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Post a LessonAnswered on 20/09/2019 Learn Miscellaneous Exercise 1
Swapna Shree
A = {x: x ∈ R and x satisfies x2 – 8x + 12 = 0}
2 and 6 are the only solutions of x2 – 8x + 12 = 0.
∴ A = {2, 6}
B = {2, 4, 6}, C = {2, 4, 6, 8 …}, D = {6}
∴ D ⊂ A ⊂ B ⊂ C
Hence, A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C
read lessAnswered on 20/09/2019 Learn Miscellaneous Exercise 1
Swapna Shree
(i) False
Let A = {1, 2} and B = {1, {1, 2}, {3}}
Now,
∴ A ∈ B
However,
(ii) False
Let
As A ⊂ B
B ∈ C
However,
(iii) True
Let A ⊂ B and B ⊂ C.
Let x ∈ A
∴ A ⊂ C
(iv) False
Let
Accordingly,and
.
However, A ⊂ C
(v) False
Let A = {3, 5, 7} and B = {3, 4, 6}
Now, 5 ∈ A and A ⊄ B
However, 5 ∉ B
(vi) True
Let A ⊂ B and x ∉ B.
To show: x ∉ A
If possible, suppose x ∈ A.
Then, x ∈ B, which is a contradiction as x ∉ B
∴x ∉ A
read lessAnswered on 20/09/2019 Learn Miscellaneous Exercise 1
Swapna Shree
Let, A, B and C be the sets such that and
.
To show: B = C
Let x ∈ B
Case I
x ∈ A
Also, x ∈ B
∴
∴ x ∈ A and x ∈ C
∴ x ∈ C
∴ B ⊂ C
Similarly, we can show that C ⊂ B.
∴ B = C
read lessTake Class 11 Tuition from the Best Tutors
Answered on 20/09/2019 Learn Miscellaneous Exercise 1
Swapna Shree
First, we have to show that (i) ⇔ (ii).
Let A ⊂ B
To show: A – B ≠ Φ
If possible, suppose A – B ≠ Φ
This means that there exists x ∈ A, x ≠ B, which is not possible as A ⊂ B.
∴ A – B = Φ
∴ A ⊂ B ⇒ A – B = Φ
Let A – B = Φ
To show: A ⊂ B
Let x ∈ A
Clearly, x ∈ B because if x ∉ B, then A – B ≠ Φ
∴ A – B = Φ ⇒ A ⊂ B
∴ (i) ⇔ (ii)
Let A ⊂ B
To show:
Clearly,
Let
Case I: x ∈ A
∴
Case II: x ∈ B
Then,
Conversely, let
Let x ∈ A
∴ A ⊂ B
Hence, (i) ⇔ (iii)
Now, we have to show that (i) ⇔ (iv).
Let A ⊂ B
Clearly
Let x ∈ A
We have to show that
As A ⊂ B, x ∈ B
∴
∴
Hence, A = A ∩ B
Conversely, suppose A ∩ B = A
Let x ∈ A
⇒
⇒ x ∈ A and x ∈ B
⇒ x ∈ B
∴ A ⊂ B
Hence, (i) ⇔ (iv).
read lessAnswered on 20/09/2019 Learn Miscellaneous Exercise 1
Swapna Shree
Let A ⊂ B
To show: C – B ⊂ C – A
Let x ∈ C – B
⇒ x ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A [A ⊂ B]
⇒ x ∈ C – A
∴ C – B ⊂ C – A
read lessAnswered on 20/09/2019 Learn Miscellaneous Exercise 1
Swapna Shree
Let A and B be two sets such that A ∩ X = B ∩ X = f and A ∪ X = B ∪ X for some set X.
To show: A = B
It can be seen that
A = A ∩ (A ∪ X) = A ∩ (B ∪ X) [A ∪ X = B ∪ X]
= (A ∩ B) ∪ (A ∩ X) [Distributive law]
= (A ∩ B) ∪ Φ [A ∩ X = Φ]
= A ∩ B … (1)
Now, B = B ∩ (B ∪ X)
= B ∩ (A ∪ X) [A ∪ X = B ∪ X]
= (B ∩ A) ∪ (B ∩ X) [Distributive law]
= (B ∩ A) ∪ Φ [B ∩ X = Φ]
= B ∩ A
= A ∩ B … (2)
Hence, from (1) and (2), we obtain A = B.
read lessTake Class 11 Tuition from the Best Tutors
Answered on 20/09/2019 Learn Miscellaneous Exercise 1
Swapna Shree
Answered on 20/09/2019 Learn Miscellaneous Exercise 1
Swapna Shree
Let U be the set of all students who took part in the survey.
Let T be the set of students taking tea.
Let C be the set of students taking coffee.
Accordingly, n(U) = 600, n(T) = 150, n(C) = 225, n(T ∩ C) = 100
To find: Number of student taking neither tea nor coffee i.e., we have to find n(T' ∩ C').
n(T' ∩ C') = n(T ∪ C)'
= n(U) – n(T ∪ C)
= n(U) – [n(T) + n(C) – n(T ∩ C)]
= 600 – [150 + 225 – 100]
= 600 – 275
= 325
Hence, 325 students were taking neither tea nor coffee.
read lessAnswered on 20/09/2019 Learn Miscellaneous Exercise 1
Swapna Shree
Let U be the set of all students in the group.
Let E be the set of all students who know English.
Let H be the set of all students who know Hindi.
∴ H ∪ E = U
Accordingly, n(H) = 100 and n(E) = 50
= 25
n(U) = n(H) + – n(H ∩ E)
= 100 + 50 – 25
= 125
Hence, there are 125 students in the group.
read lessTake Class 11 Tuition from the Best Tutors
Answered on 20/09/2019 Learn Miscellaneous Exercise 1
Swapna Shree
Let A be the set of people who read newspaper H.
Let B be the set of people who read newspaper T.
Let C be the set of people who read newspaper I.
Accordingly, n(A) = 25, n(B) = 26, and n(C) = 26
n(A ∩ C) = 9, n(A ∩ B) = 11, and n(B ∩ C) = 8
n(A ∩ B ∩ C) = 3
Let U be the set of people who took part in the survey.
(i) Accordingly,
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)
= 25 + 26 + 26 – 11 – 8 – 9 + 3
= 52
Hence, 52 people read at least one of the newspapers.
(ii) Let a be the number of people who read newspapers H and T only.
Let b denote the number of people who read newspapers I and H only.
Let c denote the number of people who read newspapers T and I only.
Let d denote the number of people who read all three newspapers.
Accordingly, d = n(A ∩ B ∩ C) = 3
Now, n(A ∩ B) = a + d
n(B ∩ C) = c + d
n(C ∩ A) = b + d
∴ a + d + c + d + b + d = 11 + 8 + 9 = 28
⇒ a + b + c + d = 28 – 2d = 28 – 6 = 22
Hence, (52 – 22) = 30 people read exactly one newspaper.
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