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If x is an integer, is |x| > 1?

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If x is an integer, is |x| > 1? [#permalink]

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If x is an integer, is |x| > 1?

(1) (1 - 2x)(1 + x) < 0
(2) (1 - x)(1 + 2x) < 0
[Reveal] Spoiler: OA
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Re: Inequality [#permalink]

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New post 06 Aug 2010, 17:57
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ssgomz wrote:
If is an integer, is |X| > 1[stopwatch]
[Reveal] Spoiler:
C
[/stopwatch]?

1.(1 - 2x)(1 + x) < 0
2.(1 - x)(1 + 2x) < 0


|x| >1 Is x< -1 or x >1 ?

1. (1-2x) (1+x) < 0 . 2 possibilities
a) 1-2x< 0 and 1+x >0 => x>1/2 and x > -1 => x >1/2
b) 1-2x> 0 and 1+x <0 => x<1/2 and x < -1 => x < -1
When x < -1, |x| >1 is true but when x>1/2 |x|>1 may or may not be true. So 1 alone is not sufficient

2. (1-x) (1+2x) < 0 . Again 2 possibilities
a) 1-x< 0 and 1+2x >0 => x>1 and x > -1/2 => x >1
b) 1-x> 0 and 1+2x <0 => x<1 and x < -1/2 => x < -1/2
When x >1, |x| >1 is true but when x<-1/2 |x|>1 may or may not be true. So 2 alone is not sufficient

Combining 1 and 2 ,

x>1/2 or x< -1 and x>1 or x<-1/2 => x<-1 or x>1
So 1 and 2 combined are sufficient. Answer C
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Re: Inequality [#permalink]

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New post 18 Aug 2010, 16:00
crack700 wrote:
ssgomz wrote:
If is an integer, is |X| > 1[stopwatch]
[Reveal] Spoiler:
C
[/stopwatch]?

1.(1 - 2x)(1 + x) < 0
2.(1 - x)(1 + 2x) < 0


|x| >1 Is x< -1 or x >1 ?

1. (1-2x) (1+x) < 0 . 2 possibilities
a) 1-2x< 0 and 1+x >0 => x>1/2 and x > -1 => x >1/2
b) 1-2x> 0 and 1+x <0 => x<1/2 and x < -1 => x < -1
When x < -1, |x| >1 is true but when x>1/2 |x|>1 may or may not be true. So 1 alone is not sufficient

2. (1-x) (1+2x) < 0 . Again 2 possibilities
a) 1-x< 0 and 1+2x >0 => x>1 and x > -1/2 => x >1
b) 1-x> 0 and 1+2x <0 => x<1 and x < -1/2 => x < -1/2
When x >1, |x| >1 is true but when x<-1/2 |x|>1 may or may not be true. So 2 alone is not sufficient

Combining 1 and 2 ,

x>1/2 or x< -1 and x>1 or x<-1/2 => x<-1 or x>1
So 1 and 2 combined are sufficient. Answer C



But x is an intiger so the ans should be B
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Re: Inequality [#permalink]

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New post 31 Oct 2010, 21:15
Bunuel, could you explain how to interpret all the diverse results of these inequalities?
Thanks!
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Re: Inequality [#permalink]

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New post 31 Oct 2010, 21:25
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1. (1 - 2x)(1 + x) < 0

Substituting both 1 and 2,satisfies the equation. Hence options A and D eliminated.

2. (1 - x)(1 + 2x) < 0

Substituting both -1 and -2,satisfies the equation. Hence option B eliminated.

1+2 , both 1 and -1 are eliminated as -1 dosen't satisfy (1) and 1 dosen't satisfy (2).

Hence x = -2,2,-3,3 and so on.

OA : C.
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Re: Inequality [#permalink]

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New post 31 Oct 2010, 21:34
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metallicafan wrote:
Bunuel, could you explain how to interpret all the diverse results of these inequalities?
Thanks!


This post might help to get the ranges for (1) and (2) - "How to solve quadratic inequalities - Graphic approach": x2-4x-94661.html#p731476

If x is an integer, is |x| > 1?

First of all: is \(|x| > 1\) means is \(x<-1\) (-2, -3, -4, ...) or \(x>1\) (2, 3, 4, ...), so for YES answer \(x\) can be any integer but -1, 0, and 1.

(1) (1 - 2x)(1 + x) < 0 --> rewrite as \((2x-1)(x+1)>0\) (so that the coefficient of x^2 to be positive after expanding): roots are \(x=-1\) and \(x=\frac{1}{2}\) --> "\(>\)" sign means that the given inequality holds true for: \(x<-1\) and \(x>\frac{1}{2}\). \(x\) could still equal to 1, so not sufficient.

(2) (1 - x)(1 + 2x) < 0 --> rewrite as \((x-1)(2x+1)>0\): roots are \(x=-\frac{1}{2}\) and \(x=1\) --> "\(>\)" sign means that the given inequality holds true for: \(x<-\frac{1}{2}\) and \(x>1\). \(x\) could still equal to -1, so not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is \(x<-1\) and \(x>1\). Sufficient.

Answer: C.
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Re: Inequality [#permalink]

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New post 31 Oct 2010, 21:36
Thanks Bunuel!

I believed that you were in a Halloween party! LOL
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Re: Inequality [#permalink]

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New post 19 Mar 2011, 23:53
Correct me if my approach is wrong:

X is an integer, and we are checking if |X|>1. It is easier to check for "NO". For us to say "No, |X| < 1", we should check for X= -1, 0 and 1.

(1) fails for X=0 and -1
(2) fails for X=0 and 1

Combining does not help either. So, "yes, |X| >1".

Does this make sense?
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Re: Inequality [#permalink]

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New post 20 Mar 2011, 02:34
Ix |x| > 1 => is x < -1 or x > 1

From 1

(1 - 2x)(1 + x) < 0

or (2x-1)(x+1) > 0

so x > 1/2 and x < -1 for this to be true.

But not sufficient.

From 2

(1 - x)(1 + 2x) < 0

(x-1)(2x+1) > 0

So X > 1 and x < -1/2 for this to be true.

Not sufficient again.

But from (1) and (2), x > 1 and x < -1,so sufficient.

Answer C.
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Re: If x is an integer, is |x| > 1? [#permalink]

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New post 14 Feb 2014, 07:41
We are asked if x>1 or x<-1.

From statement 1 we get that x>1/2 or x<-1. But note that we are told that x is an integer. Therefore x>1/2 is x>=1. But still not sufficient
From statement 2 we get that x<-1/2 and x>1. Same Here

Both together we get that

x>=1 or x<-1
x<=-1 or x>1

Both together we get that x<-1 or x>1 so our answer is YES. This is because x>1 is more restrictive than x>=1. Same for the other.

C it is

Hope its clear
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Re: If x is an integer, is |x| > 1? [#permalink]

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New post 01 Apr 2014, 12:18
ssgomz wrote:
If x is an integer, is |x| > 1?

(1) (1 - 2x)(1 + x) < 0
(2) (1 - x)(1 + 2x) < 0


Sol:

|x| > 1 i.e (+ve)-----(-1)----(-ve)------(1)--------(+ve)

since there is ">" so we choose +ve

so x<-1 or x>1 eq1

=================================

st1: (1-2x)(1+x) < 0
(2x-1)(x+1)>0 (+ve)-----(-1)-------(-ve)--------(1/2)-----(+ve)

so x<-1 or x>(1/2) after checking with eq1 it is insufficient

=======================================

St2: (1-x)(1+2x)<0
(x-1)(1+2x)>0 (+ve)----------(-1/2)-----(-ve)--------(1)-----(+ve)

so x<(-1/2) or x>1 after checking with eq1 it is still insufficient

---------------------------------------------------------------------
combining statement 1 and 2 we have

x<-1 or x>1

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Re: If x is an integer, is |x| > 1? [#permalink]

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New post 06 Jun 2014, 03:03
Arrows indicate range

Question asked:

<---------------|____________________|-------------->

----------------(-1)--------0---------------(1)----------------



From s1

<--------------------|

---------------------(-1)----------0-------1/2-----------------------


From s2

---------------------------------------------------------|------------>

--------------------------(-1/2)----0------------------ 1-------------


Combine 1+2
Answer C.
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Re: If x is an integer, is |x| > 1? [#permalink]

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Re: If x is an integer, is |x| > 1?   [#permalink] 13 Oct 2015, 02:35
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