(i) Given f₁ = {(1, 3), (2, 5), (3, 7)}; A = {1, 2, 3}, B = {3, 5, 7}

Injectivity:

f₁(1) = 3

f₁(2) = 5

f₁(3) = 7

⇒ Every element of A has different images in B.

So, f₁ is one-one.

Surjectivity:

Co-domain of f₁ = {3, 5, 7}

Range of f₁ = set of images  =  {3, 5, 7}

⇒ Co-domain = range

So, f₁ is onto.

(ii) Given f₂ = {(2, a), (3, b), (4, c)}; A = {2, 3, 4}, B = {a, b, c}

f₂ = {(2, a), (3, b), (4, c)}; A = {2, 3, 4}, B = {a, b, c}

Injectivity:

f₂(2) = a

f₂(3) = b

f₂(4) = c

⇒ Every element of A has different images in B.

So, f₂ is one-one.

Surjectivity:

Co-domain of f₂ = {a, b, c}

Range of f₂ = set of images = {a, b, c}

⇒ Co-domain = range

So, f₂ is onto.

(iii) Given f₃ = {(a, x), (b, x), (c, z), (d, z)} ; A = {a, b, c, d,}, B = {x, y, z}

Injectivity:

f₃(a) = x

f₃(b) = x

f₃(c) = z

f₃(d) = z

⇒ a and b have the same image x.

Also c and d have the same image z

Therefore, f₃ is not one-one.

Surjectivity:

Co-domain of f₁ = {x, y, z} 

Range of f₁ = set of images = {x, z}

So, the co-domain  is not same as the range.

Hence, f₃ is not onto.