Is x > x ^3 ?

(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.

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

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.)

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.

+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.

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.

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

Question : Is x > x ^3 ?

for x to be greater than x^3
Case 1: Either x < -1 or
Case 2: 0 < x < 1

Statement 1: x < 0
x may be -0.5 or x may be -2 hence
NOT SUFFICIENT

Statement 2: x^2 - x^3 > 2
For this statement to be true x < -1, (Just try some values to substitute in expression to check acceptable values of x), Hence
SUFFICIENT

Answer: Option B

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is x > x ^3 ?

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


When it comes to inequality questions, it is crucial that if range of que includes range of con, that con is sufficient.
When you modify the original condition and the question, they become x^3-x<0?, x(x-1)(x+1)<0? -> x<-1, 0<x<1?. Then, there is 1 variable(x), which should match with the number of equations. So you need 1 more equation. For 1) 1 equation, for 2) 1 equation, which is likey to make D the answer.
For 1), if x<0, the range of que doesn’t include the range of con, which is not sufficient.
For 2), in x^3-x+2<0, (x+1)(x^2-2x+2)<0, x^2-2x+2=(x-1)^2+1 is derived, which is always bigger than 0. Then, x+1<0 -> x<-1, in which the range of que includes the range of con. Thus, it is sufficient and the answer is B.


-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

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?[/quote]

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[/quote][/quote]

i could not understand with those links...
as x^2-x^3>2 we can write as X^2(1-X)>2
so roots are 0,1
So range will be x>1 or x<0
or in number line if i plot i get +ve values within 0<x<1(i m plotting with roots 0 and 1 only in consideration)
Plz help.........
secondly plz clear me also that the for selecting different roots is same for different inequality like x^2-x^3>2 and x^2-x^3>0

Thanks

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