aRb <=> 1+ab >0
a. is reflexive because a*a for any real number except 0 will be positive hence >0, and if a=0 then a*a + 1 >0.
b. if a*b + 1 > 0 then b*a + 1 will also be > 0, hence symmetric.
c. a=-2 b=0, -2*0 + 1 >0, ab+1 > 0
b=0, and if c=4 then 0*4 + 1 > 0
but a is not related to c, because a=-2, c=4, and -2*4 + 1 < 0
Hence, the given relation is reflexive and symmetric but not transitive.
(b) is correct option.
