Question 1: Find whether the following functions are one-one or not:

Answer:

Hence, is a one-one function from to itself.

Here, does not provide the unique solution but it provides Hence, is not a one-one function.

Question 2: Show that the function defined by for all is a many-one function.

Answer:

Question 3: Discuss the surjectivity of the following functions:

Answer:

(i) Let be an arbitrary element of . Then,

Hence, negative real numbers in R(co-domain) do not have their pre-images in R(domain). Hence, f is not an onto function.

Hence, negative real numbers in (co-domain) do not have their pre-images in (domain). Hence, is not an onto function.

(iii) Let y be an arbitrary element of (co-domain). Then,

Thus, does not have its pre-image in Z(domain). Hence, is not an onto function.

Question 4: Show that the function and for every , is onto but not one-one.

Answer:

It is given that

Let be an arbitrary element in (co-domain). Then,

Question 5: Show that the Signum function , given by

is neither one-one nor onto.

Answer:

Clearly, all positive real numbers have the same image equal to 1.

Hence, is a many-one function.

We observe that the range of which is not equal to the co-domain of .

Hence, is not onto.

Hence, is neither one-one nor onto

Answer:

We observe the following properties of .

Injectivity: Let be two arbitrary elements in . Then,

Hence, is an injective map.

Surjectivity: Let be an arbitrary element of . Then,

Thus, for all (co-domain) there exists (domain).

That is every element in the co-domain has its pre-image in .

Hence, is a surjection.

Answer:

We observe the following properties of

Injectivity: Let be any two elements of (domain). Then,

Hence, is an injective map.

Surjectivity: Let be an arbitrary element of (co-domain). Then,

This shows that every element in the co-domain has its pre-image in the domain. Hence, is a surjection.

Hence, is a bijection.

Question 8: Let . Show that given by is a bijection.

Answer:

We observe the following properties of .

Injectivity: Let be any two elements in . Then,

Hence, is an injective map.

Surjectivity: We have

If takes all values between 0 and 1 including these two points.

Also, if takes all values between -1 and 0 including -1. Therefore, takes every value between – 1 and 1 including – 1 and 1. Hence, range of is same as its co-domain.

Hence, it is a bijection.

Question 9: Let A be the set of all 50 students of class XII in a central school. Let be a function defined by Roll number of student . Show that is one-one but not onto.

Answer:

Here, associates each students to his (her) roll number. Since no two different students of the class can have the same roll number.

Therefore, is one-one.

We observe that i.e. range of is not same as its co-domain. Hence, is not onto.

Question 10: Prove that is one-one and onto

Answer:

We observe the following properties of .

Hence, is one-one.

Surjectivity: Let be any real number in (co-domain). Then,

Thus, for each (co-domain) has its pre-image in domain.

This means that each element in co-domain has its pre-image in domain.

Hence, is onto.

Hence, is a bijection.

Question 11: Show that the function , is neither one-one nor onto.

Answer:

We observe that i.e. there are two distinct elements in which have the same image. Hence, f is not one-one.

Since assumes only non-negative values. Hence, no negative real number in (co-domain) has its pre-image in domain of . Consequently is not onto.

Question 12: Show that , is a bijection.

Answer:

We observe the following properties of .

Injectivity: Let such that Then,

Hence, is one-one.

Surjectivity: Let (co-domain). Then,

Clearly, (domain) for all (co-domain).

Thus, for each (co-domain) there exists (domain) such that

Hence, is onto.

Hence, is a bijection.

Question 13: Show that the function is one-one onto, where is the set of all non-zero real numbers. ls the result true, if the domain is replaced by with co-domain being same as ?

Answer:

We observe the following properties of .

Injectivity: Let such that ). Then,

Hence, is one-one

Surjectivity: Let be an arbitrary element of (co-domain) such that . Then,

Hence, is onto.

Hence, is one-one onto.

For any , we find that

Hence, is one-one.

Question 14: Prove that the greatest integer function , given by , is neither one-one nor, where denotes the greatest integer less than or equal to .

Answer:

Also, does not attain non-integral values. Therefore, non-integer points in (co-domain) do not have their pre-images in the domain. Hence, is not onto.

Hence, is neither one-one nor onto.

Question 15: show that the modulus function is neither one-one nor onto.

Answer:

We observe that . Hence, f is not one-one.

Also, assumes only non-negative values. Hence, negative real numbers in (co-domain) do not have their pre-images in (domain).

Hence, is neither one-one nor onto

Question 16: Let and denote the set of all complex numbers and all real numbers respectively. Then show that given by for all is neither one-one nor onto.

Answer:

Injectivity: We find that and are two distinct complex numbers in such that

This shows that different elements in may have the same image. Hence, is not an injection.

Surjectivity: is not a surjection, because negative real numbers in do not have their pre-images in . In other words, for every negative real number a there is no complex number such that

Hence, is not a surjection.

Question 17: Show that the function is a bijection. **[CBSE 2010]**

Answer:

Injectivity: Let be any two real numbers. Then,

Hence, is an injection.

Surjectivity: Let be an arbitrary element of (co-domain). Then,

This shows that every element in co-domain has its pre-image in domain. Hence, f is a surjection.

Hence, is a bijection.

Question 18: Show that the function , is neither one-one nor onto.

Answer:

Injectivity: We know that

hence, different elements in may have the same image. Hence, is not an injection.

Surjectivity: Since the values of lie between and , it follows that the range of is not equal to its co-domain. Hence, f is not a surjection.

Answer:

Injectivity: Let be any two elements of . Then,

Hence, is an injective.

Surjectivity: Let be an arbitrary element of . Then,

Hence, is a surjective.

Hence, is a bijective.

Question 20: Let and be two sets. Show that defined by is a bijection.

Answer:

Injectivity: Let and such that

Thus,

Hence, is an injective map.

Surjectivity: Let be an arbitrary element of . Then,

Thus, for all, there exists such that

Hence, is an onto function.

Hence, is a bijection.

Question 21: Let be any non-empty set. The, prove that the identity function on set is a bijection.

Answer:

The identity function is defined as

Injectivity: Let be any two elements of . Then,

Hence, is an injective map.

Surjectivity: Let . Then, there exists such that

Hence, is a surjective map.

Hence, is a bijection.

Question 22: Let . Find all one-to-one functions from to .

Answer:

Let be a one-one function.

Then, has two choices, namely, 1 or 2.

Case 1: When

As is one-one. Therefore,

Thus, we have

Case 2: When

Since is one-one. Therefore,

Thus, in this case, we have

Hence, there are two one-one functions say from given by

Question 23: Consider the identity function defined as, for all . Show that although is onto but defined as

Answer:

We know that the identity function on a given set is always a bijection. Therefore, is onto. We have,

This means that under images of natural numbers are even natural numbers.

Hence, odd natural numbers in (co-domain) do not have their pre-images in domain .

For example,1,3,5 etc. do not have their pre-images. Hence, is not onto.

Answer:

Hence, is not one-one.

Question 25: Let be a function. Define a relation on given by .

Show that is an equivalence relation on **[CBSE 2010]**

Answer:

We observe the following properties of relation R.

Reflexivity: For any , we have

Symmetry: Let be such that Then,

Hence, is symmetric.

Transitivity: Let be such that Then,

Hence, is transitive.

Hence, is an equivalence relation.

Answer:

We have,

Hence, is an injective map.

Surjectivity: When , we have

Let be any real number. Then,

Hence, is an onto function from

Hence, is an injective map.

Surjectivity: When , we have

Let be an arbitrary real number such that . Then,

Hence, is an onto function from

Hence, is a one-one onto function.

Question 27: Show that the function is a bijection.

Answer:

Injectivity: Let such that

Hence, is an injective map.

Surjective: Let be an arbitrary element of . Then,

We know w that an odd degree equation has at least one real root. Therefore, for every real value of the equation has a real root such that

Thus, for every there exists such that . Hence, is a surjective map.

Hence, is a bijection.

Question 28: Show that defined by

is many-one onto function. **[CBSE 2009]**

Answer:

We observe that

Thus . Hence, f is a many-one function.

Surjectivity: Let be an arbitrary element of .

If n is an odd natural number, then is also an odd natural number such that

If n is an even natural number, then is also an even natural number such that

Thus, for every (whether even or odd) there exists its pre-image in . Hence, f is a surjection.

Hence, is a many-one onto function.

Answer:

lnjectivity: Let be any two even natural numbers. Then,

If are any two odd natural numbers. Then,

Thus in both the cases,

Hence, is an injective map.

Surjectivity: Let he an arbitrary natural number.

If is an odd natural number, then there exists an even natural number such that

If is an even natural number, then there exists an odd natural number such that

Thus, every has its pre-image in Hence, is a surjection.

Hence, is a bijection.

Show that is a bijection.

Answer:

is an injection : Let

If and are even, then

If and are odd, then

Thus, in both case, we have

If is odd and is even then is even and is odd. Therefore,

Similarly, if is even and is odd, then

Hence, is an injection.

is a surjection: Let be an arbitrary element of

If is an odd natural number, there exist an even natural number (domain) such that

If is an even natural number, then there exists an odd natural number (domain) such that

Also,

Thus, every element of (co-domain) has its pre-image in (domain). Hence, is an onto function.