**Exercise 2.1 Relations and Functions**

**Question and Answers**

**Class 11 – Maths**

Class | Class 11 |

Subject | Mathematics |

Chapter Name | Relations and Functions |

Chapter No. | Chapter 2 |

Exercise | Exercise 2.1 |

Category | Class 11 Maths NCERT Solutions |

**Question 1 If (x/3 + 1, y-2/3) =(5/3, 1/3) find the values of x and y.**

**Answer **As the ordered pairs are equal, the corresponding elements should also be equal.

Thus, x/3 + 1 = 5/3 and y – 2/3 = 1/3

Solving, we get

x + 3 = 5 and 3y – 2 = 1 [Taking L.C.M. and adding]

x = 2 and 3y = 3

Therefore,

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×B).**

**Answer** The number of elements in (A×B) = Number of elements in A × Number of elements in B

The number of elements in (A×B) = 3 × 3=9

So, the number of elements in (A×B) is 9.

**Question 3 If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.**

**Answer **G = {7, 8} and H = {5, 4, 2}

The Cartesian product of two non-empty sets A and B is defined as

A × B = {(a, b) :a ∈ A and b ∈ B

So, the value of G×H will be,

G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}

The value of H×G will be,

H × 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 × Q = {(m, n),(n, m)}.**

**(i)** The statement is False.

If P = {*m*, *n*} and Q = {*n*, *m*}, then

P × Q = {(*m*, *m*), (*m*, *n*), (*n,* *m*), (*n*, *n*)}

**(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.**

**(ii)** True

** (iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ.**

**(iii)** True

**Question 5 If A = {–1, 1}, find A × A × A.**

**Answer **The A × A × A for a non-empty set A is given by

A × A × A = {(*a*, *b*, *c*): *a*, *b*, *c *∈ A}

Here, it is given A = {–1, 1}

A × A × A = {(–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 × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.**

**Answer** A × B = {(a,x), (a,y), (b,x),( b,y)}

The Cartesian product of two non-empty sets A and B is defined as

A × B = {(a ,b) : a ∈ A and b ∈ B}

Hence, A is the set of all first elements, and B is the set of all second elements.

Therefore, 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 × (B ∩ C) = (A × B) ∩ (A × C).**

**(i)** A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

Verify : A × (B ∩ C) = (A × B) ∩ (A × C)

Now, B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ

Thus , L.H.S. = A × (B ∩ C) = A × Φ = Φ

A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

Thus, R.H.S. = (A × B) ∩ (A × C) = Φ

Therefore, L.H.S. = R.H.S.

Hence verified

**(ii) A × C is a subset of B × D.**

**(ii)** A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

Verify : A × C is a subset of B × D

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}

Now, it’s clearly seen that all the elements of set A × C are the elements of set B × D.

Thus, A × C is a subset of B × D.

Hence verified

**Question 8 Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.**

**Answer **A = {1, 2} and B = {3, 4}

A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}

Number of elements in A × B is *n *(A × B) = 4

We know that, If C is a set with *n*(C) = *m*, then *n*[P(C)] = 2* ^{m}*.

Thus, the set A × B has 2^{4} = 16 subsets.

And these subsets are as given below:

Φ, {(1, 3)}, {(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, 3), (2, 3), (2, 4)}, {(1, 4), (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 × B, find A and B, where x, y and z are distinct elements.**

**Answer **We are given that n(A) = 3 and n(B) = 2

(x, 1), (y, 2), (z, 1) are in A × B

A = Set of first elements of the ordered pair elements of A × B

B = Set of second elements of the ordered pair elements of A × B

So, clearly, *x*, *y*, and *z* are the elements of A; and

1 and 2 are the elements of B.

As *n*(A) = 3 and *n*(B) = 2, it is clear that set A = {*x*, *y*, *z*} and set B = {1, 2}

**Question 10 The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0,1). Find the set A and the remaining elements of A × A.**

**Answer** We are provided with, n ( A × A ) = 9

If *n*(A) = *p *and *n*(B) = *q, *then *n*(A × B) = *pq*.

Also, *n*(A × A) = *n*(A) × *n*(A)

*n*(A × A) = 9

So, *n*(A) × *n*(A) = 9

Thus, *n*(A) = 3

Also, given that the ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A × A.

And, we know in A × A = {(*a, a*): *a* ∈ A}

Thus, –1, 0, and 1 have to be the elements of A.

As *n*(A) = 3, clearly A = {–1, 0, 1}

Hence, the remaining elements of set A × A are as follows:

(–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), and (1, 1)

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