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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 16"
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<math>\textbf{(A) }</math> the <math>y</math>-axis <math>\qquad \textbf{(B) }</math> the line <math>x = 1</math> <math>\qquad \textbf{(C) }</math> the origin <math>\qquad
 
<math>\textbf{(A) }</math> the <math>y</math>-axis <math>\qquad \textbf{(B) }</math> the line <math>x = 1</math> <math>\qquad \textbf{(C) }</math> the origin <math>\qquad
 
\textbf{(D) }</math> the point <math>\left(\dfrac12, 0\right)</math> <math>\qquad \textbf{(E) }</math> the point <math>(1,0)</math>
 
\textbf{(D) }</math> the point <math>\left(\dfrac12, 0\right)</math> <math>\qquad \textbf{(E) }</math> the point <math>(1,0)</math>
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== Solution 1 ==
 
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Since <math>f(1-x)=|\lfloor 1-x \rfloor|-|\lfloor x \rfloor|=-f(x)</math>, <math>f(x)</math> is symmetric about the point <math>\left(\dfrac12, 0\right)</math>
 
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<math>\box{\textbf{(D) } } \text{the point} \left(\dfrac12, 0\right)</math>
 

Revision as of 12:07, 23 November 2021

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 $x$.)

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

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