Problem Solving

If 0<x<1, what is the median of the values x, x^-1, x^2, squareroot(x), and x^3?

x
x^-1
x^2
squareroot(x)
x^3

I figured it out after the test, but got it wrong originally. Just throwing it out for conversation..

easiest way is to plug in.
Let x = 1/4, then
x = 1/4
x^-1 = 4
x^-2 = 16
√x = 1/2
x^3 = 1/64

So, 1/64,1/4,1/2,4,16 are in ascending order. i.e x^3 < x < √x < 1/x < 1/(x^2). So the median is [spoiler]√x[/spoiler].

IMO D

As x is a fraction, the the more times you multiply the number by itself the smaller it gets.

Therefore the smallest number will be x^3 and then the next smallest is x ^ 2. The largest number will be x ^ -1 as this will be a real number greater than 1.

So we have the following till now in ascending order : x^3 , x^2 , _ , _ , x^-1

squareroot (x) > x ( as x is a fraction, the squareroot will be greater than the number itself)

So we finally have the following :

x^3 , x^2 , x , sqrrt (x) , x ^-1

So the median is x

Above is the logical way of doing it. We can ofcourse use neelgandham's method of plugging-in values as it is a much quicker option. The only mistake in my humble opinion was that neelgandham has considered x ^ -2 whereas the problem states x ^2.