got C here
(1) 1 st is valid for x=2, but it is not with in -1<x<1
or x=-1/2, and -1/2 is with in -1<x<1, insuff
(2) st 2 true for any x<0,also insuff
both give us -1<x<0 so suff

statement 1: case a: x is positive, 1<x (satisfies)
case b: x is negative, -1<x (does not satisfies in the domain -1<x<0)
Thus, 1 alone is insufficient

Statement 2 can be reduced to x<0, Now, -1<x<0 satisfies the condition, however, x<-1 does not. Thus, it alone is insufficient.

Combining both, -1<x<0, which satisfies the condition. Thus, C

Edit: NVM found -1 < x < 0 is valid for statement 1

Guys I'm getting (A)... what am I missing?

If x is not equal to 0, is |x| < 1?

(1) x/|x| < x

Multiplying by |x| on both sides won't change sign and we get x < x|x|
For this to be possibe x > 1, thus x = GT(1)
|GT(1)| < 1 ? Answer is no. Sufficient

(2) |x| > x
For this to be possible x < 0, thus x = LT(0)
|LT(0)| < 1 ? Can't answer with certainty. Insufficient

MAAJ wrote:Edit: NVM found -1 < x < 0 is valid for statement 1

Guys I'm getting (A)... what am I missing?

If x is not equal to 0, is |x| < 1?

(1) x/|x| < x

Multiplying by |x| on both sides won't change sign and we get x < x|x|
For this to be possibe x > 1, thus x = GT(1)
|GT(1)| < 1 ? Answer is no. Sufficient

(2) |x| > x
For this to be possible x < 0, thus x = LT(0)
|LT(0)| < 1 ? Can't answer with certainty. Insufficient

hi MAAJ
in st 1 you missed that it is valid not only for x>1, but also -1<x<0, try to insert x=-1/2 in the st 1 and you`ll see that it satisfies both st 1 and problem as whole