Let A={1,2,3,4, … ., 10} and B={0,1,2,3,4}. The number of elements in the relation R={(a, b) ∈ A × A: 2(a-b)²+3(a-b) ∈ B} is ___________.

Solution

Maths sir · 4 steps

Put d = a - b, find which integer d makes the expression land in B, then count the pairs.

1. One variable. Let d = a - b. Since a, b ∈ {1, …, 10}, d is an integer from -9 to 9. We need 2d² + 3d ∈ {0, 1, 2, 3, 4}.

   d	2d² + 3d	In B?
   0	0	yes
   1	5	no
   -1	-1	no
   -2	2	yes
   2	14	no
   -3	9	no

2. Larger values fail. For d ≥ 2 or d ≤ -3 the value 2d² + 3d is already above 4 and keeps growing. So only d = 0 and d = -2 work.
3. Count pairs. d = 0 means a = b, which gives 10 pairs. d = -2 means b = a + 2, and a can be 1 to 8, which gives 8 pairs.
4. So. The relation has 10 + 8 = 18 elements.

The common slip: 8. Forgetting the 10 pairs with a = b (where the expression is 0) leaves only 8.

The idea to keep. When a condition depends only on a - b, set d = a - b, solve for the allowed d, then count how many pairs inside the set give each d.

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