NCERT Solutions class 11 Mathematics 1. Sets Exercise 1.4

Exercise 1.4

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Q1. Find the union of each of the following pairs of sets :

(i) X = {1, 3, 5} Y = {1, 2, 3}
(ii) A = [ a, e, i, o, u} B = {a, b, c}
(iii) A = {x : x is a natural number and multiple of 3}
    B = {x : x is a natural number less than 6}
(iv) A = {x : x is a natural number and 1 < x ≤ 6 }
    B = {x : x is a natural number and 6 < x < 10 }
(v) A = {1, 2, 3}, B = φ

Solution: 

(i) X = {1, 3, 5} Y = {1, 2, 3}

X ∪ Y= {1, 2, 3, 5}

(ii) A = {a, e, i, o, u} B = {a, b, c}

A ∪ B = {a, b, c, e, i, o, u}

(iii) A = {x: x is a natural number and multiple of 3} = {3, 6, 9 ...}

B = {x: x is a natural number less than 6} = {1, 2, 3, 4, 5, 6}

A ∪ B = {1, 2, 4, 5, 3, 6, 9, 12 ...}

∴ A ∪ B = {x: x = 1, 2, 4, 5 or a multiple of 3}

(iv) A = {x: x is a natural number and 1 < x ≤ 6} = {2, 3, 4, 5, 6}

B = {x: x is a natural number and 6 < x < 10} = {7, 8, 9}

A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9}

∴ A∪ B = {x: x ∈ N and 1 < x < 10}

(v) A = {1, 2, 3}, B = Φ

A ∪ B = {1, 2, 3}

Q2. Let A = { a, b }, B = {a, b, c}. Is A ⊂ B ? What is A ∪ B ?

Solution:

Here, A = {a, b} and B = {a, b, c}

Yes, A ⊂ B.

As a ∈ B and b ∈ B 

A ∪ B = {a, b, c} = B

{Rule: if A ∪ B = B then A ⊂ B Or if A ∪ B = A then B ⊂ A } 

Q3. If A and B are two sets such that A ⊂ B, then what is A ∪ B ?

Solution: 

Given that: A and B are two sets such that A ⊂ B 

Then A ∪ B = B 

Illustration by example:

Let A = {1, 2, 3} and B = {1, 2, 3, 4, 5} 

Here A ⊂ B Because All elements of A 1, 2, 3 ∈ B

[B also contains 1, 2, 3]

Now, A ∪ B = {1, 2, 3, 4, 5} = B 

Therefore, A ∪ B = B

Q4. 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: 

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

(i) A ∪ B = {1, 2, 3, 4, 5, 6}

(ii) A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}

(iii) B ∪ C = {3, 4, 5, 6, 7, 8}

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

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

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

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

Q5. Find the intersection of each pair of sets of question 1 above.

(i) X = {1, 3, 5} Y = {1, 2, 3}
(ii) A = [ a, e, i, o, u} B = {a, b, c}
(iii) A = {x : x is a natural number and multiple of 3}
    B = {x : x is a natural number less than 6}
(iv) A = {x : x is a natural number and 1 < x ≤ 6 }
    B = {x : x is a natural number and 6 < x < 10 }
(v) A = {1, 2, 3}, B = φ

Solution:

(i) X = {1, 3, 5}, Y = {1, 2, 3}

   X ∩ Y = {1, 3}

(ii) A = {a, e, i, o, u}, B = {a, b, c}

   A ∩ B = {a}

(iii) A = {x: x is a natural number and multiple of 3} = (3, 6, 9 ...}

   B = {x: x is a natural number less than 6} = {1, 2, 3, 4, 5}

   ∴ A ∩ B = {3}

(iv) A = {x: x is a natural number and 1 < x ≤ 6}

Or  A = {2, 3, 4, 5, 6}

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

Or B = {7, 8, 9}

   A ∩ B = Φ

(v) A = {1, 2, 3}, B = Φ.

   So, A ∩ B = Φ

Q6. If A = { 3, 5, 7, 9, 11 }, 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 = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13},

Therefore, A ∩ B = {7, 9, 11}

Solution:

(ii) B = {7, 9, 11, 13}, and C = {11, 13, 15}

Therefore, B ∩ C = {11, 13}

Solution:

(iii) A = { 3, 5, 7, 9, 11 }, C = {11, 13, 15} and D = {15, 17};

A ∩ C ∩ D = { A ∩ C} ∩ D = {11} ∩ {15, 17} = Φ

Solution:

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

Therefore, A ∩ C = {11}

Solution:

(v) B = {7, 9, 11, 13}, and D = {15, 17};

Therefore, B ∩ D = Φ

Solution:

(vi) A ∩ (B C) = (A ∩ B) (A ∩ C)

= {7, 9, 11} {11} = {7, 9, 11}

Solution:

(vii) A ∩ D = Φ

Solution:

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

We know; A ∩ (B ∪ D) = (A ∩ B) ∪ (A ∩ D)

 = {7, 9, 11} Φ = {7, 9, 11}

Solution:

(ix) A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, and C = {11, 13, 15}

(A ∩ B) = {7, 9, 11}

(B ∪ C) = {7, 9, 11, 13, 15}

(A ∩ B) ∩ (B ∪ C)

= {7, 9, 11} ∩ {7, 9, 11, 13, 15}

= {7, 9, 11}

Solution:

(x) Given A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, C = {11, 13, 15}and D = {15, 17};

(A ∪ D) ∩ (B ∪ C)

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

= {7, 9, 11, 15}

Q7. If A = {x : x is a natural number },

B = {x : x is an even natural number},
C = {x : x is an odd natural number} and

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 7:

A = {x: x is a natural number} = {1, 2, 3, 4, 5 ...}

B = {x: x is an even natural number} = {2, 4, 6, 8 ...}

C = {x: x is an odd natural number} = {1, 3, 5, 7, 9 ...}

D = {x: x is a prime number} = {2, 3, 5, 7 ...}

Solution:

(i) A ∩ B = {2, 4, 6, 8 .....}

A ∩ B = {x : x is a even natural number}

A ∩ B = B

Solution:

(ii) A ∩ C = {1, 3, 5, 7, 9 ...}

A ∩ C = {x : x is an odd natural number}

A ∩ C = C

Solution:

(iii) A ∩ D = {2, 3, 5, 7 ...}

A ∩ D = {x : x is a prime number}

A ∩ D = D

Solution:

(iv) B ∩ C = Φ

Solution:

(v) B ∩ D = {2}

Solution:

(vi) C ∩ D = {3, 5, 7, 11 ....} 

C ∩ D = {x : x is odd prime number}

Q8. 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}

{Note : Disjoint means no elements matched in each others, in other words intersection of two sets give null set then it said to be disjoint.}

Solution:

(i) Let A = {1, 2, 3, 4} and 

   B = {x: x is a natural number and 4 ≤ x ≤ 6}

Or B = {4, 5, 6}

Now, A ∩ B 

= {1, 2, 3, 4} ∩ {4, 5, 6}

= {4}

Therefore, this pair of sets is not disjoint.

Solution:

(ii) Let X = {a, e, i, o, u} and 

       Y = (c, d, e, f}

Now, X ∩ Y

= {a, e, i, o, u} ∩ (c, d, e, f}

= {e}

Therefore, {a, e, i, o, u} and (c, d, e, f} are not disjoint.

Solution:

(iii) Let A = {x : x is an even integer}

Or     A = {2, 4, 6, 8 ... } and 

        B = {x : x is an odd integer}

Or      B = {1, 3, 5, 7 ... }

Now, A ∩ B 

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

= Φ

Therefore,

{x : x is an even integer} ∩ {x : x is an odd integer}

= Φ

Therefore, A ∩ B gives null set so this pair of sets is disjoint.

Q9. If A = {3, 6, 9, 12, 15, 18, 21}, B = { 4, 8, 12, 16, 20 },
C = { 2, 4, 6, 8, 10, 12, 14, 16 }, 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, 15, 18, 21}

(ii) A – C = {3, 9, 15, 18, 21}

(iii) A – D = {3, 6, 9, 12, 18, 21}

(iv) B – A = {4, 8, 16, 20}

(v) C – A = {2, 4, 8, 10, 14, 16}

(vi) D – A = {5, 10, 20}

(vii) B – C = {20}

(viii) B – D = {4, 8, 12, 16}

(ix) C – B = {2, 6, 10, 14}

(x) D – B = {5, 10, 15}

(xi) C – D = {2, 4, 6, 8, 12, 14, 16}

(xii) D – C = {5, 15, 20}

Q10. If X = { a, b, c, d } and Y = { f, b, d, g}, find
(i) X – Y

(ii) Y – X

(iii) X ∩ Y

Solution: 

(i) X – Y = { a, c }

(ii) Y – X = {f, g}

(iii) X ∩ Y = {b, d}

Q11. If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?

Solution: 

R = {x : x is set of real numbers}

Q = {x : x is set of rational numbers} 

We know that,

Real numbers = Rational Numbers + Irrational numbers

Real numbers - Rational Numbers = Irrational numbers
R - Q = I

Therefore, R – Q is a set of irrational numbers.

Q12. 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.

Solution: 

(i) False, Because { 2, 3, 4, 5 } ∩ { 3, 6} = {3} 

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

Solution: 

(ii) False, Because { a, e, i, o, u } ∩ { a, b, c, d } = {a}

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

Solution: 

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

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

Solution: 

(iv) True, Because { 2, 6, 10 } ∩ { 3, 7, 11} = Φ