RBSE Solutions for Class 11 Maths Chapter 1 Sets
RBSE Solutions for Class 11 Maths Chapter 1 Sets
Rajasthan Board RBSE Class 11 Maths Chapter 1 Sets Ex 1.1
Question 1.
Fill the symbols ∈ or ∉ in the blanks to make the following statements correct:
(i) 3…. {1, 2, 3, 4, 5}
(ii) 2.5…N
(iii) 0………Q
Solution:
(i) ∈
(ii) ∉
(iii) ∈
Question 2.
Fill the symbols ⊂ or ⊄ in the blanks to make following statements correct:
(i) {2, 3, 4}……….{1, 2, 3, 4, 5}
(ii) {a, e, o}……..{a, b, c}
(iii) {x : x is an equilateral triangle in a plane}……..{x : x is a triangle in a plane}
(iv) {x : x is a natural number}…….{x : x is an odd whole number}
Solution:
(i) {2, 3, 4} ⊂ {1, 2, 3, 4, 5}
(ii) {a, e, o} ⊄ {a, b, c}
(iii) Many triangles are possible in a plane.
So, {x: x is an equilateral triangle in a plane} ⊂ (x : x is a triangle in a plane}
(iv) First set in the tabular form is (1, 2, 3, 4,….} and second set in the tabular form is {…… -3, -1, 0, 1, 3, 5,..}.
So, (x : x is a natural number} ⊄ {x : x is an odd whole number}.
Question 3.
Examine the following statements :
(i) {a, b} ⊂ {b, a, c]
(ii) {a, e} ⊂ (x : x is vowel of English alphabet}
(iii) {1, 2, 3} ⊄ {1, 3, 2, 5}
(iv) {x : x is an even natural number less than 6} ⊄ {x : x is a natural number which divide 36}
Solution:
(i) {a, b} ⊂ {b, a, c}
Elements of first set (a, b) is present in second set.
Hence, the statement is true.
(ii) {a, e} ⊂ {x : x is vowel of English alphabet}
Tabular form of seconds set is {a, e, i, o, u}
Elements of first set (a, e is present is second set.
Hence, the statement is true.
(iii) {1, 2, 3} ⊄ {1, 3, 2, 5}
Elements of first set is also the elements of seconds set.
Hence, the statement is false.
(iv) {x : x is an even natural number less then 6} ⊂ (x : x is a natural number which divide 36}
{2, 4} ⊂ {1, 2, 3, 4, 6, 9, 12, 18, 36}
Hence, the statement is true.
Question 4.
Write power sets of the following:
(i) {a}
(ii) {1, 2, 3}
(iii) {a, b}
(iv) Φ
Solution:
(i) {a}
Subsets of {a} are Φ and {a}.
Hence P{a} = {Φ, {a}}
(ii) {1, 2, 3}
Subsets of {1, 2, 3} are Φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3} and {1, 2, 3}.
Hence, P{1, 2, 3} = {Φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}
(iii) {a, b}
Subsets of {a, b} are Φ, {a}, {b}, {a, b}
Hence, P{a, b} = {Φ, {a}, {b}, {a, b}}
(iv) Φ
Φ is null set, then P(Φ) = {Φ}
Question 5.
Write the following as intervals :
(i) {x : x ∈ R, -3 < x < 6}
(ii) {x : x ∈ R, -4 ≤ x ≤ 8}
(iii) {x : x ∈ R, 4 < x ≤ 9}
(iv) {x : x ∈ R, -6 ≤ x < -1}
Solution:
(i) {x : x ∈ R, -3 < x < 6}
-3 < x < 6 is open interval where -3 and 6 are not included.
Hence, {x : x ∈ R, -3 < x < 6} has interval (-3, 6).
(ii) {x : x ∈ R, -4 ≤ x ≤ 8}
-4 ≤ x ≤ 8 is closed interval in which (-4 and 8) are included.
Hence, interval is [-4, 8]
(iii) {x : x ∈ R, 4 < x ≤ 9}
4 < x ≤ 9 is semi clsoed interval where the interval is less than 4 and equal to 9.
Hence, {x : x ∈ R, 4 < x ≤ 9} has interval (4, 9].
(iv) {x : x ∈ R, -6 ≤ x < -1}
-6 ≤ x < -1 is semi closed interval where the interval is equal to -6 and less than -1.
Hence, {x : x ∈ R, – 6 ≤ x < -1} has interval [-6, -1)
Question 6.
Write the following set in builder form.
(i) (-4, 0)
(ii) [6, 8]
(iii) [-3, 7)
(iv) (3, 10]
Solution:
(i) (-4, 0) = {x : x ∈ R, -4 < x < 0}
(ii) [6, 8] = {x : x ∈ R, 6 ≤ x ≤ 8}
(iii) [- 3, 7) = {x : x ∈ R, -3 ≤ x < 7}
(iv) (3, 10] = (x : x ∈ R, 3 < x ≤ 10}
Question 7.
A = {1, 3, 5}, B = {1, 4, 6} and C = {2, 4, 6, 8}, then which of the following may be considered as in universal set:
(i) {0, 1, 2, 3, 4, 5, 6}
(ii) (1, 2, 3, 4, 5, 6, 7, 8}
(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(iv) Φ
Solution:
For set A = {1, 3, 5}, B = {2, 4, 6} and C = {2, 4, 6, 8}
we have (ii) and (iii) universal set, because it contains all the elements of the given sets A, B and C.
Hence, option (ii) and (iii) are correct.
Rajasthan Board RBSE Class 11 Maths Chapter 1 Sets Miscellaneous Exercise
Question 1.
Solution of equation x2 + x – 2 = 0 in Roster form is:
(A) {1, 2}
(B) {-1, 2}
(C) {-1, -2}
(D) {1, -2}
Solution:
Given, x2 + x – 2 = 0
x2 + 2x – x – 2 = 0
x(x + 2) – 1(x + 2) = o
(x + 2)(x – 1) = 0
when x + 2 = 0 then x = -2
when x – 1 = 0 then x = 1
Hence, roster form of equation is {-2, 1} or {1, -2}
Hence, option (D) is correct.
Question 2.
{B = Y : y is a vowel of English alphabet), its Roster form:
(A) {a, e, i, o}
(B) {a, o, u}
(C) {a, e, o, u}
(D) {a, e, i, o, u}
Solution:
Vowels of English alphabet are a, e, i, o, u.
Thus, the set of vowels is {a, e, i, o, u}.
Hence, the option (D) is correct.
Question 3.
Set builder form of set A = {1, 4, 9, 16, 25,…….} will be:
(A) {x : x is an odd natural no.)
(B) {x : x is an even natural no.}
(C) {x : x is square of natural no.}
(D) {x : x is a prime natural no.}
Solution:
Elements of set A = {1, 4, 9, 16, 25,…} are the square of 1, 2, 3, 4, 5,… respectives.
Hence, option (C) is correct.
Question 4.
Which set is infinite of the following:
(A) {x : x ∈ N and (x – 1)(x – 2) = 0}
(B) {x : x ∈ N and x2 = 4}
(C) {x : x ∈ N and 2x – 1 = 0}
(D) {x : x ∈ N and x is a prime no.}
Solution:
Option (D) is correct, because element of this set is a prime number which is infinite.
Hence, this set is infinite.
Question 5.
If A = {0}, B = {x : x > 15 and x < 5}, C = {x : x – 5 = 0}, D = {x : x2 = 25}, E = {x : x is a positive integer root of equation x2 – 2x – 15 = 0}, then pair of equal sets is:
(A) A, B
(B) B, C
(C) B, C
(D) C, E
Solution:
Option (D) is correct, as in the tabular form the set C = {5} and set E = {5}
So, C = E
Pair of equal sets is C and E.
Question 6.
For sets Φ, A = {1, 3}, B = {1, 5, 9}, C = {1, 5, 7, 9} True option is:
(A) A ⊂ B
(B) B ⊂ C
(C) C ⊂ B
(D) A ⊂ C
Solution:
Option (B) is correct, because all the elements of set B = {1, 5, 9} is present in set C = (1, 5, 7, 9}
Hence, B ⊂ C
Question 7.
If A = {2, 4, 6, 8} and B = {1, 4, 7, 8} then A – B and B – A will be respectively:
(A) {2, 6}; {1, 7}
(B) {1, 7}; {4, 8}
(C) {1, 7}; {2, 6}
(D) {4, 8}; {1, 7}
Solution:
A = {2, 4, 6, 8} and B = {1, 4, 7, 8}
A – B= {2, 6} and B – A = {1, 7}
Hence, option (A) is correct.
Question 8.
Which of the following statement is true?
(A) A ∩ B = Φ ⇒ A = Φ and B = Φ
(B) A – B = Φ ⇒ A ⊂ B
(C) A ∪ B = Φ ⇒ A ⊂ B
(D) None of these
Solution:
Option (A) is correct, because in A ∩ B = Φ, Φ is common in A and B.
Question 9.
If A ∩ B = then:
(A) A – B = Φ
(B) A – B = A
(C) A ∪ B = Φ
(D) A – B = B
Solution:
A ∩ B = Φ
Φ, is common in A and B both. So, other elements of A is not present in B.
Thus A – B = A
Hence, option (B) is correct.
Question 10.
Shaded portion of the following Venn diagram represents:
(A) A ∪ B
(B) A ∩ B
(C) A – B
(D) B – A
Solution:
Shaded portion shows that all the elements in A which is not in B i.e, A – B
Hence, option (C) is correct.
Question 11.
If A = {1, 2, 3, 4, 5, 6}, B = {2, 4, 6, 8}, then A – B will be:
(A) {1, 3, 5, 8}
(B) {1, 3, 5}
(C) {1, 2, 3, 4, 5, 6, 8}
(D) = { }
Solution:
Given, A = {1, 2, 3, 4, 5, 6}
and B = {2, 4, 6, 8}
A – B means A contains the element which is not present in B.
Thus, A – B = { 1, 3, 5}
Hence, option (B) is correct.
Question 12.
Which of the following statement is true ?
(A) {2, 3, 4, 5} and {3, 6} are disjoint sets.
(B) {a, e, i, o, u} and {a, b, c, d) are disjoint sets.
(C) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.
(D) {2, 7, 10} and {3, 7, 11} are disjoint sets.
Solution:
Option (C) is correct because in the two sets there are no common elements.
Question 13.
If U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5} then:
(A) (A ∪ B)’ = {2, 3, 4, 5}
(B) B – A = {4, 5}
(C) A – B = {2, 4, 5}
(D) (A ∪ B) = {3}
Solution:
Option (B) is correct because,
B – A means B contains the element which is not present in A.
Thus, B – A = {4, 5}
Question 14.
The shaded portion of the following Venn diagram represents:
(A) (A ∩ B) ∩ C
(B) (A ∪ B) ∩ C
(C) (A ∩ B) ∪ C
(D) (A ∩ B) ∪ (A ∩ C)
Solution:
In the figure, the first shaded part is common of A and B
i.e., A ∩ B and the second shaded part is common of A and C
i.e., A ∩ C and full shaded region means union of A ∩ B and A ∩ C
i.e.,(A ∩ B) ∪ (A ∩ C)
Hence, option (D) is correct.
Question 15.
Which of Venn diagram represents set A – (B ∩ C) is:
Solution:
A – (B ∩ C) means that all the elements in set A is not present in set (B ∩ C) which is shown in figure B.
Question 16.
Which of the following are sets? Justify your answer.
(i) Collection of even natural less than 8.
(ii) The collection of big cities in India.
(iii) The collection of various geometrical figures.
(iv) The collection of all the integers which divides 46.
(v) The collection of the best 20 cricket batsmen of the world.
(vi) The collection of all even integers.
(vii) The collection of literature written by the poet Kalidas.
(viii) The collection of Greatmen contributed to Indian culture.
Solution:
(i) Even natural numbers less than 8 are 2, 4, 6. Hence, {2, 4, 6} is a set.
(ii) The collection of big cities in India is not a set because there is no scale to measure it.
(iii) The collection of various geometrical figures is not a set because these figures are not in particular shape.
(iv) The numbers which divide the numbers 46 are 1, 2, and 23. Hence, {1, 2, 23} is a set.
(v) The collection of the best 20 cricket batsman of the world is not a set because there is no scale to measure their qualities.
(vi) The collection of all even integer is a set because all even integers are 2, 4, 6, 8, 10,……..
(vii) The collection of literature written by the poet Kalidas is a set because it is renowned literature.
(viii) The collection of Greatmen contributed in Indian culture is not a set because we cannot give proper credit for their contribution.
Hence, in above option (i), (iv), (vi) and (vii) are sets.
Question 17.
Write the following sets in roster form.
(i) A = {x : x ∈ N, 2 ≤ x ≤ 9}
(ii) B = {x : x is a two digit natural number sum of whose digits is 6}
(iii) C = {set of all the letters of word MATHEMATICS}
(iv) D = {x : x is a prime number less than 50}
Solution:
(i) A = {x : x ∈ N, 2 ≤ x ≤ 9}
Roster form A = {2, 3, 4, 5, 6, 7, 8, 9}
(ii) B = (x : x is a two digit natural number sum of whose digits is 6}
The two digit natural number of this war are 15, 24, 33, 42,51.
So, Roster form B = {15, 24, 33, 42, 51}
(iii) C = {set of all the letters of word MATHEMATICS}
Roster form C = {M, A, T, H, E, I, C, S)
(iv) D = {x : x is a prime number less than 50}
Prime number less than 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
So, Roster form D = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}.
Question 18.
If A = {1, 2, 3, 4, 5, 6}, B = {2, 3, 4} and C = {4, 6, 8, 10} then insert appropriate symbols:
(i) 4…A, 5…..B
(ii) 2..A, 4…C
(iii) B…A, A….C
(iv) A – B…C
(v) A….B = B
(vi) B – C,…{2}
(vii) B ∩ C = {….}
(viii)B ∪ C – A = {….}
Solution:
(i) 4 ∈A, 5 ∉ B
(ii) 2 ∈ A, 3 ∈ B, 4 ∈ C
(iii) B ⊂ A, A ⊄ C
(iv) A – B = {1, 5, 6}, A – B ≠ C
(v) {1, 2, 3, 4, 5, 6} ∩ {2, 3, 4} = {2, 3, 4}
A ∩ B = B
(vi) B – C….{2}
B – C = {2, 3}
B – C ≠ {2}
(vii) B ∩ C = {…} = {2, 3, 4} ∩ {4, 6, 8, 10} = {4}
(viii) B ∪ C – A = {…}
B ∪ C = {2, 3, 4} ∪ {4, 6, 8, 10} = {2, 3, 4, 6, 8, 10}
B ∪ C – A = {8, 10}
Question 19.
Give two examples of each of the following :
(i) Null set
(ii) Finite set
(iii) Infinite set
(iv) Universal set
Solution:
(i) Null Set
(a) A = {x : x is a natural number less than 1}
(b) B = {x : x is a positive natural number which lies between 2 and 3}
(ii) Finite set
(a) A = {x : x2 < 10, where x is a prime number}
(b) B = {x : x, is any month of a year}
(iii) Infinite set
(a) P = {x : x = 2n, where n is a natural number}
(b) Q = {x : x = pq, wherep and q are integer and q ≠ 0}
(iv) Universal set
(a) Set of integers for natural numbers.
(b) Set of real numbers for natural numbers, integers and rational numbers.
Question 20.
If A = {a, b, c, d}, B = {p, q, r} and C = {a, b, p, q}, then examine the validity of the following:
(i) A – B = C
(ii) B – C ≠ A
(iii) B – A = Φ
(iv) (A ∪ B) – C = {c, d, r}
(v) (A ∪ B) ∩ C = C
Solution:
Given,
A = {a, b, c, d}, B = {p, q, r}, C = {a, b, p, q}
(i) (A – B) = C
L.H.S. = A – B = {a, b, c, d} – {p, q, r} = {a, b, c, d}
⇒ A – B ≠ C (False)
(ii) B – C ≠ A
B – C = {p, q, r,} – {a, b, p, q} = {r}
≠ A (True)
(iii) B – A = Φ
B – A = {p, q, r,} – {a, b, c, d} = {p, q, r)
≠ Φ (False)
(iv) (A ∪ B) – C
A ∪ B = {a, b, c, d} ∪ {p, q, r} = {a, b, c, d, p, q, r}
(A ∪ B) – C = {a, b, c, d, p, q, r} – {a, b, p, q)
= {c, d, r,} (True)
(v) (A ∪ B) ∩ C = C
From A ∪ B = {a, b, c, d, p, q, r)
(A ∪ B) ∩ C = {a, b, c, d, p, q, r} ∩ {a, b, p, q}
= {a, b, p, q} = C (True)
Question 21.
If A = Φ i.e, A is a null set and P(A) contains only one element
Solution:
If A = Φ i.e., A is a null set and P (A) contains only one element Φ.
Question 22.
Write the following sets in interval form.
(i) {x : x ∈ R, a < x < b}
(ii) {x : x ∈ R, 3 < x ≤ 5}
(iii) {x : x ∈ R, 0 ≤ x < 8}
(iv) {x : x ∈ R, -1 ≤ x ≤ 5}
Solution:
(i) {x : x ∈ R, a < x < b} = (a, b)
(ii) {x : x ∈ R, 3 < x ≤ 5} = (3, 5]
(iii) {x : x ∈ R, 0 ≤ x < 8}= [0, 8)
(iv) {x : x ∈ R, -1 ≤ x ≤ 5} = [-1, 5]
Question 23.
Write the following intervals in set builder form.
(i) (2, 5)
(ii) [10, 7)
(iii) [2, 10]
(iv) (-5, 0]
Solution:
(i) (2, 5) = {x : x ∈ R, 2 < x < 5}
(ii) [0, 7) = {x : x ∈ R, 0 ≤ x < 7}
(iii) [2, 10) = {x : x ∈ R, 2 ≤ x < 10}
(iv) (-5, 0] = (x : x ∈ R, -5 < x ≤ 0}
Question 24.
If A = {x : x ∈ N, 2 ≤ x ≤ 9} and B = {x : x is two digits natural number, sum of which digits is 8}, then find the following sets:
(i) A ∪ B
(ii) A ∩ B
(iii) A – B
(iv) (A – B) ∪ (B – A)
Solution:
A = {x : x ∈ N, 2 ≤ x ≤ 9} = (2, 3, 4, 5, 6, 7, 8, 9}
B = {x : x two digit natural number, sum of whose digits is 8} = {17, 26, 35, 44, 53, 62, 71}
(i) A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9} ∪ {17, 26, 35, 44, 53, 62, 71}
A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9, 17, 26, 35, 44, 53, 62, 71}
(ii) A ∩ B = {2, 3, 4, 5, 6, 7, 8, 9} ∩ {17, 26, 35, 44, 53, 62, 71} = Φ
(iii) A – B = {2, 3, 4, 5, 6, 7, 8,9} – {17, 26, 35, 44, 53, 62, 71}
= All elements of A which arc not present in B.
= {2, 3, 4, 5, 6, 7, 8, 9}
= A
(iv) (A – B) ∪ (B – A)
A – B = {2, 3, 4, 5, 6, 7, 8, 9}
B – A = {17, 26, 35, 44, 53, 62, 71}
(A – B) ∪ (B – A) = {2, 3, 4, 5, 6, 7, 8, 9, 17, 26, 35, 44, 53, 62, 71} = A ∪ B
Question 25.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 4, 5, 6}, B = {2, 3, 4} and C = {4, 6, 8, 10} then find the value of following sets:
(i) (A ∪ B) ∩ B
(ii) (A ∩ B) ∪ C
(iii) A’ ∪ B’
(iv) (A ∪B)’
Solution:
(i) (A ∪ B) ∩ B
A ∪ B = { 1,2, 3, 4, 5, 6} ∪ {2, 3, 4} = {1, 2, 3, 4, 5, 6}
(A ∪ B) ∩ B = {1, 2, 3, 4, 5, 6} ∩ {2, 3, 4} = {2, 3, 4} = B
(ii) (A ∩B) ∪ C
A ∩ B = {1, 2, 3, 4, 5, 6} ∩ {2, 3, 4} = {2, 3, 4}
(A ∩ B) ∪ C = {2, 3, 4} ∪ {4, 6, 8, 10} = {2, 3, 4, 6, 8, 10}
(iii) A’ ∪ B’
A = {1, 2, 3, 4, 5, 6}
U = {1,2, 3, 4, 5, 6, 7, 8, 9, 10}
A’ = U – A = {1,2,…..10} – {1, 2,… 6} = {7, 8, 9, 10}
Similarly, B’ = {1, 5, 6, 7, 8, 9, 10}
A’ ∪ B’ = {7, 8, 9, 10} ∪ {1, 5, 6, 7, 8, 9, 10} = {1, 5, 6, 7, 8, 9, 10}
(iv) (A ∪ B)’
A ∪ B = { 1, 2, 3, 4, 5, 6} = A ∪ {2, 3, 4} = {1, 2, 3, 4, 5, 6}
(A ∪ B)’ = U – (A ∪ B) = {1, 2,… 10} – {1, 2, 3, 4, 5, 6} = {7, 8, 9, 10}
Question 26.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 4, 5, 6} and B = {2, 3, 4} then prove that:
(i) (A ∪B)’ = A’ ∩ B’
(ii) (A ∩ B)’ = A’ ∪ B’
Solution:
A’ = U – A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {1,2, 3, 4, 5, 6} = {7, 8, 9, 10}
B’ = U – B = {1,2, 3, 4, 5, 6, 7, 8, 9, 10} – {2, 3, 4} = {1, 5, 6, 7, 8, 9, 10}
(i) (A ∪B) = {1, 2, 3, 4, 5, 6} ∪ {2, 3, 4} = {1, 2, 3, 4, 5, 6}
(A ∪B)’ = U – (A ∪ B)
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {1, 2, 3, 4, 5, 6} = {7, 8, 9, 10)
and A’ ∩ B’ = {7, 8, 9, 10} ∩ {1, 5, 6, 7, 8, 9, 10} = {7, 8, 9, 10}
(A ∪B)’ = A’ ∩ B’
(ii) A ∩ B = {1, 2, 3, 4, 5, 6,} ∩ {2, 3, 4} = {2, 3, 4}
(A ∩ B)’ = U – (A ∩ B) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {2, 3, 4} = {1, 5, 6, 7, 8, 9, 10}
A’ ∪ B’ = {7, 8, 9, 10} ∪ {1, 5, 6, 7, 8, 9, 10} = {1, 5, 6, 7, 8, 9, 10}
(A ∩ B)’ = A’ ∪ B’
Question 27.
Using Venn diagrams, represent the sets given below:
(i) (A ∪ B) ∩ C
(ii) (A ∩ B) ∪ C
(iii) A’ ∪ B’
(iv) (A ∪ B)’
Solution:
(i) (A ∪ B) ∩ C
Firstly we shade the area of A and B by horizontal lines which represent A ∪ B. Then we shade the area of (A ∪ B) ∪ C by vertical lines.
The cross shaded area represents the area of (A ∪ B) ∩ C.
(ii) (A ∩ B) ∩ C
Firstly we shade the common area of A and B which represent (A ∩ B) after that we shade the area of (A ∩ B) and C which represent (A ∩ B) ∪ C.
(iii) A’ ∪ B’
A’ = U – A, B’ = U – B
Horizontal shaded region A’ vertical shaded region B’ Hence, whole shaded region = A’ ∪ B’
(iv) (A ∪ B)’
The shaded region is (A ∪ B)’
Hence Proved.
Question 28.
With the help of Venn diagram prove that:
(i) (A ∪ B)’ = A’ ∩ B’
(ii) (A ∩ B)’ = A’ ∪ B’
Solution:
(i) (A ∪ B)’ = A’ ∩ B’
Shaded region (A ∪ B)’
Horizontal shaded A’
Vertical shaded region B’
Cross shaded region A’ ∩ B’
Hence (A ∪ B)’ = A’ ∩ B’
(ii) (A ∩ B)’ = A’ ∪ B’
Shaded region (A ∩B)’
Horizontal shaded A’
Vertical shaded B’
Whole shaded region A’ ∩ B’
Hence Proved.
The Complete Educational Website