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Difference between revisions of "2011 AMC 12A Problems/Problem 21"

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== Problem ==
 
== Problem ==
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Let <math>f_{1}(x)=\sqrt{1-x}</math>, and for integers <math>n \geq 2</math>, let <math>f_{n}(x)=f_{n-1}(\sqrt{n^2 - x})</math>. If <math>N</math> is the largest value of <math>n</math> for which the domain of <math>f_{n}</math> is nonempty, the domain of <math>f_{N}</math> is <math>[c]</math>. What is <math>N+c</math>?
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Let <math>f_{1}(x)=\sqrt{1-x}</math>, and for integers <math>n \geq 2</math>, let <math>f_{n}(x)=f_{n-1}(\sqrt{n^2 - x})</math>. If <math>N</math> is the largest value of <math>n</math> for which the domain of <math>f_{n}</math> is nonempty, the domain of <math>f_{N}</math> is <math>\{c\}</math>. What is <math>N+c</math>?
  
 
<math>
 
<math>

Latest revision as of 13:47, 27 February 2025

Problem

Let $f_{1}(x)=\sqrt{1-x}$, and for integers $n \geq 2$, let $f_{n}(x)=f_{n-1}(\sqrt{n^2 - x})$. 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 $N+c$?

$\textbf{(A)}\ -226 \qquad \textbf{(B)}\ -144 \qquad \textbf{(C)}\ -20 \qquad \textbf{(D)}\ 20 \qquad \textbf{(E)}\ 144$

Solution 1

The domain of $f_{1}(x)=\sqrt{1-x}$ is defined when $x\leq1$. \[f_{2}(x)=f_{1}\left(\sqrt{4-x}\right)=\sqrt{1-\sqrt{4-x}}\]

Applying the domain of $f_{1}(x)$ and the fact that square roots must be positive, we get $0\leq\sqrt{4-x}\leq1$. Simplifying, the domain of $f_{2}(x)$ becomes $3\leq x\leq4$.

Repeat this process for $f_{3}(x)=\sqrt{1-\sqrt{4-\sqrt{9-x}}}$ to get a domain of $-7\leq x\leq0$.

For $f_{4}(x)$, since square roots must be nonnegative, we can see that the negative values of the previous domain will not work, so $\sqrt{16-x}=0$. Thus we now arrive at $16$ being the only number in the of domain of $f_4 x$ that defines $x$. However, since we are looking for the largest value for $n$ for which the domain of $f_{n}$ is nonempty, we must continue checking until we arrive at a domain that is empty.

We continue with $f_{5}(x)$ to get a domain of $\sqrt{25-x}=16 \implies x=-231$. Since square roots cannot be negative, this is the last nonempty domain. We add to get $5-231=\boxed{\textbf{(A)}\ -226}$.

Solution 2

We start with smaller values. Notice that $f_4(x) = \sqrt{1-\sqrt{4-\sqrt{9-\sqrt{16-x}}}}$. Notice that the mess after $\sqrt{1-(mess)}$ must be greater than 0, since it's a square root, and less than 1, since otherwise the inside of the larger square root on the outside would be negative.


Continuing, we get that $\sqrt{16-x} = 0$, which means $x=16$ is the only value in the domain of $f_4(x)$. Now we move on to $f_5(x)$. The only change with $f_5(x)$ is replacing the $x$ from $f_4(x)$ with $\sqrt{25-x}$. Since we had $x=16$ in $f_4(x)$, in $f_5(x)$, $\sqrt{25-x} = 16$, forcing $x=-231$.


Clearly, we can't move on from here, since $f_6(x)$ would replace $x$ with $\sqrt{36-x}$, and we would need $-231 = \sqrt{36-x}$, but a square root can never be negative, so $N=5$, $c=-231$, and the answer is $5-231 = \boxed{\textbf{(A) }-226}$.

-skibbysiggy

See also

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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