FREE Class 9 Advanced Maths Solution pdf of Chapter 2 SETS
Get FREE Class 9 Advanced Maths Solution pdf of Chapter 2 SETS. Here, we will discuss Seba class 9 advanced mathematics solutions which are explained by well-trained faculty. So dear learners get your solutions which is very helpful for your homework.
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If yes, then this article is very helpful for you, here you will get complete solutions of Class 9 Advanced Mathematics Solutions of Chapter 2 SETS. This chapter is very interesting and also useful for the higher classes if you will go Science Stream.
Basic Introduction of SETS.....
Hope you will enjoy the learning...
Also Check: CLASS 10 GENERAL MATHS IMPORTANT QUESTIONS PDF
Brief Introductions:
In real cases, very often we discuss groups or collections and work on their ideas. Some of such collection has been mentioned here are :
- Planets of the Solar System,
- Names of days in a week in Assamese etc.
A collection or a group is called a set if we are certain about that which elements belong and which
elements do not belong to that group.
You can also Check: SEBA Class 9 Advanced Mathematics Solutions pdf of all the chapter
In other words, elements belonging to the set should be well-defined.
A set is denoted by the symbol { } with the elements. For example, the collections of Planets of the Solar System may be denoted as a set
by {Venus, Mars, Earth, Mercury, Jupiter…….}.
A set has no definition. Generally, a set is understood to be
a well-defined collection of elements. Each element is called members.
A set having no element at all is called a null or an empty
set.
A null set is denoted by the Greek letter Ф(phi).
Set can be represented in two ways, such as:
- Tabular or Roster Method: Here elements are displayed physically.
- Rule( or Ser-builder) Method: Here a representative variable element is used instead of the actual elements.
Subsets:
A set A is a subset of the set B (A ⊆ B ) if and only if every element of A is in B.
This means, A ⊆ B iff x ∊ A⇒ x ∊ B
If A⊆B, then B is called the superset of A.
If A ⊆B, but B ≠ A, then A is called a Proper subset of B.
It is written as A ⊂ B ( Read A is included in B)
According to the definitions of a proper subset that we gave above, is a proper subset of any set.
Free Advanced Maths SEBA Solutions for Class 9 Chapter 2 Ex 2.2
Advanced Mathematics Solutions of Class 9 SEBA/HSLC
Chapter 2
SETS
EXERCISE-2.2
Solutions of Advanced Maths Free
1. Express the following sets both in Roster and Rule form:
a) Set of even prime numbers.
Solutions: Let,
A=Set of even prime numbers
Tabular or Roster Method: A={2}
Rule( or Ser-builder) Method: A= { x : x is an even prime number; x> 0 }
b) Set of odd numbers lying between 4 and 20.
Solutions: Let,
B = Set of odd numbers lying between 4 and 20.
Tabular or Roster Method: B={5, 7, 9, 11, 13, 15, 17, 19}
Rule( or Ser-builder) Method: B= { x : x is an odd number; 4 < x > 20 }
c) Set of the positive real roots of the equations
x3 – 2x2 – x + 2 = 0
Solutions: Let,
C = Set of the positive real roots of the equations
x3 – 2x2 – x + 2 = 0
Tabular or Roster Method: C={1,2}
Rule( or Ser-builder) Method: C = { x : x is a positive real roots of the equations
x3 – 2x2 – x + 2 = 0 }
d) Set of all multiples of 5, which are natural numbers.
Solutions: Let,
D = Set of all multiples of 5, which are natural numbers
Tabular or Roster Method: D ={ 5, 10,15, 20, 25, 30,.....}
Rule( or Ser-builder) Method: D = { x : x is a natural number divisible by 5; x> 0 }
e) Set of all integers whose square is less than 64.
Solutions: Let,
E = Set of all integers whose square is less than 64
Tabular or Roster Method: E={0, 土 1,土2,土3,土4,土5,土6,土7}
Rule( or Ser-builder) Method: E= { x : x is an integers ; x2 < 64 }
2.Write true or false.
a) If A = {a,b,c,d,e} then -
True/False
| Q. No. | a) If A = {a,b,c,d,e} then - | True/False |
|---|---|---|
| i | {a}∉ A | True |
| ii | {a,b}∈A | False |
| iii | {c,d,e}⊂A | True |
| iv | ϕϵA | False |
| v | ϕ⊂A | True |
| vi | AϵA | False |
| vii | {e}⊂A | True |
| viii | A={A} | False |
| ix | {0,a}⊂A | False |
| x | {0,a}={a} | False |
| xi | {b,c,d}={c,d,b} | True |
| xii | ∅⊂{a} | True |
| xiii | {a,b,c}={{a},{d},{c}} | False |
| xiv | φ⊂A | True |
| xv | φ⊂{φ} | True |
3. If A ⊂ B, B ⊂ C, show that A ⊂ C.
Solutions: Given,
A ⊂ B
B ⊂ C
Now, A ⊂ C can be shown with the following Venn diagram.
A ⊂ C
4. If A = {a,b,c,d }, what is P(A) and n(P(A)) ?
Solutions: Given,
A = { a,b,c,d }
Therefore, the power set of A will be
P(A) = {{a},{b},{c},{d},{a,b},{a,c},{a,d},{b,c},{b,d},{c,d},{a,b,c},{a,b,d},{a,c,d},{b,c,d},{a,b,c,d},ф}
The number of element of A= 4,
therefore, the number of elements of n (P(A) = 24 =16
5. Write all the subsets of the sets {a} and ф.
Solutions: a) All the subsets of set {a}= {{a},ф}
b)All the subsets of set ф= {ф,{}}
6. Represent the sets in the same Venn diagram:
U = { 1,2,3,....,8}
A = {1},
B = {1, 4,7} and
C = {2,4,5,8}
Solutions: Here,
U = { 1,2,3,....,8}
A = {1},
B = {1, 4,7} and
C = {2,4,5,8}
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Free Advanced Maths SEBA Solutions for Class 9 Chapter 2 Ex 2.3
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