Difference between revisions of "2011 AMC 12A Problems/Problem 21"
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\textbf{(E)}\ 144 </math> | \textbf{(E)}\ 144 </math> | ||
− | == Solution == | + | == Solution 1== |
− | The domain of <math>f_{1}(x)=\sqrt{1-x}</math> is defined when <math>x\ | + | 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>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | We start with smaller values. Notice that <math>f_4(x) = \sqrt{1-\sqrt{4-\sqrt{9-\sqrt{16-x}}}}</math>. Notice that the mess after <math>\sqrt{1-(mess)}</math> 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 <math>\sqrt{16-x} = 0</math>, which means <math>x=16</math> is the only value in the domain of <math>f_4(x)</math>. Now we move on to <math>f_5(x)</math>. The only change with <math>f_5(x)</math> is replacing the <math>x</math> from <math>f_4(x)</math> with <math>\sqrt{25-x}</math>. Since we had <math>x=16</math> in <math>f_4(x)</math>, in <math>f_5(x)</math>, <math>\sqrt{25-x} = 16</math>, forcing <math>x=-231</math>. | ||
+ | |||
+ | |||
+ | Clearly, we can't move on from here, since <math>f_6(x)</math> would replace <math>x</math> with <math>\sqrt{36-x}</math>, and we would need <math>-231 = \sqrt{36-x}</math>, but a square root can never be negative, so <math>N=5</math>, <math>c=-231</math>, and the answer is <math>5-231 = \boxed{\textbf{(A) }-226}</math>. | ||
+ | |||
+ | -skibbysiggy | ||
== 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 13:02, 4 September 2024
Contents
[hide]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 1
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 .
Solution 2
We start with smaller values. Notice that . Notice that the mess after 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 , which means is the only value in the domain of . Now we move on to . The only change with is replacing the from with . Since we had in , in , , forcing .
Clearly, we can't move on from here, since would replace with , and we would need , but a square root can never be negative, so , , and the answer is .
-skibbysiggy
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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