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

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Repeat this process for <math>f_{3}(x)=\sqrt{1-\sqrt{4-\sqrt{9-x}}}</math> to get a domain of <math>-7\leq x\leq0</math>.   
 
Repeat this process for <math>f_{3}(x)=\sqrt{1-\sqrt{4-\sqrt{9-x}}}</math> to get a domain of <math>-7\leq x\leq0</math>.   
  
−
For <math>f_{4}(x)</math>, since square roots are positive, we can exclude the negative values of the previous domain to arrive at <math>\sqrt{16-x}=0</math> as the domain of <math>f_{4}(x)</math>. We now arrive at a domain with a single number that defines <math>x</math>, however, since we are looking for the largest value for <math>n</math> for which the domain of <math>f_{n}</math> is nonempty, we must continue until we arrive at a domain that is empty. We continue with <math>f_{5}(x)</math> to get a domain of <math>\sqrt{25-x}=16</math>. Solve for <math>x</math> to get <math>x=-231</math>. Since square roots cannot be negative, this is the last nonempty domain. We add to get <math>5-231=\boxed{\textbf{(A)}\ -226}</math>.
+
For <math>f_{4}(x)</math>, since square roots must be nonnegative, we can see that the negative values of the previous domain will not work, so <math>\sqrt{16-x}=0</math>. Thus we now arrive at <math>16</math> being the only number in the of domain of <math>f_4 x</math> that defines <math>x</math>. However, since we are looking for the largest value for <math>n</math> for which the domain of <math>f_{n}</math> is nonempty, we must continue checking until we arrive at a domain that is empty.  
 +
 
 +
We continue with <math>f_{5}(x)</math> to get a domain of <math>\sqrt{25-x}=16 \implies x=-231</math>. Since square roots cannot be negative, this is the last nonempty domain. We add to get <math>5-231=\boxed{\textbf{(A)}\ -226}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=20|num-a=22|ab=A}}
 
{{AMC12 box|year=2011|num-b=20|num-a=22|ab=A}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:20, 5 October 2022

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

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 $5-231=\boxed{\textbf{(A)}\ -226}$.

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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