Question

Let Z denote 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. $3$
• C. $4$
• D. $5$
• E. $6$

Answer 1

Answer: (E)

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