Cut-n-pasting from one of my earlier replies...

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/x ... bingo, you have (I) satisfied

For x>0.7 and <1, try x = 0.8
we have x^2 < 1/x < 2x ... bingo, you have (II) satisfied

It is easy to see that for all values x > 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)

Thnkx jayhawk..

D is the OA

Great stuff. This brings up an additional question.

When looking for boundaries, it seems that even more can be found. Using couple of the above boundaries for example,

1/x = 2x
would imply
x^2 = 1/2, so x= positive and negative 0.7

and
2x = x^2
x^2 - 2x = 0
x(x-2) = 0
x = 2 and x=0

At this point, I have too many boundaries to test. Am I doing something incorrectly?

thumpin_termis wrote:Great stuff. This brings up an additional question.

When looking for boundaries, it seems that even more can be found. Using couple of the above boundaries for example,

1/x = 2x
would imply
x^2 = 1/2, so x= positive and negative 0.7

and
2x = x^2
x^2 - 2x = 0
x(x-2) = 0
x = 2 and x=0

At this point, I have too many boundaries to test. Am I doing something incorrectly?

We are given that x is positive, so -0.7 is automatically out.

Also by virtue of x being positive, our lower-limit automatically
starts at 0. So, knowing one of our boundaries is 0.7, we will
have to choose something between 0 and 0.7. So, technically
no additional boundaries

Ah-hah. Another reminder to read the question carefully

Thanks a bunch!

One concern - if we dont find the boundaries, how would we solve this problem then.....

800GMAT - I generally follow this approach.

Assume (1) to be true => x^2 < 2x < 1/x
Find the value of x using inequalities
x^2 < 1/x => x<1
2x < 1/x => x < 1/sqrt2 => x<0.70
Since x is positive (given in question stem) x can be between (0,0.7)

Take x as 0.5 , then x^2 < 2x < 1/x holds.

Apply this approach to remaining inequalities.

Hope this helps...

800GMAT wrote:One concern - if we dont find the boundaries, how would we solve this problem then.....

You have a valid concern and is one that I dread the most as well.
Basically if we cannot make an educated guess at the boundaries, we
can never be sure.

In this particular problem, there are 3 values to compare -- 1/x, x^2 and 2x.

It should be immediately apparent that we have to try a value of x which is
in the range of 0 - 1. Also, we will try a value of x > 1. So, that will
knock out 2 out of 3 choices. However, to prove II, we need to plug
a value for x in the range 0.7 - 1...unfortunately this will be apparent
only if we know the boundaries...