|x|>1

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|x|>1

by Sher1 » Mon Aug 31, 2009 8:13 am
If x is an integer, is |x|>1

(1 - 2x)(1 + x) < 0
(1 - x)(1 + 2x) < 0

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by srivas » Mon Aug 31, 2009 8:24 am
from stat I

(1 - 2x)(1 + x) < 0
1-2x<0 1<2x or x>1/2
1+x <0 1 <-x or x > -1 insufficient

from statII
(1 - x)(1 + 2x) < 0

1-x<0 1<x or x>1
1+2x<0 1< -2x or -1/2 > X insufficient
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Re: |x|>1

by fruti_yum » Tue Sep 01, 2009 1:10 pm
Sher1 wrote:If x is an integer, is |x|>1

(1 - 2x)(1 + x) < 0
(1 - x)(1 + 2x) < 0
I think its C..

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Re: |x|>1

by fruti_yum » Tue Sep 01, 2009 1:12 pm
fruti_yum wrote:
Sher1 wrote:If x is an integer, is |x|>1

(1 - 2x)(1 + x) < 0
(1 - x)(1 + 2x) < 0
I think its C.. When we take the two together we find that x is between -0.5 and 0.5. The question asked whether x is between -1 and 1... thus it qualifies for an answer combining the two together


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by navalpike » Thu Sep 03, 2009 1:33 pm
I'd go with B.

1.
(1-2x)(1+x)<0
2x^2+x>1
X(2x+1) > 1
X > 1 OR
(2x+1) > 1
X > 0 Insuff.

2.
(1-x) (1+2x) < 0
X(2x-1) > 1
X > 1 OR
(2x-1)> 1
x> 1.

In Either case, x is greater than 1. Which means |x|>1. Suff.