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ordering

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VP
Joined: 18 May 2008
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03 Feb 2009, 03:25
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SVP
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03 Feb 2009, 05:12

to make it easy.. multiple with x throug all options
I x^2 < 2x < 1/x
--> x^3<2x^2 <1
substitiute x=1/2 1/8<1/2<1
II. x^2 < 1/x <2x

--> x^3 <1 <2x^2
here assume value such that 2x^2>1 ..
X= 1/sqrt(2) 2x^2=1
above equation could be answer when x<1 and x>1/sqrt(2)

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

-->2x^2 <x^3 <1
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Joined: 17 Dec 2008
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03 Feb 2009, 07:30
I and II could be correct.

lesson:
simplify qn stem.
check for fractions.
Re: ordering   [#permalink] 03 Feb 2009, 07:30
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