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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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Is 1+x+x^2+x^3+x^4<1/(1-x)?

1) x>0
2) x<1


* A solution will be posted in two days.
[Reveal] Spoiler: OA

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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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MathRevolution wrote:
Is 1+x+x^2+x^3+x^4<1/(1-x)?

1) x>0
2) x<1


* A solution will be posted in two days.


INFO from Q stem
\(1+x+x^2+x^3+x^4<\frac{1}{(1-x)}\)..
\(1+x+x^2+x^3+x^4-\frac{1}{(1-x)}<0\)..
\(\frac{{(1+x+x^2+x^3+x^4)(1-x)-1}}{(1-x)}<0\)..
\(\frac{{1+x+x^2+x^3+x^4-x-x^2-x^3-x^4-x^5)-1}}{(1-x)}<0\)..
\(\frac{-x^5}{(1-x)}<0\)..
\(\frac{x^5}{(x-1)}<0\)..

so two conditions
1) if x>0, x<1
2) if x<0, x>1 ...... Not possible

lets check the statements
1) x>0
if x<1.. ans is YES
if x>1.. ans is NO
Insuff

2) x<1
if 0<x<1.. ans is YES
if x<0.. ans is NO
Insuff

combined..
Suff as per point 1, ans is always YES
C

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

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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is 1+x+x^2+x^3+x^4<1/(1-x)?

1) x>0
2) x<1


The question is (1-x^5)/(1-x)<1/(1-x)? and –x^5/(1-x)<0?, x^5/(x-1)<0? is derived when delete 1/(1-x) from the both equations of 1/(1-x)-x^5/(1-x)<1/(1-x)?. When you multiply (x-1)^2 to the both equations, x^5(x-1)<0? is derived. Divide it with x^4 and it becomes x(x-1)<0?. That is, 0<x<1? is derived, which makes C the answer.
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Re: Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1 [#permalink]

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New post 26 Mar 2016, 20:18
Is 1+x+x^2+x^3+x^4<1/(1-x)?

Although the series on L.H.S. is small otherwise we culd hav used G.P. for the sum ..
1(x^5-1)/(x-1) <1/(1-x)
x^5/(x-1)<0

----(+ve)----0-------(-ve)----------1--------(+ve)-----
So if x lies between 0 and 1 then we have YES for the above equation else NO.

1) x>0. Insufficient.
2) x<1. Insufficient.

Combining Sufficient. C is the answer.
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Re: Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1 [#permalink]

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New post 25 Apr 2016, 00:16
MathRevolution wrote:
Is 1+x+x^2+x^3+x^4<1/(1-x)?

1) x>0
2) x<1


* A solution will be posted in two days.


I took x=1,2 and x=1/2,0

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

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New post 05 Sep 2017, 05:24
Ans is C
is
\(1+x+x^2+x^3+x^4\) < \(\frac{1}{(1-x)}\)

Multiply and divide by 1-x on both numerator and denominator on L.H.S.

equation becomes
is
\(\frac{(1-x^5)}{(1-x)}\) < \(\frac{1}{(1-x)}\)
or
\(\frac{x^5}{(1-x)}\) > 0

Taking 3 points -1, 0 and 1 on number line
value above 1 i.e. x>1 will not hold this equation
x<1 but >0 will hold this equation
x<0 but > -1 will again not hold equation
x< -1 will again satisfy equation

so soln is x< -1 and x should be from (0,1) open brackets i.e. 0 and 1 not included.

1) x>0 is insufficient as it includes x>1 also which is not part of solution. A,D eliminated
2) x<0 is insufficient as it includes -1<x<0 which is not part of solution B eliminated

combine both
0<x<1
is our solution and Ans is C
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Re: Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1 [#permalink]

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New post 17 Sep 2017, 23:14
I have a query - If I take (C)
--> it means x = 0.5, 0.3, 0.2
With x = 0.5
1+ x + x sq + x^3 + x^4 < 1/1-x
1+ 0.5+ 0.25 + 0.125 + 0.0625 = 1/1-0.5
1.93< 2 SUFF
if x= 0.3
1+0.3+0.09 + 0.027 + 0.0081 = < 1/1 - 0.3
1.42 no less than 0.7
if x = 0.2
1+0.2+0.04 + 0.008 + 0.0016 <1/ 1-0.2
1.24 not less than 0.8
So I chose E.
Can anyone explain the issue in my reasoning here?
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Re: Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1 [#permalink]

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New post 17 Sep 2017, 23:26
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Madhavi1990 wrote:
I have a query - If I take (C)
--> it means x = 0.5, 0.3, 0.2
With x = 0.5
1+ x + x sq + x^3 + x^4 < 1/1-x
1+ 0.5+ 0.25 + 0.125 + 0.0625 = 1/1-0.5
1.93< 2 SUFF
if x= 0.3
1+0.3+0.09 + 0.027 + 0.0081 = < 1/1 - 0.3
1.42 no less than 0.7
if x = 0.2
1+0.2+0.04 + 0.008 + 0.0016 <1/ 1-0.2
1.24 not less than 0.8
So I chose E.
Can anyone explain the issue in my reasoning here?


It's \(\frac{1}{1 - 0.3}\) NOT \(\frac{1}{1}- 0.3\).

\((1 + 0.3 +0.3^2 + 0.3^3 + 0.3^4 = 1.4251)< (\approx 1.428=\frac{1}{1 - 0.3})\)

\((1 + 0.2 +0.2^2 + 0.2^3 + 0.2^4 = 1.2496)< (1.25=\frac{1}{1 - 0.3})\)
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Re: Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1 [#permalink]

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New post 17 Sep 2017, 23:41
hi, sorry I wasn not clear on my post earlier.
So after x = 0.2
I got 1.42 < 1/0.7 --> 1.42 < 10/7 which is approx 1.4
Thus
1.42 < 1.44 (TRUE)

1.24 < 1/0.8
1.24 < 5/4
1.24 < 1.25 (TRUE)

if x = 0.6
then 2.3 < 2.5 (TRUE)

Thank you, the above cleared it up for me. :)
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Re: Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1 [#permalink]

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New post 03 Dec 2017, 15:05
MathRevolution wrote:
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is 1+x+x^2+x^3+x^4<1/(1-x)?

1) x>0
2) x<1


The question is (1-x^5)/(1-x)<1/(1-x)? and –x^5/(1-x)<0?, x^5/(x-1)<0? is derived when delete 1/(1-x) from the both equations of 1/(1-x)-x^5/(1-x)<1/(1-x)?. When you multiply (x-1)^2 to the both equations, x^5(x-1)<0? is derived. Divide it with x^4 and it becomes x(x-1)<0?. That is, 0<x<1? is derived, which makes C the answer.


Thanks for such clear and precise explanation. But can this question be done by taking numbers and plugging in?
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Re: Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1 [#permalink]

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New post 04 Dec 2017, 10:22
Buttercup3 wrote:
MathRevolution wrote:
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is 1+x+x^2+x^3+x^4<1/(1-x)?

1) x>0
2) x<1


The question is (1-x^5)/(1-x)<1/(1-x)? and –x^5/(1-x)<0?, x^5/(x-1)<0? is derived when delete 1/(1-x) from the both equations of 1/(1-x)-x^5/(1-x)<1/(1-x)?. When you multiply (x-1)^2 to the both equations, x^5(x-1)<0? is derived. Divide it with x^4 and it becomes x(x-1)<0?. That is, 0<x<1? is derived, which makes C the answer.


Thanks for such clear and precise explanation. But can this question be done by taking numbers and plugging in?


Here algebra is slightly better but yes, it can be done by plugging in numbers. Let me try:

(1) x > 0

Now RHS of the question stem is 1/(1-x). Since its 1-x in the denominator, and this statement says x > 0, we should immediately look for two cases: one where 0<x<1 and one where x>1 (x cannot be 1 as for that RHS would be undefined).

So lets take a value of x between 0 and 1, say 0.5. RHS becomes 1/(1-0.5) = 2 LHS becomes 1+0.5+0.25+0.125+0.0625=1.9375. LHS is less than RHS
If we try any other value in the same range say 0.2 or 0.9, still RHS would be greater.

Now lets take another case where x > 1, in this case clearly we can see that 1-x would be negative, so RHS would be negative while LHS positive. So LHS would become greater.

So not sufficient.


(2) x < 1

We have got to take cases where 0<x<1, where -1<x<0, and where x<-1

If 0<x<1, RHS would be greater as discussed in first statement.

If -1<x<0, lets say x=-0.5. RHS = 1/(1+0.5) = 0.666.. and LHS = 1-0.5+0.25-0.125+0.0625 = 0.6875. LHS is greater

We dont even need to look at third case because we can see that this statement is not sufficient.



Combining the two statements: 0<x<1. In that case RHS would be greater as discussed. Sufficient.
Hence C answer
Re: Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1   [#permalink] 04 Dec 2017, 10:22
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