Let, $x + \frac{1}{x} + 4 = t$
$\therefore x + \frac{1}{x} = t - 4$
$ \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = {t^2} - 8t + 16$
$ \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 14 = {t^2} - 8t = t(t - 8)$
$\therefore f(t) = t(t - 8)$
So, $f(0) = 0$
If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096 $ $\therefore 2^n = 4096 =2^{12} $ $\Rightarrow n = 12 $ Greatest coefficient = ${}^{12}{C_6} = 924$