Question 1: Give an example of a function

(i) which is one-one but not onto

(ii) which is not one-one but onto

(iii) which is neither one-one nor onto

Answer:

**(i) **

**Check for Injectivity:**

**Check for Surjectivity:**

**Check for Injectivity: **

So, from definition

**Check for Surjectivity: **

(iii) which is neither one-one nor onto

Question 2: Which of the following functions from A to B are one-one and onto?

Answer:

(i) Now, As given,

Thus we can see that,

**Check for Injectivity: **

Every element of has a different image from

Hence is a One – One function

**Check for Surjectivity: **

Also, each element of is an image of some element of

Hence is Onto.

(ii) Now, As given,

Thus we can see that

**Check for Injectivity: **

Every element of has a different image from

Hence is a One – One function

**Check for Surjectivity: **

Also, each element of is an image of some element of

Hence is Onto.

(iii) Now, As given,

Thus we can clearly see that

**Check for Injectivity: **

Every element of does not have different image from

Therefore is not One – One function

**Check for Surjectivity: **

Also each element of is not image of any element of

Hence is not Onto.

Answer:

**Check for Injectivity: **

So, from definition

**Check for Surjectivity: **

Answer:

**Check for Injectivity: **

**Check for Surjectivity: **

Question 5: Classify the following functions as injection, surjection or bijection:

Answer:

**Injection condition: **

**Surjection condition: **

**Injection condition: **

**Surjection test: **

**Injection condition:**

**Surjection condition: **

**Injection condition:**

**Surjection condition: **

**Injection condition:**

**Surjection test: **

**Injection test: **

**Surjection test: **

**Injection test: **

**Surjection test: **

**Injection test: **

**Surjection test: **

Both are not same.

**Injection test: **

**Surjection test: **

**Injection test: **

**Surjection test: **

**Injection condition: **

Therefore, for all elements in the domain, the image is 1.

**Surjection condition: **

Both are not same.

**Injection test: **

**Surjection test: **

**Injection test: **

**Surjection test: **

**Injection test: **

**Surjection test: **

**Injection test: **

**Surjection test: **

**Injection condition: **

**Surjection test: **

**Injection condition: **

**Surjection test: **

Question 6: If is an injection such that range of Determine the number of elements in .

Answer:

Given that is an injection

And also given the range of

Therefore, the number of images of

Since, is an injection, there will be exactly one image for each element of

Therefore, number of elements in

Answer:

Let us show that, given function is one-one and on-to

**Injectivity: **

**Surjectivity: **

Therefore, for every element in the co-domain, there exists some pre-image in the domain.

Answer:

Let us show that the given function is one-one and on-to

**Injection test: **

**Surjection test: **

We have to show that the given function is one-one and on-to

**Injection test: **

**Surjection test: **

Let us show that the given function is one-one and on-to

**Injection test: **

**Surjection test: **

Question 9: Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective:

Answer:

(i) Let

Since, for each element in domain set, there is a unique related element y in co-domain set.

Therefore, is the function.

**Injection test: **

As, can be mother of two or more persons

Therefore, is not injective.

**Surjection test: **

For every mother defined by there exists a person for whom is mother. So, is surjective.

Therefore, is surjective function.

(ii) Let

As, the ordered map does not map person to a living person.

Therefore, is not a function.

Question 10: Let Write all one-one from to itself.

Answer:

Given as

The number of elements in

The number of one-one functions = number of ways of arranging 3 elements

Answer:

**Injectivity: **

**Surjectivity: **

Therefore, for every element in the co-domain, there exists some pre-image in the domain. is onto.

Since, is both one-to-one and onto, it is a bijection.

Question 12: Show that the exponential function given by is one-one but not onto. What happens if the co-domain is replaced by (set of all positive real numbers).

Answer:

Therefore is one-one function.

We know that range of

Both are not same.

Therefore is not onto function.

If the co-domain is replaced by , then the co-domain and range become the same and in that case, is onto and hence, it is a bijection.

Question 13: Show that the logarithmic function given by is a bijection.

Answer:

Therefore is one-one function.

Let

So for every element in the co-domain there exists some pre-image in the domain.

Therefore is onto.

Therefore f is one-one and onto, therefore it is bijection.

Question 14: . If show that a one-one function must be onto.

Answer:

A function is called an onto function if the range of is

In other words, if each there exists at least one such that then is an on-to function.

Here,

Here,

Here, as given function is one-one and for each value of there exists a value in given function is onto function.

Question 15: If show that an onto function must be one-one.

Answer:

Suppose is not one-one.

Then, there exists two elements, say 1 and 2 in the domain whose image in the co-domain is same.

Also, the image of 3 under f can be only one element.

Therefore, the range set can have at most two elements of the co-domain

i.e is not an onto function, a contradiction.

Hence, must be one-one.

Question 16: Find the number of all onto functions from the set to itself.

Answer:

Taking set

Since f is onto, all elements of have unique pre-image.

Total number of one-one function

Since f is onto, all elements of have unique pre-image.

Total number of one-one function

Answer:

Since is a constant function, it is not one-one.

Answer:

Since is a constant function, it is not surjective.

Question 19: Show that if and are one-one maps from to then the product need not be one-one.

Answer:

Question 20: Suppose and are non-zero one-one functions from to

Answer:

Question 21: Given Construct an example of each of the following:

(i) an injective map from to

(ii) a mapping from to which is not injective

(iii) a mapping from to

Answer:

Given

Question 22: Show that is neither one-one nor onto.

Answer:

Now, check for Injection:

Now, check for Surjection:

Question 23: Let be defined by

Show that is a bijection.

Answer:

We have,

Injection Test:

Case 1: If is odd

Case 2: If is even

Hence, is invective.

Surjection Test:

Hence, is surjective.

Therefore is a bijection.