Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 AMC 12A Problems/Problem 13"

m
Line 22: Line 22:
  
 
~Technodoggo
 
~Technodoggo
 +
 +
==Solution 2==
 +
 +
Consider the graphs of <math>y=e^{x+1}-1</math> and <math>y=e^{-x}-1. 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 </math>x=-1/2, giving us the axis of symmetry. 
 +
 +
~woeIsMe
  
 
==See also==
 
==See also==
 +
 +
 
{{AMC12 box|year=2024|ab=A|num-b=12|num-a=14}}
 
{{AMC12 box|year=2024|ab=A|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:35, 8 November 2024

Problem

The graph of $y=e^{x+1}+e^{-x}-2$ has an axis of symmetry. What is the reflection of the point $(-1,\tfrac{1}{2})$ over this axis?

$\textbf{(A) }\left(-1,-\frac{3}{2}\right)\qquad\textbf{(B) }(-1,0)\qquad\textbf{(C) }\left(-1,\tfrac{1}{2}\right)\qquad\textbf{(D) }\left(0,\frac{1}{2}\right)\qquad\textbf{(E) }\left(3,\frac{1}{2}\right)$

Solution 1

The line of symmetry is probably of the form $x=a$ for some constant $a$. A vertical line of symmetry at $x=a$ for a function $f$ exists if and only if $f(a-b)=f(a+b)$; we substitute $a-b$ and $a+b$ into our given function and see that we must have

\[e^{a-b+1}+e^{-(a-b)}-2=e^{a+b+1}+e^{-(a+b)}-2\]

for all real $b$. 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 $e^{a+1}-e^{-a}\neq0$, then $e^{-b}=e^b$ for all real $b$; this is clearly impossible, so let $e^{a+1}-e^{-a}=0\implies a+1=-a\implies a=-\dfrac12$. Thus, our line of symmetry is $x=-\dfrac12$, and reflecting $\left(-1,\dfrac12\right)$ over this line gives $\boxed{\textbf{(D) }\left(0,\dfrac12\right)}.$

~Technodoggo

Solution 2

Consider the graphs of $y=e^{x+1}-1$ and $y=e^{-x}-1. A rough sketch will show that they intercept somewhere between -1 and 0 and the axis of symmetry is vertical. Thus,$\boxed{\textbf{(D) }\left(0,\dfrac12\right)}.$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$ (Error compiling LaTeX. Unknown error_msg)x=-1/2, giving us the axis of symmetry.

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_12A_Problems/Problem_13&oldid=232223"