Question

If $A=\{1,2,3,4\}$ then which of the following are functions from $A$ to itself

• A. $f_1=\{(x,y):y=x+1\}$
• B. $f_2=\{(x,y):x=x+y>4\}$
• C. $f_3=\{(x,y):y<x\}$
• D. $f_4=\{(x,y):x+y=5\}$

Answer 1

The correct option is D $f_4=\{(x,y):x+y=5\}$

For function, every element of the domain must have a unique image in co-domain...

(A)Let us redefine each function as a set of ordered pairs

and see if they satisfy the above condition (A) of a function $f_1=\{(1,2),(2,3),(3,4)\}$ ; $(4,5)$ cannot be an ordered pair as $5$ does not belong to co-domain , Thus the element $4$ of domain does not have an image.

$f_2=\{(1,4),(2,3),(2,4),(3,2)(3,3),(3,4),(4,1),(4,2),(4,3),(4,4)\}$

The uniqueness is violated as the elements have not one but many images.

$f_3=\{(2,1),(3,2),(3,1),(4,1),(4,2),(4,3)\}$ Not a function as in $f_2$ uniqueness is violated.

$f_4=\{(1,4),(2,3),(3,2)(4,1)\}$ It is a function as each element has a unique image.