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1/x^3,1/x^2,1/x,x,x^2,x^3

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1/x^3,1/x^2,1/x,x,x^2,x^3 [#permalink] New post 28 Sep 2020, 02:45
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37% (02:34) correct 62% (01:22) wrong based on 27 sessions
\(\frac{1}{x^3},\frac{1}{x^2},\frac{1}{x},x,x^2,x^3\)


If \(-1<x<0\) , what is the median for the six numbers in the list above ?

A. \(\frac{1}{x}\)

B. \(x^2\)

C. \(\frac{x^2(x+1)}{2}\)

D. \(\frac{x(x^2+1)}{2}\)

E. \(\frac{x^2+1}{2x}\)
[Reveal] Spoiler: OA

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Re: 1/x^3,1/x^2,1/x,x,x^2,x^3 [#permalink] New post 28 Sep 2020, 02:45
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Re: 1/x^3,1/x^2,1/x,x,x^2,x^3 [#permalink] New post 28 Sep 2020, 04:11
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let x value be -0.1

given set will be -1000,100,-10,-0.1,0.01,-0.001

writing them in ascending order -1000,-10,-0.1,-0.001,0.01,100

median will be (-0.1-0.001)/2 =-0.0505

lets verify options
A) (1/x)=-10 not correct

B) \(x^2\)=0.01not correct

C) \(x^2(x+1)/2\)=0.0045not correct

D)\(x(x^2 +1)/2\)=-0.0505correct

E)\((x^2+1)/2x\)=5.05 not correct

so option D is correct
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Re: 1/x^3,1/x^2,1/x,x,x^2,x^3 [#permalink] New post 30 Sep 2020, 05:04
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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] New post 03 Oct 2020, 10:31
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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] New post 27 Oct 2020, 01:52
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
Re: 1/x^3,1/x^2,1/x,x,x^2,x^3   [#permalink] 27 Oct 2020, 01:52
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