Question

If $f\left(x\right)$ be a polynomial function satisfying $f\left(x\right)·f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right)$ and $f\left(4\right)=65$. Then find $f\left(6\right)$.

• A. $65$
• B. $217$
• C. $63$
• D. none of these

Answer 1

The correct option is B

$217$

Step 1: Simplify the equation

• Given: $f\left(x\right)·f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right)$ and $f\left(4\right)=65$

$$\begin{gathered} f\left(x\right)·f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right) \\ f\left(x\right)·f\left(\frac{1}{x}\right)-f\left(x\right)-f\left(\frac{1}{x}\right)+1=1 \\ f\left(x\right)\left[f\left(\frac{1}{x}\right)-1\right]-1\left[-1+f\left(\frac{1}{x}\right)\right]=1 \\ \left[f\left(x\right)-1\right]\left[f\left(\frac{1}{x}\right)-1\right]=1 \end{gathered}$$

Step 2: Let $f\left(x\right)-1=x^n$

• Now, let $f\left(x\right)-1=x^n$

$$\begin{gathered} f\left(x\right)-1=x^n\Rightarrow f\left(\frac{1}{x}\right)-1=\frac{1}{x^n} \\ f\left(x\right)=x^n+1 \end{gathered}$$

Step 3: Solve for the value of $f\left(6\right)$

$$\begin{gathered} \Rightarrow f\left(4\right)=65 \\ \Rightarrow 4^n+1=65 \\ \Rightarrow 4^n=64 \\ \Rightarrow 4^n=4^3 \\ \Rightarrow n=3 \\ \Rightarrow f\left(x\right)=x^3+1 \\ \Rightarrow f\left(6\right)={\left(6\right)}^3+1 \\ \Rightarrow f\left(6\right)=217 \end{gathered}$$

Hence, option B is correct.