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).

To answer this, I would suggest first to consider that |x| and x^2 are both ALWAYS POSITIVE.

Next, notice that this is a MUST question, so finding any good NO case (counterexample) will prove it wrong.

I. Given that |x| is defined as sqrt(x^2), then it follows that the number x must be greater than 1 -- that is, the only cases where the square is smaller than the number itself are fractions between 0 and 1. This MUST BE TRUE.

II.
>NO case make x = -2 and we see |-2| = 2 while (-2)^2 = 4

This one doesn't have to be true.

III.
>NO case x = 2 and we see |2| = 2 while 2^2 = 4

This one doesn't have to be true.

The Answer is A.
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