Explanation

(b)
Given, $f(x).f\left( \frac{1}{2} \right) = f(x) + f\left( \frac{1}{x} \right)$
Let $f(x) = x^{n} \pm 1$, where $n \in I$.
Now, $f(4) = 65$
Case I
Let $f(x) = x^{n} + 1$
$\Rightarrow \ \ \ \ \ \ \ f(4) = 4^{n} + 1$
$\Rightarrow \ \ \ 65 = 4^{n} + 1$
$\Rightarrow \ \ \ \ \ \ n = 3$
Case II
Let $f(x) = x^{n} - 1$
$\Rightarrow \ \ \ \ f(4) = 4^{n} - 1\ \Rightarrow 65 = 4^{n} - 1$
$\Rightarrow \ \ \ \ \ \ \ \ 4^{n} = 66$
The quality does not hold true for $n \in Z$.
Therefore, $f(x) = x^{3} + 1$
Now, $f(6) = 6^{3} + 1 = 216 + 1 = 217$