Sets

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

Correct! Wrong!

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

Correct! Wrong!

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 ____.

Correct! Wrong!

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 x2 + x − 2 = 0in roaster form is ____.

Correct! Wrong!

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

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

Correct! Wrong!

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 ____.

Correct! Wrong!

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.

Correct! Wrong!

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 = {2x2 − 3x +1 = 0 x∈ N} is equal to ____.

Correct! Wrong!

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}.

Given A = {1, 9, 25, 49, 81} B = {(2n - 1)2 | 1 ≤ n ≤ 5, n ∈ N} C = { n2 | 1 ≤ n ≤ 9, n ∈ N} D = {(2n - 1) | 1 ≤ n ≤ 9, n ∈ N} ____ Sets are equal.

Correct! Wrong!

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

Of the options given, ____ is an empty set.

Correct! Wrong!

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}

Correct! Wrong!

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.

Correct! Wrong!

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.]

Correct! Wrong!

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

Correct! Wrong!

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 ____.

Correct! Wrong!

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

Correct! Wrong!

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 ____.

Correct! Wrong!

Solution: B - (A ∩ B) = B ∩ (A ∩ B)C [because A - B = A ∩ B']
= B ∩ (AC U BC)[ By De Morgan's law]
= (B ∩ AC) U (B ∩ BC) [By distributive law]
= (B ∩ AC) U ϕ
= B ∩ AC
= B – A

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

Correct! Wrong!

Solution: A U (B - A) = A U (B ∩ AC)
= (A U B) ∩ (A U AC) [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 ____.

Correct! Wrong!

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 ____.

Correct! Wrong!

Solution: AC = {b, c, d, e, 1}
AC U B = {a, b, c, d, e, 1, 5}
(AC U B) ∩ C = {b, d, 1}

(A ∩ C')' U (B ∩ C) is equal to ____.

Correct! Wrong!

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 = ____.

Correct! Wrong!

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?

Correct! Wrong!

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 ____.

Correct! Wrong!

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.

Correct! Wrong!

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 ___.

Correct! Wrong!

Solution: Let A and B be such sets, i.e n (A) = m and n(B) = n
So, n(P(A)) = 2m and n(P(B)) = 2n
2m = 2n + 48(This implies m > n)
⇒ 2m - 2n = 48
i.e. 2n (2 m-n - 1) = 48 = 24 × 3
⇒ n = 4 and 2 m-n - 1 = 3
⇒ 2m-n = 4 = 22
⇒ m - n = 2
⇒ m = n + 2 = 4 + 2 = 6
So, m = 6 and n = 4

A ∩ (B U C) is equal to ____.

Correct! Wrong!

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)