Question

Let $f_1(x) = (x-2)^2$, $f_2(x) = ((x-2)^2 - 2)^2$, $f_3(x) = (((x-2)^2 - 2)^2 - 2)^2$,...... and so on ; so that $f_k(x) = \underbrace{(((...(x-2)^2 - 2)^2 - 2)^2..... - 2)^2}_{k \text{ times}} = A_k + B_k x + C_k x^2 + D_k x^3 + ......$

$B_5$ is equal to

• A. -2048
• B. -32
• C. -1024
• D. -512

Answer 1

Answer: (C)

-1024

Answer 2

Define $f_k(x) = (f_{k-1}(x) - 2)^2$ for $k \ge 2$, and $f_1(x) = (x-2)^2$. The coefficient $B_k$ is $f_k'(0)$.

Calculate $B_1$: $f_1(x) = (x-2)^2 \implies f_1'(x) = 2(x-2)$. So, $B_1 = f_1'(0) = 2(0-2) = -4$.

Find the recurrence for $B_k$: Differentiate $f_k(x) = (f_{k-1}(x) - 2)^2$ to get $f_k'(x) = 2(f_{k-1}(x) - 2)f_{k-1}'(x)$.

Evaluate at $x=0$: $f_k'(0) = 2(f_{k-1}(0) - 2)f_{k-1}'(0)$.

First, find $f_k(0)$: $f_1(0) = 4$. If $f_{k-1}(0)=4$, then $f_k(0) = (4-2)^2 = 4$. Thus, $f_k(0)=4$ for all $k$.

Substitute $f_{k-1}(0)=4$ into the recurrence for $f_k'(0)$: $B_k = 2(4-2)B_{k-1} = 4B_{k-1}$.

Solve the recurrence: $B_k = B_1 \cdot 4^{k-1} = -4 \cdot 4^{k-1} = -4^k$.

Calculate $B_5$: $B_5 = -4^5 = -1024$.