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 \(\frac{a-|x|}{(a+1)-|x|}=y\)
\(\begin{aligned}
& \Rightarrow a-|x|=y(a+1)-y|x| \quad[\because \text { assuming }|x| \neq a+1] \\
& \Rightarrow(y-1)|x|=y(a+1)-a \\
& \Rightarrow|x|=\frac{y(a+1)-a}{y-1} \geq 0, \forall x \in \text { domain of } f(x) \\
& \therefore y \in\left(-\infty, \frac{a}{a+1}\right] \cup(1, \infty) \quad(\text { as } a > 0)
\end{aligned}\)
So, range of \(f(x)=\sqrt{y} \in\left[0, \sqrt{\frac{a}{a+1}}\right] \cup(1, \infty)\)
\(\text {[as } \sqrt{y} \geq 0]\)
Hence, option (c) is correct.