Question

If Z denotes the set of all integers and $A = \left\{ {(a,b):{a^2} + 3{b^2} = 28,a,b \in Z} \right\}$ and $B = \left\{ {(a,b):a < b,a,b \in Z} \right\}$ , then the number of elements in $A \cap B$ is
a.$2$
b.$3$
c.$4$
d.$5$
e.$6$

Answer 1

Hint: In this question we have been given the set of integers such that $A = \left\{ {(a,b):{a^2} + 3{b^2} = 28,a,b \in Z} \right\}$ and $B = \left\{ {(a,b):a < b,a,b \in Z} \right\}$ . In this question we will first put the values for “b” to find the value of set A. and then we can solve this question by counting the number of sets.

Complete step by step answer:
Here we have been given the question that $A = \left\{ {(a,b):{a^2} + 3{b^2} = 28,a,b \in Z} \right\}$ .
Let us put the value of $b = 1$ in the equation and we have:
$ \Rightarrow {a^2} + 3{(1)^2} = 28$
On simplifying we have:
$ \Rightarrow {a^2} = 28 - 3$
$ \Rightarrow a = \sqrt {25} $
There it gives us value $a = \pm 5$
Now we take the second value i.e. $b = 2$ in the equation and we have:
$ \Rightarrow {a^2} + 3{(2)^2} = 28$
On simplifying we have:
$ \Rightarrow {a^2} = 28 - 12$
$ \Rightarrow a = \sqrt {16} $
There it gives us value $a = \pm 4$
Let us put the third value i.e. $b = 3$ in the equation and we have:
$ \Rightarrow {a^2} + 3{(3)^2} = 28$
On simplifying we have:
$ \Rightarrow {a^2} = 28 - 27$
$ \Rightarrow a = \sqrt 1 $
There it gives us value $a = \pm 1$ .
Similarly if we repeat the exact same method by putting the values of “a” we can find the value of set B.
It gives us $A = \left\{ {(5,1),( - 5, - 1),(5, - 1),( - 5,1),(4,2),( - 4, - 2),( - 4,2),(4, - 2),(1,3),( - 1, - 3),( - 1,3),(1, - 3)} \right\}$
We have been given the condition that $B = \left\{ {(a,b):a < b,a,b \in Z} \right\}$
Therefore it gives us:
$A \cap B = \left\{ {(1,3),( - 1,3),( - 4, - 2),( - 4,2),( - 5, - 1),( - 5,1)} \right\}$
We can see that the total number of elements is six.

Hence the correct option is (e) $6$.

Note:
We should note that if we take the value of $a = 1$ in the equation it gives us the value:
${(1)^2} + 3{b^2} = 28$
On simplifying this it gives us;
$ \Rightarrow 3{b^2} = 28 - 1$
$ \Rightarrow {b^2} = \dfrac{{27}}{3}$
It gives us value:
$b = \sqrt 9 $
We have $ \Rightarrow b = \pm 3$