Question-211029

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

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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