Practicing ML Aggarwal Class 11 Solutions Chapter 1 Sets Ex 1.1 is the ultimate need for students who intend to score good marks in examinations.
ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Ex 1.1
Question 1.
State which of the given collection of objects is a set :
(i) A collection of popular cinema actors of India.
(ii) The collection of even natural numbers less than 51.
(iii) The collection of counting numbers less than 1.
(iv) Collection of Interesting books written by Shakes peare.
(v) The collection of novels written by Munshi Prem Chand.
(vi) The collection of 10 most talented students of your school.
(vii) Collection of all rivers flowing in India.
(viii) Collection of 5 rivers flowing in India.
(ix) Collection of all rational numbers which lie between – 1 and 1.
(x) A team of eleven best cricketers of the world.
(xi) A collection of most dangerous animals of the world.
Solution:
(i) A collection of popular cinema actors of India is not a set as the collection of popular cinema actors of India is not well defined. As a particular actor is popular according to
one person might not be popular according to another person.
(ii) The collection of even natural numbers less than 51 is well defined so it represents a set.
(iii) The collection of counting numbers less than 1 is well defined so it represents a set.
(iv) The collection of interesting books written by Shakespeare is not a set as the collection of interesting books written by Shakespeare is not well defined. A particular book
considered interesting by one person might not be considered interesting by another.
(v) The collection of novels written by Munshi Premchand is well defined so it represents a set.
(vi) The collection of 10 most talented students of your school is not well defined so it does not represents a set. Since a particular student considered as talented according to
one person might not be talented according to other.
(vii) The collection of all rivers flowing in India is well defined so it represents a set.
(viii) The collection of 5 rivers flowing in India is not well defined as it is not defined that which 5 rivers are flowing in India. So given collection does not represents a set.
(ix) The collection of all rational numbers which lie between – 1 and 1 is well defined since there are infinitely many rationals between – 1 and 1. Thus given collection represents a set.
(x) A team of eleven best cricketer s of the world is not a well defined collection. Since a particular player is considered as best according to one might not be considered as best according to other. Hence given collection does not represents a set.
(xi) Collection of most dangerous animals of the world is not well defined. Since a particular animal is dangerous according to one might not considered as dangerous according to other. Hence, the given collection does not represents a set.
Question 2.
If A = {3, 5, 7, 9, 11}, then write which of the following statements are true. If a statement is not true, mention why.
(i) 3 ∈ A
(ii) 5, 9 ∈ A
(iii) 8 ∈ A
(iv) 9 ∈ A
(vi) (3) ∈ A
(vi) {5, 7} ∈ A
Solution:
(i) Given A = {3, 5, 7, 9, 11}
True.
Since 3 be an element of set A.
∴ 3 ∈ A
(ii) True, since 5 and 9 are elements of set A
∴ 5, 9 ∈A
(iii) True, since 8 is not a member of set A
∴ 8 ∉ A
(iv) False, since 9 be an element of set A
∴ 9 ∈ A
(v) False, since 3 be an element of set A but {3}
i.e. singleton set does not belong to A.
(vi) False, since 5 and 7 are members of set A but {5, 7}
i.e. set containing two elements 5 and 7 is not a member of set A.
Question 3.
Use the roster method to represent the following sets :
(i) The counting numbers which are multiples of 6 and less than 50.
(ii) The fractions whose numerator is 1, and whose denominator is a counting number less than 10.
(iii) {x : x ∈ N and x is a prime factor of 84).
(iv) The set of odd integers lying between – 4 and 8.
(v) The set of all natural numbers x for which x + 6 is greater than 10.
(vi) The set of all integers x for which x + 6 is less than 10.
(viii) The set of all integers x for which is a natural number.
(ix) [3n + 5 ; n ∈ N and n ≤ 6]. [ISC 2020]
(x) {x : x ∈ N and 4x – 3 ≤ 15).
(xi) {x : x ∈ N, x2 < 40}
(xii) {x : x ∈ Z and x2 < 16},
(xiii) The set of all digits in our number system.
(xiv) The set of all letters in the word TRIGONOMETRY.
(xv) The set of all vowels in the English alphabet which precedes k).
(xvi) {x : x is a consonant in the English alphabet which precedes k).
Solution:
(i) The counting members which are multiples of 6 and less than 50 are 6, 12, 18, 24, 30, 36, 42, 48
∴ required set in roster form be {6, 12, 18, 24, 30, 36, 42, 48)
(ii) The fractions whose numerator is 1 and whose denominator is a counting number < 10 are ;
\(\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}, \frac{1}{8}, \frac{1}{9}\)
∴ In roster form, given set = \(\left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}, \frac{1}{8}, \frac{1}{9}\right\}\)
(iii) The prime factors of 84 are 2, 3, 7.
Thus, in roster form, given set = {2, 3, 7)
(iv) We know that, odd intcgrs between – 4 and 8 are – 3, – 1, 1, 3, 5, 7
∴ reqd.set {- 3, – 1, 1, 3, 5, 7)
(v) Since x + 6 < 10
⇒ x < 10 – 6
⇒ x < 4 and x ∈ N ∴ x = 1, 2, 3 Thus, given set {1, 2, 3}
(vi) Since x + 6 > 10, x ∈ I
⇒ x > 4, x ∈ I
∴ x = 5, 6, 7, 8, 9
Thus, in roster form, given set = {5, 6, 7, ……………….}
(vii) Since x + 6 < 10, x ∈ I
⇒ x < 4, x ∈ I
∴ x = …………….. ± 3, ± 2, ± 1, 0
⇒ given set = (……………….. – 3, – 2, – 1, 0, 1, 2, 3}
(viii) Since x ∈ I and \(\frac{60}{x}\) ∈ N,
When x = 1 ;
\(\frac{60}{1}\) = 60 ∈ N
When x = 2 ;
\(\frac{60}{2}\) = 30 ∈ N
When x = 3 ;
\(\frac{60}{3}\) = 20 ∈ N
When x = 4 ;
\(\frac{60}{4}\) = 15 ∈ N
When x = 5 ;
\(\frac{60}{5}\) = 12 ∈ N
When x = 6 ;
\(\frac{60}{6}\) = 10 ∈ N
When x = 7 ;
\(\frac{60}{7}\) ∉ N
When x = 10 ;
\(\frac{60}{10}\) = 6 ∈ N
When x = 12 ;
\(\frac{60}{12}\) = 5 ∈ N
When x = 15 ;
\(\frac{60}{15}\) = 4 ∈ N
When x = 20 ;
\(\frac{60}{20}\) = 3 ∈ N
When x = 30 ;
\(\frac{60}{30}\) = 2 ∈ N
and When x = 60 ;
\(\frac{60}{60}\) = 1 ∈ N
∴ Required set in roster form = {1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60}
(ix) Given x = 3n + 5, n ∈ N and n ≤ 6
∴ n = 1, 2, 3, 4, 5, 6
When n = 1 ;
x = 3 + 5 = 8
When n = 2 ;
x = 3 × 2 + 5 = 11
When n = 3 ;
x = 3 × 3 + 5 = 14
When n = 4 ;
x = 3 × 4 + 5 = 17
When n = 5 ;
x = 3 × 5 + 5 = 20
When n = 6 ;
x = 3 × 6 + 5 = 23
∴ given set = {8, 11, 14, 17, 20, 23)
(x) Since 4x – 3 ≤ 15
⇒ 4x ≤ 18
⇒ x ≤ \(\frac{9}{2}\) and x ∈ N
∴ x = {1, 2, 3, 4}
(xi) As x2 < 40 and x ∈ N
Since 12 < 40 ;
22 < 40 ;
32 = 9 < 40 ;
42 = 16 < 40 ;
52 < 40
and 62 = 36 < 40
∴ in roster form reqd. set = {1, 2, 3, 4, 5, 6)
(xii) Since squares of integers 0, ± 1, ± 2, ± 3 are less than 16,
∴ given set in roster form = {0, ± 1, ± 2, ± 3}
(xiii) ∴ given set = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
(xiv) In roster form, given set {T, R, I, G, O, N, M, E, Y)
(xv) In roster form, given set {a, e, i, o}
(xvi) Since all the consonents in the English alphabet which precedes k are b, c, d, f, g, h, j
∴ In roster form, reqd. set = {b, c, d, f g, h, j}
Question 4.
Write the following sets in the builder form:
(i) {1, 3, 5, 7, 9, 11, 13)
(ii) {2, 4, 6, 8, ……………..}
(iii) {3, 6, 9, 12, 15}
(iv) {2, 4, 8, 16, 32, 64}
(v) {5, 25, 125, 625}
(vi) {1, 4, 9, 16, ……………., 100}
(vii) {1, 4, 9, 16, 25, ………………..}
(viii) {\(\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}\), ………………….}
Solution:
(i) Since 1, 3, 5, 7, 9, 11, 13 are all odd natural numbers less than 14.
In set builder form, given set = {x : x is an odd natural number less than 14}
(ii) Since 2, 4, 0, 8 are all even natural numbers.
∴ In set builder fonn. given set {x : x be an even natural number}
(iii) Since each member of given set is a multiple of 3 so, in set builder form,
given set = {x : x = 3n, n ∈ N and n ≤ 5)
(iv) Since given numbers 2, 4, 8, 16, 32, 64 are of the form, 21, 22, 23, 24, 25, 26.
i.e. every member of the set is of the form 2n where n ∈ N, n ≤ 6.
∴ In set builder form, given set = {x : x = 2n : x ∈ N, x ≤ 6}
(v) Since 5, 25, 125 and 625 are of the form 51, 52, 53 and 54
∴ every number of the set is of the form 5n where n ∈ N, n ≤ 4
In set builder form, given set = {x : x = 5n, n ∈ N, n ≤ 4}
(vi) Since 1, 4, 9, 16, ……………….., 100 are of the form 12, 22, 32, 42, ………………….. 102
i.e. every member of the set is of the form n2 where n ∈ N, n ≤ 10.
∴ In set-builder form, given set = {x : x = n2 ; n ∈ N, n ≤ 10)
(vii) Since every number of the set can be written as 12, 22, 32, …………………
i.e. of the form n2, n ∈ N
∴ In set builder form, given set = {x : x = n2, n ∈ N}
(viii) Here, we observe that, each member in the given set has numerator one less than the denominator.
Hence, given set in set builder form = {x : x = \(\frac{n}{n+1}\), n ∈ N}.
Question 5.
Which of the following are examples of the null set?
(i) Set of even prime numbers.
(ii) Set of odd natural numbers divisible by 2.
(iii) Set of all Indian kids 5 metres tall.
(iv) {x : x ∈ N, x < 5 and x > 8}.
(v) {x : x is a point common to any two parallel straight lines}.
(vi) {x : x is a student ofy our school presently studying in both classes XI and XII).
Solution:
(i) Since 2 he the only even prime number.
Hence the set of even prime numbers is not a null set as it contains one element 2.
(ii) Since there is no odd natural number which is divisible by 2 and hence given set be a null set as it does not contain any element.
(iii) Since there is no indian kid which is 5 metres tall and hence given set contains no element and hence be a null set.
(iv) Since there is no natural number which is > 5 and greater than 8.
Hence given set contains no element and thus given set be a null test.
(v) Since it is not possible to find a point which is common to any two parallel straight lines and hence given set contains no element and hence given set be a null set.
Question 6.
Which of the following sets arc finite or infinite?
(i) The set of days of a week.
(ii) The set of numbers which are multiples of 7.
(iii) The set of animals living on Earth.
(iv) The set of consonants in the English alphabet.
(v) The set of circles drawn in a plane.
(vi) The set of prime numbers which are less than one crore.
Solution:
(i) Since these are 7 days in a week and the given set be a finite set since 7 be a finite number.
(ii) Since numbers which are multiples of 7 are 7, 14, 21, 28, ………………..
which are clearly infinite in number hence given set be an infinite set.
(iii) Since the number of animals living on Earth are finite in number and hence given set be a finite set.
(iv) Since there are 21 consonants and 5 vowels in English alphabet and which are finite in number and hence given set be a finite set.
(v) Since there are infinite number of circles drawn in a plane and given set be an infinite set.
(vi) Since the prime numbers which are less than one crore are finite in number and hence given set be a finite set.
Question 7.
Find the cardinal number of the following sets:
(i) { }
(ii) {0}
(iii) A = {1, 2, 2, 1, 3}
(iv) The set of all Indians having 8 legs.
(v) The set of all letters in the word PRINCIPAL
(vi) The set of all vowels in the word PRINCIPAL.
Solution:
(i) { ) be an empty set and contains no element and hence cardinal number of given set be 0.
(ii) Let A = {0}, since set A contains one element 0.
∴ cardinal number of A = n(A)
(iii) Given A = {1, 2, 2, 1, 3}
Since set A contains three distinct elements 1, 2 and 3.
∴ n(A) = 3.
(iv) Since there is no Indian who have 8 legs
∴ given set contains no element.
Thus cardinal coordinal number of given set = 0
(v) A = {P, R, I, N, C, A, L)
Since set A contains 7 distinct elements.
∴ Cardinal number of A = n(A) = 7
(vi) Let B et of all vowels in the word PRINCIPAL
∴ B = {I, A}
Thus number of distinct clement in set B = n(B) = 2.
Question 8.
(i) Write the cardinal number of the set A, where A = {x : x is a two digit number, sum of whose digits is 8).
(ii) Write the cardinal number of the set of all integers x for which \(\frac{30}{x}\) is a natural number.
(iii) What is the cardinal number of the set X, where X = {x : x is a letter in the word ‘CHANDIGARH’} ?
Solution:
(i) Given A = {x : x is a two digit number, sum of whose digits is 8}
∴ A = {17, 26, 35, 44, 53, 62, 71, 80}
Thus cardinal number of A = no. of distinct elements in A = n(A) = 8.
(ii) Let A = {x ∈ I ; \(\frac{30}{x}\) ∈ N}
at x = 1 ;
\(\frac{30}{x}\) = 30 ;
at x = 5 ;
\(\frac{30}{x}\) = \(\frac{30}{5}\) = 6 ∈ N
at x = 2 ;
\(\frac{30}{x}\) = \(\frac{30}{2}\) = 15 ;
at x = 6 ;
\(\frac{30}{x}\) = \(\frac{30}{6}\) = 5 ∈ N
at x = 3 ;
\(\frac{30}{x}\) = \(\frac{30}{3}\) = 10
at x= 10 ;
\(\frac{30}{x}\) = \(\frac{30}{10}\) = 3 ∈ N
at x = 15 ;
\(\frac{30}{x}\) = \(\frac{30}{15}\) = 2 ∈ N;
at x = 30 ;
\(\frac{30}{x}\) – \(\frac{30}{30}\) = 1 ∈ N
∴ A = {1, 2, 3, 5, 6, 10, 15, 30}
Thus cardinal no. of A = n(A) = 8
(iii X = {C, H, A, N, D, I, G, A, R, H)
∴ No. of distinct elements in X = {C, H, A, N, D, I, G R} = 8
Thus, cardinal no. of X = n(X) = 8
Question 9.
State which of the following statements are true and which are false. Justify your answer.
(i) 37 ∉ {x : x has exactly two positive factors}
(ii) 35 ∈ (x : x has exactly four positive factors}
(iii) 496 ∈ {y : the sum of all positive factors of y is 2y}.
(iv) 3 ∉ {x : x4 – 5x3 + 2x2 – 112x + 6 = 0}
Solution:
(i) False, since two positive factors of 37 are 1 and 37.
(ii) True : Since 35 has exactly four positive factors 1,5, 7, 35
∴ 35 ∈ {x : x has exactly four positive factors}
(iii) By prime factorisation ; we have
496 = 2 × 2 × 2 × 2 × 31
Thus positive factors of 496 arc 1, 2, 4, 8, 16, 31, 62, 124, 248, 496
Thus, sum of positive factors
= 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 + 496
= 2 × 496
Thus, 496 ∈ {y : the sum of positive factors of y is 2y}
(iv) Since x4 – 5x3 + 2x – 112x + 6 = 0
when x = 3 ;
LH.S. = 34 – 5 × 33 + 2 × 32 – 112 × 3 + 6
= 81 – 135 + 18 – 336 + 6
= – 468 ≠ 0 = 1
= R.H.S.
Hence 3 be not the root of given eqn.
∴ 3 ∉ {x : x4 – 5x3 + 2x2 – 112x + 6 = 0}
Thus given statement is true.
Question 10.
Classify the following sets into finite set and infinite set. in casc of finite sets, mention the cardinal number.
(i) A = {x : x ∈ I, x < 5}. (ii) A = { x : x ∈ W, x is divisible by 4 and 9}. (iii) P = {x : x is an even prime number > 2}.
(iv) F = {x : x ∈ N and x is a factor of 84}.
(v) B = {x : x is a two digit number, sum of whose digits is 12}
(vi) C = {x : x ∈ W, 3x – 7 ≤ 8}.
(vii) {x : x = 5n, n ∈ N and x < 20}.
(viii) {x : x = 5n, n ∈ I and x < 20}.
(ix) {x : x = \(\frac{n}{n+1}\), n ∈ W and n ≤ 10}
(x) {x : x = \(\frac{2 n}{n+3}\), n ∈ N and 5 < n < 20}
Solution:
(i) Given A = {x : x ∈ 1, x < 5} = {………………… – 4, – 3, – 2,- 1, 0, 1, 2, 3, 4} Clearly set A contains infinite no. of elements and hence A be an infinite set. (ii) A = {x : x ∈ W, x is divisible by 4 and 9} = {x : x ∈ W, x is divisible by 4 × 9 = 36} = {36, 72, 108, …………………} Hence given set A contains infinite numbers and A be an infinite set. (iii) Given P = {x : x is an even prime number > 2)
Since there is no even prime number > 2
as 2 be the only even prime number.
Hence P be an empty set.
∴ Cardinal no. of P = no. of distinct elements of P = 0
(iv) Given F = {x : x ∈ N and x is a factor of 84
Thus the factors of84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
∴ Cardinal no. of F = {1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84} = 12
(v) Given B = {x : x is a two digit number, sum of whose digits is 12}
= (39, 48, 57, 66, 75, 84, 93}
∴ n(B) = no. of distinct elements of set B = 7
(vi) Given C = {x : x ∈ W, 3x – 7 ≤ 8}
Since 3x – 7 ≤ 8
⇒ 3x ≤ 15
⇒ x ≤ 5 and x ∈ W
∴ x = 0, 1, 2, 3, 4, 5
∴ C = {0, 1, 2, 3, 4, 5}
Thus n(C) = Cardinal no. of C = 6
(vii) Let D = {x : x = 5n, n ∈ N and x < 20}
Now x = 5n, n ∈ N
When n = 1 ;
x = 5 × 1 = 5;
When n = 2 ;
x = 5 × 2 = 10
When n = 3 ;
x = 5 × 3 = 15 < 20
When n = 4 ;
x = 5 × 4 = 20 ≥ 20
∴ D = {5, 10, 15) which contains 3 elements
and hence set D be a finite set
∴ n(D) = 3
(viii) Let D = {x : x = 5n, n ∈ I and x < 20}
i.e. D = {…………., – 15, – 10, – 5, 0, 5, 10, 15)
which are clearly infinite in number and hence set D be an infinite set.
(ix) Let E = {x : x = \(\frac{n}{n+1}\), n ∈ W and x ≤ 10}
Since n ∈ W and n ≤ 10
∴ n = 0, 1, 2, 3, ……………….., 10.
when n = 0 ;
x = \(\frac{0}{0+1}\) = 0 ;
when n = 1 ;
x = \(\frac{1}{1+1}=\frac{1}{2}\)
when n = 2 ;
x = \(\frac{2}{2+1}=\frac{2}{3}\)
when n = 3 ;
x = \(\frac{3}{3+1}=\frac{3}{4}\)
when n = 4 ;
x = \(\frac{4}{4+1}=\frac{4}{5}\)
when n = 5 ;
x = \(\frac{5}{5+1}=\frac{5}{6}\)
when n = 6 ;
x = \(\frac{6}{6+1}=\frac{6}{7}\)
when n = 7;
x = \(\frac{7}{7+1}=\frac{7}{8}\)
when n = 8 ;
x = \(\frac{8}{8+1}=\frac{8}{9}\)
when n = 9 ;
x = \(\frac{9}{9+1}=\frac{9}{10}\)
when n = 10 ;
x = \(\frac{10}{10+1}=\frac{10}{11}\)
∴ E = \(\left\{0, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}, \frac{8}{9}, \frac{9}{10}, \frac{10}{11}\right\}\)
Thus E contains finite no. of elements and hence be a finite set
∴ cardinal no. of E = No. of distinct element in E = n(E) = 11
(x) Let F = {x : x = \(\frac{2n}{n+3}\); n ∈ N and 5 < n < 20}
Since n ∈ N and 5 < n < 20
∴ n = 6, 7, 8, 9, 10, ……………….., 18, 19
when n = 6 ;
x = \(\frac{2 \times 6}{6+3}=\frac{12}{9}\) ;
when n = 7 ;
x = \(\frac{2 \times 7}{7+3}=\frac{14}{10}\)
when n = 8 ;
x = \(\frac{2 \times 8}{8+3}=\frac{16}{11}\)
…………………………………………………
…………………………………………………
When n = 18 ;
x = \(\frac{2 \times 18}{18+3}=\frac{36}{21}\)
When n = 19 ;
x = \(\frac{2 \times 19}{19+3}=\frac{38}{22}\)
∴ F = \(\left\{\frac{12}{9}, \frac{14}{10}, \frac{16}{11}, \frac{18}{12}, \frac{20}{13}, \frac{22}{14}, \frac{24}{15}, \frac{26}{16}, \frac{28}{17}, \frac{30}{18}, \frac{32}{19}, \frac{34}{20}, \frac{36}{21}, \frac{38}{22}\right\}\)
Thus F contains finite no. of elements and hence be a finite set.
∴ Cardial no. of F = No. of distinct elements in F = n(F) = 14