Modulus and inequality

[rahulvsd]

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

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

[spoiler]OA: C[/spoiler]

[killer1387]

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

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

[spoiler]OA: C[/spoiler]

need to find if
x>1 and x<-1

A)x<-1 and x>1/2
Insufficient

B) x<-1/2 and x>1
Insufficient

both: Sufficient

C

[aneesh.kg]

Statement 1: (1 - 2x)(1 + x) < 0
multiplying by -1 on both sides,
(2x - 1)(x + 1) > 0
By plotting critical points,
x < -1 , x > 1/2
So |x| can be < 1 or > 1.

Statement 2: (1 - x)(1 + 2x) < 0
multiplying by -1 on both sides,
(x - 1)(2x + 1) > 0
By plotting critical points,
x < -1/2, x > 1
So |x| can be < 1 or > 1.

Lets combine them.
The common solution is x > 1, x < -1.
and, |x| > 1.

The answer to the question is a conclusive YES.
(C) is the answer.

Note: Plot a number line to understand better. And, the Critical Points method is very useful in such questions.

[sam2304]

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

A)x < -1 and x > 1/2
x = -2, |x| = 2 YES
x = 1, |x| = 1 No
Insufficient

B) x < -1/2 and x > 1
x = -1, |x| = 1 NO
x = 2, |x| = 2 YES
Insufficient

combining both we need to take the common range which satisfies both the inequalities
So x > 1 and x < -1. Be it any value |x| is > 1. SUFF

IMO C