JEE Main 2018 — Sets And Relations Question with Solution

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