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If x 0, is x^2/IxI < 1? (1) x < 1 (2) x > 1

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VP
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If x 0, is x^2/IxI < 1? (1) x < 1 (2) x > 1 [#permalink] New post 19 Jan 2008, 05:57
If x ≠ 0, is x^2/IxI < 1?
(1) x < 1
(2) x > −1
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CEO
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Concentration: Entrepreneurship, Other
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Re: abs value [#permalink] New post 19 Jan 2008, 06:59
Expert's post
C

x^2/|x| < 1 \text{ }\to\text{ }|x| < 1\text{ }\to\text{ } x e (-1,1)

1. x e (-\infty,1) insuff.

2. x e (-1,\infty) insuff.

1&2. x e (-1,1) suff.
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Re: abs value [#permalink] New post 19 Jan 2008, 07:59
walker wrote:
C

x^2/|x| < 1 \text{ }\to\text{ }|x| < 1\text{ }\to\text{ } x e (-1,1)

1. x e (-\infty,1) insuff.

2. x e (-1,\infty) insuff.

1&2. x e (-1,1) suff.


how did you get rid of x^2?
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CEO
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Re: abs value [#permalink] New post 19 Jan 2008, 08:06
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\frac{x^2}{|x|}=\frac{|x|^2}{|x|}=|x|
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Re: abs value [#permalink] New post 27 Jan 2008, 15:12
from stem, can conclude that inequality only holds if -1<x<1

stat 1 says x<1. try something like x=-2 and youll see it doesnt work. insuff

stat 2 says x>-1. try something like x=2 and youll see it doesnt work, but for something like x=-1/2, it does. insuff.

together, we know -1<x<1. sufficient.
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Re: abs value [#permalink] New post 27 Jan 2008, 16:01
Also can solve by rephrasing the question...

Rephrase Is...

(x^2/x^1) < 1 which calculated to x < 1

and since it is abs value...

(x^2/-x^1) < 1 which calculated to -x < 1, rearrange to x > -1

So A gives us x < 1
and B gives us x > -1

So the answer is C
Re: abs value   [#permalink] 27 Jan 2008, 16:01
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If x 0, is x^2/IxI < 1? (1) x < 1 (2) x > 1

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