A) 2
B) 3
C) 4
D) 5
E) 6
Correct Answer: E
Solution :
\[\because \] \[A=\{(a,b):{{a}^{2}}+{{b}^{2}}=28,a,b\in Z\}\] \[=\{(5,1),(-5,-1)(5,-1),(-5,1)(1,3),\] \[(-1,-3),(-1,3)(1,-3),(4,2)(-4,-2)\] \[(4,-2),(-4,2)\}\] 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.You need to login to perform this action.
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