Maths Q2 – JEE-MAIN 2025

Home Maths Relations and FunctionsQuestion #378

Let

A = {1, 2, 3, . . ., 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}), . . ., (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 :

Clue 1 of 3: Understand the Chain

The problem describes a chain of relations where the second element of a pair becomes the first element of the next. This forms a decreasing sequence of numbers.

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Step 1: Establish the Recursive Relationship✦ Active

The given relation is $R = {(a, b) : a = 2 b + 1}$ . The sequence of $k$ elements is $(a_{1}, a_{2}), (a_{2}, a_{3}), . . ., (a_{k}, a_{k + 1})$ . This implies $a_{i} = 2 a_{i + 1} + 1$ for $i = 1, 2, . . ., k$ . All $a_{j}$ must belong to the set $A = {1, 2, . . ., 100}$ .

Step 2: Derive a General Formula for

a_{1}

○ Expand

From $a_{i} = 2 a_{i + 1} + 1$ , we can add 1 to both sides: $a_{i} + 1 = 2 a_{i + 1} + 2 = 2 (a_{i + 1} + 1)$ . Let $b_{i} = a_{i} + 1$ . Then $b_{i} = 2 b_{i + 1}$ , or $b_{i + 1} = \frac{b_{i}}{2}$ . This is a geometric progression. Therefore, $b_{1} = 2^{k} b_{k + 1}$ , which means $a_{1} + 1 = 2^{k} (a_{k + 1} + 1)$ . So, $a_{1} = 2^{k} (a_{k + 1} + 1) - 1$ .

Step 3: Apply Constraints to Find Maximum

k

○ Expand

Since all $a_{j} \in A$ , we have $1 \leq a_{j} \leq 100$ . To maximize $k$ , we should minimize $a_{k + 1}$ . The smallest possible value for $a_{k + 1}$ is $1$ . Substituting $a_{k + 1} = 1$ into the formula for $a_{1}$ : $a_{1} = 2^{k} (1 + 1) - 1 = 2^{k} \cdot 2 - 1 = 2^{k + 1} - 1$ . We also know $a_{1} \leq 100$ . So, $2^{k + 1} - 1 \leq 100 ⟹ 2^{k + 1} \leq 101$ . The powers of 2 are $2^{1} = 2, 2^{2} = 4, 2^{3} = 8, 2^{4} = 16, 2^{5} = 32, 2^{6} = 64, 2^{7} = 128$ . For $2^{k + 1} \leq 101$ , the maximum value for $k + 1$ is $6$ . Thus, $k + 1 = 6 ⟹ k = 5$ . For $k = 5$ , $a_{1} = 2^{5 + 1} - 1 = 2^{6} - 1 = 63$ . This value $63 \in A$ . The sequence of $a_{i}$ values would be $63, 31, 15, 7, 3, 1$ , all of which are in $A$ . If $k = 6$ , $a_{1} = 2^{6 + 1} - 1 = 2^{7} - 1 = 127$ , which is not in $A$ . Therefore, the largest integer $k$ is $5$ .

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Maths Question 2 – JEE-MAIN 2025

Clue 1 of 3: Understand the Chain

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