MCQMediumJEE 2025Relations

JEE Mathematics 2025 Question with Solution

Let A={3,2,1,0,1,2,3}A = \{-3,-2,-1,0,1,2,3\}. Let RR be a relation on AA defined by xRyxRy if and only if 0x2+2y40 \le x^2 + 2y \le 4. 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 reflexive relation. then l+ml + m is equal to

  • A

    1919

  • B

    2020

  • C

    1717

  • D

    1818

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: A={3,2,1,0,1,2,3}A = \{-3,-2,-1,0,1,2,3\} and RR is defined by 0x2+2y40 \le x^2 + 2y \le 4.

Find: ll = number of ordered pairs in RR, mm = minimum number of ordered pairs to be added to make RR reflexive, and then compute l+ml+m.

For (x,y)R(x,y) \in R,

0x2+2y40 \le x^2 + 2y \le 4

so

x22y4x2-x^2 \le 2y \le 4-x^2

and hence

x22y4x22-\frac{x^2}{2} \le y \le \frac{4-x^2}{2}

Now check each value of xAx \in A:

  • For x=3x=-3 or x=3x=3, x2=9x^2=9, so 4.5y2.5-4.5 \le y \le -2.5. Hence only y=3y=-3 is possible. This gives 22 ordered pairs.
  • For x=2x=-2 or x=2x=2, x2=4x^2=4, so 2y0-2 \le y \le 0. Hence y=2,1,0y=-2,-1,0. This gives 66 ordered pairs.
  • For x=1x=-1 or x=1x=1, x2=1x^2=1, so 0.5y1.5-0.5 \le y \le 1.5. Hence y=0,1y=0,1. This gives 44 ordered pairs.
  • For x=0x=0, x2=0x^2=0, so 0y20 \le y \le 2. Hence y=0,1,2y=0,1,2. This gives 33 ordered pairs.

Therefore,

l=2+6+4+3=15l = 2+6+4+3 = 15

To make RR reflexive, every (a,a)(a,a) for aAa \in A must belong to RR. So test

0a2+2a40 \le a^2 + 2a \le 4

for each aAa \in A.

Checking diagonal pairs:

  • a=3a=-3: (3)2+2(3)=3(-3)^2+2(-3)=3, so (3,3)R(-3,-3) \in R
  • a=2a=-2: (2)2+2(2)=0(-2)^2+2(-2)=0, so (2,2)R(-2,-2) \in R
  • a=1a=-1: (1)2+2(1)=1(-1)^2+2(-1)=-1, so (1,1)R(-1,-1) \notin R
  • a=0a=0: 02+2(0)=00^2+2(0)=0, so (0,0)R(0,0) \in R
  • a=1a=1: 12+2(1)=31^2+2(1)=3, so (1,1)R(1,1) \in R
  • a=2a=2: 22+2(2)=82^2+2(2)=8, so (2,2)R(2,2) \notin R
  • a=3a=3: 32+2(3)=153^2+2(3)=15, so (3,3)R(3,3) \notin R

Thus the missing self-pairs are (1,1)(-1,-1), (2,2)(2,2), and (3,3)(3,3). Hence

m=3m = 3

Finally,

l+m=15+3=18l+m = 15+3 = 18

Therefore, the correct option is D.

Count by fixing y

Given: 0x2+2y40 \le x^2 + 2y \le 4 with x,yA={3,2,1,0,1,2,3}x,y \in A = \{-3,-2,-1,0,1,2,3\}.

Find: l+ml+m.

Rewrite the condition as

2yx242y-2y \le x^2 \le 4-2y

Now fix each value of yy and count the possible values of xx.

  • For y=3y=-3: 6x2106 \le x^2 \le 10, so x=±3x=\pm 3
  • For y=2y=-2: 4x284 \le x^2 \le 8, so x=±2x=\pm 2
  • For y=1y=-1: 2x262 \le x^2 \le 6, so x=±2x=\pm 2
  • For y=0y=0: 0x240 \le x^2 \le 4, so x=2,1,0,1,2x=-2,-1,0,1,2
  • For y=1y=1: 2x22-2 \le x^2 \le 2, so x=1,0,1x=-1,0,1
  • For y=2y=2: 4x20-4 \le x^2 \le 0, so x=0x=0
  • For y=3y=3: 6x22-6 \le x^2 \le -2, so no such xx

Hence the total number of ordered pairs is

l=2+2+2+5+3+1=15l = 2+2+2+5+3+1 = 15

For reflexivity, all diagonal pairs (a,a)(a,a) must be present. From the condition 0a2+2a40 \le a^2+2a \le 4, the missing pairs are (1,1)(-1,-1), (2,2)(2,2), and (3,3)(3,3). Therefore

m=3m=3

So,

l+m=15+3=18l+m=15+3=18

Thus the correct option is D.

Common mistakes

  • Counting ordered pairs as unordered pairs. Here (2,1)(-2,-1) and (1,2)(-1,-2) are different, because a relation is a set of ordered pairs. Always count each valid (x,y)(x,y) separately.

  • Assuming reflexive means only checking some convenient values. A relation on AA is reflexive only if every (a,a)(a,a) for all aAa \in A is present. Check all seven elements of AA.

  • Making an algebra mistake while converting 0x2+2y40 \le x^2+2y \le 4 into bounds for yy. Divide the whole inequality correctly by 22 to get x22y4x22-\frac{x^2}{2} \le y \le \frac{4-x^2}{2}.

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