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If x is positive, which of the following could be the

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Director
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If x is positive, which of the following could be the [#permalink] New post 09 Apr 2006, 15:39
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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
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Re: DS + numbers [#permalink] New post 09 Apr 2006, 16:33
bewakoof wrote:
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.

So I and III.
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 [#permalink] New post 09 Apr 2006, 18:01
sm176811 wrote:
I and II

OA?


what set of numbers satisfied II?
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 [#permalink] New post 09 Apr 2006, 18:15
bewakoof wrote:
prof, check again


ok, now i go with I, II, and III.

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.
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 [#permalink] New post 09 Apr 2006, 18:20
Professor wrote:
bewakoof wrote:
prof, check again


ok, now i go with I, II, and III.

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
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 [#permalink] New post 09 Apr 2006, 18:42
Whats OA... I can provide the nos for II.. will post back l8r :P
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 [#permalink] New post 09 Apr 2006, 20:01
bewakoof wrote:
Professor wrote:
bewakoof wrote:
prof, check again


ok, now i go with I, II, and III.

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.

i and ii are correct.....
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 [#permalink] New post 09 Apr 2006, 20:09
We could solve the set in the following way:

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.
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 [#permalink] New post 10 Apr 2006, 01:07
I. x ^2 < 2x < 1/x
passes If x = 1/2
but fails if x = 2

ii. x ^2 < 1/x < 2x
agree with Zooroopa on this

iii. 2x < x ^2 < 1/x
This is also not possible if x = 3/2
3<>2.5<0.66
if x = 5
10<25<>0.2

so I will go with I and II
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 [#permalink] New post 10 Apr 2006, 17:17
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.
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 [#permalink] New post 10 Apr 2006, 17:48
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


II is possible
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 [#permalink] New post 10 Apr 2006, 18:27
Choosing x = 1 and x = 2 are going to violate both I and II.

For x = 1, we would get

I: 1 < 2 < 1 (this is meaningless)
II: 1 < 1 < 2 (this is meaningless as well)

Similar pitfalls exist if you choose x = 2.
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 [#permalink] New post 11 Apr 2006, 06:02
choose 3 numbers:
a)one whole number > 2 (aka 3!)
b)fraction < 1 (3/4)
c)fraction > 1 (3/2)
  [#permalink] 11 Apr 2006, 06:02
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