If $f(x)$ is a polynomial function satisfying $f(x) \cdot f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)$ and $f(4)=257$, then $f(3)=$

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Aicharya · Accepted Answer

Let $ f(x)=a_0 x^n+a_1 x^{n-1}+a_2 x^{n-2}+\ldots+a_{n-1} x+a_n $ Then, $f(x) \cdot f\left(\frac{1}{x}ight)=f(x)+f\left(\frac{1}{x}ight)$ $ \begin{aligned} \Rightarrow\left(a_0 x^n+a_1 x^{n-1}+\ldots+a_night) & \left(\frac{a_0}{x^n}+\frac{a_1}{x^{n-1}}+\ldots+a_night) \end{aligned} $ On comparing the coefficient of $x^n$, we have $ a_0 a_n=a_0 \Rightarrow a_n=1 $ Comparing the coefficient of $x^{n-1}$, we have $ \begin{aligned} & \Rightarrow a_0 a_{n-1}+a_n a_1=a_1 \ & \Rightarrow \quad a_0 a_{n-1}+a_1=a_1 \ & {\left[	ext { as } a_n=1ight]} \ & \Rightarrow \quad a_0 a_{n-1}=0 \ & \Rightarrow \quad a_{n-1}=0 \quad\left[	ext { as } a_0 
eq 0ight] \ & \end{aligned} $ Similarly, $\quad a_{n-1}=a_{n-2}=\ldots=a_1=0$ and $\quad a_0= \pm 1$ $ \begin{aligned} & 	herefore \quad f(x)=1 \pm x^n \ & f(4)=1 \pm 4^n=257 \Rightarrow 4^n=256 \ & \Rightarrow \quad 4^n=256 \quad \Rightarrow \quad 4=4 \ & f(x)=1+x^4 \ & \end{aligned} $ So, $f(3)=1+3^4=82$

If $f(x)$ is a polynomial function satisfying $f(x) \cdot…

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