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(1) \(x^3-x>0\) --> \(x(x^2-1)>0\): Two cases: A.\(x>0\), \(x^2-1>0\) --> \(x>0\) and \(x>1\) \(x<-1\)--> \(x>1\) B. \(x<0\), \(x^2-1<0\) --> \(x<0\) and \(-1<x<1\) --> \(-1<x<0\)

Two ranges \(x>1\) or \(-1<x<0\), not sufficient.

(2) \(x>0\), \(x^2>x\) --> \(x^2-x>0\)--> \(x(x-1)>0\) --> as \(x>0\), \(x-1>0\) --> \(x>1\). One range \(x>1\). Sufficient.

Answer: B.

Hi Bunuel!

I have seen already your answers in which you mention the ranges both in inequalities & in modules questions. Can you kindly explain how does it work or refer to some good resource to understand the raging concept.
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(1) \(x^3-x>0\) --> \(x(x^2-1)>0\): Two cases: A.\(x>0\), \(x^2-1>0\) --> \(x>0\) and \(x>1\) \(x<-1\)--> \(x>1\) B. \(x<0\), \(x^2-1<0\) --> \(x<0\) and \(-1<x<1\) --> \(-1<x<0\)

Two ranges \(x>1\) or \(-1<x<0\), not sufficient.

(2) \(x>0\), \(x^2>x\) --> \(x^2-x>0\)--> \(x(x-1)>0\) --> as \(x>0\), \(x-1>0\) --> \(x>1\). One range \(x>1\). Sufficient.

Answer: B.

Hi Bunuel!

I have seen already your answers in which you mention the ranges both in inequalities & in modules questions. Can you kindly explain how does it work or refer to some good resource to understand the raging concept.

Unfortunately I can not refer to any resources as I studied this staff in school. But I'll try to explain the basics of it if you specify the issue or post question(s).
_________________

x>0, x^2>x --> x^2-x>0--> x(x-1)>0 --> as x>0, x-1>0 --> x>1. One range x>1. Sufficient.

you haven't considered the second half i guess..

x(x-1)>0 --->x<0, x<1--->x<1..

if x = -2 then -2(-2-1)>0..

I think its E..correct me if am wrong..

Not so.

Statement 2 states: x^2>x>0

When we split we get x^2>x AND x>0, so there is no second part. x can not be less than zero, so the only chance x(x-1)>0 to be true is when x and x-1 is BOTH more than zero.
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