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Correct Order

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tl372
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Correct Order [#permalink] New post   26 May 2006, 10:29
If x is positive, which could be the correct order 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: Correct Order [#permalink] New post   26 May 2006, 11:09
Only I ( when X is decimal )
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Re: Correct Order [#permalink] New post   26 May 2006, 11:37
tl372 wrote:
If x is positive, which could be the correct order 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


If x=2
II) is true

If x=1/2
I) is true


So I) and II)
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Re: Correct Order [#permalink] New post   26 May 2006, 11:40
geemaat wrote:
tl372 wrote:
If x is positive, which could be the correct order 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


If x=2
II) is true

If x=1/2
I) is true


So I) and II)


if x=2

II) x^2 < 1/x < 2x

4 < 1/2 < 4

How can it be true?
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Re: Correct Order [#permalink] New post   26 May 2006, 11:43
if x is less than 1 then I is correct in any case...
x^2 < 2x < 1/x
if x = 0.2
0.04 < 0.4 < 5

But I am not sure if II is correct for x = 5

x^2 < 1/x < 2x
25 < 1/5 < 10????
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Re: Correct Order [#permalink] New post   26 May 2006, 11:49
deowl wrote:
geemaat wrote:
tl372 wrote:
If x is positive, which could be the correct order 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


If x=2
II) is true

If x=1/2
I) is true


So I) and II)


if x=2

II) x^2 < 1/x < 2x

4 < 1/2 < 4

How can it be true?


You are right... I meant to do x=3 and of course I have the I) and II) swapped...

But you raised a good point, that when x=2 or for that matter when x=1, none are true.

What is the OA? I do not see how anyone of them can be certainly true!
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Re: Correct Order [#permalink] New post   26 May 2006, 12:01
The question here is whether there is a X > 0 that the order is correct.

(I) is correct for 0 <x< 1 , others are not correct for any x > 0 ( which is given )
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Re: Correct Order [#permalink] New post   26 May 2006, 14:42
The OA is I & II.

This question is from GmatPrep. When I originally answered the question, my answer was I only.

I still can't figure out how II could be true as well. Any thoughts?
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Re: Correct Order [#permalink] New post   26 May 2006, 14:52
May be ( II ) should be rewritten as:
II) x^2 <= 1/x < 2x

So it holds true for x=1
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Re: Correct Order [#permalink] New post   26 May 2006, 14:53
tl372 wrote:
The OA is I & II.

This question is from GmatPrep. When I originally answered the question, my answer was I only.

I still can't figure out how II could be true as well. Any thoughts?


if x=3/4 then
x^2 < 1/x < 2x i.e 9/16 <4/3<1.5
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Re: Correct Order [#permalink] New post   26 May 2006, 14:58
Good one Macca! I see it now.

II works when plugging in large fractions that are close to the value of 1.
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Re: Correct Order [#permalink] New post   26 May 2006, 15:11
macca wrote:
tl372 wrote:
The OA is I & II.

This question is from GmatPrep. When I originally answered the question, my answer was I only.

I still can't figure out how II could be true as well. Any thoughts?


if x=3/4 then
x^2 < 1/x < 2x i.e 9/16 <4/3<1.5


Well done! :-D
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Re: Correct Order [#permalink] New post   07 Jun 2006, 08:20
there's a formal solution for that

1/x < 2x
1/x-2x<0
1/x(1-2x^2)<0

so 1/x>0 (as x is always positive)
we're left with
1-2x^2<0
x^2>1/2

pick this kind of fractions and the inequality will work out
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Re: Correct Order [#permalink] New post   10 Jun 2006, 19:05
Sorry to drag this one out from the past. I would like to add a point. For this kind of questions sometimes it's helpful to draw some curves. For this questions, you can easily get some crucial points, eg. x=0, x=1 and x=2. Draw the curves then you'll see their relative positions really clearly.



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Re: Correct Order [#permalink]
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