Ian Stewart wrote: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).
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. x^2 < 2x < 1/x
The 3 possible conditions are:
a) x^2<2x ==> x<2
b) x^2<1/x ==> x<1
c) 2x<1/x ==> x<1/sqrt(2)
So, x has a value less than 1/sqrt(2)
II. x^2 < 1/x < 2x
The 3 possible conditions are:
a) x^2<1/x ==> x<1
b) 1/x<2x ==> x>1/sqrt(2)
c) x^2<2x ==> x<2
So, x has a value between 1/sqrt(2) & 1
III. 2x < x^2 < 1/x
The 3 possible conditions are:
a) 2x<x^2 ==> x>2
b) 2x<1/x ==> x<1/sqrt(2)
c) x^2<1/x ==> x<1
This does not boil down to a range value for x (its either less than 1 or greater than 2).
I have almost always been unsuccessful with "Must be True" type of questions when plugging in numbers.
