Question

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______.

Answer 1

Correct Answer: 18

Given A = {1, 2, 3, . . . , 10} and B = {0, 1, 2, 3, 4}, the relation R is defined as:

R = {(a, b) ∈ A × A : 2(a − b)² + 3(a − b) ∈ B}. 

Let a − b = k. Then 2k² + 3k ∈ B. Since a, b ∈ A, a − b can take integer values from -9 to 9.

We need to find the number of pairs (a, b) such that 2(a − b)² + 3(a − b) ∈ {0, 1, 2, 3, 4}.

Solution:Click to view the solution

Let a − b = k. Then we have 2k² + 3k ∈ B. This means 0 ≤ 2k² + 3k ≤ 4. Since a, b ∈ {1, 2, ..., 10}, we have −9 ≤ k ≤ 9.

Case 1: If k = 0, then 2(0)² + 3(0) = 0 ∈ B. This corresponds to a = b. Since there are 10 possible values for a, there are 10 such ordered pairs.

Case 2: If k = −2, then 2(−2)² + 3(−2) = 8 − 6 = 2 ∈ B. This corresponds to a = b − 2. Since 1 ≤ a ≤ 10 and 1 ≤ b ≤ 10, we have 1 ≤ b − 2 ≤ 10. Therefore, 3 ≤ b ≤ 10, which gives 8 possible values for b. Thus, there are 8 such ordered pairs.

Case 3: If k = −1, then 2(−1)² + 3(−1) = 2 − 3 = −1, which is not in B.

Case 4: If k = 1, then 2(1)² + 3(1) = 2 + 3 = 5, which is not in B.

Therefore, the only valid cases are a = b and a = b − 2. When a = b, there are 10 pairs: (1,1), (2,2), ..., (10,10). When a = b − 2, there are 8 pairs: (1,3), (2,4), (3,5), ..., (8,10).

Final Answer: The total number of elements in R is 10 + 8 = 18.