We have, $f(x)=2 x+1$ and $g(x)=x^2+1$

$$\begin{aligned} \text{(i)}\quad (f+g)(x) & =f(x)+g(x) \\ & =2 x+1+x^2+1=x^2+2 x+2 \end{aligned}$$

$$\begin{aligned} \text{(i)}\quad (f-g)(x) & =f(x)-g(x)=(2 x+1)-\left(x^2+1\right) \\ & =2 x+1-x^2-1=2 x-x^2=x(2-x) \end{aligned}$$

$$\begin{aligned} \text{(iii)}\quad (f g)(x) & =f(x) \cdot g(x)=(2 x+1)\left(x^2+1\right) \\ & =2 x^3+2 x+x^2+1=2 x^3+x^2+2 x+1 \end{aligned}$$

(iv) $\frac{f}{g}(x)=\frac{f(x)}{g(x)}=\frac{2 x+1}{x^2+1}$

We have, $f: X \rightarrow R, f(x)=x^3+1$.

Where $X=\{-1,0,3,9,7\}$,

When $x=-1$, then $f(-1)=(-1)^3+1=-1+1=0$

$\begin{aligned} & x=0, \text { then } f(0)=(0)^3+1=0+1=1 \\ & x=3 \text { then } f(3)=(3)^3+1=27+1=28 \\ & x=9 \text {, then } f(9)=(9)^3+1=729+1=730 \\ & x=7, \text { then } f(7)=(7)^3+1=343+1=344 \\ & f=\{(-1,0),(0,1),(3,28),(9,730),(7,344)\}\end{aligned}$

$\therefore \quad$ Range of $f=\{0,1,28,730,344\}$

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

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

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 $28-x \neq 0$ $$ \begin{aligned} & \Rightarrow \quad x \neq 28 \\ & \therefore \quad \text { Domain of } f=R-\{28\} \end{aligned}$$