(i) (a) : Clearly (1, 1), (2, 2), (3, 3), \(\in\) R. So, R is reflexive on A.
Since, \((1,2) \in R \text { but }(2,1) \notin R\) So, R is not symmetric on A.
Since, \((2,3), \in R\) and \((3,1) \in R\) but \((2,1) \notin R\) .So, R is not transitive on A.
(ii) (b) : Since, (1,1), (2, 2) and (3, 3) are not in R. So, R is not reflexive on A.
Now, \((1,2) \in R \Rightarrow(2,1) \in R\) 
and \((1,3) \in R \Rightarrow(3,1) \in R\) 
So, R is symmetric
Clearly,\((1,2) \in R \text { and }(2,1) \in R \text { but }(1,1) \notin R\) 
So, R is not transitive on A.
(iii) (c) : We have, \(R=\{(x, y): y=x+5 \text { and } x<4\}\) ,where \(x, y \in N\) .
\(\therefore R=\{(1,6),(2,7),(3,8)\}\) 
Clearly, (1, 1), (2, 2) etc. are not in R. So, R is not reflexive.
Since, \((1,6) \in R\) but \((6,1) \notin R\) So, R is not symmetric. 
Since, \((1,6) \in R\) R and there is no order pair in R which has 6 as the first element. Same is the case for (2, 7) and (3, 8). So, R is transitive.
(iv) (d) : We have,R = {(x, y) : 3x - y = 0}, where \(x, y \in A=\{1,2, \ldots \ldots, 14\}\) .
\(\therefore\) R = {(I, 3), (2, 6), (3, 9), (4, 12)}
Clearly,\((1,1) \notin R\) So, R is not reflexive on A.
Since, \((1,3) \in R\) but \((3,1) \notin R\) .So, R is not symmetric on A.
Since, \((1,3) \in R\) and \((3,9) \in R\)  but  \((1,9) \notin R\) So, R is not transitive on A.
(v) (d) : Clearly, (1, 1), (2, 2), (3, 3) ∈  R. So, R is reflexive on A.
We find that the ordered pairs obtained by interchanging the components of ordered pairs in R are also in R. So, R is symmetric on A.
For \(1,2,3 \in A\)  such that (1, 2) and (2, 3) are in Rimplies that (1, 3) is also, in R. So, R is transitive on A. Thus, R is an equivalence relation.

(i) (a) : For f(x) to be defined \(x-2 \neq 0\) i.e.,\(x \neq 2\) 
\(\therefore\) Domain of f = R - {2}
(ii) (b) : Let y =J(x), then \(y=\frac{x-1}{x-2}\) 
\(\Rightarrow x y-2 y=x-1 \Rightarrow x y-x=2 y-1 \Rightarrow x=\frac{2 y-1}{y-1}\) 
Since, \(x \in R-\{2\}\),therefore \(y \neq 1\)
Hence, range of f = R-{1} 
(iii) (d): We have,g(x) = 2f(x) - 1
\(=2\left(\frac{x-1}{x-2}\right)-1=\frac{2 x-2-x+2}{x-2}=\frac{x}{x-2}\)
(iv) (a) : We have, \(g(x)=\frac{x}{x-2}\) ,
Let  \(g\left(x_{1}\right)=g\left(x_{2}\right) \Rightarrow \frac{x_{1}}{x_{1}-2}=\frac{x_{2}}{x_{2}-2}\) 
 \(\Rightarrow x_{1} x_{2}-2 x_{1}=x_{1} x_{2}-2 x_{2} \Rightarrow 2 x_{1}=2 x_{2} \Rightarrow x_{1}=x_{2}\) 
Thus, \(g\left(x_{1}\right)=g\left(x_{2}\right) \Rightarrow x_{1}=x_{2}\) 
Hence, g(x) is one-one.
(v) (c)