Is x > x ^3 ?

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.

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.

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.

(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.)

+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.

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

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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