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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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\(\frac{1}{x^3},\frac{1}{x^2},\frac{1}{x},x,x^2,x^3\) 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. \(\frac{x^2+1}{2x}\)
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Re: 1/x^3,1/x^2,1/x,x,x^2,x^3 [#permalink]
28 Sep 2020, 02:45
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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 (0.10.001)/2 =0.0505 lets verify options A) (1/x)=10 not correctB) \(x^2\)=0.01 not correctC) \(x^2(x+1)/2\)=0.0045 not correctD)\(x(x^2 +1)/2\)=0.0505 correctE)\((x^2+1)/2x\)=5.05 not correctso 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 Posted from my mobile device



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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





