A) \[0\]
B) \[2\]
C) \[7\]
D) \[14\]
Correct Answer: C
Solution :
Given, \[G=\{(b,b),\,(b,\,c),\,(c,c),(c,d)\}\] and \[H=\{(b,a),\,(c,b),(d,c)\}\] Now, \[(G\cup H)\,\,\oplus {{(G\cup H)}^{-1}}\] \[=(G\cup H)\oplus (G'\cap H')\] \[[\because \,\,{{(G\cup H)}^{-1}}=G'\cap H']\] \[=\{(G\cup H)-(G'\cap H')\}\cup \{(G'\cap H')-(G\cup H)\}\]\[=\{(G\cup H)\cap (G'\cap H')'\}\] \[\cup \{(G'\cap H')\cap (G\cup H)'\}\] \[[\because \,\,A-B=A\cap B']\] \[=\{(G\cup H)\cap (G\cup H)\}\] \[\cup \{(G'\cap H')\cap (G'\cap H')\}\] \[\Rightarrow \] \[(G\cup H)\cup (G'\cap H')\] \[\Rightarrow \] \[(G\cup H)\cup (G\cup H)'=G\cup H\] \[G\cup H=\{(b,b),(b,c),(c,c),(c,d),(b,a),(c,b),\] \[(d,c)\}\] \[n(G\cup H)=7\]
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