blaster wrote:Is X negative?

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


why 1) is not sufficient?
my approach like this,may be i'm missing something

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

let's give is x=2

2^3*-3= result is negative.

let's now give x=-2

-2^3*-3=result is positive , but statement indicate us that result must be negative.

from here we can conclude that , x can only be positive

Can anyone expain this?

x^3(1-x^2)<0
Means either x^3<0 OR 1-x^2<0
If x^3 <0 then x should be negative.
If 1-x^2<0, then x^2>1 implies x cannot lie between -1 and 1
Hence we are not sure if x is positive or negative.
Hence stmt1 is INSUFF

hope this helps!!

hi guys
by the way what is oa?
i got C, with 0<x<1 so, the answer is no
and what are your opinions?

x is -ve?

stmt1,

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

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

x^3<0,x is -ve and x<-1

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

x<-1 or x>1

x can be -ve or +ve

Insuff

stmt2,

x^2-1<0

x^2<1

x^2>=0

So x can be only 0.But o is neither +ve or -ve.

Combining 1 and 2,

x^2-1<0 or 1-x^2>0

-->x^3<0

x is -ve.[x<-1]
Suff

Pick C

hi selango
if i got you right, in the above problem you came that x<0
but if we insert x=-1 for example in the first st
x^3(1-x^2)=(-1)^3(1-(-1)^2)=-1*(1-1)=0.according to the st1 left part of the inequality must be less than 0, here it equals to 0, the result contradicts st 1
what i am missing?

(for sure it does not matter, as i also got C but with other outcome to x)

clock60 wrote:hi selango
if i got you right, in the above problem you came that x<0
but if we insert x=-1 for example in the first st
x^3(1-x^2)=(-1)^3(1-(-1)^2)=-1*(1-1)=0.according to the st1 left part of the inequality must be less than 0, here it equals to 0, the result contradicts st 1
what i am missing?

(for sure it does not matter, as i also got C but with other outcome to x)

x^3<0-->x is negative and x<-1 to satisfy the equation.

Now edited my post.

ok let it be x=-2, to satisfy the restriction x<-1
(-2)^3=-8
(1-(-2)^2)=1-4=-3
-8*-3=24, and 24>0?? again contradiction

clock60 wrote:ok let it be x=-2, to satisfy the restriction x<-1
(-2)^3=-8
(1-(-2)^2)=1-4=-3
-8*-3=24, and 24>0?? again contradiction

For Both Stmt together -
we know that -1 < x < 1 <-- from stmt 1

For stmt 2 , lets pick numbers

x = -1/2 --> (-1/2)^3 X (1 - (-1/2)^2) < 0
--> -1/8 X 3/4 < 0 <-- x is Negative

x = 0 --> 0 < 0 Which is not true so x cannot be zero

x = 1/2 --> 1/8 X (3/4) < 0 , Again this cannot be true, so x cannot be 1/2

From this we can safely conclude that x would be negative, -1 < x < 0 to be precise.
Hope this is right..

OA C