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Is x > x ^3 ?

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Is x > x ^3 ? [#permalink] New post 28 Jan 2011, 04:59
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Is x > x ^3 ?

(1) x < 0
(2) x^2 - x^3 > 2
[Reveal] Spoiler: OA

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

Answer: B.
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Re: Friday Algebra DS [#permalink] New post 17 Mar 2011, 21:49
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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.
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Re: Friday Algebra DS [#permalink] New post 17 Mar 2011, 23:02
(1)

If x is not a fraction, say -2 , then x > x^3

But if x = -1/2, then -1/2 < -1/8


(2) x is not a proper fraction, and is a -ve number

(-3)^2 - (-3)^3 = 9 - (-27) = 36


So 2 is sufficient. Answer B.
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Re: Friday Algebra DS [#permalink] New post 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?
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Re: Friday Algebra DS [#permalink] New post 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?


Check here:
x2-4x-94661.html#p731476,
inequalities-trick-91482.html,
everything-is-less-than-zero-108884.html?hilit=extreme#p868863,
xy-plane-71492.html?hilit=solving%20quadratic#p841486
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Friday Algebra DS [#permalink] New post 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.
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Re: Is x > x ^3 ? [#permalink] New post 07 Mar 2013, 01:03
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Expert's post
rxs0005 wrote:
Is x > x ^3 ?

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


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+2x-2]>0

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

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

as(x-1)^2+1 will always be positive, thus (x+1) has to be negative.

or x+1<0

or x<-1

Sufficient.

B.
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Last edited by mau5 on 08 Mar 2013, 06:57, edited 2 times in total.
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Re: Is x > x ^3 ? [#permalink] New post 08 Mar 2013, 05:43
vinaymimani wrote:
rxs0005 wrote:
Is x > x ^3 ?

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


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?
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Re: Is x > x ^3 ? [#permalink] New post 08 Mar 2013, 06:56
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Quote:

+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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Re: Is x > x ^3 ?   [#permalink] 08 Mar 2013, 06:56
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