Relations and Functions – Relations – JEE Main 02 April 2025 Shift 2

Solution:

From the given relation $R=\{(a,b): a=2b+1\}$, for the sequence of ordered pairs $(a_i, a_{i+1})$:
$$a_i = 2a_{i+1} + 1$$

Adding $1$ to both sides:
$$a_i + 1 = 2(a_{i+1} + 1)$$

This recursive relation implies:
$$a_1 + 1 = 2(a_2 + 1) = 2^2(a_3 + 1) = \dots = 2^k(a_{k+1} + 1)$$

Since all elements must belong to the set $A = \{1, 2, 3, \dots, 100\}$, the maximum possible value for $a_1$ is $100$.
$$a_1 + 1 \le 101$$
$$2^k(a_{k+1} + 1) \le 101$$

To maximize $k$, we must minimize $a_{k+1}$. The minimum possible values in set $A$ are $1$ and $2$.

If $a_{k+1} = 1$:
$$2^k(1 + 1) \le 101$$
$$2^{k+1} \le 101 \Rightarrow 2^{k+1} \le 64 \Rightarrow k+1 \le 6 \Rightarrow k \le 5$$
The ordered elements are: $63 \rightarrow 31 \rightarrow 15 \rightarrow 7 \rightarrow 3 \rightarrow 1$.

If $a_{k+1} = 2$:
$$2^k(2 + 1) \le 101$$
$$3 \cdot 2^k \le 101 \Rightarrow 2^k \le 33.33 \Rightarrow k \le 5$$
The ordered elements are: $95 \rightarrow 47 \rightarrow 23 \rightarrow 11 \rightarrow 5 \rightarrow 2$.

Thus, the largest integer $k$ for which such a sequence exists is $5$.

Ans. (3)