Question

Let Z be the set of all integers and A = {(a, b) : a² + 3b² = 28, a, b ∈ Z}
B = {(a, b) : a > b, a, b ∈ Z}. Then, the number of elements in
A ⋂ B, is
(a) 2 (b) 4 (c) 6 (d) 5

Answer 1

Answer:

let Z be the set of all integers and a!= {(a, b) :a² + 3b² = 28, a, b € Z}

answer 6

Answer 2

Answer :

Given -

• A = {(a, b) : a² + 3b² = 28, a, b ∈ Z}
• B =  {(a, b) : a > b, a, b ∈ Z}

To Find -

• A ⋂ B

Solution -

Equation : a² + 3b² = 28.

Putting the values in the given equation to find the values of a and b.

Substituting b = 3,

a² + 3(3)² = 28

a² + 3×9 = 28

a² = 28 - 27

a = ±1

Substituting b = 2,

a² + 3(2)² = 28

a² + 3×4 = 28

a² + 12 = 28

a² = 28 - 12

a = ±4

Substituting b,

a² + 3(1)² = 28

a² + 3×1 = 28

a² + 3 = 28

a² = 28 - 3

a = ±5

Similarly, we can find other such values.

At last :

A = {(1,3), (1,-3), (-1,-3), (5,1), (-1,3), (-5,1), (5,-1), (-5,-1), (-4,-2), (4,-2), (-4,2), (4,2)}

Now,

B = {(a, b) : a > b, a, b ∈ Z}.

So,

B = {(1,-3), (-1,-3), (5,1), (5,-1), (4,-2), (4,2)}.

Hence,

A ⋂ B = {(1,-3), (-1,-3), (5,1), (5,-1), (4,-2), (4,2)}.

Hence, number of elements is 6.

So, (c) 6.