# Plus One Math's Solution Ex 2.1 Chapter2 Relations and Functions

In this section, you’ll learn what Cartesian product of sets is and how to represent it graphically and in the form of sets.

In this chapter, you’ll get a clear understanding of Relations and Functions. Both Relations and Functions have a different meaning in mathematics; however many get confused and use these words interchangeably. A ‘relation’ means a relationship between two elements of a set.  It is a set of inputs and outputs, denoted as ordered pairs (input, output). We can also represent a relation as a mapping diagram or a graph. A relation can either be symbolised by Roster method or Set-builder method. On the other hand, a ‘function’ is a special type of relation, in which each input is related to a unique output. So, all functions are relations, but not all relations are functions.

 Board SCERT, Kerala Text Book NCERT Based Class Plus One Subject Math's Textbook Solution Chapter Chapter 2 Exercise Ex 2.1 Chapter Name Relations and Functions Category Plus One Kerala

## Kerala Syllabus Plus One Math's Textbook Solution Chapter  2 Relations and Functions Exercises 2.1

### Chapter  2  Relations and Functions Textbook Solution

Kerala plus One maths NCERT textbooks, we provide complete solutions for the exercise and answers provided at the end of each chapter. We also cover the entire syllabus given by the Board of secondary education, Kerala state.

### Chapter  2  Relations and Functions Exercise   2.1

If, find the values of x and y.

It is given that.

Since the ordered pairs are equal, the corresponding elements will also be equal.

Therefore,  and.

∴ x = 2 and y = 1

If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A × B)?

It is given that set A has 3 elements and the elements of set B are 3, 4, and 5.

⇒ Number of elements in set B = 3

Number of elements in (A × B)

= (Number of elements in A) × (Number of elements in B)

= 3 × 3 = 9

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

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

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

We know that the Cartesian product P × Q of two non-empty sets P and Q is defined as

P × Q = {(pq): p∈ P, q ∈ Q}

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

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

State whether each of the following statement are true or false. If the statement is false, rewrite the given statement correctly.

(i) If P = {mn} and Q = {nm}, then P × Q = {(mn), (nm)}.

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

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

(i) False

If P = {mn} and Q = {nm}, then

P × Q = {(mm), (mn), (n, m), (nn)}

(ii) True

(iii) True

If A = {–1, 1}, find A × A × A.

It is known that for any non-empty set A, A × A × A is defined as

A × A × A = {(abc): ab∈ A}

It is given that 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)}

If A × B = {(ax), (ay), (bx), (by)}. Find A and B.

It is given that A × B = {(ax), (a, y), (bx), (by)}

We know that the Cartesian product of two non-empty sets P and Q is defined as P × Q = {(pq): p ∈ P, q ∈ Q}

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

Thus, A = {ab} and B = {xy}

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)

(ii) A × C is a subset of B × D

(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)

We have B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Î¦

∴L.H.S. = A × (B ∩ C) = A × Î¦ = Î¦

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)}

∴ R.H.S. = (A × B) ∩ (A × C) = Î¦

∴L.H.S. = R.H.S

Hence, A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) To 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)}

We can observe that all the elements of set A × C are the elements of set B × D.

Therefore, A × C is a subset of B × D.

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

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

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

⇒ n(A × B) = 4

We know that if C is a set with n(C) = m, then n[P(C)] = 2m.

Therefore, the set A × B has 24 = 16 subsets. These are

Î¦, {(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)}

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 xy and z are distinct elements.

It is given that n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A × B.

We know that 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.

∴ xy, and z are the elements of A; and 1 and 2 are the elements of B.

Since n(A) = 3 and n(B) = 2, it is clear that A = {xyz} and B = {1, 2}.

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.

We know that if n(A) = and n(B) = q, then n(A × B) = pq.

∴ n(A × A) = n(A) × n(A)

It is given that n(A × A) = 9

∴ n(A) × n(A) = 9

⇒ n(A) = 3

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

We know that A × A = {(a, a): a ∈ A}. Therefore, –1, 0, and 1 are elements of A.

Since n(A) = 3, it is clear that A = {–1, 0, 1}.

The remaining elements of set A × A are (–1, –1), (–1, 1), (0, –1), (0, 0),

(1, –1), (1, 0), and (1, 1)

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