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when x=1/2 then 1/x is 2 ,2x is 1 and x^2 is 1/2 ,this satisfies (I) x^2<2x<1/x

when x =3 then 1/x is 1/3, 2x is 6 and x^2 is 9 ,this does not satisfy (II)

when x = 1/10 then 1/x is 10 , 2x is 1/5 and x^2 is 1/100,this again does not satify (II)

Even -ve numbers dont seem to work

when x=-3 then 1/x is -1/3 ,2x is -6 and X^2=9,this does not satisfy (II) x=-1/3 then 1/x is -3 ,2x is -2/3 and x^2=-1/9,this does not satisfy (II)

Can any one give an example which satisfies option (II)

7. 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

(A) None (B) I only (C) III only (D) I and II only (E) I II and III

First note that we are asked "which of the following COULD be the correct ordering" not MUST be. Basically we should determine relationship between \(x\), \(\frac{1}{x}\) and \(x^2\) in three areas: \(0<1<2<\).

\(x>2\)

\(1<x<2\)

\(0<x<1\)

When \(x>2\) --> \(x^2\) is the greatest and no option is offering this, so we know that x<2. If \(1<x<2\) --> \(2x\) is greatest then comes \(x^2\) and no option is offering this.

So, we are left with \(0<x<1\): In this case \(x^2\) is least value, so we are left with:

I. \(x^2<2x<\frac{1}{x}\) --> can \(2x<\frac{1}{x}\)? Can \(\frac{2x^2-1}{x}<0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)

II. \(x^2<\frac{1}{x}<2x\) --> can \(\frac{1}{x}<2x\)? The same here \(\frac{2x^2-1}{x}>0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)

Answer: D.

Second condition: \(x^2<\frac{1}{x}<2x\)

The question is which of the following COULD be the correct ordering not MUST be.

Put \(0.9\) --> \(x^2=0.81\), \(\frac{1}{x}=1.11\), \(2x=1.8\) --> \(0.81<1.11<1.8\). Hence this COULD be the correct ordering.

Each one of these gives you two inequalities. You know that x is positive, so you don't need to worry about the sign changing direction.

(1) x^2<2x<1/x

This means that x^2<2x so divide by x to get x<2. The second one tells you that 2x<1/x which simplifies to x < 1/sqrt(2). These can obviously both be satisfied at the same time, so (1) works.

(2) x^2<1/x<2x

This means that x^2<1/x which gives x^3<1, or x<1. The second half gives you 1/x<2x or 1<2(x^2) or x>1/sqrt(2). So any number that satisfies 1/sqrt(2)<x<1 will work.

(3) 2x<x^2<1/x. The first part gives 2x<x^2 or x>2. The second half gives x^2<1/x or x^3<1 or x<1. Since the regions x>2 and x<1 do not overlap, (3) can not be satisfied.

The Answer choice is (4), 1 and 2 only.
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You can see from the number of "Kudos Given" on my profile that I am generous while giving Kudos. Is it too much to ask for a similar favor???

IMO B. for 0<x<1, only statement I holds. Brunuel, if u put x=1/2: II. II. x^2<1/x<2x >>>>> will not hold true. x^2 = 1/4, 1/x=2 and 2x=1 then this expression will not hold. 1/4<2<1 [Incorrect]

If x=1/9 then: x^2=1/81, 1/x=9 and 2x=2/9 1/81<9<2/9 [Incorrect]

Let's check the III option for above values: III. 2x<x^2<1/x For x=1/2: 1<1/4<2 [Incorrect] For x=1/9: 2/9<1/81<9 [Incorrect]

So, B should be the correct answer. Please check.

OA IS D.

Algebraic approach is given in my solution. Here is number picking:

I. \(x^2<2x<\frac{1}{x}\) --> \(x=\frac{1}{2}\) --> \(x^2=\frac{1}{4}\), \(2x=1\), \(\frac{1}{x}=2\) --> \(\frac{1}{4}<1<2\). Hence this COULD be the correct ordering.

II. \(x^2<\frac{1}{x}<2x\) --> \(x=0.9\) --> \(x^2=0.81\), \(\frac{1}{x}=1.11\), \(2x=1.8\) --> \(0.81<1.11<1.8\). Hence this COULD be the correct ordering.

III. \(2x<x^2<\frac{1}{x}\) --> \(x^2\) to be more than \(2x\), \(x\) must be more than 2 (for positive \(x-es\)). But if \(x>2\), then \(\frac{1}{x}\) is the least value from these three and can not be more than \(2x\) and \(x^2\). So III can not be true.

Thus I and II could be correct ordering and III can not.

(a) none (b) I only (c) III only (d) I and II (e) I, II and III

Let's look at this question logically. There will be some key takeaways here so don't focus on the question and the (long) solution. Focus on the logic.

First of all, we are just dealing with positives so life is simpler. To compare two terms e.g. \(x^2\) and \(2x\), we should focus on the points where they are equal. \(x^2 = 2x\) holds when \(x = 2\). When \(x < 2, x^2 < 2x\) When \(x > 2, x^2 > 2x\)

Similarly \(1/x = x^2\) when \(x = 1\) When \(x < 1, 1/x > x^2\). When \(x > 1, 1/x < x^2\)

Going on, \(1/x = 2x\) when \(x = 1/\sqrt{2}\) When \(x < 1/\sqrt{2}, 1/x > 2x\) When \(x > 1/\sqrt{2}, 1/x < 2x\)

So now you know that: If \(x < 1/\sqrt{2}\), \(1/x > 2x, 1/x > x^2\) and \(x^2 < 2x\) So \(x^2 < 2x < 1/x\) is possible.

If \(1/\sqrt{2} < x < 1\) \(1/x < 2x, 1/x > x^2\) So \(x^2 < 1/x < 2x\) is possible.

If \(1 < x < 2\) \(1/x < 2x, 1/x < x^2, x^2 < 2x\) So \(1/x < x^2 < 2x\) is possible.

If \(x > 2\) \(1/x < 2x, 1/x < x^2, x^2 > 2x\) So \(1/x < 2x < x^2\) is possible.

For no positive values of x is the third relation possible.

I still dont get why D is the answer. There is a flaw in your reasoning for statement II

if x = 1/2, then 1/1/2 = 2 which is greater than 2(1/2)

can someone please use figures (fractions or decimals) to demonstrate this. Thanks

x^2 < 1/x < 2x....If I translate this into two parts, it tell me that x^3 < 1 and 2x^2 > 1...that means, x < 1 and x > 1/1.414 That means approximately, x should be between 0.7 and 1.

Take a value of x = 0.9. Here, x^2 = 0.81, 1/x = 10/9 and 2x = 1.8 and these values satisfy the inequality.

Re: If x is positive, which of the following could be correct [#permalink]

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29 Jul 2012, 01:37

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Please, refer to the attached drawing, in which the three graphs \(y=1/x,\) \(y=2x,\) and \(y=x^2\) are depicted for \(x>0\). The exact values for A, B, and C can be worked out, but they are not important to establish the order of the three algebraic expressions.

So, the correct orderings are: If \(x\) between 0 and A: \(x^2<2x<1/x\) If \(x\) between A and B: \(x^2<1/x<2x\) If \(x\) between B and C: \(1/x<x^2<2x\) If \(x\) greater than C: \(1/x<2x<x^2\)

We can see that only the first two of the above options are listed as answers (I and II).

Answer: D.

Attachments

3Graphs.jpg [ 15.15 KiB | Viewed 6391 times ]

_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

IMO the clearest way to solve the problem is to plot the three curves / line: - x^2 is a parabola tangent to the origin and passes through point (1,1) and (2,4) - 1/x is the hyperbola tangent to the x-axis and the y-axis that passes through (1,1) - 2x is the line that passes through the origin and (1,2)

The intersections among the three curves define the 4 possible cases: for x<1/\sqrt{2}: x^2 < 2x < 1/x for 1/\sqrt{2}<x<1: x^2 < 1/x < 2x for 1<x<2: 1/x < x^2 < 2x for x>2: 1/x < 2x < x^2

It's much easier when you draw the three curves and notice the intersection points (but don't know at this point how to include an image). It did not occur to me to draw them during the test and mistankenly chose answer B. With retrospect, the only way I could have figured out the four zones was by drawing.

Re: If x is positive, which of the following could be correct [#permalink]

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27 Feb 2012, 03:23

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@imhimanshu

In this kind of questions, it is best to check with numbers. Also, when fractions are involved in the question, test with 3 values: < 0.5; = 0.5 ; > 0.5

This distribution definitely helps you.
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I marked I only in Gmatprep by figuring out that x lies between 0 and 1. But how does one use mathematic logic to try out the value 0.9 since plugging 0.9 and 0.5 each gives different results

7. 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

In all the expression the common expression is x^2 < 1/x which is possible when 0<x<1 Now take expression x^2<2x x^2-2x < 0 which is possible if x< 0 & x > 2 OR x >0 and x<2 .. so using the above expression 0<x<1 this expression also could be true. now take expression x^2>2x which is possible if x<0 & x<2 OR x>0 & x>2 .. this doesnt fall under the condition of 0<x<1 .. so this expression could not be true.. III is out for this case alone now take expression 2x<1/x x^2 < 1/2 .. the range of values of x is a subset of the range 0<x<1 .. so this expression could be true for some values of x. I can be the correct ordering now take expression 1/x<2x x^2>1/2 .. the range of values of x has some values in the range 0<x<1 .. so this expression could be true for some vaues of x. II can be the correct ordering.

Re: If x is positive, which of the following could be correct [#permalink]

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27 Feb 2012, 08:23

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The answer would be D

(I) this is clear and is easily deciphered ...

but for (II) x^2 < 1/x < 2x

This means that x^2<1/x which gives x^3<1, or x<1. The second half gives you 1/x<2x or 1<2(x^2) or x>1/sqrt(2). So any number that satisfies 1/sqrt(2)<x<1.. hence (II) COULD also be true for some values..

The question asked "what could be the correct ordering" means it asked for the possibilities. What is question asked "what must be the correct ordering" ? In that case would we be required to choose an option which is true for all scenarios ?

The ordering will be different for different values of x so the question cannot ask for a single correct ordering.
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Hello Bunuel, Can you show us graphical approach to this question. I was able to draw the graph for all three equations and intersection points, however, I was not able to negate the third ordering. Would you please help me out.

Thanks imhimanshu P.S - How can I post graphs here.

Here is the graph:

Attachment:

Ques3.jpg [ 11.4 KiB | Viewed 4563 times ]

III. 2x < x^2 < 1/x

For 2x to be less than x^2, the graph of 2x should lie below the graph of x^2. This happens when the graph of 2x is the red line. For x^2 to be less than 1/x at the same time, the graph of x^2 should lie below the graph of 1/x in the region of the red line. But in the region of the red line, the graph of x^2 is never below the graph of 1/x. It will never be because graph of 1/x is going down toward y = 0 while graph of x^2 is going up toward y = infinity. Hence this inequality will not hold for any region.
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However, it didn occur to me that I must look something like 0.9. Request experts to help me understand the logic behind picking such numbers. Have my actual GMAT in 10 days, any help would be immensely valuable!

pick one number like 1/2 you will see that it is not correcy but dont rush pick another number which close to 1 but smaller then one like 9/10 and tried

you will see it is right.

so the answer is D

with that kind of guestions pick one less then 1/2 and greater then 1/2 less then 1.
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As we can see that question used term "COULD", so even if any of the choice satisfy the requirement it will be an answer. Now let look at the choices:

Choice I: Just put x=0.5, we can see that it satisfy the condition Choice II: Put x=2, we can see that it also satisfy the condition Choice III: For any value of x, it does not satisfy the condition.

So we can easily figure out that the answer is D.

Lemme me know if any one have any question on this.

So, we are left with \(0<x<1\): In this case \(x^2\) is least value, so we are left with:

I. \(x^2<2x<\frac{1}{x}\) --> can \(2x<\frac{1}{x}\)? Can \(\frac{2x^2-1}{x}<0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)

II. \(x^2<\frac{1}{x}<2x\) --> can \(\frac{1}{x}<2x\)? The same here \(\frac{2x^2-1}{x}>0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\).

Great explanation Bunuel, but for curiosity purpose, as I understand \(2x^2-1\) should be negative for this equation \(\frac{2x^2-1}{x}<0\) to be true. However if I algebraically find the values for which \(2x^2-1\) is negative, then on plugging those values in \(x^2<\frac{1}{x}<2x\) I do not find that the equation satisfies. Instead it is the value for which \(2x^2-1\) is positive, that I find the end quation satifies .... why is that ?
_________________

Please give me kudos, if you like the above post. Thanks.

So, we are left with \(0<x<1\): In this case \(x^2\) is least value, so we are left with:

I. \(x^2<2x<\frac{1}{x}\) --> can \(2x<\frac{1}{x}\)? Can \(\frac{2x^2-1}{x}<0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)

II. \(x^2<\frac{1}{x}<2x\) --> can \(\frac{1}{x}<2x\)? The same here \(\frac{2x^2-1}{x}>0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\).

Great explanation Bunuel, but for curiosity purpose, as I understand \(2x^2-1\) should be negative for this equation \(\frac{2x^2-1}{x}<0\) to be true. However if I algebraically find the values for which \(2x^2-1\) is negative, then on plugging those values in \(x^2<\frac{1}{x}<2x\) I do not find that the equation satisfies. Instead it is the value for which \(2x^2-1\) is positive, that I find the end quation satifies .... why is that ?

You are mixing I and II. If you find the values of \(x\) from the range \(0<x<1\) for which \(2x^2-1\) is negative then \(x^2<2x<\frac{1}{x}\) will hold true (not \(x^2<\frac{1}{x}<2x\)).

Below is number plugging method:

I. \(x^2<2x<\frac{1}{x}\) --> \(x=\frac{1}{2}\) --> \(x^2=\frac{1}{4}\), \(2x=1\), \(\frac{1}{x}=2\) --> \(\frac{1}{4}<1<2\). Hence this COULD be the correct ordering.

II. \(x^2<\frac{1}{x}<2x\) --> \(x=0.9\) --> \(x^2=0.81\), \(\frac{1}{x}=1.11\), \(2x=1.8\) --> \(0.81<1.11<1.8\). Hence this COULD be the correct ordering.

III. \(2x<x^2<\frac{1}{x}\) --> \(x^2\) to be more than \(2x\), \(x\) must be more than 2 (for positive \(x-es\)). But if \(x>2\), then \(\frac{1}{x}\) is the least value from these three and can not be more than \(2x\) and \(x^2\). So III can not be true.

Thus I and II could be correct ordering and III can not.

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