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X / |X| < x
<=> X / |X| - x < 0
<=> x*(1-|x|) < 0 as |x| > 0 and this inequation is not using x = 0 as a possibility, which is prohibited
<=> x*(|x| - 1) > 0
<=> sign(x) = sign(|x|-1)

In other word, we have to know
o for which x > 0, |x|-1 > 0
o for which x < 0, |x|-1 < 0

o If x > 0, then
|x|-1 > 0
<=> x - 1 > 0
<=> x > 1

o If x < 0, then
|x|-1 < 0
<=> -x - 1 < 0
<=> x > -1.... So 0 > x > -1

That means :
0 > x > -1 OR x > 1.... In other word x must be superior to -1.

(A) x > 1
Assume x = 2 then we get 2 / |2| = 1 which is LESS than 2. Hence, correct.
This would be true for any value of x > 1

(B) x > -1
Assume x = 0.5 then we get 0.5 / |0.5| = 1 which is MORE than 0.5. Hence, wrong.
So all the values of x > -1 do not satisfy X / |X| <x>-1 satisfies all the values.

B is not saying everything greater than (-1) WORKS. It just states every value of x that DOES work must be > (-1). In order to kill B you must find something smaller than (-1) that works, and that seems impossible.

(-.5) seems to kill choice A, no? It's smaller than 1 and it gives us (-1) < (-.5). No problem there. So I'm going with B.

Ok, plugging any value > -1 makes it correct, it's B.

Fig or beckee529, help me understand this. Correct me what I am doing wrong below:

For +ve X,

X/X < X yields 1 <X> 1

For -ve X,

-X/-X < -X yields 1 < -X or we can say X < -1 When I say X < -1, it means X can be -2, -3, -4.........etc.

These absolute questions are tricky.

It's in your way to replace |x|/x < x when x is negative.... U have to replace only |x| by -x. The rest remains unchanged as they do not have to be positive all time

Ok, plugging any value > -1 makes it correct, it's B.

Fig or beckee529, help me understand this. Correct me what I am doing wrong below:

For +ve X,

X/X < X yields 1 <X> 1

For -ve X,

-X/-X < -X yields 1 < -X or we can say X < -1 When I say X < -1, it means X can be -2, -3, -4.........etc.

These absolute questions are tricky.

It's in your way to replace |x|/x < x when x is negative.... U have to replace only |x| by -x. The rest remains unchanged as they do not have to be positive all time

Fig, thanks for pointing out the wrong step. It makes sense now.