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# Is 1+x+x^2+x^3+x^4 > 1/(1-x) ? 1). x > 0 2). x < 1

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VP
Joined: 06 Jun 2004
Posts: 1053
Location: CA
Is 1+x+x^2+x^3+x^4 > 1/(1-x) ? 1). x > 0 2). x < 1 [#permalink]

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04 Feb 2006, 23:13
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0% (00:00) correct 0% (00:00) wrong based on 2 sessions

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Is 1+x+x^2+x^3+x^4 > 1/(1-x) ?

1). x > 0
2). x < 1
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Director
Joined: 17 Oct 2005
Posts: 928

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04 Feb 2006, 23:21
C

I -insuff
II-insuff

Both-suff
VP
Joined: 21 Sep 2003
Posts: 1057
Location: USA

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04 Feb 2006, 23:27
Again C?

1) x > 0

x = 0.5
1+x+x^2+x^3+x^4 = 1+ 0.5+ 0.25+0.125+0.0625 < (1/(1-05) = 2 (False)
x = 2 , True

2) x < 1

x = 0.5 (False)
x = -1 (True)

Combining (1) & (2)
For 0<x<1
1+x+x^2+x^3+x^4 < 1/(1-x) Always False

So C!
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Senior Manager
Joined: 05 Jan 2006
Posts: 381

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05 Feb 2006, 01:09
Got C as combining both are always false though took real long time! I hate to substitute numbers, I like to work with equation.. is there a way of proving this by equation ! I always feels that choosing number is time consuming and may not be accurate if your sample are not representative...

Any way I tried following to combine both... lets say x = n/m where m>n so now

1 + n/m + (n/m)^2 + (n/m)^3 + (n/m)^4 < 1/(1-n/m)

now 1/(1-n/m) = m/m-n = m-n+n/m-n = 1 + n/m-n

so ineuqality will reduce to...
n/m + (n/m)^2 + (n/m)^3 + (n/m)^4 < n/m-n

1/m + n/m^2 + n^2/m^3 + n^3/m^4 < 1/m-n...

Now I can not see why this equality will be always false so unfortunately I needed to start plugging number!
Director
Joined: 13 Nov 2003
Posts: 789
Location: BULGARIA

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05 Feb 2006, 01:52
may be i am wrong but i multiplied the left part of the inequality with (1-x) and got 1-x+x-x^2+x^2-x^3+x^3-x^4+x^4-x^5>1 wich simplifies to -x^5>0 or x^5<0
now think that it should be A)
Senior Manager
Joined: 19 Feb 2005
Posts: 486
Location: Milan Italy

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05 Feb 2006, 04:11
BG wrote:
may be i am wrong but i multiplied the left part of the inequality with (1-x) and got 1-x+x-x^2+x^2-x^3+x^3-x^4+x^4-x^5>1 wich simplifies to -x^5>0 or x^5<0
now think that it should be A)

you can do this if 1-x is greater than 1 or smaller than -1
Otherwise you've to flip the inequality sign

In this case, try 1/2 as X -> 31/16 is less than 1/0.5=2 (answ=NO)
try 2 as X -> 1+2... is greater than 1/2=0.5 (ANSW=YES)
Not sufficient
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Senior Manager
Joined: 11 Jan 2006
Posts: 269
Location: Chennai,India

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05 Feb 2006, 10:47
thearch wrote:
BG wrote:
may be i am wrong but i multiplied the left part of the inequality with (1-x) and got 1-x+x-x^2+x^2-x^3+x^3-x^4+x^4-x^5>1 wich simplifies to -x^5>0 or x^5<0
now think that it should be A)

you can do this if 1-x is greater than 1 or smaller than -1
Otherwise you've to flip the inequality sign

In this case, try 1/2 as X -> 31/16 is less than 1/0.5=2 (answ=NO)
try 2 as X -> 1+2... is greater than 1/2=0.5 (ANSW=YES)
Not sufficient

thnks arch! i was having the same prob BS had.. got it now
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05 Feb 2006, 10:47
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