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

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== Solution ==
 
== Solution ==
{{Solution}}
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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>.
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From there, we get that x is either 2 or 0. 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 \textbf{(D)}
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== See Also ==
 
== See Also ==

Revision as of 12:21, 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)}


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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