The range of f(x)=sqrt{frac{a-|x|}{(a+1)-|x|} | vedclass

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(C) Given function is $f(x) = \sqrt{\frac{a-|x|}{(a+1)-|x|}}, (a > 0)$.Since $f(x) \geq 0$ for all $x$ in the domain,let $y^2 = \frac{a-|x|}{(a+1)-|x|

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$\left[0, \sqrt{\frac{a}{a+1}}\right] \cup (1, \infty)$

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(C) Given function is $f(x) = \sqrt{\frac{a-|x|}{(a+1)-|x|}}, (a > 0)$.Since $f(x) \geq 0$ for all $x$ in the domain,let $y^2 = \frac{a-|x|}{(a+1)-|x|}$ where $y \geq 0$.Then $y^2((a+1)-|x|) = a-|x|$.$y^2(a+1) - y^2|x| = a - |x|$.$|x|(1 - y^2) = a - y^2(a+1)$.$|x| = \frac{a - y^2(a+1)}{1 - y^2} = \frac{y^2(a+1) - a}{y^2 - 1}$.Since $|x| \geq 0$,we have $\frac{y^2(a+1) - a}{y^2 - 1} \geq 0$.Solving the inequality using the critical points $y^2 = \frac{a}{a+1}$ and $y^2 = 1$,we get $y^2 \in [0, \frac{a}{a+1}] \cup (1, \infty)$.Since $f(x) = y$,the range is $[0, \sqrt{\frac{a}{a+1}}] \cup (1, \infty)$.

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

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