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Is |x| < 1 ?

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Is |x| < 1 ? [#permalink] New post 05 Jan 2014, 20:28
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A
B
C
D
E

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Question Stats:

56% (02:18) correct 44% (01:41) wrong based on 34 sessions
Is |x| < 1 ?

(1) x^4-1> 0

(2) \frac{1}{1-|x|}> 0

How to solve this algebraically ?
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Jan 2014, 02:12, edited 1 time in total.
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Re: Is |x| < 1 ? [#permalink] New post 05 Jan 2014, 23:51
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gmatgambler wrote:
Is |x| < 1 ?

A) x^4-1> 0

B) \frac{1}{1-|x|> 0}

How to solve this algebraically ?


There are two possible cases here: Either x is between 1 and -1 in which case |X|<1 or X is less than -1 or greater than 1 in which case |X|>1

Statement I is sufficient:

If X^4>1 then we are sure that X cannot be between -1 and 1. It has to be greater than 1 or less than -1. Hence this statement is sufficient

Statement II is sufficient
If x is between 0 and 1 then (1/(1 - |X|) will be greater than zero.

If x is greater than 1 or -1 then this statement will not hold true as the denominator will become negative.

Note: the question should have a disclaimer that x is not equal to 1 or -1 as 1/(1 - 1) becomes undefined.
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Re: Is |x| < 1 ? [#permalink] New post 06 Jan 2014, 00:33
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gmatgambler wrote:
Is |x| < 1 ?

A) x^4-1> 0

B) \frac{1}{1-|x|>}0

How to solve this algebraically ?


Hi,

To solve algebrically you first must understand the question
Is |x| < 1 ?
rewriting question statement as
Is -1<x<1?


Using first statement,

x^4-1> 0

(x^2+1)(x^2-1)>0

(x^2+1)(x+1)(x-1)>0

x^2+1 will always be greater than 1.(As x can be anything but square of anything +1 will always be greater than 1.)

So there are two points +1 and -1.

(x+1)(x-1) >0

will be solved only if x <-1 and x>1

so No.
hence Sufficient.

Using statement 2:

\frac{1}{(1-|x|)}>0

1-|x| > 0 (if denominator is +ve than only this expression can be greater than zero.)
|x|<1
so -1<x<1. Yes Sufficient.

So the answer will be D.
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Re: Is |x| < 1 ? [#permalink] New post 06 Jan 2014, 02:18
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Is |x| < 1 ?

(1) x^4-1> 0 --> x^4>1 --> since both side are non-negative we can take the fourth root from both: |x|>1, Sufficient.

(2) \frac{1}{1-|x|}> 0 --> since the numerator is positive, then the fraction to be positive denominator must also be positive: 1-|x|>0 --> |x|<1. Sufficient.

Technically answer should be D, as EACH statement ALONE is sufficient to answer the question.

But even though formal answer to the question is D (EACH statement ALONE is sufficient), this is not a realistic GMAT question, as: on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other. But the statements above contradict each other:
From (1) we have that |x|>1 and from (2) we have that |x|<1. The statements clearly contradict each other.

So, the question is flawed. You won't see such a question on the test.
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Re: Is |x| < 1 ?   [#permalink] 06 Jan 2014, 02:18
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