Question

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.

Answer 1

Answer:

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.

Answer 2

f(a) = b, f(b) = a, f(c) = c, f(d) = d and f(a) = d, f(b) = b, f(c) = c, f(d) = d are one-to-one function and f(a) = b, f(b) = b, f(c) = d, f(d) = c is not one-to-one function. Then the correct statements are 1, 4, and 5.

What is a function?

The function is an expression, rule, or law that defines the relationship between one variable to another variable. Functions are ubiquitous in mathematics and are essential for formulating physical relationships.

One-to-one - every image has exactly one unique pre-image in the domain.

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

b has a pre image in a domian that is a

a has a pre image in a domian that is b

c has a pre image in a domian that is c

d has a pre image in a domian that is d

Here all elements have a unique pre-image in a 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 a domian that is a

a has a pre image in a domian that is b

c has a pre image in a domian that is c

d has a pre image in a domian that is d

Here all elements have a unique pre-image in a 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 a domian that is a

b has a pre image in a domian that is b

d has a pre image in a domian that is c

c has a pre image in a domian that is d

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

Therefore it is not one-to-one.

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

b has a pre image in a domian that is a

b has a pre image in a domian that is b

d has a pre image in a domian that is c

c has a pre image in a domian that is d

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

Therefore it is not one-to-one.

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

d has a pre image in a domian that is a

b has a pre image in a domian that is b

c has a pre image in a domian that is c

d has a pre image in a domian that is d

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

Therefore it is one-to-one.

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

d has a pre image in a domian that is a

b has a pre image in a domian that is b

c has a pre image in a domian that is c

d has a pre image in a domian that is d

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

Therefore it is one-to-one.

More about the function link is given below.

https://brainly.com/question/5245372