#### 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 Γ 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 β 2 = 0in roaster form is ____.

The given equation can be written as:

x^{2} + x β 2 = 0

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

2x - 3 <5, x β Z

β2x <8, x β Z

βx <4, x β Z

β 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 β€ n β€ 6,n β 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 β R and x + a = a for all a β R} D = {x: x β R and x. a = a for all a β R) Set ____ is empty.

5/ 12 β A, 0 β C, 1 β 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} β 3x +1 = 0 xβ N} is equal to ____.

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

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

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

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

Solution: 6^{2}= 36 β C, but 36 β A

2 Γ 4 β 1) β D, but 7 β A

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

B = {(2 Γ 1 - 1)^{2}, (2 Γ 2 - 1)^{2}, (2 Γ 3 - 1)^{2}, (2 Γ 4 - 1)^{2}, (2 Γ 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 β A

p β B

eβ 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 β 3}, T = {7, 2, 8, 5}, U = {β3, 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 β 3}, T = {7, 2, 8, 5}, U = {β3, 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 = Ο

P (A) = {Ο}

P (P (A)) = {Ο, {Ο}}

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

#### Which of the statement is false? A. A β A βͺ B B. A β© B β A C. B β B β A D. A β© B β A βͺ B

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

B - A = {4}

(2, 3, 4) β (4)

Hence, B β 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)(Ο) β P [PA)] (b) Ο β A (c) Ο β A (d) [(Ο)] β A

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

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

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

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

#### B - (A β© B) is equal to ____.

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

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

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

= (B β© A^{C}) U Ο

= B β© A^{C}

= B β A

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

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

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

= (A U B) β© (U) [U universal set]

= A U B [because A β© 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')β = {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) β© 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) β© C = {b, d, 1}

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

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

= (A' U C) U (B β© C)

= [(A' U C) U B] β© [(A' U C) U C]

= [A' U B U C] β© [A' U C]

= A' β© C [β΅A'βͺCβA'βͺBβͺC and AβBβAβ©B = A]

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

Solution: Given,

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

(A βͺ B) β© C

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

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

(A βͺ B) β© C = {2, 3, 4, 5, 7, 8, 9, 10} β© {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 Γ 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 Γ k + 1. Hence, the set can be represented as[x: 3k + 1 k β 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) β A

a β A

x β A

But X βͺ Y βͺ Z β 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)

β 2^{m} - 2^{n} = 48

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

β n = 4 and 2 ^{m-n} - 1 = 3

β 2^{m-n} = 4 = 2^{2}

β m - n = 2

β m = n + 2 = 4 + 2 = 6

So, m = 6 and n = 4

#### A β© (B U C) is equal to ____.

Solution: Aβ©(BβͺC) β (Aβ©B) βͺ(Aβ©C) =( Aβ©B)βͺ(Aβ©C)βAβ©(BβͺC)

Let x β A β© B βͺ C

βx β A and x β B or x β C

βx β A and x β B or x β A and x β C

βx β (Aβ©B) or x β (Aβ©C)

βx β (Aβ©B) βͺ (Aβ© C)

So, A β© ( B U C) β (A β© B) U (A β© C)

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

Now, let x β (A β©B) βͺ (Aβ©C)

βx β A β© B or x β A β© C

βx β A and x β B or x β A and x β C

βx β A and x β B or x β C

βx β A and x β B βͺ C

βx β A β©(B βͺ C)

β΄(Aβ© B)βͺ( A β©C) β A β©(B βͺ C)

β΄A β©( B βͺ C)=(A β© B) βͺ (A β©C)