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# Is x negative? 1) x^3(1-x^2)<0 2) x^2-1<0

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Manager
Joined: 25 May 2009
Posts: 145
Concentration: Finance
GMAT Date: 12-16-2011
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Kudos [?]: 59 [0], given: 2

Is x negative? 1) x^3(1-x^2)<0 2) x^2-1<0 [#permalink]  21 Jun 2009, 17:59
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Is x negative?
1) $$x^3(1-x^2)<0$$
2) $$x^2-1<0$$

Please explain and help me where I went wrong in the spoiler. Thanks.

[Reveal] Spoiler:
I got E, the OA is C.
1)Rearranging the equation:$$x^3-x^5<0$$, if x=2 or x=-1/2, they both work and one is positive, one is negative, so Not Sufficient.
2)Rearranging the equation:$$(x+1)(x-1)<0$$, then x=1 or x=-1 (this may be wrong because its less than, not equal to. How are you supposed to treat these? I got Not Sufficient.
C)Using x=1 and x=-1 from (2) I plugged it into (1) and neither fit the equation, so I concluded on E.
Senior Manager
Joined: 15 Jan 2008
Posts: 295
Followers: 2

Kudos [?]: 25 [1] , given: 3

Re: Is x negative? [#permalink]  21 Jun 2009, 21:28
1
KUDOS
Choosing X = 1 or X = -1, will make the second equation equal to Zero but not less than zero.

combining both equations,
we get that X^2 is less than 1.

and hence 1 - X^2 is positive.

So make the entire equation -ve, X has to be negative.
hence C
Manager
Joined: 28 Jan 2004
Posts: 203
Location: India
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Kudos [?]: 14 [0], given: 4

Re: Is x negative? [#permalink]  21 Jun 2009, 22:48
Stmt1 -
Is not suff. as you already determined.Just to add something here i would like to say that whenever we have a x2 its almost always insuff. (because in this case +ve and -ve values yield he same result).Again this is not a rule but something to consider while solving hese kind of problems.

Stmt 2 -
Not suff.
Same reason as above. True for both X = 1/2 and X = -1/2.

Combine....here is a trick.
we can write stmt 2 as
1-sqr(X) > 0.......................stmt 2
(multiply by -1 and flip the sign,he reason i m doing this to make this identical to stmt 1)
Stmt 1 tell us that
If cube(X) is +ve then 1-sqr(x) is -ve
or
If cube(X) is -ve then 1-sqr(x) is +ve.
However stmt. 2 tell us that 1-sqr(x) is +ve, so cube(X) has to -ve and in that case X is -ve.

Let me know if it helps.
Senior Manager
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Kudos [?]: 90 [0], given: 16

Re: Is x negative? [#permalink]  23 Jun 2009, 00:47
another/better way to solve :

1.
$$x^3(1-x^2)<0$$

for above term to be -ve , either $$x^3$$ or $$(1-x^2)$$ will be negative . Both can not have the same sign.
However there is no way to find out which one is -ve and so no way to check the sign of x
INSUFFICIENT.

2.
$$x^2-1<0$$
$$x^2<1$$
x can lie in the range -1 to 1.
INSUFFICIENT.

TOGETHER:
from 1.
either $$x^3$$ or $$(1-x^2)$$ will be negative
from 2.
x lies in the range -1 to 1 so $$(1-x^2)$$ can not be -ve.

Hance $$x^3$$ is -ve and hance x is -ve.
_________________

Lahoosaher

Senior Manager
Joined: 23 Jun 2009
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Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago
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Kudos [?]: 96 [0], given: 63

Re: Is x negative? [#permalink]  23 Jun 2009, 23:10
IMO both are needed.

First condition given; we can say that; -1<x<0 or 1<x this condition is not sufficient.

In the second condition; we can say that -1<x<1 so this condition is not sufficient too.

But gathering them together, we find that -1<x<0.
Re: Is x negative?   [#permalink] 23 Jun 2009, 23:10
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