mehrasa wrote:f x is an integer, is |x|> 1?
1. (1-2x)(1+x) < 0
2. (1-x)(1+2x) < 0
Q is saying is x>1 or x<-1?
stat 1) is insufficient bcuz it gives us x>1/2 and x<-1
stat 2) is also insufficient bcuz the range is x>1 and x<-1/2
my Q is that when we want to consider both statements together, we have to find the intersection of answers from each of statements or their union?
[spoiler]OA: C[/spoiler]
Question rephrased: Is the distance between 0 and x greater than 1?
Statement 1: (1-2x)(1+x) < 0.
The critical points are x=-1 and x=1/2.
These are the only values of x that make (1-2x)(1+x) equal to 0.
When x is any other value, (1-2x)(1+x) < 0 or (1-2x)(1+x) > 0.
To determine the range of x, plug in one value to the left and right of each critical point.
x<-1:
Let x=-2.
(1- 2(-2))(1+ (-2)) < 0
-5 < 0.
This works.
x<-1 is part of the range.
-1<x<1/2:
Let x=0.
(1- 2*0)(1+0) < 0
1 < 0.
Doesn't work.
-1<x<1/2 is not part of the range.
x>1/2:
Let x=1.
(1- 2*1))(1+1) < 0
-2 < 0.
This works.
x>1/2 is part of the range.
Thus, x<-1 or x>1/2:
The distance between 0 and x can be less than 1, equal to 1, or greater than 1.
Insufficient.
Statement 2: (1-x)(1+2x) < 0
The critical points are x=-1/2 and x=1.
These are the only values of x that make (1-x)(1+2x) equal to 0.
When x is any other value, (1-x)(1+2x) < 0 or (1-2x)(1+x) > 0.
To determine the range of x, plug in one value to the left and right of each critical point.
x<-1/2:
Let x=-1.
(1- (-1))(1+ 2(-1)) < 0
-2< 0.
This works.
x<-1/2 is part of the range.
-1/2<x<1:
Let x=0.
(1-0)(1+ 2*0) < 0
1 < 0.
Doesn't work.
-1/2<x<1 is not part of the range.
x>1:
Let x=2.
(1- 2)(1+ 2*2) < 0
-5<0.
This works.
x>1 is part of the range.
Thus, x<-1/2 or x>1:
The distance between 0 and x can be less than 1, equal to 1, or greater than 1.
Insufficient.
Statements 1 and 2 combined:
Only the values represented by the GREEN lines are included in BOTH ranges:
x<-1 or x>1.
Thus, the distance between 0 and x must be greater than 1.
Sufficient.
The correct answer is
C.
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