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

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==Problem==
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== 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>
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<math>\textbf{(A) } \{-1, 0\} \qquad\textbf{(B) } \text{The set of nonpositive integers} \qquad\textbf{(C) } \{-1, 0, 1\} \qquad\textbf{(D) } \{0\}</math> <br> <math>\textbf{(E) } \text{The set of nonnegative integers} </math>
 
 
==Solution==
 
  
There are 4 cases we need to test here:
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=== Solution 1 ===
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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>.
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'''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>.
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'''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>.
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'''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>.
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'''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>
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Thus the range of the function <math>f</math> is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>.
  
==Solution 2==
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~IronicNinja, edited by someone else hehe
  
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.
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=== Solution 2 ===
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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  
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When <math>x</math> is positive, <math>\lfloor x\rfloor \geq 0</math>, so  
<cmath>f(x)=\lfloor|x|\rfloor-|\lfloor x\rfloor|</cmath>
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<cmath>\begin{split}f(x)&=\lfloor|x|\rfloor-|\lfloor x\rfloor| \\
<cmath>=\lfloor x\rfloor-\lfloor x\rfloor</cmath>
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&=\lfloor x\rfloor-\lfloor x\rfloor \\
<cmath>=\textbf{0}</cmath>
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&=0\end{split}</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>):
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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>f(x)=\lfloor|-a-b|\rfloor-|\lfloor -a-b\rfloor|</cmath>
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<cmath>\begin{split}f(x)&=\lfloor|-a-b|\rfloor-|\lfloor -a-b\rfloor| \\
<cmath>=\lfloor a+b\rfloor-|-a-1|</cmath>
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&=\lfloor a+b\rfloor-|-a-1| \\
<cmath>=a-(a+1)=\textbf{-1}</cmath>
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&=a-(a+1)=-1\end{split}</cmath>
  
Thus, the range of f is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>
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Thus, the range of f is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>.
  
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''Note'': One could solve the case of <math>x</math> as a negative non-integer in this way:
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<cmath>\begin{split}f(x)&=\lfloor|x|\rfloor-|\lfloor x\rfloor| \\
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&=\lfloor -x\rfloor-|-\lfloor -x\rfloor-1| \\
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&=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = -1\end{split}</cmath>
  
Note: One could solve the case of <math>x</math> as a negative non-integer this way:
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=== Video Solution ===
<cmath>f(x)=\lfloor|x|\rfloor-|\lfloor x\rfloor|</cmath>
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https://youtu.be/PgqjsTkNYdc
<cmath>=\lfloor -x\rfloor-|-\lfloor -x\rfloor-1|</cmath>
 
<cmath>=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = \textbf{-1}</cmath>
 
  
==See Also==
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~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}}
SUB2PEWDS
 

Latest revision as of 00:46, 19 October 2020

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 $f$?

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

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, edited by someone else hehe

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: \[\begin{split}f(x)&=\lfloor|x|\rfloor-|\lfloor x\rfloor| \\ &=\lfloor -x\rfloor-|-\lfloor -x\rfloor-1| \\ &=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = -1\end{split}\]

Video Solution

https://youtu.be/PgqjsTkNYdc

~savannahsolver

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
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

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