Difference between revisions of "2019 AMC 10B Problems/Problem 9"
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Thus the range of function f is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math> | Thus the range of function f is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math> | ||
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==Solution 2== | ==Solution 2== | ||
Revision as of 11:16, 15 February 2019
Contents
Problem
The function 
is defined by ![\[f(x) = \lfloor|x|\rfloor - |\lfloor x \rfloor|\]](http://latex.artofproblemsolving.com/2/8/9/289baf6aac9fddce693a26364041d4d5b550c685.png)
for all real numbers 
, where 
denotes the greatest integer less than or equal to the real number 
. What is the range of ?
Solution
There are 4 cases we need to test here:
Case 1: x is a positive integer. WLOG, assume x=1. Then f(1) = 1 - 1 = 
.
Case 2: x is a positive fraction. WLOG, assume x=0.5. Then f(0.5) = 0 - 0 = .
Case 3: x is a negative integer. WLOG, assume x=-1. Then f(-1) = 1 - 1 = .
Case 4: x is a negative fraction. WLOG, assume x=-0.5. Then f(-0.5) = 0 - 1 = 
.
Thus the range of function f is 
Solution 2
It is easily verified that when is an integer, then 
is zero. We need only to consider the case when is not.
When is a positive number, 
, so
![\[f(x)=\lfloor|x|\rfloor-|\lfloor x\rfloor|\]](http://latex.artofproblemsolving.com/0/6/4/064264ddd7feea1371ee7557f34c4d4dcc016d1f.png)
![\[=\lfloor x\rfloor-\lfloor x\rfloor\]](http://latex.artofproblemsolving.com/4/5/2/452a3cd64745c70dad499c8d0d5fde09640d00b3.png)
![\[=\textbf{0}\]](http://latex.artofproblemsolving.com/d/a/c/dac37ef3271bb785001e206840ec07780c850378.png)
When is a negative number, let 
be composed of integer part 
and decimal part 
(both 
):
![\[f(x)=\lfloor|-a-b|\rfloor-|\lfloor -a-b\rfloor|\]](http://latex.artofproblemsolving.com/2/d/6/2d6104f12da05e1c6673cba10777f328e9886c40.png)
![\[=\lfloor a+b\rfloor-|-a-1|\]](http://latex.artofproblemsolving.com/8/4/9/849cda41f67e261d0b04575969655f323ab6a619.png)
![\[=a-(a+1)=\textbf{-1}\]](http://latex.artofproblemsolving.com/a/a/b/aab20fe7848413c53a3fed9951c31d977a38d06e.png)
Thus, the range of f is
>>> Intelligence_Inc
Note: One could solve the case of as a negative non-integer this way:
![\[=\lfloor -x\rfloor-|-\lfloor -x\rfloor-1|\]](http://latex.artofproblemsolving.com/0/4/4/044f9160e5698c32beb1aad5e52368f463b8e032.png)
![\[=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = \textbf{-1}\]](http://latex.artofproblemsolving.com/b/9/a/b9ac9e56fb51e1982bc648a831f2d051adb3a37c.png)
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. 
