Inequality question : PS Archive
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Inequality question

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Director
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14 Jul 2009, 09:29
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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. $$2x < x^2 < 1/x$$
O None
O I only
O III only
O I and II only
O I, II, and III
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14 Jul 2009, 09:44
bigoyal 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. $$2x < x^2 < 1/x$$
O None
O I only
O III only
O I and II only
O I, II, and III

if i do simple plugin of x=1 i get 1,2,1 and none of them fits. If I use fraction like x=1/2 then I (answer B) fits. If i use X=3 then again nothing...
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14 Jul 2009, 12:54
B because if you plug a fraction then option 1 is possible and question is asking 'could be' and not 'must be'.
Senior Manager
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14 Jul 2009, 13:45
bigoyal 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. $$2x < x^2 < 1/x$$
O None
O I only
O III only
O I and II only
O I, II, and III

D

for I: plug 1/2 and I is true

for II: plug 9/10, it is true

III is not true for any value.

so I and II can be correct
Director
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14 Jul 2009, 21:35
rashminet84 wrote:
for I: plug 1/2 and I is true

for II: plug 9/10, it is true

III is not true for any value.

Btw, how did you determine to plug 9/10 for IInd ? That value is not a very common value to determine.

I was able to prove Ist and IIrd. But somehow couldn't solve for IInd.
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15 Jul 2009, 05:25
bigoyal wrote:

Btw, how did you determine to plug 9/10 for IInd ? That value is not a very common value to determine.

I was able to prove Ist and IIrd. But somehow couldn't solve for IInd.

I could make out that one needs to check for fractions, and just like one tries with small and big integers, one needs to try with small and big fractions too.
i tried with 0.1 and 0.9, ie 1/10 and 9/10

btw, how did u prove IIIrd one?? or did u mean IInd?
Director
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15 Jul 2009, 21:59
rashminet84 wrote:
btw, how did u prove IIIrd one?? or did u mean IInd?

I mean, I was able to prove that IIIrd option was incorrect.

III. 2x < x^2 < 1/x
Taking first part:
2x < x^2 (as x is +ve, divide both sides with x)
=> 2 < x ..... (i)

2nd part:
x^2 < 1/x
=> x^3 < 1
as its already given x is positive
=> 0< x <1 ..... (ii)

As (i) and (ii) contradicts, we can say IIIrd option is not possible for any value.
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Re: Inequality question   [#permalink] 15 Jul 2009, 21:59
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