Question

Let Z denote the set of all integers and $ A=\{(a,b):{{a}^{2}}+3{{b}^{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,\text{ }b):a>b,\text{ }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.