Question

Let $Z$ denote the set of all integers where
$A=\{(a,b):a^2+3b^2=28,a,b∈Z\}$ and
$B=\{(a,b):a>b,a,b∈Z\}$, then the number of elements in $Aâˆ\copyright B$ is

• A. $2$
• B. $6$
• C. $4$
• D. $5$

Answer 1

Answer: (B)

$A=\{(a,b):a^2+3b^2=28,a,b∈Z\}$
$(a,b)$ can be $(1,3)$, $(−1,3)$, $(1,−3)$, $(−1,−3)$, $(5,1)$
$(−5,1)$, $(5,−1)$, $(−5,−1)$, $(4,2)$, $(−4,2)$, $(4,−2)$, $(−4,−2)$
$n(A)=12$, $n(B)=∞$
$Aâˆ\copyright B=\{(a,b):a^2+3b^2=28$ and $a>b,a,b∈Z\}=6$