1. The set A = {2, 6, 12, 20, 30, 42, 56} in roaster form is ___
The set A = {.....-9, -6 -3, 0, 3, 6, 9, 12,} in set-builder form is ____.
The elements in the set are of the form 3 Γ 3. Hence, the correct answer is {x: x is a multiple of 3}.
The set A = {all elements in a week except Sunday} in roaster form is ____.
The elements in the set should contain all days, except Sunday. Hence, A = {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
The solution set of the equation x2 + x β 2 = 0in roaster form is ____.
The given equation can be written as:
x2 + x β 2 = 0
The set A = {x: 2x - 3 < 5, x β Z} in the roaster form is ____.
2x - 3 <5, x β Z
β2x <8, x β Z
βx <4, x β Z
β A = {......-5, -4, -3, -2, -1, 0, 1, 2, 3)
Hence, the set in the roaster form is {......-5, -4, -3, -2, -1, 0, 1, 2, 3}.
set A = (2, 5/2,10/3,17/4,26/5,37/6) in the set-builder form is ____.
The first element is 2 = 1 + 1
The second elements is 5/2= 2 +1/2
The third element is 10/3= 3 +1/3
The sixth element is 37/6= 6 +1/6
Hence, the set in the set-builder form is x=n+1/n:1 β€ n β€ 6,n β N).
Given A = {x: x is rational number between 1/3and1/2} B = {x: x is a prime number between 7 and 10} C = {x: x β R and x + a = a for all a β R} D = {x: x β R and x. a = a for all a β R) Set ____ is empty.
5/ 12 β A, 0 β C, 1 β D - Hence, these are non-empty. There are only two elements between 7 and 10 namely 8 and 9. These are not prime. Hence, B is empty.
The set A = {2x2 β 3x +1 = 0 xβ N} is equal to ____.
The roots of the equation 2x2 - 3x + 1 = 0 are 1/2 and 1.
But as x β N, 1 is the only element of A.
The set A =x: 2x2- 3x + 1 = 0 x β N is equal to {1}.
Given A = {1, 9, 25, 49, 81} B = {(2n - 1)2 | 1 β€ n β€ 5, n β N} C = { n2 | 1 β€ n β€ 9, n β N} D = {(2n - 1) | 1 β€ n β€ 9, n β N} ____ Sets are equal.
Solution: 62= 36 β C, but 36 β A
2 Γ 4 β 1) β D, but 7 β A
A = {1, 9, 25, 49, 81}
B = {(2 Γ 1 - 1)2, (2 Γ 2 - 1)2, (2 Γ 3 - 1)2, (2 Γ 4 - 1)2, (2 Γ 5 - 1)2} = {1, 9, 25, 49, 81}, which is equal to A.
Hence, sets A and B are equal
Of the options given, ____ is an empty set.
Solution: e β A
p β B
eβ D
But as no letter can be both vowel and a consonant, C is an empty set.
From the sets given below, select equal sets. P = {1, 2, 3, 5}, Q = {2, 7, 8, 5}, R = {a, b}, S = {1 β 3}, T = {7, 2, 8, 5}, U = {β3, 1}
Solution: We know that,
Two sets are said to be equal if they have exactly the same elements.
P = {1, 2, 3, 5}, Q = {2, 7, 8, 5}, R = {a, b}, S = {1 β 3}, T = {7, 2, 8, 5}, U = {β3, 1}
Q = T and S = U are the equal sets, because all the elements of set Q are elements of set T and vice-versa and all the elements of set S are elements of set U and vice-versa.
Given A = {x: x is an even divisor of 12} B = {x: x is an even natural number less than 7} C = {x: x is a prime divisor of 6} D = {x: x is a divisor of 12} The sets ____ and ____ are equal.
Solution: A = {x: x is an even divisor of 12} = {2, 4, 6}
B = {x: x is an even natural number less than 7} = {2, 4, 6}
C = {x: x is a prime divisor of 6} = {2, 3}
D = {x: x is a divisor of 12} = {1, 2, 3, 4, 6, 12}
So, A and B are equal.
If A is empty set, then number of elements in P (P (A)) is ____ [P (A) = denotes the power set of A.]
Solution: A is an empty set
i.e. A = Ο
P (A) = {Ο}
P (P (A)) = {Ο, {Ο}}
So P (P (A)) has 2 elements.
Which of the statement is false? A. A β A βͺ B B. A β© B β A C. B β B β A D. A β© B β A βͺ B
Solution: Let's take A = {1, 2, 3}, B = {2, 3, 4}
B - A = {4}
(2, 3, 4) β (4)
Hence, B β B - A is false.
If A = {1, 2, 3, 5, 8}, B = {3, 5, 8}, then the proper subsets of A which are super sets of B are ____.
Solution: Option B is not correct as {3, 5, 8} in not included.
Option C is not correct as {1, 2, 5, 8} is not a super set of B
Option D is not correct as {1, 2, 5, 3} is not a super set of B
The correct answer is A - {3, 5, 8}, {1, 3, 5, 8}, {2, 3, 5, 8}
Which of the following statement is true for all A? (a)(Ο) β P [PA)] (b) Ο β A (c) Ο β A (d) [(Ο)] β A
Solution: Option A is not correct as Ο is not always a element of A.
Option B is not correct as {Ο} is not always a subset of A.
Option C is not correct as {Ο} is not always a element of A.
Option D is correct, as Ο is always an element of P(A) and so {Ο} is always an element of P(P(A)).
B - (A β© B) is equal to ____.
Solution: B - (A β© B) = B β© (A β© B)C [because A - B = A β© B']
= B β© (AC U BC)[ By De Morgan's law]
= (B β© AC) U (B β© BC) [By distributive law]
= (B β© AC) U Ο
= B β© AC
= B β A
A U (B - A) is equal to ____.
Solution: A U (B - A) = A U (B β© AC)
= (A U B) β© (A U AC) [Distributive law]
= (A U B) β© (U) [U universal set]
= A U B [because A β© U = A, for all sets A]
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Given the sets A = {2, 5, 6}, B = {5, 6, 7}, C = {1, 2, 5, 7, 8}, then (i) (B U C')' is ____ and (ii) (A - B')' is ____.
Solution: C' = {3, 4, 6, 9, 10}
B U C' = {3, 4, 5, 6, 7, 9, 10}
(B U C')β = {1, 2, 8}
B' = {1, 2, 3, 4, 8, 9, 10}
A - B' = {5, 6}
(A - B')' = {1, 2, 3, 4, 7, 8, 9, 10}
U = {a, b, c, d, e, 1, 2, 5} = Universal set If A = {a, 2, 5}, B = {a, b, 5} and C = {b, 2, 1, d}, then (AC U B) β© C is equal to ____.
Solution: AC = {b, c, d, e, 1}
AC U B = {a, b, c, d, e, 1, 5}
(AC U B) β© C = {b, d, 1}
(A β© C')' U (B β© C) is equal to ____.
Solution: (A β© C')' U (B β© C) = (A' U (C')') U (B β© C)[By De Morgan's law]
= (A' U C) U (B β© C)
= [(A' U C) U B] β© [(A' U C) U C]
= [A' U B U C] β© [A' U C]
= A' β© C [β΅A'βͺCβA'βͺBβͺC and AβBβAβ©B = A]
If A = {2, 4, 5, 7, 8}, B = {3, 5, 7, 9, 10} and C = {1, 2, 3, 4, 5}, then (A βͺ B) β© C = ____.
Solution: Given,
A = {2, 4, 5, 7, 8}, B = {3, 5, 7, 9, 10} and C = {1, 2, 3, 4, 5}
(A βͺ B) β© C
A βͺ B = {2, 4, 5, 7, 8} βͺ {3, 5, 7, 9, 10}
= {2, 3, 4, 5, 7, 8, 9, 10}
(A βͺ B) β© C = {2, 3, 4, 5, 7, 8, 9, 10} β© {1, 2, 3, 4, 5}
= {2, 3, 4, 5}
Which of the following statements is true?
Solution: The divisors of 28 are 1, 2, 4, 7, 14 28.
The sum of the divisors is 56 which is 2 Γ 28..
Hence, the correct answer is B..
Option A is not correct as 23 has exactly two positive divisors..
Option C is not correct as 7, 542 is not a multiple of 27..
Option D is not correct, as 513 is not a perfect cube..
513 is not a perfect cube.
but divisors of 28 are 1, 2, 4, 7, 14, 28.
1 + 2 + 4 + 7 + 14 + 28 = 56 = 2. 28.
So 'B' is true
The set A = {....-5, -2, 1, 4, 7, 10,} in set builder form is ____.
Solution: The set has elements of the form 3 Γ k + 1. Hence, the set can be represented as[x: 3k + 1 k β Z] in the set builder form.
Given the sets X = {a, b, 1, 100, 2} Y = {{1}, 2, 7, {x}} Z = {y, 5} ____ may be considered as a universal set for all the three sets X, Y, Z.
Solution: As (1) β A
a β A
x β A
But X βͺ Y βͺ Z β C
So, {a, b, y, 1, 2, 5, 7, 100, {x}, {1}} is the universal set for X, Y and Z.
Two finite sets have m and n elements respectively. The total number of subset of first set is 48 more than the total number of the second set. The values of m and n respectively are ___.
Solution: Let A and B be such sets, i.e n (A) = m and n(B) = n
So, n(P(A)) = 2m and n(P(B)) = 2n
2m = 2n + 48(This implies m > n)
β 2m - 2n = 48
i.e. 2n (2 m-n - 1) = 48 = 24 Γ 3
β n = 4 and 2 m-n - 1 = 3
β 2m-n = 4 = 22
β m - n = 2
β m = n + 2 = 4 + 2 = 6
So, m = 6 and n = 4
A β© (B U C) is equal to ____.
Solution: Aβ©(BβͺC) β (Aβ©B) βͺ(Aβ©C) =( Aβ©B)βͺ(Aβ©C)βAβ©(BβͺC)
Let x β A β© B βͺ C
βx β A and x β B or x β C
βx β A and x β B or x β A and x β C
βx β (Aβ©B) or x β (Aβ©C)
βx β (Aβ©B) βͺ (Aβ© C)
So, A β© ( B U C) β (A β© B) U (A β© C)
[Note:- To prove AβB, one has to show every element x of A belongs to B (i.e) x β Aβx β B]
Now, let x β (A β©B) βͺ (Aβ©C)
βx β A β© B or x β A β© C
βx β A and x β B or x β A and x β C
βx β A and x β B or x β C
βx β A and x β B βͺ C
βx β A β©(B βͺ C)
β΄(Aβ© B)βͺ( A β©C) β A β©(B βͺ C)
β΄A β©( B βͺ C)=(A β© B) βͺ (A β©C)