Consider the following lists.A. f(x)=frac{|x+ | vedclass

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(C) Since $f(x) = \frac{|x+2|}{x+2}, x 
eq -2$,we have $f(x) = 1$ for $x > -2$ and $f(x) = -1$ for $x < -2$. Thus,the range is $\{-1, 1\}$.$(B)$ Sinc

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$A-5, B-3, C-4, D-1$

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(C) Since $f(x) = \frac{|x+2|}{x+2}, x 
eq -2$,we have $f(x) = 1$ for $x > -2$ and $f(x) = -1$ for $x $(B)$ Since $g(x) = |[x]|$,and $[x]$ is an integer,$|[x]|$ is a non-negative integer,which is the set $W$.$(C)$ Since $h(x) = |x - [x]| = |\{x\}|$,and the fractional part $\{x\} \in [0, 1)$,the range is $[0, 1)$.$(D)$ Since $-1 \leq \sin 3x \leq 1$,we have $1 \leq 2 - \sin 3x \leq 3$. Taking the reciprocal,$\frac{1}{3} \leq \frac{1}{2 - \sin 3x} \leq 1$. Thus,the range is $[\frac{1}{3}, 1]$.Matching the results: $A-5, B-3, C-4, D-1$.

$A$. $f(x)=\frac{\|x+2\|}{x+2}, x \neq-2$	$1$. $[\frac{1}{3}, 1]$
$B$. $g(x)=\|[x]\|, x \in R$	$2$. $Z$
$C$. $h(x)=\|x-[x]\|, x \in R$	$3$. $W$
$D$. $f(x)=\frac{1}{2-\sin 3x}, x \in R$	$4$. $[0, 1)$
	$5$. $\{-1, 1\}$

Consider the following lists.
$A$. $f(x)=\frac{|x+2|}{x+2}, x \neq-2$ $1$. $[\frac{1}{3}, 1]$
$B$. $g(x)=|[x]|, x \in R$ $2$. $Z$
$C$. $h(x)=|x-[x]|, x \in R$ $3$. $W$
$D$. $f(x)=\frac{1}{2-\sin 3x}, x \in R$ $4$. $[0, 1)$
$5$. $\{-1, 1\}$

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Consider the following lists.$A$. $f(x)=\frac{|x+2|}{x+2}, x \neq-2$$1$. $[\frac{1}{3}, 1]$$B$. $g(x)=|[x]|, x \in R$$2$. $Z$$C$. $h(x)=|x-[x]|, x \in R$$3$. $W$$D$. $f(x)=\frac{1}{2-\sin 3x}, x \in R$$4$. $[0, 1)$$5$. $\{-1, 1\}$