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Is x<0 ?

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Is x<0 ?  [#permalink]

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New post 10 Dec 2011, 17:52
3
14
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

63% (02:05) correct 38% (02:04) wrong based on 319 sessions

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Is x<0 ?

(1) x^3(1-x^2)<0
(2) x^2-1<0
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Re: OG 11  [#permalink]

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New post 11 Dec 2011, 16:32
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: OG 11  [#permalink]

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New post 12 Dec 2011, 13:26
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: OG 11  [#permalink]

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New post 15 Dec 2011, 03:29
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: OG 11  [#permalink]

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New post 15 Dec 2011, 04:25
1
Lucky2783 wrote:
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


Actually just re-write with factors:

x^3(1-x^2) < 0
So: (x^3)(1-x)(1+x) < 0

so either all three or negative or any one can be negative for the expression to be true.

If (x^3) is negative, x < 0
If (1-x) is negative, x > 1
If (1+x) is negative, x < -1

When you combine statement 1 and 2, the only overlap is x < -1 so we know for sure that x is negative

Hope this answers your question.
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Re: OG 11  [#permalink]

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New post 15 Dec 2011, 04:52
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: OG 11  [#permalink]

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New post 15 Dec 2011, 17:45
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: OG 11  [#permalink]

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New post 15 Dec 2011, 19:53
Study1 you sound promising...
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Re: OG 11  [#permalink]

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New post 06 Jul 2012, 23:43
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 ?  [#permalink]

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New post 07 Jul 2012, 03:01
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Is x<0 ?

(1) x^3(1-x^2)<0 --> \(x^3(1-x)(1+x)<0\) --> the roots are -1, 0, and 1 (equate the expressions to zero to get the roots and list them in ascending order), this gives us 4 ranges: \(x<-1\), \(-1<x<0\), \(0<x<1\), and \(x>1\).

Now, test some extreme value: for example if \(x\) is very large number then \(x\) and \(1+x\) are positive, while \(1-x\) is negative, which gives negative product for the whole expression, so when \(x>1\) the expression is negative. Now the trick: as in the 4th range expression is negative then in 3rd it'll be positive, in 2nd it'll be negative again and finally in 1st it'll be positive: + - + -. So, the ranges when the expression is negative are: \(-1<x<0\) and \(x>1\).

So, \(x\) could be negative as well as positive. Not sufficient.

Or: \(x^3(1-x^2)<0\) --> \(x(1-x^2)<0\) --> \(x<x^3\) --> \(-1<x<0\) or \(x>1\). Not sufficient.

(2) x^2-1<0 --> \(x^2<1\) --> \(-1<x<1\). Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is \(-1<x<0\), hence the answer to the question whether \(x<0\) is YES. Sufficient.

Answer: C.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486

Hope it helps.
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Re: Is x<0 ?  [#permalink]

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New post 30 Jan 2014, 06:57
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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Re: Is x<0 ?  [#permalink]

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Re: Is x<0 ?   [#permalink] 26 Sep 2018, 05:44
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