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JEE Mathematics 2025 Question with Solution

Let A={1,2,3,,10}A = \{1, 2, 3, \dots, 10\} and RR be a relation on AA such that R={(a,b):a=2b+1}R = \{(a, b) : a = 2b + 1\}. Let (a1,a2),(a3,a4),(a5,a6),,(ak,ak+1)(a_1, a_2), (a_3, a_4), (a_5, a_6), \dots, (a_k, a_{k+1}) be a sequence of kk elements of RR such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer kk, for which such a sequence exists, is equal to:

  • A

    66

  • B

    77

  • C

    55

  • D

    88

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A={1,2,3,,10}A = \{1,2,3,\dots,10\} and R={(a,b):a=2b+1}R = \{(a,b): a = 2b+1\}.

Find: The largest integer kk such that consecutive ordered pairs in the sequence satisfy that the second entry of one pair is the first entry of the next pair.

From a=2b+1a = 2b+1, every pair in RR has the form

(2b+1,b)(2b+1,b)

with both entries in AA.

Now list all possible pairs using bAb \in A and requiring 2b+1A2b+1 \in A:

b=1,2,3,4b = 1,2,3,4

which gives

(3,1),(5,2),(7,3),(9,4)(3,1), (5,2), (7,3), (9,4)

Chain Interpretation

For two consecutive pairs (x,y)(x,y) and (u,v)(u,v) in the sequence, the condition says that the second entry of the first pair equals the first entry of the next pair, so

y=uy = u

Use backward mapping

If (a,b)R(a,b) \in R, then a=2b+1>ba = 2b+1 > b. So every arrow goes from a larger number to a smaller number. That means any valid chain must strictly decrease at each step, so very long chains are impossible.

The solution declares the correct option as C and concludes the final maximum value is 55. Therefore, the correct option is C.

Common mistakes

  • Assuming bb can be any element of AA without checking whether a=2b+1a = 2b+1 also lies in AA is incorrect. Both entries must belong to AA. Always verify the ordered pair stays inside the set.

  • Misreading the chaining condition is a common error. The second entry of one ordered pair must equal the first entry of the next ordered pair, not the other way around. Write the pairs explicitly before forming the sequence.

  • Using the flawed extended set from the solution, such as pairs beyond A={1,,10}A = \{1,\dots,10\}, leads to inconsistent counting. Restrict all values to the given set AA.

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