Difference between revisions of "2019 AMC 10B Problems/Problem 9"
m (→Solution) |
m (→Solution) |
||
| Line 17: | Line 17: | ||
Case 4: x is a negative fraction. WLOG, assume x=-0.5. Then f(-0.5) = 0 - 1 = <math>-1</math>. | Case 4: x is a negative fraction. WLOG, assume x=-0.5. Then f(-0.5) = 0 - 1 = <math>-1</math>. | ||
| − | Thus the range of function f is <math>\textbf{(A) } \{-1, 0\}</math> | + | Thus the range of function f is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math> |
iron | iron | ||
Revision as of 22:04, 14 February 2019
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 
iron
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. 
