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x^2, (1/x), 2x

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x^2, (1/x), 2x

by ashforgmat » Thu May 13, 2010 9:03 pm
139) If n is positive, which of the following could be the correct ordering of 1/x, 2x, and x2?
I : x2 < 2x < 1/x
II: x2 < 1/x < 2x
III: 2x < x2 < 1/x
a. None
b. I
c. III
d. I and II
e. I, II, and III


Hi Please can someone explian how II can be true.
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by krazy800 » Thu May 13, 2010 9:33 pm
X is positive (need not be an integer unless mentioned)

so X can take decimal values

Take X = .9

X^2 = .81
1/X= 1.11
2X = 1.8

Therefore, X^2< 1/x< 2x

HTH!!
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by gmatjedi » Fri May 14, 2010 2:34 pm
If x is a fraction then I : x2 < 2x < 1/x
is correct
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by gmatmachoman » Fri May 14, 2010 11:45 pm
gmatjedi wrote:If x is a fraction then I : x2 < 2x < 1/x
is correct
If X=1..None!! I am still looking for n but i see only X
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by M811 » Sat May 15, 2010 12:40 am
krazy800 wrote:X is positive (need not be an integer unless mentioned)

so X can take decimal values

Take X = .9

X^2 = .81
1/X= 1.11
2X = 1.8

Therefore, X^2< 1/x< 2x

HTH!!


if X= .5
x^2 = 0.25
1/X = 2
2X = 1

in this case X^2 < 2X < 1/X
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by tanviet » Wed Aug 03, 2011 6:34 pm
be very carefull with gmatprep questions

because x>0 we can multiple the inequalities with x and we have

for the first inequality

x^3<1<2x^2

now , we pick number. it is much easier.
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by gmatboost » Wed Aug 03, 2011 10:23 pm
Plugging in numbers works here if you happen to pick the right numbers. Often on these questions it is enough to try, say, 1/2 and 2. Why isn't that the case here?

The answer is that the "boundary points," by which I mean the values that make at least two of these expressions equal to each other, are a bit unusual.

If x^2 = 2x, we get x^2 - 2x = 0, x(x-2) = 0, so x = 0 or x = 2
If x^2 = 1/x, we get x^3 = 1, so x = 1
If 2x = 1/x, we get 2x^2 = 1, or x^2 = 1/2, so x = 1/sqrt(2) = sqrt(2)/2 approx = 1.4/2 = 0.7

So, the different boundary points, in increasing order, are 0, 1/sqrt(2), 1, 2

In order to make sure we catch all possible cases, we need to try numbers in between each set of boundary points. So, we could try 0.5, 0.8, 1.5, and 3

They will give us...
0.5: x^2 < 2x < 1/x
0.8: x^2 < 1/x < 2x
1.5: 1/x < x^2 < 2x
3: 1/x < 2x < x^2

So, there are 4 possibilities, even though only 2 are among the answer choices. Is it possible to get this right without going through all this? Of course. If you happened to pick both 0.9 and 0.5, you could have seen that both I and II work.

But I would say that it's not obvious that you need to try both a number below 0.7 and a number between 0.7 and 1. And it's definitely not obvious that different numbers between 0 and 1 will give you different answers (as 0.5 and 0.8 do).
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by shashank.ism » Thu Aug 04, 2011 12:08 am
ashforgmat wrote:139) If n is positive, which of the following could be the correct ordering of 1/x, 2x, and x2?
I : x2 < 2x < 1/x
II: x2 < 1/x < 2x
III: 2x < x2 < 1/x
a. None
b. I
c. III
d. I and II
e. I, II, and III


Hi Please can someone explian how II can be true.
So My 1000th post and I am GMAT Titan Now.. :)

Lets take x as positive and not n(as n here signifies nothing).
so first we should calculate the boundary points and then move forward to check.

1) x2 = 2x --> x(x-2) = 0 --> x=2 (taking x as +ve).
thus for 0<x<2, 2x>x2 & for x>2 x2>2x & for for x=2 x2=2x.

2)2x = 1/x --> 2x2-1 = 0 --> x2-1/2 = 0 --> x = 1/√2 (taking x as +ve).
thus for 0<x<2, 2x>x2 & for x>2 x2>2x & for for x=2 x2=2x.

3)x2 = 1/x --> x3-1 = 0 --> x = 1(taking x as +ve).
thus for 0<x<1 1/x>x2 & for x>1 x2>1/x & for x=1 x2=1/x.
Last edited by shashank.ism on Thu Aug 04, 2011 2:39 am, edited 1 time in total.
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by GMATGuruNY » Thu Aug 04, 2011 2:13 am
I posted a solution here:

https://www.beatthegmat.com/need-explanation-t85192.html
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by ozlemmetje » Sun Apr 06, 2014 3:13 am
Hi!

How can we know during the exam whether we have to look for critical points or not? I mean at first glance I would just pick 1/2 which and try and that leads the 1. answer to be correct. So I would just go for answer B- only 1.

Thanks


gmatboost wrote:Plugging in numbers works here if you happen to pick the right numbers. Often on these questions it is enough to try, say, 1/2 and 2. Why isn't that the case here?

The answer is that the "boundary points," by which I mean the values that make at least two of these expressions equal to each other, are a bit unusual.

If x^2 = 2x, we get x^2 - 2x = 0, x(x-2) = 0, so x = 0 or x = 2
If x^2 = 1/x, we get x^3 = 1, so x = 1
If 2x = 1/x, we get 2x^2 = 1, or x^2 = 1/2, so x = 1/sqrt(2) = sqrt(2)/2 approx = 1.4/2 = 0.7

So, the different boundary points, in increasing order, are 0, 1/sqrt(2), 1, 2

In order to make sure we catch all possible cases, we need to try numbers in between each set of boundary points. So, we could try 0.5, 0.8, 1.5, and 3

They will give us...
0.5: x^2 < 2x < 1/x
0.8: x^2 < 1/x < 2x
1.5: 1/x < x^2 < 2x
3: 1/x < 2x < x^2

So, there are 4 possibilities, even though only 2 are among the answer choices. Is it possible to get this right without going through all this? Of course. If you happened to pick both 0.9 and 0.5, you could have seen that both I and II work.

But I would say that it's not obvious that you need to try both a number below 0.7 and a number between 0.7 and 1. And it's definitely not obvious that different numbers between 0 and 1 will give you different answers (as 0.5 and 0.8 do).
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by ozlemmetje » Sun Apr 06, 2014 3:14 am
Hi!

How can we know during the exam whether we have to look for critical points or not? I mean at first glance I would just pick 1/2 which is the easiest and try and that works for 1. answer.So I would just go for answer B- only 1.

Is there any general rule for inequalties to look for critical points? When should we bother about them?

Thanks


gmatboost wrote:Plugging in numbers works here if you happen to pick the right numbers. Often on these questions it is enough to try, say, 1/2 and 2. Why isn't that the case here?

The answer is that the "boundary points," by which I mean the values that make at least two of these expressions equal to each other, are a bit unusual.

If x^2 = 2x, we get x^2 - 2x = 0, x(x-2) = 0, so x = 0 or x = 2
If x^2 = 1/x, we get x^3 = 1, so x = 1
If 2x = 1/x, we get 2x^2 = 1, or x^2 = 1/2, so x = 1/sqrt(2) = sqrt(2)/2 approx = 1.4/2 = 0.7

So, the different boundary points, in increasing order, are 0, 1/sqrt(2), 1, 2

In order to make sure we catch all possible cases, we need to try numbers in between each set of boundary points. So, we could try 0.5, 0.8, 1.5, and 3

They will give us...
0.5: x^2 < 2x < 1/x
0.8: x^2 < 1/x < 2x
1.5: 1/x < x^2 < 2x
3: 1/x < 2x < x^2

So, there are 4 possibilities, even though only 2 are among the answer choices. Is it possible to get this right without going through all this? Of course. If you happened to pick both 0.9 and 0.5, you could have seen that both I and II work.

But I would say that it's not obvious that you need to try both a number below 0.7 and a number between 0.7 and 1. And it's definitely not obvious that different numbers between 0 and 1 will give you different answers (as 0.5 and 0.8 do).
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by GMATGuruNY » Sun Apr 06, 2014 3:38 am
ozlemmetje wrote:Hi!

How can we know during the exam whether we have to look for critical points or not? I mean at first glance I would just pick 1/2 which and try and that leads the 1. answer to be correct. So I would just go for answer B- only 1.

Thanks
Since the question question asks which of the statements COULD be true, we have to account for ANY CASE that could make I, II or III true.
So that we don't miss a viable case, I would calculate the critical points.
Here again is the problem, along with my solution:
If x is positive, which of the following could be the correct ordering of 1/x, 2x, and x²?

I. x² < 2x < 1/x
II. x² < 1/x < 2x
III. 2x < x² < 1/x

a. None
b. I
c. III
d. I and II
e. I, II, and III
Determine the CRITICAL POINTS by setting the expressions equal to each other:

1/x = 2x
2x² = 1
x² = 1/2
x = √(1/2) = 1/√2 ≈ 1/1.4 ≈ 10/14 ≈ 5/7.

1/x = x²
x^3 = 1
x = 1.

2x = x²
x=2
(We can divide by x because x>0.)

The critical points are x=5/7, x=1, x=2.
These critical points indicate where two of the expressions are EQUAL.
Thus, to the left and right of each critical point, the value of one expression must be GREATER than the value of another.

To determine which of I, II and II could be true, plug in values to the left and right of each critical point.
Start with the range that many test-takers will fail to consider: 5/7 < x < 1.

5/7 < x < 1:
If x = 3/4, then:
1/x = 4/3.
x² = 9/16.
2x = 3/2.
Since x² < 1/x < 2x, we know that II could be true.
Eliminate A, B and C.

In statement III, 2x<x², which implies that 2<x.
But if x>2, then 1/x cannot be the greatest of the three values.
Thus, III is not possible.
Eliminate E.

The correct answer is D.
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