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1. Sets Mathematics Class 11 In English Medium Ncert Book Solutions Exercise 1.4


 NCERT Solutions Mathematics class 11

 

1. Sets : Exercise 1.4 Mathematics class 11th:English Medium NCERT Book Solutions

NCERT Books Subjects for class 11th Hindi Medium

Chapter 1. Sets Exercise 1.4: NCERT Book Solutions for class 11th. All solutions and extra or additional solved questions for 1. Sets : Exercise 1.4 Mathematics class 11th:English Medium NCERT Book Solutions.. All ncert books and cbse syllabus are solved chapter by chapter and also exercise within chapter and exercise solved by our expert in Hindi and English Medium for studends.

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Page 4 of 7

1. Sets

 

Exercise 1.4

Exercise 1.4


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, ∩ 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

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Page 4 of 7

See other sub-topics of this chapter:

1. Exercise 1.1 class 11 Chap-1. Sets

2. Exercise 1.2 class 11 Chap-1. Sets

3. Exercise 1.3 class 11 Chap-1. Sets

4. Exercise 1.4 class 11 Chap-1. Sets

5. Exercise 1.5 class 11 Chap-1. Sets

6. Exercise 1.6 class 11 Chap-1. Sets

7. Miscellaneous class 11 Chap-1. Sets

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Mathematics Chapter List

  • 1. Sets

  • 2. Relations and Functions

  • 3. Trigonometric Functions

  • 4. Principle Of Mathematical Induction

  • 5. Complex Numbers and Quadratic Equations

  • 6. Linear Inequalities

  • 7. Permutations and Combinations

  • 8. Binomial Theorem

  • 9. Sequences and Series

  • 10. Straight Lines

  • 11. Conic Sections

  • 12. Introduction to Three Dimensional Geometry

  • 13. Limits and Derivatives

  • 14. Mathematical Reasoning

  • 15. Statistics

  • 16. Probability


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