Let \(\mathrm{A}=\{1,2,3, \ldots, 100\}\) and R be a relation on A such that \(\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}=2 \mathrm{~b}+1\}\). Let \((\mathrm{a}_{1}, \mathrm{a}_{2})\), \((\mathrm{a}_{2}, \mathrm{a}_{3})\), \((\mathrm{a}_{3}, \mathrm{a}_{4})\), \(\ldots\), \((\mathrm{a}_{\mathrm{k}}, \mathrm{a}_{\mathrm{k}+1})\) 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:

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Sets Relations and Functions

Let

A = {1, 2, 3, \dots, 100}

and R be a relation on A such that

R = {(a, b) : a = 2 b + 1}

. Let

(a_{1}, a_{2})

(a_{2}, a_{3})

(a_{3}, a_{4})

\dots

(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✓ Correct
(D)8

Answer: (C)

Solution

The relation pairs

(a, b)

with

a = 2 b + 1

. A chain

(a_{1}, a_{2}), (a_{2}, a_{3}), \dots

needs

a_{i} = 2 a_{i + 1} + 1

, so each entry is twice the next plus one, giving a strictly increasing run

a_{k + 1} < a_{k} < \dots < a_{1}

with every entry in

A

. Starting from the smallest value

1

gives

1, 3, 7, 15, 31, 63

, since

2 (63) + 1 = 127

exceeds

100

. These six values yield five consecutive related pairs

(63, 31), (31, 15), (15, 7), (7, 3), (3, 1)

, and no other start gives a longer run. Hence the largest

k

5

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Sets Relations and Functions

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