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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 [#permalink] New post 17 Jun 2008, 18:05
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A
B
C
D
E

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Is x negative?

1. x^3(1-x^2)<0
2. x^2-1<0
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Re: DS: Inequalities [#permalink] New post 17 Jun 2008, 18:09
vksunder wrote:
Is x negative?

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


1) equivalent to x*(1-x^2)=x*(1-x)*(1+x) < 0 -> x < -1 or x between 0 and 1 //insufficient
2) (x-1)*(x+1) < 0 -> x is between -1 and 1 -> insufficient

1&2: x is between 0 and 1 -> x negative -> sufficient -> C
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Re: DS: Inequalities [#permalink] New post 18 Jun 2008, 03:53
Answer should be E. Will explain if it's right..actually i have typed in full explain but some problem occured in n/w... :cry:
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Re: DS: Inequalities [#permalink] New post 18 Jun 2008, 04:50
vksunder wrote:
Is x negative?

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

lets see..

1)
x^3<0 or (1-x^2)<0
if X^3<0 then x is negative.. if (1-x^2)<0 then X is a positive integer..Insuff

2) x^2<1 again x=0.5 or -2 Insuff

together..

we know that x^2<1 then it means (1-x^2) >0

which leaves x^3<0...if x^3 is less then zero then X<0

C it is
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Re: DS: Inequalities [#permalink] New post 18 Jun 2008, 05:12
fresinha12 wrote:
1)
x^3<0 or (1-x^2)<0
if X^3<0 then x is negative.. if (1-x^2)<0 then X is a positive integer..Insuff


that's not correct. it's either
a) x^3 < 0 and (1-x^2) > 0 //not just x^3 < 0
or
b) x^3 > 0 and (1-x^2) < 0 //not just 1-x^2 < 0
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Re: DS: Inequalities [#permalink] New post 18 Jun 2008, 05:20
maratikus wrote:
fresinha12 wrote:
1)
x^3<0 or (1-x^2)<0
if X^3<0 then x is negative.. if (1-x^2)<0 then X is a positive integer..Insuff


that's not correct. it's either
a) x^3 < 0 and (1-x^2) > 0 //not just x^3 < 0
or
b) x^3 > 0 and (1-x^2) < 0 //not just 1-x^2 < 0


what i mean is that Either X^3 can be neagtive or (1-x^2) can be negative but not both!
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Re: DS: Inequalities [#permalink] New post 18 Jun 2008, 06:15
Is x negative?

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

1. x (1-x) (1+x) < 0 - insufficient.

2. x^2 -1 < 0 ==> 1-x^2 > 0 => (1-x)(1+x)>0

Combining 1 & 2: x<0

Guys I'm doubtful of my math skills so any response on this approach will help me get that confidence back!
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Re: DS: Inequalities [#permalink] New post 19 Jun 2008, 06:25
I will take from what was said above but come to the E conclusion:

1) equivalent to x*(1-x^2)=x*(1-x)*(1+x) < 0 -> *** Here is where I differ. This conclusion gives us -1 < x < 1 but x /= 0. So still insufficient, but it changes the final answer.
2) (x-1)*(x+1) < 0 -> x is between -1 and 1 -> insufficient

The key to this problem is seeing that 1 and 2 are essentially the same information except 1 excludes 0.

So now we combine the two. Both answers allow for a positive or negative result. Try -1/2 or +1/2. So E is the correct answer.
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Re: DS: Inequalities [#permalink] New post 19 Jun 2008, 07:11
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jmaynardj wrote:
I will take from what was said above but come to the E conclusion:

1) equivalent to x*(1-x^2)=x*(1-x)*(1+x) < 0 -> *** Here is where I differ. This conclusion gives us -1 < x < 1 but x /= 0. So still insufficient, but it changes the final answer.
2) (x-1)*(x+1) < 0 -> x is between -1 and 1 -> insufficient

The key to this problem is seeing that 1 and 2 are essentially the same information except 1 excludes 0.

So now we combine the two. Both answers allow for a positive or negative result. Try -1/2 or +1/2. So E is the correct answer.


"So now we combine the two. Both answers allow for a positive or negative result. Try -1/2 or +1/2. So E is the correct answer."

If you plug in +1/2 for x in statement (1) the inequality would be x^3 > x^5, not x^3 < x^5. Thus the answer is C
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Factorials were someone's attempt to make math look exciting!!!

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Re: DS: Inequalities [#permalink] New post 19 Jun 2008, 07:36
jmaynardj wrote:
I will take from what was said above but come to the E conclusion:

1) equivalent to x*(1-x^2)=x*(1-x)*(1+x) < 0 -> *** Here is where I differ. This conclusion gives us -1 < x < 1 but x /= 0. So still insufficient, but it changes the final answer.
2) (x-1)*(x+1) < 0 -> x is between -1 and 1 -> insufficient

The key to this problem is seeing that 1 and 2 are essentially the same information except 1 excludes 0.

So now we combine the two. Both answers allow for a positive or negative result. Try -1/2 or +1/2. So E is the correct answer.


Ok so 1) equivalent to x*(1-x^2)=x*(1-x)*(1+x) < 0 -> *** Here is where I differ. This conclusion gives us -1 < x < 1 but x /= 0. So still insufficient, but it changes the final answer.

has to be changed to -1 < x < 0 or x > 1. So by combining the two ranges from 1 & 2 you have only -1 < x < 0. Therefore C. Great problem. Hope I don't see anything quite so hard on the real test :)
Re: DS: Inequalities   [#permalink] 19 Jun 2008, 07:36
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