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10 Jun 2010, 11:28
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
56% (01:23) correct
44%
(01:39)
wrong
based on 85
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Not Attempted Yet
(x^2 + x + 1)^x > 1
1. (-infinity, 1)
2. (-1, infinity)
3. (0, infinity)
4. (1, infinity)
5. (0, infinity)
1. (-infinity, 1)
2. (-1, infinity)
3. (0, infinity)
4. (1, infinity)
5. (0, infinity)
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10 Jun 2010, 14:46
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gmatbull
Not a GMAT question.
First let's check when \((x^2 + x + 1)^x=1\):
When the power is zero: - \(x=0\);
or
When the base is 1: \(x^2 + x + 1=1\) --> \(x=0\) or \(x=-1\).
So 2 values, 3 ranges to check (checking the values of \(x^2 + x + 1\) for different values of \(x\)):
Attachment:
graph.php.png [ 9.55 KiB | Viewed 5449 times ]
\(-1<x<0\) --> \(\frac{3}{4}\leq{x^2 + x + 1}<1\) --> the positive fraction less than one in negative power more than -1 will be more than 1 (check \((\frac{3}{4})^{(-\frac{1}{2})}=\sqrt{{\frac{4}{3}}}>1\)). So this range is OK;
\(x>0\) --> \(x^2 + x + 1>{1}\) --> the number more than 1 in positive power will be more than 1. So this range is also OK.
So we got the ranges \(-1<x<0\) and \(x>0\) (remember when \(x=0\) or \(x=-1\), then \((x^2 + x + 1)^x=1\) so these values are not ok, we should exclude them).
No correct choice in answers, maybe it's meant to be 2, but the range (-1,infinity) includes 0 and it's not ok, also it's not clear whether -1 is included in the range or not.
Anyway don't worry about this question.
