Problem Solving

Thankfully, x is positive here. That means we can divide or multiply both sides of an inequality by x without needing to worry about whether to reverse the inequality. Look at:

II. x^2 < 1/x < 2x

This is really three different inequalities:

a) x^2 < 1/x
b) 1/x < 2x
c) x^2 < 2x

Look at a):

x^2 < 1/x
x^3 < 1
x < 1

So II could only possibly be true for x < 1. We need to check the other conditions though:

Look at b):

1/x < 2x
1 < 2x^2
1/sqrt(2) < x
sqrt(2)/2 < x

So we now see that, for II to be true, x must be between sqrt(2)/2 and 1. c) doesn't tell us anything more. But if you choose any x between sqrt(2)/2 and 1, you'll see how II can hold (try x = 3/4, for example).

Vemuri wrote:
That's a good explanation Ian (as always). I have a question though. From your approach, looks like we have to check for which conditions consistently maintain. The one that gives a contradictory condition is to be eliminated. Is this right?

I have almost always been unsuccessful with "Must be True" type of questions when plugging in numbers.

Yes, everything you've done is correct. For a 'could be true' question, like the above, you might be able to make a quick decision about one or more of the items by trying out a simple value for x, just to cut down on some of the algebraic work. So for I) here, you might notice that the inequalities are true for x = 1/2, so I) can be true, and then you only need to check II) and III) algebraically.

Then when you do the algebra here, if you arrive at an answer which is logically impossible (as you did with III) - you found that x was both greater than 2 and less than 1), then the item certainly cannot be true. If, on the other hand, you arrive at a range of possible values for x, as you did with II), as long as you've completely considered all of the implications of the information provided, you will have found all of the values which x can have. And by breaking down the three-part inequality into three two-part inequalities, we can be certain that we are using all of our information here.