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: Understanding the Question:
We need to determine which of the given relations from set $A$ to set $B$ do not satisfy the definition of a mathematical function.

Step 2: Key Formula or Approach:
A relation from set $A$ to set $B$ is a function if and only if every element in the domain ($A$) maps to exactly one unique element in the codomain ($B$).

Step 3: Detailed Explanation:
Let's check each relation:
$R_1$: $a \to 1, b \to 2, c \to 1, d \to 2$. Every element in $A$ has exactly one image. It is a function.
$R_2$: $a \to 1, b \to 1, c \to 1, d \to 1$. Every element in $A$ has exactly one image (it's a constant function). It is a function.
$R_3$: $a \to 2, b \to 3, c \to 2, d \to 2$. Every element in $A$ has exactly one image. It is a function.
$R_4$: $a \to 1, b \to 2, a \to 2, d \to 3$. Here, the element '$a$' in the domain maps to two different elements ('$1$' and '$2$') in the codomain. Additionally, element '$c$' is completely missing.
Therefore, $R_4$ is not a function.

Step 4: Final Answer:
Only $R_4$ is not a function, matching option (D).