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# If 0 < x < 1, which of the following inequalities must be

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VP
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If 0 < x < 1, which of the following inequalities must be [#permalink]  08 Oct 2008, 05:48
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35% (medium)

Question Stats:

65% (02:09) correct 35% (00:56) wrong based on 106 sessions
If 0 < x < 1, which of the following inequalities must be true ?

I. $$x^5 < x^3$$
II. $$x^4 + x^5 < x^3 + x^2$$
III. $$x^4 - x^5 < x^2 - x^3$$

A. None
B. I only
C. II only
D. I and II only
E. I, II and III
[Reveal] Spoiler: OA

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Director
Joined: 01 Jan 2008
Posts: 629
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Kudos [?]: 139 [0], given: 1

Re: If 0 < x < 1, which of the following inequalities must be [#permalink]  08 Oct 2008, 06:55
amitdgr wrote:
If 0<x<1 , which of the following inequalities must be true ?

I. $$x^5 < x^3$$
II. $$x^4 + x^5 < x^3 + x^2$$
III. $$x^4 - x^5 < x^2 - x^3$$

* None
* I only
* II only
* I and II only
* I,II and III

I: x^5-x^3 = x^3*(x-1)*(x+1) < 0 since x > 0 and x < 1
II: x^4+x^5-x^3-x^2 = x^2*(x+1)*(x-1)*(x+1) < 0 for same reasons
III: x^4-x^5-x^2+x^3 = x^2*(1-x)*(x-1)*(x+1) = -x^2*(x-1)^2*(x+1) < 0
Senior Manager
Joined: 21 Apr 2008
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Schools: Kellogg, MIT, Michigan, Berkeley, Marshall, Mellon
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Kudos [?]: 30 [0], given: 13

Re: If 0 < x < 1, which of the following inequalities must be [#permalink]  08 Oct 2008, 07:36
My two cents:

I beeter if you divided by x^2: x^2+x^3<x+1 -> always
II better if you divided by x^2: x^2-x^3<1-x^2 always
III dividing by x^2: x^2-x^3<1-x^2
if x=0.1 thus 0.01-0.001<1-0.01 always
if x=0.9 thus 0.81-0.729<1-0.9 // 0.081<1 always

E

OA?

Cheers
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VP
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Kudos [?]: 348 [0], given: 1

Re: If 0 < x < 1, which of the following inequalities must be [#permalink]  08 Oct 2008, 07:47
OA is E. Thanks guys I missed the third equation !!
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Intern
Joined: 27 Sep 2008
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Kudos [?]: 2 [0], given: 0

Re: If 0 < x < 1, which of the following inequalities must be [#permalink]  08 Oct 2008, 08:01
I also believe the answer is E since even the third equation,
x^4 - x^5 < x^2 - x^3
= x^4 + x^3 < x^2 + x^5
=x^2 + x < 1+x^3
which is true since the value on the right hand side will be greater than 1.

thanks
SM
Manager
Joined: 15 Apr 2008
Posts: 166
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Kudos [?]: 10 [0], given: 1

Re: If 0 < x < 1, which of the following inequalities must be [#permalink]  08 Oct 2008, 09:33
i will also go with E

raising a decimal number to any power reduces the value of the number.
so .1^2=.01
and .1^3=.001

can be solved by picking numbers
MBA Blogger
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Location: India
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WE: Analyst (Computer Software)
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Kudos [?]: 27 [0], given: 47

Re: If 0 < x < 1, which of the following inequalities must be [#permalink]  10 Sep 2014, 07:17
People, please throw some more light!!
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Kudos [?]: 40979 [1] , given: 5654

Re: If 0 < x < 1 , which of the following inequalities must be [#permalink]  10 Sep 2014, 07:52
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
scofield1521 wrote:
People, please throw some more light!!

If 0 < x < 1, which of the following inequalities must be true ?

I. $$x^5 < x^3$$
II. $$x^4 + x^5 < x^3 + x^2$$
III. $$x^4 - x^5 < x^2 - x^3$$

A. None
B. I only
C. II only
D. I and II only
E. I, II and III

If 0 < x < 1, then x > x^2 > x^3 > x^4 > x^5 ...

I. $$x^5 < x^3$$. True.

II. $$x^4 + x^5 < x^3 + x^2$$. Each term on the left hand side is less than each term on the right hand side, thus LHS < RHS.

III. $$x^4 - x^5 < x^2 - x^3$$ --> $$x^4(1 - x) < x^2(1 - x)$$. Since 0 < x < 1, then 1 - x > 0, so we can reduce by it: $$x^4 < x^2$$. True.

Hope it's clear.
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Re: If 0 < x < 1, which of the following inequalities must be [#permalink]  10 Sep 2014, 08:15
Thanks bunuel, Its much clear now.

Bunuel wrote:
scofield1521 wrote:
People, please throw some more light!!

If 0 < x < 1, which of the following inequalities must be true ?

I. $$x^5 < x^3$$
II. $$x^4 + x^5 < x^3 + x^2$$
III. $$x^4 - x^5 < x^2 - x^3$$

A. None
B. I only
C. II only
D. I and II only
E. I, II and III

If 0 < x < 1, then x > x^2 > x^3 > x^4 > x^5 ...

I. $$x^5 < x^3$$. True.

II. $$x^4 + x^5 < x^3 + x^2$$. Each term on the left hand side is less than each term on the right hand side, thus LHS < RHS.

III. $$x^4 - x^5 < x^2 - x^3$$ --> $$x^4(1 - x) < x^2(1 - x)$$. Since 0 < x < 1, then 1 - x > 0, so we can reduce by it: $$x^4 < x^2$$. True.

Hope it's clear.

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Re: If 0 < x < 1, which of the following inequalities must be   [#permalink] 10 Sep 2014, 08:15
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