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NCERT Solutions for Class 11 Maths Chapter 2 Relations And Functions are discussed here. In the previous chapter you have learned about sets. This NCERT book chapter relation and function class 11 is the continuation of chapter 1 sets. In this article, you will get NCERT solutions for class 11 maths chapter 2 relations and functions. The answers to the exercises in the NCERT textbook are made to help students get ready for their exams and do really well. NCERT Solution are written by very knowledgeable teachers who explain each answer in an easy-to-understand way that follows the latest CBSE Syllabus 2023. Using these answers can help Class 11 students get really good at understanding Relations and Functions Important topics like domain, co-domain, and range of functions are covered in this chapter relation and function class 11.
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In NCERT solutions for class 11 maths chapter 2 relations and functions, questions related to these topics are covered. Also, you will learn different types of specific real-valued functions and their graphs in ch 2 maths class 11. NCERT solutions for class 11 maths chapter 2 relations and functions will build your fundamentals of functions which will be helpful in the 12th board exam also. Check all NCERT solutions from class 6 to 12 to learn science and Maths. Here you will get NCERT solutions for class 11 also.
Also read:
Relations:
R is a relation between sets A and B: R ⊆ A × B
Inverse of Relation R: R⁻¹ = {(b, a) : (a, b) ∈ R}
Domain of R = Range of R⁻¹
Range of R = Domain of R⁻¹
Functions:
A function f: A → B maps every element of A to one and only one element in B.
Cartesian product A × B: A × B = {(a, b) : a ∈ A, b ∈ B}
(a, b) = (x, y) implies a = x and b = y
n(A) = x, n(B) = y, then n(A × B) = xy and A × ∅ = ∅
A × B ≠ B × A
Function f: A → B can be denoted as f(x) = y.
Algebra of Functions:
For functions f: X → R and g: X → R:
(f + g)(x) = f(x) + g(x)
(f - g)(x) = f(x) - g(x)
(f * g)(x) = f(x) * g(x)
(kf)(x) = k * (f(x)), where k is a real number
{f/g}(x) = f(x)/g(x), g(x) ≠ 0
Free download NCERT Solutions for Class 11 Maths Chapter 2 Relations And Functions for CBSE Exam.
Relations and functions class 11 questions and answers - Exercise: 2.1
Question:1 If , find the values of x and y.
Answer:
It is given that
Since the ordered pairs are equal, the corresponding elements will also be equal
Therefore,
Therefore, values of x and y are 2 and 1 respectively
Question:2 If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in .
Answer:
It is given that set A has 3 elements and the elements in set B are 3 , 4 , and 5
Therefore, the number of elements in set B is 3
Now,
Number of elements in
= ( Number of elements in set A ) ( Number of elements in set B)
= 3 3
= 9
Therefore, number of elements in is 9
Question:3 If G = {7, 8} and H = {5, 4, 2}, find
Answer:
It is given that
G = {7, 8} and H = {5, 4, 2}
We know that the cartesian product of two non-empty sets P and Q is defined as
P Q = {(p,q) , where p P , q Q }
Therefore,
G H = {(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)}
And
H G = {(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)}
Answer:
FALSE
If P = {m, n} and Q = { n, m}
Then,
Answer:
It is a TRUE statement
If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that and
Answer:
This statement is TRUE
If A = {1, 2}, B = {3, 4}, then
There for
Question:5 If A = {–1, 1}, find
Answer:
It is given that
A = {–1, 1}
A is an non-empty set
Therefore,
Lets first find
Now,
Question:6 If = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.
Answer:
It is given that
= {(a, x),(a , y), (b, x), (b, y)}
We know that the cartesian product of two non-empty set P and Q is defined as
Now, we know that A is the set of all first elements and B is the set of all second elements
Therefore,
Question:7 (i) Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that .
Answer:
It is given that
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
Now,
Now,
And
Now,
From equation (i) and (ii) it is clear that
Hence,
Question:7 (ii) Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that is a subset of
Answer:
It is given that
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
Now,
And
We can clearly observe that all the elements of the set are the elements of the set
Therefore, is a subset of
Question:8 Let A = {1, 2} and B = {3, 4}. Write . How many subsets will have?
List them.
Answer:
It is given that
A = {1, 2} and B = {3, 4}
Then,
Now, we know that if C is a set with
Then,
Therefore,
The set has subsets.
Answer:
It is given that
n(A) = 3 and n(B) = 2 and If (x, 1), (y, 2), (z, 1) are in A × B.
By definition of Cartesian product of two non-empty Set P and Q:
Now, we can see that
P = set of all first elements.
And
Q = set of all second elements.
Now,
(x, y, z) are elements of A and (1,2) are elements of B
As n(A) = 3 and n(B) = 2
Therefore,
A = {x, y, z} and B = {1, 2}
Answer:
It is given that Cartesian product A × A having 9 elements among which are found (–1, 0) and (0,1).
Now,
Number of elements in (A× B) = (Number of elements in set A) × (Number of elements in B)
It is given that
Therefore,
Now,
By definition A × A = {(a, a): a ? A}
Therefore,
-1, 0 and 1 are the elements of set A
Now, because, n(A) = 3 therefore, A = {-1, 0, 1}
Therefore,
the remaining elements of set (A × A) are
(-1,-1), (-1,1), (0,0), (0, -1), (1,1), (1, -1) and (1, 0)
NCERT class 11 maths chapter 2 question answer - Exercise: 2.2
Answer:
It is given that
Now, the relation R from A to A is given as
Therefore,
the relation in roaster form is ,
Now,
We know that Domain of R = set of all first elements of the order pairs in the relation
Therefore,
Domain of
And
Codomain of R = the whole set A
i.e. Codomain of
Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
range of
Question:2 Define a relation R on the set N of natural numbers by is a natural number less than . Depict this relationship using roster form. Write down the domain and the range.
Answer:
It is given that
is a natural number less than
As x is a natural number which is less than 4.
Therefore,
the relation in roaster form is,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of
Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
the range of
Therefore, domain and the range are respectively
Question:3 A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by . Write R in roster form.
Answer:
It is given that
A = {1, 2, 3, 5} and B = {4, 6, 9}
And
Now, it is given that the difference should be odd. Let us take all possible differences.
(1 - 4) = - 3, (1 - 6) = - 5, (1 - 9) = - 8(2 - 4) = - 2, (2 - 6) = - 4, (2 - 9) = - 7(3 - 4) = - 1, (3 - 6) = - 3, (3 - 9) = - 6(5 - 4) = 1, (5 - 6) = - 1, (5 - 9) = - 4
Taking the difference which are odd we get,
Therefore,
the relation in roaster form,
Question:4 (i) The Fig2.7 shows a relationship between the sets P and Q. Write this relation in set-builder form
Answer:
It is given in the figure that
P = {5,6,7}, Q = {3,4,5}
Therefore,
the relation in set builder form is ,
OR
Question:4 (ii) The Fig2.7 shows a relationship between the sets P and Q. Write this relation roster form. What is it domain and range?
Answer:
From the given figure. we observe that
P = {5,6,7}, Q = {3,4,5}
And the relation in roaster form is ,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of
Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
the range of
Question:5 (i) Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by Write R in roster form
Answer:
It is given that
A = {1, 2, 3, 4, 6}
And
Therefore,
the relation in roaster form is ,
Question:5 (ii) Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by Find the domain of R
Answer:
It is given that
A = {1, 2, 3, 4, 6}
And
Now,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of
Question:5 (iii) Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by Find the range of R.
Answer:
It is given that
A = {1, 2, 3, 4, 6}
And
Now,
As the range of R = set of all second elements of the order pairs in the relation.
Therefore,
Range of
Question:6 Determine the domain and range of the relation R defined by
Answer:
It is given that
Therefore,
the relation in roaster form is ,
Now,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of
Now,
As Range of R = set of all second elements of the order pairs in the relation.
Range of
Therefore, the domain and range of the relation R is respectively
Question:7 Write the relation in roster form.
Answer:
It is given that
Now,
As we know the prime number less than 10 are 2, 3, 5 and 7.
Therefore,
the relation in roaster form is ,
Question:8 Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.
Answer:
It is given that
A = {x, y, z} and B = {1, 2}
Now,
Therefore,
Then, the number of subsets of the set
Therefore, the number of relations from A to B is
Question:9 Let R be the relation on Z defined by
Find the domain and range of R.
Answer:
It is given that
Now, as we know that the difference between any two integers is always an integer.
And
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
The domain of R = Z
Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
range of R = Z
Therefore, the domain and range of R is Z and Z respectively
NCERT class 11 maths chapter 2 question answer - Exercise: 2.3
Answer:
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
Now,
As Range of R = set of all second elements of the order pairs in the relation.
Therefore,
Range of
Therefore, domain and range of R are respectively
Answer:
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
Now,
As Range of R = set of all second elements of the order pairs in the relation.
Therefore,
Range of
Therefore, domain and range of R are respectively
Question:1 (iii) Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range. {(1,3), (1,5), (2,5)}.
Answer:
Since the same first element 1 corresponds to two different images 3 and 5. Hence, this relation is not a function.
Question:2 (i) Find the domain and range of the following real functions:
Answer:
Given function is
Now, we know that
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 , 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
Question:2 (ii) Find the domain and range of the following real functions:
Answer:
Given function is
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 - x 2 ≥ 0 [Since a value less than 0 will give an imaginary value]
Therefore,
The domain of f(x) is
Now,
If we put the value of x from we will observe that the value of function varies from 0 to 3
Therefore,
Range of f(x) is
Question:3 (i) A function f is defined by f(x) = 2x –5. Write down the values of f (0),
Answer:
Given function is
Now,
Therefore,
Value of f(0) is -5
Question:3 (ii) A function f is defined by f(x) = 2x –5. Write down the values of f (7)
Answer:
Given function is
Now,
Therefore,
Value of f(7) is 9
Question:3 (iii) A function f is defined by f(x) = 2x –5. Write down the values of f (-3)
Answer:
Given function is
Now,
Therefore,
Value of f(-3) is -11
Question:4(i) The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t (0)
Answer:
Given function is
Now,
Therefore,
Value of t(0) is 32
Question:4(ii) The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t (28)
Answer:
Given function is
Now,
Therefore,
Value of t(28) is
Question:4(iii) The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t (-10)
Answer:
Given function is
Now,
Therefore,
Value of t(-10) is 14
Answer:
Given function is
Now,
Therefore,
When t(C) = 212 , value of C is 100
Question:5 (i) Find the range of each of the following functions.
Answer:
Given function is
It is given that
Now,
Add 2 on both the sides
Therefore,
Range of function is
Question:5 (ii) Find the range of each of the following functions
Answer:
Given function is
It is given that x is a real number
Now,
Add 2 on both the sides
Therefore,
Range of function is
Question:5 (iii) Find the range of each of the following functions.
Answer:
Given function is
It is given that x is a real number
Therefore,
Range of function is R
Relations and functions class 11 NCERT solutions - Miscellaneous Exercise
Answer:
It is given that
Now,
And
At x = 3,
Also, at x = 3,
We can see that for , f(x) has unique images.
Therefore, By definition of a function, the given relation is function.
Now,
It is given that
Now,
And
At x = 2,
Also, at x = 2,
We can clearly see that element 2 of the domain of relation g(x) corresponds to two different images i.e. 4 and 6. Thus, f(x) does not have unique images
Therefore, by definition of a function, the given relation is not a function
Hence proved
Question:3 Find the domain of the function
Answer:
Given function is
Now, we will simplify it into
Now, we can clearly see that
Therefore, the Domain of f(x) is
Question:4 Find the domain and the range of the real function f defined by
Answer:
Given function is
We can clearly see that f(x) is only defined for the values of x ,
Therefore,
The domain of the function is
Now, as
take square root on both sides
Therefore,
Range of function is
Question:5 Find the domain and the range of the real function f defined by
Answer:
Given function is
As the given function is defined of all real number
The domain of the function is R
Now, as we know that the mod function always gives only positive values
Therefore,
Range of function is all non-negative real numbers i.e.
Question:6 Let R be a function from R into R. Determine the range of f.
Answer:
Given function is
Range of any function is the set of values obtained after the mapping is done in the domain of the function. So every value of the codomain that is being mapped is Range of the function.
Let's take
Now, 1 - y should be greater than zero and y should be greater than and equal to zero for x to exist because other than those values the x will be imaginary
Thus,
Therefore,
Range of given function is
Question:7 Let f, g : R R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and f/g
Answer:
It is given that
Now,
Therefore,
Now,
Therefore,
Now,
Therefore, values of are respectively
Answer:
It is given that
And
Now,
At x = 1 ,
Similarly,
At ,
Now, put this value of b in equation (i)
we will get,
Therefore, values of a and b are 2 and -1 respectively
Question:9 (i) Let R be a relation from N to N defined by . Are the following true?
Answer:
It is given that
And
for all
Now, it can be seen that But,
Therefore, this statement is FALSE
Question:9 (ii) Let R be a relation from N to N defined by . Are the following true?
Answer:
It is given that
And
implies
Now , it can be seen that and , But
Therefore,
Therefore, given statement is FALSE
Question:9 (iii) Let R be a relation from N to N defined by . Are the following true?
(a,b) R, (b,c) R implies (a,c) R.
Answer:
It is given that
And
implies
Now, it can be seen that because and , But
Therefore,
Therefore, the given statement is FALSE
Answer:
It is given that
and
Now,
Now, a relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B
And we can see that f is a subset of
Hence f is a relation from A to B
Therefore, given statement is TRUE
Answer:
It is given that
and
Now,
As we can observe that same first element i.e. 2 corresponds to two different images that is 9 and 11.
Hence f is not a function from A to B
Therefore, given statement is FALSE
Question:11 Let f be the subset of defined by . Is f a function from Z to Z? Justify your answer.
Answer:
It is given that
Now, we know that relation f from a set A to a set B is said to be a function only if every element of set A has a unique image in set B
Now, for value 2, 6, -2, -6
Now, we can observe that same first element i.e. 12 corresponds to two different images that are 8 and -8.
Thus, f is not a function
Answer:
It is given that
A = {9,10,11,12,13}
And
f : A → N be defined by f(n) = the highest prime factor of n.
Now,
Prime factor of 9 = 3
Prime factor of 10 = 2,5
Prime factor of 11 = 11
Prime factor of 12 = 2,3
Prime factor of 13 = 13
f(n) = the highest prime factor of n.
Hence,
f(9) = the highest prime factor of 9 = 3
f(10) = the highest prime factor of 10 = 5
f(11) = the highest prime factor of 11 = 11
f(12) = the highest prime factor of 12 = 3
f(13) = the highest prime factor of 13 = 13
As the range of f is the set of all f(n), where
Therefore, the range of f is: {3, 5, 11, 13}.
2.2 Cartesian Products of Sets
2.3 Relations
2.4 Functions
If you are interested in Relation And Function Class 11 exercises solutions then these are listed below.
relation and function class 11 Exercise 2.1 10 Questions
relation and function class 11 Exercise 2.2 9 Questions
relation and function class 11 Exercise 2.3 5 Questions
relation and function class 11 Miscellaneous Exercise 12 Questions
chapter-1 | Sets |
chapter-2 | Relations and Functions |
chapter-3 | Trigonometric Functions |
chapter-4 | Principle of Mathematical Induction |
chapter-5 | Complex Numbers and Quadratic equations |
chapter-6 | Linear Inequalities |
chapter-7 | Permutation and Combinations |
chapter-8 | Binomial Theorem |
chapter-9 | Sequences and Series |
chapter-10 | Straight Lines |
chapter-11 | Conic Section |
chapter-12 | Introduction to Three Dimensional Geometry |
chapter-13 | Limits and Derivatives |
chapter-14 | Mathematical Reasoning |
chapter-15 | Statistics |
chapter-16 | Probability |
Comprehensive Coverage: The class 11 relations and functions NCERT solutions provide a comprehensive explanation of all the topics covered in the chapter, ensuring a strong understanding of relations and functions.
Step-by-Step Solutions: Each problem and example in the NCERT textbook is solved step by step, making it easier for students to follow the logical progression of concepts.
Clear and Concise Language: The maths chapter 2 class 11 solutions are presented in clear and concise language, making complex mathematical concepts more accessible to students.
NCERT solutions for class 11 biology |
NCERT solutions for class 11 maths |
NCERT solutions for class 11 chemistry |
NCERT solutions for Class 11 physics |
Happy Reading !!!
The definition of a function in class 11 chapter 2 states that each element in the domain can be related to only one element in the range. This means that if a student draws a vertical line on a graph, it can intersect the x-axis only once. To determine if a relation is a function, one can use vertical line tests or various formulas.
Basic definitions of relations and functions, Cartesian products of sets, and domain and range of the functions are the important topic of this chapter. you can practice them to get maximum benifit. Practice problems form class 11 maths chapter 2 NCERT solutions pdf after downloading using link given in this article.
Relations are just a bunch of ordered pairs, with one object from each set. Functions can also be considered relations, but they have different concepts. The NCERT Solutions for Class 11 Maths Chapter 2 provide students with a clear definition and analysis of relations according to the CBSE Syllabus 2023. The solutions have many examples to help students solve relation-related problems with ease.
To learn how to solve problems from class 11 Maths NCERT solutions chapter 2 that are related to the Cartesian product of sets, it is essential for students to understand the first exercise of the chapter well. Before each set of problems in the exercise, solved examples are given to help students learn how to solve problems quickly. By practicing problems from the NCERT textbook, students can improve their understanding of the concepts, which is necessary for performing well in exams.
Here you will get the detailed NCERT solutions for class 11 maths by clicking on the link. Practice problems from class 11 maths chapter 2 solutions to get command and in-depth understanding of concepts.
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