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X(|x| -1 ) > 0
means x and |x| - 1 should have same sign.
if x > 0 then |x| - 1 > 0 =>|x| > 1 . Since x>0 , x > 1
if x <0 then |x| - 1 <0> -x -1 > 0 => x > -1

So x is at least not less than 1 if we have to satisfies above equations.
Suff

But could you please explain me the following:
"x > x/|x| : x e (-1,0)&(1,+∞). the expression is true for x e (-1,0) and false for x e (1,+∞). INSUFF. "

x > x/|x|: if x is -ve, then we have x > - x/x ..> x > -1, why (mathematically) do you take range (-1,0) and not (-1,+∞) excluding 0?

But could you please explain me the following: "x > x/|x| : x e (-1,0)&(1,+∞). the expression is true for x e (-1,0) and false for x e (1,+∞). INSUFF. "

x > x/|x|: if x is -ve, then we have x > - x/x ..> x > -1, why (mathematically) do you take range (-1,0) and not (-1,+∞) excluding 0?

You've answered your question
x is -ve and x > -1 give you (-1,0)
You've lost (-∞,0) condition.

Is |x| < 1?
If x > 0 then |x| = x and the question becomes is x < 1
If x < 0 then |x| = -x and the question becomes is x > -1
Combining the 2, the question becomes is -1 < x < 1?

Stat 1:
Since |x| is always +ve, we multiply each side by |x|
x|x| > x
x|x| - x > 0
x(|x| - 1) > 0
x > 0 or |x| - 1 > 0
x > 0 or |x| > 1
x > 0 or x > 1 & x < -1
Insuff.

Stat 2:
Tells us that x < 0. Insuff.

Together:
If x < 0 as per stat 2, then to be less than -1 as per stat 1. This means that x does not fall in the range of -1 < x < 1. The answer is a definite no. Suff.

Can someone confirm that above approach is correct? Thanks.

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...