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Maths Question 1 – JEE-MAIN 2025

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Let A={−3,−2,−1,0,1,2,3}. Let R be a relation on A defined by xRy if and only if 0≤x2+2y≤4. Let l be the number of elements in R and m be the minimum number of elements required to be added in R to make it a reflexive relation. Then l+m is equal to

First, determine the elements of the relation R based on the given condition. Then, recall the definition of a reflexive relation.

Step 1: Determine the elements of relation R✦ Active

The set is A={−3,−2,−1,0,1,2,3}. The relation R is defined by xRy if 0≤x2+2y≤4. We iterate through all possible pairs (x,y)∈A×A:

x=±3(x2=9):0≤9+2y≤4⟹−4.5≤y≤−2.5⟹y=−3. Pairs: (−3,−3),(3,−3) (2 elements)
x=±2(x2=4):0≤4+2y≤4⟹−2≤y≤0⟹y∈{−2,−1,0}. Pairs: (−2,−2),(−2,−1),(−2,0),(2,−2),(2,−1),(2,0) (6 elements)
x=±1(x2=1):0≤1+2y≤4⟹−0.5≤y≤1.5⟹y∈{0,1}. Pairs: (−1,0),(−1,1),(1,0),(1,1) (4 elements)
x=0(x2=0):0≤2y≤4⟹0≤y≤2⟹y∈{0,1,2}. Pairs: (0,0),(0,1),(0,2) (3 elements)

The total number of elements in R is l=2+6+4+3=15.

Step 2: Determine elements required for reflexivity○ Expand

For R to be reflexive, all pairs (a,a) for a∈A must be in R. The diagonal elements are: (−3,−3),(−2,−2),(−1,−1),(0,0),(1,1),(2,2),(3,3). We check which of these are *not* in R by substituting y=x into the condition 0≤x2+2x≤4:

(−3,−3):(−3)2+2(−3)=9−6=3.0≤3≤4. (In R)
(−2,−2):(−2)2+2(−2)=4−4=0.0≤0≤4. (In R)
(−1,−1):(−1)2+2(−1)=1−2=−1.0≤−1≤4 is false (since −1<0). (Not in R)
(0,0):(0)2+2(0)=0.0≤0≤4. (In R)
(1,1):(1)2+2(1)=1+2=3.0≤3≤4. (In R)
(2,2):(2)2+2(2)=4+4=8.0≤8≤4 is false (since 8>4). (Not in R)
(3,3):(3)2+2(3)=9+6=15.0≤15≤4 is false (since 15>4). (Not in R)

The pairs that must be added to R to make it reflexive are (−1,−1), (2,2), and (3,3). The minimum number of elements to add is m=3.

Step 3: Calculate l+m○ Expand

Using the values found for l and m:

l+m=15+3=18
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