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If $f(x) $ is polynomial and $f(x).f(\large\frac{1}{x})$$=f(x)+f(\large\frac{1}{x})$ and $f(3)=28 $ then find $f(4)$

$\begin{array}{1 1} 63 \\ 64 \\ 65 \\ 67 \end{array}$

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Given $f(x) $ is a polynomial.
$\therefore\:Let\:f(x)=a_0x^n+a_1x^{n-1}+a_2x^{n-2}+...............a_{n-1}x+a_n$ where $a_0\neq 0$
Given that $f(x).f(\large\frac{1}{x})$$=f(x)+f(\large\frac{1}{x})$.
$\Rightarrow\:(a_0x^n+a_1x^{n-1}+........a_n)$ $ (\large\frac{a_0}{x^n}+\frac{a_1}{x^{n-2}}+...........$$a_n)$
$=(a_0x^n+a_1x^{n-1}+........a_n) +$ $ (\large\frac{a_0}{x^n}+\frac{a_1}{x^{n-2}}+...........$$a_n)$
Comparing the coefficients of $x^n$ terms on either sides we get
$a_0a_n=a_0\Rightarrow\:a_0(a_n-1)=0$
$\Rightarrow\:a_n=1,$ because $a_0\neq 0$
Comparing the coefficient of $x^{n-1}$ on either sides we get
$a_1a_n+a_0a_{n-1}=a_1$
Substituting the value of $a_n$ we get
$a_1+a_0a_{n-1}=a_1$
$\Rightarrow\:a_0a_{n-1}=0\Rightarrow\:a_0=0,\:\:or\:\:a_{n-1}=0$
But since $a_0\neq 0, a_{n-1`}=0$
Similarly by comparing the coefficient of $x^{n-2},\:x^{n-3}$........x on either sides
and substituting the values of $a_n,\:a_{n-1}$........etc., we can prove that
$a_{n-2}=a_{n-3}=a_{n-3}.............=a_1=0$
Also by comparing the constant term on either sides we get
$a_0^2+a_1^2+a_2^2+.............a_n^2=2a_n$
Substituting the values of $a_n,a_{n-1},a_{n-2}............a_1,$ we get
$a_0^2+1=2$ $\Rightarrow\:a_0^2=1$
$\Rightarrow\:a_0=\pm 1$
Substituting the values of $a_0,a_1,a_2,...........a_n$ in the polynomial $f(x),$
We get $f(x)=-x^n+1$ or $f(x)=x^n+1$
given $f(3)=28$
If $f(x)=-x^n+1$, then
$f(3)=-3^n+1=28$
$\Rightarrow\:3^n=-27$ which is not possible.
$\therefore\:f(x)=x^n+1$
$f(3)=3^n+1=28$ $\Rightarrow\:3^n=27$
$\Rightarrow n=3$
$f(4)=4^3+1=65$
answered May 30, 2013 by rvidyagovindarajan_1
edited May 19, 2014 by rohanmaheshwari0831_1
 
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