Let A={1,2,3, … ., 100} and R be a relation on A such that R={(a, b): a=2 b+1}. Let (a₁, a₂),(a₂, a₃),(a₃, a₄), … .,(aₖ, aₖ₊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 :

Solution

Maths sir · 4 steps

Run the chain backwards from the smallest number, doubling and adding 1, until you cross 100.

1. Read the chain. Each pair (aᵢ, aᵢ₊₁) in R means aᵢ = 2aᵢ₊₁ + 1. So every term is built from the next one by doubling and adding 1, and the terms grow as you move left.
2. Start small. To make the chain as long as possible, let the last term aₖ₊₁ be as small as possible, which is 1. Then build backwards: 1 → 3 → 7 → 15 → 31 → 63 → 127.
3. Stop inside A. 127 > 100, so it is not in A. The longest chain is 63, 31, 15, 7, 3, 1. That is 6 numbers joined by 5 pairs: (63,31), (31,15), (15,7), (7,3), (3,1).

   a₁ = 2ᵏ(aₖ₊₁ + 1) - 1 ≤ 100 ⇒ 2ᵏ × 2 - 1 ≤ 100 ⇒ 2ᵏ ≤ 50.5

4. So. The largest k with 2ᵏ ≤ 50.5 is k = 5, since 2⁵ = 32 and 2⁶ = 64. The answer is 5, option (d).

The idea to keep. For chains of a relation like a = 2b + 1, start from the smallest allowed value and apply the rule repeatedly. Count pairs, not numbers: n numbers make n - 1 links.

Answer from the NTA final answer key. Solution by PrepWiser.