Operations on Sets MCQ Quiz in मल्याळम - Objective Question with Answer for Operations on Sets - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

• Symmetric Difference of two Sets: Let A and B be two sets. The symmetric difference of sets A and B is the set (A - B) ∪ (B - A) and is denoted as A Δ B.

          i.e A Δ B = (A - B) ∪ (B - A)

• The Venn diagram representation of symmetric difference of two sets is shown below

Calculation:

Given: X = {x ∈ R: x2 = x} and Y = {y ∈ R: y2 = 1}

∵ x2 = x

⇒ x² - x = 0

⇒ x (x - 1) = 0

⇒ x = 0 or 1

So, X = {0, 1}

∵ y² = 1

⇒ y = - 1 or 1

So, Y = {- 1, 1}

As we know that, A Δ B = (A - B) ∪ (B - A)

⇒ X - Y = {0} and Y - X = {- 1}

⇒ X Δ Y = {0, - 1}

Concept:

De Morgan's Law:

For any set A, B and U is a universal set for A and B  then we have:

• (A ∪ B)' = A' ∩ B'
• (A ∩ B)' = A' ∪ B'
• A' = U - A

Calculation:

Given: A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and U = {1, 2, 3, 4, 5, 6, 7, 8, 9} where U is the universal set for A and B.

Here, we have to find the value of A' ∪ B'

As we know that, (A ∩ B)' = A' ∪ B'

⇒ A ∩ B = {1, 2, 3, 4} ∩ {2, 4, 6, 8} = {2, 4}

As we know that, A' = U - A

⇒ (A ∩ B)' = U - (A ∩ B) = {1, 2, 3, 4, 5, 6, 7, 8, 9} -  {2, 4} = {1, 3, 5, 6, 7, 8, 9}

Calculation:

Given: n(Z) = 90 and the number of elements in Y and Z are in the ratio 4 : 5

To find: b = ?

n(Y) = 16 + 18 + 17 + b

n(Z) = 12 + 18 + 17 + c = 90 

Hence, option (3) is correct.

Concept:

• If A and B are two sets then A ∩ B = {x : x ∈ A and x ∈ B}

Calculation:

Given that: Na = {an: n ∈ N}, 

⇒ N6 = {6n: n ∈ N} = {6, 12, 24, 30, .... }

⇒ N8 = {8n: n ∈ N} = {8, 16, 24, 32, ...}

⇒ N6 ∩ N8​ = {24, 48, 72, .... }

⇒ N6 ∩ N8​ = {24n: n ∈ N}

⇒ N₆ ∩ N₈ = N₂₄.

Hence, option 2 is correct.

Concept:

Let A and B be any two sets.

A × B = {(x, y) where x in A and y in B}

Calculations:

Let A and B are two non-empty sets.

Consider, A = {1, 2, 3} and B = {2,3}

Here in A and B, there are 2 common elements.

⇒A × B = {(1, 2), (1, 3), (2, 2), (2, 3),(3, 2), (3,3)}

and B × A = {(2, 1), (2, 2), (2, 3), (3, 1),(3, 2), (3,3}

In A × B and B × A, there are 4 = 2²  common elements. = n² 

Hence, if A and B are two non-empty sets having n elements in common, then n2 common elements in the sets A × B and B × A

Concept:

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) 

Calculation:

Given:

A = { x : x ∈ N, 0

B = {x : x is a prime natural number, 0

A = { x : x ∈ N, 0

A = {1, 2, 3, 4, 5}

B = {x : x is a prime natural number, 0

B = {2, 3, 5, 7}

A - B = {1, 2, 3, 4, 5} - {2, 3, 5, 7}

= {1, 4}

∴ The value A - B is {1, 4}.

Concept:

The intersection of two sets X and Y is the set of elements that are common to both set X and set Y.

It is denoted by X ∩ Y and is read 'X intersection Y '

Cardinality is the number of elements present in the set.

Calculation:

Here, A = {x : x is a square of a natural number and x is less than 100}

So, A = {1, 4, 9, 16, 25, 36, 49, 64, 81} and

B is a set of even natural numbers.

So, B = {2, 4, 6, 8, ....., }

Now, A ∩ B = {4, 16, 36, 64}

∴ Cardinality = n(A ∩ B) = 4

Hence, option (1) is correct. 

Solution:

Given, S1 = {x ϵ R ∶ x2 + | x | - 2 = 0} and S2 = {x ϵ R ∶ x2 + x - 2 = 0}

For S₁ : x² + |x| - 2 = 0

Case 1: x ≥ 0

x² + x - 2 = 0

⇒ x² + 2x - x - 2 = 0

⇒ x(x + 2) -1(x + 2) = 0

⇒ (x - 1)(x + 2) = 0

⇒ x = 1 and -2

But according to case 1, x ≠ -2

x = 1

Case 2: x

x² - x - 2 = 0

⇒ x² + x - 2x - 2 = 0

⇒ x(x + 1) -2(x + 1) = 0

⇒ (x - 2)( x + 1) = 0

⇒ x = 2 and -1

But according to case 2, x ≠ 2

x = -1 

Hence, S₁ = {1, -1}

For S₂: x² + x - 2 = 0

⇒ x2 + 2x - x - 2 = 0

⇒ x(x + 2) -1(x + 2) = 0

⇒ (x - 1)(x + 2) = 0

⇒ x = 1 and -2

Hence, S₂ = {1, -2}

So, S₁ ∩ S₂ = {1, -1} ∩ {1, -2}

= {1}

∴ The correct option is (4)

Concept:

A ⋂ B, where ⋂ an intersection represents the objects that belong to set A and set B.

A ⋃ B where ⋃ is union, that represents the objects that belong to set A or set B

A ⊆ B, ⊆ represents a subset, A is a subset of B means set A is included in set B.

Calculation:

Given:

A = {1, 2, 3, 4}

B = {2, 4 , 6}

Now, A ∩ B = {1, 2, 3, 4} ∩ {2, 4 , 6} 

As A ⋂ B represents the objects that belong to set A and set B.

⇒ A ∩ B = {2, 4}

And A ⋃ B = {1, 2, 3, 4} ⋃ {2, 4 , 6}

A ⋃ B represents the objects that belong to set A or set B.

⇒ A ⋃ B = {1, 2, 3, 4, 6}

A ∩ B ⊆ C ⊆ A ⋃ B

{2, 4} ⊆ C ⊆ {1, 2, 3, 4, 6}

A ⊆ B represents  A is a subset of B, set A is included in set B.

C = {2, 4}, {1, 2, 3, 4, 6}, {2, 4, 1}, {2, 4, 3}, {2, 4, 6}, {2, 4, 1, 3}, {2, 4, 1, 6}, {2, 4, 3, 6}

Number of sets = 8

∴ The number of sets C is 8.

Concept:

A ⋃ B, where ⋃ is union, that represents the objects that belong to set A or set B

A ⋂ B, where ⋂ is intersection, that represents the objects that belong to set A and set B

Calculation:

Given:

A = {1, 2, 3} and B = {3, 8}

A ∪ B = {1, 2, 3} ∪ {3, 8}

⇒ A ∪ B = {1, 2, 3, 8}

And A ∩ B = {1, 2, 3} ∩ {3, 8}

⇒ A ∩ B = {3}

(A ∪ B) × (A ∩ B) = {1, 2, 3, 8} × {3}

(A ∪ B) × (A ∩ B) = {(1, 3), (2, 3), (3, 3), (8, 3)}

∴ (A ∪ B) × (A ∩ B) = {(1, 3), (2, 3), (3, 3), (8, 3)}.