Difference between revisions of "2011 AMC 12A Problems/Problem 21"
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− | 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 a square root can never be negative, so <math>N=5</math>, <math>c=-231</math>, and the answer is | + | 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 | -skibbysiggy |
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 |
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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