JEE MAIN - Mathematics (2018 - 16th April Morning Slot - No. 13)
Let N denote the set of all natural numbers. Define two binary relations on N as R = {(x, y) $$ \in $$ N $$ \times $$ N : 2x + y = 10} and R2 = {(x, y) $$ \in $$ N $$ \times $$ N : x + 2y = 10}. Then :
Range of R1 is {2, 4, 8).
Range of R2 is {1, 2, 3, 4}.
Both R1 and R2 are symmetric relations.
Both R1 and R2 are transitive relations.
Explanation
For R1; 2x + y = 10 and x, y $$ \in $$ N 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)
$$\therefore\,\,\,$$ R1 = { (1, 8), (2, 6), (3, 4), (4, 2) }
$$\therefore\,\,\,$$ Range of R1 is {2, 4, 6, 8}
R1 is not symmetric.
R1 is not transitive also as
(3, 4), (4, 2) $$ \in $$ R , but (3, 2) $$ \notin $$ R1
For R2 : x + 2y = 10 and x, y $$ \in $$ N
Possible values of 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) and
x = 2, y = 4 i.e (2, 4)
$$\therefore\,\,\,$$ R2 = {(8, 1) (6, 2) (4, 3) (2, 4)}
$$\therefore\,\,\,$$ Range of R2 = $$\left\{ {1,2,3,4} \right\}$$
R2 is not symmetric and transitive
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)
$$\therefore\,\,\,$$ R1 = { (1, 8), (2, 6), (3, 4), (4, 2) }
$$\therefore\,\,\,$$ Range of R1 is {2, 4, 6, 8}
R1 is not symmetric.
R1 is not transitive also as
(3, 4), (4, 2) $$ \in $$ R , but (3, 2) $$ \notin $$ R1
For R2 : x + 2y = 10 and x, y $$ \in $$ N
Possible values of 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) and
x = 2, y = 4 i.e (2, 4)
$$\therefore\,\,\,$$ R2 = {(8, 1) (6, 2) (4, 3) (2, 4)}
$$\therefore\,\,\,$$ Range of R2 = $$\left\{ {1,2,3,4} \right\}$$
R2 is not symmetric and transitive
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