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x is positive

Expert replies
Source: — Problem Solving |

by caprion » Mon May 25, 2009 10:47 pm
I think the question just asks for all the possible combinations...

The key could be based on values of x

For for x>1
always x2>1/x

and for all x where 0<x<1
always 1/x > x2 and 2x > x2

based on this condition option III is invalid.
2x<x2<1/x

2x is always greater then x2 when x2<1/x.

So the answer should be I and II only.
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by gmat740 » Tue May 26, 2009 11:59 am
There are two possibilities:

when x>1
Plug in x =3 to check

1/x< 2x <x^2

when 0<x<1

1/x > 2x> x^2
Plug in x =0.5 to check.

But first statement is not given in the option

correct me where am I wrong.

Thanks
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by crackgmat007 » Tue May 26, 2009 12:13 pm
How did you compute to conclude that statement II could be correct? I tried plug in, but doesnt seem to work.
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by tdadic84 » Tue May 26, 2009 12:25 pm
if you put 0.9 into statement 2 ..it is true

0.81<1.1<1.8

hence 1 and 2....
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by crackgmat007 » Tue May 26, 2009 12:44 pm
I tried with x=1/10

x^2 = 1/100
1/x = 10
2x = 1/5

Looks like it varies with which number one picks...
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by rs2010 » Tue May 26, 2009 2:58 pm
We need to test few scenarios



Write question as

1/x, 2x and x^2 --- 1, 2x^2, x^3

Now test your conditions
x<.7

x^3<2x^2<1 while for values .7<=x<1.25
x^3<1<2x^2

and when 1.25<=x<2
1<x^3<2x^2
at x=2

1<x^3=2x^2

and x>2

1<2x^2<x^3
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by caprion » Tue May 26, 2009 5:54 pm
The trick i think is in the question..which says...

What all combinations could be correct..

We found valid values where I and II could be correct. But i proved earlier that III option can never be correct.

Please comment
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