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

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

@@ Line 53: / Line 53: @@
 <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>.
-Therefore the only two possible values for <math>f(x)</math> are <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>.
+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>.
+~KingRavi
 == Video Solution ==
 https://youtu.be/PgqjsTkNYdc

2019 AMC 10B (Problems • Answer Key • Resources)
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