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

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25 Feb 2006, 21:36
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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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Re: DS inequality [#permalink]

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26 Feb 2006, 07:45
laxieqv wrote:
Is 1+x+x^2+x^3+x^4 > 1/(1-x) ?
(1) x > 0
(2) x < 1

from (1)
if x = 0.000000001==> (1+x+x^2+x^3+x^4) is not >1/(1-x).
if x = 1.00000001 ==> (1+x+x^2+x^3+x^4) is >1/(1-x). so not enough.

from (2),
if x = 0.000000001==> (1+x+x^2+x^3+x^4) is not >1/(1-x).
if x = 0.999999999 ==> (1+x+x^2+x^3+x^4) is >1/(1-x). so not suff.

both statements togather also not suff....
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26 Feb 2006, 18:25
St. 1)
If x = 1/2, then LHS = 1 + 1/2 + 1/4 + 1/8 + 1/16 = 31/16, RHS = 2 (LHS < RHS)
If x = 2, then LHS = 1 + 2 + 4 + 8 + 16 = 31, RHS = -1 (LHS > RHS)

Insufficient.

St. 2)
If x = 1/2, LHS < RHS
If x = -1, then LHS = 1 - 1 + 1 - 1 + 1 = 1 RHS = 1/2 (LHS > RHS)

Insufficient.

Using St.1 and St 2
0<x<1
x must be a positive fraction and this will always yield the inequality LHS < RHS

Ans C
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Re: DS inequality [#permalink]

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27 Feb 2006, 00:30
C with simple set of values.

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

(1) x > 0

(2) x < 1

(1)
x = 5....Main stem is true.
x = 0.1...Main stem is false since 1.1111 < 1.1111111...

(2)
x = -1...Main stem reduces to 1 > 1/2 ...and is true.
x = 0.1...Main stem is false as shown in statement I.

(1) + (2)....0<x<1.....main stem is always false...values checked x = 0.1 and x = 0.9.
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Re: DS inequality [#permalink]

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27 Feb 2006, 10:19
laxieqv wrote:
Is 1+x+x^2+x^3+x^4 > 1/(1-x) ?
(1) x > 0
(2) x < 1

i checked, even if 0<x>1, we cannot say whether (1+x+x^2+x^3+x^4) is greater or less or equal to 1/(1-x).

if x = 0.00001, (1+x+x^2+x^3+x^4) is almost equal to 1/(1-x).
if x = 0.9999, (1+x+x^2+x^3+x^4) is less than 1/(1-x).

so I still stick with E.
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27 Feb 2006, 17:46
I will second the 'C'

Whats the official answer?
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Re: DS inequality [#permalink]

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27 Feb 2006, 19:39
laxieqv wrote:
Is 1+x+x^2+x^3+x^4 > 1/(1-x) ?

(1) x > 0

(2) x < 1

(1) x>0 :
the LHS is always > 0
but the RHS can be <0 coz if x>1 ---> 1-x<0 ---> 1/(1-x) <0
----> we can't conclude ----> insuff

(2) x<1 :
that means x can come very very very close to 1 ----> 1-x comes very very very close to 0 ----> 1/(1-x) is very very very big. Whereas, LHS can't be too bigger than 1 since x<1 ---> all the power of x will be < x
---->we can't conclude ---> insuff

(1) and (2):
Transfer RHS all to LHS, we have the problem become:
(1+x+x^2+x^3+x^4) - 1/ (1-x) > 0
<==> [(1+x+x^2+x^3+x^4)* (1-x)]/ (1-x) - 1/(1-x) >0
we have : (1+x+x^2+x^3+x^4)* (1-x) = 1+x+x^2+x^3+x^4 - ( x+x^2+x^3+x^4+x^5) = 1-x^5

---> LHS= (1-x^5 -1) / (1-x) = -x^5 / (1-x)

since 0<x<1 ---> 0< 1-x<1 AND -x^5 < 0
----> -x^5/ (1-x) < 0

----> the answer is LHS is always < RHS
-----> suff.

C it is.
Re: DS inequality   [#permalink] 27 Feb 2006, 19:39
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