Difference between revisions of "2021 Fall AMC 10A Problems/Problem 16"

(Problem)
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\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)\qquad \textbf{(E) }\text{the point }(1,0)</math>
 
\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)\qquad \textbf{(E) }\text{the point }(1,0)</math>
  
==Solution==
+
==Solution 1 (Piecewise Function)==
 
<b>IN PROGRESS AND WILL FINISH SOON. NO EDIT PLEASE. A MILLION THANKS.</b>
 
<b>IN PROGRESS AND WILL FINISH SOON. NO EDIT PLEASE. A MILLION THANKS.</b>
 +
 +
~MRENTHUSIASM
 +
 +
==Solution 2 (Graphing)==
 +
The graph of <math>y=|\lfloor x \rfloor|</math> is shown below:
 +
<b>DIAGRAM WILL BE READY VERY SOON</b>
 +
 +
The graph of <math>y=|\lfloor 1 - x \rfloor|</math> is shown below:
 +
<b>DIAGRAM WILL BE READY VERY SOON</b>
 +
 +
Taking the difference, we graph <math>f(x) = |\lfloor x \rfloor| - |\lfloor 1 - x \rfloor|,</math> as shown below:
 +
<b>DIAGRAM WILL BE READY VERY SOON</b>
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 09:14, 25 November 2021

Problem

The graph of \[f(x) = |\lfloor x \rfloor| - |\lfloor 1 - x \rfloor|\] is symmetric about which of the following? (Here $\lfloor x \rfloor$ is the greatest integer not exceeding $x$.)

$\textbf{(A) }\text{the }y\text{-axis}\qquad \textbf{(B) }\text{the line }x = 1\qquad \textbf{(C) }\text{the origin}\qquad \textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)\qquad \textbf{(E) }\text{the point }(1,0)$

Solution 1 (Piecewise Function)

IN PROGRESS AND WILL FINISH SOON. NO EDIT PLEASE. A MILLION THANKS.

~MRENTHUSIASM

Solution 2 (Graphing)

The graph of $y=|\lfloor x \rfloor|$ is shown below: DIAGRAM WILL BE READY VERY SOON

The graph of $y=|\lfloor 1 - x \rfloor|$ is shown below: DIAGRAM WILL BE READY VERY SOON

Taking the difference, we graph $f(x) = |\lfloor x \rfloor| - |\lfloor 1 - x \rfloor|,$ as shown below: DIAGRAM WILL BE READY VERY SOON

Therefore, the answer is $\boxed{\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)}.$

~MRENTHUSIASM

See Also

2021 Fall AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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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