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# Is |x| < 1 ?

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Is |x| < 1 ? [#permalink]

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05 Jan 2014, 21:28
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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, 03:12, edited 1 time in total.
Edited the question.
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Re: Is |x| < 1 ? [#permalink]

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06 Jan 2014, 00:51
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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]

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06 Jan 2014, 01:33
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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]

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06 Jan 2014, 03:18
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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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Absolute Value [#permalink]

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30 Aug 2016, 09:52
Is |x| < 1 ?

(1) x^4 – 1 > 0

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

Could anyone please help with the solution.
Thanks in Advance!
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Re: Is |x| < 1 ? [#permalink]

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30 Aug 2016, 10:18
GMattest2016 wrote:
Is |x| < 1 ?

(1) x^4 – 1 > 0

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

Could anyone please help with the solution.
Thanks in Advance!

Merging topics. Please refer to the discussion above.

Also, please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Thank you.
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Re: Is |x| < 1 ?   [#permalink] 30 Aug 2016, 10:18
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