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Question Number 211029 by BaliramKumar last updated on 26/Aug/24

	Answered by Ar Brandon last updated on 26/Aug/24

	$f (2) = 5 therefore f (x) must be a non - zero polynomial .$ $Let f (x) = a_{0} + a_{1} x + a_{2} x^{2} + \cdot \cdot \cdot + a_{n} x^{n}, a_{n} \neq 0$ $Suppose f (x) + f (\frac{1}{x}) = f (x) f (\frac{1}{x}) for all x \neq 0$ $Then \underset{r = 0}{\sum^{n}} a_{r} x^{r} + \underset{r = 1}{\sum^{n}} \frac{a_{r}}{x^{r}} = (\underset{r = 0}{\sum^{n}} a_{r} x^{r}) (\underset{r = 1}{\sum^{n}} \frac{a_{r}}{x^{r}})$ $Multiplying through by x^{n},$ $\underset{r = 0}{\sum^{n}} a_{r} x^{n + r} + \underset{r = 0}{\sum^{n}} a_{r} x^{n - r} = (\underset{r = 0}{\sum^{n}} a_{r} x^{r}) (\underset{r = 0}{\sum^{n}} a_{r} x^{n - r})$ $That is,$ $(a_{0} x^{n} + a_{1} x^{n + 1} + \cdot \cdot \cdot a_{n} x^{2 n}) + (a_{0} x^{n} + a_{1} x^{n - 1} + \cdot \cdot \cdot + a_{n - 1} x + a_{n})$ $= (a_{0} + a_{1} x + \cdot \cdot \cdot a_{n} x^{n}) (a_{0} x^{n} + a_{1} x^{n - 1} + \cdot \cdot \cdot + a_{n - 1} x + a_{n})$ $Equating the corresponding coefficients of powers of x,$ $we have$ $a_{n} = a_{0} a_{n}, a_{n - 1} = a_{0} a_{n - 1} + a_{1} a_{n}$ $a_{n - 2} = a_{2} a_{n} + a_{1} a_{n - 1} + a_{n - 2} a_{0}$ $2 a_{0} = a_{0}^{2} + a_{n}^{2}$ $a_{n} = a_{0} a_{n} \Rightarrow a_{0} = 1 (since a_{n} \neq 0)$ $a_{n - 1} = a_{0} a_{n - 1} + a_{1} a_{n} \Rightarrow a_{1} a_{n} = 0 \Rightarrow a_{1} = 0$ $a_{n - 2} = a_{2} a_{n} + a_{1} a_{n - 1} + a_{n - 2} a_{0} \Rightarrow a_{n - 2} = a_{2} a_{n} + a_{n - 2} \Rightarrow a_{2} = 0$ $Continuing this process, we get a_{n - 1} = 0 and 2 = 1 + a_{n}^{2} .$ $Hence a_{n} = \pm 1 . Therefore$ $f (x) = 1 \pm x^{n}$ $But f (2) = 5 hence$ $5 = 1 \pm 2^{n}$ $Therefore f (x) cannot be 1 - x^{n} . Thus, f (x) = 1 + x^{n} and$ $5 = 1 + 2^{n} \Rightarrow n = 2 . So f (x) = 1 + x^{2} and \begin{matrix} f (3) = 10 \end{matrix} .$
	Commented by Frix last updated on 26/Aug/24
	��
	Commented by Ar Brandon last updated on 26/Aug/24
	Greetings, Sir! it's been quite a long time��
	Commented by BaliramKumar last updated on 27/Aug/24

	$T h a n k s s i r$
	Commented by Ar Brandon last updated on 27/Aug/24
	You're welcome!
	Answered by Frix last updated on 26/Aug/24

	$We know 3 things :$ $(1) f (x) f (\frac{1}{x}) = f (x) + f (\frac{1}{x}) for all x \in R$ $(2) f (2) = 5$ $((1) \Rightarrow f (\frac{1}{x}) = \frac{f (x)}{f (x) + 1}) \land (2) \Rightarrow$ $(3) f (\frac{1}{2}) = \frac{5}{4}$ $\Rightarrow we can have 3 unknowns \Rightarrow we have a$ $polynomial of degree 2 \Rightarrow$ $f (x) = {a x}^{2} + b x + c$ $(2) 4 a + 2 b + c = 5$ $(3) \frac{a}{4} + \frac{b}{2} + c = \frac{5}{4}$ $\Rightarrow b = - \frac{5 (a - 1)}{2} \land a = c$ $Insert into (1) & transform$ $c^{2} - \frac{2 (x^{2} - 5 x + 1)}{(x - 2) (2 x - 1)} c - \frac{5 x}{(x - 2) (2 x - 1)} = 0$ $\Rightarrow c = 1 \lor c = - \frac{5 x}{(x - 2) (2 x - 1)}$ $Since c is a constant which must be valid$ $for all x \in R$ $\Rightarrow c = 1$ $\Rightarrow a = 1 \land b = 0$ $\Rightarrow$ $f (x) = x^{2} + 1$ $f (3) = 10$
	Commented by BaliramKumar last updated on 27/Aug/24

	$T h a n k s s i r$
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