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Algebra: If x is positive

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by saurav.jha » Mon Mar 11, 2013 6:07 am
Hi I think the signs have been reversed.
The answer shall be x^2 > 2x > 1/x.
There are many ways to attack this problem.
One approach is if graphs of the 3 expressions are taken
y=X^2(parabola)
y=2x(straight line)
y=1/x(hyperbola)
The nature of the slopes shall be in decreasing order of the expression ie x^2 > 2x > 1/x.
The nature of first differentials shall also be 2x , 2 and -1/x^2.
hence the answer
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by psm12se » Mon Mar 11, 2013 6:25 am
Nope, its a GMAT prep question, even i wondered if the signs are reversed.
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by Anju@Gurome » Mon Mar 11, 2013 6:36 am
psm12se wrote:If x is positive, which of the following could be the correct ordering of 1/x, 2x, and x^2?
It is mentioned that x is positive NOT positive integer.
The most easiest approach will be to draw the graphs of 1/x, 2x, and x² as follows...
Image

We can see that the following options are possible...
  • Red < Blue < Green ---> x² < 2x < 1/x ----> Option I
    Red < Green < Blue ---> x² < 1/x < 2x ----> Option II
    Green < Red < Blue ---> 1/x < x² < 2x
    Green < Blue < Red ---> 1/x < 2x < x²
Last edited by Anju@Gurome on Mon Mar 11, 2013 7:34 am, edited 1 time in total.
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by GMATGuruNY » Mon Mar 11, 2013 7:13 am
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 right and left of each critical point, the value of one expression must be greater than the value of another.

To determine which answer choices are possible, plug in one value to the left and one value to the right of each critical point.

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

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 B.

In III, the largest value listed is 1/x.
For 1/x to be the largest value, x would have to be a fraction.
Having tried a fraction on each side of the critical point of 5/7, we know that there is no way that III could be true.

The correct answer is D.
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by razaul karim » Sun Apr 14, 2013 10:55 pm
Dear Anju Agarwal (Quant Expert)
Can you please explain the relation between slope (increasin or decreasing)and ordering real number?

The nature of the slopes shall be in decreasing order of the expression ie x^2 > 2x > 1/x.
The nature of first differentials shall also be 2x , 2 and -1/x^2.
Then how the followings are answer--------
x² < 1/x < 2x
x² < 2x < 1/x
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by Anju@Gurome » Sun Apr 14, 2013 11:02 pm
razaul karim wrote:Dear Anju Agarwal (Quant Expert)
Can you please explain the relation between slope (increasin or decreasing)and ordering real number?
Slope of x² and 1/x are out of scope for GMAT.
Also slope is a different thing and difficult to correlate to the values of the function attain.

What I have described is simply by drawing the graphs of 1/x, 2x, and x², we can easily check their order by identifying whether one graph is above the other or not etc.

Hope that helps.
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by saon » Sun Apr 14, 2013 11:27 pm
psm12se 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
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