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Exercise 2.1

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Comparing both side as order pairs are equal, so corresponding elements also will be equal,  

Q2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A×B).

Solution: 

Given : n(A) = 3 and set B = {3, 4, 5}

∴ n(B) = 3 

Number of elements in (A × B)

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

= n(A) × n(B) 

= 3 × 3

= 9 

Q3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.

Solution: 

Given: 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 = {(p, q): 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)} 

Q4. 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)}.

(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.

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

Solution: 

(i) If P = {m, n} and Q = { n, m}, then P × Q = {(m, n),(n, m)}.

False

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

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

Solution: 

(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

Solution: 

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

(iii) True 

Q5. If A = {–1, 1}, find A × A × A.

Solution: 

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

A × A × A = {(a, b, c): a, b, c ∈ A}

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)} 

Q6. If A × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.

Solution: 

The cartesian product of two non-empty sets P and Q is defined as P × Q = {(p, q): p ∈ P, q ∈ Q} 

Given that: A × B = {(a, x),(a , y), (b, x), (b, y)}

∴ a, b ∈ A abd x, y ∈ B 

So, A = {a, b} and B = {x, y} 

Q7. 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.

Solution: 

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

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

∴ 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)}

∴ R.H.S. = (A × B) ∩ (A × C) = Φ

∴ 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

Solution:

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

A × 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)}

All the elements of set A × C are the elements of set B × D.

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

Q8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.

Solution: 

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)] = 2^m.

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

These are 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)}

Q9. 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.

Solution: 

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.

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

Since n(A) = 3 and n(B) = 2,

∴ A = {x, y, z} and B = {1, 2}. 

Q10. 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.

Solution: 

We know that if n(A) = p 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). 

 

Exercise 2.2

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Q1. Let A = {1, 2, 3,...,14}. Define a relation R from A to A by R = {(x, y) : 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and range.

Solution: 

The relation R from A to A is given as R = {(x, y): 3x – y = 0, where x, y ∈ A} 

Given, R = {(x, y): 3x - y = 0, where x, y ∈ A} 

3x - y = 0 

∴ 3x = y 

Putting value x= 1, 2, 3, 4 from A we have y = 3, 6, 9, 12 

∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)}

The domain of R is the set of all first elements of the ordered pairs in the relation R.

∴ Domain of R = {1, 2, 3, 4}

The whole set A is the codomain of the relation R.

∴ Co-domain of R = {1, 2, 3… 14}

The range of R is the set of all second elements of the ordered pairs in the relation R.

∴Range of R = {3, 6, 9, 12} 

Q2. 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:

R = {(x, y): y = x + 5, x is a natural number less than 4, x, y ∈ N}

The natural numbers less than 4 are 1, 2, and 3.

∴ R = {(1, 6), (2, 7), (3, 8)}

The domain of R is the set of all first elements of the ordered pairs in the relation R.

∴ Domain of R = {1, 2, 3}

The range of R is the set of all second elements of the ordered pairs in the relation R.

∴ Range of R = {6, 7, 8} 

Q3. 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 ∈ A, y ∈ B}. Write R in roster form.

Solution: 

A = {1, 2, 3, 5} and B = {4, 6, 9}

R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}

∴ R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)} 

Q4. The Fig 2.7 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: From the given figure, we have

A = {5, 6, 7}, B = {3, 4, 5}

(i)R = {(x, y): y = x – 2; x ∈ A and y ∈ B} or R = {(x, y): y = x – 2 for x = 5, 6, 7}

(ii) R = {(5, 3), (6, 4), (7, 5)} Domain of R = {5, 6, 7} Range of R = {3, 4, 5}

Q5. 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: 

A = {1, 2, 3, 4, 6}, R = {(a, b): a, b ∈ A, b is exactly divisible by a}

(i) 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}

(iii) Range of R = {1, 2, 3, 4, 6} 

Q6. Determine the domain and range of the relation R defined by R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.

Solution: 

R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}

∴ R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}

∴ Domain of R = {0, 1, 2, 3, 4, 5} Range of R = {5, 6, 7, 8, 9, 10} 

Q7. Determine the domain and range of the relation R defined by R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.

Solution: 

R = {(x, x3): x is a prime number less than 10}

The prime numbers less than 10 are 2, 3, 5, and 7.

∴ R = {(2, 8), (3, 27), (5, 125), (7, 343)} 

Q8. 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}.

∴ A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}

Since n(A × B) = 6,

The number of subsets of A × B is 2⁶.

∴ the number of relations from A to B is 2⁶. 

Q9. 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: 

R = {(a, b): a, b ∈ Z, a – b is an integer}

It is known that the difference between any two integers is always an integer.

∴ Domain of R = Z

   Range of R = Z 

Exercise 2.3 

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Q1. 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) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

Let be R = {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)} 

There is no same component in the domain of given relation, so this relation is a function. 

Here, domain = {2, 5, 8, 11, 14, 17} and range = {1}

Solution: 

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

Let be R = {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)} 

There are also unique domain components or order pair in the given relation, so this relation is a function.

Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}

Solution: 

(iii) {(1, 3), (1, 5), (2, 5)}

Since the domain has the same first element i.e., 1 corresponds to two different images i.e.,

3 and 5, this relation is not a function. 

Q2. Find the domain and range of the following real functions:

(i) f(x) = – |x|

Solution2: 

Q3:  A function f is defined by f(x) = 2x – 5

(i) f (0)                      (ii) f(7)                        (iii) f(-3)

Solution 3:

The given function is f(x) = 2x – 5

Therefore,

 (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

Solution:

Solution:

(iii) f(x) = x is real number.

It is clear that the range of f is the set of all real number.

Therefore,  range of f = R

 

MISCELLANEOUS EXERCISE

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