If A=\[\{x|x\in \,N\,\]and \[({{x}^{2}}-4)\]\[({{x}^{2}}-5)\]=0} and B=\[\{x|x\in {{l}^{+}}\]and \[x(x-1)\]\[(x-2)\]=0} then \[(A\cup B)\]-\[(A\cap B)\]is

If A=\[\{x|x\in \,N\,\]and \[({{x}^{2}}-4)\]\[({{x}^{2}}-5)\]=0} and B

1.	If A=\[\{x\|x\in \,N\,\]and \[({{x}^{2}}-4)\]\[({{x}^{2}}-5)\]=0} and B=\[\{x\|x\in {{l}^{+}}\]and \[x(x-1)\]\[(x-2)\]=0} then \[(A\cup B)\]-\[(A\cap B)\]is
A.	{1, 2}
B.	{1}
C.	{2}
D.	\[\phi \]
Answer» C. {2}

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If A=\[\{x|x\in \,N\,\]and \[({{x}^{2}}-4)\]\[({{x}^{2}}-5)\]=0} and B=\[\{x|x\in {{l}^{+}}\]and \[x(x-1)\]\[(x-2)\]=0} then \[(A\cup B)\]-\[(A\cap B)\]is

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