Carcass wrote:
Given: |x|<x2;
Reduce by |x|: 1<|x| (side note: we can safely do this as absolute value is non-negative and in this case we know it's not zero too, as if x would be zero inequality wouldn't hold true, so |x|>0);
So we have that x<−1 or x>1.
I. x2>1 --> always true;
II. x>0 --> may or may not be true;
III. x<−1 --> may or may not be true.
Answer: A (I only).
Could you please explain the case for III part? below in the reply the case for third part has been explained but it takes x as 2 whereas x is supposed to be a negative number here