NCERT Solutions for Class 11 Maths Chapter 1 - Sets Exercise 1.5

The NCERT Solutions for Class 11 Maths Chapter 1 Exercise 1.5 are created by subject experts according to the latest CBSE Syllabus 2023-24. Here, we have provided the solutions for the questions in the fifth exercise of the chapter. Exercise 1.5 of NCERT Solutions for Class 11 Maths Chapter 1 – Sets are based on the following topics:

1. Complement of a Set: If U is the universal set and A a subset of U. Then the complement of A is the set of all elements of U which are not the elements of A.
   1. De Morgan’s laws: The complement of the union of two sets is the intersection of their complements, and the complement of the intersection of two sets is the union of their complements.
   2. Some Properties of Complement Sets:
      1. Complement laws
      2. De Morgan’s law
      3. Law of double complementation
      4. Laws of empty set and universal set

Students can access these NCERT Solutions for Class 11 Maths Chapter 1 and kickstart their exam preparations.

NCERT Solutions for Class 11 Maths Chapter 1 – Sets Exercise 1.5

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Class 11 Maths Chapter 1 – Sets Exercise 1.5 Solutions

1. Let U = {1, 2, 3; 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8}, and C = {3, 4, 5, 6}. Find

(i) A’

(ii) B’

(iii) (A U C)’

(iv) (A U B)’

(v) (A’)’

(vi) (B – C)’

Solution:

It is given that

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

A = {1, 2, 3, 4}

B = {2, 4, 6, 8}

C = {3, 4, 5, 6}

(i) A’ = {5, 6, 7, 8, 9}

(ii) B’ = {1, 3, 5, 7, 9}

(iii) A U C = {1, 2, 3, 4, 5, 6}

So, we get

(A U C)’ = {7, 8, 9}

(iv) A U B = {1, 2, 3, 4, 6, 8}

So, we get

(A U B)’ = {5, 7, 9}

(v) (A’)’ = A = {1, 2, 3, 4}

(vi) B – C = {2, 8}

So, we get

(B – C)’ = {1, 3, 4, 5, 6, 7, 9}

2. If U = {a, b, c, d, e, f, g, h}, find the complements of the following sets:

(i) A = {a, b, c}

(ii) B = {d, e, f, g}

(iii) C = {a, c, e, g}

(iv) D = {f, g, h, a}
Solution:

(i) A = {a, b, c}

So, we get

A’ = {d, e, f, g, h}

(ii) B = {d, e, f, g}

So, we get

B’ = {a, b, c, h}

(iii) C = {a, c, e, g}

So, we get

C’ = {b, d, f, h}

(iv) D = {f, g, h, a}

So, we get

D’ = {b, c, d, e}

3. Taking the set of natural numbers as the universal set, write down the complements of the following sets:

(i) {x: x is an even natural number}

(ii) {x: x is an odd natural number}

(iii) {x: x is a positive multiple of 3}

(iv) {x: x is a prime number}

(v) {x: x is a natural number divisible by 3 and 5}

(vi) {x: x is a perfect square}

(vii) {x: x is perfect cube}

(viii) {x: x + 5 = 8}

(ix) {x: 2x + 5 = 9}

(x) {x: x ≥ 7}

(xi) {x: x ∈ N and 2x + 1 > 10}

Solution:

We know that

U = N: Set of natural numbers

(i) {x: x is an even natural number}´ = {x: x is an odd natural number}

(ii) {x: x is an odd natural number}´ = {x: x is an even natural number}

(iii) {x: x is a positive multiple of 3}´ = {x: x ∈ N and x is not a multiple of 3}

(iv) {x: x is a prime number}´ ={x: x is a positive composite number and x = 1}

(v) {x: x is a natural number divisible by 3 and 5}´ = {x: x is a natural number that is not divisible by 3 or 5}

(vi) {x: x is a perfect square}´ = {x: x ∈ N and x is not a perfect square}

(vii) {x: x is a perfect cube}´ = {x: x ∈ N and x is not a perfect cube}

(viii) {x: x + 5 = 8}´ = {x: x ∈ N and x ≠ 3}

(ix) {x: 2x + 5 = 9}´ = {x: x ∈ N and x ≠ 2}

(x) {x: x ≥ 7}´ = {x: x ∈ N and x < 7}

(xi) {x: x ∈ N and 2x + 1 > 10}´ = {x: x ∈ N and x ≤ 9/2}

4. If U = {1, 2, 3, 4, 5,6,7,8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Verify that

(i) (A U B)’ = A’ ∩ B’

(ii) (A ∩ B)’ = A’ U B’

Solution:

It is given that

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

A = {2, 4, 6, 8}

B = {2, 3, 5, 7}

(i) (A U B)’ = {2, 3, 4, 5, 6, 7, 8}’ = {1, 9}

A’ ∩ B’ = {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9} = {1, 9}

Therefore, (A U B)’ = A’ ∩ B’.

(ii) (A ∩ B)’ = {2}’ = {1, 3, 4, 5, 6, 7, 8, 9}

A’ U B’ = {1, 3, 5, 7, 9} U {1, 4, 6, 8, 9} = {1, 3, 4, 5, 6, 7, 8, 9}

Therefore, (A ∩ B)’ = A’ U B’.

5. Draw an appropriate Venn diagram for each of the following:

(i) (A U B)’

(ii) A’ ∩ B’

(iii) (A ∩ B)’

(iv) A’ U B’

Solution:

(i) (A U B)’

(ii) A’ ∩ B’

(iii) (A ∩ B)’

(iv) A’ U B’

6. Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A’?

Solution:

A’ is the set of all equilateral triangles.

7. Fill in the blanks to make each of the following a true statement.

(i) A U A’ = ……..

(ii) Φ′ ∩ A = …….

(iii) A ∩ A’ = …….

(iv) U’ ∩ A = …….

Solution:

(i) A U A’ = U

(ii) Φ′ ∩ A = U ∩ A = A

So, we get

Φ′ ∩ A = A

(iii) A ∩ A’ = Φ

(iv) U’ ∩ A = Φ ∩ A = Φ

So, we get

U’ ∩ A = Φ

Access other exercise solutions of Class 11 Maths Chapter 1 – Sets

The NCERT Class 11 Solutions of Chapter 1 can be accessed using the links below.

Exercise 1.1 Solutions 6 Questions

Exercise 1.2 Solutions 6 Questions

Exercise 1.3 Solutions 9 Questions

Exercise 1.4 Solutions 12 Questions

Exercise 1.6 Solutions 8 Questions

Miscellaneous Exercise on Chapter 1 Solutions 16 Questions