Question

If $f\left(x\right)$ is a polynomial in $x$ $(>0)$ satisfying the equation $f\left(x\right)+f\left(\frac{1}{x}\right)=f\left(x\right)f\left(\frac{1}{x}\right)$ and $f\left(2\right)=9,$ then $f\left(3\right)=$

• A. $26$
• B. $27$
• C. $28$
• D. $29$

Answer 1

The correct option is C $28$

Suppose $f\left(x\right)$ is a polynomial of degree 'n'.
$f\left(x\right)=a_0+a_{1}x+........a_{n-1}{x}^{n-1}+a_n{x}^n\Rightarrow f\left(\frac{1}{x}\right)=a_0+\frac{a_1}{x}+\frac{a_2}{x^2}+........+\frac{a_{n-1}}{x^{n-1}}+\frac{a_n}{x^n}$
Given that $f\left(x\right)+f\left(\frac{1}{x}\right)=f\left(x\right).f\left(\frac{1}{x}\right)$
$\Rightarrow (a_0+a_{1}x+........a_{n-1}{x}^{n-1}+a_n{x}^n)+(a_0+\frac{a_1}{x}+\frac{a_2}{x^2}+........+\frac{a_{n-1}}{x^{n-1}}+\frac{a_n}{x^n})=(a_0+a_{1}x+........a_{n-1}{x}^{n-1}+a_n{x}^n)(a_0+\frac{a_1}{x}+\frac{a_2}{x^2}+........+\frac{a_{n-1}}{x^{n-1}}+\frac{a_n}{x^n})$
.
Comparing the coefficients of $x^n$ on both sides, we get $a_n=a_n{a}_0\Rightarrow a_0=1$ ( $\because a_n\ne 0$)
Similarly comapring the coefficients of $x^{n-1}$ on both sides, we get $a_{n-1}=a_n{a}_1+a_{n-1}{a}_0\Rightarrow a_n{a}_1=0\Rightarrow a_1=0$
($\because a_0=1$)
Similarly comparing the coefficients of $x^{n-2},x^{n-3},.....x$, we get
$a_2=a_3=a_4=..........a_{n-1}=0$
$\Rightarrow f\left(x\right)=a_n{x}^n+1$
$\Rightarrow f\left(\frac{1}{x}\right)=\frac{a_n}{x^n}+1\Rightarrow (a_n{x}^n+1)+(\frac{a_n}{x^n}+1)=(a_n{x}^n+1)(\frac{a_n}{x^n}+1)$
Comparing the constant terms on both sides we get $2=a_n^2+1\Rightarrow a_n=\pm 1$
$\Rightarrow f\left(x\right)=\pm x^n+1$
Given $f\left(2\right)=9\Rightarrow f\left(2\right)=\pm 2^n+1=9$
$\Rightarrow 2^n=8$ ( $\because -ve$ is not possible)
$\Rightarrow n=3$
$f\left(x\right)=x^3+1$
$\therefore f\left(3\right)=27+1=28$