With these kinds of questions, it is always useful to find the boundaries

We have 3 equations giving us 3 boundaries --

1/x = x^2 implies x = 1
1/x = 2x implies x = 0.7 approx
2x = x^2 implies x = 2

Our 3 boundaries are 0.7, 1 and 2

Take 1 sample for each range

For x<0.7, try x = 0.5
We have x^2 < 2x <1>0.7 and <1, try x = 0.8
we have x^2 < 1/x <2x> 1, 1/x will give you the
least of the 3. So, (III) can never be true.

So, choose D (i.e. I and II alone)

Jayhawk,

Can you please elaborate this part :

Take 1 sample for each range

For x<0.7, try x = 0.5
We have x^2 < 2x <1>0.7 and <1, try x = 0.8
we have x^2 < 1/x <2x> 1, 1/x will give you the
least of the 3. So, (III) can never be true.

So, choose D (i.e. I and II alone)

raulverde wrote:Jayhawk,

Can you please elaborate this part :

Take 1 sample for each range

For x<0.7, try x = 0.5
We have x^2 < 2x <1>0.7 and <1, try x = 0.8
we have x^2 < 1/x <2x 1, 1/x will give you the
least of the 3. So, (III) can never be true.

So, choose D (i.e. I and II alone)

We have 3 boundaries here i.e. 0.7, 1 and 2

To find out which one can never be true, try out values <0> 2

For x = 0.5, we get x^2 < 2x < 1/x

For x = 0.8, we get x^2 < 1/x < 2x

For values of x greater than 2, 1/x will be the least of the 3

So, you can safely conclude that III will never be true

Hence D

ri2007 wrote:If X is positive, which of the following could be the correct ordering of x, 1/x and 2x

1) X^2 < 2x < 1/x
2) X^2 < 1/x < 2x
3) 2x < X^2 < 1/x

None
1 only
2 only
1 & 2 only
all 3

Hi
Is there a quick way of doing this? Also can any one show me a number set which satisfys the order in statement 2?

Thanks

OA [spoiler]1 & 2 only[/spoiler]

In the questions is it x^2 instead of x, is x a typo? I meant ordering of x, 1/x and 2x, should it be x^2,1/x,2x??