Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 1 Sets Ex 1.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 1 Sets Ex 1.4

Question 1.

Find the union of each of the following pairs of sets:

1. X = {1, 3, 5} Y = {1, 2, 3}

2. A = {a, e, i, o, u} B = {a, b, c}

3. A = {x : x is a natural number and multiple of 3},

B = {x : x is a natural number less than 6}.

4. A = {x : x is a natural number and 1 < x ≤ 6},

B = {x : x is a natural number and 6 < x < 10}.

5. A = {1, 2, 3}, B = ϕ

Solution:

- X ∪ Y = {1, 3, 5} ∪ (1, 2, 3} = {1, 2, 3, 5}
- A ∪ B = [a, e, i, o, u} ∪ {a, b, c} = {a, b, c, e, i, o, u}
- A ∪ B = {3, 6, 9, …} ∪ {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5, 6, 9, 12, 15, …}
- A ∪ B = {2, 3, 4, 5, 6} ∪ {7, 8, 9} = {2, 3, 4, 5, 6, 7, 8, 9}.
- A ∪ B = (1, 2, 3} ∪ϕ = {1, 2, 3}.

Question 2.

Let A = {a, b} and B = {a, b, c). Is A ⊂ B? What is A ∪ B?

Solution:

Yes A ⊂B, because every element of A is also an element of B. Therefore, A is a subset of B, i.e., A ⊂ B.

A ∪ B = (a, b} ∪ {a, b, c} = {a, b, c}.

Question 3.

If A and B are two sets such that A ⊂ B, then what is A ∪ B?

Solution:

Since A is subset of B, therefore every element of set A is contained in the set B and hence A ∪ B = B.

Question 4.

If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10},

Find:

(i) A ∪ B

(ii) A ∪ C

(iii) B ∪ C

(iv) B ∪ D

(v) A ∪ B ∪ C

(vi) A ∪ B ∪ D

(vii) B ∪ C ∪ D.

Solution:

(i) A ∪ B = {1, 2, 3, 4} ∪ {3, 4, 5, 6} = {1, 2, 3, 4, 5, 6}

(ii) A ∪ C = (1, 2, 3, 4} ∪ {5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8}

(iii) B ∪ C = {3, 4, 5, 6} ∪ {5, 6, 7, 8} = {3, 4, 5, 6, 7, 8}

(iv) B ∪ D = {3, 4, 5, 6} ∪{7, 8, 9, 10} = {3, 4, 5, 6, 7, 8, 9, 10}

(v) A ∪ B ∪ C = {1, 2, 3, 4} ∪{3, 4, 5, 6} ∪ {5, 6, 7, 8}

= {1, 2, 3, 4, 5, 6} ∪ {5, 6, 7, 8}

= {1, 2, 3, 4, 5, 6, 7, 8}.

(vi) A ∪ B ∪ D = (1, 2, 3, 4} ∪ {3, 4, 5, 6} ∪{7, 8, 9, 10}

= {1, 2, 3, 4, 5, 6} u {7, 8, 9, 10}

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

(vii) B ∪ C ∪ D = {3, 4, 5, 6} ∪{5, 6, 7, 8} ∪{7, 8, 9, 10}

= {3, 4, 5, 6, 7, 8} ∪ {7, 8, 9, 10}

= {3, 4, 5, 6, 7, 8, 9, 10}.

Question 5.

Find the intersection of each pair of sets of question 1 above?

Solution:

- X ∩ Y = {1, 3, 5} ∩ {1, 2, 3} = (1, 3}.
- A ∩ B = {a, e, i, o, u) ∩ [a, b, c] = {a}.
- A ∩ B = {3, 6, 9, ….,} ∩ {1, 2, 3, 4, 5} = {3}.
- A ∩ B = {2, 3, 4, 5, 6} ∩ {7, 8, 9} = ϕ.
- A ∩ B = {1, 2, 3} ∩ ϕ = ϕ.

Question 6.

If A = {3, 5, 7, 9, 11} and B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}, find:

(i) A ∩ B

(ii) B ∩ C

(iii) A ∩ C ∩ D

(iv) A ∩ C

(v) B ∩ D

(vi) A ∩(B ∪ C)

(vii) A ∩ D

(viii) A ∩ (B ∪D)

(ix) (A ∩B) ∩ (B ∪ C)

(x) (A ∪ D) ∩ (B ∪ C).

Solution:

(i) A ∩ B = {3, 5, 7, 9, 11} ∩ {11, 9, 11, 13} = {7, 9,11}

(ii) {7, 9, 11, 13} ∩ {11, 13, 15} = {11, 13}.

(iii) {3, 5, 7, 9,11} ∩ {11, 13, 15} ∩ [15, 17} = {11} ∩ {15, 17} = ϕ.

(iv) A ∩ C = {3, 5, 7, 9, 11} ∩ {11, 13, 15} = {11}

(v) B ∩ D = {7, 9, 11, 13} ∩ {15, 17} = ϕ.

(vi) A ∩(B ∪C) = {3, 5, 7, 9, 11} ∩ ({7, 9, 11, 13} ∪ {11, 13, 15})

= {3, 5, 7, 9, 11} ∩ (7, 9, 11, 13, 15}

= {7, 9, 11}.

(vii) A ∩ D = {3, 5, 7, 9, 11} ∩ {15, 17} = ϕ.

(viii) A ∩ (B ∪ D) = {3, 5, 7, 9, 11} ∩ ({7, 9, 11, 13} ∪ {15, 17})

= {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15, 17}

= {7, 9, 11}.

(ix) A ∩ B = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13}

= {7, 9, 11}

B ∪ C = {7, 9, 11, 13} ∪ {11, 13, 15}

= {7, 9, 11, 13, 15}.

∴ (A ∩ B) ∩ (B ∪ C) = {7, 9, 11} ∪ {7, 9, 11, 13, 15}

= {7, 9, 11}.

(x) A ∪ D = {3, 5, 7, 9, 11} ∪ {15, 17}

= {3, 5, 7, 9, 11, 15, 17}.

B ∪ C = {7, 9, 11, 13, 15} [From part (ix)]

∴ (A ∪ D) ∩ (B ∪ C) = {3, 5, 7, 9, 11, 15, 17} ∩ {7, 9, 11, 13, 15}

= {7, 9, 11, 15}.

Question 7.

Let A = {x : x is a natural number}, B = {x : x is an even natural number}, C = {x : x is an odd natural number}, D = {x : x is a prime number}. Find:

(i) A ∩ B

(ii) A ∩ C

(iii) A ∩ D

(iv) B ∩ C

(v) B ∩ D

(vi) C ∩ D

Solution:

Given:

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

C = (1, 3, 5, 7, …}, D = {2, 3, 5, 7, 11, 13, …}

(i) A ∩ B = {1, 2, 3, 4, …} ∩ {2, 4, 6, 8, …}

= {2, 4, 6, 8, …} = B.

(ii) A ∩ C = {1, 2, 3, 4, …} ∩ (1, 3, 5, 7, …}

= {1, 3, 5, 7, …} = C.

(iii) A ∩ D = {1, 2, 3, 4, …} ∩ {2, 3, 5, 7, 11, 13, …)

= 12, 3, 5, 7, 11, 13, …} = D.

(iv) B ∩ C = {2, 4, 6, 8, …} ∩ {1, 3, 5, 7, …}

= ϕ.

(v) B ∩ D = (2, 4, 6, 8, …} ∩ {2, 3, 5, 7, 11, 13, …}

= {2}.

(vi) C ∩ D = {1, 3, 5, 7, …) ∩ (2, 3, 5, 7, 11, 13, …}

= (3, 5, 7, 11, 17, 19, …)

= (x : x is an odd prime number}.

Question 8.

Which of the following pairs of sets are disjoint?

(i) {1, 2, 3, 4} and x : x is a natural number and 4 ≤ x ≤ 6}

(ii) {a, e, i, o, u) and {c, d, e, f}

(iii) {x : x is an even integer} and {x : x is an odd integer}.

Solution:

(i) {1, 2, 3, 4} and {x : x is a natural number and 4 ≤ x ≤ 6} i.e., {4, 5, 6} are not disjoint sets as they have 4 as a common element.

(ii) {a, e, i, o, u} and {c, d, e, f} are not disjoint sets because they have common element e.

(iii) {x : x is an even integer} and {x : x is an odd integer} are disjoint sets because they have no common element.

Question 9.

Let A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16} and D = {5, 10, 15, 20}. Find:

(i) A – B

(ii) A – C

(iii) A – D

(iv) B – A

(v) C – A

(vi) D – A

(vii) B – C

(viii) B – D

(ix) C – B

(x) D – B

(xi) C – D

(xii) D – C

Solution:

(i) A – B = {3, 6, 9, 12, 15, 18, 21} – {4, 8, 12, 16, 20}

= {3 ,6, 9, 15, 18, 21}.

(ii) A – C = {3 ,6, 9, 12, 15, 18, 21} – {2, 4, 6, 8, 10, 12, 14, 16}

= {3, 9, 15, 18, 21}.

(iii) A – D = (3, 6, 9, 12, 15, 18, 21} – {5, 10, 15, 20}

= {3, 6, 9, 12, 18, 21}.

(iv) B – A = {4, 8, 12, 16, 20} – {3, 6, 9, 12, 15, 18, 21}

= {4, 8, 16, 20}.

(v) C – A = {2, 4, 6, 8, 10, 12, 14, 16} – (3, 6, 9, 12, 15, 18, 21}

= {2, 4, 8, 10, 14, 16}.

(vi) D – A = {5, 10, 15, 20} – {3, 6, 9, 12, 15, 18, 21}

{5, 10, 20}.

(vii) B – C = {4, 8, 12, 16, 20} – {2, 4, 6, 8, 10, 12, 14, 16}

= {20}.

(viii) B – D = {4, 8, 12, 16, 20} – {5, 10, 15, 20}

= {4, 8, 12, 16}.

(ix) C – B = {2, 4, 6, 8, 10, 12, 14, 16} – {4, 8, 12, 16, 20}

= {2, 6, 10, 14}.

(x) D – B = {5, 10, 15, 20} – {4, 8, 12, 16, 20}

= (5, 10, 15}.

(xi) C – D = {2, 4, 6, 8, 10, 12, 14, 16} – {5, 10, 15, 20}

= {2, 4, 6, 8, 12, 14, 16}.

(xii) D – C = {5, 10, 15, 20} – {2, 4, 6, 8, 10, 12, 14, 16}

= {5, 15, 20}.

Question 10.

If X = {a, b, c, d} and Y = {f, b, d, g}, find:

- X – Y
- Y – X
- X ∩ Y

Solution:

- X – Y = {a, b, c, d} – {f, b, d, g} = {a, c}.
- Y – X = {f, b, d, g) – {a, b, c, d} = {f, g).
- X ∩ Y = {a, b, d, d} ∩ {f, b, d, g} = {b, d}.

Question 11.

If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?

Solution:

Since set of real numbers contains the set of rational numbers and the set of irrational numbers, therefore R – Q = the set of irrational numbers.

Question 12.

State whether each of the following statement is true or false. Justify your answer?

(i) {2, 3, 4, 5} and {3, 6} are disjoint sets.

(ii) {a, e, i, o, u} and {a, b, c, d} are disjoint sets.

(iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.

(iv) {2, 6, 10} and {3, 7, 11} are disjoint sets.

Solution:

(i) False, A ∩ B = {2, 3, 4, 5} ∩ {3, 6} = {3} ≠ ϕ

∴ The given sets are not disjoint.

(ii) False, because {a, e, i, o, u} ∩ {a, b, c, d} – {a} ≠ ϕ

(iii) True, because {2, 6, 10, 14} ∩ {3, 7, 11, 15} = ϕ

(iv) True, because {2, 6, 10} ∩ {3, 7, 11} = ϕ