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

Solution

Maths sir · 3 steps

Build R row by row for each x, then spot the missing diagonal pairs and the missing reverse pairs.

1. Rewrite the rule. xRy means y ≤ 2 - 2x, so for each x keep every y in A below that bound. A diagonal pair (x, x) is in R only when 3x ≤ 2, that is x ≤ 0. A reverse pair (y, x) is in R only when 2y + x ≤ 2.

   x	Allowed y	Pairs	(x,x) in R?	Pairs whose reverse is missing
   -2	y ≤ 6: all of A	7	yes	y = 3, 4: 2
   -1	y ≤ 4: all of A	7	yes	y = 2, 3, 4: 3
   0	y ≤ 2: -2 to 2	5	yes	y = 2: 1
   1	y ≤ 0: -2, -1, 0	3	no	none
   2	y ≤ -2: only -2	1	no	none
   3	y ≤ -4: none	0	no	none
   4	y ≤ -6: none	0	no	none

2. Count everything. The Pairs column gives l = 7 + 7 + 5 + 3 + 1 = 23. The diagonal pairs (1,1), (2,2), (3,3), (4,4) are missing, so m = 4. The last column gives n = 2 + 3 + 1 = 6, from the pairs (3,-2), (4,-2), (2,-1), (3,-1), (4,-1), (2,0).
3. So. l + m + n = 23 + 4 + 6 = 33, which is option (c).

The idea to keep. For counting questions on a relation, build one table with a row for each x. The same table then gives the size of R, the missing (x,x) pairs and the missing reverse pairs.

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