MCQMediumJEE 2026Relations

JEE Mathematics 2026 Question with Solution

Let A={2,3,5,7,9}A = \{2, 3, 5, 7, 9\}. Let RR be the relation on AA defined by xRyxRy if and only if 2x3y2x \le 3y. Let ll be the number of elements in RR, and mm be the minimum number of elements required to be added in RR to make it a symmetric relation. Then l+ml + m is equal to :

  • A

    2323

  • B

    2121

  • C

    2525

  • D

    2727

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A={2,3,5,7,9}A = \{2, 3, 5, 7, 9\} and RR is defined by xRy    2x3yxRy \iff 2x \le 3y.

Find: The value of l+ml + m, where ll is the number of elements in RR and mm is the minimum number of elements to be added to make RR symmetric.

A relation is symmetric if whenever (x,y)R(x,y) \in R, then (y,x)R(y,x) \in R must also hold.

We first count all ordered pairs in RR.

For x=2x=2:

2(2)=43y    y432(2) = 4 \le 3y \implies y \ge \frac{4}{3}

Possible yy are {2,3,5,7,9}\{2,3,5,7,9\}, so there are 55 pairs.

For x=3x=3:

2(3)=63y    y22(3) = 6 \le 3y \implies y \ge 2

Possible yy are {2,3,5,7,9}\{2,3,5,7,9\}, so there are 55 pairs.

For x=5x=5:

2(5)=103y    y1032(5) = 10 \le 3y \implies y \ge \frac{10}{3}

Possible yy are {5,7,9}\{5,7,9\}, so there are 33 pairs.

For x=7x=7:

2(7)=143y    y1432(7) = 14 \le 3y \implies y \ge \frac{14}{3}

Possible yy are {5,7,9}\{5,7,9\}, so there are 33 pairs.

For x=9x=9:

2(9)=183y    y62(9) = 18 \le 3y \implies y \ge 6

Possible yy are {7,9}\{7,9\}, so there are 22 pairs.

Therefore,

l=5+5+3+3+2=18l = 5 + 5 + 3 + 3 + 2 = 18

Now check which pairs need their reverse ordered pair to make the relation symmetric.

Pairs with $$x

Counting and symmetry check

Given: R={(x,y)A×A:2x3y}R = \{(x,y) \in A \times A : 2x \le 3y\} on A={2,3,5,7,9}A = \{2,3,5,7,9\}.

Find: First compute l=Rl = |R|, then compute the minimum additional pairs mm needed for symmetry.

To count ll, fix each value of xx and determine all allowed values of yy.

  1. For x=2x=2, the inequality becomes 43y4 \le 3y, so every element of AA works.
  2. For x=3x=3, the inequality becomes 63y6 \le 3y, so every element of AA works.
  3. For x=5x=5, the inequality becomes 103y10 \le 3y, so only 5,7,95,7,9 work.
  4. For x=7x=7, the inequality becomes 143y14 \le 3y, so only 5,7,95,7,9 work.
  5. For x=9x=9, the inequality becomes 183y18 \le 3y, so only 7,97,9 work.

Thus,

l=5+5+3+3+2=18l = 5 + 5 + 3 + 3 + 2 = 18

For symmetry, diagonal pairs such as (2,2),(3,3),(5,5),(7,7),(9,9)(2,2), (3,3), (5,5), (7,7), (9,9) do not create any issue because their reverse is the same pair.

So we only inspect non-diagonal pairs. Every time (x,y)R(x,y) \in R but (y,x)R(y,x) \notin R, one new pair must be added.

The missing reverse pairs are exactly:

  • from (2,5)(2,5), add (5,2)(5,2)
  • from (2,7)(2,7), add (7,2)(7,2)
  • from (2,9)(2,9), add (9,2)(9,2)
  • from (3,5)(3,5), add (5,3)(5,3)
  • from (3,7)(3,7), add (7,3)(7,3)
  • from (3,9)(3,9), add (9,3)(9,3)
  • from (5,9)(5,9), add (9,5)(9,5)

Therefore,

m=7m = 7

And hence,

l+m=18+7=25l+m = 18+7 = 25

Therefore, the value of l+ml+m is 2525.

Common mistakes

  • Counting only pairs with x<yx<y while finding ll is incorrect because RR is a relation on A×AA \times A and includes ordered pairs with x=yx=y and also pairs with x>yx>y whenever 2x3y2x \le 3y holds. Count all valid ordered pairs row-wise for each fixed xx.

  • Assuming every pair needs a distinct reverse pair, including diagonal pairs, is wrong. For symmetry, diagonal pairs such as (a,a)(a,a) already satisfy their own reflection condition. Only non-diagonal pairs need checking.

  • Using the inequality in the reverse direction, such as checking 2y3x2y \le 3x incorrectly or forgetting to test (y,x)(y,x) separately, leads to wrong values of mm. For each pair (x,y)(x,y) in RR, explicitly test whether (y,x)(y,x) also satisfies 2y3x2y \le 3x.

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