Question

If $Z$ denotes the set of all integers and $A=\left\{\left(a,b\right):a^2+3b^2=28,a,b\in Z\right\}$ and $B=\left\{\left(a,b\right):a>b,a,b\in Z\right\}$. Then the number of elements in $A$ intersection $B$ is

• A. $2$
• B. $3$
• C. $4$
• D. $5$
• E. $6$

Answer 1

The correct option is E

$6$

Explanation for the correct option.

Find the number of elements in $A\cap B$.

The set $A$ is defined as: $A=\left\{\left(a,b\right):a^2+3b^2=28,a,b\in Z\right\}$.

So the elements in set $A$ are all integer pair $\left(a,b\right)$ satisfying $a^2+3b^2=28$.

$A=\left\{\left(1,3\right),\left(1,-3\right),\left(4,2\right),\left(4,-2\right),\left(5,1\right),\left(5,-1\right),\left(-1,3\right),\left(-1,-3\right),\left(-4,2\right),\left(-4,-2\right),\left(-5,1\right),\left(-5,-1\right)\right\}$

Now set $B$ is defined as: $B=\left\{\left(a,b\right):a>b,a,b\in Z\right\}$.

So $A\cap B$ will have all elements of $A$ in which the first element of the ordered pair is greater than the second element.

So, $A\cap B=\left\{\left(1,-3\right),\left(4,2\right),\left(4,-2\right),\left(5,1\right),\left(5,-1\right),\left(-1,-3\right)\right\}$

So the number of elements in $A$ intersection $B$ is $6$.

Hence, the correct option is E.