Find the Domain and range of the following real valued functions.

$\text{ i) }\log \left|4-x^2\right|$

 $\text{ ii) }\frac{x^2-4}{x-2}$

$\text{ iii) }f\left(x\right)=\sqrt{9-x^2}$

$\text{ iv) }f\left(x\right)=\frac{x}{2-3x}$

(i) let $y=f\left(x\right)=\log \left|4-x^2\right|$

$f\left(x\right)\in R\Rightarrow 4-x^2\ne 0$

$\Rightarrow x^2\ne 4\Rightarrow x\ne \pm 2$

Domain is $R-\{-2,2\}$

$\because y=\log \left|4-x^2\right|\Rightarrow \left|4-x^2\right|=e^y$

$\because e^y>0,\mathrm{\forall }y\in R\therefore$ Range of f is R

(ii)$\text{ Let }y=f\left(x\right)=\frac{x^2-4}{x-2}$

$\Rightarrow y=\frac{(x+2)(x-2)}{x-2}\Rightarrow x-2\ne 0\Rightarrow x\ne 2$

$\therefore$ domain of f is $R-\left\{2\right\}$

Then $y=x+2\Rightarrow y\ne 4[\because x\ne 2]$

Then its range $R-\left\{4\right\}$

(iii)$\Rightarrow 9-x^2\ge 0\Rightarrow (3-x)(3+x)\ge 0$

$\Rightarrow (x-3)(x+3)\le 0$

$\therefore$ Domain of f is $[-3,3]$

If $-3\le x\le 3\Rightarrow 0\le x^2\le 9$

$\Rightarrow 0\ge -x^2\ge 9\Rightarrow 9\ge 9-x^2\ge 0$

$\Rightarrow 0\le \sqrt{9-x^2}\le 3\Rightarrow 0\le f\left(x\right)\le 3$

$\Rightarrow f\left(x\right)\in [0,3]$

$\therefore$ Range of $f=[0,3]$

(iv) $f\left(x\right)=\frac{x}{2-3x}$ is defined when $2-3x\ne 0$

$=x\ne \frac{2}{3}$    $\therefore$ Domain is $R-\left\{\frac{2}{3}\right\}$

let $f\left(x\right)=y\Rightarrow \frac{x}{2-3x}=y$

$\Rightarrow x=2y-3xy\Rightarrow x+3xy=2y$

$\Rightarrow x(1+3y)=2y$

$\Rightarrow x=\frac{2y}{1+3y}$ is defined when $1+3y\ne 0$

$\Rightarrow y\ne \frac{-1}{3}$

$\therefore$ Range is $R-\left\{\frac{-1}{3}\right\}$