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.