The range of the function f(x) = frac{sqrt{1 | vedclass

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(A) Given the function $f(x) = \frac{\sqrt{1 - x^2}}{1 + |x|}$.Since the domain of $\sqrt{1 - x^2}$ is $x \in [-1, 1]$,we can substitute $x = \sin 	h

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$[0, 1]$

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(A) Given the function $f(x) = \frac{\sqrt{1 - x^2}}{1 + |x|}$.Since the domain of $\sqrt{1 - x^2}$ is $x \in [-1, 1]$,we can substitute $x = \sin 	heta$ where $	heta \in [-\pi/2, \pi/2]$.Then $|x| = |\sin 	heta|$ and $\sqrt{1 - x^2} = \sqrt{1 - \sin^2 	heta} = \sqrt{\cos^2 	heta} = |\cos 	heta|$.So,$f(x) = \frac{|\cos 	heta|}{1 + |\sin 	heta|}$.Let $t = |\sin 	heta|$,where $t \in [0, 1]$. Then $|\cos 	heta| = \sqrt{1 - t^2}$.Thus,$y = \frac{\sqrt{1 - t^2}}{1 + t} = \frac{\sqrt{(1 - t)(1 + t)}}{1 + t} = \sqrt{\frac{1 - t}{1 + t}}$.As $t$ increases from $0$ to $1$,the expression $\frac{1 - t}{1 + t}$ decreases from $1$ to $0$.Therefore,the range of $y = \sqrt{\frac{1 - t}{1 + t}}$ is $[0, 1]$.

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

The range of the function $f(x) = \frac{\sqrt{1 - x^2}}{1 + |x|}$ is

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Consider the following lists.
$A$. $f(x)=\frac{|x+2|}{x+2}, x \neq-2$ $1$. $[\frac{1}{3}, 1]$
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