## Identify the correct statements for the given functions from the set

Advanced MathIdentify the correct statements for the given functions from the set 11 months ago

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.