Class JEE Mathematics Sets, Relations, and Functions Q #1136
COMPETENCY BASED
APPLY
4 Marks 2024 JEE Main 2024 (Online) 9th April Morning Shift NUMERICAL
5 Let $A=\{2,3,6,7\}$ and $B=\{4,5,6,8\}$. Let $R$ be a relation defined on $A \times B$ by $(a_1, b_1) R(a_2, b_2)$ if and only if $a_1+a_2=b_1+b_2$. Then the number of elements in $R$ is __________.

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Step-by-Step Solution

  1. First, list all possible pairs of elements from set A and set B: $A = \{2, 3, 6, 7\}$ and $B = \{4, 5, 6, 8\}$.

  2. We are given the relation $R$ defined on $A \times B$ such that $(a_1, b_1) R (a_2, b_2)$ if and only if $a_1 + a_2 = b_1 + b_2$. We need to find the number of elements in $R$.

  3. Let's consider all possible pairs $(a_1, b_1)$ and $(a_2, b_2)$ from $A \times B$ and check the condition $a_1 + a_2 = b_1 + b_2$. Since we are looking for the number of elements in $R$, we need to find pairs of pairs that satisfy the given condition.

  4. Instead of checking all possible pairs of pairs, let's analyze the possible values of $a_1 + a_2$ and $b_1 + b_2$.

    The possible values of $a_1 + a_2$ are: $2+2=4, 2+3=5, 2+6=8, 2+7=9, 3+2=5, 3+3=6, 3+6=9, 3+7=10, 6+2=8, 6+3=9, 6+6=12, 6+7=13, 7+2=9, 7+3=10, 7+6=13, 7+7=14$.

    The possible values of $b_1 + b_2$ are: $4+4=8, 4+5=9, 4+6=10, 4+8=12, 5+4=9, 5+5=10, 5+6=11, 5+8=13, 6+4=10, 6+5=11, 6+6=12, 6+8=14, 8+4=12, 8+5=13, 8+6=14, 8+8=16$.

  5. Now, we need to find the number of pairs $(a_1, b_1), (a_2, b_2)$ such that $a_1 + a_2 = b_1 + b_2$.

    Let's consider all pairs $(a, b) \in A \times B$. $A \times B = \{(2,4), (2,5), (2,6), (2,8), (3,4), (3,5), (3,6), (3,8), (6,4), (6,5), (6,6), (6,8), (7,4), (7,5), (7,6), (7,8)\}$. There are $4 \times 4 = 16$ such pairs.

  6. We need to find how many pairs of pairs satisfy the condition. This is equivalent to counting the number of pairs $(a_1, b_1), (a_2, b_2)$ such that $a_1 + a_2 = b_1 + b_2$.

  7. Instead, let's consider the possible values of $S = a_1 + a_2 = b_1 + b_2$.

    The minimum value of $a_1 + a_2$ is $2+2=4$ and the maximum value is $7+7=14$. The minimum value of $b_1 + b_2$ is $4+4=8$ and the maximum value is $8+8=16$.

    So, we are looking for values of $S$ that are possible for both $a_1 + a_2$ and $b_1 + b_2$. The possible values of $S$ are $8, 9, 10, 12, 13, 14$.

  8. Now, let's count the number of ways to obtain each value of $S$:

    • $S=8$: $a_1+a_2=8 \implies (2,6), (6,2)$. $b_1+b_2=8 \implies (4,4)$. Number of pairs: $2 \times 1 = 2$
    • $S=9$: $a_1+a_2=9 \implies (2,7), (3,6), (6,3), (7,2)$. $b_1+b_2=9 \implies (4,5), (5,4)$. Number of pairs: $4 \times 2 = 8$
    • $S=10$: $a_1+a_2=10 \implies (3,7), (7,3)$. $b_1+b_2=10 \implies (4,6), (6,4), (5,5)$. Number of pairs: $2 \times 3 = 6$
    • $S=12$: $a_1+a_2=12 \implies (6,6)$. $b_1+b_2=12 \implies (4,8), (6,6), (8,4)$. Number of pairs: $1 \times 3 = 3$
    • $S=13$: $a_1+a_2=13 \implies (6,7), (7,6)$. $b_1+b_2=13 \implies (5,8), (8,5)$. Number of pairs: $2 \times 2 = 4$
    • $S=14$: $a_1+a_2=14 \implies (7,7)$. $b_1+b_2=14 \implies (6,8), (8,6)$. Number of pairs: $1 \times 2 = 2$
  9. Total number of elements in $R$ is $2 + 8 + 6 + 3 + 4 + 2 = 25$.

Correct Answer: 20

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to apply the definition of a relation and the given condition to find the number of elements that satisfy the relation.
Knowledge Dimension: CONCEPTUAL
Justification: The question tests the understanding of the concept of relations and how they are defined on Cartesian products. It requires applying the given condition to identify the elements that belong to the relation.
Syllabus Audit: In the context of JEE, this is classified as COMPETENCY. It requires the student to apply the definition of a relation to a specific problem, rather than just recalling the definition.

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