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10 Jul 2013, 00:40
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
52% (01:53) correct
48%
(02:19)
wrong
based on 88
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Given x not equal to -1, is \(\frac{1}{1+x} > 1-x + x^2\)?
(1) \(x < 0\)
(2) \(x > -1\)
(1) \(x < 0\)
(2) \(x > -1\)
10 Jul 2013, 00:50
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Given X not equal to -1
Is \(\frac{1}{1+x} > 1-x + x^2\)
I would rewrite the expression as: \(\frac{1}{1+x}-\frac{(1+x)(-1+x-x^2)}{(1+x)}>0\)
\(\frac{-x^3}{1+x}>0\)
The D is >0 if \(x>-1\).
The N is >0 if \(x<0\).
So overall the expression is positive if \(-1<x<0\)
(1) \(x < 0\)
(2) \(x > -1\)
Both are necessary C
Is \(\frac{1}{1+x} > 1-x + x^2\)
I would rewrite the expression as: \(\frac{1}{1+x}-\frac{(1+x)(-1+x-x^2)}{(1+x)}>0\)
\(\frac{-x^3}{1+x}>0\)
The D is >0 if \(x>-1\).
The N is >0 if \(x<0\).
So overall the expression is positive if \(-1<x<0\)
(1) \(x < 0\)
(2) \(x > -1\)
Both are necessary C
10 Jul 2013, 00:58
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Zarrolou
We can write another step \(\frac{X^3}{1+x} < 0\)
We also know X^2 will always be positive so the question becomes
Is \(\frac{x}{1+x} < 0\)
I usually have trouble picking numbers any tips for that? Test values in different sectors on the time line? ---- -1----0-----1------
