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(B) Given $f(x) = \begin{cases} \frac{x+5}{x-2}, & x 
eq 2 \ 1, & x=2 \end{cases}$.First,$f(x)$ is discontinuous at $x=2$. Therefore,$f(f(x))$ is di

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at exactly two values of $x$

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(B) Given $f(x) = \begin{cases} \frac{x+5}{x-2}, & x 
eq 2 \ 1, & x=2 \end{cases}$.First,$f(x)$ is discontinuous at $x=2$. Therefore,$f(f(x))$ is discontinuous at $x=2$.Next,we find the points where $f(f(x))$ might be discontinuous by considering the points where $f(x)$ is discontinuous or where $f(x)$ takes the value $2$ (since $f(x)$ is discontinuous at $2$).For $x 
eq 2$,$f(x) = \frac{x+5}{x-2}$.We set $f(x) = 2$ to find other points of discontinuity:$\frac{x+5}{x-2} = 2$$x+5 = 2(x-2)$$x+5 = 2x-4$$x = 9$.Thus,$f(f(x))$ is discontinuous at $x=2$ and $x=9$.Hence,$f(f(x))$ is discontinuous at exactly two values of $x$.

Consider the function $f(x) = \begin{cases} \frac{x+5}{x-2}, & \text{if } x \neq 2 \\ 1, & \text{if } x=2 \end{cases}$. Then,$f(f(x))$ is discontinuous

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