1958 AHSME Problems/Problem 46
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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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