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Students in class 12 will have to concentrate on their studies and prepare well in advance for their upcoming boards. RD Sharma class 12th exercise 2.1 will be of great help to students. With RD Sharma class 12th mathematics solutions, you will get the best answers all in one place. RD Sharma Solution It will provide you with a clear understanding of all the concepts to ace your mathematics paper.

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Functions exercise 2.1 question1(i)

The example for a function which is one-one but not onto.

One-one function means every element in the domain has a distinct image in the co-domain

., then

Where, domain of

Onto function defined as every element in the co-domain has at least one pre image in the domain of function.

Here, we need to give an example of a function for one-one but not onto.

Let we consider the function,

, given by

Let us consider two elements and in the domain of f, so we get

and

Now we know the condition for finding a one-one function.

Hence, the function is one-one.

Now, we need to proveis not onto

Let , such that x,

We get,

If we put

, which is cannot be true as supposed in solution.

Hence, the given function is not onto.

So, is an example of one-one but not onto function.

Functions exercise 2.1 question 1 (ii)

The example for a function which is not one-one but onto.

One-one function means every element in the domain has a distinct image in the co-domain. If one-one is given for any function., then

Where, domain of

Onto function defined as every element in the co-domain has at least one pre image in the domain of function.

Solution:

Here, we need to give an example of a function for a not one-one but onto function.

Let the function, given by

Here,

Since different elements have same image 1,

is not one-one.

Let, such that

Here,y is a natural number and for every y there is a value of x which is a natural number.

Hence,f is onto.

So, the function , given by is not one-one but onto.

Functions exercise 2.1 question 1 (iii)

The example for a function which is neither one-one nor onto.One-one function means every element in the domain has a distinct image in the co-domain. If one-one is given for any function as if, then

Where, domain of

Onto function means every element in the co-domain has at least one pre image in the domain of function.

Let function , given by

Calculate and

Since, doesn’t have unique image,

Now,

Let such that

Since y is real number, then it can be negative also.

If which is not possible as the root of a negative number is not real.

Hence,x is not real, so f is not onto.

Function given byis neither one-one nor onto.

Functions exercise 2.1 question 2 (i)

is one-one and onto function.One-one function means every element in the domain has a distinct image in the co-domain.

Onto function means every element in the codomain has at least one pre image in the domain of function.

We need to find the one-one function

Injectivity: (one-one)

Every element of A has different images in B.

Function is one-one.

Surjectivity: (onto)

Co-domain of

Range of set of image

Co-domain=Range

So, is onto.

Functions exercise 2.1 question 2 (ii)

is one-one and onto function.One-one function means every element in the domain has a distinct image in the co-domain.

Onto function means every element in the codomain has at least one pre image in the domain of function.

We need to find the one-one function

Injectivity:

Every element of A has different images in B.

Function is one-one.

Surjectivity:

Co-domain of

Range of set of image

Codomain = Range

So, is onto

Functions exercise 2.1 question 2 (iii)

is not one-one and onto function.One-one function means every element in the domain has a distinct image in the co-domain.

Onto function means every element in the codomain has at least one pre image in the domain of function.

We need to find the one-one function

Injectivity: (one-one)

Here have the same image x

Also have the same image z

Therefore, is not one-one.

Functions exercise 2.1 question 3

, given byis one-one but not onto.Let x and y be any two elements in the domain (N), such that

So,f is one-one.

, defined by

Let us prove that the given function is one-one.

We have … (i)

Calculate … (ii)

Now equate equation (i) and (ii)

From (iii) equation, we can write

, because if we substitute

Hence, for any

Where as,

So,f is a one-one function.

But, for all

So, doesn’t assume values of1 and 2

is not an onto function.

Hence, we proved that, is one-one function not an onto function.

Functions exercise 2.1 question 4

is neither one-one nor onto.One-one function means every element in the domain has a distinct image in the co-domain.

Onto function means every element in the co-domain has at least one pre image in the domain of function.

We need to prove that is neither one-one nor onto.

First of all we check the function with one-one,

So,f is not one-one.

Now, we go for onto function,

Co-domain of

Range of

Codomain of Range of . (both are not same)

Hence, f is not onto

Functions exercise 2.1 question 5 (i)

One-one but not onto.given by .

One-one function means every element in the domain has a distinct image in the co-domain.

Onto function means every element in the co-domain has at least one pre image in the domain of function.

For any function to be a bijective, the given function be one-one and onto.

Let's start with the injection test.

Let x and y be any two elements in the domain(N) such that

Where x and y are in N , so we don’t get 1 value.

So,f is an injection.

Surjection test:

Let y be any element in the codomain (N) , such that for some element x in N .

which may not be in N

Example:

If which is not in N .

So,f is not a surjection.

The condition of bijection is as we know the function should be one-one and onto.

Hence for this, f is not bijection

Functions exercise 2.1 question 5 (ii)

Neither one-one nor onto.given by

One-one function means every element in the domain has a distinct image in the co-domain.

Onto function means every element in the co-domain has at least one pre image in the domain of function.

For any function to be a bijective, the given function be one-one and onto.

Let's start with the injection test.

Let x and y be any two elements in the domain (Z) such that

Therefore,f is not an injection

Surjection test:

Let y be any element in the codomain (Z) , such that for some element x in Z .

which may not be in N

Example:

If which is not in N .

So,f is not a surjection.

The condition of bijection is as we know the function should be one-one and onto.

Hence for this, f is not bijection.

Functions exercise 2.1 question 5 (iii)

Injective but not surjective.given by .

One-one function means every element in the domain has a distinct image in the co-domain.

Onto function means every element in the co-domain has at least one pre image in the domain of function.

For any function to be bijective, the given function be one-one and onto.

Let us check for the given function is injection, surjection and bijection.

Let's start with the injection test.

Let x and y be any two elements in the domain (N) such that

So,f is an injection

Surjection test:

Let y be any element in the codomain (N) , such that for some element x in N .

which may not be in N

So,f is not a surjection.

Here f is injective but not surjective, so the f is not a bijective.

The condition of bijection is as we know the function should be one-one and onto.

Hence for this, f is not bijection.

Functions exercise 2.1 question 5 (iv)

Injective but not surjective.given by

One-one function means every element in the domain has a distinct image in the co-domain.

Onto function means every element in the co-domain has at least one pre image in the domain of function.

For any function to be a bijective, the given function be one-one and onto.

Let us check for the given function is injection, surjection and bijection.

Let's start with the injection test.

Let x and y be any two elements in the domain (N) such that

So,f is an injection

Surjection test:

Let y be any element in the codomain (N) , such that for some element x in N .

which may not be in N

So,f is not a surjection.

Here f is injective but not surjective, so the f is not a bijection.

The condition of bijection is as we know the function should be one-one and onto.

Hence for this, f is not bijection.

Functions exercise 2.1 question 5 (v)

Neither an injection nor a surjection., defined by .

Injective function means every element in the domain has a distinct image in the co-domain.

Surjective

function means every element in the co-domain has at least one pre image in the domain of function.

Bijection ⇒ Function should fulfill the injection, surjection condition.

Let us check if the given function is injective, surjective, or bijective.

Let x and y be any two elements in the domain (R) such that

Therefore f is not an injection.

Surjection test:

Let y be any element in the codomain (R) , such that for some element x in R .

Therefore, f is surjective.

Since f is not an injective , but surjective. So the f is not bijective.

Functions exercise 2.1 question 5 (vi)

Neither injective nor surjective.defined by .

One-one function means every element in the domain has a distinct image in the co-domain.

Onto function means every element in the co-domain has at least one pre image in the domain of function.

Bijective ⇒ Function should fulfill the injective, surjective condition.

Let us check if the given function is an injective, surjective, bijective.

Let x,y be any two elements in domain .

Here we cannot say that x=y.

For example,

and

Then

Therefore, we have two number 2 and -3 in the domain z whose image is same as 6.

So f is not an injective.

Surjection:

Let y be any element in co-domain (R) such that for some element x in (R)

Here we can’t say .

So, f is not surjective and f is not a bijective.

Functions exercise 2.1 question 5 (vii)

Bijective.As defined by

One-one function means every element in the domain has a distinct image in the co-domain.

Onto function means every element in the co-domain has at least one pre image in the domain of function.

Bijection ⇒ Function should fulfill the injection, surjection condition.

Injection test:

Let be any two elements in domain(Z) such that

So, f is an injection.

Surjection test:

Let y be any element in the co-domain(Z) such that for some element x in Z( domain)

which is in Z.

So, Z is a surjection and f is a bijection.

Functions exercise 2.1 question 5 (viii)

Neither injective nor surjective.defined by .

One-one function means every element in the domain has a distinct image in the co-domain.

Onto function means every element in the co-domain has at least one pre image in the domain of function.

Injection test

Let x and y be any two elements in the domain (R) such that

Here x may not be equal to y because

So, f is not an injection.

Surjection test:

Range of

Co-domain of

Both are not the same.

So f is not a surjection and f is not a bijection

Functions exercise 2.1 question 5 (ix)

Bijectivedefined by .

Bijection: function should be one-one and onto function.

Injection test:

Let x and y be any two elements in the domain (R) such that

So, f is an injection.

Surjection test:

Let y be any element in the codomain (N) , such that for some element x in R .

So, f is a surjection.

So, f is a bijection.

Functions exercise 2.1 question 5 (x)

Surjective but not injective, defined by

Injective function means every element in the domain has a distinct image in the co-domain.

surjective function means every element in the co-domain has at least one pre image in the domain of function.

Injective test:

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

Here we cannot say

For example

So, -1 and 1 have the same image 0.

Thus f is not an injection.

Surjection test:

Let x and y be any two elements in the domain (R) such that for some element (R) in R.

In order to identity surjection,

Therefore, f is a surjective.

Therefore, f is not a bijective.

Functions exercise 2.1 question 5 (xi)

Neither injective nor surjective.defined by .

Injective (one-one): function means every element in the domain has a distinct image in the co-domain.

Surjective(Onto): function means every element in the co-domain has at least one pre image in the domain of function.

Injection condition:

( by formula)

So, for every x in R.

For all elements in the domain the image is 1. So,f is not an injective.

Surjection:

Range of , co-domain of f=R

Both are not the same.

So, f is not a surjection and f is not a bijection.

Functions exercise 2.1 question 5 (xii)

Injective but not surjectiveInjective (one-one): function means every element in the domain has a distinct image in the co-domain.

Surjective(Onto): function means every element in the co-domain has at least one pre image in the domain of function.

Injection:

let x,y be any two elements in the domain, such that .

So, f is an injection.

Subjection test:

Let y be any element in the co-domain Q, such that for some element x in ( (domain).

So, f is not surjective.

Therefore, f is not bijective

Functions exercise 2.1 question 5 (xiii)

Injectivedefined by

Injective (one-one): function means every element in the domain has a distinct image in the co-domain.

Surjective(Onto): function means every element in the co-domain has at least one pre image in the domain of function.

Injection: Let x,y be any two element in the domain Q,

such that .

Therefore, f is an injection.

Surjection: Let y be any two element in the co-domain Q, such that for some element x in Q.

which may not be Q.

So, f is not surjective.

Therefore, f is not bijective.

Functions exercise 2.1 question 5 (xvi)

Bijective, defined by

Bijective function should be one-one and onto function.

Let x and y be any two elements in the domain(R), such that .

So, f is an injection.

Subjection test:

Let y be any element in the co-domain(R), such that .

So, f is a surjective.

Hence, f is a bijective.

Functions exercise 2.1 question 5 (xv)

Bijective, defined by .

Bijective function should be one-one and onto function.

Let

Hence f is an injective.

Subjection test:

Let

is onto

Hence, f is bijection.

Functions exercise 2.1 question 5 (xvi)

Neither injective nor surjective.defined by

Injective (one-one): function means every element in the domain has a distinct image in the co-domain.

Surjective(Onto): function means every element in the co-domain has at least one pre image in the domain of function.

Injection condition:

Let x,y be any two elements in domain R such that

Therefore, f is not an injection.

Surjection test:

Let y be any two elements in domain (R) such that for some element x in R.

So, f is not a surjection and f is not a bijection

Functions exercise 2.1 question 5 (xvii)

Neither injective nor surjective.defined by

Injective (one-one): function means every element in the domain has a distinct image in the co-domain.

Surjective(Onto): function means every element in the co-domain has at least one pre image in the domain of function.

Let x and y be any two elements in domain (R), such that f(x) = f(y).

xy

xy

−xy(−y + x) + 1(x − y) = 0

(x − y) (1 – xy) = 0

x = y or x =

So, f is not an injection.

Let y be any element in co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

yx

x((-1) ± √(1 - 4x

= (1 ± √(1 - 4y

For example, if y = 1, then (1 ± √(1-4))/(2y) = (1 ± i√3)/2, which is not in R Therefore, f is not surjection and f is not bijection.

Functions exercise 2.1 question 6

is an injection such that the range of

For an injection there will be exactly one image for each element.

Here

⇒ Since F is an injection, there will be exactly one image for each element of f.

⇒ So, the number of images of f is one.

Number of element in A is one.

Functions exercise 2.1 question 7

**Given:**

**Hint: **

**For a bijection, we have to prove that this function is one-one and onto**

**Solution:**

**Let and**

**Consider the function**

**Here **

**For one-one function, we have **

** **

** **

** **

** **

** **

** **

**So,is a one-one function.**

**For onto function, we have**

** **

** **

** **

** **

** **

** **

**Again here**

**Substituting the value ofin**

** **

** **

** **

**is onto function.**

**Hence is bijective.**

**Note:**

**If any function is bijective, then it is invertible.**

Functions exercise 2.1 question 8 (i)

f is not bijective., given by

Here we have to show that the given function is one-one and onto.

For one-one:

Let and be any two elements in the domain A, such that

So,f is one-one.

For onto:

Let y be any element in the codomain A, such that for some element x in domain A

Then,

, which may not be in A=[-1, 1]

So f is not onto.

Hence,f is not bijective.

Functions exercise 2.1 question 8 (ii)

f is not bijective, given by

Function f is one-one, onto or bijective.

For any function to be a bijective, the given function should be one-one and onto.

First we will check whether the given function is one-one or not.

We know for one-one function

So, g is not one-one.

Let such that

Hence, the value of y is defined only if y is positive,

But y is a real number

Hence, if y is negative, there is not corresponding element of x

i.e; , there is no value of x inA

Hence, f is not onto.

So, g is not onto.

Hence g is not bijective.

Functions exercise 2.1 question 8 (iii)

f is not bijective., given by

Function h is one-one, onto or bijective.

For any function to be bijective, the given function should be one-one and onto.

Solution:

First we will check whether the given function is one-one or not.

We know for function h to be one-one

Let such that

Hence, the value of y is defined only if y is positive,

But y is a real number

Hence, if y is negative, there is not corresponding element of x

i.e; for , there is no value of x in A.

Hence, h is not onto.

Hence, h is not bijective.

Functions exercise 2.1 question 9 (i)

f is surjective.Let is a person, y is the mother of x

As for each element x in the domain set, there is a unique element y in the co-domain set.

So, f is a function.

First we check for injection or one-one.

We know: For one-one,for each element x in the domain set, there should be a unique element y in the co-domain set.

But here,y can be the mother of two or more persons.

So, f is not injective.

Again for surjective,

For every mother y defined by (x,y) , there exists a person x for whom y is mother.

So, f is surjective.

Hence, f is surjective function.

Functions exercise 2.1 question 9 (ii)

Let is person and b is an ancestor of aThe mapping is injective, surjective or not.

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

So, f is a function.

Since the ordered map (a,b) is not a map because "a" can have many ancestors. So the codomain set is not unique.

So, g is not a function.

If the function does not exist then we can’t show that the function is an injective or surjective.

Functions exercise 2.1 question 10

GivenNumber of element in A =3

We know that,

Number of one-one functionnumber of ways of arranging 3 elements.

=3! = 6

Hence, the number of one-one function as following:

Functions exercise 2.1 question 11

f is bijection.is a function defined by .

f is bijection.

Any function to be bijection. The given function should be one-one and onto.

First we will check whether the function is one-one or not.

Let x and y be any two element in the domain R such that

So, f is one-one.

Now, we will check if the given function is onto or not.

Let y be any element in the co-domain R such that for some element x in R (domain).

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

So, f is onto.

Hence f is bijection.

Functions exercise 2.1 question 12

, given by.

f is one-one but not onto.

If different element in A have distinct images in B then the function is one-one and if every element in B has at least one pre-image in A then the function is onto.

First we will check whether the function is one-one or not.

Here, let such that

is one-one

Now, we check onto

Let such that

clearly the range of but the given range isR

co-domain≠ Range

is not onto.

When co-domain is replaced by, i.e. then f become onto function.

Functions exercise 2.1 question 13

given by .

is bijection.

For any function to be bijective, the function should be one-one and onto.

Let such that

is one-one.

Now, let be arbitrary then

Thus for all , there exist such that .

is onto

is one-one and onto.

Hence f is bijective.

Functions exercise 2.1 question 14

One-one function must be onto.

Here and

We know that, since f is one-one, so the three elements of must be taken to different element of the co-domain under f.

Hence f must be onto.

Functions exercise 2.1 question 15

, given by.

Onto function must be one-one.

Let f is not one-one then there exists two elements say 1 and 2 in the domain having same image in the co-domain.

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

The range set will have at most two element of the co-domain, but we know that in the range set we have three elements .

i.e. There arises a contradiction that f is not onto function.

Hence f must be one-one.

Functions exercise 2.1 question 16

Here given that is the set.

Here we have find all onto function from the given set.

Since if f is onto then all element have a unique pre image.

Taking the set

Total number of one-one function

As example since f is onto, all elements of have unique pre-image

Element | Number of possible pairs |

1 | 3 |

2 | 2 |

3 | 1 |

So as the diagram, we get

Elements | Number of possible pairs |

1 | n |

2 | n-1 |

3 | n-2 |

. | . |

. | . |

n-1 | . |

n | 2 |

1 |

Functions exercise 2.1 question 17

Let and be two function given byWe can easily verify that and are one-one functions.

Now,

is a function given by

Since is a constant function.

Hence it is not one-one.:

Functions exercise 2.1 question 18

Then and are surjective functions.

Now,

is given by

Since is a constant function.

Hence it is not surjective.

Functions exercise 2.1 question 19

and are one-one mapping.

The product is not one-one.

Let be defined by

and be defined by

Clearly and are one-one functions.

Now, is defined by

…(i)

Clearly

since the two elements have the same image.

is not one-one.

Hence is not one-one.

Functions exercise 2.1 question 20

Let and are two function defined by and .Clearly and are one-one functions.

Now, given by

Let

defined by

Now,

since, the two elements have the same image.

is not one-one.

Hence, is not one-one.

Functions exercise 2.1 question 21 (i)

Answer:

Here given that

and

Here we have to construct an example of injective.

For injective, we have to construct the example of one-one function.

Here and

Hence for injective (i.e) one-one.

Hence the result.

Functions exercise 2.1 question 21 (ii)

Here given that

and

Here we have to construct an example as the mapping from A to B which is not injective.

In this question we construct an example of many one function.

Here and

Hence we have

It is many-one.

Functions exercise 2.1 question 21 (iii)

Answer:

and

Here we have to find a mapping from A to B.

Here and

This is the required mapping.

Functions exercise 2.1 question 22

The given function is neither one-one nor onto.

First we will check whether the given function is one-one or not.

For one-one; we have,

for all

is not one-one where as many-one.

Again, Range

is not onto.

Hence f is neither one-one nor onto

Functions exercise 2.1 question 23

Answer:

**Given:****To prove:**

Function f is bijection.**Hint:**

For any function to be a bijection, the given function should be one-one and onto.**Solution:**

Here

First we will check whether the given function is one-one or not.

For one-one,

Case I : If n is odd

Let such that

As

Case II : If n is even

Let such that

As

So, f is injective (one-one).

Again we will check whether the given function is onto or not.

For onto,

Case I : If n is odd

As for every, , there exists

Case II : If n is even

As for every , there exists

So f is surjective (onto).

Hence f is a bijection.

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The class 12 RD Sharma chapter 1 exercise 2.1 solution has the Functions chapter. It is the second chapter in the book and needs special attention from students due to its complex nature. The concepts discussed include identity function, constant function, polynomial function, signum function, etc. The exercise will explore questions from these parts and contains a total of 45 questions on two levels.

**Chapter-wise RD Sharma Class 12 Solutions**

- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable

1. How can class 12 RD Sharma chapter 1 exercise 2.1 solution be helpful for students?

The second chapter in the Mathematics book of class 12 deals with Functions that you have learned in the previous class. The class 12 RD Sharma chapter 1 exercise 2.1 solution will help you to solve these questions and provide you with a clear understanding of how you must do the calculations.It will be of great help to students in exam preparations for both school and board exams.

2. Is RD Sharma class 12 solutions Functions ex 2.1 set according to the latest syllabus?

The RD Sharma 12 solutions Functions ex 2.1 is always kept up to date with the latest syllabus in the NCERT book. With every new edition of the NCERT book, the pdf of the RD Sharma 12 solutions chapter 1 ex 2.1 is also updated so that students will find answers to all the questions in the book.

3. Can I use RD Sharma class 12th exercise 2.1 for my exam preparations?

The RD Sharma Sharma class 12 chapter 1 exercise 2.1 is a highly acclaimed NCERT solution for class 12 students. The answers and explanations provided in the book have some unique tips and tricks that will help you solve questions easier and faster than others.

4. Do I have to buy the RD Sharma class 12 chapter 1 exercise 2.1 book?

You don't need to buy the RD Sharma class 12 chapter 1 exercise 2.1 book at all. The pdf of the book is easily available online. You can download it anytime you want and save it on your electronic devices. Moreover, the pdf is available completely free of cost, so you don't have to make any financial investment.

5. How can the Class 12 RD Sharma chapter 1 exercise 2.1 solution be beneficial for me?

The Class 12 RD Sharma chapter 1 exercise 2.1 solution book will be quite helpful for students who want to score well in their exams and have started preparing in advance. It can also be used for self-testing, self-practice, and understanding your weak points. The book is trusted by many students and teachers, which again shows that it can help you with exam preparation

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