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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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Updated on: 04 Feb 2012, 09:30
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Question Stats:

66% (01:35) correct 34% (01:48) wrong based on 182 sessions

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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

Originally posted by rxs0005 on 04 Feb 2012, 09:23.
Last edited by Bunuel on 04 Feb 2012, 09:30, edited 1 time in total.
Edited the question
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Joined: 02 Sep 2009
Posts: 59588
Re: x > x^2  [#permalink]

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04 Feb 2012, 09:29
1
2
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.

Answer: B.

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is-x-between-0-and-1-1-x-x-3-2-x-x-107401.html
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Check this for more on solving this kind of inequalities:
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xy-plane-71492.html?hilit=solving%20quadratic#p841486

Hope it helps.
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Re: x > x^2  [#permalink]

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04 Feb 2012, 09:58
1
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 3187 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 3192 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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24 Apr 2014, 22:31
1
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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Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t  [#permalink]

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22 Jun 2015, 13:53
1
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
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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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10 Jul 2017, 00:04
1
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.
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Re: x > x^2  [#permalink]

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04 Feb 2012, 09:30
thanks B

nice to have u back
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Re: x > x^2  [#permalink]

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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 ?
Math Expert
Joined: 02 Sep 2009
Posts: 59588
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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13 Mar 2016, 08:31
If the question really wants to know if x lies in the range (0,1)
hence B is sufficient
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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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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
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Joined: 28 Mar 2017
Posts: 1192
Location: India
GMAT 1: 730 Q49 V41
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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]

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12 Sep 2019, 20:32
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Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t   [#permalink] 12 Sep 2019, 20:32
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