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

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Senior Manager
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1+X+X^2+X^3+X^4+X^5<1/(1-X)? [#permalink] New post 08 Jul 2004, 07:55
00:00
A
B
C
D
E

Difficulty:

(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
1 1+X+X^2+X^3+X^4+X^5<1/(1-X)?
a. X>0
b. X<1
Director
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 [#permalink] New post 08 Jul 2004, 08:24
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)
Director
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 [#permalink] New post 09 Jul 2004, 05:16
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.
Senior Manager
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 [#permalink] New post 10 Jul 2004, 10:12
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

C is the answer
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 [#permalink] New post 11 Jul 2004, 05:49
Agree L. :beat
  [#permalink] 11 Jul 2004, 05:49
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