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Re: Is x^2 greater than x ? [#permalink]
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 \(x > 1\) suff

In less than 50 seconds. Is fine or I'm wrong ??

Thanks a lot :)
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Re: Is x^2 greater than x ? [#permalink]
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carcass wrote:
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 \(x > 1\) suff

In less than 50 seconds. Is fine or I'm wrong ??

Thanks a lot :)


Everything is correct except the red part above.

Is x^2 greater than x ?

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.

Answer: D.
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Re: Is x^2 greater than x ? [#permalink]
Quote:

Everything is correct except the red part above.

Is x^2 greater than x ?

Is \(x^2 > x\)? --> is \(x(x-1)>0\)? Is \(x<0\) or \(x>1\)? So, as Karishma correctly noted above \(x^2>x\), for all negative value 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.

Answer: D.



grrrrrrrrrrrrrrrrrrrrrrr always this silly stupid dumb mistake :wall: I'm not so far from a good score ( \(>=48\) )

and always if you do not know the sign of x you can't square both sides, simply.


thanks both of you :roll:
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Is x^2 greater than x ? [#permalink]
carcass wrote:
Is x^2 greater than x ?

(1) x is less than -1.

(2) x^2 is greater than 1.


to answer this question, we need to know whether:
x<0
or x>1.
if 0<x<1, x^2 will be less than x.

1. sufficient.
2. it means that |x|>1, sufficient.
the answer is D.
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Is x^2 greater than x ? [#permalink]
VeritasKarishma 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)

VeritasKarishma

Is it not the case that
\(x<x^3 \implies\) \(-1<x<0\) OR \(1<x\) ?

And is it not also the case that

\(x^3<x \implies x<-1\) OR \(0<x<1\)?
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Re: Is x^2 greater than x ? [#permalink]
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