2011 AMC 12A Problem 21

Problem 21 of 25HarderAlgebra

Let

f_1(x) = \sqrt{1 - x},

and for integers

n \ge 2,

let

f_n(x) = f_{n-1}\!\left(\sqrt{n^2 - x}\right).

If

N

is the largest value of

n

for which the domain of

f_n

is nonempty, the domain of

f_N

is

\{c\}.

What is

N + c?

Show solution

Solution

Each step requires

\sqrt{n^2 - x}

to lie in the domain of

f_{n-1}.

Tracking the domains:

f_1: (-\infty, 1].

f_2: \sqrt{4 - x} \in (-\infty, 1] \Rightarrow [3, 4].

f_3: \sqrt{9 - x} \in [3, 4] \Rightarrow [-7, 0].

f_4: \sqrt{16 - x} \in [-7, 0] \Rightarrow \{16\}

(only the value

0

is possible).

f_5: \sqrt{25 - x} = 16 \Rightarrow \{-231\}.

For

f_6

we would need

\sqrt{36 - x} = -231,

impossible, so the domain is empty. Hence

N = 5,

c = -231,

and

N + c = -226.

Thus, the correct answer is A.

More practice

Concepts: function · radical · recursion

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.