Consider $\mathrm{f}:\mathrm{ℝ}-\left\{-\frac{4}{3}\right\}\to \mathrm{ℝ}-\left\{\frac{4}{3}\right\}$ given by $\mathrm{f}\left(\mathrm{x}\right)=\frac{4\mathrm{x}+3}{3\mathrm{x}+4}$ Show that f is bijective. Find the inverse of f and hence find  ${\mathrm{f}}^{-1}\left(0\right)$ and x such that ${\mathrm{f}}^{-1}\left(\mathrm{x}\right)=2$.

OR

Let $\mathrm{A}=\mathrm{\mathbb{Q}}\times \mathrm{\mathbb{Q}}$ and let * be a binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for (a, b), (c, d) 𝜖 A. Determine, whether * is commutative and associative. Then, with respect to * on A
(i) Find the identity element in A.
(ii) Find the invertible elements of A.

Let ${\mathrm{x}}_1,{\mathrm{x}}_2\in \mathrm{R}-\left\{-\frac{4}{3}\right\}$ and $\mathrm{f}\left({\mathrm{x}}_1\right)=\mathrm{f}\left({\mathrm{x}}_2\right)$
$$\begin{gathered} \Rightarrow \frac{4{\mathrm{x}}_1+3}{3{\mathrm{x}}_1+4}=\frac{4{\mathrm{x}}_2+3}{3{\mathrm{x}}_2+4}\Rightarrow \left(4{\mathrm{x}}_1+3\right)\left(3{\mathrm{x}}_2+4\right)=\left(4{\mathrm{x}}_2+3\right)\left(3{\mathrm{x}}_1+4\right) \\ \begin{array}{l}\Rightarrow 12{\mathrm{x}}_1{\mathrm{x}}_2+16{\mathrm{x}}_1+9{\mathrm{x}}_2+12=12{\mathrm{x}}_1{\mathrm{x}}_2+16{\mathrm{x}}_2+9{\mathrm{x}}_2+12\\ \Rightarrow 16\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)-9\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)=0\Rightarrow 7\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)=0\\ \Rightarrow {\mathrm{x}}_1-{\mathrm{x}}_2=0\Rightarrow {\mathrm{x}}_1={\mathrm{x}}_2\end{array} \end{gathered}$$
$\Rightarrow \mathrm{f}$ is a one one function.
Let $\mathrm{y}=\frac{4\mathrm{x}+3}{3\mathrm{x}+4}$for $\mathrm{y}\in \mathrm{R}-\left\{\frac{4}{3}\right\}$
$\begin{array}{l}\Rightarrow 3\mathrm{xy}+4\mathrm{y}=4\mathrm{x}+3\Rightarrow 4\mathrm{x}-3\mathrm{xy}=4\mathrm{y}-3\Rightarrow \mathrm{x}(4-3\mathrm{y})=4\mathrm{y}-3\\ \Rightarrow \mathrm{x}=\frac{4\mathrm{y}-3}{4-3\mathrm{y}}\Rightarrow \mathrm{\forall }\mathrm{y}\in \mathrm{R}-\left\{\frac{4}{3}\right\},\mathrm{x}\in \mathrm{R}-\left\{-\frac{4}{3}\right\}\end{array}$

$\Rightarrow \mathrm{f}$ is onto function and hence it is bijective.
Now, 
$\because \mathrm{x}=\frac{4\mathrm{y}-3}{4-3\mathrm{y}}\Rightarrow {\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\frac{4\mathrm{x}-3}{4-3\mathrm{x}}$

$f^{-1}\left(0\right)=\frac{-3}{4}$
for ${\mathrm{f}}^{-1}\left(\mathrm{x}\right)=2\Rightarrow \frac{4\mathrm{x}-3}{4-3\mathrm{x}}=2\Rightarrow 4\mathrm{x}-3=8-6\mathrm{x}$
$\Rightarrow 10\mathrm{x}=11\Rightarrow \mathrm{x}=\frac{11}{10}$

OR

A =ℚ×ℚ....... [Given]
For any (a, b), (c, d) ∈A, ∗ is defined by
(a, b) ∗ (c, d) = (ac, b + ad) ..... [Given]
To check ∗ is commutative i.e. to check (a, b) ∗ (c, d) = (c, d) ∗ (a, b) for any (a, b), (c, d) ∈ A
Now, (a, b) ∗ (c, d) = (ac, b + ad)
(c, d) ∗ (a, b) = (ca, d + cb) = (ac, d + bc) = (ac, b + ad)
∴(a, b) ∗ (c, d) = (c, d) ∗ (a, b)
Thus, ∗ is not commutative ....... (1)
To check associativity
Let (a, b), (c, d), (e, f) ∈ A
Now, (a, b) ∗ ((c, d) ∗ (e, f)) = (a, b) ∗ (ce, d + cf)
= (ace, b + a(d + cf))
= (ace, b + ad + acf) ...... (2)
∴ (a, b) ∗ ((c, d) ∗ (e, f)) = (ace, b + ad + acf)
((a, b) ∗ (c, d)) ∗ (e, f) = (ac, b + ad) ∗ (e, f)
= (ace, b + ad + acf)
= (a, b) ∗ ((c, d) ∗ (e, f)) ..... From (2)
∴ (a, b) ∗ ((c, d) ∗ (e, f)) = ((a, b) ∗ (c, d)) ∗ (e, f)
Thus, ∗ is associative ....... (3)
(i) To find identity element
Let e = (a′, b′) be identity element of A
⟹ (a, b) ∗ (a′, b′) = (a, b) = (a′, b′) ∗ (a, b)
As (a, b) ∗ (a′, b′) = (a, b)
⟹ (aa′, b + ab′) = (a, b) ........ Using definition of ∗
⟹ aa′ = a and b + ab′=b
⟹ a′ = 1 and b′ = 0
We can verify it as follows
(a′, b′) ∗ (a, b) = (a′a, b′ + a′b) = (1⋅a, 0+1⋅b) = (a, b)
Similarly, (a, b) ∗ (a′, b′) = (a, b)
Hence, e= (1, 0) is the identity element in A.
(ii) To find inverse element
Let f = (c′, d′) be inverse element of (a, b) ∈ A
⟹ (a, b) ∗ (c′, d′) = (1, 0) = (c′, d′) ∗ (a, b) ...... Using definition of inverse element
Now, (a, b) ∗ (c′, d′) = (1,0)
⟹ (ac′, b + ad′) = (1,0) ........ [Using definition of ∗]
⟹ ac′ = 1 and b + ad′ =0
$\Rightarrow {\mathrm{c}}^{\mathrm{\prime }}=\frac{1}{\mathrm{a}}\text{ and }{\mathrm{d}}^{\mathrm{\prime }}=-\frac{\mathrm{b}}{\mathrm{a}}$
We can verify it as follows
$\left({\mathrm{c}}^{\mathrm{\prime }},{\mathrm{d}}^{\mathrm{\prime }}\right)\ast (\mathrm{a},\mathrm{b})=\left({\mathrm{c}}^{\mathrm{\prime }}\mathrm{a},{\mathrm{d}}^{\mathrm{\prime }}+{\mathrm{c}}^{\mathrm{\prime }}\mathrm{b}\right)=\left(\frac{1}{\mathrm{a}}\times \mathrm{a},-\frac{\mathrm{b}}{\mathrm{a}}+\frac{1}{\mathrm{a}}\times \mathrm{b}\right)=(1,0)$
Hence, $\mathrm{f}=\left(\frac{1}{\mathrm{a}},-\frac{\mathrm{b}}{\mathrm{a}}\right)$ is the inverse element of A