Bunuel wrote:

Good question. +1.

Is x> 1?

(1) (x+1) (|x| - 1) > 0. Consider two cases:

If x>0 then |x|=x so (x+1) (|x| - 1) > 0 becomes (x+1) (x - 1) > 0 --> x^2-1>0 --> x^2>1 --> x<-1 or x>1. Since we consider range when x>0 then we have x>1 for this case;

If x\leq{0} then |x|=-x so (x+1) (|x| - 1) > 0 becomes (x+1) (-x - 1) > 0 --> -(x+1) (x+1) > 0 --> -(x+1)^2>0 --> (x+1)^2<0. Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that (x+1) (|x| - 1) > 0 holds true only when x>1. Sufficient.

(2) |x| < 5 --> -5<x<5. Not sufficient.

Answer: A.

Hope it's clear.

Thanks Bunnel.

I have a question regarding the modulus of X.

|X| = -X when X<= 0

|X| = X when X> 0

Which range does zero belong to? Since we know the result in the end solely depends on whether the absolute value of x (whether it is positive or negative) and zero is rather inconsequential in deciding whether |X| is to become -X or +X.

So, is

|X| = -X when X< 0

|X| = X when X >= 0 wrong? (if you observe I have swapped the equal to zero sign from the negative to the positive case)

So where exactly does zero fall in a modulus scenario? It is important because it decides whether the values on the verge of the range are considered in the solution set or not.

Hope the description of my question is clear enough!

Thank you in advance!

My guess is that we will need to plug in the values on the edges of the range and then decide where the zero falls (or rather which values are to be considered)