Identify the correct statements for the given functions from the set {a, b, c, d} to itself.

a) f(a) = b, f(b) = a, f(c) = c, f(d) = d

b) f(a) = b, f(b) = a, f(c) = c, f(d) = d

c) f(a) = b, f(b) = b, f(c) = d, f(d) = c

d) f(a) = b, f(b) = b, f(c) = d, f(d) = c

e) f(a) = d, f(b) = b, f(c) = c, f(d) = d

f) f(a) = d, f(b) = b, f(c) = c, f(d) = d

Step 1

One to one: Every image has exactly one unique pre-image in domain.

a) f(a) = b, f(b)=a, f(c)=c, f(d)=d

b has a pre-image in domain i.e a

a has a pre-image in domain i.e b

c has a pre-image in domain i.e c

d has a pre-image in domain i.e d

Here all elements have a unique pre- image in domain.

Srep 2

Therefore it is one to one.

b) f(a) = b, f(b)=a, f(c)=c, f(d)=d

b has a pre-image in domain i.e a

a has a pre-image in domain i.e b

c has a pre-image in domain i.e c

d has a pre-image in domain i.e d

Here all elements have a unique pre- image in domain.

Step 3

Therefore it is one to one.

c) f(a) = b, f(b)=b, f(c)=d, f(d)=c

b has a pre-image in domain i.e a

b has a pre-image in domain i.e b

d has a pre-image in domain i.e c

c has a pre-image in domain i.e d

b has two pre image.

Here all elements have not a unique pre- image in domain.

Step 4

Therefore it is not a one to one mapping.

d) f(a) = b, f(b)=b, f(c)=d, f(d)=c

b has a pre-image in domain i.e a

b has a pre-image in domain i.e b

d has a pre-image in domain i.e c

c has a pre-image in domain i.e d

b has two pre image.

Here all elements have not a unique pre- image in domain.

Step 5

Therefore it is not a one to one mapping.

e) f(a) = d, f(b)=b, f(c)=d, f(d)=c

d has a pre-image in domain i.e a

b has a pre-image in domain i.e b

d has a pre-image in domain i.e c

c has a pre-image in domain i.e d

Here all elements have a unique pre- image in domain.

Step 6

Therefore it is a one to one mapping.

f) f(a)=d, f(b)=b, f(c)=c, f(d)=d

d has a pre-image in domain i.e a

b has a pre-image in domain i.e b

c has a pre-image in domain i.e c

d has a pre-image in domain i.e d

Here all elements have a unique pre- image in domain.

Therefore it is a one to one mapping.

Step 1

Definition f from A to B has the property that each element of A has been assigned to exactly one element of B.

The function f is onto if and only if for every element bin B there exist an element ain A such that f(a)=b

Step 2

Solution

A={a, b, c, d}

B={a, b, c, d}

a) Given: f(a)=b, f(b)=a, f(c)=c, f(d)=d

The function f is onto, because every element of B={a, b, c, d} is the image of an element.

Step 3

b) Given: f(a)=b, f(b)=b, f(c)=d, f(d)=c

The function is not onto, because the element a is not image of any element.

Step 4

c) Given: f(a)=d, f(b)=b, f(c)=c, f(d)=d

The function is not onto, because the element a is not the image of any element.

Step 1

In order for a function to be a one-to-one function, for each y-value, there is exactly one corresponding x-value. Therefore,

a) Yes since for each y-value, there is exactly one corresponding x-value.

b) No since for the y-value b, there are two corresponding x-value: a and b.

c) No since for the y-value d, there are two corresponding x-value: a and d.