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# Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t

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Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t [#permalink]

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04 Feb 2012, 09:23
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Is x > x^2 ?

(1) x is greater than x^3

(2) x is greater than x^4

hi can you solve this Agebraically instead of picking numbers
[Reveal] Spoiler: OA

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Last edited by Bunuel on 04 Feb 2012, 09:30, edited 1 time in total.
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04 Feb 2012, 09:29
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rxs0005 wrote:
hi can you solve this Agebraically instead of picking numbers

Is x > x^2 ?

S1 x is greater than x^3

S2 x is greater than x^4

Is $$x > x^2$$ --> is $$x(x-1)<0$$? --> is $$0<x<1$$?

(1) x > x^3 --> $$x*(x^2-1)<0$$ --> $$(x+1)*x*(x-1)<0$$ --> $$0<x<1$$ or $$x<-1$$. Not sufficient.

(2) x > x^4 --> $$0<x<1$$ (only in this range x will be more than x^4). Sufficient.

Similar questions (by the way all posted by you):
is-x-0-x-x-2-x-x-3-i-did-this-algebraically-i-108110.html
is-x-between-0-and-1-1-x-x-3-2-x-x-107401.html
is-x-0-s1-x-x-2-s2-x-2-x-108207.html
is-x-x-3-s1-x-0-s2-x-2-x-108395.html

Check this for more on solving this kind of inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
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04 Feb 2012, 09:30
thanks B

nice to have u back
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04 Feb 2012, 09:39
B

what I have been realizing ( i feel its wrong now ) is that

x ( x - 1 ) < 0 Means

either x< 0 or x - 1 < 0 or x < 1

so x < 0 or x < 1 how did u get 0 < x < 1 ?
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04 Feb 2012, 09:58
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rxs0005 wrote:
B

what I have been realizing ( i feel its wrong now ) is that

x ( x - 1 ) < 0 Means

either x< 0 or x - 1 < 0 or x < 1

so x < 0 or x < 1 how did u get 0 < x < 1 ?

$$x(x-1)<0$$ --> roots are $$x=0$$ and $$x=1$$ --> "<" sign indicates that the solution lies between the roots: $$0<x<1$$:
Attachment:

1.gif [ 891 Bytes | Viewed 2491 times ]

If it were: $$x(x-1)>0$$ --> the same roots: $$x=0$$ and $$x=1$$ --> ">" sign indicates that the solution lies to the left of the smaller root and to the right of the larger root: x<0 or x>1:
Attachment:

2.gif [ 997 Bytes | Viewed 2494 times ]

Follow the links in my previous post for more on this subject.

Hope it helps.
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Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t [#permalink]

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23 Feb 2014, 05:57
Bumping for review and further discussion.
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Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t [#permalink]

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24 Apr 2014, 06:54
Good that you bumped onto this thread Bunuel

Here's my question

On second statement how do we do critical points?

So we have x>x^4

Therefore x^4-x<0

Now then factorizing we have: x^2 (x-1)(x+1) , we only need to take care of (x-1)(x+1) since x^2 is always >=0

But then I get the range -1<x<1 which is not correct given that for instance when x=0 clearly inequality does not hold

Would you mind showing me where I went wrong here?

Thanks!
Cheers
J

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Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t [#permalink]

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24 Apr 2014, 22:31
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jlgdr wrote:
Good that you bumped onto this thread Bunuel

Here's my question

On second statement how do we do critical points?

So we have x>x^4

Therefore x^4-x<0

Now then factorizing we have: x^2 (x-1)(x+1) , we only need to take care of (x-1)(x+1) since x^2 is always >=0

But then I get the range -1<x<1 which is not correct given that for instance when x=0 clearly inequality does not hold

Would you mind showing me where I went wrong here?

Thanks!
Cheers
J

Note that the inequality is
$$x^4 - x < 0$$

So on factoring you get
$$x (x^3 - 1) < 0$$
$$x(x - 1)(x^2 + x + 1) < 0$$

Now, (x^2 + x + 1) is always positive. It has no real roots. Hence we might as well assume it to be a positive number and ignore it for the sake of getting our transition points.
Transition points are 0 and 1. Since we want the negative region, the inequality will be negative between 0 and 1.

0 < x < 1

Though, you should try to understand the relation between x, x^2 and x^3 in the ranges < -1, -1 < x < 0, 0 < x <1 and x > 1 so that you don't need to use much algebra for such questions.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17845 [1], given: 235 Non-Human User Joined: 09 Sep 2013 Posts: 15565 Kudos [?]: 283 [0], given: 0 Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t [#permalink] ### Show Tags 21 Jun 2015, 02:31 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Kudos [?]: 283 [0], given: 0 SVP Joined: 06 Nov 2014 Posts: 1905 Kudos [?]: 542 [1], given: 23 Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t [#permalink] ### Show Tags 22 Jun 2015, 13:53 1 This post received KUDOS Is x > x^2 ? This would hold true if x<1 (1) x is greater than x^3 x can be a negative integer. For example x = -2. -2^3 = -8. But -2 is not greater than -2^2 = 4. Not sufficient. (2) x is greater than x^4 For this to hold true, x would have to be less than 1. sufficient Kudos [?]: 542 [1], given: 23 Retired Moderator Joined: 12 Aug 2015 Posts: 2213 Kudos [?]: 876 [0], given: 602 Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t [#permalink] ### Show Tags 13 Mar 2016, 08:31 If the question really wants to know if x lies in the range (0,1) hence B is sufficient _________________ Give me a hell yeah ...!!!!! Kudos [?]: 876 [0], given: 602 Director Joined: 28 Mar 2017 Posts: 584 Kudos [?]: 145 [0], given: 132 Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t [#permalink] ### Show Tags 08 Jul 2017, 03:45 VeritasPrepKarishma wrote: jlgdr wrote: Good that you bumped onto this thread Bunuel Here's my question On second statement how do we do critical points? So we have x>x^4 Therefore x^4-x<0 Now then factorizing we have: x^2 (x-1)(x+1) , we only need to take care of (x-1)(x+1) since x^2 is always >=0 But then I get the range -1<x<1 which is not correct given that for instance when x=0 clearly inequality does not hold Would you mind showing me where I went wrong here? Thanks! Cheers J Note that the inequality is $$x^4 - x < 0$$ So on factoring you get $$x (x^3 - 1) < 0$$ $$x(x - 1)(x^2 + x + 1) < 0$$ Now, (x^2 + x + 1) is always positive. It has no real roots. Hence we might as well assume it to be a positive number and ignore it for the sake of getting our transition points. Transition points are 0 and 1. Since we want the negative region, the inequality will be negative between 0 and 1. 0 < x < 1 Though, you should try to understand the relation between x, x^2 and x^3 in the ranges < -1, -1 < x < 0, 0 < x <1 and x > 1 so that you don't need to use much algebra for such questions. Hello VeritasPrepKarishma, Its been a while since i studies all such formulaes like (a+1)^3 or a^3-1 or (a+b+c)^2 etc, so should I revisit each and every of these formulaes? Coming back to the question, Can I say that since x>x^4 and since x^4>0 always therefore x MUST be >0 Thus, x>0 and x^4-x<0 => x(x^3-1)<0 => x^3-1<0 => x^3<1 Now, since x >0 then x^3 will be less than 1 only when 0<x<1; since for any real number greater than 1, the cube is greater than the number itself. Thus i can say x<0<1. Hence sufficient. Is my reasoning correct? Regards _________________ Kudos if my post helps! Helpful links: 1. e-GMAT's ALL SC Compilation Kudos [?]: 145 [0], given: 132 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7745 Kudos [?]: 17845 [1], given: 235 Location: Pune, India Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t [#permalink] ### Show Tags 10 Jul 2017, 00:04 1 This post received KUDOS Expert's post gmatexam439 wrote: VeritasPrepKarishma wrote: jlgdr wrote: Good that you bumped onto this thread Bunuel Here's my question On second statement how do we do critical points? So we have x>x^4 Therefore x^4-x<0 Now then factorizing we have: x^2 (x-1)(x+1) , we only need to take care of (x-1)(x+1) since x^2 is always >=0 But then I get the range -1<x<1 which is not correct given that for instance when x=0 clearly inequality does not hold Would you mind showing me where I went wrong here? Thanks! Cheers J Note that the inequality is $$x^4 - x < 0$$ So on factoring you get $$x (x^3 - 1) < 0$$ $$x(x - 1)(x^2 + x + 1) < 0$$ Now, (x^2 + x + 1) is always positive. It has no real roots. Hence we might as well assume it to be a positive number and ignore it for the sake of getting our transition points. Transition points are 0 and 1. Since we want the negative region, the inequality will be negative between 0 and 1. 0 < x < 1 Though, you should try to understand the relation between x, x^2 and x^3 in the ranges < -1, -1 < x < 0, 0 < x <1 and x > 1 so that you don't need to use much algebra for such questions. Hello VeritasPrepKarishma, Its been a while since i studies all such formulaes like (a+1)^3 or a^3-1 or (a+b+c)^2 etc, so should I revisit each and every of these formulaes? Coming back to the question, Can I say that since x>x^4 and since x^4>0 always therefore x MUST be >0 Thus, x>0 and x^4-x<0 => x(x^3-1)<0 => x^3-1<0 => x^3<1 Now, since x >0 then x^3 will be less than 1 only when 0<x<1; since for any real number greater than 1, the cube is greater than the number itself. Thus i can say x<0<1. Hence sufficient. Is my reasoning correct? Regards Some important algebraic identities: $$(x + y)^2 = x^2 + y^2 + 2xy$$ $$(x - y)^2 = x^2 + y^2 - 2xy$$ $$x^2 - y^2 = (x + y)*(x - y)$$ Others, if needed, you can calculate there and then. $$(x + y)^3 = (x + y)(x + y)(x + y) = (x^2 + y^2 + 2xy)(x + y) = x^3 + xy^2 + 2x^2y ...$$ As for this question, yes you can arrive at 0 < x< 1 through that method too. No problem in the reasoning though I would suggest you to look at the other methods discussed above too. They are more generic. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t [#permalink]

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10 Jul 2017, 07:09
VeritasPrepKarishma wrote:
Some important algebraic identities:

$$(x + y)^2 = x^2 + y^2 + 2xy$$

$$(x - y)^2 = x^2 + y^2 - 2xy$$

$$x^2 - y^2 = (x + y)*(x - y)$$

Others, if needed, you can calculate there and then.

$$(x + y)^3 = (x + y)(x + y)(x + y) = (x^2 + y^2 + 2xy)(x + y) = x^3 + xy^2 + 2x^2y ...$$

As for this question, yes you can arrive at 0 < x< 1 through that method too. No problem in the reasoning though I would suggest you to look at the other methods discussed above too. They are more generic.

Thank you Karishma You guyz are doing a very good job. Will look at the other methods described above as well.
Regards
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Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t   [#permalink] 10 Jul 2017, 07:09
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