Difference between revisions of "2005 AMC 12A Problems/Problem 20"
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== Solution == | == Solution == | ||
− | For the two functions <math>f(x)=2x,0\le x\le \frac{1}{2}</math> and <math>f(x)=2-2x,\frac{1}{2}\le x\le 1</math>, | + | For the two functions <math>f(x)=2x,0\le x\le \frac{1}{2}</math> and <math>f(x)=2-2x,\frac{1}{2}\le x\le 1</math>,as long as <math>f(x)</math> is between <math>0</math> and <math>1</math>, <math>x</math> will be in the right domain, so we don't need to worry about the domain of <math>x</math>. |
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+ | Also, every time we change <math>f(x)</math>, the expression for the final answer in terms of <math>x</math> will be in a different form, so we get a different value of x. Every time we have two choices for <math>f(x</math>) and altogether we have to choose <math>2005</math> times. Thus, <math>2^{2005}\Rightarrow\boxed{E}</math>. | ||
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==See Also== | ==See Also== | ||
{{AMC12 box|ab=A|num-b=19|num-a=21|year=2005}} | {{AMC12 box|ab=A|num-b=19|num-a=21|year=2005}} |
Revision as of 18:41, 9 November 2013
Problem
For each in
, define
Let
, and
for each integer
. For how many values of
in
is
?
Solution
For the two functions and
,as long as
is between
and
,
will be in the right domain, so we don't need to worry about the domain of
.
Also, every time we change , the expression for the final answer in terms of
will be in a different form, so we get a different value of x. Every time we have two choices for
) and altogether we have to choose
times. Thus,
.
See Also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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.