Which of the following is a value of x for which x^11-x^30

Question

Which of the following is a value of x for which x^11-x^3>0

A. -2
B. -1
C. -1/2
D. 1/2
E. 1

Bunuel · Accepted Answer

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 --> -1x^3 when x>1 or -1x2-4x-94661.html#p731476 inequalities-trick-91482.html data-suff-inequalities-109078.html range-for-variable-x-in-a-given-inequality-109468.html everything-is-less-than-zero-108884.html graphic-approach-to-problems-with-inequalities-68037.html All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184 All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189 700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html Hope this helps.

Bunuel · Answer

x^{11}>x^3 to hold true either -11 . Only -1/2 is in either of the range. Answer: C.

prasannajeet · Answer

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

JusTLucK04 · Answer

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

KarishmaB · Answer

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.

HarveyS · Answer

Bunuel, Thus we have that x^{11}>x^3 when x>1 or -1

KarishmaB · Answer

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

saadis87 · Answer

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

EMPOWERgmatRichC · Answer

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:

Show Spoiler

C

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

sairam595 · Answer

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 --> -1x^3 when x>1 or -1x2-4x-94661.html#p731476 inequalities-trick-91482.html data-suff-inequalities-109078.html range-for-variable-x-in-a-given-inequality-109468.html everything-is-less-than-zero-108884.html graphic-approach-to-problems-with-inequalities-68037.html All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184 All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189 700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html Hope this helps. Can Someone explain how the below are deduced. X^8>1-->X<-1 OR X>1 X^8<1--->-1

chetan2u · Answer

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

sairam595 · Answer

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

AndyNeedsGMAT · Answer

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

KarishmaB · Answer

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

AndyNeedsGMAT · Answer

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

Vinayprajapati · Answer

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

Attachments

x11-x3 inequality.png [ 5.36 KiB | Viewed 19151 times ]

TheUltimateWinner · Answer

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 --> -1x^3 when x>1 or -1

Jaya6 · Answer

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

kelly_jacques · Answer

KarishmaB  I prefer to use the graphical method and normally always do. But how do you know to keep expanding the exponents like this. I have seen problems where you keep the higher powers, but here, clearly you needed to keep simplifying to get the third root. (e.g. if you had stopped at x^8(x^3-1) you only would have had 0 and 1 as roots

Which of the following is a value of x for which x^11-x^3>0

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GMAT Problem Solving (PS) Questions

Which of the following is a value of x for which x^11-x^3>0

Prep Toolkit

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