Given function is
$f(x)=\sqrt{\frac{a-|x|}{(a+1)-|x|}},(a>\,0)$
$\because\, f(x) \geq 0, \forall x \in$ domain of $f(x)$
Now, let $\frac{a-|x|}{(a+1)-|x|}=y$
$\Rightarrow \,a-|x|=y(a+1)-y|x|$ $[\because$ assuming $|x| \neq a+1]$
$\Rightarrow \, (y-1)|x|=y(a+1)-a$
$\Rightarrow \, |x|=\frac{y(a+1)-a}{y-1} \geq 0, \forall x \in$ domain of $f(x)$
$\therefore \, y \in\left(-\infty, \frac{a}{a+1}\right] \cup(1, \infty)($ as $a>\,0)$
So, range of $f(x)=\sqrt{y} \in\left[0, \sqrt{\frac{a}{a+1}}\right] \cup(1, \infty)$
[ as $\sqrt{y} \geq\, 0]$