**Miscellaneous Exercise**

Chapter 1 Sets

**Question and Answers**

**Class 11 – Maths**

Chapter 1 Sets

Class | Class 11 |

Subject | Mathematics |

Chapter Name | Sets |

Chapter No. | Chapter 1 |

Exercise | Miscellaneous Exercise |

Category | Class 11 Maths NCERT Solutions |

**Question 1 Decide, among the following sets, which sets are subsets of one and another: **

**A = { x : x ∈ R and x satisfy x ^{2} – 8x + 12 = 0 }, **

**B = { 2, 4, 6 },**

** C = { 2, 4, 6, 8, . . . },**

** D = { 6 }.**

**Answer** A = {x:x ∈ R satisfy x^{2}−8x+12=0}

x^{2}−8x+12=0

(x−2)(x−6)=0

x = 2, 6

A set A is said to be a subset of B if every element of A is also an element of B

Hence, A = {2, 6}

B = {2, 4, 6}, C = {2, 4, 6, 8 …}, D = {6}

Hence, D ⊂ A ⊂ B ⊂ C

Hence, A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C

**Question 2 In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example. **

**(i) If x ∈ A and A ∈ B , then x ∈ B**

**(i)** Let A = {1,2},

B = {1,{1,2},{3}}

Now 2 ∈ {1,2 }, {1,2} ∈ {1, {1,2}, {3}}

But 2∉ {1, {1,2} ,{3}} 2 ∉ {1, {1,2}, {3}}

Hence the given statement is false.

**(ii) If A ⊂ B and B ∈ C , then A ∈ C**

**(ii)** Let us assume that,

A = {2}

B = {0, 2}

And, C = {1, {0, 2}, 3}

A ⊂ B, Hence, B ∈ C

But, we know,

A ∉ C

The given statement is false.

** (iii) If A ⊂ B and B ⊂ C , then A ⊂ C**

**(iii)** Let A ⊂ B, B ⊂ C

Let x ∈ A

x ∈ B so A ⊂ B

x ∈ C so B ⊂ C

The given statement is true.

**(iv) If A ⊄ B and B ⊄ C, then A ⊄ C**

**(iv)** Let A = {1,2}

B = {0,6,8}

C = {0,1,2,6,9}

Now by the statement,

A ⊄ B and B ⊄ C

But A ⊂ C

The given statement is false

**(v) If x ∈ A and A ⊄ B , then x ∈ B**

**(v)** Let A = {3, 5, 7}

B = {3, 4, 6}

Now, 5 ∈ A and A ⊄ B

But 5 ∉ B

The given statement is false

** (vi) If A ⊂ B and x ∉ B , then x ∉ A**

**(vi)** Let A ⊂ B and x ∉ B

To show that x ∉ A

Suppose x∈A

Then x ∈ B which is a contradiction

∴ x ∉ A

The given statement is true

**Question 3 Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. **

**Show that B = C.**

**Answer** Let x ∈ B

x ∈ A ∪ B since [B ⊂ A ∪ B]

x ∈ A ∪ C since [A ∪ B=A ∪ C]

x ∈ A or ∈ C

Also x ∈ B

x ∈ A ∩ B since [B ⊂ A ∩ B]

x ∈ A ∩ C since [A ∩ B=A ∩ C]

x ∈ A or x ∈ C

∴ x ∈ C

∴ B ⊂ C

Similarly, we can show that C ⊂ B

∴ It has been proved that B=C

**Question 4 Show that the following four conditions are equivalent :**

** (i) A ⊂ B**

**(ii) A – B = φ **

**(iii) A ∪ B = B **

**(iv) A ∩ B = A**

**Answer** First showing that, (i)⬄(ii)

Here, (i) = A ⊂ B and (ii) = A – B ≠ ϕ

Let us assume that A ⊂ B

To prove, A – B ≠ ϕ, Let A – B ≠ ϕ

Hence, there exists X ∈ A, X ≠ B, but since A⊂ B, it is not possible

∴ A – B = ϕ

And A ⊂ B ⇒ A – B ≠ ϕ

Let us assume that A – B ≠ ϕ

To prove: A ⊂ B , Let X ∈ A

So, X ∈ B (if X ∉ B, then A – B ≠ ϕ)

Hence, A – B = ϕ => A ⊂ B

∴ (i) ⬌ (ii)

Let us assume that A ⊂ B

To prove, A ∪ B = B

⇒ B ⊂ A ∪ B

Let us assume that, x ∈ A∪ B

⇒ X ∈ A or X ∈ B

Taking Case I : X ∈ B , A ∪ B = B

Taking Case II: X ∈ A

⇒ X ∈ B (A ⊂ B)

⇒ A ∪ B ⊂ B

Let A ∪ B = B

Let us assume that X ∈ A

⇒ X ∈ A ∪ B (A ⊂ A ∪ B)

⇒ X ∈ B (A ∪ B = B)

∴A ⊂ B

Hence, (i) ⬌ (iii)

To prove (i) ⬌ (iv)

Let us assume that A ⊂ B

A ∩ B ⊂ A

Let X ∈ A

To prove, X ∈ A∩ B

Since, A ⊂ B and X ∈ B

Hence, X ∈ A ∩ B

⇒ A ⊂ A ∩ B

⇒ A = A ∩ B

Let us assume that A ∩ B = A

Let X ∈ A

⇒ X ∈ A ∩ B

⇒ X ∈ B and X ∈ A

⇒ A ⊂ B

∴ (i) ⬌ (iv)

∴ (i) ⬌ (ii) ⬌ (iii) ⬌ (iv)

Hence, proved

**Question 5 Show that if A ⊂ B, then C – B ⊂ C – A.**

**Answer** Given that, A ⊂ B

To show that, C−B ⊂ C−A

Let x ∈ C−B

x ∈ C and x ∈ B

x ∉ A [A⊂B] and x ∈ C

x ∈ C−A

∴C− B ⊂ C − A

Hence it has been showed that C−B ⊂ C−A

**Question 6 Show that for any sets A and B,**

** A = ( A ∩ B ) ∪ ( A – B ) and A ∪ ( B – A ) = ( A ∪ B )**

**Answer **(A∩B)∪(A−B)

= (A∩B) ∪ (A∩B′)

= A ∩ (B∪B′) (By distributive law)

= A ∩ U=A

Hence A=(A ∩ B) ∪ (A−B)

Also A ∪ (B−A)

= A ∪ (B ∩ A′)

= (A ∪ B )∩ (A ∪ A′) (By distributive law)

= (A ∪ B) ∩ U

= A ∪ B

Hence A ∪ (B−A)= A ∪ B.

**Question 7 Using properties of sets, show that **

**(i) A ∪ ( A ∩ B ) = A **

**(ii) A ∩ ( A ∪ B ) = A.**

**Answer **

**(i)** To show: A ∪ (A ∩ B) = A

We know that, A ⊂ A

A ∩ B ⊂ A

∴ A ∪ (A ∩ B) ⊂ A ….(i)

We have:

A ⊂ A ∪ (A ∩ B) ….(ii)

Hence, from equation (i) and (ii), We have

A ∪ (A ∩ B) = A

**(ii)** To show, A ∩ (A ∪ B) = A

A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B)

= A ∪ (A ∩ B) = A

**Question 8 Show that A ∩ B = A ∩ C need not imply B = C.**

**Answer **Let us assume,

A = {0, 1} B = {0, 2, 3} C = {0, 4, 5}

According to the question,

A ∩ B = {0}

A ∩ C = {0}

∴ A ∩ B = A ∩ C = {0}

But, 2 ∈ B and 2 ∉ C

Therefore, B ≠ C

**Question 9) Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B.**

** (Hints A = A ∩ ( A ∪ X ) , B = B ∩ ( B ∪ X ) and use Distributive law )**

**Answer ** Let A and B be two sets such that A ∩ X = B ∩ X = ϕ and A ∪ X = B ∪ X for some set X.

To show, A = B

A = A ∩ (A ∪ X) = A ∩ (B ∪ X) [A ∪ X = B ∪ X]

A = (A ∩ B) ∪ (A ∩ X) [Distributive law]

A = (A ∩ B) ∪ Φ [A ∩ X = Φ]

A = A ∩ B ….(i)

Now, B = B ∩ (B ∪ X)

B = B ∩ (A ∪ X) [A ∪ X = B ∪ X]

B = (B ∩ A) ∪ (B ∩ X) … [Distributive law]

B = (B ∩ A) ∪ Φ [B ∩ X = Φ]

B = A ∩ B ….(ii)

Hence, from equations (i) and (ii), we obtain A = B.

**Question 10) Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = φ.**

**Answer**

Let us assume, A = {0, 1} B = {1, 2} C = {2, 0}

According to the question,

A ∩ B = {1}

B ∩ C = {2}

And,

A ∩ C = {0}

∴ A ∩ B, B ∩ C and A ∩ C are not empty sets

Hence, we get,

A ∩ B ∩ C = Φ

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