Poly fun

Algebra Level 5

f(x) is a polynomial function with positive integral coefficients degree of f is greater than or equal to 1 f ( f ( [ x ] ) + 1 ) f ( [ x ] ) = k \frac{f(f([x])+1)}{f([x])}=k x ϵ [ a + 1 , b ) x\epsilon[a+1,b) k takes 3 value a,b,c The equation (x-a)(x-b)(x-c)=1 whose roots are given by x i x_{i} then i = 1 3 l i m m l i m n ( c o s ( π n ! x i ) ) m = a \sum_{i=1}^{3}lim_{m\rightarrow\infty}lim_{n\rightarrow\infty}(cos(\pi n!x_{i}))^{m}=a

find (a+b) find a+b


The answer is 2.

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1 solution

Milun Moghe
Mar 1, 2014

f ( x ) = a n x n + a n 1 x n 1 + . . . + a 0 f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{0}

[x]=m

f ( m ) = a n m n + a n 1 m n 1 + . . . + a 0 f(m)=a_{n}m^{n}+a_{n-1}m^{n-1}+...+a_{0}

f ( f ( m ) + 1 ) = a n ( f ( m ) + 1 ) n + a n 1 ( f ( m ) + 1 ) n 1 + . . . + a 0 = C f ( m ) + f ( 1 ) f(f(m)+1)=a_{n}(f(m)+1)^{n}+a_{n-1}(f(m)+1){}^{n-1}+...+a_{0}=Cf(m)+f(1)

f ( f ( m ) + 1 ) f ( m ) = C + f ( 1 ) f ( m ) = i n t e g e r \frac{f(f(m)+1)}{f(m)}=C+\frac{f(1)}{f(m)}=integer

m=1

( x a ) ( x b ) ( x c ) 1 = p ( x ) q ( x ) (x-a)(x-b)(x-c)-1=p(x)q(x) ......(1)

let p(x) is linear and q(x) is quadratic with integral coefficients

p ( a ) q ( a ) = p ( b ) q ( b ) = p ( c ) q ( c ) = 1 p(a)q(a)=p(b)q(b)=p(c)q(c)=-1

as p(a) is has integer coefficients and a is an integer

similarly with b and c

p ( a ) = 1 o r 1 p(a)=1or-1

similar with b and c

a linear polynomial being one one cannot attain one- value for two numbers

therefore (1) as irrational roots

therefore the given limit =0 as x i i r r a t i o n a l x_{i}\rightarrow irrational

a=0 b=2

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