How would you solve?
If x is not equal to 0, is |x| less than 1?
1) x/|x| is less than x
2) |x| is greater than x
ween from i, x/lxl < x refers to x as -ve and but grater than -1. so its sufficient to say that lxl is <1> x only implies that x is ve. nsf.
(1) (A) is also insuff because if x/|x| <x
then -1 < x <0> 1
(2) |x| > x => x must be -ve => insuff
(1) & (2) => suff