Answer: $\mathbf A$
Explanation:
$\mathrm {f_1(x) = (x-2)^2\;;\;f_{n+1} = (f_n-2)^2} \;\;\; \text {[Given]}$
$\mathrm {f_1(x) = (x-2)^2 = 4 - 4 x +x^2}$
$\mathrm {f_2(x) = (f_1(x)-2)^2 = (2-4x+x^2)^2 = 2^2 + 2.2(-4x) +\cdots=4-4^2x+\cdots}$
$\mathrm {f_3(x) = (f_2(x)-2)^2 = (4-2^4x+\cdots-2)^2 = (2-2^4x+\cdots)^2\cdots =2^2-2.2.4^2x+\cdots=2^2-4^3x+\cdots}$
$\mathrm {f_4(x) = (f_3(x)-2)^2=2^2-4^4x+\cdots}$
$\mathrm {f_5(x) = (f_4(x)-2)^2 = 2^2-4^5x+\cdots}$
So, $\mathrm {a_n = 4\;,\;b_n = -4^{(n^{th}\;term \;\;of\; 1,2,3,4,5)}}$
$\therefore \mathbf A $ is the correct option.