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

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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 ==

Revision as of 23:22, 13 March 2015

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

$\fbox{}$

See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 45
Followed by
Problem 47
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All AHSME Problems and Solutions

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