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Difference between revisions of "2019 AMC 10B Problems/Problem 9"
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== Solution 4 ==
 
== Solution 4 ==
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We have 2 cases: <math>x</math> is positive and <math>x</math> is negative.
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We have 2 cases: either <math>x</math> is positive or <math>x</math> is negative.
  
  
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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  
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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 (CREATIVE PROBLEM SOLVING!!!)==
 
==Video Solution (CREATIVE PROBLEM SOLVING!!!)==

Revision as of 16:12, 29 May 2023

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

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 (CREATIVE PROBLEM SOLVING!!!)

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

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