Bunuel wrote:
You are correct.
Statement 1: Positive fractions will never make this statement true. The square of a positive fraction is always smaller than the original fraction itself.
Statement 2: Only positive numbers greater than 1 make this statement true. Therefore, by combining them, there is only one possibility to make both true: Numebrs greater than 1, which allows us to answer the original question.
No your ranges for (2) are not correct, it should be: \(-1<x<0\) or \(x>1\). Negative fractions from the range (-1,0) also make statement (2) true. Below is complete solution:
Is \(x>0\)?(1) \(x<x^2\) --> \(x*(1-x)>0\) --> \(x<0\) or \(x>1\) -->
---------0----1----. Not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO.
(2) \(x<x^3\) --> \(x(x-1)(x+1)>0\) --> \(-1<x<0\) or \(x>1\) -->
----(-1)----(0)----(1)----. Not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO.
(1)+(2) Intersection of the ranges from (1) and (2) is:
----(-1)----(0)----(1)----Again not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO.
Answer: E.
Hope it helps.[/quote]
Hi Bunuel,
Could you please explain why you're considering only the negative factors while considering \(x<x^2\) --> \(x*(1-x)>0\) --> \(x<0\) or \(x>1\).. From \(x*(1-x)>0\) can't we consider both x and (x-1) to be positive also ?
Similarly for \(x(x-1)(x+1)>0\) --> \(-1<x<0\) or \(x>1\), I seem to be missing your logic.
Thanks