Difference between revisions of "2019 AMC 10B Problems/Problem 9"
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<math>\textbf{(A) } \{-1, 0\} \qquad\textbf{(B) } \text{The set of nonpositive integers} \qquad\textbf{(C) } \{-1, 0, 1\} \qquad\textbf{(D) } \{0\}</math> <br> <math>\textbf{(E) } \text{The set of nonnegative integers} </math> | <math>\textbf{(A) } \{-1, 0\} \qquad\textbf{(B) } \text{The set of nonpositive integers} \qquad\textbf{(C) } \{-1, 0, 1\} \qquad\textbf{(D) } \{0\}</math> <br> <math>\textbf{(E) } \text{The set of nonnegative integers} </math> | ||
− | + | == Solution 1 == | |
There are four cases we need to consider here. | There are four cases we need to consider here. | ||
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~IronicNinja | ~IronicNinja | ||
− | + | == Solution 2 == | |
It is easily verified that when <math>x</math> is an integer, <math>f(x)</math> is zero. We therefore need only to consider the case when <math>x</math> is not an integer. | It is easily verified that when <math>x</math> is an integer, <math>f(x)</math> is zero. We therefore need only to consider the case when <math>x</math> is not an integer. | ||
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&=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = -1\end{split}</cmath> | &=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = -1\end{split}</cmath> | ||
− | + | == Video Solution == | |
https://youtu.be/PgqjsTkNYdc | https://youtu.be/PgqjsTkNYdc | ||
Latest revision as of 20:26, 17 April 2021
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
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.