The range of $f(x)=\sqrt{\frac{a-| x|}{(a+1)-| x|}},(a0)$ is

Question

The range of $f(x)=\sqrt{\frac{a-| x|}{(a+1)-| x|}},(a>0)$ is

Aicharya · Accepted Answer

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.

The range of \(f(x)=\sqrt{\frac{a-| x|}{(a+1)-| x|}},(a>0)\) is

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