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

1.

f(a) = b, f(b) = a, f(c) = c, f(d) = d is a one-to-one function as each element is an image of exactly one element.

2.

f(a) = b, f(b) = a, f(c) = c, f(d) = d is not a one-to-one function as a is mapped to b and b is mapped to a.

3.

f(a) = b, f(b) = b, f(c) = d, f(d) = c is a one-to-one function as each element is mapped to only one element.

4.

f(a) = b, f(b) = b, f(c) = d, f(d) = c is not a one-to-one function as b is an image of more than one elements.

5.

f(a) = d, f(b) = b, f(c) = c, f(d) = d is a one-to-one function as each element is mapped to only one element.

6.

f(a) = d, f(b) = b, f(c) = c, f(d) = d is not a one-to-one function as d is an image of more than one element.

The correct answers are

option(1), option (4),option (5)

Step-by-step explanation:

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

(1)

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.

Therefore it is one to one.

(2)

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.

Therefore it is one to one.

(3)

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.

Therefore it is not a one to one mapping.

(4)

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.

Therefore it is not a one to one mapping.

(5)

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.

Therefore it is a one to one mapping.

(6)

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.

idk what the answer is but for the points

step-by-step explanation:

the first one and the second one that's the answer