Question

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

• A. $[0,a]$
• B. $[0,\infty )-\left[-\sqrt{\frac{a}{a+1}},\sqrt{\frac{a}{a+1}}\right]$
• C. $\left[0,\sqrt{\frac{a}{a+1}}\right]\cup (1,\infty )$
• D. $\left[0,\sqrt{\frac{a}{a+1}}+1\right]$

Answer 1

Answer: (C)

Given function is
$f(x)=\sqrt{\frac{a-|x|}{(a+1)-|x|}},(a>0)$
$\because f(x)\ge 0,\forall x\in$ domain of $f(x)$
Now, let $\frac{a-|x|}{(a+1)-|x|}=y$
$\Rightarrow a-|x|=y(a+1)-y|x|$ $[\because$ assuming $|x|\ne a+1]$
$\Rightarrow (y-1)|x|=y(a+1)-a$
$\Rightarrow |x|=\frac{y(a+1)-a}{y-1}\ge 0,\forall x\in$ 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 $\sqrt{y}\ge 0]$