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If x is positive, which of the following could be correct ordering of

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

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If x is positive, which of the following could be correct ordering of \(\frac{1}{x}\), \(2x\), and \(x^2\)?

I. \(x^2 < 2x < \frac{1}{x}\)

II. \(x^2 < \frac{1}{x} < 2x\)

III. \(2x < x^2 < \frac{1}{x}\)


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

Originally posted by Vavali on 03 Oct 2008, 14:37.
Last edited by Bunuel on 18 Jun 2017, 02:09, edited 5 times in total.
Renamed the topic and edited the question.
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If x is positive, which of the following could be correct ordering of  [#permalink]

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New post 24 Jan 2010, 01:47
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gautamsubrahmanyam wrote:
I am not sure how the OA is D

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)


If x is positive, which of the following could be the correct ordering of 1/x, 2x and x^2 ?

I. \(x^2<2x<\frac{1}{x}\)

II. \(x^2<\frac{1}{x}<2x\)

III. \(2x<x^2<\frac{1}{x}\)

(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.

Hope it's clear.
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Re: If x is positive, which of the following could be correct ordering of  [#permalink]

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New post 07 Feb 2012, 13:51
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Copied from an other forum. Thought it might help someone. This is a great explanation...

--------------------------------------------------

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

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New post 04 Oct 2008, 23:18
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Vavali wrote:
If x is positive, which of the following could be 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
(e) I, II and III


could be correct ordering

So if we can find any example that satisfy the inequation, that statement will be correct

(I) x = 0.1 => 0.01 < 0.2 < 10
(II) x= 1/2 => 1/4 < 1/2 < 1

(III)
2x < x^2 <=> x ( 2 -x) < 0, x > 0 then x > 2

with x > 2 ==> x^2 < 1/x <=> x^3 < 1 <=> x < 1

So (III) can't happen

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

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New post 05 Oct 2008, 13:22
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Vavali wrote:
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.
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If x is positive, which of the following could be correct ordering of  [#permalink]

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New post 18 Apr 2010, 10:45
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If x is positive, which of the following could be correct ordering of \(\frac{1}{x}\), \(2x\), and \(x^2\)?

I. \(x^2 < 2x < \frac{1}{x}\)

II. \(x^2 < \frac{1}{x} < 2x\)

III. \(2x < x^2 < \frac{1}{x}\)

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


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.

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

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New post 03 Oct 2010, 14:28
Bunuel wrote:
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 ?
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Re: If x is positive, which of the following could be correct ordering of  [#permalink]

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New post 03 Oct 2010, 14:47
devashish wrote:
Bunuel wrote:
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.

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

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New post 26 Dec 2011, 11:03
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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.

Hope this helps :)
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Re: If x is positive, which of the following could be correct ordering of  [#permalink]

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New post 07 Feb 2012, 23:49
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Vavali wrote:
If x is positive, which of the following could be correct ordering of \(\frac{1}{x}\), \(2x\), and \(x^2\)?

(I) \(x^2 < 2x < \frac{1}{x}\)
(II) \(x^2 < \frac{1}{x} < 2x\)
(III) \(2x < x^2 < \frac{1}{x}\)

(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.


*Edited
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If x is positive, which of the following could be correct ordering of  [#permalink]

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New post 27 Feb 2012, 02:52
Hi Bunnel,

Could you please provide a reasoning to the below text... how did you find the range...Pls help

I am not sure how the OA is D
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Re: If x is positive, which of the following could be correct ordering of  [#permalink]

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New post 27 Feb 2012, 03:07
imhimanshu wrote:
Hi Bunnel,

Could you please provide a reasoning to the below text... how did you find the range...Pls help


The reasoning is that in these ranges x (2x), 1/x and x^2 are ordered differently:

For \(x>2\) --> \(x^2\) has the largest value. Since no option offers this we know that \(x\) cannot be more that 2;
For \(1<x<2\) --> \(2x\) has the largest value, then comes \(x^2\). Since no option offers this we know that \(x\) cannot be from this range either;

So, we are left with last range: \(0<x<1\). In this case \(x^2\) has the least value. Options, I and II offer this, so we should concentrate on them and test the values of x from 0 to 1.

Hope it's clear.
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Re: If x is positive, which of the following could be correct ordering of  [#permalink]

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

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

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New post 27 Jun 2012, 14:11
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Bunuel, please check my reasoning about statement III:

Statement III says \(2x < x^2 < \frac{1}{x}\)

So, we analyze part by part of this compound inequality:

a) \(2x < x^2\)
then, \(x>2\)

b) \(x^2 < \frac{1}{x}\)

then,\(x^3<1\) ==> \(x < 1\)

So, we have \(x>2\) and \(x<1\).
That's impossible! if \(x>2\), x cannot be less than 1 at the same time.
Statement III could not be correct!

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

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New post 28 Jun 2012, 02:15
metallicafan wrote:
Bunuel, please check my reasoning about statement III:

Statement III says \(2x < x^2 < \frac{1}{x}\)

So, we analyze part by part of this compound inequality:

a) \(2x < x^2\)
then, \(x>2\)

b) \(x^2 < \frac{1}{x}\)

then,\(x^3<1\) ==> \(x < 1\)

So, we have \(x>2\) and \(x<1\).
That's impossible! if \(x>2\), x cannot be less than 1 at the same time.
Statement III could not be correct!

Please, confirm.


Yes, your reasoning for option III is correct.
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Re: If x is positive, which of the following could be correct ordering of  [#permalink]

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New post 29 Jul 2012, 01:37
5
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.
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Re: If x is positive, which of the following could be correct ordering of  [#permalink]

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New post 29 Jul 2012, 22:58
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smartmanav wrote:
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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Re: If x is positive, which of the following could be correct ordering of  [#permalink]

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New post 11 May 2013, 02:05
Hey Karishma,

I feel below highlighted part is not correct. Please check. If I am wrong, please explain. thanks.

VeritasPrepKarishma wrote:
Vavali wrote:
If x is positive, which of the following could be correct ordering of \(\frac{1}{x}\), \(2x\), and \(x^2\)?

(I) \(x^2 < 2x < \frac{1}{x}\)
(II) \(x^2 < \frac{1}{x} < 2x\)
(III) \(2x < x^2 < \frac{1}{x}\)

(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 \(x > 1, 1/x > x^2\) Are you sure this is correct? I think.. we can use x=4 here, then 1/4>16 .. which is not correct.

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 \(x > 1\)
\(1/x < 2x, 1/x > x^2\)
So \(x^2 < 1/x < 2x\) is possible. (Same as above)

For no positive values of x is the third relation possible.
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Re: If x is positive, which of the following could be correct ordering of  [#permalink]

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New post 11 May 2013, 03:12
yogirb8801 wrote:
Hey Karishma,

I feel below highlighted part is not correct. Please check. If I am wrong, please explain. thanks.

VeritasPrepKarishma wrote:
Vavali wrote:
If x is positive, which of the following could be correct ordering of \(\frac{1}{x}\), \(2x\), and \(x^2\)?

(I) \(x^2 < 2x < \frac{1}{x}\)
(II) \(x^2 < \frac{1}{x} < 2x\)
(III) \(2x < x^2 < \frac{1}{x}\)

(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 \(x > 1, 1/x > x^2\) Are you sure this is correct? I think.. we can use x=4 here, then 1/4>16 .. which is not correct.

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 \(x > 1\)
\(1/x < 2x, 1/x > x^2\)
So \(x^2 < 1/x < 2x\) is possible. (Same as above)

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


That is a typo. If you notice, for every case, the relation is opposite on the opposite sides of the equality value. So the relation that holds in x < 1 will be opposite to the relation that holds when x > 1.
That did mess up the entire explanation. Good you pointed it out. I have edited the original post.
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