# NCERT Solutions | Class 11 Maths Chapter 2 | Functions and relations

## CBSE Solutions | Maths Class 11

Check the below NCERT Solutions for Class 11 Maths Chapter 2 Functions and relations Pdf free download. NCERT Solutions Class 11 Maths  were prepared based on the latest exam pattern. We have Provided Functions and relations Class 11 Maths NCERT Solutions to help students understand the concept very well.

### NCERT | Class 11 Maths

Book: National Council of Educational Research and Training (NCERT) Central Board of Secondary Education (CBSE) 11th Maths 2 Functions and relations English

#### Functions and relations | Class 11 Maths | NCERT Books Solutions

You can refer to MCQ Questions for Class 11 Maths Chapter 2 Functions and relations to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

## 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 non-empty sets, then Ax B is a non-empty 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 24 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 set-builder 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, x3): x is a prime number less than 10} in roster form.
Solution.
Given relation is R = {(x, x3): x is a prime number less than 10)
= {(x, x3): 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 = 26 = 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)=x2+ 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) = x2 + 2, x is a real number
We know x2≥ 0 ⇒ x2 + 2 ≥ 0 + 2
⇒ x2 + 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.