Difference between revisions of "1958 AHSME Problems/Problem 46"
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From there, we get that <math>x</math> is either <math>2</math> or <math>0</math>. 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 <math>\textbf{(D)}</math> | From there, we get that <math>x</math> is either <math>2</math> or <math>0</math>. 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 <math>\textbf{(D)}</math> | ||
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+ | Solution by: the_referee | ||
== See Also == | == See Also == |
Revision as of 12:23, 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 is either or . 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
Solution by: the_referee
See Also
1958 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 45 |
Followed by Problem 47 | |
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