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"

(→‎Solution)
(Improved grammar, readability, and LaTeX, and removed an irrelevant attribution)
Line 5: Line 5:
 
<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==
+
==Solution 1==
  
−
There are 4 cases we need to test here:
+
There are four cases we need to consider here.
  
−
Case 1: x is a positive integer. WLOG, assume x=1. Then f(1) = 1 - 1 = <math>0</math>.
+
'''Case 1''': <math>x</math> is a positive integer. Without loss of generality, assume <math>x=1</math>. Then <math>f(1) = 1 - 1 = 0</math>.
  
−
Case 2: x is a positive fraction. WLOG, assume x=0.5. Then f(0.5) = 0 - 0 = <math>0</math>.
+
'''Case 2''': <math>x</math> is a positive fraction. Without loss of generality, assume <math>x=\frac{1}{2}</math>. Then <math>f\left(\frac{1}{2}\right) = 0 - 0 = 0</math>.
  
−
Case 3: x is a negative integer. WLOG, assume x=-1. Then f(-1) = 1 - 1 = <math>0</math>.
+
'''Case 3''': <math>x</math> is a negative integer. Without loss of generality, assume <math>x=-1</math>. Then <math>f(-1) = 1 - 1 = 0</math>.
  
−
Case 4: x is a negative fraction. WLOG, assume x=-0.5. Then f(-0.5) = 0 - 1 = <math>-1</math>.
+
'''Case 4''': <math>x</math> is a negative fraction. Without loss of generality, assume <math>x=-\frac{1}{2}</math>. Then <math>f\left(-\frac{1}{2}\right) = 0 - 1 = -1</math>.
  
−
Thus the range of function f 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>.
−
 
 
−
iron
 
  
 
==Solution 2==
 
==Solution 2==
  
−
It is easily verified that when <math>x</math> is an integer, then <math>f(x)</math> is zero. We need only to consider the case when <math>x</math> is not.
+
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.
−
 
 
−
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>=\lfloor x\rfloor-\lfloor x\rfloor</cmath>
 
−
<cmath>=\textbf{0}</cmath>
 
  
−
When <math>x</math> is a negative number, let <math>x=-a-b</math> be composed of integer part <math>a</math> and decimal part <math>b</math> (both <math>\geq 0</math>):
+
When <math>x</math> is positive, <math>\lfloor x\rfloor \geq 0</math>, so
−
<cmath>f(x)=\lfloor|-a-b|\rfloor-|\lfloor -a-b\rfloor|</cmath>
+
<cmath>\begin{split}f(x)&=\lfloor|x|\rfloor-|\lfloor x\rfloor| \\
−
<cmath>=\lfloor a+b\rfloor-|-a-1|</cmath>
+
&=\lfloor x\rfloor-\lfloor x\rfloor \\
−
<cmath>=a-(a+1)=\textbf{-1}</cmath>
+
&=0\end{split}</cmath>
  
−
Thus, the range of f is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>
+
When <math>x</math> is negative, let <math>x=-a-b</math> be composed of integer part <math>a</math> and fractional part <math>b</math> (both <math>\geq 0</math>):
 +
<cmath>\begin{split}f(x)&=\lfloor|-a-b|\rfloor-|\lfloor -a-b\rfloor| \\
 +
&=\lfloor a+b\rfloor-|-a-1| \\
 +
&=a-(a+1)=-1\end{split}</cmath>
  
 +
Thus, the range of f is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>.
  
−
Note: One could solve the case of <math>x</math> as a negative non-integer this way:
+
''Note'': One could solve the case of <math>x</math> as a negative non-integer in this way:
−
<cmath>f(x)=\lfloor|x|\rfloor-|\lfloor x\rfloor|</cmath>
+
<cmath>\begin{split}f(x)&=\lfloor|x|\rfloor-|\lfloor x\rfloor| \\
−
<cmath>=\lfloor -x\rfloor-|-\lfloor -x\rfloor-1|</cmath>
+
&=\lfloor -x\rfloor-|-\lfloor -x\rfloor-1| \\
−
<cmath>=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = \textbf{-1}</cmath>
+
&=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = -1\end{split}</cmath>
  
 
==See Also==
 
==See Also==

Revision as of 19:54, 17 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 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 $\boxed{\textbf{(A) } \{-1, 0\}}$.

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 f is $\boxed{\textbf{(A) } \{-1, 0\}}$.

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

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=103207"