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

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

23 Aug 2007, 19:12

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Is x negative?

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

(2) x^2 - 1 < 0

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Re: Is x negative? - DS [#permalink]

23 Aug 2007, 19:47

above720 wrote:

Is x negative?

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

(2) x^2 - 1 < 0

from i, x^3 (1 - x)(1 + x) < 0
from ii, x^2 - 1 < 0
(x - 1) (x + 1) < 0

from i and ii: 1 + x < 0.
x < - 1

so its C.

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23 Aug 2007, 20:20

I come up with A ..

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

X(cube) - X ( power of 5) < 0

checking for values..

when X is positive - the equation is satisified

when X is negative

the equation is not satisfied....

(2)

X(square) -1 < 0

X(square) < 1

X could be a negative fraction or positive fraction

Is the answer A?

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23 Aug 2007, 20:34

Got B:
I: Ins: plug in x = -1/2 and x = 2
II: Suff: (x-1)(x+1)<0
x < 1 and x <1> Common solution: x < -1. To check - again plug in.

UPD: Sorry, I made a mistake in II: x = -1/2 and x = 1/2 also satisfy the exuation. I change my answer to E.

Last edited by Whatever on 23 Aug 2007, 23:54, edited 1 time in total.

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23 Aug 2007, 23:44

i come up with E.

All the guys above-did u try positive and negative fraction values for X.
With 1/2 and -1/2 both the eqn are satisfied so can't say for sure if it is positive or -ve.

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24 Aug 2007, 09:35

I got C. Nic to see all the options in play though :-)

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24 Aug 2007, 10:39

Is x negative ?

Statement 1

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

x^3-x^5 < 0

x^3 < x^5

can be false for x=1/2 or true for x=2
can be true for x=-1/2 or false for x=-2

insufficient

Statement 2

x^2 - 1 < 0

x^2 < 1

can be true for x=1/2 or false for x=2
can be true for x=-1/2 or false for x=-2

insufficient

Statements 1&2

x has to be negative - the only way for both formulas to work !

the answer is (C)

:-D

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25 Aug 2007, 02:12

I will go with C.

From St1:
x^3 < 0 hence x < 0 or 1-x^2 < 0 i.e. -x^2 <1> 1 i.e. x>1 or x < -1
Therefore A & D are ruled out.

From St1:
x^2 - 1 < 0 hence x^2 < 1
-1 < x < 1

if St1 and St 2 are combined, then -1 < x < 0
Hence C
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Subhen

Last edited by subhen on 25 Aug 2007, 04:21, edited 1 time in total.

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25 Aug 2007, 03:39

subhen wrote:

x^2 - 1 < 0 hence x^2 < 1
-1 < x < 1

Even if St1 and St 2 are combined, we do not know if x is +ve or -ve.

Hence E.

If X^2 < 1
combining 1 and 2.
1-x^2 > 0
so x^3 <0
=> x <0

C is SUFF

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Re: Is x negative? - DS [#permalink]

25 Aug 2007, 04:59

above720 wrote:

Is x negative?

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

(2) x^2 - 1 < 0

(1) x^3(1-x)(1+x) < 0

The above statement is false if x < -1, true if -1< x < 0 , false if 0<x<1> 1.

Thus (1) tells is that x is either a negative number greater than - 1 or a postive number greater than 1.
NOT SUFF

(2) -1 < x < 1
NOT SUFF

(T) x is either a negative number greater than - 1
SUFF

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Re: Is x negative? - DS [#permalink]

25 Aug 2007, 11:42

above720 wrote:

Is x negative?

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

(2) x^2 - 1 < 0

we need to solve inequalities system of 1 and 2 combined then it gives us the range of negative x.s so it is C

Re: Is x negative? - DS [#permalink] 25 Aug 2007, 11:42

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

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

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