Telangana SCERT Class 10 Maths Chapter 2 Sets Exercise 2.2 Maths Problems and Solution Here in this Post. Telangana SCERT Class 10 Maths Solution Chapter 2 Sets Exercise 2.2
Telangana SCERT Solution Class X (10) Maths Chapter 2 Sets Exercise 2.2
Exercise – 2.2
(Q1) IF A = {1, 2, 3, 4} and B = {1, 2, 3, 5, 6}
Then find A∩B and B∩A. Are they equal?
=> Solution:
Given: A = {1, 2, 3, 4} and
B = {1, 2, 3, 5, 6}
A∩B = {1, 2, 3, 4} ∩ {1, 2, 3, 5, 6}
= {1, 2, 3} —– (1)
[∴∩= intersection means both the sets of common elements}
B∩ A = {1, 2,3, 5, 6} ∩ {1, 2, 3, 4}
= {1, 2, 3} —— (2)
From equation (1) & (2) A∩B = B∩A
(Q2) IF A = {0, 2, 4}, find A∩∅ and A∩A. What did you notice from the result?
= Solution: A = {0, 2, 4}
A∩∅ = {0, 2, 4} ∩ { }
[∴∅ empty set]
= { }
A∩∅ = ∅ {A & ∅ sets are not common elements}
A∩A = {0, 2, 4} ∩ {0, 2, 4}
= {0, 2, 4}
A∩A = A (from given)
∴ A∩∅ and A∩A = A
(Q3) If A = {2, 4, 6, 8, 10} and B = {3, 6, 9, 12} find A-B and B-A
= Solution:
Given: A = {2, 4, 6, 8, 10} and
B = {3, 6, 9, 12, 15}
To find A – B
A – B = {2, 4, 6, 8, 10} – {3, 6, 9, 12, 15}
First we have to cancle common elements in both sets. A & B
And the write elements present in Set A only. Because we have to find A-B
A-B = {2, 4, 8, 10}
To find B – A
B – A = {3, 6, 9, 12, 15} – {2, 4, 6, 8, 10}
In same way we have to cancle common elements in both
Sets A & B and then write elements
Present in set B only because we have to find B – A
B – A = {3, 9, 12, 15}
We see that A – B ≠ B – A.
(Q4) IF A and B are two sets such that A⊂B then what is A∪B?
Explain by giving an example.
=> solution:
We have to take one example of A⊂B.
Let, A = {1, 2, 3; 4} and
B = {0, 1, 2, 3, 4, 5}
A∪B = {1, 2, 3, 4} ∪ {0, 1, 2, 3, 4, 5}
{Union means we have to take all the elements of A & B}
A∪B = {0, 1, 2, 3, 4, 5}
A∪B = B {∴B = {0, 1, 2, 3, 4, 4, 5}
∴ If A⊂B then A∪B = B
(Q5) Let A = {x : x is a natural number}
B = {x : x is an even natural number}
C = {x : x is an odd natural number}
And D = {x : x is a prime number}
Find A∩B, A∩C, A∩D, B∩C, B∩D and C∩D
=> First we have to write set builder form into roaster form.
∴ The roaster form of the given sets are.
A = {1, 2, 3, 4, 5, 6, 7, 8, —–}
B = {2, 4, 6, 8, 10, —–}
C = {1, 3, 5, 7, 9, 11, —-}
D = [2, 3, 5, 7, 11, ——}
To find A∩B
(i) A∩B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, —-}
∩ {2, 4, 6, 8, 10, —–}
A∩B = {2, 4, 6, 8, 10, —-}
(ii) A∩C = {1, 2, 3, 4, 5, 6, 7, 8, —–} ∩
{1, 3, 5, 7, 9, 11, —–}
A∩C = {1, 3, 5, 7, —–}
(iii) A∩D = {1, 2, 3, 4, 5, 7, 8, —–} ∩
{2, 3, 5, 7, 11, —–}
A∩D = {2, 3, 5, 7, —-}
(iv) B∩C = {2, 4, 6, 8, 10, —-} ∩
{1, 3, 5, 7, 9, 11, —–}
B∩C = { } = ∅
B∩C there is no common elements. So B∩C = ∅.
(v) B∩D = {2, 4, 6, 8, 10 —-} ∩
{2, 3, 5, 7, 11, —-]
B∩D = {2}
(vi) C∩D = {1, 3, 5, 7, 9 ——} ∩
{ 2, 3, 5, 7, 11, —-}
C∩D = {3, 5, 7, —-}
[∩ = intersection means we have to take common elements in both the sets]
(Q5) IF A = {3, 6, 9, 12, 15, 18, 21}
B = {4, 8, 12, 16, 20},
C = {2, 4, 6, 8, 10, 12, 14, 16} and
D = {5, 10, 15, 20},
find
(i) A – B
(ii) A – C
(iii) A – D
(iv) B – A
(v) C – A
(vi) D – A
(vii) B – C
(viii) B – D
(ix) C – B
(x) D – B
=> Solution:
The given sets are –
A = {3,6,9,12,18,21}
B = {4,8,12,16,20}
C = {2,4,6,8,10,12,14,16}
D = {5,10,15,20}
(i) A – B = {3, 6, 9, 12, 15, 18, 21} –
{4, 8, 12, 16, 20}
[We don’t have to take common elements in both the sets. Then after we have to take only elements sets A because we have to find A – B]
A – B = {3, 6, 9, 15, 18, 21}
(ii) A – C = {3, 6, 9, 12, 18, 15, 21} –
{2, 4, 6, 8, 10, 12, 14, 16}
[We have to cancled common elements in both the sets and then after we have to take remaining elements in set A only]
A – C = {3, 6, 9, 18, 15, 21}
(iii) A – D = {3, 6, 9, 12, 18, 15, 21} –
{5, 10, 15, 20}
A – D = [3, 6, 9, 12, 18, 21}
(iv) B – A = {4, 8, 12, 16, 20} – {3, 6, 9, 12, 15, 18, 21}
[We have to cancle common elements in both the sets then after we have take remaining elements in set B only because we have to find B – A]
B – A = {4, 8, 16, 20}
(v) C – A = {2, 4, 6, 8, 10, 12, 14, 16} – {3, 6, 9, 12, 15, 18, 21}
[We have to cancle common elements in both the sets then after we have to take remaining elements in get C only because we have to find C – A]
C – A = {2, 4, 6, 8, 10, 14, 16}
(vi) D – A {5, 10, 15, 20} – {3, 6, 9, 12, 15, 18, 21}
[We have to cancle common elements in both the sets]
D – A = {5, 10, 20}
[Then after we have take remaining elements in set D only because we have to find D – A]
(vii) B – C = {4, 8, 12, 16, 20} – {2, 4, 6, 8, 10, 12, 14, 16}
B – C = {20}
(viii) B – D = {4, 8, 12, 16, 20} – {5, 10, 15, 20}
B – D = {4, 8, 12, 16}
(ix) C – B = {2, 4, 6, 8, 10, 12, 14, 16}
{4, 8, 12, 16, 20}
C – B = {2, 6, 10, 14}
[We have to take remaining elements in set ‘C’ only]
(x) D – B = {5, 10, 15, 20} – {4, 8, 12, 16, 20}
D – B = {5, 10, 15}
(Q7) State whether each of the following statements is true or false, justify your answers.
(i) {2, 3, 4, 5} and {3, 6} are disjoint sets.
=> Solution:
[The sets which have no common elements are called disjoint sets”]
[If there intersection is null get or empty set we can say are disjoint set]
False
Since {2, 3, 4, 5} ∩ {3, 6} = {3} ≠ ∅
[We can see here common element is 3 so these are not disjoint sets]
(ii) {a, e, I, o, u} and {a, b, c, d} are disjoint sets.
=> False
Since, {a, e, I, o, u} ∩ {a, b, c, d}
= {a} ≠ ∅
[These are not disjoint sets because of common element is ‘a’]
(iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets
=> True
Since, {2, 6, 10, 14} ∩ {3, 7, 11, 15}
= { } = ∅
There is no common elements so these are disjoint sets.
(iv) {2, 6, 10} and {3, 7, 11} are disjoint sets.
=> True.
Since, {2, 6, 10} ∩ {3, 7, 11}
= { } = ∅
These is no common elements so these are disjoint sets.