Question

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

• A. $2$
• B. $4$
• C. $6$
• D. $5$

Answer 1

The correct option is C $6$

$\because A=\{(a,b):a^2+3b^2=28,a,b\in Z\}$
$=\{(5,1),(-5,-1),(5,-1),(-5,1),(4,2),(-4,-2),(4,-2),(-4,2),(1,3),(-1,-3),(1,-3),(-1,3)\}$
and $B=\{(a,b):a<b,a,b\in Z\}$
$\therefore A\cap B=\{(1,3),(-1,3),(-4,-2),(-4,2),(-5,-1),(-5,1)\}$
$\therefore$ The number of elements in $A\cap B$ is $6$.