NVAMediumJEE 2025Relations

JEE Mathematics 2025 Question with Solution

Let A={1,2,3}A = \{1,2,3\}. The number of relations on AA, containing (1,2)(1,2) and (2,3)(2,3), which are reflexive and transitive but not symmetric, is _____.

Answer

Correct answer:7

Step-by-step solution

Standard Method

Given: A={1,2,3}A = \{1,2,3\}. The relation on AA contains (1,2)(1,2) and (2,3)(2,3), and it must be reflexive, transitive, but not symmetric.

Find: The number of such relations.

A reflexive relation on AA must contain

(1,1),(2,2),(3,3)(1,1), (2,2), (3,3)

Since (1,2)(1,2) and (2,3)(2,3) are given, transitivity forces

(1,3)(1,3)

because

(1,2) and (2,3)    (1,3)(1,2) \text{ and } (2,3) \implies (1,3)

So the minimal relation is

R0={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}R_0 = \{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}

Now the only possible additional reverse pairs are

(2,1),(3,2),(3,1)(2,1), (3,2), (3,1)

We check which subsets of these can be added while preserving transitivity.

Valid transitive extensions are:

  1. No additional pair
  2. +(2,1)+(2,1)
  3. +(3,2)+(3,2)
  4. +(3,1)+(3,1)
  5. +(2,1),(3,1)+(2,1),(3,1)
  6. +(3,2),(3,1)+(3,2),(3,1)
  7. +(2,1),(3,2),(3,1)+(2,1),(3,2),(3,1)

The pair combination (2,1)(2,1) and (3,2)(3,2) together forces (3,1)(3,1) by transitivity, so that case is already included only in the seventh relation.

Each of these 77 relations is not symmetric, because at least one of (1,2),(2,3),(1,3)(1,2), (2,3), (1,3) does not have its reverse pair present in full.

Therefore, the number of required relations is 77.

Casewise Counting

Given: A={1,2,3}A = \{1,2,3\} and the relation contains (1,2)(1,2) and (2,3)(2,3).

Find: Count all relations that are reflexive, transitive, and not symmetric.

Reflexive means all diagonal pairs must be present:

(1,1),(2,2),(3,3)(1,1), (2,2), (3,3)

Transitivity with the given pairs gives:

(1,2) and (2,3)    (1,3)(1,2) \text{ and } (2,3) \implies (1,3)

Hence every valid relation must contain

R0={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}R_0 = \{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}

Now consider the optional reverse pairs:

(2,1),(3,2),(3,1)(2,1), (3,2), (3,1)

Check all transitive possibilities:

  • Add none: valid.
  • Add only (2,1)(2,1): valid.
  • Add only (3,2)(3,2): valid.
  • Add only (3,1)(3,1): valid.
  • Add (2,1)(2,1) and (3,1)(3,1): valid.
  • Add (3,2)(3,2) and (3,1)(3,1): valid.
  • Add (2,1)(2,1) and (3,2)(3,2): then transitivity requires (3,1)(3,1), so this is not a separate case.
  • Add all three: valid.

Thus the total number of valid relations is

77

Also, none of these 77 relations is symmetric, because symmetry would require all reverse pairs of the non-diagonal elements, which does not happen here.

Therefore, the answer is 77.

Common mistakes

  • Including only the reflexive pairs and the given pairs, but forgetting that transitivity from (1,2)(1,2) and (2,3)(2,3) forces (1,3)(1,3). This makes the relation incomplete. Always add every pair required by transitivity before counting cases.

  • Treating (2,1)(2,1) and (3,2)(3,2) as independently optional without checking transitivity. If both are included, then (3,2)(3,2) and (2,1)(2,1) force (3,1)(3,1). So those two cannot be counted as a separate valid case unless (3,1)(3,1) is also present.

  • Misreading 'not symmetric' as meaning that no reverse pair can be present at all. That is incorrect. 'Not symmetric' only means at least one pair lacks its reverse. Some reverse pairs may still be included, provided the whole relation does not become symmetric.

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