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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((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 (D) 127

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

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 +∙∙∙+a_n x^n , a_n ≠0 Suppose that f(x)+f((1/x))=f(x)f((1/x)) for all x≠0 Then Σ_(r=0) ^n a_r x^r +Σ_(r=1) ^n (a_r /x^r )=(Σ_(r=0) ^n a_r x^r )(Σ_(r=1) ^n (a_r /x^r )) Multiplying through by x^n , we get that Σ_(r=0) ^n a_r x^(n+r) +Σ_(r=0) ^n a_r x^(n−r) =(Σ_(r=0) ^n a_r x^r )(Σ_(r=0) ^n a_r x^(n−r) ) That is, (a_0 x^n +a_1 x^(n+1) +∙∙∙a_n x^(2n) )+(a_0 x^n +a_1 x^(n−1) +∙∙∙+a_(n−1) x+a_n ) =(x^n +a_1 x^n +∙∙∙a_n x^n )(a_0 x^n +a_1 x^(n−1) +∙∙∙+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 2a_0 =a_0 ^2 +a_n ^2 a_n =a_0 a_n ⇒a_0 =1 (since a_n ≠0) a_(n−1) =a_0 a_(n−1) +a_1 a_n ⇒a_1 a_n =0⇒a_1 =0 a_(n−2) =a_2 a_n +a_1 a_(n−1) +a_(n−2) a_0 ⇒a_(n−2) =a_2 a_n +a_(n−2) ⇒a_2 =0 Continuing this process, we get that a_(n−1) =0 and 2=1+a_n ^2 . Hence a_n =±1. Therefore f(x)=1±x^n Since we are given that f(2)=9 we have 9=1±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.

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

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