Contents

NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 11 |

Subject |
Maths |

Chapter |
Chapter 2 |

Chapter Name |
Relations and Functions |

Exercise |
Ex 2.1, Ex 2.2, Ex 2.3 |

Number of Questions Solved |
24 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions

**NCERT Exercises**

### Chapter 2 Relations and Functions Exercise – 2.1

**Question 1.**

If , find the values of x and y.

**Solution.**

Since the ordered pairs are equal. So, the corresponding elements are equal

∴ and

⇒ and ⇒ x = 2 and y = 1.

**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).

**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)}.

**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.

**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)}

**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}.

**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}

**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)}.

**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}.

**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).

### Chapter 2 Relations and Functions Exercise – 2.2

**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, ∴ ⇒ 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}.

**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}.

**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)}.

**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.

**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.

**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}.

**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)}.

**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.

**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.

### Chapter 2 Relations and Functions Exercise – 2.3

**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.

**Question 2.**

Find the domain and range of the following real functions:

(i) f(x) =

(ii) f(x) =

**Solution.**

**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.

**Question 4.**

The function T which maps temperature in degree Celsius into temperature in degree by

Find

(i) t (0)

(ii) t (28)

(iii) t (-10)

(iv) The value of C, when t (C = 212

**Solution.**

**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.

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