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Solutions ⇒ Class 11th ⇒ Mathematics ⇒ 1. Sets

Solutions 1. Sets - Exercise 1.4 | Class 11 Mathematics - Toppers Study

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1. Sets

| Exercise 1.4 |

Solutions 1. Sets - Exercise 1.4 | Class 11 Mathematics - Toppers Study


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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Mathematics Class - 11th

NCERT Maths book for CBSE Students.

books

Study Materials List:

Solutions ⇒ Class 11th ⇒ Mathematics
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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