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JEE Mathematics 2023 Question with Solution

Let A={2,3,4}A = \{2, 3, 4\} and B={8,9,12}B = \{8, 9, 12\}. Then the number of elements in the relation R={((a1,b1),(a2,b2))(A×B,A×B):a1 divides b2 and a2 divides b1}R = \{((a_1, b_1), (a_2, b_2)) \in (A \times B, A \times B) : a_1 \text{ divides } b_2 \text{ and } a_2 \text{ divides } b_1 \} is:

  • A

    1818

  • B

    2424

  • C

    1212

  • D

    3636

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: A={2,3,4}A = \{2, 3, 4\} and B={8,9,12}B = \{8, 9, 12\}.

Find: The number of elements in the relation R={((a1,b1),(a2,b2)):a1 divides b2 and a2 divides b1}R = \{((a_1,b_1),(a_2,b_2)) : a_1 \text{ divides } b_2 \text{ and } a_2 \text{ divides } b_1\}.

The solution states that for each element of AA, there are exactly 22 elements of BB satisfying the divisibility condition.

Check divisibility from AA to BB:

  • 22 divides 8,128, 12
  • 33 divides 9,129, 12
  • 44 divides 8,128, 12

So there are 66 valid ordered pairs (a,b)A×B(a,b) \in A \times B such that aa divides bb.

Now the relation consists of ordered pairs ((a1,b1),(a2,b2))((a_1,b_1),(a_2,b_2)) with conditions:

  • a1a_1 divides b2b_2
  • a2a_2 divides b1b_1

By the same counting, there are 66 choices for one compatible pair and 66 choices for the other compatible pair.

Total number of elements in R=6×6=36\text{Total number of elements in } R = 6 \times 6 = 36

Therefore, the number of elements in the relation is 3636. The correct option is D.

Pair Counting View

Given: A={2,3,4}A = \{2, 3, 4\} and B={8,9,12}B = \{8, 9, 12\}.

Find: The cardinality of the relation RR.

First count all ordered pairs (a,b)(a,b) with aAa \in A, bBb \in B, and aba \mid b.

28,1239,1248,12\begin{aligned} 2 &\mid 8, 12 \\ 3 &\mid 9, 12 \\ 4 &\mid 8, 12 \end{aligned}

Hence the valid pairs are:

(2,8),(2,12),(3,9),(3,12),(4,8),(4,12)(2,8), (2,12), (3,9), (3,12), (4,8), (4,12)

So the number of such pairs is 66.

In ((a1,b1),(a2,b2))R((a_1,b_1),(a_2,b_2)) \in R, the conditions are cross-linked:

  • a1b2a_1 \mid b_2
  • a2b1a_2 \mid b_1

This means choosing ((a1,b2))((a_1,b_2)) as one valid divisible pair and ((a2,b1))((a_2,b_1)) as another valid divisible pair.

Each choice has 66 possibilities, so

R=66=36|R| = 6 \cdot 6 = 36

Therefore, the correct option is D.

Common mistakes

  • Counting only pairs (a,b)(a,b) with aba \mid b and stopping at 66. This is wrong because the relation is on (A×B)×(A×B)(A \times B) \times (A \times B), so you must count ordered pairs of such pairs. After finding 66 compatible pairs, multiply by 66 again.

  • Ignoring the cross-conditions a1b2a_1 \mid b_2 and a2b1a_2 \mid b_1. This is wrong because the divisibility is not between coordinates of the same ordered pair. Read the indices carefully and count the compatible cross-pairings.

  • Assuming each element of AA divides all elements of BB. This is wrong because, for example, 22 does not divide 99 and 44 does not divide 99. List the valid divisibility cases explicitly before counting.

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