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PS: Correct ordering

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Senior Manager
Senior Manager
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Concentration: Technology, Marketing
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WE: Sales (Telecommunications)
PS: Correct ordering [#permalink]

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New post 07 Jun 2009, 11:27
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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
C.III
D.I and II
E.All
D
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Re: PS: Correct ordering [#permalink]

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New post 07 Jun 2009, 11:53
I. x^2 < 2x < 1/x
lets evaluate both parts seperately
x^2 < 2x
or x < 2 (as x is +ve)

2x < 1/x
or x^2 < 0.5
or x < 0.7

Together for x <0.7

II. x^2 < 1/x < 2x
x^2 < 1/x
or x^3 < 1
or x < 1

1/x < 2x
or 1/2 < x^2
or 0.7 < x

Together, possible for 0.7 < x < 1

III. 2x < x^2 < 1/x
2x < x^2
or 2 < x (dividing x from both sides, since x is +ve)

x^2 < 1/x
or x^3 < 1
or x < 1

together, x>2 and x<1, not possible

Hence answer I and II or "D"
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Re: PS: Correct ordering [#permalink]

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New post 08 Jun 2009, 03:30
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amolsk11 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
C.III
D.I and II
E.All
D


I.\(x^2<2x<1/x\)
\(1: \longrightarrow x^2 < 2x \longrightarrow x<2\)
\(2: \longrightarrow x^2 < 1/x \longrightarrow x^3 < 1 \longrightarrow x < 1\)
\(3: \longrightarrow 2x < 1/x \longrightarrow x^2 < 1/2 \longrightarrow x < \frac{1}{\sqrt{2}}\)
\(1&2&3 \longrightarrow x < \frac{1}{\sqrt{2}} < 1 < 2\)

Works (ie X = 1/2).

II.\(x^2<1/x<2x\)
\(1: \longrightarrow x^2<1/x \longrightarrow x^{3}<1 \longrightarrow x<1\)
\(2: \longrightarrow x^2 < 2x \longrightarrow x < 2\)
\(3: \longrightarrow 1/x<2x \longrightarrow 1/2 < x^{3} \longrightarrow x>\frac{1}{\sqrt[3]{2}}\)
\(1&2&3 \longrightarrow \frac{1}{\sqrt[3]{2}} < x < 1\)
Works -- X=9/10

III.\(2x<x^2<1/x\)
\(1: \longrightarrow 2x<x^{2} \longrightarrow x>2\)
\(2: \longrightarrow x^2<1/x \longrightarrow x^3<1 \longrightarrow x<1\)

Impossible


D 1&2
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Re: PS: Correct ordering   [#permalink] 08 Jun 2009, 03:30
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