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 6)
(→‎Solution 5a (cosh(x) image ))
Line 91: Line 91:
  
  
−
image by ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
+
image by ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] inspired from Kevin_liu
  
 
==Solution 6 ==
 
==Solution 6 ==

Revision as of 17:02, 9 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?

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

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 cheese)

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.

~woeIsMe

[asy] unitsize(2cm); real e = 2.71828; real f1(real x) {return e^(x+1)-1;} real f2(real x) {return e^(0-x)-1;} draw(graph(f1,-1.5,0.5)); draw(graph(f2,-1.5,0.5)); xaxis(-2,1,Ticks()); yaxis(f2(0.5),f1(0.5),Ticks()); draw((-0.5,f2(0.5))--(-0.5,f1(0.5)),red+dashed); /*graph by Technodoggo, 9 November 2024*/ [/asy] (graph by Technodoggo)

Solution 3(Derivatives)

For an axis of symmetry there must be a local maximum or minimum. I mean, think about it. Would it make sense for a function to be larger than to the left of its axis of symmetry? So we do a little derivativing and set said derivative equal to 0. $\frac{d[e^{x+1}+e^{-x}-2]}{dx} = 0 = e^{x+1}-e^{-x}$ Now in order to solve for e we do a lil subbing hehehe $u = e^x$ $0 = e*u - \frac{1}{u}$ -> $0 = \frac{e*u^2 - 1}{u}$ -> $0 = e*u^2 - 1$ -> $e*u^2 = 1$ -> $(e^x)^2 = \frac{1}{e}$ -> $e^{2x} = e^{-1}$ -> $2x = -1$ -> $x = -\frac{1}{2}$ Now we just flippidy flap (-1, $\frac{1}{2}$) over $x = -\frac{1}{2}$ and blammo, we got ourselves the answer choice $\boxed{\textbf{(D) }\left(0,\dfrac12\right)}.$

Note: "an axis of symmetry there must be a local maximum or minimum" is not true. consider the graph of 1/x

Solution 4

\[f(0) = e-1\] \[f(1) = e^2 + e^{-1} - 2\] \[f(-1) = 1+ e -2 = e-1\] \[f(-2) = e^{-1} + e^2 -2\]


so f(0) = f(-1) , f(1) =f(-2) then f(x) is symmetric about x=-1/2 point (-1, 1/2) reflects over axis x=-1/2 is point ( 0, 1/2) answer choice D ~luckuso

Solution 5

Notice that this is simply a transformation of $\cosh(x)$, specifically $2e^\frac{1}{2}\cosh(x + \frac{1}{2}) + 2$. We know that $\cosh(x)$ has a parabolic shape with an axis of symmetry $x=0$. Since this is simple a transformation of $\frac{1}{2}$ to the left of the origin, we have the axis of symmetry is $x = -\frac{1}{2}$ giving us $\boxed{\textbf{(D) }\left(0,\dfrac12\right)}$

~KEVIN_LIU

Solution 5a (cosh(x) image )

2024 amc12A p13.PNG


image by ~luckuso inspired from Kevin_liu

Solution 6

\[f(-x-1) = e^{-x-1+1} + e^{-(-x-1)} - 2 = e^{-x} + e^{x+1} - 2 = f(x)\]

so the axis is middle point of (x , -x-1 ) = -1/2

point (-1, 1/2) reflects over axis x=-1/2 is point ( 0, 1/2) answer choice D

~luckuso

Solution 7

symmetric on axis a means f(2a-x) = f(x)

plug in \[f(2a-x) = e^{2a-x+1} + e^{-(2a-x)} - 2 = e^{x+1} + e^{-x} - 2 = f(x)\] 2a+1 = 0 solve a= -1/2

point (-1, 1/2) reflects over axis x=-1/2 is point ( 0, 1/2) answer choice D ~luckuso

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=232802"