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"

(→‎Solution 2 (Graphing))
m (→‎Solution 2 (Graphing))
Line 27: Line 27:
 
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 the sum of both functions will be the same from one side of the interception to the other. Setting both equations equal to each other, it is trivial to see <math>x=-1/2</math>, giving us the axis of symmetry.   
+
Note: You can more rigorously think about the solution by noting that since the derivative of the power that e is raised to in one equation is equal to the derivative of the power that e is raised to multiplied by <math>-1</math>, and both equations are subtracted by 1, then the sum of both equations will be the same from one side of the interception to the other. Setting both equations equal to each other, it is trivial to see <math>x=-1/2</math>, giving us the axis of symmetry.   
  
 
(someone insert the graph pls)
 
(someone insert the graph pls)

Revision as of 19:47, 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 (Graphing)

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 the derivative of the power that e is raised to in one equation is equal to the derivative of the power that e is raised to multiplied by $-1$, and both equations are subtracted by 1, then the sum of both equations will be the same from one side of the interception to the other. 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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_12A_Problems/Problem_13&oldid=232241"