Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 5 Sets and Relations Ex 5.2 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 5 Sets and Relations Ex 5.2

Question 1.

If (x – 1, y + 4) = (1, 2), find the values of x and y.

Solution:

(x – 1, y + 4) = (1, 2)

By the definition of equality of ordered pairs, we have

x – 1 = 1 and y + 4 = 2

∴ x = 2 and y = -2

Question 2.

If \(\left(x+\frac{1}{3}, \frac{y}{3}-1\right)=\left(\frac{1}{2}, \frac{3}{2}\right)\), find x and y.

Solution:

\(\left(x+\frac{1}{3}, \frac{y}{3}-1\right)=\left(\frac{1}{2}, \frac{3}{2}\right)\)

By the definition of equality of ordered pairs, we have

x + \(\frac{1}{3}\) = \(\frac{1}{2}\) and \(\frac{y}{3}\) – 1 = \(\frac{3}{2}\)

∴ x = \(\frac{1}{2}\) – \(\frac{1}{3}\) and \(\frac{y}{3}\) = \(\frac{3}{2}\) + 1

∴ x = \(\frac{1}{6}\) and y = \(\frac{15}{2}\)

Question 3.

If A = {a, b, c}, B = {x, y}, find A × B, B × A, A × A, B × B.

Solution:

A = (a, b, c}, B = {x, y}

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

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

A × A = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}

B × B = {(x, x), (x, y), (y, x), (y, y)}

Question 4.

If P = {1, 2, 3} and Q = {1, 4}, find sets P × Q and Q × P.

Solution:

P = {1, 2, 3}, Q = {1, 4}

∴ P × Q = {(1, 1), (1, 4), (2, 1), (2, 4), (3, 1), (3, 4)}

and Q × P = {(1, 1), (1, 2), (1, 3), (4, 1), (4, 2), (4, 3)}

Question 5.

Let A = {1, 2, 3, 4}, B = {4, 5, 6}, C = {5, 6}. Verify,

(i) A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) A × (B ∪ C) = (A × B) ∪ (A × C)

Solution:

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

(i) B ∩ C = {5, 6}

A × (B ∩ C) = = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}

A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}

A × C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}

∴ (A × B) ∩ (A × C) = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}

∴ A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) B ∪ C = {4, 5, 6}

A × (B ∪ C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3,4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}

A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}

A × C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}

∴ (A × B) ∪ (A × C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}

∴ A × (B ∪ C) = (A × B) ∪ (A × C)

Question 6.

Express {(x, y) / x^{2} + y^{2} = 100, where x, y ∈ W} as a set of ordered pairs.

Solution:

{(x, y) / x^{2} + y^{2} = 100, where x, y ∈ W}

We have, x^{2} + y^{2} = 100

When x = 0 and y = 10,

x^{2} + y^{2} = 0^{2} + 10^{2} = 100

When x = 6 and y = 8,

x^{2} + y^{2} = 6^{2} + 8^{2} = 100

When x = 8 and y = 6,

x^{2} + y^{2} = 8^{2} + 6^{2} = 100

When x = 10 and y = 0,

x^{2} + y^{2} = 10^{2} + 0^{2} = 100

∴ Set of ordered pairs = {(0, 10), (6, 8), (8, 6), (10, 0)}

Question 7.

Let A = {6, 8} and B = {1, 3, 5}. Show that R_{1} = {(a, b) / a ∈ A, b ∈ B, a – b is an even number} is a null relation, R_{2} = {(a, b) / a ∈ A, b ∈ B, a + b is an odd number} is a universal relation.

Solution:

A = {6, 8}, B = {1, 3, 5}

R_{1} = {(a, b)/ a ∈ A, b ∈ B, a – b is an even number}

a ∈ A

∴ a = 6, 8

b ∈ B

∴ b = 1, 3, 5

When a = 6 and b = 1, a – b = 5, which is odd

When a = 6 and b = 3, a – b = 3, which is odd

When a = 6 and b = 5, a – b = 1, which is odd

When a = 8 and b = 1, a – b = 7, which is odd

When a = 8 and b = 3, a – b = 5, which is odd

When a = 8 and b = 5, a – b = 3, which is odd

Thus, no set of values of a and b gives a – b as even.

∴ R_{1} has a null relation from A to B.

A × B = {(6, 1), (6, 3), (6, 5), (8, 1), (8, 3), (8, 5)}

When a = 6 and b = 1, a + b = 7, which is odd

When a = 6 and b = 3, a + b = 9, which is odd

When a = 6 and b = 5, a + b = 11, which is odd

When a = 8 and b = 1, a + b = 9, which is odd

When a = 8 and b = 3, a + b = 11, which is odd

When a = 8 and b = 5, a + b = 13, which is odd

∴ R_{2} = {(6, 1), (6, 3), (6, 5), (8, 1), (8, 3), (8, 5)}

Here, R_{2} = A × B

∴ R_{2} has a universal relation from A to B.

Question 8.

Write the relation in the Roster form. State its domain and range.

(i) R_{1} = {(a, a^{2}) / a is a prime number less than 15}

(ii) R_{2} = {(a, \(\frac{1}{a}\)) / 0 < a ≤ 5, a ∈ N}

(iii) R_{3} = {(x, y / y = 3x, y ∈ {3, 6, 9, 12}, x ∈ {1, 2, 3}}

(iv) R_{4} = {(x, y) / y > x + 1, x = 1, 2 and y = 2, 4, 6}

(v) R_{5} = {(x, y) / x + y = 3, x, y ∈ {0, 1, 2, 3}}

(vi) R_{6} = {(a, b) / a ∈ N, a < 6 and b = 4}

(vii) R_{7} = {(a, b) / a, b ∈ N, a + b = 6}

(viii) R_{8} = {(a, b)/ b = a + 2, a ∈ Z, 0 < a < 5}

Solution:

(i) R_{1} = {(a, a^{2}) / a is a prime number less than 15}

∴ a = 2, 3, 5, 7, 11, 13

∴ a^{2} = 4, 9, 25, 49, 121, 169

∴ R_{1} = {(2, 4), (3, 9), (5, 25), (7, 49), (11, 121), (13, 169)}

∴ Domain (R_{1}) = {a/a is a prime number less than 15}

= {2, 3, 5, 7, 11, 13}

Range (R_{1}) = {a^{2}/a is a prime number less than 15}

= {4, 9, 25, 49, 121, 169}

(iii) R_{3} = {(x, y) / y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}}

Here y = 3x

When x = 1, y = 3(1) = 3

When x = 2, y = 3(2) = 6

When x = 3, y = 3(3) = 9

∴ R_{3} = {(1, 3), (2, 6), (3, 9)}

∴ Domain (R_{3}) ={1, 2, 3}

∴ Range (R_{3}) = {3, 6, 9}

(iv) R_{4} = {(x, y) / y > x + 1, x = 1, 2 and y = 2, 4, 6}

Here, y > x + 1

When x = 1 and y = 2, 2 ≯ 1 + 1

When x = 1 and y = 4, 4 > 1 + 1

When x = 1 and y = 6, 6 > 1 + 1

When x = 2 and y = 2, 2 ≯ 2 + 1

When x = 2 and y = 4, 4 > 2 + 1

When x = 2 and y = 6, 6 > 2 + 1

∴ R_{4} = {(1, 4), (1, 6), (2, 4), (2, 6)}

Domain (R_{4}) = {1, 2}

Range (R_{4}) = {4, 6}

(v) R_{5} = {{x, y) / x + y = 3, x, y ∈ (0, 1, 2, 3)}

Here, x + y = 3

When x = 0, y = 3

When x = 1, y = 2

When x = 2, y = 1

When x = 3, y = 0

∴ R_{5} = {(0, 3), (1, 2), (2, 1), (3, 0)}

Domain (R_{5}) = {0, 1, 2, 3}

Range (R_{5}) = {3, 2, 1, 0}

(vi) R_{6} = {(a, b)/ a ∈ N, a < 6 and b = 4}

a ∈ N and a < 6

∴ a = 1, 2, 3, 4, 5 and b = 4

R_{6} = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4)}

Domain (R_{6}) = {1, 2, 3, 4, 5}

Range (R_{6}) = {4}

(vii) R_{7} = {(a, b) / a, b ∈ N, a + b = 6}

Here, a + b = 6

When a = 1, b = 5

When a = 2, b = 4

When a = 3, b = 3

When a = 4, b = 2

When a = 5, b = 1

∴ R_{7} = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}

Domain (R_{7}) = {1, 2, 3, 4, 5}

Range (R_{7}) = {5, 4, 3, 2, 1}

(viii) R_{8} = {(a, b) / b = a + 2, a ∈ Z, 0 < a < 5}

Here, b = a + 2

When a = 1, b = 3

When a = 2, b = 4

When a = 3, b = 5

When a = 4, b = 6

∴ R_{8} = {(1, 3), (2, 4), (3, 5), (4, 6)}

Domain (R_{8}) = {1, 2, 3, 4}

Range (R_{8}) = {3, 4, 5, 6}

Question 9.

Identify which of the following relations are reflexive, symmetric, and transitive.

Solution:

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

Let a ∈ Z, then a – a ∈ Z

∴ (a, a) ∈ R

∴ R is reflexive.

Let (a, b) ∈ R

∴ a – b ∈ Z

∴ -(a – b) ∈ Z, i.e., b – a ∈ Z

∴ (b, a) ∈ R

∴ R is symmetric.

Let (a, b) and (b, c) ∈ R

∴ a – b ∈ Z and b – c ∈ Z

∴ (a – b) + (b – c) ∈ Z

∴ a – c ∈ Z

∴ (a, c) ∈ R

∴ R is transitive.

(ii) Given, R = {(a, b) : a, b ∈ N, a + b is even}

Let a ∈ N, then a + a = 2a, which is even.

∴ (a, a) ∈ R

∴ R is reflexive.

Let (a, b) ∈ R

∴ a + b is even

∴ b + a is even

∴ (b, a) ∈ R

∴ R is symmetric.

Let (a, b) and (b, c) ∈ R

∴ a + b and b + c is even

Let a + b = 2x and b + c = 2y for x, y ∈ N

∴ (a + b) + (b + c) = 2x + 2y

∴ a + 2b + c = 2(x + y)

∴ a + c = 2(x + y) – 2b = 2(x + y – b)

∴ a + c is even ……..[∵ x, y, b ∈ N, x + y – b ∈ N]

∴ (a, c) ∈ R

∴ R is transitive.

(iii) Given, R = {(a, b) : a, b ∈ N, a divides b}

Let a ∈ N, then a divides a.

∴ (a, a) ∈ R

∴ R is reflexive.

Let a = 2 and b = 8, then 2 divides 8

∴ (a, b) ∈ R

But 8 does not divide 2.

∴ (b, a) ∉ R

∴ R is not symmetric.

Let (a, b) and (b, c) ∈ R

∴ a divides b and b divides c.

Let b = ax and c = by for x, y ∈ N.

∴ c = (ax) y = a(xy)

i.e., a divides c.

∴ (a, c) ∈ R

∴ R is transitive.

(iv) Given, R = {(a, b) : a, b ∈ N, a^{2} – 4ab + 3b^{2} = 0}

Let a ∈ N, then a^{2} – 4aa + 3a^{2} = a^{2} – 4a^{2} + 3a^{2} = 0

∴ (a, a) ∈ R

∴ R is reflexive.

Let a = 3 and b = 1,

then a^{2} – 4ab + 3b^{2} = 9 – 12 + 3 = 0

∴ (a, b) ∈ R

Consider, b^{2} – 4ba + 3a^{2} = 1 – 12 + 9 = -2 ≠ 0

∴ (b, a) ∉ R

∴ R is not symmetric.

Let a = 3, b = 1 and c = \(\frac{1}{3}\),

then a^{2} – 4ab + 3b^{2} = 9 – 12 + 3 = 0 and

b^{2} – 4bc + 3c^{2} = 1 – \(\frac{4}{3}\) + \(\frac{1}{3}\) = 1 – 1 = 0

∴ we get (a, b) and (b, c) ∈ R.

Consider, a^{2} – 4ac + 3c^{2} = 9 – 4 + \(\frac{1}{3}\) = \(\frac{16}{3}\) ≠ 0

∴ (a, c) ∉ R

∴ R is not transitive.

(v) Given, R = {(a, b) : a is sister of b and a, b ∈ G = Set of girls}

Let a ∈ G, then ‘a’ cannot be a sister of herself.

∴ (a, a) ∉ R

∴ R is not reflexive.

Let (a, b) ∈ R

∴ ‘a’ is a sister of ‘b’.

∴ ‘b’ is a sister of ‘a’.

∴ (b, c) ∈ R

∴ R is symmetric.

Let (a, b) and (b, c) ∈ R

∴ ‘a’ is a sister of ‘b’ and ‘b’ is a sister of ‘c’

∴ ‘a’ is a sister of ‘c’.

∴ (a, c) ∈ R

∴ R is transitive.

(vi) Given, R = {(a, b) : Line a is perpendicular to line b in a plane}

Let a be any line in the plane, then a cannot be perpendicular to itself.

∴ (a, a) ∉ R

∴ R is not reflexive.

Let (a, b) ∈ R

∴ a is perpendicular to b.

∴ b is perpendicular to a.

∴ (b, a) ∈ R.

∴ R is symmetric.

Let (a, b) and (b, c) ∈ R.

∴ a is perpendicular to b and b is perpendicular to c.

∴ a is parallel to c.

∴ (a, c) ∉ R

∴ R is not transitive.

(vii) Given, R = {(a, b) : a, b ∈ R, a < b}

Let a ∈ R, then a ≮ a.

∴ (a, a) ∉ R

∴ R is not reflexive.

Let a = 1 and b = 2, then 1 < 2

∴ (a, b) ∈ R

But 2 ≮ 1

∴ (b, a) ∉ R

∴ R is not symmetric.

Let (a, b) and (b, c) ∈ R

∴ a < b and b < c

∴ a < c

∴ (a, c) ∈ R

∴ R is transitive.

(viii) Given, R = {(a, b) : a, b ∈ R, a ≤ b^{3}}

Let a = -3, then a^{3} = -27.

Here, a ≮ a

∴ (a, a) ∉ R

∴ R is not reflexive.

Let a = 2 and b = 9, then b^{3} = 729

Here, a < b^{3}

∴ (a, b) ∈ R

Consider, a3 = 8

Here, b ≮ a^{3}

∴ (b, a) ∉ R

∴ R is not symmetric.

Let a = 10, b = 3, c = 2,

then b^{3} = 27 and c^{3} = 8

Here, a < b^{3} and b < c^{3}.

∴ (a, b) and (b, c) ∈ R

But a ≮ c^{3}

∴ (a, c) ∉ R.

∴ R is not transitive.