(i) Since, every element in B has pre-image in A, so f1 is onto function.
(ii) Since, element in B has pre-image of A. So f2 is onto function.
(iii) Since 3 ∈ B does not have pre-image A. So f3 is not onto function.
(iv) Since, 6 ∈ B does not have pre-image of A. So f4 is not onto function.