Correct Answer – A
We have, (a, b) `in R_(1)` iff |a-b| `le 7`, where `a, b in z`
Reflexivity Let `a in z`
implies a – a = 0
`implies |a – a| le 7`
`implies 0 le 7`
`implies (a, a) in R_(1)`
`therefore` The relation `R_(1)` is reflexive.
Symmetry
`(a, b) in R_(1)`
`implies |a-b|le7implies|-(b-a)|le7`
`implies |b-a|le7implies(b, a)in R_(1)`
`therefore` The relation `R_(1)` is symmetric.
Transitivity We have (2, 6), (6, 10) `in R_(1)` because
`|2-6|=4le7 and |6-10|=4le7`
Also, |2-10| = `8cancelle7`
`therefore (2, 10) cancelinR_(1)`
Hence, the relation `R_(1)` is not transitive.

Correct Answer – A
We have, (a, b) `in R_(3)` iff `a^(2) – 4ab + 3b^(2) = 0`
where a, b `in R`
Reflexivity
`therefore a^(2)-4a.a+3d^(2)=4a^(2)-4a^(2)=0`
`therefore (a, a) in R_(3)`
`therefore` The relation `R_(3)` is reflexive.
Symmetry
`(a, b) in R_(3)`
`implies a^(2)-4ab+3b^(2)=0`, we get a = b and a = 3b
and `(b, a) in R_(3)`
implies `b^(2) – 4ab + 3a^(2) = 0`
we get b = a and b = 3a
`therefore (a, b) in R_(3) cancelimplies(b, a)in R_(3)`
`therefore` The relation `R_(3)` is not symmetric.
Transitivity We have `(3, 1), (1, (1)/(3))inR_(3)`
because `(3)^(2)-4(3)(1)+3(1)^(2)=9-12+3=0`
and `(1)^(2)-4(1)((1)/(3))+3((1)/(3))^(2)=1-(4)/(3)+(1)/(3)=0`
Also, `(3, (1)/(3))cancelinR_(3)`, because
`(3)^(2)-4.(3)((1)/(3))+3((1)/(3))^(2)=9-4+(1)/(3)=(16)/(3)ne0`
`therefore` The relation `R_(3)` is not transitive.