Explanation

(a)
Given that, $R = {(2 + \sqrt{3})}^{2n}$ and $f = R - \lbrack R\rbrack$
As $0 < 2 - \sqrt{3} < 1$, we get $0 < F = {(2 - \sqrt{3})}^{2n} < 1$
We have, $R + F = {(2 + \sqrt{3})}^{2n} + {(2 - \sqrt{3})}^{2n}$
$= 2\left\lbrack \ ^{2n}C_{0}2^{2n} + \ ^{2n}C_{2}2^{2n - 2}\left( \sqrt{3} \right)^{2} + \ ^{2n}C_{4}\left( 2^{2n - 4} \right)\left( \sqrt{3} \right)^{4} + ... + \ ^{2n}C_{2n}\left( \sqrt{3} \right)^{2n} \right\rbrack$
$\Rightarrow R + F$ is an even integer
$\Rightarrow \lbrack R\rbrack + f + F$ is an even integer
$\Rightarrow f + F$ is an integer
But, $0 \leq f < 1\ and\ 0 < F < 1$
$\Rightarrow 0 < f + F\ < 2$
But the only integer between 0 and 2 is 1. Thus,
$f + F = 1\ \Rightarrow \ 1 - f = F$
Now, $R(1 - f) = RF = {(2 + \sqrt{3})}^{2n}{(2 - \sqrt{3})}^{2n}$
$= {(4 - 3)}^{2n} = 1^{2n} = 1$