Concept:
Union of the sets:
Union of two given sets is the set that contains those elements that are either in A or in B, or in both.
The union of the sets A and B, denoted by A U B
Intersection of Sets:
The intersection of two given sets is the largest set which contains all the elements that are common to both the sets.
The intersection of Sets A and B, denoted by A ∩ B
Formula: A ∪ B = A + B  A ∩ B
Calculation:
Given: A ∪ B = A
To Find: A ∩ B
As we know,
A ∪ B = A + B  A ∩ B
⇒ A = A + B  A ∩ B
∴ A ∩ B = B
Directions: Consider the following Venn diagram, where X, Y and Z are three sets. Let the number of elements in Z be denoted by n(Z). which is equal to 90.
Calculation:
Given: n(Z) = 90 and the number of elements in Y and Z are in the ratio 4 : 5
To find: b = ?
n(Y) = 16 + 18 + 17 + b
n(Z) = 12 + 18 + 17 + c = 90
\(\rm \frac{n(y)}{n(z)}=\frac{16+18+17+b}{12+18+17+c}=\frac{4}{5}\\ \Rightarrow \frac{51+b}{90}=\frac{4}{5}\\ \Rightarrow (51+b)\times 5=360 \\\Rightarrow255+5b=360\\ \Rightarrow5b=105\\ \Rightarrow b=21\)
Hence, option (3) is correct.
Directions: Consider the following Venn diagram, where X, Y and Z are three sets. Let the number of elements in Z be denoted by n(Z). which is equal to 90.
Calculation:
Given: n(Z) = 90,
From above diagram, we get
n(X) = a + 46, n(Y) = b + 51, n(X ∩ Y) = 16 + 18 = 34, n(Y ∩ Z) = 17 + 18 =35, n(X ∩ Z) = 12 + 18 = 30, n(X ∩ Y ∩ Z) = 18
n(X) + n(Y) + n(Z)  n(X ∩ Y)  n(Y ∩ Z)  n(X ∩ Z) + n(X ∩ Y ∩ Z) =
a + 46 + b + 51 + 90  34  35  30 + 18
= a + b + 106
Hence, option (4) is correct.
Concept:
(A ⋂ B) are the elements present in both the sets (A and B)
Calculation:
Here, A = {x: x is a natural number and 1 < x ≤ 4},
A = {2, 3, 4}
B = {x: x is a natural number and 4 < x ≤ 7}
B = {5, 6, 7}
Here, not a single element is common in both sets.
So, (A ∩ B) = ϕ
Hence, option (4) is correct.
Directions: Consider the following Venn diagram, where X, Y and Z are three sets. Let the number of elements in Z be denoted by n(Z). which is equal to 90.
Concept:
Complement of X = U  X, where U is a universal set.
Calculation:
We know, n(X) + n(Y) + n(Z)  n(X ∩ Y)  n(Y ∩ Z)  n(X ∩ Z) + n(X ∩ Y ∩ Z) = n(X ∪ Y ∪ Z)
Here U  n(X ∪ Y ∪ Z) = p, whre U is a universal set
⇒ U  (a + b + 106) = p
⇒ U = p + (a + b + 106)
Now the number of elements in the complement of X, U  n(X) = p + (a + b + 106)  (a + 16 + 18 + 12)
= p + b + 60
Hence, option (1) is correct.
Consider to subsets of R^{3} given as,
S_{1} = {[2, 3, 1], [1, 0, 5], [0, 1, 0], [0, 0, 1]} and S_{2} = {[1, 0, 0], [0, 1, 1], [0, 0, 0]}. Which of the following statements is true?
Given:
S_{1} = [2, 3, 1], [1, 0, 5], [0, 1, 0], [0, 0, 1]
S_{2} = [1, 0, 0], [0, 1, 1], [0, 0, 0]
For subset S_{1}
Let,
α_{1 }= [2, 3, 1],
α_{2} = [1, 0, 5],
α_{3} = [0, 1, 0],
α_{4} = [0, 0,1]
For linearly independent:
aα_{1 }+ bα_{2} + cα_{3} + dα_{4} = 0
a [2, 3, 1] + b [1, 0, 5] + c [0, 1, 0] + d [0, 0, 1] = [0, 0, 0, 0]
2a + b = 0 ⇒ a= b/2
3a + c = 0 ⇒ c = 3a = 3b/2
a + 5b + d = 0 ⇒ d = a 5b = 5b + b/2 = 9b/2
value of a, c, d depend upon ‘b’ Hence It is linearly dependent
for subset S_{2},
α_{1} = [1, 0, 0], α_{2 }= [0, 1, 1], α_{3} = [0, 0, 0]
α_{3} is [0,0,0] So, Determinant of matrix =0.
We will solve this type of question by Matrix method:
If, Rank of matrix ≥ No. of unknown variables, then System is linearly Independent.
Else, System is linearly dependent.
With the help of α_{1}, α_{2}, and α_{3}, matrix A is formed as;
\(A = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&1\\ 0&0&0 \end{array}} \right]\)
\(\left A \right = 0\)
Rank = 2 < (No. of unknown variable = 3)
Hence It is linearly dependent.
Concept:
Let A and B denote two sets of elements.
Calculations:
Let A be the set of students who read "Hitavad" and B the set of students who read "Hindustan".
From the given information:
n(A ⋃ B) = 50  10 = 40.
n(A) = 30.
n(B) = 35.
By using n(A ⋃ B) = n(A) + n(B)  n(A ⋂ B), we get:
40 = 30 + 35  n(A ⋂ B)
⇒ n(A ⋂ B) = 25.
∴ The number of students who read both "Hitavad" and "Hindustan" newpapers is n(A ⋂ B) = 25.
Open set:
A subset S of a metric space (X, d) is open if it contains an open ball about each of its points, i.e.,
if ∀ x ∈ S : ∃ ϵ > 0: B(x, ϵ) ⊆ S.
If S_{1}, S_{2}, . . . , S_{n} are open sets, then ∩_{n}^{i=1}{S_{i}} is an open set.
In other words, the intersection of the finite number of open sets is open.
If S_{α} is an open set for each α ∈ A, then ∪_{α∈A}S_{α} is an open set. Here, A be an arbitrary set
In other words, the union of any collection of open sets is open.
Closed set:
A subset S of a metric space (X, d) is closed if it is the complement of an open set.
If S1, S2, . . . , Sn are closed sets, then ∪_{n}^{i=1}{S_{i}} is a closed set.
In other words, the union of the finite number of closed sets is close.
If S_{α} is a closed set for each α ∈ A, then ∩_{α∈A}S_{α} is a closed set.
In other words, the intersection of any collection of closed sets is closed.
Conclusion:
1. An arbitrary (finite, countable, or uncountable) union of open sets is open. (statement 1 is correct)
2. The intersection of a finite number of open sets is open. (statement 3 is correct)
3. An arbitrary (finite, countable, or uncountable) intersection of closed sets is closed. (but in statement 2 instead of the intersection, the union is mentioned, hence statement 2 is incorrect).
4. The union of a finite number of closed sets is closed.
Concept:
Union of the sets:
Union of two given sets is the set that contains those elements that are either in A or in B, or in both.
The union of the sets A and B, denoted by A U B
Intersection of Sets:
The intersection of two given sets is the largest set which contains all the elements that are common to both the sets.
The intersection of Sets A and B, denoted by A ∩ B
Formula: n(A ∪ B) = n(A) + n(B)  n(A ∩ B)
Calculation:
Given: n(A) = 20, n(B) = 35 and n(A ∪ B) = 45
To Find: n (A ∩ B)
We know that,
n(A ∪ B) = n(A) + n(B)  n(A ∩ B)
⇒ n(A ∩ B) = n(A) + n(B)  n(A ∪ B)
⇒ n(A ∩ B) = 20 + 35  45
∴ n(A ∩ B) = 10
CONCEPT:
For any two sets A and B, we have: n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
CALCULATION:
Given: Out of 50 students in a school, 25 have scooters and 35 have cycles for coming to school.
Let A be the number of students who have scooter for coming to school and B be the number of students who are having cycles for coming to school
⇒ n(A) = 25, n(B) = 35 and n(A ∪ B) = 50
As we know that, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
⇒ 50 = 25 + 35  n(A ∩ B)
⇒ n(A ∩ B) = 60  50 = 10
Hence, there are 10 students in school who have both scooter and cycle for coming to school.
Concept:
P  Q = The set of elements which belong to P but not to Q
P U Q = The set which consists of all elements of P and all elements of Q, the common elements being taken only once.
P ∩ Q = The set of all elements which are common to both A and B.
Calculation:
let us suppose P = {1, 2, 3, 4} and Q = {3, 4, 5, 6}
⇒ P  Q = {1, 2}
⇒ Q  P = {5, 6}
⇒ P U Q = {1, 2, 3, 4, 5, 6}
⇒ P ∩ Q = {3, 4}
⇒ (P  Q) ∪ (Q  P) ∪ (P ∩ Q) = {1, 2} U {5, 6} U {3, 4} = {1, 2, 3, 4, 5, 6} = P U Q
Concept:
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
Calculation:
Given: A = {a, b, c, d, e}, B = {a, c, e, g} and C = {b, e, f, g}
As we know that, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
Here, we have to find the value of (A ∩ B) ∪ (A ∩ C)
i.e we have to find the value of A ∩ (B ∪ C)
⇒ B ∪ C = {a, b, c, e, f, g}
⇒ A ∩ (B ∪ C) = {a, b, c, d, e} ∩ {a, b, c, e, f, g} = {a, b, c, e}
Hence, (A ∩ B) ∪ (A ∩ C) = {a, b, c, e}
Given
n(Q) = 36, n(R) = 40 and n(Q U R) = 50
Calculation
n(Q U R) = n(Q) + n(R)  n(Q ∩ R)
⇒ 50 = 36 + 40  n(Q ∩ R)
⇒ 50 = 76  n(Q ∩ R)
n(Q ∩ R) = 26
A, B, C and D are four sets such that A ∩ B = C ∩ D = ϕ. Consider the following:
1. A ∪ C and B ∪ D are always disjoint.
2. A ∩ C and B ∩ D are always disjoint.
Which of the above statements is/are correct?Concept:
Two sets are said to be disjoint when they have no common element.
Two sets A and B are disjoint sets if the intersection of two sets is a null set or an empty set.
A ∩ B = ϕ
Calculation:
Given: A, B, C and D are four sets such that A ∩ B = C ∩ D = ϕ
Statement 1: A ∪ C and B ∪ D are always disjoint.
⇒ (A ∪ C) ∩ (B ∪ D) = [(A ∪ C) ∩ B] ∪ [(A ∪ C) ∩ D]
= [(A ∩ B) ∪ (C ∩ B)] ∪ [(A ∩ D) ∪ (C ∩ D)]
= [ϕ ∪ (C ∩ B)] ∪ [(A ∩ D) ∪ ϕ]
= (C ∩ B)] ∪ [(A ∩ D) (∵ A ∪ ϕ = A)
(A ∪ C) and (B ∪ D) is not disjoint set.
Statement 2: A ∩ C and B ∩ D are always disjoint.
⇒ (A ∩ C) ∩ (B ∩ D) = [(A ∩ C) ∩ B] ∩ [(A ∩ C) ∩ D]
= [(A ∩ B) ∩ C] ∩ [(A ∩ (C ∩ D)]
= [ϕ ∩ C] ∩ [(A ∩ ϕ] (∵ A ∩ ϕ = ϕ)
= (ϕ ∩ ϕ)
= ϕ
(A ∩ C) and (B ∩ D) are always disjoint.
Hence statement 2 is correct.
Concept:
Calculation:
Let,
A = Number divisible by 10,
B = Number divisible by 15,
C = Number divisible by 25
\({\rm{n}}\left( {\rm{A}} \right) = \frac{{1000}}{{10}} = 100,{\rm{\;}}\)
\({\rm{n}}\left( {\rm{B}} \right) = \frac{{1000}}{{15}} = 66,{\rm{\;}}\)
\({\rm{n}}\left( {\rm{C}} \right) = \frac{{1000}}{{25}} = 40\)
Now, number divisible by A and B, i.e., n (A ∩ B)
For that we need to find out the least number which is divisible by both 10 and 15
And that is 30
\(\therefore {\rm{n}}\left( {{\rm{A}} \cap {\rm{B}}} \right) = \frac{{1000}}{{30}} = 33\)
Similarly,
\({\rm{n}}\left( {{\rm{A}} \cap {\rm{C}}} \right) = \frac{{1000}}{{50}} = 20\) (∵ The least number which is divisible by both 10 and 25 is 50)
Also,
\({\rm{n}}\left( {{\rm{B}} \cap {\rm{C}}} \right) = \frac{{1000}}{{75}} = 13\) (∵ The least number which is divisible by both 15 and 25 is 75)
And
The least number which is divisible by 10, 15 and 25 is 150
\({\rm{n}}\left( {{\rm{A}} \cap {\rm{B}} \cap {\rm{C}}} \right) = \frac{{1000}}{{150}} = 6\)
Number divisible by 10, 15 or 25
I.e., \({\rm{n}}\left( {{\rm{A\;U\;B\;U\;C}}} \right) = {\rm{\;n}}\left( {\rm{A}} \right) + {\rm{\;n}}\left( {\rm{B}} \right) + {\rm{\;n}}\left( {\rm{C}} \right)  {\rm{\;n}}\left( {{\rm{A}} \cap {\rm{B}}} \right)  {\rm{\;n}}\left( {{\rm{A}} \cap {\rm{C}}} \right)  {\rm{\;n}}\left( {{\rm{B}} \cap {\rm{C}}} \right) + {\rm{n}}\left( {{\rm{A}} \cap {\rm{B}} \cap {\rm{C}}} \right)\)
\(\therefore {\rm{n}}\left( {{\rm{A\;U\;B\;U\;C}}} \right) = 100 + 66 + 40  33  20  13 + 6\)
\( \Rightarrow {\rm{n}}\left( {{\rm{A\;U\;B\;U\;C}}} \right) = 146\)
Now, the number divisible neither by 10 nor 15, 25
= 1000 – 146
= 854
Hence, option (2) is correct.
If A = {x ∈ R : x^{2} + 6x – 7 < 0} and B = {x ∈ R : x^{2} + 9x + 14 > 0}, then which of the following is/are correct?
1. A ∩ B = {x ∈ R :  2 < x < 1}
2. A \ B = {x ∈ R :  7 < x <  2}
Select the correct answer using the code given below:Calculation:
Given:
A = {x ∈ R : x^{2} + 6x – 7 < 0}
⇒ x^{2} + 6x – 7 < 0
⇒ x^{2} + 7x  x – 7 < 0
⇒ x (x + 7) – 1 (x + 7) < 0
⇒ (x – 1) (x + 7) < 0
∴ A ∈ (7, 1)
Now, B = {x ∈ R : x^{2} + 9x + 14 > 0}
⇒ x^{2} + 9x + 14 > 0
⇒ x^{2} + 7x + 2x + 14 > 0
⇒ x (x + 7) + 2(x + 7) > 0
⇒ (x + 7) (x + 2) > 0
∴ B ∈ (∞, 7) ∪ (2, ∞)
Now, A ∩ B
∴ A ∩ B = (2, 1) ⇔ A ∩ B = {x ∈ R :  2 < x < 1}
So, statement 1 is true,
Now, A ∪ B = R – {7}
So, statement 2 is wrong,
Concept:
Let set A has n(A) elements and set B has n(B) elements.
Number of distinct relations from A to B = 2^{n(A) × n(B)}
Calculation:
Number of elements in set A = p
Number of elements in set B = q
∴ Number of distinct relations from B to A = 2n(A) × n (B)
= 2^{pq}
Calculation:
Given: The four sets have 150, 180, 210 and 240 elements respectively
n(A) = 150
n(B) = 180
n(C) = 210
n(D) = 240
Each pair of sets has 15 elements
n(A ∩ B) = 15
n(A ∩ C) = 15
n(A ∩ D) = 15
n(B ∩ C) = 15
n(B ∩ D) = 15
n(C ∩ D) = 15
Each triple of sets has 3 elements
n(A ∩ B ∩ C) = 3
n(A ∩ B ∩ D) = 3
n(A ∩ C ∩ D) = 3
n(B ∩ C ∩ D) = 3
A ∩ B ∩ C ∩ D = ϕ
n(A ∩ B ∩ C ∩ D) = 0
Now, number of elements in the union of 4 sets A, B, C and D
n(A ∪ B ∪ C ∪ D) = n(A) + n(B) + n(C) + n(D)  n(A ∩ B)  n(A ∩ C)  n(A ∩ D)  n(B ∩ C)  n(B ∩ D)  n(C ∩ D) + n(A ∩ B ∩ C) + n(A ∩ B ∩ D) + n(A ∩ C ∩ D) + n(B ∩ C ∩ D)  n(A ∩ B ∩ C ∩ D)
= 150 + 180 + 210 + 240  6 × 15 + 4 × 3  0
= 702
CONCEPT:
Intersection:
Let A and B be two sets. The intersection of A and B is the set of all those elements which are present in both the sets A and B.
The intersection of A and B is denoted by A ∩ B i.e A ∩ B = {x : x ∈ A and x ∈ B}
CALCULATION:
Given: A = {1, 2, 3, 4} and B = {x ∈ N : x ≤ 5}
The set B can be rewritten as B = {1, 2, 3, 4, 5}
As we know that, A ∩ B = {x : x ∈ A and x ∈ B}
⇒ A ∩ B = {1, 2, 3, 4} = A
Hence, correct option is 2.
Concept:
Properties of relation:
Let R be a relation on Z, and let x, y, z ∈ Z.

xRx 

xRy implies yRx 

xRy and yRz implies xRz 
Let (a, b) ∈ R
Then (a, b) ∈ R ⇒ (b, a) ∈ R^{1}
(b, a) ∈ R [Because R = R^{1}]
(a, b) ∈ R ⇒ (b, a) ∈ R, for all (a, b) ∈ A
aRb ⇒ bRa
∴ R is symmetric