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PS Problem #7. Anyone. Thx!!

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PS Problem #7. Anyone. Thx!! [#permalink] New post 26 Jul 2009, 22:33
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I was able to solve it, but it took a while, a bit of handwaving/guessing. I wish I could find a more structured and efficient way of attacking this problem:

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
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Re: PS Problem #7. Anyone. Thx!! [#permalink] New post 27 Jul 2009, 00:51
sudeep wrote:
I and II only


I know the answer. I am seeking an explanation/thought process/efficient mode of attack/methodology/etc. Could you help in that regard?
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Re: PS Problem #7. Anyone. Thx!! [#permalink] New post 27 Jul 2009, 01:06
there are three f(x), make comparisons between each to get the relative division in the real line x.

1) 1/x < 2x ==> (x>1/sqrt(2))

2) 1/x < x^2 ==> (x>1)

3) 2x < x^2 ==> (x>2)

so you have possibilities of changes

for x
(0, 1/sqrt(2)) ==> get order from above comparison results.
(1/sqrt(2), 1)
(1,2)
(2,inf)
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Re: PS Problem #7. Anyone. Thx!! [#permalink] New post 27 Jul 2009, 01:29
sudeep wrote:
there are three f(x), make comparisons between each to get the relative division in the real line x.

1) 1/x < 2x ==> (x>1/sqrt(2))

2) 1/x < x^2 ==> (x>1)

3) 2x < x^2 ==> (x>2)

so you have possibilities of changes

for x
(0, 1/sqrt(2)) ==> get order from above comparison results.
(1/sqrt(2), 1)
(1,2)
(2,inf)



This is interesting. Never seen it solved this way. So, after you have found the relative divisions, what do you do with them that would help you figure out the ones (I, II, or III) that could work. You're a pretty smart guy/lady, but please break this down as simple as possible.
Re: PS Problem #7. Anyone. Thx!!   [#permalink] 27 Jul 2009, 01:29
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PS Problem #7. Anyone. Thx!!

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