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If x is positive, which of the following could be the correct ordering [#permalink]
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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. \(x^2 < 2x < \frac{1}{x}\) 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
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Originally posted by butterfly on 01 Nov 2010, 17:28.
Last edited by Bunuel on 12 Oct 2017, 00:08, edited 2 times in total.
Edited the question.



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If x is positive, which of the following could be the correct ordering [#permalink]
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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. \(x^2 < 2x < \frac{1}{x}\) 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 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: 012. \(x>2\) \(1<x<2\) \(0<x<1\) 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 \(\frac{2x^21}{x}<0\), the expression \(2x^21\) 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 \(\frac{2x^21}{x}>0\), the expression \(2x^21\) 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 \(xes\)). 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.
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Re: If x is positive, which of the following could be the correct ordering [#permalink]
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02 Nov 2010, 05:43
Bunuel wrote: 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: 012.
\(x>2\)
\(1<x<2\)
\(0<x<1\)
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 \(\frac{2x^21}{x}<0\), the expression \(2x^21\) 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 \(\frac{2x^21}{x}>0\), the expression \(2x^21\) 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 \(xes\)). 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



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Re: If x is positive, which of the following could be the correct ordering [#permalink]
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02 Nov 2010, 05:54
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butterfly wrote: 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 '01' 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)
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Re: If x is positive, which of the following could be the correct ordering [#permalink]
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02 Nov 2010, 06:00
vitamingmat wrote: Bunuel wrote: 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: 012.
\(x>2\)
\(1<x<2\)
\(0<x<1\)
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 \(\frac{2x^21}{x}<0\), the expression \(2x^21\) 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 \(\frac{2x^21}{x}>0\), the expression \(2x^21\) 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 \(xes\)). 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. Also: how is III correct for x=3? Hope it helps.
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Re: If x is positive, which of the following could be the correct ordering [#permalink]
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07 Dec 2010, 09:26
Bunuel wrote: [b] Basically we should determine relationship between \(x\), \(\frac{1}{x}\) and \(x^2\) in three areas: 012.
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



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Re: If x is positive, which of the following could be the correct ordering [#permalink]
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07 Dec 2010, 11:54
sameerdrana wrote: Bunuel wrote: [b] Basically we should determine relationship between \(x\), \(\frac{1}{x}\) and \(x^2\) in three areas: 012.
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 xes 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.
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Re: If x is positive, which of the following could be the correct ordering [#permalink]
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05 Jun 2013, 17:35
Bunuel wrote: Basically we should determine relationship between , and in three areas: 012.Buneul, Could you please explain as to how you came to pick these ranges ? The rest of it is perfectly fine. Thanks d
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Re: If x is positive, which of the following could be the correct ordering [#permalink]
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06 Jun 2013, 01:27
dataman wrote: Bunuel wrote: Basically we should determine relationship between , and in three areas: 012.
Buneul,
Could you please explain as to how you came to pick these ranges ? The rest of it is perfectly fine.
Thanks d For each range the ordering of 1/x, 2x and x^2 is different. If \(0<x<1\), the least value is x^2; If \(1<x<2\), the greatest value is 2x (\(\frac{1}{x}<x^2<2x\)) > no option has such ordering; If \(2<x\), the greatest value is x^2 (\(\frac{1}{x}<2x<x^2\)) no option has such ordering. So, we should consider \(0<x<1\) range (where x^2 is the smallest) and find whether x^2<2x<1/x and x^2<1/x<2x could be true.
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Re: If x is positive, which of the following could be the correct ordering [#permalink]
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14 Aug 2016, 09:36
if i take value 9/10 it satisfies 2nd ordering but if i take 2/3 it doesnt satisfies 2nd ordering. why?
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Re: If x is positive, which of the following could be the correct ordering [#permalink]
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16 Aug 2016, 02:27
sananoor wrote: if i take value 9/10 it satisfies 2nd ordering but if i take 2/3 it doesnt satisfies 2nd ordering. why? The question says "...could be the correct ordering of..." So if even 1 positive value of x satisfies the ordering, it is included. Since 9/10 satisfies the second ordering, it is included in the ordering. A transition point here is 1/sqrt(2). That is why values less than 1/sqrt(2) (such as 2/3) behave differently from values more than 1/sqrt(2) (such as 9/10). Check here: http://www.veritasprep.com/blog/2013/05 ... onpoints/
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If x is positive, which of the following could be the correct ordering [#permalink]
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10 Feb 2018, 21:10
Bunuel wrote: 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
ALGEBRAIC APPROACH:
can 2x<1x2x<1x? Can 2x2−1x<02x2−1x<0
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: 012.
\(x>2\)
\(1<x<2\)
\(0<x<1\)
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 \(\frac{2x^21}{x}<0\), the expression \(2x^21\) 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 \(\frac{2x^21}{x}>0\), the expression \(2x^21\) 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 \(xes\)). 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. How did you get from can 2x< 1/x? to Can (2x^2  1)/x < 0? Shouldn't it be 2x^2  1 < 0



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If x is positive, which of the following could be the correct ordering [#permalink]
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11 Feb 2018, 01:20
jasonfodor wrote: Bunuel wrote: 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
ALGEBRAIC APPROACH:
can 2x<1x2x<1x? Can 2x2−1x<02x2−1x<0
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: 012.
\(x>2\)
\(1<x<2\)
\(0<x<1\)
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 \(\frac{2x^21}{x}<0\), the expression \(2x^21\) 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 \(\frac{2x^21}{x}>0\), the expression \(2x^21\) 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 \(xes\)). 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. How did you get from can 2x< 1/x? to Can (2x^2  1)/x < 0? Shouldn't it be 2x^2  1 < 0 \(2x< \frac{1}{x}\); \(2x  \frac{1}{x}<0\); \(\frac{2x^2  1}{x}<0\).
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Re: If x is positive, which of the following could be the correct ordering [#permalink]
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16 Feb 2018, 10:30
Really tough one for me. I tried these three ranges: a) A value between 0 and 1 (I took 0.5) b) A value between 1 and 2 (I took 1.5) c) A value > 2 (I took 3)
By taking these values, I was only getting I correct.
How are we supposed to realize that we should also be taking a value of x around 0.9?



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If x is positive, which of the following could be the correct ordering [#permalink]
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17 Feb 2018, 09:11
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Nived wrote: Really tough one for me. I tried these three ranges: a) A value between 0 and 1 (I took 0.5) b) A value between 1 and 2 (I took 1.5) c) A value > 2 (I took 3)
By taking these values, I was only getting I correct.
How are we supposed to realize that we should also be taking a value of x around 0.9? Hi NivedAs we know \(x>0\), and if \(x>1\), then there would be two possibilities \(x^2>2x>\frac{1}{x}\) or \(x^2=2x>\frac{1}{x}\) (if \(x=2\)) but these options are not given, so we need not worry about this range. Now you need to check only one range which is \(0<x<1\)  (1) square both sides of equation (1) you will get \(0<x^2<1\) (2) Now multiply both sides of equation (1) by \(2\), you will get \(0<2x<2\) (3) So from equation (2) & (3) we could get \(x^2<2x\), and this scenario is present in statements I & II now we know \(x<1\), so \(1<\frac{1}{x}\), clearly \(x^2<\frac{1}{x}\) always for this range. so Statement II is definitely one of the possibility Now let's test \(2x\) & \(\frac{1}{x}\), we know that in the given range \(2x\) will be less than \(2\) but \(\frac{1}{x}\) can be greater than \(2\) for e.g if \(x=0.1\), then \(\frac{1}{x}=10\). so if \(x\) moves closer to \(1\), then \(\frac{1}{x}\) will be closer to \(1\) but if \(x\) is closer to \(0\), then \(\frac{1}{x}\) will be closer to infinity, i.e a very large number (\(\frac{1}{0}\) is infinity), Hence \(\frac{1}{x}\) can definitely be greater than \(2\) So we have a possibility that \(2x<\frac{1}{x}\). Hence Statement I could be possible. So if you are aware of these concepts, then testing number might not be a difficult task.



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Re: If x is positive, which of the following could be the correct ordering [#permalink]
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19 Feb 2018, 04:19
Nived wrote: Really tough one for me. I tried these three ranges: a) A value between 0 and 1 (I took 0.5) b) A value between 1 and 2 (I took 1.5) c) A value > 2 (I took 3)
By taking these values, I was only getting I correct.
How are we supposed to realize that we should also be taking a value of x around 0.9? If you insist on solving by plugging in, it is required to take care of transition points. This concept is discussed in detail here: http://www.veritasprep.com/blog/2013/05 ... onpoints/
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19 Feb 2018, 06:22
Thanks niks18 and VeritasPrepKarishma. It is very clear now and a great learning for me.



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Re: If x is positive, which of the following could be the correct ordering [#permalink]
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21 Feb 2018, 14:16
butterfly wrote: 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 If 0 < x < 1, for example, x = 1/2, then 1/x = 2, 2x = 1 and x^2 = 1/4. We see that x^2 < 2x < 1/x. Roman numeral I could be true. Also, if x = 3/4; then 1/x = 4/3, 2x = 3/2 and x^2 = 9/16. We see that x^2 < 1/x < 2x. Roman numeral II could also be true. For Roman numeral III, observe that x^2 > 2x is equivalent to x^2  2x > 0 and we can factor out the x to obtain x(x  2) > 0. Since x is positive, we can divide each side by x and we will get x  2 > 0; in other words, x > 2. On the other hand, if 1/x > 2x, then 1/x  2x > 0; or, equivalently, (1  2x^2)/x > 0. Since x is positive, we can multiply each side by x to obtain 1  2x^2 > 0, which is equivalent to 2x^2 < 1. Then, x^2 < 1/2 and we know that this is only possible when x < 1. Thus, the inequalities x^2 > 2x and 1/x > 2x cannot hold simultaneously for a positive x. Therefore, Roman numeral III is not possible. In conclusion, I and II could be true. Answer: D
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