NCERT Solutions  Class 11 Maths Chapter 2  Functions and relations
CBSE Solutions  Maths Class 11
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NCERT  Class 11 Maths
Book:  National Council of Educational Research and Training (NCERT) 

Board:  Central Board of Secondary Education (CBSE) 
Class:  11th 
Subject:  Maths 
Chapter:  2 
Chapters Name:  Functions and relations 
Medium:  English 
Functions and relations  Class 11 Maths  NCERT Books Solutions
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1
Ex 2.1 Class 11 Maths Question 1.
If \(\left( \frac { x }{ 3 } +1,y\frac { 2 }{ 3 } \right) =\left( \frac { 5 }{ 3 } ,\frac { 1 }{ 3 } \right) \), find the values of x and y.
Solution.
Since the ordered pairs are equal. So, the corresponding elements are equal
∴ \(\frac { x }{ 3 } +1=\frac { 5 }{ 3 } \) and \(y\frac { 2 }{ 3 } =\frac { 1 }{ 3 } \)
⇒ \(\frac { x }{ 3 } =\frac { 5 }{ 3 } 1\) and \(y=\frac { 1 }{ 3 } +\frac { 2 }{ 3 } \) ⇒ x = 2 and y = 1.
Ex 2.1 Class 11 Maths Question 2.
If the set A has 3 elements and the set B {3, 4, 5}, then find the number of elements in (A x B).
Solution.
According to question, n(A) = 3 and n(B) = 3.
∴ n(A x B) = n(A) x n(B) = 3 x 3 = 9
∴ There are total 9 elements in (A x B).
Ex 2.1 Class 11 Maths Question 3.
If G = {7, 8} and H = {5, 4, 2}, find G x H and H x G.
Solution.
We have G = {7, 8} and H = {5, 4, 2} Then, by the definition of the cartesian product, we have
G x H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}
H x G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}.
Ex 2.1 Class 11 Maths Question 4.
State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
(i) If P = {m, n} and Q = {n, m}, then P x Q = {(m, n), (n, m)}.
(ii) If A and B are nonempty sets, then Ax B is a nonempty set of ordered pairs (x, y) such
that x ∈ A and y ∈ B.
(iii) If A = {1, 2}, B = {3, 4}, then A x (B ∩φ) = φ
Solution.
(i) False, if P = {m, n} and Q = {n, m}
Then P x Q = {(m, n), (m, m), (n, n), (n, m)}.
(ii) True, by the definition of cartesian product.
(iii) True, We have A = {1, 2} and B = {3, 4}
Now, B ∩ φ = φ ∴ A x (B ∩ φ) = A x φ = φ.
Ex 2.1 Class 11 Maths Question 5.
If A = {1, 1},find A x A x A.
Solution.
A = {1, 1}
Then, A x A = {1, 1} x {1, 1} = {(1, 1), (1,1),(1,1), (1,1)}
A x A x A = ((1,1),(1,1),(1,1),(1,1)} x {1,1}
= {(1, 1, 1), (1, 1, 1), (1, 1, 1), (1, 1,1), (1, 1, 1), (1, 1,1), (1,1,1), (1,1,1)}
Ex 2.1 Class 11 Maths Question 6.
If A x B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B.
Solution.
Given, A x B = {(a, x), (a, y), (b, x), (b, y)}
If {p, q) ∈ A x B, then p ∈ A and q ∈ B
∴ A = {a, b} and B = {x, y}.
Ex 2.1 Class 11 Maths Question 7.
Let A = {1, 2}, B = (1, 2, 3, 4), C = {5, 6} and D = {5, 6, 7, 8}. Verify that
(i) A x (B ∩ C = (A x B) ∩ (AxC)
(ii) A x C is a subset of B x D.
Solution.
Given, A = {1, 2}, B ={1, 2, 3, 4}, C = {5, 6}, D = (5, 6, 7, 8}
Ex 2.1 Class 11 Maths Question 8.
Let A = {1, 2} and B = {3, 4}. Write 4 x B. How many subsets will 4 x B have? List them.
Solution.
Given, A = {1, 2} and B = {3, 4}
Then, A x B = {(1, 3), (1,4), (2, 3), (2, 4)}
i. e., A x B has 4 elements. So, it has 2^{4} i.e. 16 subsets.
The subsets of A x B are as follows :
φ, {(1, 3)1, ((1, 4)), {(2, 3), {(2, 4)}, {(1, 3), (1,4)}, {(1,3), (2,3)},{(1,3), (2,4)), ((1,4), (2,3)},
{(1, 4), (2, 4)},{(2, 3), (2, 4)},{(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 4), (2, 3), (2,4)},{(1, 3), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2,4)}.
Ex 2.1 Class 11 Maths Question 9.
Let A and B be two sets such that n (A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A x B, find A and B, where x, y and z are distinct elements.
Solution.
Given, n(A) = 3 and n(B) = 2
Now (x, 1) ∈ A x B ⇒ x ∈ A and 1 ∈ B,
(y, 2) ∈ A x B ⇒ y ∈ A and 2 ∈ B
(z, 1) ∈ A x B ⇒z ∈ A and 1 ∈ B
∴ x, y, z ∈ A and 1, 2 ∈ B
Hence, A = {x, y, z} and B = {1, 2}.
Ex 2.1 Class 11 Maths Question 10.
The Cartesian product 4×4 has 9 elements among which are found (1, 0) and (0, 1). Find the set 4 and the remaining elements of 4 x 4.
Solution.
Since, we have n(A x A) = 9
⇒ n(A) x n(A) = 9 [ ∵ n (A x B) = n(A) x n(B)]
⇒ (n(A))^{2} = 9 ⇒ n(A) = 3
Also, given (1, 0) ∈ A x A ⇒ 1, 0 ∈ A ,
and (0,1) ∈ A x A ⇒ 0, 1 ∈ A
∴ 1, 0,1 ∈ A
Hence, A = {1, 0, 1} (∵ n(A) = 3)
and the remaining elements of A x A are (1, 1), (1,1), (0, 1), (0,0), (1, 1), (1,0), (1,1).
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.2
Ex 2.2 Class 11 Maths Question 1.
Let 4 = {1,2,3, ,14}. Define a relation R from A to A by R = {(x,y): 3x – y = 0, where x, y ∈ 4}.
Write down its domain, codomain and range.
Solution.
We have A = (1, 2, 3,……..,14)
Given relation R = {(x, y) : 3x – y = 0, where x, y ∈ A}
= {(x, y): y = 3x, where x, y ∈ A)
= {(x, 3x), where x, 3x ∈ A}
= {(1, 3), (2, 6), (3, 9), (4,12)}
[∵ 1 ≤ 3x ≤ 14, ∴ \(\frac { 1 }{ 3 } \le x\le \frac { 14 }{ 3 } \) ⇒ x = 1, 2, 3, 4 ]
Domain of R = {1, 2, 3, 4}
Codomain of R = {1, 2,……, 14}
Range of R = {3, 6, 9, 12}.
Ex 2.2 Class 11 Maths Question 2.
Define a relation R on the set N of natural numbers by R = {(x, y): y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using roster form. Write down the domain and the range.
Solution.
Given relation R = {(x, y): y = x + 5, x < 4 and x, y ∈ N)
= {(x, y): y = x + 5, x ∈ (1, 2, 3) & y ∈ N}
= {(x, x + 5): x = 1, 2, 3}
Thus, R = {(1, 6), (2, 7), (3, 8)}.
Domain of R = {1, 2, 3}, Range of R = {6, 7, 8}.
Ex 2.2 Class 11 Maths Question 3.
A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ 4, y ∈B}. Write R in roster form.
Solution.
We have, A = {1, 2, 3, 5} and B = {4, 6, 9} R = {(x, y): difference between x and y is odd;
x ∈ A, y ∈B}
= {(x, y): y – x = odd; x ∈ A, y ∈ B}
Hence R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}.
Ex 2.2 Class 11 Maths Question 4.
The figure shows a relationship between the sets P and Q. Write this relation
(i) in setbuilder form
(ii) roster form.
What is its domain and range?
Solution.
(i) Its set builder form is
R = {(x, y): x – y = 2; x ∈ P, y ∈ Q}
i.e., R = {(x, y): y = x – 2 for x = 5, 6, 7)}
(ii) Roster form is R = {(5, 3), (6, 4), (7, 5)}
Domain of R = {5, 6, 7} = P,
Range of R = {3, 4, 5} = Q.
Ex 2.2 Class 11 Maths Question 5.
Let A = {1,2,3,4,6}. Let R be the relation on A defined by {(a, b): a, b ∈ A, b is exactly divisible by a}.
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R?.
Solution.
Given A = {1, 2, 3, 4, 6}
Given relation is R = {(a, b):a,b ∈ A, b is exactly divisible by a}
(i) Roster form of R = {(1,1), (1,2), (1,3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6) (4, 4), (6, 6)}.
(ii) Domain of R = {1, 2, 3, 4, 6} = A.
(iii) Range of R = {1, 2, 3, 4, 6} = A.
Ex 2.2 Class 11 Maths Question 6.
Determine the domain and range of the relation R defined by R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.
Solution.
Given relation is R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5)}
= {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5,10)}
∴Domain of R = {0,1, 2, 3, 4, 5} and
Range of R = {5, 6, 7, 8, 9, 10}.
Ex 2.2 Class 11 Maths Question 7.
Write the relation R = {(x, x^{3}): x is a prime number less than 10} in roster form.
Solution.
Given relation is R = {(x, x^{3}): x is a prime number less than 10)
= {(x, x^{3}): x ∈ {2, 3, 5, 7}}
= {(2, 23), (3, 33), (5, 53), (7, 73)}
= {(2, 8), (3, 27), (5, 125), (7, 343)}.
Ex 2.2 Class 11 Maths Question 8.
Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.
Solution.
Given A = {x, y, z} and B = {1, 2}
∴ n(A) = 3 & n(B) = 2
Since n(A x B) = n(A) x n(B)
∴ n(A x B) = 3 x 2 = 6
Number of relations from A to B is equal to the number of subsets of A x B.
Since A x B contains 6 elements.
⇒ Number of subsets of A x B = 2^{6} = 64
So, there are 64 relations from A to B.
Ex 2.2 Class 11 Maths Question 9.
Let R be the relation on Z defined by R = {{a, b): a, b ∈ Z, a – b is an integer}. Find the domain and range of R.
Solution.
Given relation is R = {(a, b): a, b ∈ Z, a – b is an integer}
If a, b ∈ Z, then a b ∈ Z ⇒ Every ordered pair of integers is contained in R.
R = {(a, b) :a,b ∈ Z}
So, Range of R = Domain of R = Z.
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.3
Ex 2.3 Class 11 Maths Question 1.
Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
(ii) {{2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
(iii) {(1, 3), (1, 5), (2, 5)}.
Solution.
(i) We have a relation R = {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)} Since 2, 5, 8, 11, 14, 17 are the elements of domain of R having their unique images.
∴ The given relation is a function.
Hence domain = {2, 5, 8, 11, 14, 17) and Range = {1}.
(ii) We have a relation
R = {(2,1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
Since 2, 4, 6, 8, 10, 12, 14 are the elements of domain of R having their unique images.
∴ The given relation is a function.
Hence domain = {2, 4, 6, 8, 10, 12, 14} and Range = {1, 2, 3, 4, 5, 6, 7}.
(iii) We have a relation R = {(1, 3), (1, 5), (2, 5)}
Since the distinct ordered pairs (1, 3) and (1, 5) have same first element i.e., 1 does not have a unique image under R.
∴ It is not a function.
Ex 2.3 Class 11 Maths Question 2.
Find the domain and range of the following real functions:
(i) f(x) = \(\left x \right \)
(ii) f(x) = \(\sqrt { 9{ x }^{ 2 } } \)
Solution.
Ex 2.3 Class 11 Maths Question 3.
A function f is defined by f (x) = 2x – 5. Write down the values of
(i) f (0)
(ii) f (7)
(iii) f (3)
Solution.
We are given f (x) = 2x – 5
(i) f (0) = 2(0) – 5 = 0 5 = 5
(ii) f (7) = 2(7) – 5 = 14 5 = 9
(iii) f (3) = 2(3) – 5 = 6 – 5 = 11.
Ex 2.3 Class 11 Maths Question 4.
The function T which maps temperature in degree Celsius into temperature in degree by
\(t(C)=\frac { 9C }{ 5 } +32\)
Find
(i) t (0)
(ii) t (28)
(iii) t (10)
(iv) The value of C, when t (C = 212
Solution.
Ex 2.3 Class 11 Maths Question 5.
Find the range of each of the following functions.
(i) f(x) = 2 – 3x, x ∈ R, x>0.
(ii) f(x)=x^{2}+ 2, x is a real number.
(iii) f (x) = x, x is a real number.
Solution.
(i) Given f (x) = 2 – 3x, x ∈ R, x > 0
∵ x > 0 ⇒ 3x < 0 ⇒ 2 – 3x < 2 + 0 ⇒ f (x) < 2
∴ The range of f (x) is (2).
(ii) Given f (x) = x^{2} + 2, x is a real number
We know x^{2}≥ 0 ⇒ x^{2} + 2 ≥ 0 + 2
⇒ x^{2} + 2 > 2 ∴ f (x) ≥ 2
∴ The range of f (x) is [2, ∞).
(iii) Given f (x) = x, x is a real number.
Let y =f (x) = x ⇒ y = x
∴ Range of f (x) = Domain of f (x)
∴ Range of f (x) is R.
NCERT Class 11 Maths
Class 11 Maths Chapters  Maths Class 11 Chapter 2
Chapterwise NCERT Solutions for Class 11 Maths

NCERT Solutions For Class 11 Maths Chapter 1 Sets
NCERT Solutions For Class 11 Maths Chapter 2 Functions and relations
NCERT Solutions For Class 11 Maths Chapter 3 Trigonometric functions
NCERT Solutions For Class 11 Maths Chapter 4 Principle of mathematical induction
NCERT Solutions For Class 11 Maths Chapter 5 Quadratic equations and complex numbers
NCERT Solutions For Class 11 Maths Chapter 6 Linear Inequalities
NCERT Solutions For Class 11 Maths Chapter 7 Permutations and combinations
NCERT Solutions For Class 11 Maths Chapter 8 Binomial Theorem
NCERT Solutions For Class 11 Maths Chapter 9 Sequences and series
NCERT Solutions For Class 11 Maths Chapter 10 Straight lines
NCERT Solutions For Class 11 Maths Chapter 11 Conic sections
NCERT Solutions For Class 11 Maths Chapter 12 Introduction to threedimensional geometry
NCERT Solutions For Class 11 Maths Chapter 13 Limits and derivatives
NCERT Solutions For Class 11 Maths Chapter 14 Mathematical Reasoning
NCERT Solutions For Class 11 Maths Chapter 15 Statistics
NCERT Solutions For Class 11 Maths Chapter 16 Probability
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