Difference between revisions of "2019 AMC 10B Problems/Problem 9"
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&=a-(a+1)=-1\end{split}</cmath> | &=a-(a+1)=-1\end{split}</cmath> | ||
− | Thus, the range of | + | Thus, the range of x is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>. |
''Note'': One could solve the case of <math>x</math> as a negative non-integer in this way: | ''Note'': One could solve the case of <math>x</math> as a negative non-integer in this way: |
Revision as of 18:19, 22 December 2020
Problem
The function is defined by
for all real numbers
, where
denotes the greatest integer less than or equal to the real number
. What is the range of
?
Solution 1
There are four cases we need to consider here.
Case 1: is a positive integer. Without loss of generality, assume
. Then
.
Case 2: is a positive fraction. Without loss of generality, assume
. Then
.
Case 3: is a negative integer. Without loss of generality, assume
. Then
.
Case 4: is a negative fraction. Without loss of generality, assume
. Then
.
Thus the range of the function is
.
~IronicNinja, edited by someone else hehe
Solution 2
It is easily verified that when is an integer,
is zero. We therefore need only to consider the case when
is not an integer.
When is positive,
, so
When is negative, let
be composed of integer part
and fractional part
(both
):
Thus, the range of x is .
Note: One could solve the case of as a negative non-integer in this way:
Video Solution
~savannahsolver
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.