Question 1: State with reasons whether following functions have inverse:

Answer:

Therefore, does not have an inverse.

Therefore, does not have an inverse.

Here, different elements of the domain have different images in the co-domain.

Also, each element in the co-domain has a pre-image in the domain.

So,

Answer:

Here, the different elements of the domain have different images in the co-domain.

Thus, this is one-one.

Here, the different elements of the domain have different images in the co-domain.

Thus,

This is not onto because the element 0 in the co-domain (B) has no pre-image in the domain (A)

Answer:

Clearly, and are bijections.

Therefore, and are invertible.

Clearly, is a bijection.

Therefore, is invertible.

Answer:

So, and are bijections and,

Question 5: Show that the function defined by is invertible. Also, find

Answer:

Let us show that the given function is invertible.

Injection of

Let and be any two elements of the domain

Therefore, is one-one.

Surjection of

Let be in the co-domain

Therefore, is onto.

Therefore, is a bijection and, hence, it is invertible.

Let us find

Question 6: Consider given by . Show that is invertible. Find the inverse of

Answer:

Let us show that the given function is invertible.

Consider injection of

Let and be any two elements of domain

Therefore, is one-one.

Surjection of

Let be in the co-domain

Therefore, is a bijection and, hence, is invertible.

Question 7: Consider given by . Show that is invertible with inverse of f given by , where is the set of all non-negative real numbers.

Answer:

Let us show that is invertible,

Consider injection of

Let and be any two elements of the domain

Therefore, is one-one

Now surjection of

Let be in the co-domain

Therefore, is a bijection and, hence, it is invertible.

Answer:

So, the given function is invertible and the inverse of is itself.

Answer:

is onto.

Since function is one-one and onto, so it is invertible.

Answer:

Let us prove that exists

Injectivity of

Let and be two elements in domain

Therefore, is one-one.

Surjectivity of

Let be in the co-domain

Therefore is onto.

Therefore, is a bijection and, hence, it is invertible.

Find

Question 11: A function is defined as . Is it a bijection or not? In case it is a bijection, find

Answer:

Injectivity of

Let and be two elements of domain

Therefore, is one-one.

Surjectivity of

Let be in the co-domain

Therefore, is onto.

Therefore, is a bijection and, hence, is invertible.

Answer:

From ( 2 ),

Now,

From (4)

From (3) and (5)

Answer:

Given;

Hence is onto function.

Thus is one=one onto function.

Answer:

First we shall show that is one-one.

Therefore is both one-one and onto

Answer:

Answer:

For one-one

Answer:

Injectivity of

Let and be two elements of domain (R), such that

Hence, f is one-one.

Surjectivity of f:

Let y be in the co-domain such that f(x) = y.

Hence, is onto

Finding

Answer:

Injectivity of f:

Let x and y be two elements of domain (R), such that

f(x) = f(y)

Hence, f is one-one.

Surjectivity of f:

Let y be in the co-domain (0, 2) such that f(x) = y.

Hence, is onto

Therefore is bijection and hence, it is invertible.

Finding

Answer:

Injectivity test:

Therefore is an injection.

Surjectivity test:

Therefore is a surjection.

Since, is an injection and surjection, then it is bijection.

Hence, is a invertible.

From (1)

Answer:

Answer:

Injectivity:

Let and be two elements in the domain (R), such that

Hence, we cannot say that

Therefore is not one-one.

Therefore is not a bijection.

Thus, is not invertible.

Answer:

(a)

(b)

(c)

(d)

Question 23: Let A and B be two sets each with finite number of elements. Assume that there is an injective map from A to B and that there is an injective map from B to A. Prove that there is a bijection from A to B.

Answer:

Since, there is an injective mapping from A to B, each element of A has unique image in B.

Similarly, there is also an injective mapping from B to A, each element of B has unique image in A or in other words there is one to one onto mapping from A to B.

Thus, there is bijective mapping from A to B.

(i) is an injection

(ii) is a surjection.

Answer:

(i) is an injection

Injectivity of

Let and be two elements of the domain (A), such that

( As, is one-one)

( As is one-one)

Hence, is an injection.

(ii) is a surjection.

Surjectivity of

Let be an element in the co-domain of

Now, (co-domain of ) and is a surjection.

Now (co-domain of ) and is a surjection.

Hence, is a surjection.