\(x^{11}>x^3\) --> \(x^{3}(x^8-1)>0\). 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.
_________________

\(x^{11}>x^3\) to hold true either \(-1<x<0\) or \(x>1\). Only -1/2 is in either of the range.

Answer: C.
_________________

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

Why go into such depths..

x^11 > x^3 in absolute terms for -ve or +ve value of x ...the same is reversed when x is replaced by 1/x
so options with either a positive value of x^11 or a negative value of 1/x^11..
we see the second case in option C and the question is solved in under 10 secs

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/
_________________

Bunuel,

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

From Red part, why can't x=1/2?...

What am I missing here

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
_________________

plugged in and worked from wrong to right...seemed to work

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
_________________

\(x^{11}>x^3\) --> \(x^{3}(x^8-1)>0\). 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.[/quote]


Can Someone explain how the below are deduced.

X^8>1-->X<-1 OR X>1

X^8<1--->-1<X<1
_________________

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

yes it is 100% clear.. Thank you so much
_________________

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

When do the relations behave differently? They are different in case of even-odd powers.

A higher odd power is greater than a lower odd power when x > 1 or -1 < x < 0
So in these ranges, x^3 > x, x^5 > x, x^9 > x^5 etc all hold.

A higher even power is greater than a lower even power when x > 1 or x < -1
So in these ranges, x^8 > x^2, x^4 > x^2 etc all hold

Similarly, a higher even power is greater than a lower odd power when x > 1 or x < 0
So in these ranges, x^2 > x, x^4 > x^3 etc all hold
_________________

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

X^11-X^3>0
=>X^3(X^8-1)>0
=>x^3(x^4-1)(x^4+1)>0
=>x^3(x^2-1)(x^2+1)(x^4+1)>0
=>x^3(x-1)(x+1)(x^2+1)(x^4+1)>0

plotting the three roots -1, 0, 1 along with the signs of the inequality in the respective zones:
only -1<x<0 and x>1 satisfy the inequality
So answer is C

\(x^{11}>x^3\) --> \(x^{3}(x^8-1)>0\). 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...

Hey,
how can you say that x^11>x^3 is similar to x^3>X?