Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎See also)
Line 15: Line 15:
 
== 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}}

Revision as of 20:52, 3 July 2013

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 $x\leq1$. $f_{2}(x)=f_{1}(\sqrt{4-x})=\sqrt{1-\sqrt{4-x}}$. Applying the domain of $f_{1}(x)$ and the fact that square roots must be positive, we get $0\leq\sqrt{4-x}\leq1$. Simplify this to arrive at the domain for $f_{2}(x)$, which is defined when $3\leq x\leq4$. 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 are positive, we can exclude the negative values of the previous domain to arrive at $\sqrt{16-x}=0$ as the domain of $f_{4}(x)$. We now arrive at a domain with a single number 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 until we arrive at a domain that is empty. We continue with $f_{5}(x)$ to get a domain of $\sqrt{25-x}=16$. Solve for $x$ to get $x=-231$. 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_21&oldid=53843"