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

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If x 0, is |x| <1? (1) x^2<1 (2) |x| < 1/x [#permalink]

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03 Aug 2008, 19:53
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
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Re: modulus inequality [#permalink]

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03 Aug 2008, 20:23
D

statement 1: x^2 must be positive, and for it to be less than 1, x has to be -1<x<1

statement 2: |x| is positive, and for 1/x to be bigger than |x|, x has to be 0<x<1
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Re: modulus inequality [#permalink]

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03 Aug 2008, 20:26
arjtryarjtry wrote:
. If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x

S1. This is possible only if x is a fraction; hence -1< x <1. For any of these values |x| <1 will hold. Hence BCE.

S2. If x is positive then equation becomes same as in S1. x^2<1
If x is -negative then = -x * x < 1 =>
when you multiply by a -ve teh sign changes; you get -x^2 > 1; this can be again written as x^2 < 1; which is the same as S1.

Hence ans D
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Re: modulus inequality [#permalink]

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05 Aug 2008, 06:01
answer is A in my view.

from the second statement, x |x|<1, it cannot be concluded if 0<x<1.
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Re: modulus inequality [#permalink]

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05 Aug 2008, 18:38
arjtryarjtry wrote:
. If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x

clearly IMO D
here we are concerned about value of x not the sign

(1) x^2 <1 => x<1 => |x|<1 SUFFI
(2)|x|<1/x => x<1 => |x|<1 SUFFI

IMO D
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Re: modulus inequality [#permalink]

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12 Aug 2008, 02:55
kaushikb wrote:
answer is A in my view.

from the second statement, x |x|<1, it cannot be concluded if 0<x<1.

You are mistaken. x|x|<1 is only equivalent to |x|<1/x when x>0
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Re: modulus inequality [#permalink]

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13 Aug 2008, 11:16
arjtryarjtry wrote:
. If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x

1) x^2<1 sufficient.
x*x-1<0 --> (x-1) (x+1)<0
two solutions
a) x-1<0 and x+1>0 --> x<1 and x>-1
-1<x<1
b) x-1>0 and x+1<0 --> x>1 and x<-1 Not possible..

only one solution (a)

2) |x| < 1/x

|x| < 1/x --> x<1/x and -x<1/x

x*x-1<0 and -x*x<1 --. Not possible.

x*x-1<0 --> leads to -1<x<1

Sufficient.

D

what is OA.
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Re: modulus inequality   [#permalink] 13 Aug 2008, 11:16
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# If x 0, is |x| <1? (1) x^2<1 (2) |x| < 1/x

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