Difference between revisions of "1958 AHSME Problems/Problem 46"
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== Solution == | == Solution == | ||
− | {{ | + | 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 11:21, 23 October 2018
Problem
For values of less than but greater than , the expression has:
Solution
From , we can further factor and then and finally . Using , we can see that . From there, we can get that .
From there, we get that x is either 2 or 0. Substituting both of them in, you get that if , then the value is . If you plug in the value of , you get the value of . So the answer is \textbf{(D)}
See Also
1958 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 45 |
Followed by Problem 47 | |
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All AHSME Problems and Solutions |
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