$\because \quad f(x)=g(x)$

$$\begin{aligned} & \Rightarrow \quad 3 x^2-1=3+x \\ & \Rightarrow \quad 3 x^2-x-4=0 \\ & \Rightarrow \quad 3 x^2-4 x+3 x-4=0 \\ & \Rightarrow \quad x(3 x-4)+1(3 x-4)=0 \\ & \Rightarrow \quad(3 x-4)(x+1)=0 \\ & \therefore \quad x=-1, \frac{4}{3} \end{aligned}$$

We have, $$g=\{(1,1),(2,3),(3,5),(4,7)\}$$

Since, every element has unique image under $g$. So, $g$ is a function.

Now, $g(x)=\alpha x+\beta$

When $x=1$, then $g(1)=\alpha(1)+\beta\quad \text{.... (i)}$

$\Rightarrow \quad 1=\alpha+\beta$

When $x=2$, then $$g(2)=\alpha(2)+\beta$$

$$\Rightarrow \quad 3=2 \alpha+\beta\quad \text{.... (ii)}$$

On solving Eqs. (i) and (ii), we get

$$\alpha=2, \beta=-1$$

(i) We have, $f(x)=\frac{1}{\sqrt{1-\cos x}}$

$$\begin{array}{ll} \because & -1 \leq \cos x \leq 1 \\ \Rightarrow & -1 \leq-\cos x \leq 1 \\ \Rightarrow & 0 \leq 1-\cos x \leq 2 \end{array}$$

So, $f(x)$ is defined, if $1-\cos x \neq 0$

$$\begin{array}{rlrl} \cos x & \neq 1 \\ x & \neq 2 n \pi-\forall n \in Z \\ \therefore \quad \text { Domain of } f & =R-\{2 n \pi: n \in Z\} \end{array}$$

(ii) We have, $$f(x)=\frac{1}{\sqrt{x+|x|}}$$

$\begin{aligned} \because\quad x+|x| & =x-x=0, x<0 \\ & =x+x=2 x, x \geq 0\end{aligned}$

Hence, $f(x)$ is defined, if $x>0$.

$\therefore \quad$ Domain of $f=R^{+}$

(iii) We have, $\quad f(x)=x|x|$

Clearly, $f(x)$ is defined for any $x \in R$.

$\therefore \quad$ Domain of $f=R$

(iv) We have, $$f(x)=\frac{x^3-x+3}{x^2-1}$$

$$\begin{aligned} & f(x) \text { is not defined, if } \\ & x^2-1=0 \\ & \Rightarrow \quad(x-1)(x+1)=0 \\ & \Rightarrow \quad x=-1,1 \\ & \therefore \quad \text { Domain of } f=R-\{-1,1\} \end{aligned}$$

(v) We have, $$f(x)=\frac{3 x}{28-x}$$

Clearly, $f(x)$ is defined, $\quad$ if $28-x \neq 0$ $$ \begin{aligned} & \Rightarrow \quad x \neq 28 \\ & \therefore \quad \text { Domain of } f=R-\{28\} \end{aligned}$$

(i) We have, $$f(x)=\frac{3}{2-x^2}$$

$$\begin{aligned} \text{Let}\quad & y=f(x) \\ \text{Then,}\quad & y=\frac{3}{2-x^2} \Rightarrow 2-x^2=\frac{3}{y} \end{aligned}$$

$\Rightarrow \quad x^2=2-\frac{3}{y} \Rightarrow x=\sqrt{\frac{2 y-3}{y}}$

$x$ assums real values, if $2 y-3 \geq 0$ and $y>0 \Rightarrow y \geq \frac{3}{2}$

$\therefore \quad$ Range of $f=\left[\frac{3}{2}, \infty\right)$

$$\begin{aligned} \text{(ii) We know that,}\quad |x-2| & \geq 0 \Rightarrow-|x-2| \leq 0 \\ 1-|x-2| & \leq 1 \Rightarrow f(x) \leq 1 \\ \therefore \text { Range of } f & =(-\infty, 1] \end{aligned}$$

(iii) We know that, $$|x-3| \geq 0 \Rightarrow f(x) \geq 0$$

$$\therefore \quad \text { Range of } f=[0, \infty)$$

(iv) We know that,

$$\begin{aligned} & -1 \leq \cos 2 x \leq 1 \Rightarrow-3 \leq 3 \cos 2 x \leq 3 \\ & \Rightarrow \quad 1-3 \leq 1+3 \cos 2 x \leq 1+3 \Rightarrow-2 \leq 1+3 \cos 2 x \leq 1+3 \end{aligned}$$

$$\Rightarrow \quad-2 \leq f(x) \leq 4$$

$\therefore \quad$ Range of $f=[-2,4]$

$$\begin{aligned} \text{Since,}\quad |x-2| & =-(x-2), x<2 \\ x-2, x & \geq 2 \\ \text{and}\quad |2+x| & =-(2+x), x<-2 \\ (2+x), x & \geq-2 \end{aligned}$$

$\therefore \quad f(x)=|x-2|+|2+x|,-3 \leq x \leq 3$

$\begin{aligned} & =\left\{\begin{array}{cc}-(x-2)-(2+x), & -3 \leq x<-2 \\ -(x-2)+2+x, & -2 \leq x<2 \\ x-2+2+x, & 2 \leq x \leq 3\end{array}\right. \\ & =\left\{\begin{array}{cc}-2 x, & -3 \leq x<-2 \\ 4, & -2 \leq x<2 \\ 2, & 2 \leq x \leq 3\end{array}\right.\end{aligned}$