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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]  28 Sep 2020, 02:45
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Question Stats:

38% (02:34) correct 61% (01:19) wrong based on 26 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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Founder
Joined: 18 Apr 2015
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Kudos [?]: 3531 [0], given: 12627

Re: 1/x^3,1/x^2,1/x,x,x^2,x^3 [#permalink]  28 Sep 2020, 02:45
Expert's post
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
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Re: 1/x^3,1/x^2,1/x,x,x^2,x^3 [#permalink]  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]  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]  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.
Re: 1/x^3,1/x^2,1/x,x,x^2,x^3   [#permalink] 03 Oct 2020, 10:31
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