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

(Solution 3 (Formal))
(Solution 3 (Formal))
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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>.
 
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>.
  
For <math>x \geq 0</math>, the above expression is equal to <math> \lfloor |\lfloor x \rfloor + </math>{<math>x</math>}<math>| \rfloor \implies </math>
+
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 </math>
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 23:06, 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 $\lfloor |\lfloor x \rfloor +${$x$}$| \rfloor - | \lfloor \lfloor x \rfloor +$ {$x$}$\rfloor | \implies$

Video Solution

https://youtu.be/PgqjsTkNYdc

~savannahsolver

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
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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