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:
17
18
(Correct Answer)19
20
18
The relation condition is given by the inequality:
0≤x2+2y≤4
We can rewrite this inequality to easily check values for 2y:
−x2≤2y≤4−x2
Since x,y∈A, we will test each possible value of x from the set A={−3,−2,−1,0,1,2,3} to find the valid values for y.
Case 1: When x=0⟹x2=0
−0≤2y≤4−0⟹0≤2y≤4⟹0≤y≤2
Valid values for y∈A are {0,1,2}.
Pairs: (0,0),(0,1),(0,2) ⟹ 3 pairs
Case 2: When x=1 or x=−1⟹x2=1
−1≤2y≤4−1⟹−1≤2y≤3⟹−0.5≤y≤1.5
Valid values for y∈A are {0,1}.
Pairs for x=1: (1,0),(1,1) ⟹ 2 pairs
Pairs for x=−1: (−1,0),(−1,1) ⟹ 2 pairs
Case 3: When x=2 or x=−2⟹x2=4
−4≤2y≤4−4⟹−4≤2y≤0⟹−2≤y≤0
Valid values for y∈A are {−2,−1,0}.
Pairs for x=2: (2,−2),(2,−1),(2,0) ⟹ 3 pairs
Pairs for x=−2: (−2,−2),(−2,−1),(−2,0) ⟹ 3 pairs
Case 4: When x=3 or x=−3⟹x2=9
−9≤2y≤4−9⟹−9≤2y≤−5⟹−4.5≤y≤−2.5
The only integer value in this range is y=−3.
Valid values for y∈A are {−3}.
Pairs for x=3: (3,−3) ⟹ 1 pair
Pairs for x=−3: (−3,−3) ⟹ 1 pair
l=3+2+2+3+3+1+1=15
For a relation on set A to be reflexive, it must contain the pair (x,x) for every x∈A.
The required reflexive pairs for set A are:
{(−3,−3),(−2,−2),(−1,−1),(0,0),(1,1),(2,2),(3,3)}
Let's check which of these reflexive pairs are already present in our relation R:
From Case 4: (−3,−3) and (3,−3) are present ⟹ (−3,−3) is Present
From Case 3: (−2,−2) and (2,−2) are present ⟹ (−2,−2) is Present
From Case 2: (−1,1) and (1,1) are present ⟹ (1,1) is Present
From Case 1: (0,0) is present ⟹ (0,0) is Present
The pairs that are missing from R are:
{(−1,−1),(2,2),(3,3)}
Thus, the minimum number of elements required to be added (m) is:
m=3
l+m=15+3=18
(b) 18
The relation condition is given by the inequality:
0≤x2+2y≤4
We can rewrite this inequality to easily check values for 2y:
−x2≤2y≤4−x2
Since x,y∈A, we will test each possible value of x from the set A={−3,−2,−1,0,1,2,3} to find the valid values for y.
Case 1: When x=0⟹x2=0
−0≤2y≤4−0⟹0≤2y≤4⟹0≤y≤2
Valid values for y∈A are {0,1,2}.
Pairs: (0,0),(0,1),(0,2) ⟹ 3 pairs
Case 2: When x=1 or x=−1⟹x2=1
−1≤2y≤4−1⟹−1≤2y≤3⟹−0.5≤y≤1.5
Valid values for y∈A are {0,1}.
Pairs for x=1: (1,0),(1,1) ⟹ 2 pairs
Pairs for x=−1: (−1,0),(−1,1) ⟹ 2 pairs
Case 3: When x=2 or x=−2⟹x2=4
−4≤2y≤4−4⟹−4≤2y≤0⟹−2≤y≤0
Valid values for y∈A are {−2,−1,0}.
Pairs for x=2: (2,−2),(2,−1),(2,0) ⟹ 3 pairs
Pairs for x=−2: (−2,−2),(−2,−1),(−2,0) ⟹ 3 pairs
Case 4: When x=3 or x=−3⟹x2=9
−9≤2y≤4−9⟹−9≤2y≤−5⟹−4.5≤y≤−2.5
The only integer value in this range is y=−3.
Valid values for y∈A are {−3}.
Pairs for x=3: (3,−3) ⟹ 1 pair
Pairs for x=−3: (−3,−3) ⟹ 1 pair
l=3+2+2+3+3+1+1=15
For a relation on set A to be reflexive, it must contain the pair (x,x) for every x∈A.
The required reflexive pairs for set A are:
{(−3,−3),(−2,−2),(−1,−1),(0,0),(1,1),(2,2),(3,3)}
Let's check which of these reflexive pairs are already present in our relation R:
From Case 4: (−3,−3) and (3,−3) are present ⟹ (−3,−3) is Present
From Case 3: (−2,−2) and (2,−2) are present ⟹ (−2,−2) is Present
From Case 2: (−1,1) and (1,1) are present ⟹ (1,1) is Present
From Case 1: (0,0) is present ⟹ (0,0) is Present
The pairs that are missing from R are:
{(−1,−1),(2,2),(3,3)}
Thus, the minimum number of elements required to be added (m) is:
m=3
l+m=15+3=18
(b) 18
