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doubt..
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Nikkita
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Jun 23, 2017; 8:45am
doubt..
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Nikkita
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Jun 23, 2017; 8:46am
Re: doubt..
133 posts
This concept was randomly used in a ques..
Asd1995
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Jun 23, 2017; 1:55pm
Re: doubt..
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by Nikkita
Define g(x) =1+f(x) and solve. It's easier that way.
Alternatively write down general form of n degree polynomial and compare coefficients on lhs and rhs of x^k, where 0<=k<=n
Asd1995
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Jun 23, 2017; 2:37pm
Re: doubt..
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by Nikkita
You can also factorise it as (1-g(x))(1-g(1/x))=1 for all x.
If f(x)=1-g(x), we get f(x) f(1/x)=1, so that f(x)=+- x^n
Nikkita
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Jun 24, 2017; 2:26am
Re: doubt..
133 posts
Thankyou Asd1995
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