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 _ _ _ _ _ _ _ _ _ _ .

Step-by-step solution

Given, A = 1 , 2 , 3 , … , 10    B = 0 , 1 , 2 , 4 ( a , b ) ∈ A × A such that 2 a - b 2 + 3 a - b - k = 0 where k ∈ 0 , 1 , 2 , 3 , 4 Now finding discriminant D = 9 - 4 × 2 - k And 9 - 4 × 2 - k a perfect square for any possible a , b as a - b will be a integer, So, 9 + 8 k is a perfect square ⇒ k = 0 or k = 2 Now for k = 0 , 2 a - b 2 + 3 a - b = 0 ⇒ a - b 2 a - b 2 + 3 = 0 ⇒ a - b = 0 ⇒ ( a , b ) ∈ ( 1 , 1 ) , ( 2 , 2 ) … . ( 10 , 10