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Difference between revisions of "1958 AHSME Problems/Problem 46"

(Created page with "== Problem == For values of <math> x</math> less than <math> 1</math> but greater than <math> \minus{}4</math>, the expression <math>\frac{x^2 \minus{} 2x \plus{} 2}{2x \minus{} ...")
 
(→‎Solution)
 
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== Problem ==
 
== Problem ==
−
For values of <math> x</math> less than <math> 1</math> but greater than <math> \minus{}4</math>, the expression
+
For values of <math> x</math> less than <math> 1</math> but greater than <math> -4</math>, the expression
−
<math>\frac{x^2 \minus{} 2x \plus{} 2}{2x \minus{} 2}</math>
+
<math>\frac{x^2 - 2x + 2}{2x - 2}</math>
 
has:
 
has:
  
 
<math> \textbf{(A)}\ \text{no maximum or minimum value}\qquad \\
 
<math> \textbf{(A)}\ \text{no maximum or minimum value}\qquad \\
−
\textbf{(B)}\ \text{a minimum value of }{\plus{}1}\qquad \\
+
\textbf{(B)}\ \text{a minimum value of }{+1}\qquad \\
−
\textbf{(C)}\ \text{a maximum value of }{\plus{}1}\qquad \\
+
\textbf{(C)}\ \text{a maximum value of }{+1}\qquad \\
−
\textbf{(D)}\ \text{a minimum value of }{\minus{}1}\qquad \\
+
\textbf{(D)}\ \text{a minimum value of }{-1}\qquad \\
−
\textbf{(E)}\ \text{a maximum value of }{\minus{}1}</math>
+
\textbf{(E)}\ \text{a maximum value of }{-1}</math>
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{}</math>
+
From <math>\frac{x^2 - 2x + 2}{2x - 2}</math>, we can further factor <math>\frac{x^2 - 2x + 2}{2(x - 1)}</math> and then <math>\frac{(x-1)^{2}+1}{2(x - 1)}</math> and finally <math>\frac{x-1}{2}+\frac{1}{2x-2}</math>. Using <math>AM-GM</math>, we can see that <math>\frac{x-1}{2}=\frac{1}{2x-2}</math>. From there, we can get that <math>2=2 \cdot (x-1)^{2}</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{(E)}</math>
 +
 
 +
Solution by: the_referee
  
 
== See Also ==
 
== See Also ==

Latest revision as of 00:35, 12 October 2021

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 $\textbf{(E)}$

Solution by: the_referee

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

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