f(x) is an n^{text{th}} degree polynomial sat | vedclass

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(C) Given the determinant equation:$f(x) = \frac{1}{2} [f(x) \cdot f(\frac{1}{x}) - (f(\frac{1}{x}) - f(x)) \cdot 1]$$2f(x) = f(x)f(\frac{1}{x}) - f(\

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$244$

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(C) Given the determinant equation:$f(x) = \frac{1}{2} [f(x) \cdot f(\frac{1}{x}) - (f(\frac{1}{x}) - f(x)) \cdot 1]$$2f(x) = f(x)f(\frac{1}{x}) - f(\frac{1}{x}) + f(x)$$f(x) + f(\frac{1}{x}) = f(x)f(\frac{1}{x})$Let $f(x) = ax^n + c$. Then $ax^n + c + a(\frac{1}{x})^n + c = (ax^n + c)(a(\frac{1}{x})^n + c)$$ax^n + a x^{-n} + 2c = a^2 + acx^n + acx^{-n} + c^2$Comparing coefficients,we get $a = ac$,so $c = 1$ (assuming $a 
eq 0$).Then $a^2 + c^2 = 2c \implies a^2 + 1 = 2 \implies a^2 = 1$. Since $f(2) = 33$,$a(2^n) + 1 = 33 \implies a(2^n) = 32$.If $a = 1$,$2^n = 32 \implies n = 5$. Thus $f(x) = x^5 + 1$.Then $f(3) = 3^5 + 1 = 243 + 1 = 244$.

$f(x)$ is an $n^{\text{th}}$ degree polynomial satisfying $f(x) = \frac{1}{2} \begin{vmatrix} f(x) & f(\frac{1}{x}) - f(x) \\ 1 & f(\frac{1}{x}) \end{vmatrix}$. If $f(2) = 33$,then the value of $f(3)$ is

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