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
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

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

m (→‎Solution 1)
(→‎See Also)
 
(26 intermediate revisions by 11 users not shown)
Line 1: Line 1:
−
==Problem==
+
== 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>?
 
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\} \qquad\textbf{(E) } \text{The set of nonnegative integers} </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\} \qquad \textbf{(E) } \text{The set of nonnegative integers} </math>
−
 
 
−
==Solution 1==
 
  
 +
== Solution 1 ==
 
There are four cases we need to consider here.
 
There are four cases we need to consider here.
  
Line 19: Line 17:
 
Thus the range of the function <math>f</math> is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>.
 
Thus the range of the function <math>f</math> is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>.
  
−
~IronicNinja, edited by someone else hehe
+
~IronicNinja
−
 
 
−
==Solution 2==
 
  
 +
== 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.
 
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.
  
Line 35: Line 32:
 
&=a-(a+1)=-1\end{split}</cmath>
 
&=a-(a+1)=-1\end{split}</cmath>
  
−
Thus, the range of f is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>.
+
Thus, the range of x 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:
 
''Note'': One could solve the case of <math>x</math> as a negative non-integer in this way:
Line 42: Line 39:
 
&=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = -1\end{split}</cmath>
 
&=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = -1\end{split}</cmath>
  
−
==See Also==
+
== Solution 3 (Formal) ==
 +
Let {<math>x</math>} denote the fractional part of <math>x</math>; for example, {<math>2.7</math>}<math>= 0.7</math>, and {<math>-1.3</math>}<math>= 0.3</math>.
 +
Then for <math> x \geq 0</math>, <math> x = \lfloor x \rfloor +</math> {<math>x</math>} and for <math> x < 0</math>, <math> x = \lfloor x \rfloor + 1 - </math>{<math>x</math>}.
 +
 
 +
Now we can rewrite <math>\lfloor |x| \rfloor - |\lfloor x \rfloor|</math>, breaking the expression up based on whether <math> x \geq 0 </math> or <math> x < 0</math>.
 +
 
 +
For <math>x \geq 0</math>, the above expression is equal to <math> \lfloor |\lfloor x \rfloor + </math>{<math>x</math>}<math>| \rfloor - | \lfloor \lfloor x \rfloor + </math> {<math>x</math>}<math> \rfloor | \implies \lfloor \lfloor x \rfloor + </math>{<math>x</math>}<math> \rfloor - | \lfloor x \rfloor | </math>
 +
 
 +
<math> \implies \lfloor x \rfloor - \lfloor x \rfloor = \mathbf{0} </math>.
 +
 
 +
For <math> x < 0</math>, the expression is equal to <math> \lfloor |\lfloor x \rfloor + 1 - </math>{<math>x</math>}<math>| \rfloor - | \lfloor \lfloor x \rfloor + 1 - </math> {<math>x</math>}<math> \rfloor |</math>
 +
 
 +
<math> \implies \lfloor - \lfloor x \rfloor - 1 + </math>{<math>x</math>}<math> \rfloor - | \lfloor x \rfloor | \implies - \lfloor x \rfloor  - 1 - (- \lfloor x \rfloor) = \mathbf{-1}</math>.
 +
 
 +
Therefore the only two possible values for <math>f(x)</math>, and thus the range of the function, is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>.
 +
 
 +
~KingRavi
 +
 
 +
== Solution 4 ==
 +
We have 2 cases: either <math>x</math> is positive or <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==
 +
https://youtu.be/LffjyNNqf14
 +
 
 +
~Education, the Study of Everything
 +
 
 +
== Video Solution ==
 +
https://youtu.be/PgqjsTkNYdc
 +
 
 +
~savannahsolver
  
 +
== See Also ==
 
{{AMC10 box|year=2019|ab=B|num-b=8|num-a=10}}
 
{{AMC10 box|year=2019|ab=B|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
[[Category: Introductory Algebra Problems]]

Latest revision as of 16:16, 16 October 2025

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 1

There are four cases we need to consider here.

Case 1: $x$ is a positive integer. Without loss of generality, assume $x=1$. Then $f(1) = 1 - 1 = 0$.

Case 2: $x$ is a positive fraction. Without loss of generality, assume $x=\frac{1}{2}$. Then $f\left(\frac{1}{2}\right) = 0 - 0 = 0$.

Case 3: $x$ is a negative integer. Without loss of generality, assume $x=-1$. Then $f(-1) = 1 - 1 = 0$.

Case 4: $x$ is a negative fraction. Without loss of generality, assume $x=-\frac{1}{2}$. Then $f\left(-\frac{1}{2}\right) = 0 - 1 = -1$.

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

~IronicNinja

Solution 2

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

When $x$ is positive, $\lfloor x\rfloor \geq 0$, so \[\begin{split}f(x)&=\lfloor|x|\rfloor-|\lfloor x\rfloor| \\ &=\lfloor x\rfloor-\lfloor x\rfloor \\ &=0\end{split}\]

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

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

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

Solution 3 (Formal)

Let {$x$} denote the fractional part of $x$; for example, {$2.7$}$= 0.7$, and {$-1.3$}$= 0.3$. Then for $x \geq 0$, $x = \lfloor x \rfloor +$ {$x$} and for $x < 0$, $x = \lfloor x \rfloor + 1 -${$x$}.

Now we can rewrite $\lfloor |x| \rfloor - |\lfloor x \rfloor|$, breaking the expression up based on whether $x \geq 0$ or $x < 0$.

For $x \geq 0$, the above expression is equal to $\lfloor |\lfloor x \rfloor +${$x$}$| \rfloor - | \lfloor \lfloor x \rfloor +$ {$x$}$\rfloor | \implies \lfloor \lfloor x \rfloor +${$x$}$\rfloor - | \lfloor x \rfloor |$

$\implies \lfloor x \rfloor - \lfloor x \rfloor = \mathbf{0}$.

For $x < 0$, the expression is equal to $\lfloor |\lfloor x \rfloor + 1 -${$x$}$| \rfloor - | \lfloor \lfloor x \rfloor + 1 -$ {$x$}$\rfloor |$

$\implies \lfloor - \lfloor x \rfloor - 1 +${$x$}$\rfloor - | \lfloor x \rfloor | \implies - \lfloor x \rfloor - 1 - (- \lfloor x \rfloor) = \mathbf{-1}$.

Therefore the only two possible values for $f(x)$, and thus the range of the function, is $\boxed{\textbf{(A) } \{-1, 0\}}$.

~KingRavi

Solution 4

We have 2 cases: either $x$ is positive or $x$ is negative.


Case 1 - x is positive:

Let $x = n + f$, where $n$ is a positive integer and $f$ is a positive real number between 0 and 1. We have \[\lfloor |x| \rfloor = \lfloor |n+f| \rfloor = \lfloor n+f \rfloor = n\] and \[|\lfloor x \rfloor| = |\lfloor n+f \rfloor| = |n| = n.\] $n-n=0$, so the possible value of $f(x)$ if $x$ is positive is $0$.


Case 2 - x is negative:

Let $x = -n - f$, where $n$ is a positive integer and $f$ is a positive real number between 0 and 1. We have \[\lfloor |x| \rfloor = \lfloor |-n-f| \rfloor = \lfloor n+f \rfloor = n\] and \[|\lfloor x \rfloor| = |\lfloor -n-f \rfloor| =|-n|\:or\: |-n-1|= n \:or\: n+1.\]

$n-n=0$ and $n-(n+1) = -1$, so the possible values of $f(x)$ if $x$ is negative are $0$ and $-1.$


Hence, the possible values of $f(x)$ are $0$ and $-1$, so the answer is $\boxed{\textbf{(A) } \{-1, 0\}}$. ~azc1027

Video Solution

https://youtu.be/LffjyNNqf14

~Education, the Study of Everything

Video Solution

https://youtu.be/PgqjsTkNYdc

~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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_9&oldid=263275"