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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

i. x^2 < 2x < 1/x is a correct order if x <0.5
ii. x^2 < 1/x < 2x is not a correct order no matter what the value of x is.
iii. 2x < x^2 < 1/x is a correctone if 2>x >1.

i. x^2 < 2x < 1/x is a correct order if x <0.5
ii. x^2 < 1/x < 2x is also a correct order if 0.70 < x < 1.00.
iii. 2x < x^2 < 1/x is a correctone if 2>x >1.

i. x^2 < 2x < 1/x is a correct order if x <0.5
ii. x^2 < 1/x < 2x is also a correct order if 0.70 < x < 1.00.
iii. 2x < x^2 < 1/x is a correctone if 2>x >1.[/quote]

thats not right
_________________

Success is my only option, failure is not -- Eminem

i. x^2 < 2x < 1/x is a correct order if x <0.5 ii. x^2 < 1/x < 2x is also a correct order if 0.70 < x < 1.00. iii. 2x < x^2 < 1/x is a correctone if 2>x >1.[/quote]

thats not right

oh, sorry buddy. i overlooked as to the reverse order.

Case I: Solve for x^^2 < 2x and 2x < 1/x.
====
You would get x < 2 and x < 1/sqrt(2).
So, x < 1/sqrt(2) satisfies the set.

Case II: Solve for x^^2 < 1/x and 1/x < 2x.
=====
You would get x^^3 < 1 and 2x^^2 > 1
So, for all real x, x <1 and x > 1/sqrt(2).
So, 1/sqrt(2) < x < 1 satsifies the set.

Case III: Solve for 2x < x^^2 and x^^2 <1/x.
=====
You would get x >2 and x^^3 < 1.
For all real x, you must have x < 1 (from the second equation).
Now x < 1 and x >2 are incompatible.
So III does not hold good for all real x.

So I & II are correct.
Caveat: =====

Be cautious not to select values such as x =1, x = 2.
For these values, some of the functions yield the same value e.g. for x =1, x^^2 = 1/x = 1, for x = 2, 2x = x^^2 = 4.
So you would not get the desired relationships.
_________________

Whats the trick here to save time?
Should we try different values like 1/2, 1, 3/2, 4 for x and then figure out what satisfies.
I guess these kind of questions are included so that we spend maximum time on them and end up in a mess.