Difference between revisions of "2021 Fall AMC 10A Problems/Problem 16"
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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 is symmetric about which of the following? (Here is the greatest integer not exceeding .)
Solution 1 (Piecewise Function)
IN PROGRESS AND WILL FINISH SOON. NO EDIT PLEASE. A MILLION THANKS.
~MRENTHUSIASM
Solution 2 (Graphing)
The graph of is shown below: DIAGRAM WILL BE READY VERY SOON
The graph of is shown below: DIAGRAM WILL BE READY VERY SOON
Taking the difference, we graph as shown below: DIAGRAM WILL BE READY VERY SOON
Therefore, the answer is
~MRENTHUSIASM
See Also
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.