Cut-n-pasting from one of my earlier replies...
With these kinds of questions, it is always useful to find the boundaries
We have 3 equations giving us 3 boundaries --
1/x = x^2 implies x = 1
1/x = 2x implies x = 0.7 approx
2x = x^2 implies x = 2
Our 3 boundaries are 0.7, 1 and 2
Take 1 sample for each range
For x<0.7, try x = 0.5
We have x^2 < 2x < 1/x ... bingo, you have (I) satisfied
For x>0.7 and <1, try x = 0.8
we have x^2 < 1/x < 2x ... bingo, you have (II) satisfied
It is easy to see that for all values x > 1, 1/x will give you the
least of the 3. So, (III) can never be true.
So, choose D (i.e. I and II alone)
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jayhawk2001
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thumpin_termis
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Great stuff. This brings up an additional question.
When looking for boundaries, it seems that even more can be found. Using couple of the above boundaries for example,
1/x = 2x
would imply
x^2 = 1/2, so x= positive and negative 0.7
and
2x = x^2
x^2 - 2x = 0
x(x-2) = 0
x = 2 and x=0
At this point, I have too many boundaries to test. Am I doing something incorrectly?
When looking for boundaries, it seems that even more can be found. Using couple of the above boundaries for example,
1/x = 2x
would imply
x^2 = 1/2, so x= positive and negative 0.7
and
2x = x^2
x^2 - 2x = 0
x(x-2) = 0
x = 2 and x=0
At this point, I have too many boundaries to test. Am I doing something incorrectly?
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jayhawk2001
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We are given that x is positive, so -0.7 is automatically out.thumpin_termis wrote:Great stuff. This brings up an additional question.
When looking for boundaries, it seems that even more can be found. Using couple of the above boundaries for example,
1/x = 2x
would imply
x^2 = 1/2, so x= positive and negative 0.7
and
2x = x^2
x^2 - 2x = 0
x(x-2) = 0
x = 2 and x=0
At this point, I have too many boundaries to test. Am I doing something incorrectly?
Also by virtue of x being positive, our lower-limit automatically
starts at 0. So, knowing one of our boundaries is 0.7, we will
have to choose something between 0 and 0.7. So, technically
no additional boundaries
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thumpin_termis
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f2001290
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800GMAT - I generally follow this approach.
Assume (1) to be true => x^2 < 2x < 1/x
Find the value of x using inequalities
x^2 < 1/x => x<1
2x < 1/x => x < 1/sqrt2 => x<0.70
Since x is positive (given in question stem) x can be between (0,0.7)
Take x as 0.5 , then x^2 < 2x < 1/x holds.
Apply this approach to remaining inequalities.
Hope this helps...
Assume (1) to be true => x^2 < 2x < 1/x
Find the value of x using inequalities
x^2 < 1/x => x<1
2x < 1/x => x < 1/sqrt2 => x<0.70
Since x is positive (given in question stem) x can be between (0,0.7)
Take x as 0.5 , then x^2 < 2x < 1/x holds.
Apply this approach to remaining inequalities.
Hope this helps...
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jayhawk2001
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You have a valid concern and is one that I dread the most as well.800GMAT wrote:One concern - if we dont find the boundaries, how would we solve this problem then.....
Basically if we cannot make an educated guess at the boundaries, we
can never be sure.
In this particular problem, there are 3 values to compare -- 1/x, x^2 and 2x.
It should be immediately apparent that we have to try a value of x which is
in the range of 0 - 1. Also, we will try a value of x > 1. So, that will
knock out 2 out of 3 choices. However, to prove II, we need to plug
a value for x in the range 0.7 - 1...unfortunately this will be apparent
only if we know the boundaries...
