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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

ALGEBRAIC APPROACH:

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.

NUMBER PLUGGING APPROACH:

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.

I guess sometimes picking the "easy" numbers is not the best strategy.

Thanks again!

When picking a number, the most important thing is that we should try all possible numbers that could give us different answers. Let me explain by telling you how I would put in numbers and check.

I see \(\frac{1}{x}\), 2x and \(x^2\). I know I have to try numbers from two ranges at least '0-1' and '>1' since numbers in these ranges behave differently. Also, \(x^2\) is greater than 2x if x > 2 e.g. 3x3 > 2x3 but if x < 2, then \(x^2\) is less than 2x e.g 1.5 x 1.5 < 2 x 1.5. So, I need to try a number in the range 1 to 2 as well. Also, 0 and 1 are special numbers, they give different results sometimes so I have to try those as well. Let's start:

0- Since x is positive, I don't need to try it.

1/2 - I get 2, 1 and 1/4. I get the order \(x^2\) < 2x < \(\frac{1}{x}\)

1 - I get 1, 2, 1 Now, what you need to notice here is that 2x > \(\frac{1}{x}\) whereas in our above result we got \(\frac{1}{x}\) > 2x. This means there must be some value between 0 and 1 where \(\frac{1}{x}\) = 2x. Anyway, that doesn't bother me but what I have to do now is take a number very close to 1 but still less than it. I take 15/16 (random choice). I get 16/15, 15/8 and \((\frac{15}{16})^2\). The first two numbers are greater than 1 and the last one is less than 1. I get the order \(x^2 < \frac{1}{x} < 2x\)

Now I need to try 3/2. I get 2/3, 3 and 9/4. So the order is \(\frac{1}{x} < 2x < x^2\)

I try 3 - I get 1/3, 6, 9 For these numbers, \(x^2\) will be greatest but none of the options have it as the greatest term.

Only I. and II. match hence answer is (D) _________________

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

ALGEBRAIC APPROACH:

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.

NUMBER PLUGGING APPROACH:

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.

Hope it's clear.

Hi Bunuel,

I have doubt !!!

Lets submit the values in the equations...lets take x= 3

then

I. x^2<2x<1/x ===> 9<6<1/3 which is not true II. x^2<1/x<2x ===> 9<1/3<6 which is not true again III. 2x<x^2<1/x ===> 6< 9> 1/3 which is true....

so, i believe only III is the ans

First of all we are asked "which of the following COULD be the correct ordering" not MUST be.

"MUST BE TRUE" questions: These questions ask which of the following MUST be true, or which of the following is ALWAYS true no matter what set of numbers you choose. Generally for such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

As for "COULD BE TRUE" questions: The questions asking which of the following COULD be true are different: if you can prove that a statement is true for one particular set of numbers, it will mean that this statement could be true and hence is a correct answer.

Thanks Bunuel! I did exactly what you did, splitting the numbers. I then picked 1/2 and 1/16 and got the wrong answer. I guess sometimes picking the "easy" numbers is not the best strategy.

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

ALGEBRAIC APPROACH:

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.

NUMBER PLUGGING APPROACH:

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.

Hope it's clear.

Hi Bunuel,

I have doubt !!!

Lets submit the values in the equations...lets take x= 3

then

I. x^2<2x<1/x ===> 9<6<1/3 which is not true II. x^2<1/x<2x ===> 9<1/3<6 which is not true again III. 2x<x^2<1/x ===> 6< 9> 1/3 which is true....

[b] Basically we should determine relationship between \(x\), \(\frac{1}{x}\) and \(x^2\) in three areas: 0------1------2-------.

Bunuel I have reasonably implemented the key values approach in all my inequalities problems but I couldn't understand how 0, 1, 2 can be inferred to be the keys in this problem. Can you elaborate why you chose 0 1 2 ?

[b] Basically we should determine relationship between \(x\), \(\frac{1}{x}\) and \(x^2\) in three areas: 0------1------2-------.

Bunuel I have reasonably implemented the key values approach in all my inequalities problems but I couldn't understand how 0, 1, 2 can be inferred to be the keys in this problem. Can you elaborate why you chose 0 1 2 ?

Regards, Sameer

We should check which of the 3 statements COULD be the correct ordering.

Now, the same way as x and x^2 have different ordering in the ranges 0<x<1 and x>1, 2x and x^2 have different ordering in the ranges 1<x<2 (1/x<x^2<2x) and x>2 (1/x<2x<x^2). Next, you can see that no option is offering such ordering thus if there is correct ordering listed then it must be for the x-es from the range 0<x<1. So, if we want to proceed by number plugging we know from which range to pick numbers. Also as in this range x^2 is the least value we can quickly discard option III and concentrate on I and II. _________________

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

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04 Feb 2013, 01:35

this can take a long time

use both methods, picking the number and solving unequality

for the third we can solve unequality, remember one thing make it easy, x is positive so we can multiple 2 sides of un equality with x and do not change the mark of unequality

gmat is simple but is enough to kill us when we are nervous on the test.

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

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04 Feb 2013, 10:13

Took 3 minutes. Did it by plugging in. Finding algebraic method a little laborious. _________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

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16 Nov 2014, 03:08

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22 Nov 2015, 19:57

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