Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 AMC 10B Problems/Problem 9"

(→‎Solution 2)
(→‎Solution 2)
Line 27: Line 27:
  
 
When <math>x</math> is a positive number, <math>\lfloor x\rfloor \geq 0</math>, so  
 
When <math>x</math> is a positive number, <math>\lfloor x\rfloor \geq 0</math>, so  
−
<cmath>f(x)=\lfloor|x|\rfloor-|\lfloor x\rfloor</cmath>
+
<cmath>f(x)=\lfloor|x|\rfloor-|\lfloor x\rfloor|</cmath>
 
<cmath>=\lfloor x\rfloor-\lfloor x\rfloor</cmath>
 
<cmath>=\lfloor x\rfloor-\lfloor x\rfloor</cmath>
 
<cmath>=\textbf{0}</cmath>
 
<cmath>=\textbf{0}</cmath>

Revision as of 22:29, 14 February 2019

Problem

The function $f$ is defined by \[f(x) = \lfloor|x|\rfloor - |\lfloor x \rfloor|\]for all real numbers $x$, where $\lfloor r \rfloor$ denotes the greatest integer less than or equal to the real number $r$. What is the range of $f$?

$\textbf{(A) } \{-1, 0\} \qquad\textbf{(B) } \text{The set of nonpositive integers} \qquad\textbf{(C) } \{-1, 0, 1\} \qquad\textbf{(D) } \{0\} \qquad\textbf{(E) } \text{The set of nonnegative integers}$

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 = $0$.

Case 2: x is a positive fraction. WLOG, assume x=0.5. Then f(0.5) = 0 - 0 = $0$.

Case 3: x is a negative integer. WLOG, assume x=-1. Then f(-1) = 1 - 1 = $0$.

Case 4: x is a negative fraction. WLOG, assume x=-0.5. Then f(-0.5) = 0 - 1 = $-1$.

Thus the range of function f is $\boxed{\textbf{(A) } \{-1, 0\}}$

iron


Solution 2

It is easily verified that when $x$ is an integer, then $f(x)$ is zero. We need only to consider the case when $x$ is not.

When $x$ is a positive number, $\lfloor x\rfloor \geq 0$, so \[f(x)=\lfloor|x|\rfloor-|\lfloor x\rfloor|\] \[=\lfloor x\rfloor-\lfloor x\rfloor\] \[=\textbf{0}\]

When $x$ is a negative number, let $x=-a-b$ be composed of integer part $a$ and decimal part $b$ (both $\geq 0$): \[f(x)=\lfloor|-a-b|\rfloor-|\lfloor -a-b\rfloor|\] \[=\lfloor a+b\rfloor-|-a-1|\] \[=a-(a+1)=\textbf{-1}\]

Thus, the range of f is $\boxed{\textbf{(A) } \{-1, 0\}}$

>>> Intelligence_Inc


Note: One could solve the case of $x$ as a negative non-integer this way: \[f(x)=\lfloor|x|\rfloor-|\lfloor x\rfloor|\] \[=\lfloor -x\rfloor-|-\lfloor -x\rfloor-1|\] \[=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = \textbf{-1}\]

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_9&oldid=102769"