Maths Solve the following Question.(1 Marks)

B ∩ C = {}

U = {x/x ∈ N, 1 ≤ x ≤ 12} = {1, 2, 3, …., 12}

A = {1, 4, 7, 10}, B = {2, 4, 6, 7, 11}, C = {3, 5, 8, 9, 12}

(i) A ∪ B = {1, 2, 4, 6, 7, 10, 11}

Let C = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday} ∴ C = {x/x is a day of a week}

Let B = {a, e, i, o, u} ∴ B = {x/x is a vowel of English alphabets}

Let A = {10, 20, 30, 40, 50}

∴ A = {x/x = 10n, n ∈ N and n ≤ 5}

Here, (1, 2), (2, 3) ∈ R,

But (1, 3) ∉ R.

∴ R is not transitive.

Here, (1, 2) ∈ R, but (2, 1) ∉ R. ∴ R is not symmetric.

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}

(i) Here, (x, x) ∈ R, for x ∈ {1, 2, 3}

∴ R is reflexive.

R = {(a, b) / a ∈ N, a < 5, b = 4} ∴ Domain (R) = {a / a ∈ N, a < 5} = {1, 2, 3, 4} Range (R) = {b / b = 4} = {4}

B ∪ C = {2, 3, 4, 5, 6, 7, 8, 9, 11, 12} ∴ A ∩ (B ∪ C) = {4, 7}

A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

C’ = {1, 2, 4, 6, 7, 10, 11} ∴ B ∩ C’ = {2, 4, 6, 7, 11}

A – B = {1, 10}

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)}

(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

A – B = (-7, 2)

B – C = [2, 4)

B’ = (-∞, 2) ∪ (6, ∞) C’ = (-∞, 4) ∪ (9, ∞) ∴ B’ ∩ C’ = (-∞, 2) ∪ (9, ∞)

A’ = (-∞, – 7] ∪ (3, ∞) ∴ A’ ∩ B = (3, 6]

A ∩ C = { }

B ∩ C = [4, 6]

A ∩ B = [2, 3]

A ∪ C = (-7, 3] ∪ [4, 9]

B ∪ C = [2, 9]

A = (-7, 3], B = [2, 6], C = [4, 9]

A ∪ B = (-7, 6]

-9 < 2x + 7 ≤ 19 ∴ -16 < 2x ≤ 12 ∴ -8< x ≤ 6 ∴ x ∈ (-8, 6]

[-3, 4) = {x / x ∈ R, -3 ≤ x < 4}

(2, 5] = {x / x ∈ R, 2 < x ≤ 5}

(-∞, 5] = {x / x ∈ R, x ≤ 5}

(6, ∞) = {x / x ∈ R, x > 6}

[6, 12] = {x / x ∈ R, 6 ≤ x ≤ 12}

(-3, 0) = {x / x ∈ R, -3 < x < 0}

A = {1, 2, 3}

The power set of A is given by

P(A) = {{Φ}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}

n(A’ ∩ B) = n(B) – n(A ∩ B) = 20 – 10 = 10

n(A’ ∩ B) = n(B) – n(A ∩ B) = 20 – 10 = 10

n(A ∩ B) = n(A) + n(B) – n(A ∪ B) = 35 + 20 – 45 = 10

n(X) = 50, n(A) = 35, n(B) = 20, n(A’ ∩ B’) = 5 (i) n(A ∪ B) = n(X) – [n(A ∪ B)’] = n(X) – n(A’ ∩ B’) = 50 – 5 = 45

B = {3, 4, 5, 6} ∴ n(B) = 4 …..(i) A’ = {5, 6, 7, 8, 9, 10} A’ ∩ B = {5, 6} ∴ n(A’ ∩ B) = 2 A ∩ B = {3, 4} ∴ n(A ∩ B) = 2 ∴ n(A’ ∩ B) + n(A ∩ B) = 2 + 2 = 4 …..(ii) From (i) and (ii), we get n(B) = n(A’ ∩ B) + n (A ∩ B)

A = {1, 2, 3, 4}, B = {3, 4, 5, 6} A ∩ B = {3, 4}, A ∪ B = {1, 2, 3, 4, 5, 6} ∴ n(A) = 4, n(B) = 4, n(A ∩ B) = 2, n(A ∪ B) = 6 ……(i) ∴ n(A) + n(B) – n(A ∩ B) = 4 + 4 – 2 ∴ n(A) + n(B) – n(A ∩ B) = 6 …..(ii) From (i) and (ii), we get n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

B – C = {3} C – B = {7, 8} B Δ C = (B – C) ∪ (C – B) = {3, 7, 8} ∴ A ∩ (B Δ C) = {3} ……(i) A ∩ B = {3, 4} A ∩ C = {4} ∴ (A ∩ B) Δ (A ∩ C) = [(A ∩ B) – (A ∩ C)] ∪ [(A ∩ C) – (A ∩ B)] = {3} …..(ii) From (i) and (ii), we get A ∩ (B Δ C) = (A ∩ B) Δ (A ∩ C)

A ∪ B = {1, 2, 3, 4, 5, 6} …….(i) A – B = {1, 2} A ∩ B = {3, 4} B – A = {5, 6} ∴ (A – B) ∪ (A ∩ B) ∪ (B – A) = {1, 2, 3, 4, 5, 6} ……(ii) From (i) and (ii), we get A ∪ B = (A – B) ∪ (A ∩ B) ∪ (B – A)

B = {3, 4, 5, 6} …..(i) A ∩ B = {3, 4} A’ = {5, 6, 7, 8, 9, 10} A’ ∩ B = {5, 6} ∴ (A ∩ B) ∪ (A’ ∩ B) = {3, 4, 5, 6} …..(ii) From (i) and (ii), we get B = (A ∩ B) ∪ (A’ ∩ B)

A ∩ B = {3, 4} (A ∩ B)’= {1, 2, 5, 6, 7, 8, 9, 10} …….(i) A’ = {5, 6, 7, 8, 9, 10} B’ = {1, 2, 7, 8, 9, 10} ∴ A’ ∪ B’ = {1, 2, 5, 6, 7, 8, 9, 10} …….(ii) From (i) and (ii), we get (A ∩ B)’ = A’ ∪ B’

A ∪ B = {1, 2, 3, 4, 5, 6} ∴ (A ∪ B)’ = {7, 8, 9, 10} ………(i) A’ = {5, 6, 7, 8, 9, 10}, B’ = {1, 2, 7, 8, 9,10} ∴ A’ ∩ B’ = {7, 8, 9, 10} …….(ii) From (i) and (ii), we get (A ∪ B)’ = A’ ∩ B’

B ∪ C = {3, 4, 5, 6, 7, 8} ∴ A ∩ (B ∪ C) = {3, 4} ………(i) A ∩ B = {3, 4} A ∩ C = {4} ∴ (A ∩ B) ∪ (A∩ C) = {3, 4} ……..(ii) From (i) and (ii), we get A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8}, X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} (i) B ∩ C = {4, 5, 6} ∴ A ∪ (B ∩ C) = {1, 2, 3, 4, 5, 6} …..(i) A ∪ B = {1, 2, 3, 4, 5, 6} A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8} ∴ (A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6} …….(ii) From (i) and (ii), we get A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

C = {3, 5, 7, 9, … }

B = {-1, 0, 1, 2, 3, 4}

A = {M, O, V, E, N, T}