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

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==Problem==
 
The graph of <math>y=e^{x+1}+e^{-x}-2</math> has an axis of symmetry. What is the reflection of the point <math>(-1,\tfrac{1}{2})</math> over this axis?
 
The graph of <math>y=e^{x+1}+e^{-x}-2</math> has an axis of symmetry. What is the reflection of the point <math>(-1,\tfrac{1}{2})</math> over this axis?
  
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~Technodoggo
 
~Technodoggo
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==See also==
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{{AMC12 box|year=2024|ab=A|num-b=12|num-a=14}}
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{{MAA Notice}}

Revision as of 18:49, 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

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

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