Difference between revisions of "2024 AMC 12A Problems/Problem 13"
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Consider the graphs of <math>y=e^{x+1}-1</math> and <math>y=e^{-x}-1</math>. A rough sketch will show that they intercept somewhere between -1 and 0 and the axis of symmetry is vertical. Thus, <math>\boxed{\textbf{(D) }\left(0,\dfrac12\right)}.</math> is the only possible answer. | Consider the graphs of <math>y=e^{x+1}-1</math> and <math>y=e^{-x}-1</math>. A rough sketch will show that they intercept somewhere between -1 and 0 and the axis of symmetry is vertical. Thus, <math>\boxed{\textbf{(D) }\left(0,\dfrac12\right)}.</math> is the only possible answer. | ||
Note: You can more rigorously think about the solution by noting that since that the absolute value of the derivative of the power that e is raised to is the same, and they are both subtracted by 1, then for all x greater than the point of interception, both equations will grow by the same amount. Setting both equations equal to each other, it is trivial to see $x=-1/2, giving us the axis of symmetry. | Note: You can more rigorously think about the solution by noting that since that the absolute value of the derivative of the power that e is raised to is the same, and they are both subtracted by 1, then for all x greater than the point of interception, both equations will grow by the same amount. Setting both equations equal to each other, it is trivial to see $x=-1/2, giving us the axis of symmetry. | ||
| + | |||
| + | (someone insert the graph pls) | ||
~woeIsMe | ~woeIsMe | ||
Revision as of 19:36, 8 November 2024
Problem
The graph of 
has an axis of symmetry. What is the reflection of the point 
over this axis?
Solution 1
The line of symmetry is probably of the form 
for some constant 
. A vertical line of symmetry at for a function 
exists if and only if 
; we substitute 
and 
into our given function and see that we must have
![\[e^{a-b+1}+e^{-(a-b)}-2=e^{a+b+1}+e^{-(a+b)}-2\]](http://latex.artofproblemsolving.com/0/1/8/018cef12e62b1b356b5d14697e020baa5c746914.png)
for all real 
. Simplifying:
\begin{align*} e^{a-b+1}+e^{-(a-b)}-2&=e^{a+b+1}+e^{-(a+b)}-2 \\ e^{a-b+1}+e^{b-a}&=e^{a+b+1}+e^{-a-b} \\ e^{a-b+1}-e^{-a-b}&=e^{a+b+1}-e^{b-a} \\ e^{-b}\left(e^{a+1}-e^{-a}\right)&=e^b\left(e^{a+1}-e^{-a}\right). \\ \end{align*}
If 
, then 
for all real ; this is clearly impossible, so let 
. Thus, our line of symmetry is 
, and reflecting 
over this line gives 
~Technodoggo
Solution 2 (Graphing)
Consider the graphs of 
and 
. A rough sketch will show that they intercept somewhere between -1 and 0 and the axis of symmetry is vertical. Thus, is the only possible answer.
Note: You can more rigorously think about the solution by noting that since that the absolute value of the derivative of the power that e is raised to is the same, and they are both subtracted by 1, then for all x greater than the point of interception, both equations will grow by the same amount. Setting both equations equal to each other, it is trivial to see $x=-1/2, giving us the axis of symmetry.
(someone insert the graph pls)
~woeIsMe
See also
| 2024 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
