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Given sets 

A = {6, 9, 11} and ∅ = { } 

A ∪ ∅ = {6, 9, 11} ∪ { } 

= {6, 9, 11} = A

∴ A ∪ ∅ = A 

A ∩ ∅ = {6,9,11} ∩ { } = { } = ∅

 ∴ A ∩ ∅ = ∅

(i) A’= {d,e,f,g,h}

(ii) B’ = {a,b,c,h}

(iii) C = {b,d,f,h]

(iv) D’ = {b,c,d,e}.

U = {a, b, c, d, e, f, g, h}

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

A' = {d,e,f,g,h}

(ii) B = {d, e, f, g}

B' = {a,b,c,d}

(iii) C = {a, c, e, g}

C' = {b,d,f,h}

(iv) D = {f, g, h, a}

D' = {b,c,d,e}

Given A∪B=A∪C 

⇒(A∪B)∩C = (A∪C)∩C 

⇒(A∩C)∪(B∩C) = C 

⇒(A∩B)∪(B∩C) = C 

∵ A∩B = A∩C 

Now, A∪B = A∪C 

⇒ (A∪B)∩B = (A∪C)∩B 

⇒ (A∩B)u(B∩B) = (A∩B)∪(C∩B) 

⇒ B = (A∩B)∪(B∩C) B = C

(i) X – Y = {a, c)

(ii) Y- X = (f, g)

(iii) X ∩ Y = {b, d}

(i) Let A = {2, 3, 4, 5} and B = {3, 6}

∴ A ∩ B = {3} ∴ A ∩ B ≠ φ

∴Given sets are disjoint sets.

(ii) Let A = {a, e, i, o, u} and B = {a, b, c, d}

A∩B = {a}.

Given sets are disjoint sets.

(iii) Given sets are A = {2, 6, 10, 14} and B = {3, 7, 11, 15}.

∴ A ∩ B =φ

(iv) Given sets are disjoint sets.

Let A = {2, 6, 10} and B = {3, 7, 11}

∴ A ∩ B = φ

∴ Given sets are disjoint sets.

(i) The factors of 35 are 1, 5, 7 and 35. So, 35 is an element of the set. Hence, statement is true.

(ii) The factors of 128 hre 1,2,4, 8, 16, 32, 64 and 128.

Sum of factors = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255 * 2 x 128 Hence, statement is false.

(iii) We have, x⁴ – 5x³ + 2x² – 112x + 6 = 0 Forx = 3, we have

(3)⁴ – 5(3)³ + 2(3)² – 112(3) + 6 = 0

=> 81 – 135 + 18-336 + 6 = 0

=> -346 = 0, which is not true.

So 3 is not an element of the set

Hence, statement is true

iv) 496 = 2⁴ x 31

So, the factors of 496 are 1, 2, 4, 8, 16, 31, 62,124, 248 and 496.

Sum of factors = 1 +2 + 4 + 8+ 16 + 31 + 62 + 124 + 248 + 496 = 992 = 2(496)

So, 496 is the element of the set Hence, statement is false

According to the question,

L = {1, 2, 3, 4}, M = {3, 4, 5, 6} and N = {1, 3, 5}

To verify:

L – (M ∪ N) = (L – M) ∩ (L – N)

M = {3, 4, 5, 6}, N = {1, 3, 5} ⇒ M ∪ N = {1, 3, 4, 5, 6}

L = {1, 2, 3, 4} and M ∪ N = {1, 3, 4, 5, 6}

⇒ L – (M ∪ N) = {2}………………(i)

L = {1, 2, 3, 4} and M = {3, 4, 5, 6} ⇒ L – M = {1, 2}

L = {1, 2, 3, 4} and N = {1, 3, 5} ⇒ L – N = {2, 4}

L – M = {1, 2} and L – N = {2, 4}

⇒ (L – M) ∩ (L – N) = {2}………………(ii)

From equations (i) and (ii),

We have,

L – (M ∪ N) = (L – M) ∩ (L – N)

Hence verified

(i) False 

Since, 37 has exactly two positive factors, 1 and 37, 37 belongs to the set. 

(ii) True 

Since, the sum of positive factors of 28 = 1 + 2 + 4 + 7 + 14 + 28 = 56 = 2(28) 

(iii) False 

7,747 is not a multiple of 37

(i) X – Y = {a, c}

(ii) Y – X = {f, g}

(iii) X ∩ Y = {b, d}

(i) True

According to the question,

35 ∈ {x | x has exactly four positive factors}

The possible positive factors of 35 = 1, 5, 7, 35

35 belongs to given set

Since, 35 has exactly four positive factors

⇒ The given statement 35 ∈ {x | x has exactly four positive factors} is true.

(ii) False

According to the question,

128 ∈ {y | the sum of all the positive factors of y is 2y}

The possible positive factors of 128 are 1, 2, 4, 8, 16, 32, 64, 128

The sum of them

= 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128

= 255

2y = 2 × 128 = 256

Since, the sum of all the positive factors of y is not equal to 2y

128 does not belong to given set

⇒ The given statement 128 ∈ {y | the sum of all the positive factors of y is 2y} is false.

(iii) True

According to the question,

3 ∉ {x | x⁴ – 5x³ + 2x² – 112x + 6 = 0}

x⁴ – 5x³ + 2x² – 112x + 6 = 0

On putting x = 3 in LHS:

(3)⁴ – 5(3)³ + 2(3)² – 112(3) + 6

= 81 – 135 + 18 – 336 + 6

= –366

≠ 0

So, 3 does not belong to given set

⇒ The given statement 3 ∉ {x | x⁴ – 5x³ + 2x² – 112x + 6 = 0} is true.

(iv) False

According to the question,

496 ∉ {y | the sum of all the positive factors of y is 2y}

The possible positive factors of 496 are 1, 2, 4, 8, 16, 31, 62, 124, 248, 496

The sum of them

= 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 + 496

= 996

2y = 2 × 496 = 992

Since, the sum of all the positive factors of y is equal to 2y

496 belongs to given set

⇒ The given statement 496 ∉ {y | the sum of all the positive factors of y is 2y} is false.

Here, A = {a, b} and B = {a, b, c}

Yes, A ⊂ B.

A ∪ B = {a, b, c} = B

(i) A – B = {3, 6, 9, 15, 18, 21}

(ii) A – C = {3, 9, 15, 18, 21}

(iii) A – D = {3, 6, 9, 12, 18, 21}

(iv) B – A = {4, 8, 16, 20}

(v) C – A = {2, 4, 8, 10, 14, 16}

(vi) D – A = {5, 10, 20}

(vii)B – C = {20}

(viii) B – D = {4, 8, 12, 16}

(ix) C – B = {2, 6, 10, 14}

(x) D – B = {5, 10, 15}

(xi) C – D = {2, 4, 6, 8, 12, 14, 16}

(xii)D – C = {5, 15, 20}

(i) Incorrect, ∵ 3 ∈ {3, 4} but 3 ∉ A. 

(ii) Correct, ∵ {3, 4} is a member of A. 

(iii) Correct, ∵ {3, 4} ∈ A. 

(iv) Correct, ∵ 1 is a member of A. 

(v) Incorrect, ∵ 1 is A but not 1⊂ A. 

(vi) Correct, ∵ 1, 2, 5, ∈ A. 

(vii) Incorrect, ∵ {1, 2, 5} ∉ A. 

(viii) Incorrect, ∵ 3 ∈ {1, 2, 3} but 3 ∉ A 

(ix) Incorrect, ∵ φ ∉ A 

(x) Correct, ∵ φ is a subset of every set.

(xi) Incorrect, ∵ φ ∈ {φ} and φ ∈ A.

R – Q = set of irrational numbers.

(i) {1, 2, 3, 4}

{x: x is a natural number and 4 ≤ x ≤ 6} = {4, 5, 6}

Now, {1, 2, 3, 4} ∩ {4, 5, 6} = {4}

Therefore, this pair of sets is not disjoint.

(ii) {a, e, i, o, u} ∩ (c, d, e, f} = {e}

Therefore, {a, e, i, o, u} and (c, d, e, f} are not disjoint.

(iii) {x: x is an even integer} ∩ {x: x is an odd integer} = Φ

Therefore, this pair of sets is disjoint.

A = All natural numbers i.e. {1, 2, 3…..} 

B = All even natural numbers i.e. {2, 4, 6, 8…} 

C = All odd natural numbers i.e. {1, 3, 5, 7……} 

D = All prime natural numbers i.e. {1, 2, 3, 5, 7, 11, …} 

i. A ∩ B 

A contains all elements of B. 

∴ B ⊂ A 

∴ A ∩ B = B 

ii. A ∩ C 

A contains all elements of C. 

∴ C ⊂ A 

∴ A ∩ C = C 

iii. A ∩ D 

A contains all elements of D. 

∴ D ⊂ A 

∴ A ∩ D = D 

iv. B ∩ C 

B ∩ C = ϕ 

There is no natural number which is both even and odd at same time. 

v. B ∩ D 

B ∩ D = 2 

2 is the only natural number which is even and a prime number. 

vi. C ∩ D 

C ∩ D = {1, 3, 5, 7…} 

Every prime number is odd except 2

A = {1, 2, 3, 4], B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}

(i) A ∪ B = {1, 2, 3, 4, 5, 6}

(ii) A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}

(iii) B ∪ C = {3, 4, 5, 6, 7, 8}

(iv) B ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}

(v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}

(vi) A ∪ B ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(vii) B ∪ C ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}

The collection of all subsets of a set A is called the power set of A and is denoted by 

P(A). If n(A) = m, then n[P(A)] = 2^m.

A = {x: x is a natural number} = {1, 2, 3, 4, 5 …}

B ={x: x is an even natural number} = {2, 4, 6, 8 …}

C = {x: x is an odd natural number} = {1, 3, 5, 7, 9 …}

D = {x: x is a prime number} = {2, 3, 5, 7 …}

(i) A ∩B = {x: x is a even natural number} = B

(ii) A ∩ C = {x: x is an odd natural number} = C

(iii) A ∩ D = {x: x is a prime number} = D

(iv) B ∩ C = Φ

(v) B ∩ D = {2}

(vi) C ∩ D = {x: x is odd prime number}

A – B is defined as {x ϵ A : x ∉ B} 

i. A – B is defined as {x ϵ A : x ∉ B} 

A = {3, 6, 12, 15, 18, 21} 

B = {4, 8, 12, 16, 20} 

A – B = {3, 6, 15, 18, 21} 

ii. A – C is defined as {x ϵ A : x ∉ C} 

A = {3, 6, 12, 15, 18, 21} 

C = {2, 4, 6, 8, 10, 12, 14, 16} 

A – C = {3, 15, 18, 21} 

iii. A – D is defined as {x ϵ A : x ∉ D} 

A = {3, 6, 12, 15, 18, 21} 

D = {5, 10, 15, 20}. 

A – D = {3, 6, 12, 18, 21} 

iv. B – A is defined as {x ϵ B : x ∉ A} 

A = {3, 6, 12, 15, 18, 21} 

B = {4, 8, 12, 16, 20} 

B – A = {4, 8, 16, 20} 

v. C – A is defined as {x ϵ C : x ∉ A} 

A = {3, 6, 12, 15, 18, 21} 

C = {2, 4, 6, 8, 10, 12, 14, 16} 

C – A = {2, 4, 8, 10, 14, 16} 

vi. D – A is defined as {x ϵ D : x ∉ A} 

A = {3, 6, 12, 15, 18, 21} 

D = {5, 10, 15, 20}. 

D – A = {5, 10, 20}. 

vii. B – C is defined as {x ϵ B : x ∉ C} 

B = {4, 8, 12, 16, 20} 

C = {2, 4, 6, 8, 10, 12, 14, 16} 

B – C = {4, 8, 20} 

viii. B – D is defined as {x ϵ B : x ∉ D} 

B = {4, 8, 12, 16, 20} 

D = {5, 10, 15, 20} 

B – D = {4, 8, 12, 16}

(î) A – B = {3,6,9,15,18,21) 

(ii) A – C = (3,9,15,18,21) 

(iii) A – D = (3,6,9,12,18,21) 

(iv) B – A = {4,8,16,20) 

(v) C – A = {2, 4, 8,10,14,16} 

(vi) D – A = (5,10,20} 

(vii) B – C = {20) 

(viii) B – D = (4,8,12,16) 

(ix) C – B = {2,6,10,14) 

(x) D – B={5,10,15) 

(xi) C – D={2,4,6,812,14,16} 

(xii) D – C={5,15,20).

If there are some sets under consideration, then there happen to be a set which is a super set of each one of the given sets. Such a set is known as the universal set for those sets. The universal set is denoted by U.

(i) Subsets of {a} are {a}, φ

(ii) Subsets of {a, b} are {a, b},φ,{a},{b}

(iii) Subsets of {1, 2, 3} are {1, 2, 3}, are {1}, {2}, {3}, {1,2}, {2, 3}, {3,1}

(iv) Subsets of {-1, 0, 1} are {-1, 0, 1}, φ, {-1}, {0}, {1}, {-1,0}, {0,1}, {-1,1}

(i) A ∩ B = {7, 9, 11}

(ii) B ∩ C = {11, 13}

(iii) A ∩ C ∩ D = { A ∩ C} ∩ D = {11} ∩ {15, 17} = Φ

(iv) A ∩ C = {11}

(v) B ∩ D = Φ

(vi) A ∩ (B∪ C) = (A ∩ B)∪ (A ∩ C)

= {7, 9, 11}∪ {11} = {7, 9, 11}

(vii) A ∩ D = Φ

(viii) A ∩ (B∪ D) = (A ∩ B) (A ∩ D)

= {7, 9, 11} ∪Φ = {7, 9, 11}

(ix) (A ∩ B) ∩ (B∪ C) = {7, 9, 11} ∩ {7, 9, 11, 13, 15} = {7, 9, 11}

(x) (A∪ D) ∩ (B ∪C) = {3, 5, 7, 9, 11, 15, 17) ∩ {7, 9, 11, 13, 15}

= {7, 9, 11, 15}

P(A) has only one element, namely φ.

∴ PW = {φ}

Answer is P(A)=1

Given sets are 

A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ……} 

B = {2, 4, 6, 8, 10, …….} 

C = {1, 3, 5, 7, 9, …….} 

D = {2, 3, 5, 7, 11, …….} 

A ∩ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, …….} ∩ {2, 4, 6, 8, 10, ……} 

= {2, 4, 6, 8, 10, ……} 

A ∩ C = {1, 2, 3,4, 5, 6, 7, 8, 9, 10, …} ∩ {1, 3, 5, 7, 9 } 

= {1, 3, 5, 7, 9, ……} 

A ∩ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, …} ∩ {2, 3, -5, 7, 11,….} 

= {2, 3, 5, 7, 11, ……} 

B ∩ C = {2, 4, 6, 8, 10, ……} ∩ {1, 3, 5, 7, 9, …….} =

 { } = φ 

B ∩ D = {2, 4, 6, 8, 10, ……} ∩ {2, 3, 5, 7, 11, ……} 

= {2} 

C ∩ D = {1, 3, 5, 7, 9, ……} ∩ {2, 3, 5, 7, 11, 13, ……} 

= {3, 5, 7, …..}

Given A = {1,2,3,4,………………}

B = {2,4,6,8, …………..}

C = {1,3,5,7,……… }

D = {2,3,5,7,11,13, …………..}.

(i) A∩B = {2,4,6,8,……….. } = B

(ii) A∩C = {1,3,5,7,……….. } = C

(iii) A∩D = {2,3,5,7,11,13,………….}

(iv) B∩C = φ

(v) B∩D = {2}

(vi) C∩D = {3,5,7,11,13,…………}

(i) X = {1, 3, 5}, Y = {1, 2, 3}

X ∩ Y = {1, 3}

(ii) A = {a, e, i, o, u}, B = {a, b, c}

A ∩ B = {a}

(iii) A = {x: x is a natural number and multiple of 3} = (3, 6, 9 …}

B = {x: x is a natural number less than 6} = {1, 2, 3, 4, 5}

∴ A ∩ B = {3}

(iv) A = {x: x is a natural number and 1 < x ≤ 6} = {2, 3, 4, 5, 6}

B = {x: x is a natural number and 6 < x < 10}

= {7, 8, 9}

A ∩ B = Φ

(v) A = {1, 2, 3}, B = Φ. So, A ∩ B = Φ

(i) A∩B = {7,9,11} 

(ii) B∩C = {11,13} 

(iii) A∩C∩D = φ 

(iv) A∩C ={11} 

(v) B∩D ={ } =φ 

(vi ) A ∩ {B ∪ C) = {3, 5, 7,9,11}∩{7,9,11,13,15}={7,9,11} A∩D = φ 

(viii) A∩(B∪D) = {3,5,7,9,11} ∩ {7,9,11,13,15,17} = {7,9,11} 

(A∩B)∩(B∪C) = {7,9,11}∩{7,9,11,13,15}= {7,9,11} 

(A∪D)∩(B∪C) = {3,5,7,9,11,15,17} ∩ {7,9,11,13,15}. 

= {7,9,11,15}

(i) A ∪ B = {1, 2, 3, 4, 5, 6}

(ii) A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8} 

(iii) B ∪ C = {3, 4, 5, 6, 7, 8} 

(iv) B ∪ D = {3, 4, 5, 6, 7, 8, 9, 10} 

(v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8} 

(vi) A ∪ B ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 

(vii) B ∪ C ∪ O = {3, 4, 5, 6, 7, 8, 9, 10}

Given A ⊂ B i.e., every element of A is contained in the set B and hence A ∪ B = B