If x is positive, which of the following could be the correct ordering of 1/x, 2x and x^2 ?

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

(A) None
(B) I only
(C) III only
(D) I and II only
(E) I II and III

Thanks Bunuel! I did exactly what you did, splitting the numbers. I then picked 1/2 and 1/16 and got the wrong answer. I guess sometimes picking the "easy" numbers is not the best strategy.

Thanks again!

When picking a number, the most important thing is that we should try all possible numbers that could give us different answers. Let me explain by telling you how I would put in numbers and check.

I see \frac{1}{x}, 2x and x^2.
I know I have to try numbers from two ranges at least '0-1' and '>1' since numbers in these ranges behave differently.
Also, x^2 is greater than 2x if x > 2 e.g. 3x3 > 2x3 but if x < 2, then x^2 is less than 2x e.g 1.5 x 1.5 < 2 x 1.5. So, I need to try a number in the range 1 to 2 as well.
Also, 0 and 1 are special numbers, they give different results sometimes so I have to try those as well. Let's start:

0- Since x is positive, I don't need to try it.

1/2 - I get 2, 1 and 1/4. I get the order x^2 < 2x < \frac{1}{x}

1 - I get 1, 2, 1 Now, what you need to notice here is that 2x > \frac{1}{x} whereas in our above result we got \frac{1}{x} > 2x. This means there must be some value between 0 and 1 where \frac{1}{x} = 2x. Anyway, that doesn't bother me but what I have to do now is take a number very close to 1 but still less than it.
I take 15/16 (random choice). I get 16/15, 15/8 and (\frac{15}{16})^2. The first two numbers are greater than 1 and the last one is less than 1. I get the order x^2 < \frac{1}{x} < 2x

Now I need to try 3/2. I get 2/3, 3 and 9/4. So the order is \frac{1}{x} < 2x < x^2

I try 3 - I get 1/3, 6, 9
For these numbers, x^2 will be greatest but none of the options have it as the greatest term.

Only I. and II. match hence answer is (D)
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Bunuel I have reasonably implemented the key values approach in all my inequalities problems but I couldn't understand how 0, 1, 2 can be inferred to be the keys in this problem.
Can you elaborate why you chose 0 1 2 ?

Regards,
Sameer

If x is positive, which of the following could be correct ordering of \frac{1}{x}, 2x, and x^2?

(I) x^2 < 2x < {1}/{x}
(II) x^2 < {1}/{x} < 2x
(III) 2x < x^2 < {1}/{x}

(a) none
(b) I only
(c) III only
(d) I and II
(e) I, II and III
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this can take a long time

use both methods, picking the number and solving unequality

for the third we can solve unequality, remember one thing make it easy, x is positive so we can multiple 2 sides of un equality with x and do not change the mark of unequality

gmat is simple but is enough to kill us when we are nervous on the test.