$$\begin{aligned} \text{Given that,} \quad f(x) & =2 x-3, \forall x \in R \\ \text{Now, let }\quad y & =2 x-3 \\ 2 x & =y+3 \\ x & =\frac{y+3}{2} \\ \therefore\quad f^{-1}(x) & =\frac{x+3}{2} \end{aligned}$$

$$\begin{aligned} \text{Given that,}\quad A & =\{a, b, c, d\} \\ \text{and }\quad f & =\{(a, b),(b, d),(c, a),(d, c)\} \\ f^{-1} & =\{(b, a),(d, b),(a, c),(c, d)\} \end{aligned} $$

$$\begin{aligned} \text{Given that,}\quad f(x) & =x^2-3 x+2 \\ \therefore\quad f\{f(x)\} & =f\left(x^2-3 x+2\right) \\ & =\left(x^2-3 x+2\right)^2-3\left(x^2-3 x+2\right)+2 \\ & =x^4+9 x^2+4-6 x^3-12 x+4 x^2-3 x^2+9 x-6+2 \\ & =x^4+10 x^2-6 x^3-3 x \\ f\{f(x)\} & =x^4-6 x^3+10 x^2-3 x \end{aligned}$$

Given that, $g=\{(1,1),(2,3),(3,5),(4,7)\}$.

Here, each element of domain has unique image. So, $g$ is a function.

Now given that,

$$\begin{aligned} g(x) & =\alpha x+\beta \\ g(1) & =\alpha+\beta \\ \alpha+\beta & =1 \quad\text{.... (i)}\\ g(2) & =2 \alpha+\beta \\ 2 \alpha+\beta & =3\quad\text{.... (ii)} \end{aligned}$$

From Eqs. (i) and (ii),

$$\begin{aligned} 2(1-\beta)+\beta & =3 \\ \Rightarrow\quad 2-2 \beta+\beta & =3 \\ \Rightarrow\quad 2-\beta & =3 \\ \beta & =-1 \\ \text{If}\quad \beta & =-1, \text { then } \alpha=2 \\ \alpha & =2, \beta=-1 \end{aligned}$$

(i) Given set of ordered pair is $\{(x, y): x$ is a person, $y$ is the mother of $x\}$.

It represent a function. Here, the image of distinct elements of $x$ under $f$ are not distinct, so it is not a injective but it is a surjective.

(ii) Set of ordered pairs $=\{(a, b)$ : $a$ is a person, $b$ is an ancestor of $a\}$

Here, each element of domain does not have a unique image. So, it does not represent function.