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Is |x| < 1

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Veritas Prep GMAT Instructor
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Re: Is |x| < 1 [#permalink] New post 02 Feb 2014, 20:15
Expert's post
jlgdr wrote:
Bunuel wrote:
PiyushK wrote:
For second option I tried to solve it with handy trick.
x^2(x-1)<0

critical points on number line --(+)---0--(-)--1---(+)-- as equation is < 0 trick suggest to pick range between 0-1, this trick mostly worked in all cases even in option 1. But in option 2 it is not working.

You can refer trick on link : inequalities-trick-91482.html#p804990

I am not sure where I am making mistake with this trick.


The squared term, x^2, should be omitted because it's always positive and thus doesn't affect the sign of the left side.


Ahhh now I get it, that's awesome thanks. Still I don't quite know what's wrong with including x^2 in the key points, I know it will give you a different valid range but I would like to understando the logic behind it.

Cheers
J


First understand why we have transition points at all - at the transition point, the sign of one factor changes from positive to negative (starting from right to left). That reverses the sign of the entire inequality. Therefore, we plot the number line with the transition points and make that wave to give different signs to adjoining regions.
Detailed explanation of transition points: http://www.veritasprep.com/blog/2012/06 ... e-factors/

Now what happens when you have x^2? Does x^2 change sign? No, it is ALWAYS positive (or 0). It never becomes negative and hence never reverses the sign of the inequality.
Detailed explanation of even powers: http://www.veritasprep.com/blog/2012/07 ... s-part-ii/
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Re: Is |x| < 1 [#permalink] New post 02 Feb 2014, 22:19
gmatblues wrote:
Is |x| < 1

(1) x^5 > x^3
(2) x^2 > x^3


Statement I is insufficient:

x = 2 (NO), x = -0.5 (YES)

Statement II is insufficient:

x = 0.5 (NO), x = -3 (YES)

Combining is sufficient:
X has to be a negative fraction as x cannot be a positive number or negative number less than -1.

Hence answer is C
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Re: Is |x| < 1   [#permalink] 02 Feb 2014, 22:19
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