac{x-1}{2}=\frac{1}{2x-2}$. From there, we can get that $2=2 \cdot (x-1)^{2}$.

From there, we get that $x$ is either $2$ or $0$. Substituting both of them in, you get that if $x=2$, then the value is $1$. If you plug in the value of $x=0$, you get the value of $-1$. So the answer is $\textbf{(D)}$

Solution by: [b]the_referee[b]

See Also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 45 		Followed by
Problem 47
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME 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=1958_AHSME_Problems/Problem_46&oldid=98242"