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

(Solution 1)
(Solution 3 (Formal))
 
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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 1==
 
  
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== Solution 1 ==
 
There are four cases we need to consider here.
 
There are four cases we need to consider here.
  
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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
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~IronicNinja
 
 
==Solution 2==
 
  
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== 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.
  
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&=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>.
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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:
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&=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = -1\end{split}</cmath>
 
&=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = -1\end{split}</cmath>
  
==See Also==
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== Solution 3 (Formal) ==
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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>}.
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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>.
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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>
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<math> \implies \lfloor x \rfloor - \lfloor x \rfloor = \mathbf{0} </math>.
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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>
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<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>.
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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>.
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 +
~KingRavi
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== Video Solution ==
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https://youtu.be/PgqjsTkNYdc
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 +
~savannahsolver
  
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== 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}}

Latest revision as of 23:36, 31 October 2021

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

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

Video Solution

https://youtu.be/PgqjsTkNYdc

~savannahsolver

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AMC 10 Problems and Solutions

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