A) Both\[{{R}_{1}}\]and\[{{R}_{2}}\]are transitive relations
B) Both\[{{R}_{1}}\]and\[{{R}_{2}}\] are symmetric relations
C) Range of\[{{R}_{2}}\] is\[\{1,2,3,4\}\]
D) Range of\[{{R}_{1}}\] is\[\{2,4,8\}\]
Correct Answer: C
Solution :
Define two binary relations on N as \[{{R}_{1}}=\{(x,y)\in N\times N:2x+y=10\}\]and \[{{R}_{2}}=\{(x,y)\in N\times N:x+2y=10\}\] From, \[{{R}_{1}},2x+y=10\]and \[x,y\in N\] So, possible values for\[x\]and\[y\]are: \[x=1,y=8\]i.e (1, 8) \[x=2,y=6\]i.e \[(2,6)\] \[x=3,y=4\]i.e (3, 4) \[x=4,y=2\]i.e \[(4,2)\] \[{{R}_{1}}=\{(1,8),(2,6),(3,4),(4,2)\}\] Therefore, Range of\[{{R}_{1}}\]is \[\{2,4,6,8\}\] \[{{R}_{1}}\] is not symmetric Also,\[{{R}_{1}}\] is not transitive because \[(3,4),(4,2)\in {{R}_{1}}\]but\[(3,2)\cancel{\in }{{R}_{1}}\] Thus, options A, B and D are incorrect. From \[{{R}_{2}},x+2y=10\], and\[x,y\in N\] So, possible values for\[x\]and\[y\]are: \[x=8,y=1\,\,i.e\,(8,1)\] \[x=6,y=2\,\,i.e\,(6,2)\] \[x=4,y=3\,\,i.e\,(4,3)\] \[x=2,y=4\,\,i.e\,(2,4)\]\[{{R}_{2}}=\{(8,1),(6,2),(4,3),(2,4)\}\] Therefore, Range of \[{{R}_{2}}\]is \[\{1,2,3,4\}\] \[{{R}_{2}}\]is not symmetric Hence, option C is correct.
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