Difference between revisions of "2011 AMC 12A Problems/Problem 21"
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== Problem == | == Problem == | ||
+ | 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> | ||
+ | \textbf{(A)}\ -226 \qquad | ||
+ | \textbf{(B)}\ -144 \qquad | ||
+ | \textbf{(C)}\ -20 \qquad | ||
+ | \textbf{(D)}\ 20 \qquad | ||
+ | \textbf{(E)}\ 144 </math> | ||
+ | |||
== Solution == | == Solution == | ||
+ | |||
+ | The domain of <math>f_{1}(x)=\sqrt{1-x}</math> is defined when <math>x\leq1</math>. | ||
+ | <cmath>f_{2}(x)=f_{1}\left(\sqrt{4-x}\right)=\sqrt{1-\sqrt{4-x}}</cmath> | ||
+ | |||
+ | Applying the domain of <math>f_{1}(x)</math> and the fact that square roots must be positive, we get <math>0\leq\sqrt{4-x}\leq1</math>. Simplifying, the domain of <math>f_{2}(x)</math> becomes <math>3\leq x\leq4</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 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}} |
Latest revision as of 21:20, 5 October 2022
Problem
Let , and for integers
, let
. If
is the largest value of
for which the domain of
is nonempty, the domain of
is
. What is
?
Solution
The domain of is defined when
.
Applying the domain of and the fact that square roots must be positive, we get
. Simplifying, the domain of
becomes
.
Repeat this process for to get a domain of
.
For , since square roots must be nonnegative, we can see that the negative values of the previous domain will not work, so
. Thus we now arrive at
being the only number in the of domain of
that defines
. However, since we are looking for the largest value for
for which the domain of
is nonempty, we must continue checking until we arrive at a domain that is empty.
We continue with to get a domain of
. Since square roots cannot be negative, this is the last nonempty domain. We add to get
.
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.