Difference between revisions of "G285 2021 Summer Problem Set Problem 2"

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==Solution==
 
==Solution==
Note for <math>x \in \{-1,1 \}</math>, our function approaches an infinite recursion. Now, for <math>-1<x<1</math>, we have the function approaches <math>-1</math>, or <math>1</math>, which also is an infinite recursion. The answer is <math>\boxed{\textbf{(C)}\ 0}</math>
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Note for <math>x \in \{ -1,1 \}</math>, our function approaches an infinite recursion. Now, for <math>-1<x<1</math>, we have the function approaches <math>-1</math>, or <math>1</math>, which also is an infinite recursion. The answer is <math>\boxed{\textbf{(C)}\ 0}</math>

Latest revision as of 23:15, 28 June 2021

Problem

Let \[f(x,y) = \begin{cases}x^y & \text{ if } x^2>y \text{ and } |x|<y\\f(f(\sqrt{|x|},y),y) & \text{ otherwise} \end{cases}\] If $y$ is a positive integer, find the sum of all values of $x$ such that $f(x,y) \neq k$ for some constant $k$.

$\textbf{(A)}\ -1 \qquad\textbf{(B)}\ -\frac{1}{2} \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ \frac{3}{8} \qquad\textbf{(E)}\ 1$

Solution

Note for $x \in \{ -1,1 \}$, our function approaches an infinite recursion. Now, for $-1<x<1$, we have the function approaches $-1$, or $1$, which also is an infinite recursion. The answer is $\boxed{\textbf{(C)}\ 0}$