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# Is x greater than x^3 ?

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Senior Manager
Joined: 25 Nov 2006
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Is x greater than x^3 ?  [#permalink]

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Updated on: 20 Apr 2014, 12:26
2
00:00

Difficulty:

75% (hard)

Question Stats:

52% (02:18) correct 48% (02:23) wrong based on 255 sessions

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Is x greater than x^3 ?

(1) x is negative

(2) x^2 - x^3 > 2

Originally posted by lumone on 09 Feb 2009, 13:43.
Last edited by Bunuel on 20 Apr 2014, 12:26, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: Is x greater than x^3 ?  [#permalink]

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19 Jul 2016, 22:44
2
1
lumone wrote:
Is x greater than x^3 ?

(1) x is negative

(2) x^2 - x^3 > 2

Is x > x^3?

Statement 1: x is negative

If -1 < x < 0, x is less than x^3.
If x < -1, x is greater than x^3.
Not sufficient alone.

Statement 2: $$x^2 - x^3 > 2$$
$$x^3 - x^2 + 2 < 0$$
$$(x + 1)(x^2 - 2x + 2) < 0$$

(x^2 - 2x + 2) has no real roots so it doesn't cut the x axis. It is positive for all real values of x.
So for (x+1)(x^2 - 2x + 2) to be negative, (x+1) must be negative.
So x + 1 < 0
x < -1
If x < -1, x is greater than x^3.

Sufficient

Note: Know the relation between x, x^2 and x^3 on the 4 sections of the number line: x > 1, 0 < x< 1, -1 < x < 0 and x < -1
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Is x greater than x^3 ?  [#permalink]

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19 Jul 2016, 13:43
1
Is x greater than $$x^3$$ ?

(1) x is negative.

(2) $$x^2- x^3 > 2$$

Bunuel, can you explain Stat 2...I did half way ..$$x^2$$ ( 1 - x ) > 2 and taken $$x^2$$ > 0 as it is square root and tried with 1-x >2...and struck here.
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09 Feb 2009, 13:57
lumone wrote:
Is x greater than x^3

(1) x is negative

(2) x^2 - x^3 > 2

1)

x =-2 x^3=-8 x>x^3
x=-1/2 x^3 = -1/8 x<x^3

not sufficient

2)

x^2 (1 - x) > 2

x^2 is always postive.|.. so (1-x) need to be positve and x can't be fractional values between -1 and 1

x must be -ve value < -1 i.e

x > x^3

sufficient.

B.
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20 Apr 2014, 10:15
x2suresh wrote:
lumone wrote:
Is x greater than x^3

(1) x is negative

(2) x^2 - x^3 > 2

1)

x =-2 x^3=-8 x>x^3
x=-1/2 x^3 = -1/8 x<x^3

not sufficient

2)

x^2 (1 - x) > 2

x^2 is always postive.|.. so (1-x) need to be positve and x can't be fractional values between -1 and 1

x must be -ve value < -1 i.e

x > x^3

sufficient.

B.

This is a really old question, and I found it recently in Jeff Sackmann's Data Sufficiency Challenge question bank. I am afraid I don't understand why Statement (2) above is sufficient, particularly the highlighted part above.

My interpretation is this: In order to keep x^2 > x^3, (1-x) > 0 (i.e. positive). This implies that 1 > x. x could therefore be negative integer, negative fraction or positive fraction. For instance:

x = -2 => x^2 = 4 and x^3 = -8 => x^2 > x^3 and 1-(-2) > 0. So, x > x^3

x = -1/2 => x^2 = 1/4 and x^3 = -1/8 => x^2 > x^3 and 1-(-1/2) > 0. However, x < x^3

x = 1/2 => x^2 = 1/4 and x^3 = 1/8 => x^2 > x^3 and 1-(1/2) > 0. So, x > x^3

Clearly, x CAN be fractional values between -1 and 1 for x^2 > x^3. But each time, the relation between x and x^3 may change. So B is not sufficient.

Can somebody please explain where I am going wrong? Or provide an alternative explanation. Thanks a lot!
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Joined: 02 Sep 2009
Posts: 59588

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21 Apr 2014, 05:23
DrFawkes wrote:
x2suresh wrote:
lumone wrote:
Is x greater than x^3

(1) x is negative

(2) x^2 - x^3 > 2

1)

x =-2 x^3=-8 x>x^3
x=-1/2 x^3 = -1/8 x<x^3

not sufficient

2)

x^2 (1 - x) > 2

x^2 is always postive.|.. so (1-x) need to be positve and x can't be fractional values between -1 and 1

x must be -ve value < -1 i.e

x > x^3

sufficient.

B.

This is a really old question, and I found it recently in Jeff Sackmann's Data Sufficiency Challenge question bank. I am afraid I don't understand why Statement (2) above is sufficient, particularly the highlighted part above.

My interpretation is this: In order to keep x^2 > x^3, (1-x) > 0 (i.e. positive). This implies that 1 > x. x could therefore be negative integer, negative fraction or positive fraction. For instance:

x = -2 => x^2 = 4 and x^3 = -8 => x^2 > x^3 and 1-(-2) > 0. So, x > x^3

x = -1/2 => x^2 = 1/4 and x^3 = -1/8 => x^2 > x^3 and 1-(-1/2) > 0. However, x < x^3

x = 1/2 => x^2 = 1/4 and x^3 = 1/8 => x^2 > x^3 and 1-(1/2) > 0. So, x > x^3

Clearly, x CAN be fractional values between -1 and 1 for x^2 > x^3. But each time, the relation between x and x^3 may change. So B is not sufficient.

Can somebody please explain where I am going wrong? Or provide an alternative explanation. Thanks a lot!

I think you misread the second statement: it's x^2 - x^3 > 2 not (2) x^2 - x^3 > 0.
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Is x greater than x^3 ?  [#permalink]

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19 Jul 2016, 09:49
Is x greater than $$x^3$$ ?

(1) x is negative.

(2) $$x^2- x^3 > 2$$

I find such questions challenging. Please explain.
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Joined: 02 Sep 2009
Posts: 59588
Re: Is x greater than x^3 ?  [#permalink]

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19 Jul 2016, 09:54
Is x greater than $$x^3$$ ?

(1) x is negative.

(2) $$x^2- x^3 > 2$$

I find such questions challenging. Please explain.

Merging topics. please refer to the discussion above.
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Re: Is x greater than x^3 ?  [#permalink]

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22 Aug 2017, 06:17
lumone wrote:
Is x greater than x^3 ?

(1) x is negative

(2) x^2 - x^3 > 2

DS : x>x^3?

Statement 1 : x<0
-1<x<0 , x^3 > x
x<-1, x<x^3

NOT SUFFICIENT

Statement 2 : x^2 - x^3 > 2
x^2> 2+x^3
For x>1, x^3 > x^2 So, x^3 +2 > x^2. This is not the case
For 0<x<1, x^2 <1 So, x^2 <2 +x^3 This is also not the case.
For -1<x<0, x^2<1 , -1<x^3<0 So, x^2 <2 +x^3 This is also not the case.
For x = -1 , x^2=2+x^3 This is also not the case.
For x<-1, x^2>2+x^3 This is the case.

Now if x<-1, Clearly x>x^3

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Re: Is x greater than x^3 ?  [#permalink]

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11 Sep 2017, 04:29
Responding to a pm:
Quote:
I didn't quite understand how you factored the cubic polynomial x^3 - x^2 + 2
Would appreciate it if you could walk me through the steps. Thanks in advance!

Here is how you will do it:
https://www.veritasprep.com/blog/2013/0 ... rd-degree/
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Re: Is x greater than x^3 ?   [#permalink] 11 Sep 2017, 04:29
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