Question

Let $ A = \{1, 2, 3, ..., 10\} $ and $ R $ be a relation on $ A $ such that $ R = \{(a, b) : a = 2b + 1\} $. Let $ (a_1, a_2), (a_3, a_4), (a_5, a_6), ..., (a_k, a_{k+1}) $ be a sequence of $ k $ elements of $ R $ such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer $ k $, for which such a sequence exists, is equal to:

• A. 6
• B. 7
• C. 5
• D. 8

Hint:

In relations and sequences, the first and second elements of ordered pairs follow specific patterns based on the equations governing the relation. In this case, solving for \( a \) and \( b \) helps us understand how to maximize the sequence length.

Answer 1

The Correct Option is C

Given \( a = 2b + 1 \), we can solve for \( b \) as follows: \[ b = \frac{a - 1}{2} \] The set \( R \) is given by \( \{(3, 1), (5, 2), \dots, (99, 49)\} \). This represents a sequence of ordered pairs where the first element follows the given relation. Let \( (2m + 1, m) \), \( (2n - 1, n) \), etc., be such ordered pairs. From the condition, we have: \[ m = 2a - 1 \quad \Rightarrow \quad m \, \text{is odd number} \] The first element of ordered pair \( (a, b) \) is: \[ a = 2(2a - 1) + 1 = 4a - 1 \] Hence, \( a = \{3, 7, 11, \dots, 99\} \). For maximum number of ordered pairs in such a sequence, we need to solve for \( \lambda \). This gives us the largest sequence length. \[ \lambda = 2a - 1 \] The number of terms in this sequence satisfies: \[ \lambda \in \{1, 2, 3, \dots, 25\} \] Thus, for maximum ordered pairs, we evaluate cases for various values of \( \lambda \). The final maximum value of \( r \) for \( \lambda = 16 \) is 5.