SN Dey Class 11 Solutions | SN Dey Class 11 Set Theory Solutions | Class 11 Set Theory MCQ | SN Dey Solutions | S N Dey Math Solutions | Set Theory S N Dey Solutions | WBCHSE Class 11 Math Solutions

**SN Dey Class 11 Set Theory MCQ Solutions**

**Choose the correct option:**

**Ex 1: The number of subsets in a set consisting of four distinct elements is-**

**(A) 4 (B) 8 (C) 16 (D) 64**

**Answer:**

We know that the number of subsets in a set consisting of n distinct elements is equal to 2^{n}.

So the answer is

= 2^{4 }= 16.

∴ option (C) is correct.

**Ex 2: The number of proper subsets in a set consisting of five distinct elements is-**

**(A) 5 (B) 10 (C) 32 (D) 31**

**Answer:**

We know that the number of proper subsets in a set consisting of n distinct elements is equal to 2^{n}-1.

So the answer is

= 2^{5}-1= 32-1 = 31.

∴ option (D) is correct.

**Ex 3: If x ∈ A ⇒ x ∈ B then –**

**(A) A=B (B) A⊂B (C) A⊆B (D) B⊆A**

**Answer:**

As x ∈ A ⇒ x ∈ B, then we must have that either A is a proper subset of B or they are equal. So the answer is A ⊆ B.

∴ option (C) is correct.

**Ex 4: If A⊆B and B⊆A then –**

**(A) A=Φ (B) A∩B=Φ (C) A=B (D) None of these**

**Answer:**

As A⊆B and B⊆A, then it follows that A=B.

So option (C) is correct.

**Ex 5: For two sets A and B, if A∪B = A∩B then –**

**(A) A⊆B (B) B⊆A (C) A=B (D) None of these**

**Answer:**

Let $x∈A$ then $x∈A∪B.$

As $A∪B=A∩B, we have that$$x∈A∩B$

⇒ $x∈B$

$∴A⊂B$…..(i)

Similarly, if $y∈B$ then, $y∈A∪B$ = $A∩B.$

⇒ $y∈A$

Thus, we get that B$⊂A$…..(ii)

So from (i) and (ii), we obtain that

$A=B$

∴ option (C) is correct.

**Ex 6: A – B = Φ iff – [Council Sample Question ’13]**

**(A) A≠B (B) A⊂B (C) B⊂A (D) A∩B=Φ**

**Answer:**

We know that A-B ={x: x ∈ A but x ∉B}. Thus A-B=Φ means that A is fully contained in B, that is, A⊂B.

∴ option (B) is correct.

**Ex 7: If A∩B = B then –**

**(A) A⊆B (B) B⊆A (C) A=B (D) A=Φ**

**Answer:**

As B=A∩B, we have B ⊆ A∩B.

⇒ B ⊆ A and B ⊆ B.

⇒ B ⊆ A is true.

∴ option (B) is correct.

**Ex 8: If A and B are two disjoint sets then n(A∪B)=**

**(A) n(A)+n(B) (B) n(A)-n(B) (C) 0 (D) None of these**

**Answer:**

As A and B are disjoint sets, so we have A∩B = Φ. Thus n(A∩B)=0.

Now, we know that

n(A∪B) = n(A)+n(B)-n(A∩B)

= n(A)+n(B)+0

= n(A)+n(B)

∴ option (A) is correct.

**Ex 9: For any two sets A and B, n(A)+n(B)-n(A∩B)=**

**(A) n(A∪B) (B) n(A)-n(B) (C) Φ (D) None of these**

**Answer:**

As n(A∪B) = n(A)+n(B)+n(A∩B), the option (A) is correct.

**Ex 10: The dual of A∪U = U is-**

**(A) A∪U=U (B) A∪Φ=Φ (C) A∪Φ=A (D) A∩Φ=Φ**

**Answer:**

The dual of A∪U = U is A∩Φ=Φ.

∴ option (D) is correct.

**Ex 14: State which of the is the set of factors of the number 12-**

**(A) {2, 3, 4, 6} (B) {2, 3, 4, 6, 12} **

**(C) {2, 3, 4, 8, 6} (D) {1, 2, 3, 4, 6, 12}**

**Answer:**

The factors of 12 are 1, 2, 3, 4, 6, and 12.

∴ option (D) is correct.