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# If x≠0, is |x| <1? (1) x^2<1 (2) |x| < 1/x

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Intern
Joined: 08 Jan 2006
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If x≠0, is |x| <1? (1) x^2<1 (2) |x| < 1/x [#permalink]

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03 Jul 2006, 12:01
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x

From 1) -1<x<1. Therefore, |x| is <1.

How does one solve 2)?

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Senior Manager
Joined: 09 Aug 2005
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03 Jul 2006, 13:15
geemaat wrote:
If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x

From 1) -1<x<1. Therefore, |x| is <1.

How does one solve 2)?

From 2 ---- 1/x is +ve also x=! 0

there fore 1/x got to be greater than one which means x is less than 1 therefore |x|<1

D

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CEO
Joined: 20 Nov 2005
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008

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03 Jul 2006, 16:40
geemaat wrote:
If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x

From 1) -1<x<1. Therefore, |x| is <1.

How does one solve 2)?

D

St1: Its easy. -1<x<1 : SUFF

St2: I just plugged in the number and found 0<x<1 : SUFF
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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VP
Joined: 25 Nov 2004
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03 Jul 2006, 17:50
geemaat wrote:
If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x

From 1) -1<x<1. Therefore, |x| is <1.

How does one solve 2)?

from 1, we know x<1.
From 2) 1/x must be +ve since lxl is positive. only values for x work are 1>x>0.

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Director
Joined: 07 Jun 2006
Posts: 513

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03 Jul 2006, 17:59
geemaat wrote:
If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x

From 1) -1<x<1. Therefore, |x| is <1.

How does one solve 2)?

(1) x^2 < 1 => x < 1 so OK
(2) |x| < 1/x => x < 1/x or x < -1/x, then,
x < 1/x is possible only if x < 1, but it can be any number 0<x<1 and x < -1, so it does not tell us if |x| is < 1,

if we take x < -1/x, is possible if x <= -1 ... can't say.

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Director
Joined: 07 Jun 2006
Posts: 513

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03 Jul 2006, 18:07
aargh!! I always get these !@***#@\$\$@#!# inequalities wrong. Anyone got a good suggestion where I can firm up my the fundamentals?

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SVP
Joined: 30 Mar 2006
Posts: 1728

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05 Jul 2006, 03:32
D

1) suff. -1<x<1

2) plug in values and you will find that 0<x<1

hence D

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Senior Manager
Joined: 07 Jul 2005
Posts: 399

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Location: Sunnyvale, CA

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05 Jul 2006, 15:19
one more for (D)..
as explained by others.

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Intern
Joined: 08 Jan 2006
Posts: 27

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14 Jul 2006, 09:07
old_dream_1976 wrote:
geemaat wrote:
If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x

From 1) -1<x<1. Therefore, |x| is <1.

How does one solve 2)?

From 2 ---- 1/x is +ve also x=! 0

there fore 1/x got to be greater than one which means x is less than 1 therefore |x|<1

D

Thanks guys. The OA is D.

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Re: solving absolute inequality   [#permalink] 14 Jul 2006, 09:07
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# If x≠0, is |x| <1? (1) x^2<1 (2) |x| < 1/x

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