If-f-x-is-a-polynomial-function-satisfying-the-relation-f-x-f-1-x-f-x-f-1-x-for-all-0-x-R-and-if-f-2-9-then-f-6-is-A-216-B-217-C-126-

Question Number 117984 by Ar Brandon last updated on 14/Oct/20

If f (x) is a polynomial function satisfying the relation

f (x) + f (\frac{1}{x}) = f (x) f (\frac{1}{x})

for all 0 \neq x \in R and if f (2) = 9, then f (6) is

(A) 216 (B) 217 (C) 126 (D) 127

Answered by Ar Brandon last updated on 14/Oct/20

f (2) = 9 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 that 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}, we get that

\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})

= (x^{n} + a_{1} x^{n} + \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 that a_{n - 1} = 0 and 2 = 1 + a_{n}^{2} .

Hence a_{n} = \pm 1 . Therefore

f (x) = 1 \pm x^{n}

Since we are given that f (2) = 9 we have

9 = 1 \pm 2^{n}

Therefore f (x) cannot be 1 - x^{n} . Thus, f (x) = 1 + x^{n} and

9 = 1 + 2^{n} and hence n = 3 . So f (x) = 1 + x^{3} and f (6) = 217 .