1958 AHSME Problems/Problem 46

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)}

1958 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 45	Followed by Problem 47
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1958 AHSME Problems/Problem 46

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