ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Ex 1.3

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ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Ex 1.3

Question 1.
If A = {1, 2, 3, 4, 5, 6) and B = {2, 4, 6, 8}, find
(i) A ∪ B
(ii) A ∩ B
(iii) A – B
(iv) B – A
Solution:
Given A = {1, 2, 3, 4, 5, 6}
and B = {2, 4, 6, 8}
(i) A ∪ B = {1, 2, 3, 4, 5, 6, 8}
(ii) A ∩ B = {2, 4, 6}
(iii) A – B = {1, 3, 5}
(iv) B – A = {8}

Question 2.
If A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 5, 6, 8, 9, 10), then find A Δ B.
Solution:
Given A = {1, 2, 3, 4, 5, 6}
and B = {2, 4, 5, 6, 8, 9, 10}
∴ A – B = {1, 3)
and B – A = {8, 9, 10}
Thus A Δ B = (A – B) ∪ (B – A)
= {1, 3} ∪ {8, 9, 10}
= {1, 3, 8, 9, 10}

Question 3.
If A = {x : x is a natural number and 1 < x ≤ 6} and B = {x : x is a natural number and 6 < x < 10}, find
(i) A ∪ B
(ii) A ∩ B
(iii) A – B
(iv) B – A.
Solution:
Given
A= {x : x is a natural number and 1 < x ≤ 6}
∴ A = {2, 3, 4, 5, 6}
and B = {x : x is a natural number and 6 < x < 10}
= {7, 8, 9)

(i) A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9}
(ii) A ∩ B = { } or Φ
(iii) A – B = {2, 3, 4, 5, 6} = A
(iv) B – A = {7, 8, 9} = B

Question 4.
Which of the following pairs of sets are disjoint?
(i) {a, e, i, o, u) and {c, d, e, f}
(ii) {2, 6, 10, 14, 18} and {3, 7, 11, 15}
(iii) {1, 2, 3, 4} and {x : x is a natural number and 4 ≤ x ≤ 6}
(iv) {x : x is an even integer} and {x : x is an odd integer}.
Solution:
(i) Let A = (a, e, i, o, u}
and B = {c, d, e, f}
∴ A ∩ B = {e} ≠ Φ
Thus A and B are not disjoint sets.

(ii) Let A = {2,6, 10, 14, 18}
and B = {3, 7, 11, 15}
∴ A ∩ B = Φ
Thus, A and B are disjoint sets.

(iii) Let A = {1, 2, 3, 4)
and B = {4, 5, 6)
∴ A ∩ B = {4} ≠ Φ
Thus sets A and B are disjoint sets.

(iv) Let A = {x : x is an even integer}
= {……………….., – 4, – 2, 0, 2, 4, 6, …………….}
and B = {x : x is an odd integer
= {………………., – 3, – 1, 3, 5, ……………………}
Clearly A ∩ B = Φ
Thus A and B are disjoint sets.

Question 5.
If ξ = {1, 2, 3, 4, 5, 6, 7, 8}, A = {1, 2, 3, 5, 6} and B = {2, 3, 4, 7, 8}, find the following:
(i) A ∪ B
(ii) A ∩ B
(iii) A – B
(iv) B – A
(v) A ∩ B
(vi) (A ∪ B)’
(vii) A’ ∩ B’
(viii) A’ ∪ B
(ix) (A – B)’
Solution:
Given ξ = {1, 2, 3, 4, 5, 6, 7, 8}
A = {1, 2, 3, 5, 6}
and B = {2, 3, 4, 7, 8}

(i) A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8} = ξ

(ii) A ∩ B = {2, 3}

(iii) A – B = {1, 5, 6}

(iv) B – A = {4, 7, 8}

(v) Now B’ = ξ – B = {1, 5, 6}
∴ A – B’ = {1, 2, 3, 5, 6) ∪ {1, 5, 6}
= {1, 5, 6}

(vi) A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8}
(A ∪ B)’ = ξ – (A ∪ B) = Φ

(vii) Now, A’ = ξ – A
= {4, 7, 8}
and B’ – B = {1, 5, 6}
∴ A’ ∩ B’ = {4, 7, 8} ∩{1, 5, 6} = Φ

(viii) A’ ∪ B = {4, 7, 8} ∪ {2, 3, 4, 7, 8}
= {2, 3, 4, 7, 8}

(ix) A – B = {1, 5, 6}
∴ (A – B)’ = ξ – (A – B) = {2, 3, 4, 7, 8}

Question 6.
If A and B are two sets such that Ac B, then what is
(i) A ∪ B?
(ii) A ∩ B?
Solution:
Given A ⊂ B
(i) A ∪ B = B
[∵ B be the larger set and all members of A are members of B]

(ii) A ∩ B = A
[∵ A be the smaller set]

Question 7.
Taking the set of natural numbers as the universal set., write the complements of the following sets :
(i) A = {x : x is an odd natural number}
(ii) B = {x : x is an even natural number}
(iii) C = {x : x is a multiple of 5}
(iv) D = {x : x is divisible by 2 and 3}
(v) E = {x : x is a perfect cube}
(vi) F = {x : x + 5 = 8}
(vii) G = {x : x ≥ 7}
(viii) H = {x : 3x – 1 < 14}.
Solution:
Here ξ = N
= {1, 2, 3, 4, ………………….}

(i) A = {x : x is an odd natural number}
= {1, 3, 5, ……………..}
∴ A’ = ξ – A
= {2, 4, 6, ………………….}
= {x : x be an even natural number)

(ii) Given B = {x : x is an even number}
= {2, 4, 6……………..}
∴ B’ = ξ – B
= {1, 3, 5, …………………}
= {x : x is an odd natural number}

(iii) Given C = {x : x is a multiple of 5}
∴ C’ = ξ – C
= {x : x ∈ N, x is not a multiple of 5}

(iv) Given D = {x : x is divisible by 2 and 3}
= {x : x is divisible by 2 × 3 = 6}
∴ D’ = ξ – D
= {x : x ∈ N, x is not divisible by 6}

(v) Given E = {x : x is a perfect cube}
∴ E’ = – E
= {x : x ∈ N, x is not a perfect cube}

(vi) Given F = {x : x + 5 = 8}
Since x + 5 = 8
⇒ x = 8 – 5 = 3
∴ F = {3) and {x : x ∈ N and x ≠ 3} = F’

(vii) Given G = {x : x ≥ 7}
∴ G’ = ξ – G
= {x : x ∈ N, x < 7}
= {1, 2, 3, 4, 5, 6}

(viii) Given H’ = {x : 3x – 1 < 14}
since 3x – 1 < 14
⇒ 3x < 15
⇒ x < 5
and x ∈ N
∴ H = {1, 2, 3, 4)
∴ H’ = {5, 6, 7, 8, 9}
Thus, H’ = ξ – H = {x : x ∈ N ; x ≥ 5}

Question 8.
From the adjoining Venn diagram, determine the following sets :

(i) A ∪ B
(it) A ∩ B
(iii) A – B
(iv) A – B
(v) (A ∩ B)’.
Solution:
Here ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Here A = {1, 2, 4, 7}
and B = {0, 1, 3, 7, 8}
(i) A ∪ B = {0, 1, 2, 3, 4, 7, 8}

(ii) A ∩ B = {1, 7}

(iii) A – B = {2 ,4}

(iv) (A ∩ B)’ = – (A ∩ B)
= {0, 2, 3, 4, 5, 6, 8, 9}

Question 9.
From the adjoining Venn diagram, write the following:
(i) A’
(ii) B’
(iii) (A ∩ B)’

Solution:
Here ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {0, 2, 3, 4, 8)
and B = {0, 3, 5, 6, 8}

(i) A’ = {1, 5, 6, 7, 19}

(ii) B’ = {1, 2, 4, 7, 9}

(iii) A ∩ B = {3, 0, 8}
∴ (A ∩ B)’ = {1, 2, 4, 5, 6, 7, 9}

Question 10.
Let ξ = {x : x ∈ N and x < 10}, A = {x : x = 2y + 1 and y ∈ N} and B = {x : x = 3y- 1 and y ∈ N), list the elements of A’ ∪ B.
Solution:
Given ξ = {x : x ∈ N, x < 10}
= {1, 2, 3, 4, 5, 6, 7, 8, 9}
and A = {x : x = 2y + 1 and y ∈ N}
when y = 1
⇒ x = 2 + 1 = 3
when y = 2
⇒ x = 4 + 1 = 5
when y = 3
⇒ x = 6 + 1 = 7
when y = 4
⇒ x = 8 + 1 = 9
when y = 5
⇒ x = 10 + 1 = 11 ∉ ξ
∴ A = {3, 5, 7, 9}
and B = {x : x = 3y – 1 and y ∈ N}
when y = 1
⇒ x = 3 – 1 = 2
when y = 2
⇒ x = 6 – 1 = 5
when y = 4
⇒ x = 12 – 1 = 11
when y = 3
⇒ x = 9 – 1 = 8
∴ B = {2, 5, 8}
∴ A’ = ξ – A = {1, 2, 4, 6, 8}
Thus, A’ ∪ B = {1, 2, 4, 6, 8} ∪ {2, 5, 8}
= {1, 2, 4, 5, 6, 8}

Question 11.
If ξ = {1, 2, 3, ……………., 10}, A = {x : x is prime} and B = {x : x is even integer}, then write the following :
(i) A – B
(ii) A ∩ B’.
Solution:
Given ξ = {1, 2, 3, ……………….., 10}
A = {x : x is prime} = {2, 3, 5, 7}
B = {x : x is even integer) = {2, 4,6, 8, 10}
(i) A – B = {3, 5, 7}
Here B’ – B = {1, 3, 5, 7, 9}

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

Question 12.
If ξ = {all digits in our decimal system}, A = {x : x is prime}, and B = {x : x² < 25}, then write the following:
(i) B – A
(ii) A ∪ B
(iii) (A ∪ B)’.
Solution:
Given ξ = {0, 1, 2, 3 ,9}
A = {x : x is prime} = {2, 3, 5, 7}
and B = {x : x² < 25} = {1, 2, 3, 4}
[Since 1² = 1 < 25;
2² = 4 < 25 ;
3² = 9 < 25
and 4² = 16 < 25]

(i) B – A = {1, 4}

(ii) A ∪ B = {1, 2, 3, 4, 5, 7}

(iii) (A ∪ B)’ = ξ – (A ∪ B) = {0, 6, 8, 9}

Question 13.
On the real axis, if A = (0, 3] and B = [2, 6], then write the following:
(i) A ∪ B
(ii) A ∩ B
(iii) A – B
(Iv) B – A
Solution:
Given A = (0, 3]
and B = [2, 6]
(i) ∴ λ ⊥ ⇒ σ
A ∪ B = (0, 3] ∪ [2, 6] = (0, 6]

(ii) A ∩ B = [2, 3]

(iii) A – B = (0, 2)

(iv) B – A = (3, 6]

Question 14.
On the real axis, if A = – [ 1, 1) and B = [0, 4], then find the following:
(i) A’
(ii) B’
(iii) A ∪ B
(iv) A ∩ B
(v) A – B
(vi) B – A
Solution:

Given A = [- 1, 1)
and B = [0, 4]

(i) A’ = R – A
= R – [- 1, 1]
= (- ∞, – 1) ∪ [1, ∞)

(ii) B’ = R – B
= (- ∞, ∞) – [0, 4]
= (- ∞, 0) ∪ (4, ∞)

(iii) A ∪ B = [- 1, 1) ∪ [0, 4]
= [- 1, 4]

(iv) A ∩ B = [- 1, 1) ∩ [0, 4]
= [0, 1)

(v) A – B = [- 1, 1) – [0, 4]
= [- 1, 0)

(vi) B – A = [0, 4] – [- 1, 1)
= [1, 4]

Question 15.
If A = the set of letters in the word ‘JAIPUR’ and B = the set of letters in the word ‘JODHPUR’, find the following.
(i) A ∪ B
(ii) A ∩ B
(iii) A – B
(iv) B – A.
Also verify the following results :
(a) n (A ∪ B) = n (A) + n (B) – n (A ∩ B)
(b) n(A – B) = n (A) – n (A ∩ B)
= n (A ∪ B) – n (B)
Solution:
A = {J, A, I, P, U, R}
B = {J, O, D. H, P, U, R}

(i) A ∪ B = (J, A, J. P, U, R, O, D, H)

(ii) A ∩ B = {J, P, U, R}

(iii) A – B = {A, 1}

(iv) B – A = {O, D, H}
Here n(A) = no. of distinct elements in A = 6
n (B) = 7 ;
n (A ∪ B) = 9 ;
n (A ∩ B) = 4

(a) n (A ∪ B) = 9 = L.H.S.
and R.H.S. = n (A) + n (B) – n (A ∩ B)
= 6 + 7 – 4 = 9
∴ n (A ∪ B) = n (A) + n (B) – n(A ∩ B)

(b) Here n (A – B) = 2 ;
n (B – A) = 3
Now n (A) – n (A ∩ B) = 6 – 4
= 2 = n (A – B)
and n (A ∪ B) – n (B) = 9 – 7
= 2 = n (A – B)

Question 16.
If ξ = {0, 1, 2, 3, 4, 5, 6, 7), A = {2, 5, 7}, B = {0, 2, 3, 7} and C = {0, 3, 4, 6}, form the following sets :
(i) (A ∪ B)’
(ii) A – C
(iii) A ∩ (B ∪ C).
Solution:
Given ξ = {0, 1, 2, 3, 4, 5, 6, 7} ;
A = {2, 5, 7} ;
B = {0, 2, 3, 7}
and C = {0, 3, 4, 6}

(i) A ∪ B = {2, 5, 7, 0, 3}
∴ (A ∪ B)’ = ξ – AuB
= {1, 4, 6}

(ii) A – C = {2, 5, 7)

(iii) B ∪ C = {0, 2, 3, 4, 6, 7}
∴ A ∩ (B ∪ C) = {2, 7}

Question 17.
Let A = {2x : x ∈ N and 1 ≤ x < 4}, B = {x + 2 ; x ∈ N and 2 ≤ x < 5) and C = {x : x ∈ N and 3 < x < 6}, determine the following :
(i) A ∩ B
(ii) A ∪ B
(iii) (A ∪ B) ∩ C
Also verify that n (A ∪ B) = n (A) – n (A ∩ B)
Solution:
Given A = {2x : x ∈ N and 1 ≤ x < 4}
since 1 ≤ x < 4
∴ x = 1, 2, 3 as x ∈ N
∴ A = {2, 4, 6}
B = {x + 2 ; x ∈ N and 2 ≤ x < 5}
since x ∈ N and 2 ≤ x < 5
∴ x = 2, 3, 4
∴ B = {2 + 2, 3 + 2, 4 + 2} = {4, 5, 6}
and C = {x : x ∈ N and 3 < x < 6} = {4, 5}

(i) A ∩ B = {2, 4, 6) ∩ (4, 5, 6) = {4, 6}

(ii) A ∪ B = (2, 4, 6) ∪ {4, 5, 6) = {2, 4, 5, 6}

(iii) (A ∪ B) ∩ C = {2, 4, 5, 6} ∩ {4, 5} = {4, 5}
∴ n(A ∪ B) = 4 ;
n (A) = 3 ;
n (B) = 3 ;
n (A ∩ B) = 2
∴ R.H.S. = n (A) + n (B) – n (A ∩ B)
= 3 + 3 – 2
= 4 = n (A ∪ B)

Question 18.
If ξ = the set of all digits in our decimal system, A = {x : x is prime} and B = {x : x is a perfect square}, then verify the following results :
(i) A ∪ A’ = ξ
(ii) A ∩ A’= Φ
(iii) A-B = A ∩ B’
(iv) (A ∩ B)’ = A’ ∪ B’
(v) (A ∪ B)’ = A’ ∩ B’
(vi) (A’)’ = A
Solution:
Given ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {x : x is prime}
= {2, 3, 5, 7}
and B = {x : x is perfect square}
= {1, 4, 9}

(i) A’ = ξ – A
= {0, 1, 4, 6, 8, 9}
∴ A ∪ A’ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} = ξ

(ii) A ∩ A’ = {2, 3, 5, 7} ∩ {0, 1, 4, 6, 8, 9} = Φ

(iii) A – B = {2, 3, 5, 7}
Also B’ = ξ – B’
= {0, 2, 3, 5, 6, 7, 8}
∴ A ∩ B’ = {2, 3, 5, 7} ………………(2)
From (1) and (2) ; we have
A – B = A ∩ B’

(iv) A ∩ B = Φ
∴ (A ∩ B)’ = Φ’ = ξ ………………..(1)
∴ A’ ∪ B’ = {0, 1, 4, 6, 8, 9} ∪ {0, 2, 3, 5, 6, 7, 8}
= {0, 1, 2, 3, 4. 5, 6, 7, 8, 9}
= ξ ………………….(2)
From (1) and (2) ; we have
(A ∩ B)’ = A’ ∪ B’

(v) L.H.S. = (A ∪ B)’ – (A ∪ B)
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 5, 7, 9}
= {0, 6, 8} ……………………..(1)
R.H.S. = A’ ∩ B’
= {0, 1, 4, 6, 8, 9} ∩ {0, 2, 3, 5, 6, 7, 8}
= {0, 6, 8}
From (1) and (2) ; we have
(A ∪ B)’ = A’ ∩ B’

(vi) A’ = ξ – A
= {0, 1, 4, 6, 8, 9}
∴ (A’)’ = ξ A’
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 0} – {0, 1, 4, 6, 8, 9}
= {2, 3, 5, 7}
= A