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

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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>
 
<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>
  
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=== 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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~IronicNinja
 
~IronicNinja
  
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=== 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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&=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = -1\end{split}</cmath>
 
&=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = -1\end{split}</cmath>
  
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=== Video Solution ===
+
== Video Solution ==
 
https://youtu.be/PgqjsTkNYdc
 
https://youtu.be/PgqjsTkNYdc
  

Revision as of 19:26, 17 April 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: \[\begin{split}f(x)&=\lfloor|x|\rfloor-|\lfloor x\rfloor| \\ &=\lfloor -x\rfloor-|-\lfloor -x\rfloor-1| \\ &=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = -1\end{split}\]

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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