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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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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

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.

Re: If x is an integer, is |x| > 1? [#permalink]
14 Feb 2014, 06:41

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.