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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 :

JEE · Math · previous-year question

  1. A.Range of R1 is {2, 4, 8).
  2. B.Range of R2 is {1, 2, 3, 4}.correct
  3. C.Both R1 and R2 are symmetric relations.
  4. D.Both R1 and R2 are transitive relations.

Answer

B. Range of R2 is {1, 2, 3, 4}.

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

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