Difference between revisions of "2019 AMC 10B Problems/Problem 9"
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− | + | == Problem == | |
+ | The function <math>f</math> is defined by <cmath>f(x) = \lfloor|x|\rfloor - |\lfloor x \rfloor|</cmath>for all real numbers <math>x</math>, where <math>\lfloor r \rfloor</math> denotes the greatest integer less than or equal to the real number <math>r</math>. What is the range of <math>f</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. | ||
+ | |||
+ | '''Case 1''': <math>x</math> is a positive integer. Without loss of generality, assume <math>x=1</math>. Then <math>f(1) = 1 - 1 = 0</math>. | ||
+ | |||
+ | '''Case 2''': <math>x</math> is a positive fraction. Without loss of generality, assume <math>x=\frac{1}{2}</math>. Then <math>f\left(\frac{1}{2}\right) = 0 - 0 = 0</math>. | ||
+ | |||
+ | '''Case 3''': <math>x</math> is a negative integer. Without loss of generality, assume <math>x=-1</math>. Then <math>f(-1) = 1 - 1 = 0</math>. | ||
+ | |||
+ | '''Case 4''': <math>x</math> is a negative fraction. Without loss of generality, assume <math>x=-\frac{1}{2}</math>. Then <math>f\left(-\frac{1}{2}\right) = 0 - 1 = -1</math>. | ||
+ | |||
+ | Thus the range of the function <math>f</math> is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>. | ||
+ | |||
+ | ~IronicNinja, edited by someone else hehe | ||
+ | |||
+ | === 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. | ||
+ | |||
+ | When <math>x</math> is positive, <math>\lfloor x\rfloor \geq 0</math>, so | ||
+ | <cmath>\begin{split}f(x)&=\lfloor|x|\rfloor-|\lfloor x\rfloor| \\ | ||
+ | &=\lfloor x\rfloor-\lfloor x\rfloor \\ | ||
+ | &=0\end{split}</cmath> | ||
+ | |||
+ | When <math>x</math> is negative, let <math>x=-a-b</math> be composed of integer part <math>a</math> and fractional part <math>b</math> (both <math>\geq 0</math>): | ||
+ | <cmath>\begin{split}f(x)&=\lfloor|-a-b|\rfloor-|\lfloor -a-b\rfloor| \\ | ||
+ | &=\lfloor a+b\rfloor-|-a-1| \\ | ||
+ | &=a-(a+1)=-1\end{split}</cmath> | ||
+ | |||
+ | Thus, the range of f 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: | ||
+ | <cmath>\begin{split}f(x)&=\lfloor|x|\rfloor-|\lfloor x\rfloor| \\ | ||
+ | &=\lfloor -x\rfloor-|-\lfloor -x\rfloor-1| \\ | ||
+ | &=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = -1\end{split}</cmath> | ||
+ | |||
+ | === Video Solution === | ||
+ | https://youtu.be/PgqjsTkNYdc | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | == See Also == | ||
+ | {{AMC10 box|year=2019|ab=B|num-b=8|num-a=10}} | ||
+ | {{MAA Notice}} |
Revision as of 01:46, 19 October 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 f 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.