NCERT Solutions for Exercise 1.2 Class 12 Maths Chapter 1 - Relations and Functions

NCERT Solutions for Exercise 1.2 Class 12 Maths Chapter 1 - Relations and Functions

Edited By Ramraj Saini | Updated on Dec 03, 2023 01:18 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.2

NCERT Solutions for Exercise 1.2 Class 12 Maths Chapter 1 Relations and Functions are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT Solutions for Class 12 Maths chapter 1 exercise 1.2 is the most important exercise of chapter Relations and Functions and it includes topics like types of relations, functions, binary operations etc. Exercise 1.2 Class 12 Maths exposes students to questions like proving one to one functions etc. Such questions are generally asked in Board examinations. Solving NCERT syllabus for Class 12 Maths chapter 1 exercise 1.2 is recommended to students to score well in CBSE Class 12 board exam. The contribution of chapter Relations and Functions is high in competitive exams also like JEE main and NEET. There are many questions asked in previous years which are based on concepts of Class 12 Maths chapter 1 exercise 1.2.

12th class Maths exercise 1.2 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all NCERT exercise together using the link provided below.

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NCERT Solutions For Class 12 Maths Chapter 1 Relations And Functions: Exercise 1.2

Given, $f: R_* \longrightarrow R_{*}$ is defined by $f(x) = \frac{1}{x}$ .

One - One :

$f(x)=f(y)$

$\frac{1}{x}=\frac{1}{y}$

$x=y$

$\therefore$ f is one-one.

Onto:

We have $y \in R_*$ , then there exists $x=\frac{1}{y} \in R_*$ ( Here $y\neq 0$ ) such that

$f(x)= \frac{1}{(\frac{1}{y})} = y$

$\therefore f is \, \, onto$ .

Hence, the function is one-one and onto.

If the domain R is replaced by N with co-domain being same as R ∗ i.e. $g: N \longrightarrow R_{*}$ defined by

$g(x)=\frac{1}{x}$

$g(x_1)=g(x_2)$

$\frac{1}{x_1}=\frac{1}{x_2}$

$x_1=x_2$

$\therefore$ g is one-one.

For $1.5 \in R_*$ ,

$g(x) = \frac{1}{1.5}$ but there does not exists any x in N.

Hence, function g is one-one but not onto.

$f : N\rightarrow N$

$f(x) = x^2$

One- one:

$x,y \in N$ then $f(x)=f(y)$

$x^{2}=y^{2}$

$x=y$

$\therefore$ f is one- one i.e. injective.

For $3 \in N$ there is no x in N such that $f(x)=x^{2}=3$

$\therefore$ f is not onto i.e. not surjective.

Hence, f is injective but not surjective.

$f : Z \rightarrow Z$

$f(x) = x^2$

One- one:

For $-1,1 \in Z$ then $f(x) = x^2$

$f(-1)= (-1)^{2}$

$f(-1)= 1$ but $-1 \neq 1$

$\therefore$ f is not one- one i.e. not injective.

For $-3 \in Z$ there is no x in Z such that $f(x)=x^{2}= -3$

$\therefore$ f is not onto i.e. not surjective.

Hence, f is neither injective nor surjective.

$f: R \rightarrow R$

$f(x) = x^2$

One- one:

For $-1,1 \in R$ then $f(x) = x^2$

$f(-1)= (-1)^{2}$

$f(-1)= 1$ but $-1 \neq 1$

$\therefore$ f is not one- one i.e. not injective.

For $-3 \in R$ there is no x in R such that $f(x)=x^{2}= -3$

$\therefore$ f is not onto i.e. not surjective.

Hence, f is not injective and not surjective.

$f : N\rightarrow N$

$f(x) = x^3$

One- one:

$x,y \in N$ then $f(x)=f(y)$

$x^{3}=y^{3}$

$x=y$

$\therefore$ f is one- one i.e. injective.

For $3 \in N$ there is no x in N such that $f(x)=x^{3}=3$

$\therefore$ f is not onto i.e. not surjective.

Hence, f is injective but not surjective.

$f : Z \rightarrow Z$

$f(x) = x^3$

One- one:

For $(x,y) \in Z$ then $f(x) = f(y)$

$x^{3}=y^{3}$

$x=y$

$\therefore$ f is one- one i.e. injective.

For $3 \in Z$ there is no x in Z such that $f(x)=x^{3}= 3$

$\therefore$ f is not onto i.e. not surjective.

Hence, f is injective but not surjective.

$f : R\longrightarrow R$

$f (x) = [x]$

One- one:

For $1.5,1.7 \in R$ then $f(1.5)=\left [ 1.5 \right ] = 1$ and $f(1.7)=\left [ 1.7 \right ] = 1$

but $1.5\neq 1.7$

$\therefore$ f is not one- one i.e. not injective.

For $0.6 \in R$ there is no x in R such that $f(x)=\left [ 0.6 \right ]$

$\therefore$ f is not onto i.e. not surjective.

Hence, f is not injective but not surjective.

$f : R \rightarrow R$

$f (x) = | x |$

$f (x) = | x | = x \, if\, x\geq 0 \,\, and \, \, -x\, if\, x< 0$

One- one:

For $-1,1 \in R$ then $f (-1) = | -1 |= 1$

$f (1) = | 1 |= 1$

$-1\neq 1$

$\therefore$ f is not one- one i.e. not injective.

For $-2 \in R$ ,

We know $f (x) = | x |$ is always positive there is no x in R such that $f (x) = | x |=-2$

$\therefore$ f is not onto i.e. not surjective.

Hence, $f (x) = | x |$ , is neither one-one nor onto.

$f : R \rightarrow R$ is given by

$f (x) = \left\{\begin{matrix} 1 & if\;x>0 \\ 0& if\;x=0 \\ -1& if\;x<0 \end{matrix}\right.$

As we can see $f(1)=f(2)=1$ , but $1\neq 2$

So it is not one-one.

Now, f(x) takes only 3 values (1,0,-1) for the element -3 in codomain $R$ ,there does not exists x in domain $R$ such that $f(x)= -3$ .

So it is not onto.

Hence, signum function is neither one-one nor onto.

$A = \{1, 2, 3\}$

$B = \{4, 5, 6, 7\}$

$f = \{(1, 4), (2, 5), (3, 6)\}$

$f : A \rightarrow B$

$\therefore$ $f(1)=4,f(2)=5,f(3)=6$

Every element of A has a distant value in f.

Hence, it is one-one.

$f: R\rightarrow R$

$f(x) = 3 -4x$

Let there be $(a,b) \in R$ such that $f(a)=f(b)$

$3-4a = 3 -4b$

$-4a = -4b$

$a = b$

$\therefore$ f is one-one.

Let there be $y \in R$ , $y = 3 -4x$

$x = \frac{(3-y)}{4}$

$f(x) = 3 -4x$

Puting value of x, $f(\frac{3-y}{4}) = 3 - 4(\frac{3-y}{4})$

$f(\frac{3-y}{4}) = y$

$\therefore$ f is onto.

f is both one-one and onto hence, f is bijective.

$f : R\rightarrow R$

$f(x) = 1 + x^2$

Let there be $(a,b) \in R$ such that $f(a)=f(b)$

$1+a^{2} = 1 +b^{2}$

$a^{2}=b^{2}$

$a = \pm b$

For $f(1)=f(-1)=2$ and $1\neq -1$

$\therefore$ f is not one-one.

Let there be $-2 \in R$ (-2 in codomain of R)

$f(x) = 1 + x^2 = -2$

There does not exists any x in domain R such that $f(x) = -2$

$\therefore$ f is not onto.

Hence, f is neither one-one nor onto.

$f : A \times B \rightarrow B \times A$

$f (a, b) = (b, a)$

Let $(a_1,b_1),(a_2,b_2) \in A\times B$

such that $f (a_1, b_1) = f(a_2, b_2)$

$(b_1,a_1)=(b_2,a_2)$

$\Rightarrow$ $b_1= b_2$ and $a_1= a_2$

$\Rightarrow$ $(a_1,b_1) = (a_2,b_2)$

$\therefore$ f is one- one

Let, $(b,a) \in B\times A$

then there exists $(a,b) \in A\times B$ such that $f (a, b) = (b, a)$

$\therefore$ f is onto.

Hence, it is bijective.

$f : N \rightarrow N$ , $n\in N$

$f(n) = \left\{\begin{matrix} \frac{n+1}{2} & if\;n\;is\;odd \\ \frac{n}{2} & if\;n\;is\;evem \end{matrix}\right.$

Here we can observe,

$f(2)=\frac{2}{2}=1$ and $f(1)=\frac{1+1}{2}=1$

As we can see $f(1)=f(2)=1$ but $1\neq 2$

$\therefore$ f is not one-one.

Let, $n\in N$ (N=co-domain)

case1 n be even

For $r \in N$ , $n=2r$

then there is $4r \in N$ such that $f(4r)=\frac{4r}{2}=2r$

case2 n be odd

For $r \in N$ , $n=2r+1$

then there is $4r+1 \in N$ such that $f(4r+1)=\frac{4r+1+1}{2}=2r +1$

$\therefore$ f is onto.

f is not one-one but onto

hence, the function f is not bijective.

$A = R - \{3\}$

$B = R - \{1\}$

$f : A\rightarrow B$

$f(x) = \left (\frac{x-2}{x-3} \right )$

Let $a,b \in A$ such that $f(a)=f(b)$

$\left (\frac{a-2}{a-3} \right ) = \left ( \frac{b-2}{b-3} \right )$

$(a-2)(b-3)=(b-2)(a-3)$

$ab-3a-2b+6=ab-2a-3b+6$

$-3a-2b=-2a-3b$

$3a+2b= 2a+3b$

$3a-2a= 3b-2b$

$a=b$

$\therefore$ f is one-one.

Let, $b \in B = R - \{1\}$ then $b\neq 1$

$a \in A$ such that $f(a)=b$

$\left (\frac{a-2}{a-3} \right ) =b$

$(a-2)=(a-3)b$

$a-2 = ab -3b$

$a-ab = 2 -3b$

$a(1-b) = 2 -3b$

$a= \frac{2-3b}{1-b}\, \, \, \, \in A$

For any $b \in B$ there exists $a= \frac{2-3b}{1-b}\, \, \, \, \in A$ such that

$f(\frac{2-3b}{1-b}) = \frac{\frac{2-3b}{1-b}-2}{\frac{2-3b}{1-b}-3}$

$=\frac{2-3b-2+2b}{2-3b-3+3b}$

$=\frac{-3b+2b}{2-3}$

$= b$

$\therefore$ f is onto

Hence, the function is one-one and onto.

$f : R \rightarrow R$

$f(x) = x^4$

One- one:

For $a,b \in R$ then $f(a) = f(b)$

$a^{4}=b^{4}$

$a=\pm b$

$\therefore f(a)=f(b)$ does not imply that $a=b$

example: and $2\neq -2$

$\therefore$ f is not one- one

For $2\in R$ there is no x in R such that $f(x)=x^{4}= 2$

$\therefore$ f is not onto.

Hence, f is neither one-one nor onto.

Option D is correct.

$f : R\rightarrow R$

$f(x) = 3x$

One - One :

Let $\left ( x,y \right ) \in R$

$f(x)=f(y)$

$3x=3y$

$x=y$

$\therefore$ f is one-one.

Onto:

We have $y \in R$ , then there exists $x=\frac{y}{3} \in R$ such that

$f(\frac{y}{3})= 3\times \frac{y}{3} = y$

$\therefore f is \, \, onto$ .

Hence, the function is one-one and onto.

The correct answer is A .

More About NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2

The NCERT Class 12 Maths chapter Relations and Functions has a total of 5 exercises including miscellaneous. Exercise 1.2 Class 12 Maths covers solutions to 12 main questions and their sub-questions. Most of the questions are related to proving a function one to one. Hence NCERT Solutions for Class 12 Maths chapter 1 exercise 1.2 can be referred for learning the concepts related to proof etc.

Benefits of NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2

• The Class 12th Maths chapter 1 exercise provided here is solved by subject matter experts having rich experience in the domain of competitive exam preparation.
• Students can practice Exercise 1.2 Class 12 Maths to prepare various concepts like signus functions, one to one functions etc, many direct questions are asked in Board exams from this chapter.
• These Class 12 Maths chapter 1 exercise 1.2 solutions can be referred by students to revise before the exam and clarify any doubt regarding solution of exercise questions.
• NCERT Solutions for Class 12 Maths chapter 1 exercise 1.2 provided here are most recommended solutions for students aspiring to score well in examinations.
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Key Features Of NCERT Solutions for Exercise 1.2 Class 12 Maths Chapter 1

• Comprehensive Coverage: The solutions encompass all the topics covered in ex 1.2 class 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 maths ex 1.2, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 ex 1.2 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this 12th class maths exercise 1.2 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for ex 1.2 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for class 12 maths ex 1.2 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions Subject Wise

Subject wise NCERT Exemplar solutions

Happy learning!!!

1. Which concepts are covered in Exercise 1.2 Class 12 Maths?

Concepts related to one to one functions, reflexive functions etc, are discussed in the Exercise 1.2 Class 12 Maths

2. What is a set?

In Mathematics, A set is a collection of distinct or well-defined numbers or elements

3. How much weightage is given to chapter NCERT book chapter Relations and Functions for the CBSE board exam ?

Weightage of the chapters 'relation and function' is around 5 % weightage in the CBSE final board exam.

4. What are the various ways of representing a set?

There are 3 ways to represent a set:

a. Statement form.

b. Roaster form .

c. Set Builder form.

5. What is the meaning of the empty set in Class 12 Maths?

A set with no elements is called an empty set. Also known by Null set or void set.

6. What is the meaning of Relation discussed in Exercise 1.2 Class 12 Maths

A relation is the set of ordered pair numbers.

7. How many questions are covered in Exercise 1.2 Class 12 Maths ?

12 questions  are there  in Exercise 1.2 Class 12 Maths

8. What is the total number of exercises there are in the NCERT Class 12 Maths chapter 1 Relations and Functions ?

5 exercises are there including a miscellaneous exercise in the NCERT class 12 maths chapter 1.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Hello Aspirant , Hope your doing great . As per your query , your eligible for JEE mains in the year of 2025 , Every candidate can appear for the JEE Main exam 6 times over three consecutive years . The JEE Main exam is held two times every year, in January and April.

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A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9