Let A={-3,-2,-1,0,1,2,3}. Let R be a relation on A defined by x R y if and only if 0 ≤ x²+2 y ≤ 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

Solution

Maths sir · 4 steps

Fix each x, find which y keep x²+2y between 0 and 4, then check the diagonal pairs.

1. Turn it into y. For a fixed x, the condition 0 ≤ x² + 2y ≤ 4 becomes -x²/2 ≤ y ≤ (4 - x²)/2. Since x² is the same for x and -x, both get the same list of y values.

   x	Range for y	Allowed y in A	Is (x, x) in R?
   0	0 ≤ y ≤ 2	0, 1, 2 (3 pairs)	yes
   1	-0.5 ≤ y ≤ 1.5	0, 1 (2 pairs)	yes
   -1	-0.5 ≤ y ≤ 1.5	0, 1 (2 pairs)	no
   2	-2 ≤ y ≤ 0	-2, -1, 0 (3 pairs)	no
   -2	-2 ≤ y ≤ 0	-2, -1, 0 (3 pairs)	yes
   3	-4.5 ≤ y ≤ -2.5	-3 (1 pair)	no
   -3	-4.5 ≤ y ≤ -2.5	-3 (1 pair)	yes

2. Count l. Add the pairs: l = 3 + 2 + 2 + 3 + 3 + 1 + 1 = 15.
3. Count m. A reflexive relation needs all 7 pairs (x, x). R has only 4 of them, so (-1,-1), (2,2) and (3,3) must be added. That gives m = 3.
4. So. l + m = 15 + 3 = 18, which is option (a).

The idea to keep. To make a relation reflexive, you only add the missing (a, a) pairs. So m is the size of the set minus the number of diagonal pairs already in the relation.

Answer from the NTA final answer key. Solution by PrepWiser.