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# Is x > 1?

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Senior Manager
Joined: 18 Aug 2009
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01 Nov 2009, 03:56
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48% (01:52) correct 52% (00:58) wrong based on 107 sessions

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Is x > 1?

(1) x^3 > x
(2) x^2 > x > 0
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 39720

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01 Nov 2009, 04:23
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KUDOS
Expert's post
Is $$x>1$$?

(1) $$x^3-x>0$$ --> $$x(x^2-1)>0$$:
Two cases:
A.$$x>0$$, $$x^2-1>0$$ --> $$x>0$$ and $$x>1$$ $$x<-1$$--> $$x>1$$
B. $$x<0$$, $$x^2-1<0$$ --> $$x<0$$ and $$-1<x<1$$ --> $$-1<x<0$$

Two ranges $$x>1$$ or $$-1<x<0$$, not sufficient.

(2) $$x>0$$, $$x^2>x$$ --> $$x^2-x>0$$--> $$x(x-1)>0$$ --> as $$x>0$$, $$x-1>0$$ --> $$x>1$$. One range $$x>1$$. Sufficient.

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01 Nov 2009, 04:38
Bunuel wrote:
Is $$x>1$$?

(1) $$x^3-x>0$$ --> $$x(x^2-1)>0$$:
Two cases:
A.$$x>0$$, $$x^2-1>0$$ --> $$x>0$$ and $$x>1$$ $$x<-1$$--> $$x>1$$
B. $$x<0$$, $$x^2-1<0$$ --> $$x<0$$ and $$-1<x<1$$ --> $$-1<x<0$$

Two ranges $$x>1$$ or $$-1<x<0$$, not sufficient.

(2) $$x>0$$, $$x^2>x$$ --> $$x^2-x>0$$--> $$x(x-1)>0$$ --> as $$x>0$$, $$x-1>0$$ --> $$x>1$$. One range $$x>1$$. Sufficient.

bingo! and very clear derivation..
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01 Nov 2009, 05:07
If we put values in option 1, we can consider it sufficient also.

x^3>x ==>> (-2)^3> -2 ==>> wrong
==>> (2)^3>2===> Correct

So x^3>x is true for all values of x greater than 1.

Where did I go wrong???
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01 Nov 2009, 05:24
Hussain15 wrote:
If we put values in option 1, we can consider it sufficient also.

x^3>x ==>> (-2)^3> -2 ==>> wrong
==>> (2)^3>2===> Correct

So x^3>x is true for all values of x greater than 1.

Where did I go wrong???

Don't forget that 0 is also a possible value. So, x^3>x is true for all values of x greater than 0.
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01 Nov 2009, 05:27
Hussain15 wrote:
If we put values in option 1, we can consider it sufficient also.

x^3>x ==>> (-2)^3> -2 ==>> wrong
==>> (2)^3>2===> Correct

So x^3>x is true for all values of x greater than 1.

Where did I go wrong???

consider the values between -1 and 0.
x^3>x ==>> (-1/2)^3> -1/2
Math Expert
Joined: 02 Sep 2009
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01 Nov 2009, 05:33
Hussain15 wrote:
If we put values in option 1, we can consider it sufficient also.

x^3>x ==>> (-2)^3> -2 ==>> wrong
==>> (2)^3>2===> Correct

So x^3>x is true for all values of x greater than 1.

Where did I go wrong???

We are not told that x is an integer, so if you consider x=-0.5: -0.5^3>-0.5 and -0.5<1. So x^3>x not only true when x>1 but also when -1<x<0.

Watch out for ZIP trap.
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01 Nov 2009, 05:36
Shelen wrote:
Hussain15 wrote:
If we put values in option 1, we can consider it sufficient also.

x^3>x ==>> (-2)^3> -2 ==>> wrong
==>> (2)^3>2===> Correct

So x^3>x is true for all values of x greater than 1.

Where did I go wrong???

Don't forget that 0 is also a possible value. So, x^3>x is true for all values of x greater than 0.

That's not correct: x^3>x ONLY in the following ranges: x>1 and -1<x<0.
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01 Nov 2009, 05:43
Thanks!!! I got it now.

Conclusion: The range of -1 to 1 is the deadly one.

-1<x<0 :::Too dangerous
0<x<1 ::: Dangerous but well known due to square powers
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01 Nov 2009, 05:58
Bunuel wrote:
Is $$x>1$$?

(1) $$x^3-x>0$$ --> $$x(x^2-1)>0$$:
Two cases:
A.$$x>0$$, $$x^2-1>0$$ --> $$x>0$$ and $$x>1$$ $$x<-1$$--> $$x>1$$
B. $$x<0$$, $$x^2-1<0$$ --> $$x<0$$ and $$-1<x<1$$ --> $$-1<x<0$$

Two ranges $$x>1$$ or $$-1<x<0$$, not sufficient.

(2) $$x>0$$, $$x^2>x$$ --> $$x^2-x>0$$--> $$x(x-1)>0$$ --> as $$x>0$$, $$x-1>0$$ --> $$x>1$$. One range $$x>1$$. Sufficient.

Hi Bunuel!

I have seen already your answers in which you mention the ranges both in inequalities & in modules questions. Can you kindly explain how does it work or refer to some good resource to understand the raging concept.
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01 Nov 2009, 06:20
Hussain15 wrote:
Bunuel wrote:
Is $$x>1$$?

(1) $$x^3-x>0$$ --> $$x(x^2-1)>0$$:
Two cases:
A.$$x>0$$, $$x^2-1>0$$ --> $$x>0$$ and $$x>1$$ $$x<-1$$--> $$x>1$$
B. $$x<0$$, $$x^2-1<0$$ --> $$x<0$$ and $$-1<x<1$$ --> $$-1<x<0$$

Two ranges $$x>1$$ or $$-1<x<0$$, not sufficient.

(2) $$x>0$$, $$x^2>x$$ --> $$x^2-x>0$$--> $$x(x-1)>0$$ --> as $$x>0$$, $$x-1>0$$ --> $$x>1$$. One range $$x>1$$. Sufficient.

Hi Bunuel!

I have seen already your answers in which you mention the ranges both in inequalities & in modules questions. Can you kindly explain how does it work or refer to some good resource to understand the raging concept.

Unfortunately I can not refer to any resources as I studied this staff in school. But I'll try to explain the basics of it if you specify the issue or post question(s).
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02 Nov 2009, 07:39
@Bunuel

x>0, x^2>x --> x^2-x>0--> x(x-1)>0 --> as x>0, x-1>0 --> x>1. One range x>1. Sufficient.

you haven't considered the second half i guess..

x(x-1)>0 --->x<0, x<1--->x<1..

if x = -2 then -2(-2-1)>0..

I think its E..correct me if am wrong..
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02 Nov 2009, 07:52
FedX wrote:
@Bunuel

x>0, x^2>x --> x^2-x>0--> x(x-1)>0 --> as x>0, x-1>0 --> x>1. One range x>1. Sufficient.

you haven't considered the second half i guess..

x(x-1)>0 --->x<0, x<1--->x<1..

if x = -2 then -2(-2-1)>0..

I think its E..correct me if am wrong..

Not so.

Statement 2 states: x^2>x>0

When we split we get x^2>x AND x>0, so there is no second part. x can not be less than zero, so the only chance x(x-1)>0 to be true is when x and x-1 is BOTH more than zero.
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02 Nov 2009, 08:07

Thanks Bunuel...
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Re: Is x > 1? [#permalink]

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03 Dec 2013, 10:10
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Re: Is x > 1? [#permalink]

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05 Apr 2014, 09:12
gmattokyo wrote:
Is x > 1?

(1) x^3 > x
(2) x^2 > x > 0

sol:

st1:

x^3 >x (subtracting same value from both sides do not change the sign)

x^3-x >0 => x(x-1) (x+1)>0

<-----(-ve)----(-1)---(+ve)----(0)---(-ve)----(1)----(+ve)------>

-1<x<0 or x>1 two sol hence not sufficient

st2:

x^2 -x>0 (also x>0)
x(x-1)>0 (also x>0)

----(+ve)---(0)----(-ve)------1-----(+ve)

x>1 or x<0 <----this can be retired since x>0 which is actual condition

therefore x>1

Re: Is x > 1?   [#permalink] 05 Apr 2014, 09:12
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