Set MCQs

The first element of the set is 2 = 1(1 + 1)
The second element of the set is 6 = 2(2 + 1)
The third element of the set is 12 = 3(3 + 1)
The seventh element of the set is 56 = 7(7 + 1)
The set A = {2, 6, 12, 20, 30, 42, 56} in roaster form is nn+1:n <8,n∈N.

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

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

The given equation can be written as:
x² + x − 2 = 0

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}.

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).

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 roots of the equation 2x² - 3x + 1 = 0 are 1/2 and 1.
But as x ∈ N, 1 is the only element of A.
The set A =x: 2x²- 3x + 1 = 0 x ∈ N is equal to {1}.

Solution: 6²= 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

Solution: e ∈ A
p ∈ B
e∈ D
But as no letter can be both vowel and a consonant, C is an empty set.

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.

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.

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.

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.

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}

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)).

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

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]

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}

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}

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]

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}

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

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.

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.

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)) = 2ⁿ
2^m = 2ⁿ + 48(This implies m > n)
⇒ 2^m - 2ⁿ = 48
i.e. 2ⁿ (2 ^m-n - 1) = 48 = 2⁴ × 3
⇒ n = 4 and 2 ^m-n - 1 = 3
⇒ 2^m-n = 4 = 2²
⇒ m - n = 2
⇒ m = n + 2 = 4 + 2 = 6
So, m = 6 and n = 4

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)