Question

Let $A=[a, b, c, d], B=[1,2,3]$. Relation $R_1, R_2, R_3, R_4$ are as follows :
$R_1=[(a, 1), (b, 2), (c, 1), (d, 2)]$
$R_2=[(a, 1), (b, 1), (c, 1), (d, 1)]$
$R_3=[(a, 2), (b, 3), (c, 2), (d, 2)]$
$R_4=[(a, 1), (b, 2), (a, 2), (d, 3)]$, then

• A. only $R_3$ and $R_4$ are not functions
• B. only $R_1$ and $R_2$ are not functions
• C. only $R_3$ is not a function
• D. only $R_4$ is not a function

Hint:

To quickly spot a non-function in a list of ordered pairs, look for any repeating first coordinate. In $R_4$, $(a, 1)$ and $(a, 2)$ share the same input '$a$', immediately disqualifying it!

Answer 1

The Correct Option is D

Step 1: Recall what makes a function.
A relation from set $A$ to set $B$ is a function only if every element of $A$ is paired with exactly one element of $B$. No element of $A$ may be left out, and none may have two partners.

Step 2: Note the sets.
Here $A=\{a,b,c,d\}$ and $B=\{1,2,3\}$. So each of $a,b,c,d$ must appear once and only once on the left.

Step 3: Check $R_1$.
$R_1=\{(a,1),(b,2),(c,1),(d,2)\}$. Each of $a,b,c,d$ appears exactly once. It is a function.

Step 4: Check $R_2$.
$R_2=\{(a,1),(b,1),(c,1),(d,1)\}$. Each input has one image (all map to $1$). That is allowed, so it is a function.

Step 5: Check $R_3$.
$R_3=\{(a,2),(b,3),(c,2),(d,2)\}$. Each of $a,b,c,d$ appears once. It is a function.

Step 6: Check $R_4$.
$R_4=\{(a,1),(b,2),(a,2),(d,3)\}$. Here $a$ is paired with both $1$ and $2$, which breaks the rule. Also $c$ is missing. So $R_4$ is not a function.

Step 7: Pick the answer.
Only $R_4$ fails, which is option (4).
\[ \boxed{\text{Only } R_4 \text{ is not a function}} \]