Sets Class 11 OP Malhotra Exe-1D ISC Maths Solutions Ch-1 Latest editions. In this article you would learn about Union of sets and its properties. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.
Sets Class 11 OP Malhotra Exe-1D ISC Maths Solutions Ch-1 Latest editions
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-1 | Sets |
| Writer | OP Malhotra |
| Exe-1(D) | Union of sets and its properties |
Exercise- 1D
Sets Class 11 OP Malhotra Exe-1D ISC Maths Solutions Ch-1 Latest editions.
Find A ∪ B if
Que-1: (i) A = {a, b, c}, B = {x, y}
(ii) A = {c, a, t}, B = {s, a, t}
(iii) A = Set of letters in the word ‘INDIA’
B = Set of letters is the word ‘DRAIN’.
Sol: (i) A = {a, b, c}, B = {x, y}
A∪B = {a, b, c, x, y}
(ii) A = {c, a, t}, B = {s, a, t}
A∪B = {c, a, t , s}.
(iii) A = Set of letters in the word ‘INDIA’
B = Set of letters is the word ‘DRAIN’.
A∪B = {I, N, D, A, R}
Que-2: (i) A = {factor of 12}, B = {factors of 16}
(ii) A = {Prime number < 10} , B = {odd number < 10}
(iii) A = {x | x = 3p, p ∈ N}, B = {y | y = 4n, n ∈ N}
(iv) A = {x | x is a positive integer}, B = {x | x is a negative number}.
Sol: (i) A = {factor of 12}, B = {factors of 16}
A∪B = {1, 2, 3, 4, 6, 8, 12, 16}
(ii) A = {Prime number < 10} , B = {odd number < 10}
A∪B = {1, 2, 3, 5, 7, 9}
(iii) A = {x | x = 3p, p ∈ N}, B = {y | y = 4n, n ∈ N}
A is the subset of all positive multiple of 3.
A = {3, 6, 9, 12, 15, 18……}
B is the subset of all positive multiple of 4.
B = {4, 8, 12, 16, 20, 24…….}
A∪B = {3, 6, 8, 9, 12, 15, 16, 18, 21, 24……}
(iv) A = {x | x is a positive integer}, B = {x | x is a negative number}.
A = {1, 2, 3, 4, …..}
B = {….., -3, -2, -1,}
A∪B = {……., -3, -2, -1, 0, 1, 2, 3,……}
This is the set of all non-zero real numbers that are either negative numbers or positive integers.
Thus,
A∪B = {x ∈ R | x < 0}.
Que-3: 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
Sol: The given sets are
A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}
The union of two sets A and B is a set that is obtained by writing all the elements of sets A and B by avoiding duplicates.
Then
(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}
Que-4: If A = {2, 3, 5, 7, 11}, B = {1, 3, 5, 7, 9}, C = {0, 1, 2, 3}, then find (A∪B)∪C and A∪(B∪C) and verify that (A∪B)∪C = A∪(B∪C).
Sol: Given that :
A = {2, 3, 5, 7, 11}, B = {1, 3, 5, 7, 9}, C = {0, 1, 2, 3}
Taking L.H.S.
A∪B = {1, 2, 3, 5, 7, 9, 11}
Now take union of A∪B with C = {0, 1, 2, 3}
(A∪B)∪C = {0, 1, 2, 3, 5, 7, 9, 11}
Taking R.H.S.
B∪C = {0, 1, 2, 3, 5, 7, 9}
Now take union with A = {2, 3, 5, 7, 11}
A∪(B∪C) = {0, 1, 2, 3, 5, 7, 9, 11}.
Hence proved
R.H.S = L.H.S
A∪(B∪C) = (A∪B)∪C = {0, 1, 2, 3, 5, 7, 9, 11}.
Que-5: Let L = {Letters of CRICKET}, M = {Letters of CATERPILLAR} and N = {Letters of CARE TAKER} find : (i) L∪M (ii) M ∪ N (iii) L∪N
Sol: Given L = {C, R, I, K, E, T}
M = {C, A, T, E, R, P, I, L} and N = {C, A, R, E, T, K}
(i) L∪M = {C,R, I, K, E, T} ∪ {C, A, T, E, R, P, I, L}
= {C, A, T, E, I, K, R, P, L}
(ii) M∪N = {C, A, T, E, R, P, I, L} ∪ {C, A, R, E, T, K}
= {C, A, T, E, R, P, I, L, K}
(iii) L∪N = {C,R, I, K, E, T} ∪ {C, A, R, E, T, K} = {C, R, I, K, E, T, A}.
–: End Sets Class 11 OP Malhotra Exe-1D ISC Maths Ch-1 Latest editions :–
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