Difference between revisions of "1988 AHSME Problems/Problem 30"
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==Solution== | ==Solution== | ||
− | + | Note that <math>x_{0} = 0</math> gives the constant sequence <math>0, 0, ...</math>, since <math>f(0) = 4 \cdot 0 - 0^2 = 0</math>. Because <math>f(4)=0, x_{0} = 4</math> gives the sequence <math>4, 0, 0, ...</math> with two different values. Similarly, <math>f(2) = 4</math>, so <math>x_{0} = 2</math> gives the sequence <math>2, 4, 0, 0, ...</math> with three values. In general, if <math>x_{0} = a_{n}</math> gives the sequence <math>a_{n}, a_{n-1}, ... , a_{2}, a_{1}, ...</math> with <math>n</math> different values, and <math>f(a_{n+1}) = a_{n}</math>, then <math>x_{0} = a_{n+1}</math> gives a sequence with <math>n+1</math> different values. (It is easy to see that we could not have <math>a_{n+1} = a_{i}</math> for some <math>i < n + 1</math>.) Thus, it follows by induction that there is a sequence with <math>n</math> distinct values for every positive integer <math>n</math>, as long as we can verify that there is always a real number <math>a_{n+1}</math> such that <math>f(a_{n+1}) = a_{n}</math>. This makes the answer <math>\boxed{\text{E}}</math>. The verification alluded to above, which completes the proof, follows from the quadratic formula: the solutions to <math>f(a_{n+1}) = 4a_{n+1} - a_{n+1}^{2} = a_{n}</math> are <math>a_{n+1} = 2 \pm \sqrt{4 - a_{n}}</math>. Hence if <math>0 \leq a_{n} \leq 4</math>, then <math>a_{n+1}</math> is real, since the part under the square root is non-negative, and in fact <math>0 \leq a_{n+1} \leq 4</math>, since <math>4-a_{n}</math> will be between <math>0</math> and <math>4</math>, so the square root will be between <math>0</math> and <math>2</math>, and <math>2 \pm</math> something between <math>0</math> and <math>2</math> gives something between <math>0</math> and <math>4</math>. Finally, since <math>a_{1} = 0 \leq 4</math>, it follows by induction that all terms satisfy <math>0 \leq a_{n} \leq 4</math>; in particular, they are all real. | |
== See also == | == See also == |
Latest revision as of 15:16, 27 February 2018
Problem
Let . Give
, consider the sequence defined by
for all
.
For how many real numbers
will the sequence
take on only a finite number of different values?
Solution
Note that gives the constant sequence
, since
. Because
gives the sequence
with two different values. Similarly,
, so
gives the sequence
with three values. In general, if
gives the sequence
with
different values, and
, then
gives a sequence with
different values. (It is easy to see that we could not have
for some
.) Thus, it follows by induction that there is a sequence with
distinct values for every positive integer
, as long as we can verify that there is always a real number
such that
. This makes the answer
. The verification alluded to above, which completes the proof, follows from the quadratic formula: the solutions to
are
. Hence if
, then
is real, since the part under the square root is non-negative, and in fact
, since
will be between
and
, so the square root will be between
and
, and
something between
and
gives something between
and
. Finally, since
, it follows by induction that all terms satisfy
; in particular, they are all real.
See also
1988 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
Followed by Last Question | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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