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Re: 1/x^3,1/x^2,1/x,x,x^2,x^3 [#permalink]
2
From the given condition we can pick our x = -1/2
Then the set become 4,1/4, -1/2,-1/8,-2,-8
Our median is x+x^3/2 =x(x^2+1)/2
Hence D

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Re: 1/x^3,1/x^2,1/x,x,x^2,x^3 [#permalink]
4
Given the inequality -1 <x< 0, the value of x has to be a negative fraction.

Therefore, the values 1/(x^2) and x^2 have to be the greatest values as they will result in positive values.

The values 1/x and 1/x^3 have to be the smallest values as they will result in negative integer values.
Say, the value of x = -1/3, then 1/x will be -3 and the value for 1/x^3 will be (-3)^3. Giving the value as integer values.

The two middle values of this list will therefore be x^3 and x. Dividing these two values by x to get the median will be-
x^3+x/2 which gives x(x^2+1)/2

This gives us the answer D

The most conclusive way to do it would be to take a value for x, for instance -1/3 and plug in the values to get the median.
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Re: 1/x^3,1/x^2,1/x,x,x^2,x^3 [#permalink]
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for -1<x<0
x^2>x
x^3>x
=> x^2>x^3>x
1/x^2 > x^2
1/x^3 is the least value
and 1/x > 1/x^3

Finally, 1/x^2> x^2 > x^3> x > 1/x > 1/x^2


notice that the middle terms are: x^3 and x

=> median = x^3+x/2
pick D
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Re: 1/x^3,1/x^2,1/x,x,x^2,x^3 [#permalink]
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Re: 1/x^3,1/x^2,1/x,x,x^2,x^3 [#permalink]
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