Difference between revisions of "2011 AMC 12A Problems/Problem 21"
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== Solution == | == Solution == | ||
− | The domain of <math>f_{1}(x)=\sqrt{1-x}</math> is defined when <math>x\leq-1</math>. <math>f_{2}(x)=f_{1}(\sqrt{4-x})=\sqrt{1-\sqrt{4-x}}</math>. 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>. Simplify this to arrive at the domain for <math>f_{2}(x)</math>, which is defined when <math>3\ | + | The domain of <math>f_{1}(x)=\sqrt{1-x}</math> is defined when <math>x\leq-1</math>. <math>f_{2}(x)=f_{1}(\sqrt{4-x})=\sqrt{1-\sqrt{4-x}}</math>. 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>. Simplify this to arrive at the domain for <math>f_{2}(x)</math>, which is defined when <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\leqx\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>. |
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== 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}} |
Revision as of 02:35, 12 February 2011
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
. Simplify this to arrive at the domain for
, which is defined when
. Repeat this process for
to get a domain of $-7\leqx\leq0$ (Error compiling LaTeX. Unknown error_msg). For
, since square roots are positive, we can exclude the negative values of the previous domain to arrive at
as the domain of
. We now arrive at a domain with a single number that defines
, however, since we are looking for the largest value for
for which the domain of
is nonempty, we must continue until we arrive at a domain that is empty. We continue with
to get a domain of
. Solve for
to get
. 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 |