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

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Re: If x is positive, which of the following could be correct [#permalink] New post 27 Jun 2012, 13:11
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 [#permalink] New post 28 Jun 2012, 01:15
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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 [#permalink] New post 28 Jul 2012, 19:24
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 [#permalink] New post 29 Jul 2012, 00:37
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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 [#permalink] New post 29 Jul 2012, 21:58
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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 [#permalink] New post 07 Aug 2012, 23:50
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 [#permalink] New post 13 Dec 2012, 06:28
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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 [#permalink] New post 13 Dec 2012, 22:41
if x = 1/2, then 1/1/2 = 2 which is greater than 2(1/2)
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Re: If x is positive, which of the following could be correct [#permalink] New post 13 Dec 2012, 23:30
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 [#permalink] New post 11 May 2013, 01:05
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 [#permalink] New post 11 May 2013, 02:12
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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 [#permalink] New post 05 Jul 2013, 01:26
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Re: If X is positive [#permalink] New post 15 Sep 2013, 09:47
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 [#permalink] New post 02 Oct 2013, 08:40
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 [#permalink] New post 02 Oct 2013, 20:21
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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 930 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 [#permalink] New post 14 Feb 2014, 07:49
Bunuel wrote:
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.


Hi Bunuel.

How do you know which numbers to pick? I mean for example 0.9 for the second statement. Any clues?

Thanks
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Re: If x is positive, which of the following could be correct [#permalink] New post 22 Feb 2014, 07:48
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 [#permalink] New post 24 Feb 2014, 01:12
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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   [#permalink] 24 Feb 2014, 01:12
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