MCQMediumJEE 2026Relations

JEE Mathematics 2026 Question with Solution

Let A={0,1,2,...,9}A = \{0, 1, 2, ..., 9\}. Let RR be a relation on AA defined by (x,y)R(x, y) \in R if and only if xy|x - y| is a multiple of 33.

Statement I: n(R)=36n(R) = 36.

Statement II: RR is an equivalence relation.

  • A

    Both Statement I and Statement II are correct

  • B

    Both Statement I and Statement II are incorrect

  • C

    Statement I is incorrect but Statement II is correct

  • D

    Statement I is correct but Statement II is incorrect

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A={0,1,2,,9}A = \{0,1,2,\dots,9\} and (x,y)R(x,y) \in R iff xy|x-y| is a multiple of 33.

Find: Which of the two statements is correct.

The condition that xy|x-y| is a multiple of 33 means exactly that

xy(mod3)x \equiv y \pmod{3}

So the relation groups elements according to their remainders when divided by 33.

The equivalence classes are:

C0={0,3,6,9},C1={1,4,7},C2={2,5,8}C_0 = \{0,3,6,9\}, \quad C_1 = \{1,4,7\}, \quad C_2 = \{2,5,8\}

Since congruence modulo 33 is reflexive, symmetric, and transitive, RR is an equivalence relation. Therefore, Statement II is correct.

Now count the ordered pairs in RR. Every pair of elements from the same class belongs to RR.

n(R)=C02+C12+C22n(R) = |C_0|^2 + |C_1|^2 + |C_2|^2

Substituting the class sizes:

n(R)=42+32+32=16+9+9=34n(R) = 4^2 + 3^2 + 3^2 = 16 + 9 + 9 = 34

So Statement I is incorrect because it claims 3636, whereas the actual value is 3434.

Therefore, the correct option is C.

Using equivalence classes

Given: The relation is defined by xy|x-y| being divisible by 33.

Find: The truth values of Statement I and Statement II.

First rewrite the condition in modular form:

xy is a multiple of 3    xy0(mod3)    xy(mod3)|x-y| \text{ is a multiple of } 3 \iff x-y \equiv 0 \pmod{3} \iff x \equiv y \pmod{3}

Hence two elements are related exactly when they have the same remainder modulo 33.

Partition AA into residue classes modulo 33:

  • C0={0,3,6,9}C_0 = \{0,3,6,9\}
  • C1={1,4,7}C_1 = \{1,4,7\}
  • C2={2,5,8}C_2 = \{2,5,8\}

A relation defined by equality of remainders modulo 33 is an equivalence relation because:

  • Reflexive: every element has the same remainder as itself.
  • Symmetric: if xx and yy have the same remainder, then yy and xx also have the same remainder.
  • Transitive: if xx and yy have the same remainder, and yy and zz have the same remainder, then xx and zz have the same remainder. So Statement II is true.

To count n(R)n(R), count ordered pairs within each class.

  • From C0C_0: 42=164^2 = 16 pairs
  • From C1C_1: 32=93^2 = 9 pairs
  • From C2C_2: 32=93^2 = 9 pairs

Hence,

n(R)=16+9+9=34n(R) = 16 + 9 + 9 = 34

So Statement I is false.

Therefore, Statement I is incorrect but Statement II is correct, so the correct option is C.

Common mistakes

  • Counting unordered pairs instead of ordered pairs. A relation is a set of ordered pairs, so for a class of size kk, the contribution is k2k^2, not (k2)\binom{k}{2}.

  • Forgetting that pairs of the form (x,x)(x,x) are also included. Reflexive pairs belong to the relation because xx=0|x-x| = 0, and 00 is a multiple of 33.

  • Thinking that divisibility of xy|x-y| by 33 does not define an equivalence relation. It is equivalent to xy(mod3)x \equiv y \pmod{3}, which is a standard equivalence relation.

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