$f(x)=\frac{4x+3}{6x-4}$$f \circ f(x)=\frac{4\left(\frac{4x+3}{6x-4}\right)+3}{6\left(\frac{4x+3}{6x-4}\right)-4}$$=\frac{16x+12+18x-12}{24x+18-24+16}$$=\frac{34x}{34}=x$$y=\frac{4x+3}{6x-4}$$6xy-4y=4x+3$$6xy-4x=4y+3$$x=\frac{4y+3}{6y-4}$$\therefore f(x)=\frac{4x+3}{6x-4}$ OR Given : $R=\{(a,b): a < b\}$Let $A$ be the set of real numbers.(i) For symmetric : Let $a, b \in A$Then, $(a,b) \in R$ such that $a < b$$\Rightarrow b < a$ which is false$\Rightarrow (b,a) \notin R$$\therefore R$ is not symmetric.(ii) For transitive : Let $a, b, c \in A$Then, $(a,b) \in R$ and $(b,c) \in \text{bo} R$$\Rightarrow a < b \text{ and } b < c$$\Rightarrow a < c$$\Rightarrow (a,c) \in R$ $\therefore R$ is transitive.