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Re: Is x < 0 ? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
IMO it is C.

(1) By itself is not sufficient.
(2) tells us x is a +ve or -ve fraction. Not sufficient.

(Together)
If x is +ve fraction it does not satisfy st. (1) (plug in x = 1/2 for testing)
If x is -ve fraction it satifies (1). so x is a -ve fraction and hence x < 0. SUFFICIENT.
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Re: Is x < 0 ? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
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agree C

stmt 1: x<0 or x>1
stmt 2: x>-1 x<1

Together:
x>0 and x>-1
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Re: Is x < 0 ? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
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How does 1) X^3(1-X^2)<0 led to x<0 or x>1
eg: X=-2
-2^3 (1-(-2)^2) = -8*-3 = 24
eg: X=-1 , result 0<0 , not true.
eg:X=0 result 0<0, not true
X=1 , 0<0, not true.

i feel it is only when X>1 , X^3(1-X^2)<0
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Re: Is x < 0 ? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
i think we need to consider the three cases in parallel eg: if X<0 what will happen to other two expressions

If (x^3) is negative, x < 0
say X is -0.5, X^3 is -ive , 1-X is +ive and 1+X too is positive. expression holds.
Now say X=-2, X^3 is -ive , 1-X is +ive but 1+X is -ive . expression doesnt hold .
so we cannot say that if X<0 , X^3 (1-X) (1+X) < 0

1+X will be -ive only when X <-1 , in that case 1-X will be +ive and X^3 will be -ive , over all expression will be +ive
so this set of values of X also ruled out.

now
1-X will be -ive for all values of X greater than 1 and the other two X^3 and 1+X will be positive in this case.
we can safely say that for all X>1 the expression will hold.


may be i am wrong though in my thought process , pls advise.
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Re: Is x < 0 ? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
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Is X<0 ?

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

C

From 1)

-1 < x < 0 and 1 < x

From 2)

-1 < x < 1

Combining,

-1 < x < 0
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Re: Is x < 0 ? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
Study1 you sound promising...
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Re: Is x < 0 ? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
ashiima wrote:
Is X<0 ?

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


Question asks if X is negative

Statement-1
If X is -ve: x^3 will be -ve, then (1-x^2) has to be positive, can't say
If X is +ve: x^3 will be +ve then (1-x^2) has to be negative, can't say
Insufficient

Statement-2
X^2-1<0, this is possible only when x is fraction. But fraction can be negative or positive, Insufficient

Both statements together

we get
when x is -ve, x^3 is negative, and 1-x^2 is positive ( as x is a fraction now eg. 1/2)
when x is +ve, not valid
Hence x is -ve, when both statements combined together.

Hence C

Hope I am clear.
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Re: Is x < 0 ? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
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Answer is C.

(1): x³(1-x²) < 0 --> x³ < 0 or 1-x² < 0. If x³ < 0 then x < 0, BUT x² < 1 x could be both. IS.

(2): Clearly IS. x² < 1, x could be < or > 0.

Together: From 1 we know EITHER x³ <0 OR 1-x² < 0. From 2 we know that 1-x² < 0 hence x³ is > 0, thus x is > 0.
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Is x < 0 ? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
Bunuel can you help me with what is wrong in my approach?

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

--> X^3-X^5<0
--> X^3<X^5 : Thus, X!=1,0,-1

positive integer value can have X^3<X^5 eg. 8<32
But if x<0, -8/<-32. Thus X>0 ----------- 1 Sufficient

x^2-1<0
X^2<1
|X|<1 i.e. x <1 or -x<1=> x>-1. Thus -------------2 Not Sufficient.
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Re: Is x < 0 ? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
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sidchandan wrote:
Bunuel can you help me with what is wrong in my approach?

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

--> X^3-X^5<0
--> X^3<X^5 : Thus, X!=1,0,-1

positive integer value can have X^3<X^5 eg. 8<32
But if x<0, -8/<-32. Thus X>0 ----------- 1 Sufficient

x^2-1<0
X^2<1
|X|<1 i.e. x <1 or -x<1=> x>-1. Thus -------------2 Not Sufficient.


What if -1 < x < 0 ? For example, if x = -1/2, then \((x^3 = -\frac{1}{8}) < (x = -\frac{1}{32})\). So, as you can see, from (1) x can be negative as well as positive.
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Re: Is x < 0 ? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
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