In absolute value questions, testing cases in the different points of interest can be a good way to tackle a problem.
1. |x + 1| = 2*|x - 1|
If x > 1, values inside absolute sign will be positive.
=> x + 1 = 2*(x - 1)
=> x + 1 = 2x - 2
=> x = 3
If -1 < x < 1, lhs sign is positive, rhs sign is negative
=> x + 1 = 2*(1 - x)
=> x + 1 = 2 - 2x
=> 3x = 1
=> x = 1/3
If x < -1, both lhs and rhs would be negative qtys.
=> -(x + 1) = 2(1 - x)
=> -x - 1 = 2 - 2x
=> x = 3
EDIT: Made a math error here before, corrected it. Gmatmachoman is correct, this is the same as case 1 since we are multiplying both sides by -1.
Thus |x| may be 3 or 1/3 ....INSUFFICIENT
2. |x - 3| ≠0
=> x ≠3
But x could be any value other than 3, i.e. less than 1 or greater than 1.... INSUFFICIENT
Both 1 and 2:
If x cannot be 3, then x MUST be 1/3
Thus |x| < 1
SUFFICIENT
Pick C.
Last edited by
albatross86 on Thu Jul 08, 2010 10:53 am, edited 1 time in total.
~Abhay
Believe those who are seeking the truth. Doubt those who find it. -- Andre Gide
