Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 2 Sets Ex 2(a) Textbook Exercise Questions and Answers.
CHSE Odisha Class 11 Math Solutions 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) .