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

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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 Jul 2012, 20:24
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 ?
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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
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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.
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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 08 Aug 2012, 00:50
given that x>0
1 ) x^2 < 2X<1/X lets just check x^2 < 2X => 0 <x<2 ; 2X<1/X => 0<x<1/\sqrt{2}
these 2 inequities do not conradict each other. so, 1) is ok

2) x^2 <1/X< 2X check them - x^2 <1/X => 0<x<1 ; 1/X< 2X => x> 1/\sqrt{2}
these 2 inequities do not conradict each other. so, 2) is ok

3) 2X< x^2 <1/X check them - 2X< x^2 => x>2 ; x^2 <1/X => x<1
these 2 inequities conradict each other. so, 3) is not ok


p.s. dont know why such symbols as sqroot , fraction ets dont work. :oops:
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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 13 Dec 2012, 07:28
picking numbers, both integers and not and only positive.

In all cases only the 3 case doesn't work, so pretty fast you can reach D

in this question youhave to be really comfortable with theory to solve it, otherwise is best and safe picking number .
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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 14 Dec 2012, 00:30
positiveness or negativeness is important to inequalit

because x is positive we can multiple both sides of inequality with x and keep the same mark.for example

x^2<2x<1/x

is the same as

x^3<2x^2<1

(if x is negative we have to change the mark of the inequality. this question is not relevant to that cases)

now solve 2 inequality independently . this can be done quick.

this questions can be done in less than 3 minutes. Other method takes longer time and in fact is not good.

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

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lylya4 wrote:
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


if x = 1/2, 1/x = 2 .. wrong example used
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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 02 Oct 2013, 09:40
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.
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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 02 Oct 2013, 21:21
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imhimanshu wrote:
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
Ques3.jpg [ 11.4 KiB | Viewed 4695 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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Re: If x is positive, which of the following could be correct ordering of [#permalink]

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New post 22 Feb 2014, 08:48
I picked numbers: 1/2, 1, 3/2, 2, 3

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!

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

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abdb wrote:
I picked numbers: 1/2, 1, 3/2, 2, 3

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!

Thanks!


I have answered your query using this very question here: http://www.veritasprep.com/blog/2013/05 ... on-points/
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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 30 Jul 2014, 20:40
What a horrific problem?! I think a fast and simple way is to graph all functions and compare all vertically. Doing this, it can be easy to see that III is impossible.
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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 13 Aug 2014, 11:39
lylya4 wrote:
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


double check your calculation for x= 1/2 the equation won't hold true
1/0.5 = 2
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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 Aug 2015, 00:41
This is how I approached it :)

Try breaking the eqn and solve

Eq 1) x^2 < 2x < 1/x
x^2 < 2x and 2x < 1/x
or, x < 2 and x^2 < 1/2 (As X is positive, we can multiply)
x < ( 1/root 2 ) satisfies both eqn.

2) x^2 < 1/x < 2x
x^2 < 1/x and 1/x < 2x
or, x^3 < 1 and x2 > 1/2
1 > x > ( 1/root 2 ) satisfies both eqn (Eg: Root 3 / 2 )

3) 2x < x^2 < 1/x
2x < x^2 and x^2 < 1/x
or, 2 < x and x^3 < 1
Not possible

So, 4) I 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 29 Dec 2015, 16:01
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


first, let's get rid of at least one x in the given expressions, as x is positive just multiply the expressions by x and we'll get
(I) \(x^3 < 2x^2 < 1\)
as \(2x^2\) < 1 we must pick a value which is < 1, let's pick 1/2 and it works. COULD BE

(II) \(x^3 < 1< 2x^2\)
Here we can see that \(x^3\)<1 so we must pick a value <1 BUT which will make \(2x^2\) >1 if possible. Let's pick 0.9 and it works also here. COULD BE

(III) \(2x^2 < x^3 < 1\)
We must pick a value < 1 BUT as we've already seen, if we pick a fraction < 1 we cannot make \(2x^2 < x^3\), in the above cases it \(2x^2 was > x^3\) each time we picked a fraction < 1

Hope it 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 29 Dec 2015, 22:13
Alternatively, we can just divide each term by x (since x > 0) and arrive at: 1/x^2 , 2 and x
Much easier to pick number this way :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 Jan 2016, 03:02
Hi Experts,
In solving x^2<2x=>x^2-2x<0=>x(x-2)<0=>x<0 or x<2

why did we ignore x<0 is it bcz we are told in question that x is positive number...it is so then why can't it be ignored in statement 3...btw I understood the number picking methodology, just little bit curious...Thanks in advance.
Re: If x is positive, which of the following could be correct ordering of   [#permalink] 03 Jan 2016, 03:02

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