Is 1+x+x^2+x^3+x^4>1/(1-x)? 1). x>0 2). x<1 : DS Archive
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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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Manager
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Is 1+x+x^2+x^3+x^4>1/(1-x)? 1). x>0 2). x<1 [#permalink]

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04 Jun 2009, 10:34
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Is 1+x+x^2+x^3+x^4>1/(1-x)?
1). x>0
2). x<1
Senior Manager
Joined: 16 Jan 2009
Posts: 359
Concentration: Technology, Marketing
GMAT 1: 700 Q50 V34
GPA: 3
WE: Sales (Telecommunications)
Followers: 4

Kudos [?]: 199 [0], given: 16

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04 Jun 2009, 13:55
Is 1+x+x^2+x^3+x^4>1/(1-x)?

1). x>0
1+x+x^2+x^3+x^4=1.X
Where X= x+x^2+x^3+x^4

Case 1:
1/(1-x) will be –ve for x>1
So , 1+x+x^2+x^3+x^4>1/(1-x)

Case 2:
1/(1-x) will be +ve for x<1

INSUFFICIENT

2). x<1

1/(1-x) will be +ve for x<1
INSUFFICIENT

TOGETHER :
For 1>x>0
1/(1-x) will be +ve
INSUFFICIENT
IMO E
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Lahoosaher

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09 Jun 2009, 02:02
IMO C
1) with x=4 YES
With x=0.5 NO
2) with x=-3 YES
with x=0.5 NO
Together YES
Hence C
What 's the OA?
Manager
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09 Jun 2009, 22:11
E
Agree with amolsk11.
Senior Manager
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10 Jun 2009, 00:26
IMO E too..

1+x+x2+x3+X4 < 1 / (1-x) for x = 0.1

1/(1-x) = 1/ 0.9 = 1.1111111

1+x+x2+x3+x4 = 1+ .1 + .01+.001 = 1.1111

but for x = 0.5, the equation doesnt hold good.

1/(1-x) = 1/0.5 = 2

1+0.5+0.25+0.625 = 2.15 > 2

hence E
Re: DS   [#permalink] 10 Jun 2009, 00:26
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