Practicing OP Malhotra Class 11 Solutions Chapter 1 Sets Ex 1(a) is the ultimate need for students who intend to score good marks in examinations.
S Chand Class 11 ICSE Maths Solutions Chapter 1 Sets Ex 1(a)
Question 1.
Which of the collections are sets?
(i) The collection of all months of a year, beginning with letter J
(ii) The collection of most talented writers of India.
(iii) The collection of all natural numbers less than 100.
(iv) A collection of most dangerous animals of the world.
Solution:
(i) Clearly it is a well defined collection of objects so it represents a set.
(ii) As it is not a well defined collection of objects as the word the most talented is vague. As according to some, writers are talented and according to some people, those WTiters are not talented. Hence given collection does not represents a set.
(iii) The collection of all natural numbers less than 100 ={1,2, 3,…, 99} clearly it is a well defined collection so it represents a set.
(iv) given collection is not a well defined collection since people may have different opinions.
Hence given collection does not represents a set.
Question 2.
Let A = {1, 2, 3, 4, 5, 6}. Insert the appropriate symbol ∈ and ∉ in the blank spaces.
(i) 5 …. A,
(ii) 8 …. A,
(iii) 0 …. A
(iv) 4 …. A
Solution:
Given A = {1, 2, 3, 4, 5, 6}
(i) Since 5 is a member of set A ∴ 5 ∈ A
(ii) 8 is not a member of set A ∴ 8 ∉ A
(iii) 0 is not an element of set A ∴ 0 ∉ A
(iv) 4 is an element of set A ∴ 4 ∈ A
Question 3.
Write down a description of each of the following sets. (There could be different suitable descriptions.)
(i) {2, 4, 6, 8}
(ii) {7, 14, 21, 28, 35}
(iii) {1, 2, 3, 4, 6, 12}
Solution:
(i) {2, 4, 6, 8} = set of all even numbers between 1 and 9
(ii) {7, 14, 21, 28, 35} = set of all multiples of 7 between 1 and 36.
(iii) {1, 2, 3, 4, 6, 12}= set of all factors of 12
Question 4.
List the following sets in roster form.
(i) The set of square numbers less than 40.
(ii) The set of colours of the rainbow.
(iii) (a) The set of factors of 144.
(b) The set of prime factors of 144.
(iv) The set of natural numbers less than 50.
(v) The set of consonants before i in the English alphabet.
(vi) The set of letters in the word ‘Satellite’.
Solution:
(i) {1, 4, 9, 6, 16, 25, 36}
(ii) Indigo, Violet, Brown, Green, Yellow, Orange, Red}
(iii) (a) 144 = 24 x 3²
∴ factors of 144 are
1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 72, 144
∴ Required set in Roster form = {1,2, 3,4, 6, 8, 9, 12, 16, 18, 24, 36, 72, 144}
(b) {2, 3}
(iv) (1, 2, 3, 4, 5, …….., 49}
(v) {b, c, d, f g, h)
(vi) {s, a, t, e, l, i}
Question 5.
Rewrite the following sets in the indicated notation.
(i) {- 2, – 4, – 6, – 8} ;
(a) Words
(b) Set-builder notation
(ii) Positive multiple of 11 ; Roster form
(iii) {- 9, – 7, – 5, – 3, – 11} ; Set-builder notation
(iv) Even numbers between 27 and 39
(a) Roster form
(b) Set-builder notation
(v) {x: | 0 < x < 1} ;
(a) Words
(b) Roster form
(vi)
(vii) Negative multiples of 3, Set-builder form
(viii)
(ix) Numbers more than 2 units from 8, Set-builder form.
(x) x ≠ 5 and x ≤ 10 ; Set-builder form
(xi) {y | y = 5x – 2, y ∈ N} ; Roster form
(xii)
Solution:
(i) (a) set of all even numbers between – 9 and 0.
(b) {x : x < 0 and x is even}
(ii) {11, 22, 33, 44, }
(iii) {x: – 9 ≤ x ≤ – 1, x is odd}
(iv) (a) {28, 30, 32, 34, 36, 38}
(b) {x : 27 ≤ x ≤ 39 ; x is even}
(v) (a) number greater than 0 but less than 1
(b) since there are infinite numbers between 0 and 1 so the given set can’t be written in Roster form.
(vi) Since there are dark dots at x = – 3, 1, 5 on real line
∴ In roster form, solution be {1, – 3, 5}
(vii) {x : x = – 3n and n ∈ N}
(viii) Since there is only one dark dot at x = 2 on real line.
∴ Solution in set builder form is given by { x | x = 2}
(ix) {x : x = 8 + 2}
(x) {x : x < 5 or 5 < x ≤ 10}
(xi) Given y = 5x – 2, x ∈ N
When x = 1 ⇒ y = 3 ∈ N
When x = 2 ⇒ y = 10 – 2 = 8 ∈ N
When x = 3 ⇒ y = 15 – 2 = 13 ∈ N
and so on
Thus in roster form, given set
= {3, 8, 13, ….}
(xii) Given p ∈ W and p ≤ 5
∴ p = {0, 1,2, 3, 4, 5}
Also, x = \(\frac{3 p+1}{2 p-1}\)
When p = 0 ; x = \(\frac { 1 }{ -1 }\) = – 1
When p = 1 ; x = \(\frac { 4 }{ 2-1 }\)= 4
When p = 2 ; x = \(\frac { 7 }{ 3 }\)
When p = 3; x = \(\frac { 10 }{ 5 }\) = 2
When p = 4 ; x = \(\frac { 13 }{ 7 }\)
When p = 5 ; x = \(\frac { 16 }{ 9 }\)
In roster form,
given set = \(\left\{-1,4, \frac{7}{3}, 2, \frac{13}{7}, \frac{16}{9}\right\}\)
Question 6.
State whether each of the following sets is finite or infinite :
(i) The set of lines which are parallel to the x-axis
(ii) The set of letters in the English alphabet.
(iii) The set of number which are multiple of 5.
(iv) The set of animals living on earth.
(v) The set of circles through the origin
(vi) The set of whole numbers greater than 5.
(vii) The set of natural numbers less than one billion.
(viii) The set of integers between – 4 and 4.
(ix) The set of rational numbers between 0 and 1.
Solution:
(i) Since there are number of lines which are parallel to x-axis and this counting never comes to end, hence given set is an infinite set.
(ii) Since there are 26 English alphabets and hence the given set is a finite set.
(iii) {5, 10, 15, }
Here counting of multiples of 5 cannot comes to end and hence given set represents an infinite set.
(iv) Since the counting of animals living on earth comes to an end and hence given set is a finite set.
(v) Since there are infinite no. of circles that pass through the origin (0, 0). Hence given set represents an infinite set.
(vi) {6, 7, 8, 9, 10, }
Since the counting of whole numbers which are greater than 5 never comes to end.
Hence given set represents an infinite set.
(vii) Since the natural numbers which are less than one billion are finite and hence given set represents a finite set.
(viii) {- 3, – 2, – 1, 0, 1, 2, 3} since there are 7 integers between – 4 and 4.
Thus given set represents a finite set.
(ix) Since there are infinite number of rational numbers between 0 and 1. Hence given set represents an infinite set.
Question 7.
Which of the following sets are empty sets?
(i) A = (x : x is a human being living on Mars}
(ii) B = {x: x is an odd number divisible by 2}
(iii) C = {x : x is a point common to any two parallel lines}
(iv) D = {0}
(v) E = {x : x is a natural number, x < 5 and simultaneously x > 7}
Solution:
(i) Since there are no human being living on Mars and hence set A contains no element. Thus given set A be an empty set.
(ii) Since there is no odd number which is divisible by 2. given set has no element. ∴ set B be an empty set.
(iii) When two lines are parallel so they can’t be intersect each other and hence there is no common point between the parallel lines and hence given set C has no element and hence set C be an empty set.
(iv) Given D = {0}. Thus set D contains one element namely ‘0′. ∴ set D contains one element and hence given set be a non-empty set.
(v) Since there is no natural number which is less than 5 but greater than 7. ∴ set E contains no element and hence set E be an empty set.
Question 8.
Are the following sets equal? Give reasons.
(i) A = {2, 3} ; B = (x: x is a solution of x² + 5x + 6 = 0}
(ii) A = (x : x is a letter in the word FOLLOW}
B = (y’ y is a letter in the word WOLF}
Solution:
(i) Given A = (2, 3} ∴ set A contains two elements 2 and 3
Now x² + 5x + 6 = 0
⇒ (x + 2) (x + 3) = 0
⇒ x = – 2, – 3
∴ B = {- 2, – 3} set B contains two elements – 2 and – 3
Since every element of A is not an element of B. Thus A and B are not equal sets.
(ii) A = {F, O, L, W}
B = {W, O, L, F}
Since A and B have same elements and hence both sets are equal sets.
Question 9.
Which of the following are singleton sets?
(i) A = (x; | x | = 5, x ∈ N}
(ii) A = (Planets of our solar system}
(iii) C = (x : x³ = – 125, x ∈ Z}
Solution:
(i) Given A = {x : | x | = 5, x ∈ N}
since |x| = 5 ⇒ x = ±5, x∈N } ∴ x = 5
Thus, set A contains only one element ‘5’
∴ set A is a singleton set.
(ii) Given B = (x: x² – 11x + 24 = 0; x ∈ N}
Since x² – 11x + 24 = 0
x² – 3x – 8x + 24 = 0
⇒ x (x – 3) – 8 (x – 3) = 0
⇒ (x – 3) (x – 8) = 0
⇒ x = 3, 8 ∈ N
∴ In roster form, B = {3, 8}
So set B contains two elements and hence set B is not a singleton set.
(iii) C = {x : x³ = – 125 ; x ∈ Z}
since x³ = – 125 = (- 5)³
⇒ x = – 5 ∈ Z
Thus set C contains only one element and hence set C is a singleton set.
Question 10.
State the value of n (A) for each of the following sets.
(i) A = (Months of the year}
(ii) A = (Planets of our solar system}
(iii) A = {x: x is an integer and – 8 ≤ x ≤ 3}
(iv) A = (x : x is an even number}
Solution:
(i) since there are 12 months in a year
∴ n (A) = no. of elements in set A = 12
(ii) Since there are 8 planets in our solar system.
∴ n (A) = 8
(iii) A = {- 8, – 7, – 6, – 5, – 4, – 3, – 2, – 1, 0, 1, 2, 3}
which are 12 in total.
∴ n(A) = 12
(iv) A = {…… -4,-2, 0, 2, 4, -}
Since counting of even numbers is never comes to an end and hence A be an infinite set.
∴ cardinal number i.e. n (A) cannot be determined.
Use interval notation to represent each set of numbers.
Question 11.
(i) – 17
(ii) 6 ≤ x ≤ 12
(iii) -1 < x ≤ 4
(iv) – 4 ≤ x < 7
(v) x ≤ 3 or 5 < x ≤ 9
(vi) {x | x ≥ 99}
(vii) {x | x ≠ 1}
(viii) (1, 3, 5, 7, …}
(ix) x ≠ 3
Solution:
(i) – 17 < x < 0 = (- 17, 0)
since – 17 and 0 both are not included.
(ii) 6 ≤ x ≤ 12 = [6, 12], since both numbers 6 and 12 are included.
(iii) – 1 < x ≤ 4 = (- 1, 4] since 4 is included and – 1 is not included.
(iv) – 4 ≤ x < 7 = [- 4, 7) Here – 4 is included but 7 is excluded.
(v) x ≤ 3 or 5 < x ≤ 9
⇒ x ∈ (- ∞, 3] ∪ (5, 9]
(vi) {x | x ≥ 99} ⇒ x ∈ [99, ∞)
(vii) {x|x ≠ 1} = {x|x < 1 or x > 1}
= (- ∞, 1) ∪ (1, ∞)
(viii) {1, 3,5, 7,…} it can’t be expressible in interval potation.
(ix) {x | x ≠ 3} = {x | x < 3 or x > 3}
= (- ∞, 3) ∪ (3, ∞)
Question 12.
Solution:
(i) since there is a dark line start at x = – 1 and continuously goes on left side of – 1
∴ Solution set = (- ∞, – 1]
(ii) since dark line starts from 2 and conti-nuously going on right side of x = 2 but there is a hollow dot at x = 2 so 2 is not included.
∴ given representation = (2, ∞)
(iii) There is a line between x = 1 and x = 4 and hollow dots at x = 1 and x = 4 shows that 2 and 4 are not included.
∴ given representation = (1, 4)
(iv) There is a dark line from x = – 1 to x = 3 and dark dots at x = – 1 and x = 3 shows that – 1, 3 are included
∴ given representation = [- 1, 3]
(v) There is a dark line from x = – 4 to x = – 1
and dark dots at x = – 4 and – 1 shows that – 4 and – 1 included.
∴ given representation = [- 4, – 1]