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# 1+X+X^2+X^3+X^4+X^5<1/(1-X)? a. X>0 b. X<1

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1+X+X^2+X^3+X^4+X^5<1/(1-X)? a. X>0 b. X<1 [#permalink]

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17 Feb 2006, 11:11
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1 1+X+X^2+X^3+X^4+X^5<1/(1-X)?
a. X>0
b. X<1
Manager
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17 Feb 2006, 12:10
C

stmt 1:
x>0

when x = 2,3 etc
1+x+x^2+x^3+x^4+X^5 > 1/1-x

when x= 1
there is a division by 0 which is undefined

when x = 1/2 ,1/3 etc
1+x+..... < 1/1-x
INSUFF

stmt 2:
x<1

when x =0
1+x+..... = 1

when x= 1/2, 1/3 etc
1+x+.... < 1/1-x

when x = -1, -2 etc
1+x+..... < 1/1-x
INSUFF

combining:
0<x<1
so 1+x+x^2+.... < 1/1-x
SUFF

What's the OA?
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18 Feb 2006, 08:43
My answer is E. The reason being

in the case where 0<x<1

when x = 1/2

Then 1+X+X^2+X^3+X^4+X^5<1/(1-X) holds true
63/32<2

but when x=1/10

then the equation doesn't hold true
111111/100000 > 11/10

Hence E. what is the OA?
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18 Feb 2006, 09:57

(1-X^6)=(1-X).(1+X+X^2+X^3+X^4+X^5)

a) X>0 insufficient:
(1-X^6)=(1-X).(1+..+X^5)<1
for X>1: (1+...+X^5)>1/(1-X)
for X<1: (1+...+X^5)<1/(1-X)

b)X<1 insufficient:
for X<>0: (1+...+X^5)<1/(1-X)
for X=0 (1+...+X^5)=1/(1-X)=1

a+b) sufficient
We always have (1+...+X^5)<1/(1-X)
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19 Feb 2006, 13:41
Let S=1+x+x^2 + ... +x^6
S=(1-x^6)/(1-x)
Compare this with 1/(1-x)

(1-x^6)/(1-x)-1/(1-x)=-x^6/(1-x)=x^6/(x-1)

a) x>0
x^6>=0, (x-1) could be positive or negative
insufficient

b) x<1
x^6>=0
x-1<0

S<=1/(1-x)
insufficient due to that equal sign.

Combined
x^6>0
x-1<0
Sufficient

Therefore C
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Last edited by HongHu on 19 Feb 2006, 20:22, edited 1 time in total.
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19 Feb 2006, 17:54
a> x>0. Assume x = 2, then LHS = 1+2+4+8+16+32 = 63, while RHS = 1/1 = 1 so 1+X+X^2+X^3+X^4+X^5 > 1/(1-X). But if x = 1/2, then LHS = 1+1/2+1/4+1/8+1/16+1/32 = 63/32 while RHS = 2 and in this case 1+X+X^2+X^3+X^4+X^5 < 1/(1-X).

Insufficient.

b) x<1. If x = 1/2, then we have 1+X+X^2+X^3+X^4+X^5 < 1/(1-X) but if x = 0, then we have LHS = RHS. Insufficient.

Using a and b, we know x ie between 0 and 1 and must be fraction leading to 1+X+X^2+X^3+X^4+X^5 < 1. Sufficient.

Ans C
19 Feb 2006, 17:54
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