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Re: Friday Algebra DS [#permalink]
28 Jan 2011, 05:54

Expert's post

rxs0005 wrote:

Is x > x ^3

S1 x < 0

S2 x^2 - x^3 > 2

Is x> x^3?

Is x>x^3? --> is x^3-x<0? --> is (x+1)x(x-1)<0? is x<-1 or 0<x<1

(1) x<0. Not sufficient.

(2) x^2-x^3>2 --> x^2(1-x)>2 --> only true for x<-1 (note that if x>1 then x^2(1-x) is negative so this range is not good and if -1\leq{x}\leq{1} then x^2(1-x)\leq{2} so this range is also not good). Sufficient.

Re: Friday Algebra DS [#permalink]
17 Mar 2011, 21:49

1

This post received KUDOS

I goofed on this because I tricked myself but this is easy...think about it as always before putting pen to paper.. When is x>x^3? ONLY when x = negative integer OR a positive fraction. 1) x= -ve, ok but is it a fraction? Insuff 2) x^2 - x^3 >2 x^2 (x-1)<2 Use gurpreets method to draw the number line. You will see that Statement is positive for all: x>1 = positive, x>0 = positive Therefore between 0 and 1 the statement is negative..hence the original statement holds true. Because X is a positive fraction, B is sufficient.

Re: Friday Algebra DS [#permalink]
05 Mar 2013, 05:49

Bunuel wrote:

rxs0005 wrote:

Is x > x ^3

(2) x^2-x^3>2 --> x^2(1-x)>2 --> only true for x<-1 (note that if x>1 then x^2(1-x) is negative so this range is not good and if -1\leq{x}\leq{1} then x^2(1-x)\leq{2} so this range is also not good). Sufficient.

Answer: B.

Hi Bunuel,

For x^2-x^3>2 how did you directly arrive at the intervals <-1 between -1 and 1 and >1. When we deal with (1-x) >2 we get x <-1. But how did you chose the other point 1?

Re: Friday Algebra DS [#permalink]
06 Mar 2013, 01:55

Expert's post

Jaisri wrote:

Bunuel wrote:

rxs0005 wrote:

Is x > x ^3

(2) x^2-x^3>2 --> x^2(1-x)>2 --> only true for x<-1 (note that if x>1 then x^2(1-x) is negative so this range is not good and if -1\leq{x}\leq{1} then x^2(1-x)\leq{2} so this range is also not good). Sufficient.

Answer: B.

Hi Bunuel,

For x^2-x^3>2 how did you directly arrive at the intervals <-1 between -1 and 1 and >1. When we deal with (1-x) >2 we get x <-1. But how did you chose the other point 1?

Re: Friday Algebra DS [#permalink]
06 Mar 2013, 23:08

Bunuel wrote:

Check here:

Hi Bunuel, Thanks for your reply and reference to some wonderful materials around quadratic inequalities.

May I ask you one more question - going back to the basics now. What would be the roots of the eqn we have in hand in this post: x^2 - x^3 >2. I solved the roots to be 0 and -1 by following the below steps: x^2(1-x)>2 X^2 implies 0 is a root. 1-x>2 implies x<-1, so -1 is a root. From your explanations the roots seem to be 1 and -1. Where am I going wrong? (I also saw another post of yours where I was not able to solve the correct roost when x^3 was involved. Please help.)

Last edited by Jaisri on 07 Mar 2013, 05:28, edited 1 time in total.

Nothing to add after Bunuel's explanation. But just writing down how to get the inequality x<-1 from F.S 2 for those who couldn't get it.

F.S 1 clearly not sufficient. Take x= -1,you get a NO for stem, but for -0.5, you get a YES.

F.S 2, states x^2-x^3-2>0

or( x^2-1)-(x^3-1)>0

or(x-1)[(x+1) - (x^2+1+x)]>0

or(x-1)(-x^2)>0. Thus, (x-1) has to be negative

or x-1<0

or x<-1.

Sufficient.

B.

+1 Kudos. Thanks for showing how to solve this inequality! There is a minor change though... x-1 >0 gives x<1 and not x <-1. Bunuel seems to have considered -1 and 1 as the roots. Any reasons for that?

+1 Kudos. Thanks for showing how to solve this inequality! There is a minor change though... x-1 >0 gives x<1 and not x <-1. Bunuel seems to have considered -1 and 1 as the roots. Any reasons for that?

thanks for pointing out the mistake. It was a splendidly foolish mistake. Apologies.
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