**(i) One-one but not onto**

Consider A = {1, 2, 3} and B = {a, b, c, d}

So we get f = {(1, a), (2, b), (3, c}

**(ii) One-one and onto**

We know that f(x) = 2x

**Infectivity:**

Consider x_{1}, x_{2} ∈ R where f(x1) = f(x_{2})

So we get

2x_{1} = 2x_{2}

x_{1} = x_{2}

**Hence, f: R → R is one-one**

**Subjectivity:**

Consider y be any real number in R which is the co-domain

f(x) = y

We get

2x = y

It can be written as

x = y/2

We know that y/2 ∈ R for y ∈ R where

f(y/2) = 2(y/2) = y

For y ∈ R(co-domain) there exists x = y/2 ∈ R (domain) where f(x) = y

Here, each element in co-domain has pre-image in domain

**Thus, f: R → R is bijective.**

**(iii) Neither one-one nor onto**

Consider A = {1, 2, 3} and B = {4, 5, 6}

We get

f = {(1, 4), (2, 4), (3, 5)}

**(iv) Onto but not one-one**

Consider A = {1, 2, 3} and B = {4, 5, 6}

We get

f = {(1, 2), (3, 2), (5, 4)}