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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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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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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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+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] New post 09 Oct 2014, 07:57
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Re: Is x > x ^3 ? [#permalink] New post 09 Oct 2014, 20:50
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rxs0005 wrote:
Is x > x ^3 ?

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


You can either use inequalities here or the big picture approach.

When is x greater than x^3? It is when x < -1 or 0 < x <1.
(Recall that we should know the relations of x, x^2 and x^3 in the ranges 'less than -1', '-1 to 0', '0 to 1' and 'greater than 1')

(1) x < 0
When x is between -1 and 0, x^3 is greater than x. When x < -1, then x is greater than x^3. So this statement alone is not sufficient.

(2) x^2 - x^3 > 2
Now, this is not very easy to handle using inequalities. Without the cube, we would have just taken 2 to the other side and solved the quadratic. But this will be more easily solved using the big picture. Think of a number line.
What does x^2 - x^3 > 2 imply? It means x^2 is greater than x^3 and is 2 units to the right of x^3 on the number line. x^2 is never negative so it must be to the right of 0. Now there are two cases: Either x^2 is between 0 and 1 or it is greater than 1.
If x^2 were between 0 and 1, x would be between -1 and 1 and x^3 would be between 0 and -1. In this case, difference between x^3 and x^2 cannot be greater than 2. Hence this is not possible. So x^2 must be to the right of 1 and hence, x would be greater than 1 or less than -1. If x were greater than 1, x^3 would be greater than x^2 so x cannot be greater than 1. Hence x must be less than -1.
When x is less than -1, then x IS greater than x^3. Sufficient alone.

Answer (B)

Again, spend some time checking out the relations of x, x^2 and x^3 on the number line.
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Re: Is x > x ^3 ? [#permalink] New post 10 Oct 2014, 04:37
Bunuel
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

when (x+1)x(x-1)<0,, doesnt it gives us the rage as x>-1 or x<1 ... because ..as you explained in an another question that

Intersection points are the roots of the equation x^2-4x+3=0, which are x_1=1 and x_2=3. "<" sign means in which range of x the graph is below x-axis. Answer is 1<x<3 (between the roots).
From this i understood that "<" sign means roots to the RIGHT of the smaller root and to the LEFT of the bigger root).

I am getting it wrong??
please explain
Re: Is x > x ^3 ?   [#permalink] 10 Oct 2014, 04:37
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