Anand Classes offers detailed and reliable Sets NCERT Solutions Exercise 1.5 Class 11 Maths Chapter 1 to help students strengthen their understanding of key concepts in set theory. These solutions are prepared according to the latest NCERT and CBSE syllabus, providing step-by-step explanations for each question to ensure complete conceptual clarity. Perfect for Class 11 students revising for exams, these solutions make learning smooth, systematic, and exam-oriented. Click the print button to download study material and notes.
NCERT Question 1. Let U =ย {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}. ย Find:
(i) A’ $\quad $ (ii) Bโฒ $\quad $ (iii) (A โช C)โฒ $\quad $ (iv) (A โช B)โฒ$\quad $
(v) (Aโฒ)โฒ $\quad $ (vi) (B โ C)โฒ.
Solution :
(i)
Aโฒ = U โ A = { 1, 2, 3, 4, 5, 6, 7, 8, 9 } โ { 1, 2, 3, 4 } = { 5, 6, 7, 8, 9 }
(ii)
Bโฒ = U โ B = { 1, 2, 3, 4, 5, 6, 7, 8, 9 } โ { 2, 4, 6, 8 } = { 1, 3, 5, 7, 9 }
(iii)
A โช C = { 1, 2, 3, 4 } โช { 3, 4, 5, 6 } = { 1, 2, 3, 4, 5, 6 }
(A โช C)โฒ= U โ (A โช C) = { 1, 2, 3, 4, 5, 6, 7, 8, 9 } โ { 1, 2, 3, 4, 5, 6 } = { 7, 8, 9 }
(iv)
A โช B = { 1, 2, 3, 4 } โช { 2, 4, 6, 8 } = { 1, 2, 3, 4, 6, 8 }
(A โช B)โฒ= U โ (A โช B) = { 1, 2, 3, 4, 5, 6, 7, 8, 9 } โ { 1, 2, 3, 4, 6, 8 } = { 5, 7, 9 }
(v)
Aโฒ = U โ A = { 1, 2, 3, 4, 5, 6, 7, 8, 9 } โ { 1, 2, 3, 4 } = { 5, 6, 7, 8, 9 }
(Aโฒ)โฒ = U โ Aโฒ = { 1, 2, 3, 4, 5, 6, 7, 8, 9 } โ { 5, 6, 7, 8, 9 } = { 1, 2, 3, 4 } = A
(vi)
B โ C = { 2, 4, 6, 8 } โ { 3, 4, 5, 6 } = { 2, 8 }
(B โ C)โฒ= U โ (B โ C) = { 1, 2, 3, 4, 5, 6, 7, 8, 9 } โ { 2, 8 } = { 1, 3, 4, 5, 6, 7, 9 }.
NCERT Question 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}
Complement of set A = A
A’ = U – A
A’ = {a, b, c, d, e, f, g, h} – {a, b, c}
A’ = {d, e, f, g, h}
(ii)
(ii) B = { d, e, f, g }
Complement of set B = B’
B’ = U – B
B’ = {a, b, c, d, e, f, g, h} – {d, e, f, g}
B’ = {a, b, c, h}
(iii) C = {a, c, e, g}
Complement of set C = C’
C’ = U – C
C’ = {a, b, c, d, e, f, g, h} – {a, c, e, g}
C’ = {b, d, f, h}
(iv) D = {f, g, h, a}
Complement of set D = D’
D’ = U – D
D’ = {a, b, c, d, e, f, g, h} – {f, g, h, a}
D’ = {b, c, d, e}
NCERT Question 3.ย Taking the set of natural numbers as the universal set, write down the complements of the following sets:
(a) {x : x is an even natural number}
(b) {x : x is an odd natural number}
(c) {x : x is a positive multiple of 3}
(d) {x : x is a prime number}
(e) {x : x is a natural number divisible by 3 and 5}
(f) {x : x is a perfect square}
(g) {x : x is a perfect cube}
(h) {x : x + 5 = 8}
(i) {x : 2x + 5 = 9}
(j) {x : x โฅ 7}
(k) {x : x โ N and 2x + 1 > 10}
Solution : Let
U = N: set of natural numbers
(a) {x : x is an even natural number}
=> {x : x is an odd natural number}
(b) {x : x is an odd natural number}
=>{x : x is an even natural number}
(c) {x : x is a positive multiple of 3}
=>{x : x โ N and x is not a multiple of 3}
(d) {x : x is a prime number}
=> {x : x is a positive composite number and x = 1}
Def: Composite number: A natural number > 1 is said to be composite if it is not prime, i.e., if it has at least one divisor other than 1 and itself. For example, 4, 6, 8, 9, โฆ are composite.
Note. In fact N is the union of (set of primes, set of composites and {1})
(e) {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}
(f) {x : x is a perfect square}
=> {x : xโN and x is not perfect square}
(g) {x : x is a perfect cube}
=> {x : x โ N and x is not perfect cube}
(h) {x : x + 5 = 8}
=> {x : x โ N ย and x โ 3}
(i) {x : 2x + 5 = 9}
=> {x : x โ N ย and x โ 2}
(j) {x : x โฅ 7}
=> {x : xโN ย and x < 7}
(k) {x : x โ N and 2x + 1 > 10}
=> {x : xโN and x โค 9/2} = {1, 2, 3, 4}
Question 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
(a)ย (A โช B)’= A’ โฉ B’
(b) (A โฉ B)โฒ = Aโฒ โช Bโฒ
Solution:
(a)ย (A โช B)’= A’ โฉ B’
=> (A โช B)’= U – (AโชB)
=> {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 3, 4, 5, 6, 7, 8}
=> (AโชB)’ = {1, 9}
A’ โฉ B’ = (U – A) โฉ (U – B)
=> {1, 3, 5, 7, 9} โฉ {1, 4, 6, 8, 9}
=> A’ โฉ B’ = {1, 9}
Hence, Verified!!! (Aโช B)’ = A’ โฉ B’
(b) (A โฉ B)โฒ = Aโฒ โช Bโฒ
=> (A โฉ B)’ = U – (A โฉ B)
=> {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2}
=> (A โฉ B)โฒ = {1, 3, 4, 5, 6, 7, 8, 9}
A’ โช B’= (U – A) โช (U – B)
=> {1, 3, 5, 7, 9} โช {1, 4, 6, 8, 9}
=> Aโฒ โช Bโฒ = {1, 3, 4, 5, 6, 7, 8, 9}
Hence, Verified!!! (A โฉ B)โฒ = Aโฒ โช Bโฒ
NCERT Question 5.ย Draw appropriate Venn diagram for each of the following:ย
(i) (A โช B)โฒ (ii) Aโฒ โฉ Bโฒ (iii) (A โฉ B)โฒ (iv) Aโฒ โช Bโฒ.
(a) (A โช B)โฒ =ย
(b) A’ โฉ B’ =
(c) (A โฉ B)โฒ =
(d) Aโฒ โช Bโฒ =
NCERT Question 6. Let U be the set of all triangles in a plane. If A is the set of all triangles with atleast one angle different from 60ยฐ, what is Aโฒ?
Solution:
U = set of all triangles in plane
Aโฒ = U โ A
A = set of all triangles with at least one angle different from 60ยฐ
A’ = set of all triangles with no angle different from 60ยฐ i.e, set of all triangles with all angles 60ยฐ
A’ is the set of all equilateral triangle.
NCERT Question 7. Fill in the blanks to make each of the following a true statement :
(a) A โช Aโฒ = โฆ
(b) ฯโฒ โฉ A = โฆ
(c) A โฉ Aโฒ = โฆ
(d) Uโฒ โฉ A = โฆ
Solution:
(a) A โช Aโฒ = U
(b) โ โฒ โฉ A = A
(c) A โฉ Aโฒ = โ
(d) Uโฒ โฉ A = โ
