Test: Sets - 1 - JEE MCQ

Concept:

The null set is a subset of every set. (ϕ ⊆ A)

Every set is a subset of itself. (A ⊆ A)

The number of subsets of a set with n elements is 2ⁿ.

The number of proper subsets of a given set is 2n - 1

Calculation:

Statement I: If A = {x: x is an even natural number} and B = {y: y is a natural number}, A subset B.

A = {2, 4, 6, 8, 10, 12, ...} and A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, ...}.

It is clear that all the elements of set A are included in set B.

So, set A is the subset of set B.

Statement I is correct.

Statement II:  Number of subsets for the given set A = {5, 6, 7, 8) is 15.

Given: A = {5, 6, 7, 8}

The number of elements in the set is 4

We know that,

The formula to calculate the number of subsets of a given set is 2ⁿ

 = 2⁴ = 16

Number of subsets is 16

Statement II is incorrect.

Statement III: Number of proper subsets for the given set A = {5, 6, 7, 8) is 15.

The formula to calculate the number of proper subsets of a given set is 2ⁿ - 1

 = 2⁴ - 1

 = 16 - 1 = 15

The number of proper subsets is 15.

Statement III is correct.

∴ Statements I and III are correct.

Using the Venn Diagram,
(i) A - B = A - (A ∩ B) is true

(ii) A = (A ∩ B) ∪ (A - B) is true

(iii) A - (B ∪ C) = (A - B) ∪ (A - C) is false

A = {(a, b) ∈ R × R : |a − 5| < 1 and |b − 5| < 1}

⇒ a ∈ (4, 6) and b ∈ (4, 6)

Therefore, A = {(a, b): a ∈ (4, 6) and b ∈ (4, 6)}

Now, B = {a, b) ∈ R × R : 4 (a − 6) ² + 9(b−5) ² ≤ 36}

From the conditions above for set A, the maximum value of B is:

4 (a − 6) ² + 9(b−5) ² = 4 (4 − 6)² + 9(6−5)² = 25 ≤ 36

We can check for other values as well.

Elements of set A satisfy the conditions in Set B.

Hence, A is a subset of B.

n(A) = 4 and n(B) = 2

Therefore, n(A x B) = 8

Number of subsets of A x B having atleast 3 elements = 2⁸ – ⁸C₀ – ⁸C₁ – ⁸C₂

= 256 – 1 – 8 – 28

= 219

The correct answer is a) A

• B' gives us all the elements in U other than 6 and 7 i.e., B' = {1, 2, 3, 4, 5, 8, 9, 10}
• The intersection of this set with A will be the common elements in both of these (A and B') i.e., = {1, 2, 5} which is set A itself.

• This method involves specifying a rule or condition which can be used to decide whether an object can belong to the set.
• This rule is written inside a pair of curly braces and can be written either as a statement or expressed symbolically or written using a combination of statements and symbols.
• {x :  x ≠ x}
  This implies a null set.

If A ∪ B = A ∩ B then A = B.
Example: Let A = {1, 2, 3, 4} & B = {1, 2, 3, 4}
(AUB) = (A⋂B)
⇒{1,2,3,4} = {1,2,3,4}

From Venn-Euler's Diagram

∴ (A ∪ B)′ ∪ (A′ ∩ B) = A′

If A is a subset of the universal set U, then its complement A′ is also a subset of U. We have, A′={2,4,6,8,10}
Hence, (A′)′ = {x : x ∈ U and x ∉ A′} = {1,3, 5,7,9} = A
It is clear from the definition of the complement that for any subset of the universal set U, we have (A′)′=A

Understanding the Power Set P(A):

• The power set of A, written as P(A), is the set of all subsets of A.

• This includes the empty set {} and the set A itself.

Statement 1: A ∪ P(A) = P(A)

• The union A ∪ P(A) means combining all elements of A and all elements of P(A).

• Elements of A are individual elements like 1, 2, etc.

• Elements of P(A) are sets like {1}, {2}, {1,2}, etc.

• Since elements of A are not subsets of A, they are not in P(A).

• Example:
  If A = {1, 2},
  then P(A) = { {}, {1}, {2}, {1, 2} }
  A ∪ P(A) = {1, 2, {}, {1}, {2}, {1, 2}}
  So, A ∪ P(A) ≠ P(A)

• Therefore, Statement 1 is incorrect.

Statement 2: {A} ∩ P(A) = A

• {A} is a set containing A as a single element.
  For example, if A = {1, 2}, then {A} = { {1, 2} }

• P(A) contains all subsets of A, including A itself.
  So A is in P(A), and {A} ∩ P(A) = {A}

• But the statement says {A} ∩ P(A) = A
  That’s incorrect because:

  • Left side is a set containing A

  • Right side is the actual contents of A

  • Example:
    A = {1, 2}
    Then, {A} ∩ P(A) = { {1, 2} }
    But A = {1, 2}
    So they are not equal.

• Therefore, Statement 2 is also incorrect.

Statement 3: P(A) − {A} = P(A)

• P(A) includes A as one of its subsets.

• If we remove A from P(A), the result cannot be equal to P(A) itself.

• Example:
  A = {1}
  Then P(A) = { {}, {1} }
  {A} = { {1} }
  So P(A) − {A} = { {} }
  Which is not equal to P(A)

• Therefore, Statement 3 is incorrect.

Final Answer:

None of the statements are correct.

Difference of two Sets:
Let A and B be two sets. The difference of A and B is denoted as (A - B) It is the set of all those elements of A which are not present in B i.e {x:x∈A and x∈B} The Venn diagram representation of the difference of two sets is shown below

n(A−B)=n(A)−n(A∩B)
Similarly; n(B−A)=n(B)=n(A∩B)
Calculations:
Given: A and 8 are two disjoint sets.
So, (A∩B)=(B∩A)=ϕ
A−B={x:x∈A,x∈B}=A
A−(A∩B)=A−ϕ=A
So, The statement " A⋅B=A⋅(A∩B)′′ is true.
Consider the statement −B−A′=B∩A′
⇒B−A′=ϕ
Now, B∩A=∅
From (3) and (4), we have
B−A′=B∩A
The statement −B⋅A′=B∩A′′ is true.
Consider the statement "A∩B=(A−B)∩B"
Let, A−B={x:x∈A,x∉B}
⇒A−B=A
⇒(A−B)∩B=A∩B
The statement "A∩B=(A−B)∩B" is true.
Hence, the statement A∩B=(A−B)∩B is true.

Let A = {n² : n ∈ N} and B = {n³ : n ∈ N}
A = {1, 4, 9,16,…..}
and B = {1, 8, 27, 64, ……}
Now, A ∩ B = {1} which is a finite set.
Also, A ∪ B = {1, 4, 8, 9, 27, …..}
So, complement of A ∪ B is infinite set. Hence, A ∪ B ≠ N

The correct answer is c) 9

Given A=[2,4,5],B=[7,8,9]

To find: n(A×B)=?

A×B={(2,7),(2,8),(2,9),(4,7),(4,8),(4,9),(5,7),(5,8),(5,9)}

∴n(A×B)=9

As n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
But in case of disjoint sets, n(A ∩ B) = 0
∴ n(A ∪ B) = n(A) + n(B)

N Δ NP = (N − N_p) ∪ (N_P − N)
= {1,2,…} ∪ {…−2,−1} = I − {0}

The number of non-empty subsets = 2ⁿ − 1 = 2⁴ − 1 = 16 − 1 = 15
The subsets are:
[{1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,4}, {1,2,3}, {1,3,4} ,{2,3,4}, {1,2,3,4}]

Let A = {1,2}
B = {3, 4, 0}
C = {5, 6, 0}
D = {7, 8}
Such that (A ∩ B) = (C ∩ D) = ϕ
⇒ (A ∪ C) = {1, 2, 5, 6, 0}
⇒ (B ∪ D) ={3, 4, 7, 8, 0}
⇒ (A ∪ C) ∩ (B ∪ D) = {0}
So (A ∪ C) and (B ∪ D) are not always disjoint ⇒ (A ∩ C) = ϕ and (B ∩ D) = ϕ
So (A ∩ C) and (B ∩ D) are always disjoint.

If A⊂B and B⊂C, then these sets 15 represented in Venn diagram as
Clearly, A∪B=B
and B∩C=B
Hence, A∪B=B∩C.

A ∩ (B − A) = ψ,[∵ x ∈ B − A ⇒ x∉ A]

Suppose a ∈ X and a ∈ A ⇒ a ∈ X ∪ A ⇒ a∈Y∪A
⇒a∈Y and a∈A(∵X∪A=Y∪A)
⇒a∈Y∩A⇒Y∩A is non-empty
This contradicts that Y∩A=ϕ So X=Y

Explanation for the correct option: Power set is the collection of all the subsets of the set A.
Let A = {1, 2} and {A} = {1, 2} ⇒ P(A) = {{1}, {2}, {1,2}, {ϕ}}
Now,
Option (B): {A} ∩ P(A) = A{A} ∩ P(A) = {{1, 2} ∩ {1}, {2}, {1, 2}, {φ}} = {1, 2} = A
Explanation for the incorrect option :
Option (A):A ∪ P(A) = P(A)
∵A U P(A) ≠ P(A) because Ais subset of Powerset P
Option (C): P(A) − {A} = P(A)
P(A) − {A} = {{1}, {2}, {1, 2}, {ϕ}} − {1, 2} = {{1}, {2}, {ϕ}} ≠ P(A)
Therefore statement (2) is correct.
Hence, Option 'B' is correct.

For subsets A and B of U, If(A − B) ∪ (B − A) = A
⇒ B = ϕ

n( A′ ∩ B′) = n( A ∪ B)′ = n(U) − n( A ∪ B)
= n(U) − [n( A) + n( B) − n( A ∩ B)]
= 700 − [200 + 300 − 100] = 300

–6 < n² – 10n + 19 < 6

⇒ n² – 10n + 25 > 0 and n² – 10n + 13 < 0

(n – 5)² > 0 ⇒ 5 – 3√2 < n < 5 + 3√2

N ∈ Z \ {5}
⇒ n = {2, 3, 4, 5, 6, 7, 8} 

Taking intersection, we get

N = {2, 3, 4, 6, 8}

Number of values of n = 6