VeritasPrepKarishma wrote:

carcass wrote:

Is \(x^2\) greater than x ?

(1) \(x\) is less than -1.

(2) \(x^2\) is greater than 1.

For 700+ level questions, you should know these properties of numbers very well:

When is x^2 > x?

For all negative values of x (since the square will be positive) or whenever x > 1 (Square will be more than the number).

When is x^2 < x?

When 0 < x < 1

When is x^3 > x?

When x > 1 or -1 < x < 0

When is x^3 < x?

When 0 < x < 1 or x < -1

Draw them on the number line and mark the corresponding regions. Take examples from each region to convince yourself why the squares and cubes behave this way. Then, memorize both the diagrams!

Coming back to this question:

(1) \(x\) is less than -1.

For negative numbers, x^2 is more than x. So answer is 'Yes'. Sufficient.

(2) \(x^2\) is greater than 1

implies mod x is greater than 1 or we can say, x is greater than 1 or less than -1. In either case, x^2 is greater than x. So answer is 'Yes'. Sufficient.

Answer (D)

Thanks Karishma

I know: I 'm working to memorize these rules and tackle the question with more proficiency.

Infact at the moment I 'm quite comfortable with this question but I'm working on how attack a question from an odd angle i.e. trying different strategy.

This is why I posted here this question. Please see if I did correct

\(x^2 > x\) or \(x^2 - x > 0\)

This imply that \(x < 0\) and \(x > 1\)

1) \(x < - 1\) suff

2)

\(x^2 > 1\) basically says \(x > 1\) suffIn less than 50 seconds. Is fine or I'm wrong ??

Thanks a lot

Everything is correct except the red part above.

Is \(x^2 > x\)? --> is \(x(x-1)>0\)? Is \(x<0\) or \(x>1\)? So, as Karishma correctly noted above \(x^2>x\) hods true for all negative values of x as well as for values of x which are more than 1.

(1) x is less than -1. If x is negative, then \(x^2=positive>x=negative\). Sufficient.

(2) x^2 is greater than 1 --> \(x^2>1\) --> \(|x|>1\) --> \(x<-1\) or \(x>1\) --> for both case \(x^2>x\). Sufficient.