Let A = {a,b,c,d} and B = {c,d,e,f,g}. Let R1 = {(a,c), (b,d), (c,e)} R2 = {(a,c), (a,g), (b,d), (c,e), (d,f)} R3 = {(a,c), (b,d), (c,e), (d,f)} Justify which of the given relation is a function from A to B. (c) Let f be a real valued function defined by f(x) = 1 x2−9 . (i) What is the domain of f? (ii) What is the range of f? (iii) Represent f as a set of ordered pairs.

Given $A = \{a,b,c,d\}$ and $B = \{c,d,e,f,g\}$ .

$R_1 = \{(a,c), (b,d), (c,e)\}, R_2 = \{(a,c), (a,g), (b,d), (c,e), (d,f)\}, \\ R_3 = \{(a,c), (b,d), (c,e), (d,f)\}$

A relation is a function when every element of set A has image in B and a element of set A can-not have more than one image in set B.

So, Relation $R_3$ is a function.

(c) Given $f$ is a real valued function defined by $f(x) = x^2 - 9$ .

(I) Function is defined for all real values of $x$. Hence,

Domain of$f = \R$

(ii) Now, as $x^2 \geq 0 \implies x^2-9 \geq -9$

Hence, Range of $f = [-9,\infin)$

(iii) Representation of $f$ as a set of ordered pair = $\{ (x,x^2-9) : x \in \R\}$