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 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
