#### 1. The set A = {2, 6, 12, 20, 30, 42, 56} in roaster form is ___

#### The set A = {.....-9, -6 -3, 0, 3, 6, 9, 12,} in set-builder form is ____.

The elements in the set are of the form 3 × 3. Hence, the correct answer is {x: x is a multiple of 3}.

#### The set A = {all elements in a week except Sunday} in roaster form is ____.

The elements in the set should contain all days, except Sunday. Hence, A = {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}

#### The solution set of the equation x^{2} + x − 2 = 0in roaster form is ____.

The given equation can be written as:

x^{2} + x − 2 = 0

#### The set A = {x: 2x - 3 < 5, x ∈ Z} in the roaster form is ____.

2x - 3 <5, x ∈ Z

⇒2x <8, x ∈ Z

⇒x <4, x ∈ Z

⇒ A = {......-5, -4, -3, -2, -1, 0, 1, 2, 3)

Hence, the set in the roaster form is {......-5, -4, -3, -2, -1, 0, 1, 2, 3}.

#### set A = (2, 5/2,10/3,17/4,26/5,37/6) in the set-builder form is ____.

The first element is 2 = 1 + 1

The second elements is 5/2= 2 +1/2

The third element is 10/3= 3 +1/3

The sixth element is 37/6= 6 +1/6

Hence, the set in the set-builder form is x=n+1/n:1 ≤ n ≤ 6,n ∈ N).

#### Given A = {x: x is rational number between 1/3and1/2} B = {x: x is a prime number between 7 and 10} C = {x: x ∈ R and x + a = a for all a ∈ R} D = {x: x ∈ R and x. a = a for all a ∈ R) Set ____ is empty.

5/ 12 ∈ A, 0 ∈ C, 1 ∈ D - Hence, these are non-empty. There are only two elements between 7 and 10 namely 8 and 9. These are not prime. Hence, B is empty.

#### The set A = {2x^{2} − 3x +1 = 0 x∈ N} is equal to ____.

The roots of the equation 2x^{2} - 3x + 1 = 0 are 1/2 and 1.

But as x ∈ N, 1 is the only element of A.

The set A =x: 2x^{2}- 3x + 1 = 0 x ∈ N is equal to {1}.

#### Given
A = {1, 9, 25, 49, 81}
B = {(2n - 1)^{2 }| 1 ≤ n ≤ 5, n ∈ N}
C = { n^{2} | 1 ≤ n ≤ 9, n ∈ N}
D = {(2n - 1) | 1 ≤ n ≤ 9, n ∈ N}
____ Sets are equal.

Solution: 6^{2}= 36 ∈ C, but 36 ∉ A

2 × 4 – 1) ∈ D, but 7 ∈ A

A = {1, 9, 25, 49, 81}

B = {(2 × 1 - 1)^{2}, (2 × 2 - 1)^{2}, (2 × 3 - 1)^{2}, (2 × 4 - 1)^{2}, (2 × 5 - 1)^{2}} = {1, 9, 25, 49, 81}, which is equal to A.

Hence, sets A and B are equal

#### Of the options given, ____ is an empty set.

Solution: e ∈ A

p ∈ B

e∈ D

But as no letter can be both vowel and a consonant, C is an empty set.

#### From the sets given below, select equal sets. P = {1, 2, 3, 5}, Q = {2, 7, 8, 5}, R = {a, b}, S = {1 − 3}, T = {7, 2, 8, 5}, U = {−3, 1}

Solution: We know that,

Two sets are said to be equal if they have exactly the same elements.

P = {1, 2, 3, 5}, Q = {2, 7, 8, 5}, R = {a, b}, S = {1 − 3}, T = {7, 2, 8, 5}, U = {−3, 1}

Q = T and S = U are the equal sets, because all the elements of set Q are elements of set T and vice-versa and all the elements of set S are elements of set U and vice-versa.

#### Given A = {x: x is an even divisor of 12} B = {x: x is an even natural number less than 7} C = {x: x is a prime divisor of 6} D = {x: x is a divisor of 12} The sets ____ and ____ are equal.

Solution: A = {x: x is an even divisor of 12} = {2, 4, 6}

B = {x: x is an even natural number less than 7} = {2, 4, 6}

C = {x: x is a prime divisor of 6} = {2, 3}

D = {x: x is a divisor of 12} = {1, 2, 3, 4, 6, 12}

So, A and B are equal.

#### If A is empty set, then number of elements in P (P (A)) is ____ [P (A) = denotes the power set of A.]

Solution: A is an empty set

i.e. A = ϕ

P (A) = {ϕ}

P (P (A)) = {ϕ, {ϕ}}

So P (P (A)) has 2 elements.

#### Which of the statement is false? A. A ⊂ A ∪ B B. A ∩ B ⊂ A C. B ⊂ B – A D. A ∩ B ⊂ A ∪ B

Solution: Let's take A = {1, 2, 3}, B = {2, 3, 4}

B - A = {4}

(2, 3, 4) ⊈ (4)

Hence, B ⊂ B - A is false.

#### If A = {1, 2, 3, 5, 8}, B = {3, 5, 8}, then the proper subsets of A which are super sets of B are ____.

Solution: Option B is not correct as {3, 5, 8} in not included.

Option C is not correct as {1, 2, 5, 8} is not a super set of B

Option D is not correct as {1, 2, 5, 3} is not a super set of B

The correct answer is A - {3, 5, 8}, {1, 3, 5, 8}, {2, 3, 5, 8}

#### Which of the following statement is true for all A? (a)(ϕ) ∈ P [PA)] (b) ϕ ∈ A (c) ϕ ⊂ A (d) [(ϕ)] ⊂ A

Solution: Option A is not correct as ϕ is not always a element of A.

Option B is not correct as {ϕ} is not always a subset of A.

Option C is not correct as {ϕ} is not always a element of A.

Option D is correct, as ϕ is always an element of P(A) and so {ϕ} is always an element of P(P(A)).

#### B - (A ∩ B) is equal to ____.

Solution: B - (A ∩ B) = B ∩ (A ∩ B)^{C} [because A - B = A ∩ B']

= B ∩ (AC U B^{C})[ By De Morgan's law]

= (B ∩ A^{C}) U (B ∩ B^{C}) [By distributive law]

= (B ∩ A^{C}) U ϕ

= B ∩ A^{C}

= B – A

#### A U (B - A) is equal to ____.

Solution: A U (B - A) = A U (B ∩ A^{C})

= (A U B) ∩ (A U A^{C}) [Distributive law]

= (A U B) ∩ (U) [U universal set]

= A U B [because A ∩ U = A, for all sets A]

#### U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Given the sets A = {2, 5, 6}, B = {5, 6, 7}, C = {1, 2, 5, 7, 8}, then (i) (B U C')' is ____ and (ii) (A - B')' is ____.

Solution: C' = {3, 4, 6, 9, 10}

B U C' = {3, 4, 5, 6, 7, 9, 10}

(B U C')’ = {1, 2, 8}

B' = {1, 2, 3, 4, 8, 9, 10}

A - B' = {5, 6}

(A - B')' = {1, 2, 3, 4, 7, 8, 9, 10}

#### U = {a, b, c, d, e, 1, 2, 5} = Universal set If A = {a, 2, 5}, B = {a, b, 5} and C = {b, 2, 1, d}, then (AC U B) ∩ C is equal to ____.

Solution: A^{C }= {b, c, d, e, 1}

A^{C }U B = {a, b, c, d, e, 1, 5}

(A^{C} U B) ∩ C = {b, d, 1}

#### (A ∩ C')' U (B ∩ C) is equal to ____.

Solution: (A ∩ C')' U (B ∩ C) = (A' U (C')') U (B ∩ C)[By De Morgan's law]

= (A' U C) U (B ∩ C)

= [(A' U C) U B] ∩ [(A' U C) U C]

= [A' U B U C] ∩ [A' U C]

= A' ∩ C [∵A'∪C⊂A'∪B∪C and A⊂B⇒A∩B = A]

#### If A = {2, 4, 5, 7, 8}, B = {3, 5, 7, 9, 10} and C = {1, 2, 3, 4, 5}, then (A ∪ B) ∩ C = ____.

Solution: Given,

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

(A ∪ B) ∩ C

A ∪ B = {2, 4, 5, 7, 8} ∪ {3, 5, 7, 9, 10}

= {2, 3, 4, 5, 7, 8, 9, 10}

(A ∪ B) ∩ C = {2, 3, 4, 5, 7, 8, 9, 10} ∩ {1, 2, 3, 4, 5}

= {2, 3, 4, 5}

#### Which of the following statements is true?

Solution: The divisors of 28 are 1, 2, 4, 7, 14 28.

The sum of the divisors is 56 which is 2 × 28..

Hence, the correct answer is B..

Option A is not correct as 23 has exactly two positive divisors..

Option C is not correct as 7, 542 is not a multiple of 27..

Option D is not correct, as 513 is not a perfect cube..

513 is not a perfect cube.

but divisors of 28 are 1, 2, 4, 7, 14, 28.

1 + 2 + 4 + 7 + 14 + 28 = 56 = 2. 28.

So 'B' is true

#### The set A = {....-5, -2, 1, 4, 7, 10,} in set builder form is ____.

Solution: The set has elements of the form 3 × k + 1. Hence, the set can be represented as[x: 3k + 1 k ∈ Z] in the set builder form.

#### Given the sets X = {a, b, 1, 100, 2} Y = {{1}, 2, 7, {x}} Z = {y, 5} ____ may be considered as a universal set for all the three sets X, Y, Z.

Solution: As (1) ∉ A

a ∉ A

x ∉ A

But X ∪ Y ∪ Z ⊆ C

So, {a, b, y, 1, 2, 5, 7, 100, {x}, {1}} is the universal set for X, Y and Z.

#### Two finite sets have m and n elements respectively. The total number of subset of first set is 48 more than the total number of the second set. The values of m and n respectively are ___.

Solution: Let A and B be such sets, i.e n (A) = m and n(B) = n

So, n(P(A)) = 2^{m }and n(P(B)) = 2^{n}

2^{m} = 2^{n} + 48(This implies m > n)

⇒ 2^{m} - 2^{n} = 48

i.e. 2^{n} (2 ^{m-n} - 1) = 48 = 2^{4} × 3

⇒ n = 4 and 2 ^{m-n} - 1 = 3

⇒ 2^{m-n} = 4 = 2^{2}

⇒ m - n = 2

⇒ m = n + 2 = 4 + 2 = 6

So, m = 6 and n = 4

#### A ∩ (B U C) is equal to ____.

Solution: A∩(B∪C) ⊂ (A∩B) ∪(A∩C) =( A∩B)∪(A∩C)⊂A∩(B∪C)

Let x ∈ A ∩ B ∪ C

⇒x ∈ A and x ∈ B or x ∈ C

⇒x ∈ A and x ∈ B or x ∈ A and x ∈ C

⇒x ∈ (A∩B) or x ∈ (A∩C)

⇒x ∈ (A∩B) ∪ (A∩ C)

So, A ∩ ( B U C) ⊂ (A ∩ B) U (A ∩ C)

[Note:- To prove A⊂B, one has to show every element x of A belongs to B (i.e) x ∈ A⇒x ∈ B]

Now, let x ∈ (A ∩B) ∪ (A∩C)

⇒x ∈ A ∩ B or x ∈ A ∩ C

⇒x ∈ A and x ∈ B or x ∈ A and x ∈ C

⇒x ∈ A and x ∈ B or x ∈ C

⇒x ∈ A and x ∈ B ∪ C

⇒x ∈ A ∩(B ∪ C)

∴(A∩ B)∪( A ∩C) ⊂ A ∩(B ∪ C)

∴A ∩( B ∪ C)=(A ∩ B) ∪ (A ∩C)