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1+X+X^2+X^3+X^4+X^5<1/(1-X)?

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Senior Manager
Joined: 11 Nov 2003
Posts: 356

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Location: Illinois

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08 Jul 2004, 08:55
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1 1+X+X^2+X^3+X^4+X^5<1/(1-X)?
a. X>0
b. X<1

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Director
Joined: 01 Feb 2003
Posts: 844

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08 Jul 2004, 09:24
Answer should be C, as for x<1 the equation is not totally satisfied.

However, if I change this into a PS problem to solve the given equation, then we need to consider 2 scenarios:
(1) x<1 (2) x>1

for x<1 we can cross multiply, without change in sign and get the solution as x^6 > 0 which is true for any x not equal to zero

and for x>1, we will have x^6-1>1 => x^6>2 => x > 2^(1/6)

the solution seems to be x> 2^(1/6)

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Director
Joined: 05 Jul 2004
Posts: 894

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09 Jul 2004, 06:16
I vote for (B)

inequality can be written as : (1-x^6)/(1-x) < (1/(1-x)
(A) Insufficient :
x<1 : Satisfies
x>1 : does not satisfy

(B) Sufficient :
Condition: x<1
x --> [0, 1] Satifies
x --> [-1, 0] Satisfies
x --> (inf, -1] Satisfies (1-x^6) will be negative.

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Senior Manager
Joined: 07 Oct 2003
Posts: 350

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Location: Manhattan

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10 Jul 2004, 11:12
jpv wrote:
I vote for (B)

inequality can be written as : (1-x^6)/(1-x) < (1/(1-x)
(A) Insufficient :
x<1 : Satisfies
x>1 : does not satisfy

(B) Sufficient :
Condition: x<1
x --> [0, 1] Satifies
x --> [-1, 0] Satisfies
x --> (inf, -1] Satisfies (1-x^6) will be negative.

Regarding (B)
try x = -1, you get 0<.5 ok
x = 0, you then get 1<1, not ok
hence B is insufficient

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Director
Joined: 05 Jul 2004
Posts: 894

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11 Jul 2004, 06:49
Agree L.

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11 Jul 2004, 06:49
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