Difference between revisions of "2019 AMC 10B Problems/Problem 9"
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+ | == Solution 4 == | ||
+ | We have 2 cases: <math>x</math> is positive and <math>x</math> is negative. | ||
+ | |||
+ | |||
+ | '''Case 1: x is positive.''' | ||
+ | |||
+ | Let <math>x = n + f</math>, where <math>n</math> is a positive integer and <math>f</math> is a positive real number between 0 and 1. We have | ||
+ | <cmath>\lfloor |x| \rfloor = \lfloor |n+f| \rfloor = \lfloor n+f \rfloor = n</cmath> and | ||
+ | <cmath>|\lfloor x \rfloor| = |\lfloor n+f \rfloor| = |n| = n.</cmath> | ||
+ | <math>n-n=0</math>, so the possible value of <math>f(x)</math> if <math>x</math> is positive is <math>0</math>. | ||
+ | |||
+ | |||
+ | '''Case 2: x is negative. ''' | ||
+ | |||
+ | Let <math>x = -n - f</math>, where <math>n</math> is a positive integer and <math>f</math> is a positive real number between 0 and 1. We have | ||
+ | <cmath>\lfloor |x| \rfloor = \lfloor |-n-f| \rfloor = \lfloor n+f \rfloor = n</cmath> and | ||
+ | <cmath>|\lfloor x \rfloor| = |\lfloor -n-f \rfloor| =|-n|\:or\: |-n-1|= n \:or\: n+1.</cmath> | ||
+ | |||
+ | <math>n-n=0</math> and <math>n-(n+1) = -1</math>, so the possible values of <math>f(x)</math> if <math>x</math> is negative are <math>0</math> and <math>-1.</math> | ||
+ | |||
+ | |||
+ | Hence, the possible values of <math>f(x)</math> are <math>0</math> and <math>-1</math>, so the answer is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>. ~azc1027 | ||
==Video Solution (CREATIVE PROBLEM SOLVING!!!)== | ==Video Solution (CREATIVE PROBLEM SOLVING!!!)== |
Revision as of 16:11, 29 May 2023
Contents
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:
Solution 3 (Formal)
Let {} denote the fractional part of ; for example, {}, and {}. Then for , {} and for , {}.
Now we can rewrite , breaking the expression up based on whether or .
For , the above expression is equal to {} {}{}
.
For , the expression is equal to {} {}
{}.
Therefore the only two possible values for , and thus the range of the function, is .
~KingRavi
Solution 4
We have 2 cases: is positive and is negative.
Case 1: x is positive.
Let , where is a positive integer and is a positive real number between 0 and 1. We have and , so the possible value of if is positive is .
Case 2: x is negative.
Let , where is a positive integer and is a positive real number between 0 and 1. We have and
and , so the possible values of if is negative are and
Hence, the possible values of are and , so the answer is . ~azc1027
Video Solution (CREATIVE PROBLEM SOLVING!!!)
~Education, the Study of Everything
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.