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If x is positive, which of the following could be the [#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 < 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
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Last edited by Bunuel on 30 Jul 2012, 11:38, edited 1 time in total.
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Re: 1/x, 2x, x^2 [#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<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.
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Re: 1/x, 2x, x^2 [#permalink]
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01 Nov 2010, 19:53
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.
Thanks again!



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Re: 1/x, 2x, x^2 [#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: 1/x, 2x, x^2 [#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: 1/x, 2x, x^2 [#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: 1/x, 2x, x^2 [#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: 1/x, 2x, x^2 [#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 [#permalink]
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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.
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04 Feb 2013, 10:13
Took 3 minutes. Did it by plugging in. Finding algebraic method a little laborious.
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Re: If x is positive, which of the following could be the [#permalink]
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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 [#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 [#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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If x is positive, which of the following could be the [#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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