Let $∆ = \left|\begin{array}{cc}f\left(x\right)& f\left(\frac{1}{x}\right)+f\left(x\right)\\ 1& f\left(\frac{1}{x}\right)\end{array}\right| = 0$ so that f(x) is polynomial and f(2) = 17 then f(5) =

$$\begin{gathered} ∆ = \left|\begin{array}{cc}\mathrm{f}\left(\mathrm{x}\right)& \mathrm{f}\left(\frac{1}{\mathrm{x}}\right)+\mathrm{f}\left(\mathrm{x}\right)\\ 1 & \mathrm{f}\left(\frac{1}{\mathrm{x}}\right)\end{array}\right| = 0 \\ \mathrm{f}\left(\mathrm{x}\right) \mathrm{f}\left(\frac{1}{\mathrm{x}}\right)-\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)-\mathrm{f}\left(\mathrm{x}\right)=0 \\ \mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\mathrm{f}\left(\mathrm{x}\right)-\mathrm{f}\left(\frac{1}{\mathrm{x}}\right) \\ \therefore \mathrm{f}\left(\mathrm{x}\right)=1+{\mathrm{x}}^{\mathrm{n}}; \mathrm{f}\left(2\right)=17 \\ =1+2^4 \\ \therefore \mathrm{f}\left(\mathrm{x}\right)=1+{\mathrm{x}}^4 \\ \mathrm{f}\left(5\right)=62 \mathrm{b} \end{gathered}$$