Carcass wrote:
Given: \(|x|<x^2\);
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. \(x^2>1\) --> 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