# 2019 AMC 10B Problems/Problem 9

## 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

~savannahsolver

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 