solutions for CHSE Odisha Plus 2 1st Year Elements of Mathematics Chapter 2 Sets. Chapter exercise answers, formulas, and PDF guide for Class 11 Math.
Overview of CHSE Odisha Class 11 Mathematics Chapter 2: Sets
Key Concepts & Formulas Summary
1. Set:
- 'Set is an undefined term in mathematics. But we understand set as "a collection of well defined objects".
- Set is a collection.
- The objects (called elements) in a set must be well defined.
2. Set notesion :
- We denote sets as capital alphabets like A,B,C,D....and the elements by the small alaphabets like x,y,z
- If x is an element of set A we say "x belongs to A" and write x ∈ A'.
- If x is not an element of A we say "x does not belong to A" and we write "x ∉ A".
3. Set Representation:
(a) Extension or tabular or Roster Method:
In this method we describe a set by listing the elements, separated by commas within curly brackets.
Note: While listing out the elements the repetition of objects have no effect. Thus, we don't do this.
(b) Intention or set builder or set selector method:
In this method a set is described by a characterising property p(x) of element x. In this case the set is described as {x: p(x) holds}
4. Types of set :
(a) Empty of null or void set:
It is a set with no element.
• We denote empty set by 'Ï•'
There is only one empty set.
(b) Singleton set :
It is a set with only one element.
(c) Finite set:
A set is finite if it has finite number of elements.
(d) Infinite Set:
A set which is not finite is called an infinite set.
(e) Equal sets:
Two sets A and B are equal if they have same elements.
• Two sets A and B are equal if all elements of A are also elements of B and all elements of B are also elements of A.
(f) Equivalent set:
Two finite sets A and B are equivalent if they have same number of elements.
(g) Sub sets:
Let A and B are two sets.
If every elements of A is an element of B then A is called a subset of B (we write A ⊂ B) and B is called a super set of A (We write B⊃A)
Thus A⊂ B is x ∈ A⇒x∈ B
Note.
i) A set is a subset of itself.
ii)Empty set 'Ï•' is a subset of every set.
iii) A is called a proper subset of B if B contains at least one element that is not in A.
(iv) If A has n elements then total number of subsets of A = 2".
6. Universal set:
A set 'U' that contains all sets in a given context is called the universal set.
7. Power set:
Let A is any set. The collection (or set) of all subsets of A is called the power
set of A. We denote it as P (A)
P(A)={S: S⊂A}
8. Set operations:
(a) Union of sets :
The union of two sets A and B is the set of all elements of A or B or both.
AUB = {x ∈ A or x ∈ B}
(b) Intersection of sets:
Intersection of two sets A and B is the set of all those elements that belong to both A and B. (or all common elements of A and B)
AnB = {x: x ∈ A and x ∈ B}
Two sets A and B are disjoint if
A n B = Ï• . Otherwise A and B intersecting or overlaping sets.
(c) Difference of sets:
The difference of two sets 'A and B' is the set of all elements of A which do not belong to B.
A-B={x: x ∈ A and x ∉ B)
(d) Symmetric difference of two sets:
Symmetric difference of two sets A and B is the set
(A-B) U (BA)
A Δ B
=(A-B) U (B-A)
= (A U B) - (A n B)
(e) Complement of a set:
Let complement of a set A (denoted as A' or AC) is defined as U - A
• A'= {x ∈ U: x ∈ A}
X ∈ A' ⟺ X∈ A
9. Laws of set Algebra :
(a) Idempotent law :
For any set A we have
(i) AUA=A
(ii) A n A = A
(b) Identity laws :
For any set A we have
(i) A U Ï• = A and
(ii) A n U = A
(c) Commutative laws:
For any three sets A, B and C
(i) A U B=B U A
(ii) A n B=B n A
Elements of Mathematics Chapter 2 Sets Exercise 2(a)
Question 1.
Construct five different examples of sets. Describe each with the help of a proposition.
Solution:
(i) Collection of all the days of a week.
i.e., D = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
= {x: x is a day of a week}
(ii) Collection of writing instruments
i.e., A = {pen, pencil, paper, ink}
= {x: x is a writing instrument}
(iii) Collection of all kings having more than one queen.
i.e., B = {x: x is a king having more than one queen.}
(iv) Collection of all nationalized political parties i.e., C = {BJP, Congress, CPI, CPI(M), JD(U), JD(S), JP}
= {x: x is a set of nationalized political parties}
(v) Collection of all integers of multiples of 3.
i.e., D = {0, ± 3, ± 6}
= {x: x is a set of all integers of multiples of 3}
(vi) Collection of all fingers of a hand
i.e., E = {x: x is a finger of hand}
Question 2.
Give an example of a set that has exactly 10 elements and express it through a defining property.
Solution:
Collection of all positive prime numbers less than 30 i.e„ A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} = {x: x is a prime number less than 30}
Question 3.
It is possible to express every set through a defining property? Justify your answer.
Solution:
Yes, it is possible to express every set through a defining property, as every set can be written in the set builder form
Question 4.
If {x: p1 (x)} = {x: p2 (x)}, show for each x, p1 (x) and p2 (x) have the same truth value.
Solution:
Given that {x : p1 (x)} = {x : p2 (x)}
∴ The two sets have the same element.
So p1 (x) ≠ p2 (x) i.e., they have the same truth value.
Question 5.
For each of the following words, write down the set of letters forming that word :
(i) Administration,
(ii) Misrepresentation,
(iii) Mathematics,
(iv) Concurrence,
(v) Demonstration
Solution:
(i) {a, d, i, m, n, o, r, s, t }
(ii) {a, e, i, m, n,o,p, r, s, t }
(iii) {a, c, e, h, i, m, s, t }
(iv) {c, e, n, n, o, r, u }
(v) {a, d, e, i, m, n, o, r, s, t
Question 6.
State with reason, which of the following are sets and which are not :
(i) All big rivers of India.
(ii) All natural numbers having at least one prime factor.
(iii) All sincere students of Ravenshaw college during the academic year 1998 -99.
(iv) All real numbers with negative squares.
(v) All citizens of India earning more than Rs. 10,000/- per month.
(vi) All college teachers who are citizens of India.
(vii) All finite subsets of the set Z of integers.
(viii)Collection of all possible sets.
(ix) Collection of all winged horses.
(x) Collection of all residents of Odisha who will live for more than 100 years.
Solution:
(i) It is not a set, as the word ‘big’ is not properly defined.
(ii) It is a set, as it is properly defined.
(iii) It is not a set, as the word ‘sincere’ is not properly defined.
(iv) It is a set, as it is well-defined.
(v) It is a set, as it is well-defined.
(vi) It is a set, as it is well-defined.
(vii) It is a set, as it is well-defined.
(viii)It is a set, as it is well-defined.
(ix) It is a set, as is it properly defined.
(x) It is not a set, as we do not know, who will live for more than 100 years.
Question 7.
Write the following sets in the form of lists:
(i) {x: x is a prime number and 1 ≤ x ≤ 100}
(ii) {x: x = is an odd integer}
(iii) {x: x = 1 or x = 2 or x = 3 }
(iv) {x: x = can be written as a sum of two odd integers}
(v) Set of all natural numbers that are divisible by 5.
Solution:
(i) {2, 3, 5, 7, …….. 97}
(ii) {± 1, ± 3, ± 5, ……..}
(iii) {1, 2, 3}
(iv) {2, 4, 6, 8, 10, ………}
(v) {5, 10, 15, 20, ……..
Question 8.
Write the following sets in the intention (or specification form)
(i) {a}
(ii) Φ
(iii) {1, 2}
(iv) {1, 2, 3, 4, 5}
(v) P (Φ)
(vi) {1, 3, 9, 27}
Solution:
(i) {x: x = a}
(ii) {x: x ≠ x}
(iii) {x: x = 1 or x = 2}
(iv) {x: x ∈ N, 1 ≤ x ≤ 5}
(v) {x: x = Φ}
(vi) {x: x =3n n ∈ Z, 0 ≤ n ≤ 3 }
Question 9.
Determine if set A is a proper subset of the set B where A and B are as given below:
(i) A = { 1, 2, 3, …….}
B = {x: x is a rational number}
Solution:
A is a proper subset of B as all the elements of A are in B.
(ii) A = {x: x is a prime number}
B = {2n – 1: n = 1, 2, 3, …… }
Solution:
A = {x : x is a prime number}
= {2, 3, 5, 7, 11, ……}
B = {2n – 1: n = 1, 2, 3, …. }
= {1, 3, 5, 7, ….}
∴ A ⊄ B, because of 2 ∈ A but 2 ∉.
(iii) A = {-l, 1, 3}
B = {x: x ∈ R and x3 – 2x3 -x + 2 = 0}
Solution:
Solving x3 – 2x2 – x + 2 = 0
we have x2 (x – 2) – 1 (x – 2) = 0
or, (x – 2) (x2– 1) = 0
∴ x = 1, -1, 2
∴ B ={ -1, 1, 2}
∴ A is not a subset of B.
(iv) A= {1, 2, 3, 4}
B = [n ∈ N, n is a divisor of 60}
Solution:
B = {1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60}
∴ A ⊂ B
Question 10.
For each of the following pairs of sets A and B, determine if A ⊂ B or A ⊄ B:
Solution:
(i) A = Φ, B = {Φ}
Here A ⊂ B, as Φ is the subset of every set
(ii) A = {x : x is an integer}
= 0, ± 1, ± 2……..
B = {3x : x is an integer}
= {0, ± 3, ± 6,……} A ⊄ B.
(iii) A = {x: x is an odd integer}
B = {x: x is real and not an even integer}
∴ A = {± 1, ± 3, ± 5, …….} and B is the set containing all the real numbers except for even integers.
∴ A ⊂ B
(iv) A = {x : x is an integer which is both even and odd}
B = {x: x is an integer, x ≠ x}
∴ A ⊂ B, … A = Φ and B = Φ)
(v) A = |a, b, c|, B = {|a|, |b|, |c|}
∴ A ⊄ B
Question 11.
Determine the truth of the falsity of the following propositions with reasons.
(a) {1, 2} € {1,2,3}
(b) A ⊂ A for any set A.
(c) Every set has a proper subset.
(d) Every set in a proper subset of the same set
(e) For any object x, there is a set A such that, x ∈ A.
(f) For any object x, there is a set A such that, x ∉ A.
(g) If A, B, C sets, then either A = B or A ⊂ B or B ⊂ A.
(h) a ∈ {(a)}
(i) a ∈ {(a, b), b}, a ≠ b.
(j) If A is a proper subset of B and B is a subset of C, then A is a proper subset of C.
(k) A ⊂ Φ if and only if A = Φ.
Solution:
(a) {1,2} ∈ {1,2,3} is false, as {1, 2} is not an element of { 1, 2, 3} but it is a subset.
(B) A ⊂ A is false, as A is an improper subset of itself, not a proper subset.
(C) ‘Every set has a proper subset is false as Φ has no proper subset.
(d) Every set is a proper subset of the same set is false.
(e) For any object x, there is a set A such that, x ∈ A. It is true.
(f) For every object x, there is a set A such that, x ∉ A. It is true.
(g) If A, B, C sets, then either A = B or A ⊂ B or B ⊂ A. It is false, as there is a possibility for A ≠ B.
e.g. a = {1, 2}, b = {a, b}
(h) a ∈ {{a}}. It is false as ‘a’ is not an element of the set {(a)}.
(i) as ∈ {{a, b}, b}, a ≠ b It is false, as ‘a’ is not an element of the set {{a, b}, b}
(j) If A is a proper subset of B and B is a subset of C, then A is a proper subset of C.
It is true, as
x ∈ A ⇒ x ∈ B ∴ (A ⊂ B)
and x ∈ B ⇒ x ∈ C ∴ (B ⊂ A)
∴ x ∈ A ⇒ x ∈ C means A ⊂ C.
(k) A ⊂ Φ if and only if A = Φ is true.
Proof:
Necessary part: Given A ⊂ Φ and prove A = Φ.
Now A ⊂ Φ ….(1)
And Φ is the subset of every set.
So Φ ⊂ A ….(2)
Combining (1) and (2) we get A = Φ.
Sufficient part: Given A = Φ and to prove A ⊂ C.
A = Φ ⇒ A ⊂ Φ and Φ ⊂ A
Thus A ⊂ Φ
Question 12.
Write down the power set of
(i) {a, b, c}
(ii) {a, {a}}
(iii) Φ
(iv) {Φ}
(v) {a, {a}, {a,b}}
(vi) {{Φ}}
Solution:
(i) Let A = {a, b, c}
P(A) = {{a}, {b}, {c}, {a, b}, {a, c}, {b, c}, A, Φ}
(ii) Let A = {a, {a}} then P(A) = {{a}, {{a}}, Φ, A}
(iii) P (Φ) = {Φ}
(iv) P({Φ})-= {{Φ}, Φ}
(v) Let A = {a, {a}, {a, b}}
∴ P(A) = {{a}, {{a}}, {{a, b}}, {a}, {a}
{{a}, {a,b}}, {a, {a,b}, A, Φ)}
(vi) Let A = {{Φ}}
∴ P(A) = {A, Φ}
Question 13.
Prove that P(A) ⊂ P(B) if and only if A ⊂ B. When is the inclusion P(A) ⊂ P(B) proper?
Solution:
Necessary Part :
Let P(A) ⊂ P(B),
we shall prove that A ⊂ B.
∴ Let x ∈ A
⇒ {x} ∈ P(A)
⇒ {x} ∈ P(B) (∴ P(A) ⊂P(B)
⇒ x ∈ B
Sufficient Part:
Let A ⊂ B
Let x ∈ P(A)
⇒ x ⊂ A
⇒x ⊂ B (∴ A ⊂ B)
⇒ x ∈ P(B)
Thus P(A) ⊂ P(B)
The inclusion P(A) ⊂ P(B) is proper when A is a proper subset of B.
Question 14.
A set can be finite or infinite (as understood in an informal way). For instance, {1, 2, 3, 4} is a finite set whereas Z is an infinite set. The number of elements of a set A, denoted by |A|, is called its cardinal number. Without going into the necessary technicalities, we may just observe that
|Φ| = 0, |{x1, x2………,xn }| = n
Two sets A and B are called similar if they have the same cardinal number. Thus, the sets {1, 2, 3} and {2, 4, 6} are similar. We write A ~ B to express the fact that A and B are similar. Now, answer the following questions.
(i) What are the cardinal numbers of the following sets?
{Φ}, {a, {a, b}}, {Z} {0.5}, {0, {5}}, {a, b, {a, b} }, {{Φ}}, {Φ, {Φ}}
(ii) For any natural number n, give an example of a set A such that |A = n|.
(iii) Determine the cardinal number of the set {x: x is real and x3 – x2 + x- 1 = 0 }
Solution:
(i) Cardinal number of the set {Φ} is 1.
Cardinal number of the set {a, {a, b} } is 2.
Cardinal number of the set {z} is 1.
Cardinal number of the set {0,5} is 2.
Cardinal number of {0, {5 } } is 2.
Cardinal number of the set {a, b {a, b}) is 3.
Cardinal number of the set {{Φ}} is 1.
Cardinal number of {Φ, {Φ}} is 2
(ii) Let A = {1, 2, 3}
∴ | A | = 4 where 4 ∈ N for n = 4.
(iii) Let A = {x : x is real and
x3 – x2 + x – 1 = 0 }
Solving, x3 – x2 + x – 1 = 0
or, x2 ( x – 1) + 1 (x – 1 ) = 0
or, (x- 1) (x2 + 1) = 0
x – 1 = 0, x2 + 1 = 0
x = 1, ± i
As x is real, we have A = {1}
| A | = 1
Question 15.
Which sets are finite and which are infinite?
(i) The set N of positive natural numbers.
(ii) The set Z of integers.
(iii) The set Q of rational numbers.
(iv) The set R of real numbers.
(v) The set of prime numbers.
(vi) The set of even integers.
(vii) The set of human beings.
(viii)The set of integers less than 10.
Solution:
(i) “The set N of positive natural numbers” is an infinite set.
(ii) “The set Z of integers” is an infinite set.
(iii) “The set Q of rational numbers” is an infinite set.
(iv) “The set R of real numbers” is an infinite set.
(v) “The set of prime numbers” is an infinite set.
(vi) “The set of even integers” is an infinite set.
(vii) “The set of human beings” is a finite set.
(viii)“The set of integers less than 10” is an infinite set.
Question 16.
Verify that
| P (Φ) | = 20
| P ({a}) | = 21
| P ({a, b}) | = 22
| P({a, b, c}) | = 23
Solution:
(i) Let A = Φ, then P(A) = {Φ}
∴ P(A) = 1 = 20
(ii) Let A = {a} then P(A) = {{a}, Φ}
∴ P(A) = 2 = 21
(iii) Let A = {a, b)
then P(A) = {{a}, {b}, {a, b}, Φ}
∴ P(A) = 4 = 22
(iv) Let A = {a, b, c}
P(A) = {{a}, {b}, {c}, {a, b}, {b, c}, A, Φ}
∴ P(A) = 8 = 23
Question 17.
Find the number of elements of
(i) P(P(Φ))
(ii) P(P(P(Φ)))
(iii) P(P(P(P(Φ))))
Solution:
(i) We have | P (Φ) | = 20 = 1
∴ | P(P(Φ)) | = 21 = 2
(ii) | P(P(P(Φ))) | = 22 = 4
(iii) | P(P(P(P(Φ))) | =24 = 16
Question 18.
Prove by the method of induction that if A has n number of elements, then | P (A) | = 2n.
Solution:
When n = 1, we have, let A = {a1}
P (A) = {{ a1 }, Φ} ⇒ | P(A) | = 2 = 21
∴ p1 is true.
Let pk be true i.e., if A = {a1, a2, ……….. ak }
then | A | = k and | P(A) | = 2k
Now let B = {a1, a2, ……….. ak, ak+1 }
∴ | B | =k + 1.
Here A has 2k subsets. These 2k subsets are also in B. Which includes the additional elements ak+1 ∈ B.
So there are 2k. 2 = 2k+1 subsets in B.
∴ | P(B) | = 2k+1
i.e., if | A| = k +1 then | P(A) | = 2k+1
∴ pk+1 is true.
∴ pn is true for all values of n ∈ N.
∴ | P(A) | =2n if | A | = n.
Question 19.
Can you say how many elements P(P(A)) if A has n elements?
Solution:
If | A | = n then | p(A) | =2n
and | P(P(A)) | = 2(2n) .
