\(x^{11}>x^3\) --> \(x^{3}(x^8-1)>0\). So, far correct.

Next, for \(x^{3}(x^8-1)\) to be positive \(x^3\)and \(x^8-1\) must have the same sign, so both of them must be positive or both of them must be negative.

\(x^3>0\) and \(x^8-1>0\):
\(x^3>0\) gives \(x>0\);
\(x^8-1>0\) --> \(x^8>1\) --> \(x<-1\) or \(x>1\).
Both to hold true \(x\) must be greater than 1. So, \(x^3>0\) and \(x^8-1>0\) when \(x>1\).

\(x^3<0\) and \(x^8-1<0\):
\(x^3<0\) gives \(x<0\);
\(x^8-1<0\) --> \(x^8<1\) --> \(-1<x<1\).
Both to hold true \(x\) must be from -1 to 1. So, \(x^3<0\) and \(x^8-1<0\) when \(-1<x<1\).

Thus we have that \(x^{11}>x^3\) when \(x>1\) or \(-1<x<1\).



Hope this helps.
_________________

Hi Bununel
Could you explain how to proceed as i stuck up after few steps???

X^11-X^3>0
=>X^3(X^8-1)>0
=>Now X has 2 solution
such as X^3>0 then X>0...1
Now X^8-1>0
=>X^8>=1
=>X>1.....2

We got 2 solution in this case I think D satisfies

Help...........

Rgds
Prasannajeet

The analysis of \(x^{11} > x^3\) will be similar to that of \(x^3 > x\). We know the ranges where this holds: x > 1 or -1 < x < 0.

Hence for x = -1/2, this will hold.

For a question discussing ranges and the properties of numbers in these ranges, check: http://www.veritasprep.com/blog/2011/08 ... -question/
_________________

This is the range given by Bunuel. \(-1<x<0\) or \(x>1\)
So x cannot be 1/2.

The post you are quoting has a typo. In the second part, since x^3 < 0, we get x < 0.
So when you get -1 < x< 1, the only possible range is -1 < x< 0
_________________

Hi All,

Since the answer choices are numbers, we can TEST THE ANSWERS to find the one that "fits." This approach is made all the easier since the answers are so simple. There are also some great Number Property shortcuts that we can take advantage of.

So, which answer would fit X^11 - X^3 > 0?

Since X^11 and X^3 both have odd exponents, plugging in 1 or -1 would yield the same result, so X^11 - X^3 = 0..... NOT > 0 like we need. Eliminate B and E.

With -2, X^11 will be CONSIDERABLY MORE NEGATIVE than X^3, so X^11 - X^3 < 0. Eliminate A.

With 1/2, X^11 will be CONSIDERABLY SMALLER than X^3, so X^11 - X^3 < 0. Eliminate D.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
_________________

Hi,
when you raise anything less than 1 to an integer power, it becomes smaller as the power increases..
example 1/2.. 1/2 ^2=1/4.. 1/2^3=1/8...

lets see the two inequalities now..

X^8>1..
since the power of x is even number 8, x^8 will always be positive..
so if x is any negative integer less than -1, x^8 will >1..
example x=-2, (-2)^8 will be greater than 1..
but as we have seen earlier anything less than 1 when raised to a power will become even smaller, so it will always be<1..
that is why x cannot take any value between -1 and 1, both inclusive, for x^8 to be >1..
so x<-1 or x>1..

x^8<1..
Opposite of the above , for x^8 to be less than 1, x has to be between 1 and -1..

hope it helps
_________________

Hi Karishma ,
Can you kindly explain , how \(x^{11} > x^3\) will be similar to that of \(x^3 > x\) ?

Hey Karishma - Thanks a lot for clarifying my doubts.
+1 kudos

\(x^{11}>x^3\) --> \(x^{3}(x^8-1)>0\). So, far correct.

Next, for \(x^{3}(x^8-1)\) to be positive \(x^3\)and \(x^8-1\) must have the same sign, so both of them must be positive or both of them must be negative.

\(x^3>0\) and \(x^8-1>0\):
\(x^3>0\) gives \(x>0\);
\(x^8-1>0\) --> \(x^8>1\) --> \(x<-1\) or \(x>1\).
Both to hold true \(x\) must be greater than 1. So, \(x^3>0\) and \(x^8-1>0\) when \(x>1\).

\(x^3<0\) and \(x^8-1<0\):
\(x^3<0\) gives \(x<0\);
\(x^8-1<0\) --> \(x^8<1\) --> \(-1<x<1\).
Both to hold true \(x\) must be from -1 to 1. So, \(x^3<0\) and \(x^8-1<0\) when \(-1<x<1\).

Thus we have that \(x^{11}>x^3\) when \(x>1\) or \(-1<x<1\).
/quote]
Hi Bunuel,
Hope you're well brother.
For the first green part we do not consider the first red part. But, why do we consider second red part (1) after knowing the second green part (x<0)
Thank you...