NCERT Solutions for Exercise 2.3 Class 11 Maths Chapter 2 - Relations and Functions

NCERT Solutions for Exercise 2.3 Class 11 Maths Chapter 2 - Relations and Functions

Edited By Vishal kumar | Updated on Nov 03, 2023 08:18 AM IST

NCERT Solutions for Class 11 Maths Chapter 2 - Relations And Functions Exercise 2.3- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 2: Relations and Functions Exercise 2.3- In the previous exercise, you have learned about the relations, domain of relations, co-domain of relations, and range of relations. In the NCERT solutions for Class 11 Maths chapter 2 exercise 2.3, you will learn about the definition of a function, the domain of the function, range of the function, etc. Some different types of functions and their graph is given before exercise 2.3 Class 11 Maths. You must go through these examples and definitions given before this ex 2.3 class 11. These definitions are very important to understand the Class 11 Maths chapter 2 exercise 2.3.

The questions related to the algebra of real functions such as the addition of two real functions, subtraction of a real function from another function, multiplication of scalar with real function, multiplication of two real functions, and quotient of two real functions are also covered in Class 11 Maths chapter 2 exercise 2.3 solutions. There are some questions related to some special functions like greatest integer function, signum function, modulus function, rational functions in the Class 11 Maths chapter 2 exercise 2.3. You must try to solve all the NCERT problems by yourself. If you find difficulties while solving them, you can go through the Class 11 Maths, Class 11 Maths chapter 2 exercise 2.3 solutions. Also, if you looking for NCERT Solutions, click on the given link to get descriptive NCERT solutions.

The 11th class maths exercise 2.3 answers, prepared by subject experts at Careers360, maintain the same approach as in Exercises 2.1 and 2.2. These exercise 2.3 class 11 maths solutions continue to offer detailed, step-by-step explanations to help students understand mathematical concepts effectively. Furthermore, a PDF version of the class 11 math Exercise 2.3 solutions is accessible for convenient use, and, just like previous resources, they are provided free of charge. This resource remains a valuable and cost-effective tool for students looking to improve their understanding of mathematics and excel in their studies.

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Question:1 (i) Which of the following relations are functions? Give reasons. If it is a function,determine its domain and range. {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}

Since, 2, 5, 8, 11, 14 and 17 are the elements of domain R having their unique images. Hence, this relation R is a function.
Now,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R =\left \{ {2, 5, 8, 11, 14, 17} \right \}$

Now,
As Range of R = set of all second elements of the order pairs in the relation.
Therefore,
Range of $R =\left \{ {1} \right \}$

Therefore, domain and range of R are $\left \{ {2, 5, 8, 11, 14, 17} \right \} \ and \ \left \{ 1 \right \}$ respectively

Since, 2, 4, 6, 8, 10,12 and 14 are the elements of domain R having their unique images. Hence, this relation R is a function.

Now,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R =\left \{ {2, 4, 6, 8, 10,12, 14} \right \}$

Now,
As Range of R = set of all second elements of the order pairs in the relation.
Therefore,
Range of $R =\left \{ {1, 2, 3, 4, 5,6, 7} \right \}$

Therefore, domain and range of R are $\left \{ {2, 4, 6, 8, 10,12, 14} \right \} \ and \ \left \{ {1, 2, 3, 4, 5,6, 7} \right \}$ respectively

Since the same first element 1 corresponds to two different images 3 and 5. Hence, this relation is not a function.

$f (x ) = - |x|$

Given function is
$f (x ) = - |x|$
Now, we know that

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

Now, for a function f(x),
Domain: The values that can be put in the function to obtain real value. For example f(x) = x, now we can put any value in place of x and we will get a real value. Hence, the domain of this function will be Real Numbers.
Range: The values that we obtain of the function after putting the value from domain. For Example: f(x) = x + 1, now if we put x = 0, f(x) = 1. This 1 is a value of Range that we obtained.

Since f(x) is defined for $x \ \epsilon \ R$, the domain of f is R.

It can be observed that the range of f(x) = -|x| is all real numbers except positive real numbers. Because will always get a negative number when we put a value from the domain.
Therefore, the range of f is $(-\infty , 0]$

Question:2 (ii) Find the domain and range of the following real functions:

$f ( x ) = \sqrt { 9- x ^2 }$

Given function is
$f ( x ) = \sqrt { 9- x ^2 }$
Now,
Domain: These are the values of x for which f(x) is defined.
for the given f(x) we can say that, f(x) should be real and for that,9 - x2 ≥ 0 [Since a value less than 0 will give an imaginary value]
$\Rightarrow 3^2-x^2= (3-x)(3+x)\geq 0$
$\Rightarrow -3\leq x\leq 3$
Therefore,
The domain of f(x) is $[-3,3]$
Now,
If we put the value of x from $[-3,3]$ we will observe that the value of function $f ( x ) = \sqrt { 9- x ^2 }$ varies from 0 to 3
Therefore,
Range of f(x) is $[0,3]$

Given function is
$f(x) = 2x-5$
Now,
$f(0) = 2(0)-5=0-5 = -5$
Therefore,
Value of f(0) is -5

Given function is
$f(x) = 2x-5$
Now,
$f(7) = 2(7)-5=14-5 = 9$
Therefore,
Value of f(7) is 9

Given function is
$f(x) = 2x-5$
Now,
$f(-3) = 2(-3)-5=-6-5 = -11$
Therefore,
Value of f(-3) is -11

Given function is
$t ( C ) = \frac{9 C }{5} + 32$
Now,
$t ( 0 ) = \frac{9 (0) }{5} + 32= 0+ 32 = 32$
Therefore,
Value of t(0) is 32

Given function is
$t ( C ) = \frac{9 C }{5} + 32$
Now,
$t ( 28 ) = \frac{9 (28) }{5} + 32= \frac{252}{5}+ 32 = \frac{252+160}{5}= \frac{412}{5}$
Therefore,
Value of t(28) is $\frac{412}{5}$

Given function is
$t ( C ) = \frac{9 C }{5} + 32$
Now,
$t ( -10 ) = \frac{9 (-10) }{5} + 32= \frac{-90}{5}+ 32 = -18+32= 14$
Therefore,
Value of t(-10) is 14

Given function is
$t ( C ) = \frac{9 C }{5} + 32$
Now,
$212 = \frac{9 (C) }{5} + 32$
$212 \times 5= {9 (C) } + 160$
${9 (C) } =1060-160$
$C = \frac{900}{9} = 100$
Therefore,
When t(C) = 212 , value of C is 100

Question:5 (i) Find the range of each of the following functions.

$f (x) = 2 - 3x, x \epsilon R, x > 0.$

Given function is

$f (x) = 2 - 3x, x \epsilon R, x > 0.$
It is given that $x > 0$
Now,
$\Rightarrow 3x > 0$
$\Rightarrow -3x < 0$
Add 2 on both the sides
$\Rightarrow -3x+2 < 0+2$
$\Rightarrow 2-3x < 2$
$\Rightarrow f(x) < 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because f(x) = 2-3x)$
Therefore,
Range of function $f(x) = 2 -3x$ is $(-\infty,2)$

Question:5 (ii) Find the range of each of the following functions

$f ( x ) = x ^2 +2$ , x is a real number.

Given function is

$f ( x ) = x ^2 +2$
It is given that x is a real number
Now,
$\Rightarrow x^2 \geq 0$
Add 2 on both the sides
$\Rightarrow x^2+2 \geq 0+2$
$\Rightarrow f(x) \geq 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because f(x) = x^2+2)$
Therefore,
Range of function $f ( x ) = x ^2 +2$ is $[2,\infty)$

Question:5 (iii) Find the range of each of the following functions.

f (x) = x, x is a real number

Given function is

$f ( x ) = x$
It is given that x is a real number
Therefore,
Range of function $f ( x ) = x$ is R

More About NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.3:-

Class 11 Maths Chapter 2 exercise 2.3 consists of questions related to identifying the functions, finding the domain and range of the functions, etc. There are eight examples in ex 2.3 class 11, some graphs of different types of functions, and definitions are given before the Class 11 Maths chapter 2 exercise 2.3. You must go through these examples and definitions before solving the NCERT exercise problems.

Also Read| Relations And Functions Class 11th Notes

Benefits of NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.3:-

• NCERT solutions for class 11 maths ex 2.3 are not only important for CBSE exams but you can use these solutions to conceptual clarity about the functions and their properties.
• The last two exercises of this chapter are very important and useful in other chapters like trigonometric functions, inverse trigonometric functions, etc.
• Class 11 Maths chapter 2 exercise 2.3 solutions can be used for the revision of important concepts before the CBSE exam,

Key Features of 11th Class Maths Exercise 2.3 Answers

1. Comprehensive Solutions: The class 11 maths ex 2.3 solutions offer comprehensive and detailed explanations for each problem, making it easier for students to understand and apply mathematical concepts.

2. Clarity and Precision: The class 11 ex 2.3 solutions are presented with clarity and accuracy, ensuring that students can confidently prepare for exams and gain a deep understanding of the material.

3. Curriculum Alignment: The 11th class maths exercise 2.3 answers solutions are closely aligned with the NCERT curriculum, covering the topics and concepts according to the official syllabus.

4. Varied Practice Problems: Exercise 2.3 includes a range of practice problems to help students reinforce their knowledge and develop their problem-solving skills.

5. Accessibility: These class 11 ex 2.3 solutions are typically available for free, ensuring easy access for students from various backgrounds.

6. Format Options: PDF versions of the ex 2.3 class 11 solutions is provided, allowing students to download and use them at their convenience, both online and offline.

Also see-

Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. What is an identity function ?

The identity function is a function that always returns the same value as the input you have given to the function.

2. What is a constant function ?

The identity function is a function that always returns the constant value independent of the input you have given to the function.

3. What is a polynomial function ?

A polynomial function is a function that involves only positive integer exponents of a variable in an equation.

4. Write an example of polynomial function ?

Example of a polynomial function - p(x) = x^2 + 2x + 3

5. Whats is the modulus function ?

Modulus function is a function that gives us the absolute value of the variable or number that has been given as the input to the function.

6. What is the greatest integer function ?

The greatest integer function is a function that gives the greatest integer, less than or equal to x where x is the input of the greatest integer function.

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