gmatbull wrote:
(x^2 + x + 1)^x > 1
1. (-infinity, 1)
2. (-1, infinity)
3. (0, infinity)
4. (1, infinity)
5. (0, infinity)
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 \(x\)):
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\(x\leq{-1}\) --> \(x^2 + x + 1\geq{1}\) --> the number more than one in power less than -1 (remember power is \(x\), which is \(\leq{-1}\)) will be less than one (check \(3^{(-2)}=\frac{1}{9}<1\)). So this range is not OK;
\(-1<x<0\) --> \(\frac{3}{4}\leq{x^2 + x + 1}<1\) --> 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.
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