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

$$\mathrm{If}f\left(x\right)\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{function}\mathrm{satisfying}\mathrm{the}\mathrm{relation}$$$$f\left(x\right)+f\left(\frac{1}{x}\right)=f\left(x\right)f\left(\frac{1}{x}\right)$$$$\mathrm{for}\mathrm{all}0\ne x\in \mathbb{R}\mathrm{and}\mathrm{if}f\left(2\right)=9,\mathrm{then}\mathrm{f}\left(6\right)\mathrm{is}$$$$\left(\mathrm{A}\right)216\left(\mathrm{B}\right)217\left(\mathrm{C}\right)126\left(\mathrm{D}\right)127$$

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

$$f\left(2\right)=9\mathrm{therefore}f\left(x\right)\mathrm{must}\mathrm{be}\mathrm{a}\mathrm{non}-\mathrm{zero}\mathrm{polynomial}.$$$$\mathrm{Let}f\left(x\right)=a_0+a_{1}x+a_2{x}^2+\cdot \cdot \cdot +a_n{x}^n,a_n\ne 0$$$$\mathrm{Suppose}\mathrm{that}f\left(x\right)+f\left(\frac{1}{x}\right)=f\left(x\right)f\left(\frac{1}{x}\right)\mathrm{for}\mathrm{all}x\ne 0$$$$\mathrm{Then}\underset{\mathrm{r}=0}{\sum ^n}{a}_{\mathrm{r}}{x}^{\mathrm{r}}+\underset{\mathrm{r}=1}{\sum ^n}\frac{a_{\mathrm{r}}}{x^{\mathrm{r}}}=\left(\underset{\mathrm{r}=0}{\sum ^n}{a}_{\mathrm{r}}{x}^{\mathrm{r}}\right)\left(\underset{\mathrm{r}=1}{\sum ^n}\frac{a_{\mathrm{r}}}{x^{\mathrm{r}}}\right)$$$$\mathrm{Multiplying}\mathrm{through}\mathrm{by}{x}^n,\mathrm{we}\mathrm{get}\mathrm{that}$$$$\underset{\mathrm{r}=0}{\sum ^n}{a}_{\mathrm{r}}{x}^{n+\mathrm{r}}+\underset{\mathrm{r}=0}{\sum ^n}{a}_{\mathrm{r}}{x}^{n-\mathrm{r}}=\left(\underset{\mathrm{r}=0}{\sum ^n}{a}_{\mathrm{r}}{x}^{\mathrm{r}}\right)\left(\underset{\mathrm{r}=0}{\sum ^n}{a}_{\mathrm{r}}{x}^{n-\mathrm{r}}\right)$$$$\mathrm{That}\mathrm{is},$$$$(a_0{x}^n+a_1{x}^{n+1}+\cdot \cdot \cdot a_n{x}^{2n})+(a_0{x}^n+a_1{x}^{n-1}+\cdot \cdot \cdot +a_{n-1}x+a_n)$$$$=(x^n+a_1{x}^n+\cdot \cdot \cdot a_n{x}^n)(a_0{x}^n+a_1{x}^{n-1}+\cdot \cdot \cdot +a_{n-1}x+a_n)$$$$\mathrm{Equating}\mathrm{the}\mathrm{corresponding}\mathrm{coefficients}\mathrm{of}\mathrm{powers}\mathrm{of}x,$$$$\mathrm{we}\mathrm{have}$$$$a_n=a_0{a}_n,a_{n-1}=a_0{a}_{n-1}+a_1{a}_n$$$$a_{n-2}=a_2{a}_n+a_1{a}_{n-1}+a_{n-2}{a}_0$$$$2a_0=a_0^2+a_n^2$$$$a_n=a_0{a}_n\Rightarrow a_0=1(\mathrm{since}{a}_n\ne 0)$$$$a_{n-1}=a_0{a}_{n-1}+a_1{a}_n\Rightarrow a_1{a}_n=0\Rightarrow a_1=0$$$$a_{n-2}=a_2{a}_n+a_1{a}_{n-1}+a_{n-2}{a}_0\Rightarrow a_{n-2}=a_2{a}_n+a_{n-2}\Rightarrow a_2=0$$$$\mathrm{Continuing}\mathrm{this}\mathrm{process},\mathrm{we}\mathrm{get}\mathrm{that}{a}_{n-1}=0\mathrm{and}2=1+a_n^2.$$$$\mathrm{Hence}{a}_n=\pm 1.\mathrm{Therefore}$$$$f\left(x\right)=1\pm x^n$$$$\mathrm{Since}\mathrm{we}\mathrm{are}\mathrm{given}\mathrm{that}f\left(2\right)=9\mathrm{we}\mathrm{have}$$$$9=1\pm 2^{\mathrm{n}}$$$$\mathrm{Therefore}f\left(x\right)\mathrm{cannot}\mathrm{be}1-x^n.\mathrm{Thus},f\left(x\right)=1+x^n\mathrm{and}$$$$9=1+2^n\mathrm{and}\mathrm{hence}n=3.\mathrm{So}f\left(x\right)=1+x^3\mathrm{and}f\left(6\right)=217.$$

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