Answers: (i) This statement is True Reason - Neither {2} and nor {1} is a subset of set A. (ii) This statement is True Reason - All three {ϕ, {ϕ}, {1, ϕ}} are subset of set A. (iii) True Reason -...
Answers: (i) This statement is False Reason - 2 is not a subset of set A, it is an element of set A. (ii) This statement is True Reason - {2, {1}} is not a subset of set A.
Answers: (i) This statement is False Reason - 1 is not an element of A. (ii) This statement is True Reason - {2, Φ} is a subset of A.
Answers: (i) This statement is True Reason - Φ belongs to set A. (ii) This statement is True Reason - {Φ} is an element of set A.
Answers: (i) This statement is False Reason - Φ is a subset of A, not an element of A. (ii) This statement is True Reason - Φ is a subset of every set, so it is a subset of A.
Answers: (i) This statement is True. Reason = {6, 7, 8} ∈ A. (ii) This statement is True Reason = {{4, 5}} is a subset of A.
Answers: (i) This statement is False Reason - 1 is not an element of A. (ii) This statement is True Reason - Correct Form = {1,2,3} ∈ A
Answers: (i) This statement is False Reason - ϕ ⊂ A (ii) This statement is False Reason - ϕ ⊂ A
Answers: (i) This statement is False Reason - {a, b, e} does not belong to A. Correct Form = {a, b, e} ⊂ A (ii) This statement is False Reason - {a, b, c} is not a subset of A
Answers: (i) This statement is False Reason - a is not a subset of A and belongs to A. (ii) This statement is True Reason - {a, b, e} is a subset of A
Answers: (i) This statement is True Reason - {c, d} is a subset of A. (ii) This statement is True Reason - a belongs to A
Answers: (i) This statement is False Reason - {c, d} is not a subset of A but it belong to A. (ii) This statement is True Reason - {c, d} belongs to A
Answers: (i) This statement is incorrect Correct Form = ϕ ⊂ {a, b} (ii) This statement is the correct form. (iii) This statement is incorrect Correct Form = {x: x + 3 = 3} ≠ ϕ
Answers: (i) This statement is incorrect Correct Form = {b, c} ∈ {a,{b, c}} (ii) {a, b} is not a subset of given set. Correct Form = {a, b}⊄{a,{b, c}}
Answers: (iii) This statement is incorrect Correct Form = {a} ∈ {{a}, b} (iv) This statement is incorrect Correct Form = {a} ∈ {{a}, b}
Answers: (i) This statement is incorrect Correct Form = a ∈{a,b,c} (ii) This statement is incorrect Correct Form = {a} ⊂ {a, b, c}
Answers: (i) This statement is False P = {a} B = {{a}} But {a} = P B = {P} P and B are not equal (ii) This statement is True A = For “LITTLE” A = {L, I, T, E} = {E, I, L, T} B = For “TITLE” B = {T,...
Answers: (i) The statement is True Reason - A rational number is represented by the form p/q where p and q are integers and (q not equal to 0) keeping q = 1 we can place any number as p. Which then...
Answers: (i) The statement is False Reason - Every square can be a rectangle, but every rectangle cannot be a square. (ii) The statement is False Reason - Every square can be a rectangle, but every...
Answer: A = x2 – 8x + 12=0 (x–6) (x–2) =0 x = 2 or 6 Then, A = {2, 6} B = {2, 4, 6} C = {2, 4, 6, 8} D = {6} Thus, D⊂A⊂B⊂C
Answers: (iii) The statement is False Reason - a is a subset of the set. So, it cannot be an element. (iv) The statement is True Reason - Repetition of elements is not allowed in a set. (v) The...
Answers: (i) The statement is True Reason - 1 is present in the given set. (ii) The statement is False Reason - a is an element of a set and not a subset.
Answers: (v) The statement is False. Reason - Repetition is not allowed in the elements of a set. (vi) The statement is True. Reason - The equivalent sets have same number of elements. (vii) The...
Answers: (iii) The statement is True Reason - Basically, the smaller part of something which is finite can never be infinite. (iv) The statement is False Reason - Null set or empty set does not have...
Answers: (i) The statement is False. Reason - It is not required for the two sets A and B to be A B or B A. (ii) The statement is False. Reason - It is finite subset of infinite set which belongs to...
Solution: (i) The statement given in the question is true. Now here {2, 6, 10, 14} ∩ {3, 7, 11, 15} $=$ Φ (ii) The statement given in the question is true. Now here {2, 6, 10} ∩ {3, 7, 11} $=$...
Solution: (i) The statement given in the question is false. So if 3 ∈ {2, 3, 4, 5}, 3 ∈ {3, 6} Therefore, we obtain {2, 3, 4, 5} ∩ {3, 6} $=$ {3} (ii) The statement given in the question is false....
Solution: It is known that Set of real numbers - R Set of rational numbers - Q As a result, R – Q is a set of irrational numbers.
{a, b, c, d} and Y {f, b, d, g}, find (i) X ∩ YSolution: (i) X ∩ Y $=$ {b, d}
{a, b, c, d} and Y {f, b, d, g}, find (i) X – Y (ii) Y – XSolution: (i) X – Y $=$ {a, c} (ii) Y – X $=$ {f, g}
{3, 6, 9, 12, 15, 18, 21}, B {4, 8, 12, 16, 20}, C {2, 4, 6, 8, 10, 12, 14, 16}, D {5, 10, 15, 20}; find (i) C – D (ii) D – CSolution: (i) C – D $=$ {2, 4, 6, 8, 12, 14, 16} (ii) D – C $=$ {5, 15, 20}
{3, 6, 9, 12, 15, 18, 21}, B {4, 8, 12, 16, 20}, C {2, 4, 6, 8, 10, 12, 14, 16}, D {5, 10, 15, 20}; find (i) C – B (ii) D – BSolution: (i) C – B = {2, 6, 10, 14} (ii) D – B = {5, 10, 15}
{3, 6, 9, 12, 15, 18, 21}, B {4, 8, 12, 16, 20}, C {2, 4, 6, 8, 10, 12, 14, 16}, D {5, 10, 15, 20}; find (i) B – C (ii) B – DSolution: (i) B – C $=$ {20} (ii) B – D $=$ {4, 8, 12, 16}
{3, 6, 9, 12, 15, 18, 21}, B {4, 8, 12, 16, 20}, C {2, 4, 6, 8, 10, 12, 14, 16}, D {5, 10, 15, 20}; find (i) C – A (ii) D – ASolution: (i) C – A $=$ {2, 4, 8, 10, 14, 16} (ii) D – A $=$ {5, 10, 20}
{3, 6, 9, 12, 15, 18, 21}, B {4, 8, 12, 16, 20}, C {2, 4, 6, 8, 10, 12, 14, 16}, D {5, 10, 15, 20}; find (i) A – D (ii) B – ASolution: (i) A – D $=$ {3, 6, 9, 12, 18, 21} (ii) B – A $=$ {4, 8, 16, 20}
{3, 6, 9, 12, 15, 18, 21}, B {4, 8, 12, 16, 20}, C {2, 4, 6, 8, 10, 12, 14, 16}, D {5, 10, 15, 20}; find (i) A – B (ii) A – CSolution: (i) A – B $=$ {3, 6, 9, 15, 18, 21} (ii) A – C $=$ {3, 9, 15, 18, 21}
is an even integer} and { is an odd integer}Solution: (i) {$x: x$ is an even integer} ∩ {$x: x$ is an odd integer} $=$ Φ As a result, this pair of sets is disjoint.
is a natural number and 4 ≤ x ≤ 6} (ii) {a, e, i, o, u} and {c, d, e, f}Solution: (i) {1, 2, 3, 4} {$x: x$ is a natural number and 4 ≤ x ≤ 6} $=$ {4, 5, 6} As a result, we obtain {1, 2, 3, 4} ∩ {4, 5, 6} $=$ {4} As a result, this pair of sets is not disjoint. (ii) {a,...
{ is a natural number}, B { is an even natural number} C { is an odd natural number} and D { is a prime number}, find (i) B ∩ D (ii) C ∩ DSolution: This can be written as A $=$ {$x: x$ is a natural number} $=$ {1, 2, 3, 4, 5 …} B $=$ {$x: x$ is an even natural number} $=$ {2, 4, 6, 8 …} C $=$ {$x: x$ is an odd natural number} $=$ {1,...
{ is a natural number}, B { is an even natural number} C { is an odd natural number} and D { is a prime number}, find (i) A ∩ D (ii) B ∩ CSolution: This can be written as A $=$ {$x: x$ is a natural number} $=$ {1, 2, 3, 4, 5 …} B $=$ {$x: x$ is an even natural number} $=$ {2, 4, 6, 8 …} C $=$ {$x: x$ is an odd natural number} $=$ {1,...
{ is a natural number}, B { is an even natural number} C { is an odd natural number} and D { is a prime number}, find (i) A ∩ B (ii) A ∩ CSolution: This can be written as A $=$ {$x: x$ is a natural number} $=$ {1, 2, 3, 4, 5 …} B $=$ {$x: x$ is an even natural number} $=$ {2, 4, 6, 8 …} C $=$ {$x: x$ is an odd natural number} $=$ {1,...
{3, 5, 7, 9, 11}, B {7, 9, 11, 13}, C {11, 13, 15} and D {15, 17}; find (i) (A ∩ B) ∩ (B ∪ C) (ii) (A ∪ D) ∩ (B ∪ C)Solution: (i) (A ∩ B) ∩ (B ∪ C) $=$ {7, 9, 11} ∩ {7, 9, 11, 13, 15} $=$ {7, 9, 11} (ii) (A ∪ D) ∩ (B ∪ C) $=$ {3, 5, 7, 9, 11, 15, 17) ∩ {7, 9, 11, 13, 15} $=$ {7, 9, 11,...
{3, 5, 7, 9, 11}, B {7, 9, 11, 13}, C {11, 13, 15} and D {15, 17}; find (i) A ∩ D (ii) A ∩ (B ∪ D)Solution: (i) A ∩ D $=$ Φ (ii) A ∩ (B ∪ D) $=$ (A ∩ B) ∪ (A ∩ D) $=$ {7, 9, 11} ∪ Φ $=$ {7, 9, 11}
{3, 5, 7, 9, 11}, B {7, 9, 11, 13}, C {11, 13, 15} and D {15, 17}; find (i) B ∩ D (ii) A ∩ (B ∪ C)Solution: (i) B ∩ D $=$ Φ (ii) A ∩ (B ∪ C) $=$ (A ∩ B) ∪ (A ∩ C)
{3, 5, 7, 9, 11}, B {7, 9, 11, 13}, C {11, 13, 15} and D {15, 17}; find (i) A ∩ C ∩ D (ii) A ∩ CSolution: (i) A ∩ C ∩ D $=$ {A ∩ C} ∩ D $=$ {11} ∩ {15, 17} $=$ Φ (ii) A ∩ C $=$ {11}
{3, 5, 7, 9, 11}, B {7, 9, 11, 13}, C {11, 13, 15} and D {15, 17}; find (i) A ∩ B (ii) B ∩ CSolution: (i) A ∩ B $=$ {7, 9, 11} (ii) B ∩ C $=$ {11, 13}
{1, 2, 3}, B ΦSolution: (i) A $=$ {1, 2, 3}, B $=$ Φ Therefore the intersection of the given set can be re-written as A ∩ B $=$ Φ
{ is a natural number and multiple of 3} B { is a natural number less than 6} (ii) A { is a natural number and 1 < x ≤ 6} B { is a natural number and 6 < x < 10}Solution: (i) A $=$ {$x: x$ is a natural number and multiple of 3} $=$ {3, 6, 9 …} B $=$ {$x: x$ is a natural number less than 6} $=$ {1, 2, 3, 4, 5} Therefore the intersection of the given set can...
{1, 3, 5} Y {1, 2, 3} (ii) A {a, e, i, o, u} B {a, b, c}Solution: (i) X $=$ {1, 3, 5}, Y $=$ {1, 2, 3} Therefore the intersection of the given set can be re-written as X ∩ Y $=$ {1, 3} (ii) A $=$ {a, e, i, o, u}, B $=$ {a, b, c} Therefore the...
{1, 2, 3, 4}, B {3, 4, 5, 6}, C {5, 6, 7, 8} and D {7, 8, 9, 10}; find (i) B ∪ C ∪ DSolution: Provided that A $=$ {1, 2, 3, 4}, B $=$ {3, 4, 5, 6}, C $=$ {5, 6, 7, 8} and D $=$ {7, 8, 9, 10} (i) B ∪ C ∪ D $=$ {3, 4, 5, 6, 7, 8, 9, 10}
{1, 2, 3, 4}, B {3, 4, 5, 6}, C {5, 6, 7, 8} and D {7, 8, 9, 10}; find (i) A ∪ B ∪ C (ii) A ∪ B ∪ DSolution: Provided that A $=$ {1, 2, 3, 4}, B $=$ {3, 4, 5, 6}, C $=$ {5, 6, 7, 8} and D $=$ {7, 8, 9, 10} (i) A ∪ B ∪ C $=$ {1, 2, 3, 4, 5, 6, 7, 8} (ii) A ∪ B ∪ D $=$ {1, 2, 3, 4, 5, 6, 7, 8, 9,...
{1, 2, 3, 4}, B {3, 4, 5, 6}, C {5, 6, 7, 8} and D {7, 8, 9, 10}; find (i) B ∪ C (ii) B ∪ DSolution: Provided that A $=$ {1, 2, 3, 4}, B $=$ {3, 4, 5, 6}, C $=$ {5, 6, 7, 8} and D $=$ {7, 8, 9, 10} (i) B ∪ C $=$ {3, 4, 5, 6, 7, 8} (ii) B ∪ D $=$ {3, 4, 5, 6, 7, 8, 9, 10}
{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 ∪ CSolution: Provided that A $=$ {1, 2, 3, 4}, B $=$ {3, 4, 5, 6}, C $=$ {5, 6, 7, 8} and D $=$ {7, 8, 9, 10} (i) A ∪ B $=$ {1, 2, 3, 4, 5, 6} (ii) A ∪ C $=$ {1, 2, 3, 4, 5, 6, 7, 8}
Solution: If it is given that A and B are two sets such that A ⊂ B, then the value of A ∪ B = B.
{a, b}, B {a, b, c}. Is A ⊂ B? What is A ∪ B?Solution: Provided that: A $=$ {a, b} and B $=$ {a, b, c} Yes, A ⊂ B Therefore the union of the pairs of set can be re-written as A ∪ B $=$ {a, b, c} $=$ B
{1, 2, 3}, B ΦSolution: (i) A $=$ {1, 2, 3}, B $=$ Φ Therefore the union of the pairs of set can be re-written as A ∪ B $=$ {1, 2, 3}
is a natural number and multiple of 3} B = { is a natural number less than 6} (ii) A = { is a natural number and 1 < x ≤ 6} B = { is a natural number and 6 < x < 10}Solution: (i) A $=$ {$x: x$ is a natural number and multiple of 3} $=$ {3, 6, 9 …} B $=$ {$x: x$ is a natural number less than 6} $=$ {1, 2, 3, 4, 5, 6} Therefore the union of the pairs of set can...
Solution: (i) X = {1, 3, 5} Y = {1, 2, 3} Therefore the union of the pairs of set can be re-written as X ∪ Y= {1, 2, 3, 5} (ii) A = {a, e, i, o, u} B = {a, b, c} Therefore the union of the pairs of...