Miscellaneous Exercise on Chapter 1 Class 11 from the NCERT book contains 10 questions. The questions in this exercise are based on conceptual understanding of the previous exercises from 1.1 to 1.5.
Miscellaneous Exercise on Chapter 1 Class 11 NCERT
1. Decide among the following sets, which sets are subsets of one and another.
A =
B = {2, 4, 6}, C = {2, 4, 6, 8, . . .}, D = {6}.
Answer
The given sets are
A =
B = {2, 4, 6},
C = {2, 4, 6, 8, . . .},
D = {6}.
From the set A, we have,
So, the set A in roster form is
A = {2,6}
We have to find that sets which are subsets of one another. Clearly,
A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C
2. In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
(i) If ∈ A and A ∈ B, then ∈ B.
Answer
False
Example:
A = {2, 7} and B = {1, {2, 7}, 8, 9}
Here, 2 ∈ A and A ∈ B but 2 ∉ B
That's why it is false.
(ii) If A ⊂ B and B ∈ C, then A ∈ C
Answer
False
Example:
A = {1, 2}, B = {1, 2, 3} and C = {4, {1, 2, 3}, 5}
Here, A ⊂ B and B ∈ C but A ∉ C
(iii) If A ⊂ B and B ⊂ C, then A ⊂ C
Answer
True
Proof: Let ∈ A.
⇒ B (∵ A ⊂ B)
⇒ C (∵ B ⊂ C)
So, we have shown that
If
A ⊂ C
Hence, proved!
(iv) If A ⊄ B and B ⊄ C, then A ⊄ C
Answer
False
Example:
A = {0, 1, 2}
B = {1, 2, 3}
C = {0, 1, 2, 4}
Here, A ⊄ B and B ⊄ C but A ⊂ C
That's why it is false.
(v) If ∈ A and A ⊄ B, then ∈ B
Answer
False
Example:
A = {1, 2, 3}
B = {0, 4, 5}
Here, 2 ∈ A and A ⊄ B but 2 ∉ B
That's why it is false.
(vi) If A ⊂ B and ∉ B, then ∉ A.
Answer
True
Proof: We know that if A ⊂ B, then all elements of A is in B. And if there is an element ∉ B, it definitely means that ∉ A.
3. Let A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C.
Answer
Given:
A ∪ B = A ∪ C,
A ∩ B = A ∩ C
To show: B = C
Proof: To show B = C, we have to show that all elements of B is in C and all elements of C is in B, i.e. B ⊆ C and C ⊆ B
First, we will show B ⊆ C
let ∈ B
⇒ ∈ A ∪ B (∵ A ∪ B contains elements of both A and B)
⇒ ∈ A ∪ C (∵ A ∪ B = A ∪ C)
⇒ ∈ A or ∈ C
Two cases arise.
Case I: When ∈ A
Case II: When ∈ C
Case I: When ∈ A
In this case, we have,
∈ A and ∈ B
⇒ ∈ A ∩ B
⇒ ∈ A ∩ C (∵ A ∩ B = A ∩ C)
⇒ ∈ A and ∈ C
⇒ ∈ C
So, when ∈ B ⇒ ∈ C
⇒ B ⊆ C . . . (1)
Case II: When ∈ C
In this case,
When ∈ B ⇒ ∈ C
⇒ B ⊆ C . . . (1)
Now, we will show C ⊆ B
let ∈ C
⇒ ∈ A ∪ C (∵ A ∪ C contains elements of both A and C)
⇒ ∈ A ∪ B (∵ A ∪ B = A ∪ C)
⇒ ∈ A or ∈ B
Two cases arise.
Case I: When ∈ A
Case II: When ∈ B
Case I: When ∈ A
In this case, we have,
∈ A and ∈ C
⇒ ∈ A ∩ C
⇒ ∈ A ∩ B (∵ A ∩ B = A ∩ C)
⇒ ∈ A and ∈ B
⇒ ∈ B
So, when ∈ C ⇒ ∈ B
⇒ C ⊆ B . . . (2)
Case II: When ∈ B
In this case,
When ∈ C ⇒ ∈ B
⇒ C ⊆ B . . . (2)
From (1) and (2),
B ⊆ C and C ⊆ B
⇒ B = C
Hence, proved!
4. Show that the following four conditions are equivalent:
(i) A ⊂ B
(ii) A − B = ϕ
(iii) A ∪ B = B
(iv) A ∩ B = A
Answer
The given conditions are
(i) A ⊂ B
(ii) A − B = ϕ
(iii) A ∪ B = B
(iv) A ∩ B = A
To show that the above four conditions are equivalent, we have to show that from one condition, we can derive other three conditions.
(i) A ⊂ B ⇒ All elements of A are in B.
⇒ If we exclude or remove all elements of B from A, we will be left with an empty set, i.e. ϕ . It means (ii) A − B = ϕ condition is equivalent to (i).
Also, if all elements of A are in B, then union of A and B will be B itself. So,(iii) A ∪ B = B condition is equivalent to (i)
Lastly, if all elements of A are in B, then intersection of sets A and B i.e. A ∩ B is equal to A itself.
So, the condition (iv) A ∩ B = A is equivalent to (i).
Similarly, we can show for other conditions that they are equivalent to three others.
5. Show that if A ⊂ B, then C − B ⊂ C − A.
How to prove
When you are given to prove something like question 5 is asking, you have to do the following.
Let us take C − B ⊂ C − A, as given in the question. For proving this, we first suppose an element ∈ C − B and then somehow using our understanding of sets and all, we try to show that ∈ C − A. Having shown it, we would have stood proved. Now you can see the solution for practical proof.
Answer
Given: A ⊂ B
To prove: C − B ⊂ C − A
Proof: Let ∈ C − B
⇒ ∈ C and ∉ B
∵ ∉ B and A ⊂ B (given)
⇒ ∉ A
Now,
∵ ∈ C and ∉ A
⇒ ∈ C − A
Clearly, we have shown that
whenever ∈ C − B ⇒ ∈ C − A
⇒ C − B ⊂ C − A
Hence proved.
6. Show that for any sets A and B,
A = (A ∩ B) ∪ (A − B) and A ∪ (B − A) = (A ∪ B)
Answer
First part
Given: Any two sets A and B.
To prove: A = (A ∩ B) ∪ (A − B)
Method: To prove A = (A ∩ B) ∪ (A − B), we will follow two step method.
In the first step, we will prove A ⊆ (A ∩ B) ∪ (A − B)
In the second step, we will prove (A ∩ B) ∪ (A − B) ⊆ A
We know that if P ⊆ Q and Q ⊆ P ⇒ P = Q.
Proof : Let ∈ A,
⇒ ∈ (A ∩ B) (∵ A ⊆ A ∩ B) . . . (1)
and ∈ (A − B) (∵ A ⊆ (A − B)) . . . (2)
⇒ ∈ (A ∩ B) ∪ (A − B) . . . From (1) & (2).
∵ Whenever ∈ A ⇒ ∈ (A ∩ B) ∪ (A − B)
⇒ A ⊆ (A ∩ B) ∪ (A − B) . . . (3)
Conversely,
let ∈ (A ∩ B) ∪ (A − B)
⇒ ∈ (A ∩ B) or ∈ (A − B)
∵ ∈ (A ∩ B)
⇒ ∈ A and ∈ B
⇒ ∈ A
So,
∵ Whenever ∈ (A ∩ B) ∪ (A − B) ⇒ ∈ A
⇒ (A ∩ B) ∪ (A − B) ⊆ A . . . (4)
From (3) & (4)
A = (A ∩ B) ∪ (A − B)
Hence proved.
Second part
Given: Any two sets A and B.
To prove: A ∪ (B − A) = (A ∪ B)
Proof:
Let ∈ A ∪ (B − A)
⇒ ∈ A or ∈ (B − A)
⇒ ∈ A or ( ∈ B and ∉ A)
⇒ ( ∈ A or ∈ B) and ( ∈ A or ∉ A)
Neglecting ( ∈ A or ∉ A), as it has no affect over the result.
⇒ ( ∈ A or ∈ B)
⇒ ∈ A ∪ B
So,
Whenever ∈ A ∪ (B − A) ⇒ ∈ A ∪ B
⇒ A ∪ (B − A) ⊆ (A ∪ B) . . . (1)
Conversely,
Let ∈ (A ∪ B)
⇒ ∈ A or ∈ B
(∵ belongs to A already, we can remove elements of A from B without affecting the meaning)
⇒ ∈ A or ∈ B − A
⇒ ∈ A ∪ (B − A)
Thus, whenever ∈ (A ∪ B) ⇒ ∈ A ∪ (B − A)
⇒ (A ∪ B) ⊆ A ∪ (B − A) . . . (2)
From (1) and (2), we can write.
A ∪ (B − A) = (A ∪ B)
Hence proved.
7. Using properties of sets, show that
(i) A ∪ (A ∩ B) = A
(ii) A ∩ (A ∪ B) = A.
Answer
(i) A ∪ (A ∩ B) = A
Taking LHS, we have,
A ∪ (A ∩ B)
= (A ∪ A) ∩ (A ∪ B) (∵ Distributive law of union)
= A ∩ (A ∪ B) (∵ Idempotent Law)
= A (∵ The common elements between A and (A ∪ B) are the elements of A)
= RHS
Hence proved.
(ii) A ∩ (A ∪ B) = A
Taking LHS, we have,
A ∩ (A ∪ B)
= (A ∩ A) ∪ (A ∩ B) (∵ Distributive law of intersection)
= A ∪ (A ∩ B) (∵ Idempotent Law)
= A (∵ A ∩ B ⊆ A, the elements of A ∩ B are actually the common elements of A and B, so, it can be equal to A or less than A)
= RHS.
Hence proved.
8. Show that A ∩ B = A ∩ C need not imply B = C
Answer
Let us take an example to show it.
Let A = {1, 2}, B = {1, 2, 3, 4} and C = {1, 2, 5, 6}
Here,
A ∩ B = {1, 2}
A ∩ C = {1, 2}
So, A ∩ B = A ∩ C, but B ≠ C
Thus, (A ∩ B) = (A ∩ C) need not imply B = C.
Hence, proved.
9. Let A and B be sets. If A ∩ X = B ∩ X = ϕ and A ∪ X = B ∪ X for some set X, show that A = B.
Answer
Given:
A ∩ X = B ∩ X = ϕ
A ∪ X = B ∪ X
To prove: A = B
Proof:
Taking LHS, we have,
A = A ∪ ϕ
= A ∪ (A ∩ X) ... (∵ A ∩ X = ϕ)
= (A ∪ A) ∩ (A ∪ X) ... (∵ Distributive law)
= A ∩ (A ∪ X) ... (∵ A ∪ A = A)
= A ∩ (B ∪ X) ... (∵ A ∪ X = B ∪ X)
= (A ∩ B) ∪ (A ∩ X) ... (∵ Distributive law)
= (B ∩ A) ∪ (B ∩ X) ... (∵ A ∩ X = B ∩ X)
= B ∩ (A ∪ X) ... (∵ Distributive law)
= B ∩ (B ∪ X) ... (∵ A ∪ X = B ∪ X)
= (B ∩ B) ∪ (B ∩ X) ... (∵ Distributive law)
= B ∪ ϕ ... (∵ (B ∩ B) = B & (B ∩ X) = ϕ)
= B ... (∵ An empty set has no elements)
Thus, A = B
Hence proved.
10. Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = ϕ.
Answer
The following sets can be chosen as per the given conditions.
A = {1, 2, 3, 4},
B = {3, 4, 5, 6},
C = {1, 2, 5, 6}
Here,
A ∩ B = {3, 4}
B ∩ C = {5,6}
A ∩ C = {1, 2}
But, A ∩ B ∩ C = ϕ
