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ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Ex 1.2
Question 1.
In the following, state whether A = B or not :
(î) A = {x : x + 2 = 3), B = {x : x ∈ N and is less than 2}
(ii) A = {x : x ∈ N and 3x – 1 < 2}, B = {x : x ∈ W and 3x – 1 < 2}
(iii) A = (x : x ∈ N and a prime factor of 36), B = {1, 2, 3, 4, 6, 9, 12}
(iv) A = {x : x ∈ I and x2 ≤ 4), B = {x : x ∈ R and x2 – 3x + 20}
(v) A = {2, 3} and B = {x : x is a solution of x2 + 5x + 6 = 0}
(vi) A = {x : x is a letter in the word LOYAL} and B = {y : y is a letter in the world ALLOY}
(vii) A = {x : x is a letter in the word WOLF} and B = {y : y is a letter in the world FOLLOW)
(viii) A = {10, 20, 30, 40, 50, ………………} and B = {x : x is a multiple of 10}.
Solution:
(i) Given A = {x : x + 2 = 3}
Since x + 2 = 3
⇒ x = 3 – 2 = 1
∴ A{1}
and B = {x : x ∈ N and is less than 2} = {1}
Since sets A and B have same element
∴ A and B are equal sets i.e. A = B.
(ii) Given A = {x : x ∈ N and 3x – 1 < 2}
Since 3x – 1 < 2
⇒ 3x < 3
⇒ x < 1, x ∈ N
∴ A = Φ i.e. A be an empty set.
and B = {x : x ∈ W and 3x – 1 < 2}
Since 3x – 1 < 2
⇒ 3x < 3
⇒ x < 1, x ∈ W
∴ x = 0
∴ B = {0} and B be a singleton set.
Hence A and B do not have same elements.
∴ A ≠ B.
(iii) Given A = {x : x ∈ N and is a prime factor of 36)
Since prime factors of 36 are 2, 3
∴ x = 2, 3
∴ A = {2, 3}
and B = {1, 2, 3, 4, 6, 9, 12}
Clearly n(A) = 2
and n(B) = 7
∴ n(A) ≠ n(B)
Moreover A and B does not have same elements
∴ A and B are not equal sets.
Thus A ≠ B.
(iv) Given A = {x : x ∈ I and x2 ≤ 4}
Since x2 ≤ 4
⇒ |x| ≤ 2
⇒ – 2 ≤ x ≤ 2, x ∈ I
∴ x = – 2, – 1, 0, 1, 2
Thus, A = {- 2, – 1, 0, 1, 2}
and B = {x : x ∈ R and x2 – 3x + 20}
Since x2 – 3x + 2 = 0
⇒ (x – 1) (x – 2) = 0
⇒ x = 1, 2.
Thus. B = {1, 2)
Clearly the elements oíA and B are not same.
∴ A and B are not equal sets
∴ A ≠ B.
(v) Given A = {2, 3)
and B = {x : x is a solution of x2 + 5x + 6 = 0)
Since x2 + 5x + 6 = 0
⇒ (x + 2) (x + 3) = 0
⇒ x = – 2, – 3
Thus B = {- 2, – 3}
Clearly the elements of set A and B are not same
∴ A ≠ B.
(vi) Given A = {x : x is a letter in the word LOYAL)
= {L, O, Y, A}
and B = {y : y is a letter in the word ALLOY)
= {A, L, O, Y}
Clearly sets A and B have same elements and hence A and B are equal sets
∴ A = B
(vii) Given A = {x : x is a letter in the word WOLF)
= {W, O, L, F)
and B = {y : y is a letter in the word FOLLOW)
= {W, O, L, F)
Clearly sets A and B have same elements and hence both sets are equal
∴ A = B
(viii) Given A = {10, 20, 30, 40, 50, …………..)
B = {x : x is a multiple of 10}
Clearly multiple of 10 are 0, 10, 20, 30, 40, 50.
∴ B = {0, 10, 20, 30, 40, 50}
Thus set A and set B does not have same elements and hence Aand B are not equal sets
∴ A ≠ B.
Question 2.
If A = {1, 2, 3, 4, 5), B = {2, 3, 4) and C = {2, 4, 5); state whether the following statements are true or false.
(i) A ⊂ B
(ii) A ⊂ C
(iii) B ⊂ A
(iv) B ⊂ C
(y) C ⊂ A
(vi) C ⊂ B
(vii) B = C
(viii) Φ ⊂ B
(ix) A ↔ B
(x) B ↔ C
(xi) A ↔ C
Solution:
Given, A = {1, 2, 3, 4, 5} ;
B = {2, 3, 4}
and C = {2, 4, 5}
(i) since 1 ∈ A but 1 ∉ B
∴ A ⊄ B
∴ given statement is false
(ii) False, since 1 ∈ A but 1 ∉ C
∴ A ⊄ C
(iii) True, since every member of B is a member of A.
(iv) False. since 3 ∈ B and 3 ∉ C
∴ B ⊄ C
(v) True, since every member of C is a member of A
∴ C ⊂ A
(vi) False, As 5 ∈ C and 5 ∉ B
∴ C ⊄ B
(vii) False, since 3 ∈ B but 3 ∉ C
Thus sets B and C does not have same elements
∴ B and C are not equal seis.
∴ B ≠ C
(viii) True, since empty set is a subset of every set.
∴ C ⊂ A
(ix) False, since cardinal number of A = no. of distinct elements in A = n(A) = 5
while n(B) = cardinal number of set B = 3
∴ n(A) ≠ n(B)
∴ both sets A and B are not equivalent sets.
∴ A ↮ B.
(x) True; since cardinal number of B = n(B) = 3
and cardinal number of C = n(C) = 3
∴ n(A) = n(B)
⇒ B ↔ C
(xi) False, since cardinal no. of A = n(A) = 5
and cardinal no. of C = n(C) = 3
∴ n(A) ≠ n(C)
Thus, A and C are not equivalent sets.
∴ A ↮ C.
Question 3.
Use appropriate symbol ‘∈, ∉, ⊂, ⊃, =‘ to fill in the blanks:
(i) 4 …………… {1, 2, 3, 4}
(ii) – 5 ……………. (2, 3, 4, 5, 6)
(iii) (2) ……………. (2, 3, 4)
(iv) [a, b, c] …………. {a, b, b, a, c}
(vi) {a, i, u} ……………. {a, a, i}
(vii) MUMBAI …………………. {x : x is a capital city of countries in Asia}.
Solution:
(i) Since 4 be a member of {1, 2, 3, 4}
∴ 4 ∈ {1, 2, 3, 4}
(ii) Since – 5 is not a member of {2, 3, 4, 5, 6)
∴ – 5 ∈ {2, 3, 4, 5, 6}
(iii) Since {2} is not a member of {2, 3, 4}
∴ {2} ∉ {2, 3, 4}
also {2} be a subset of {2, 3, 4}
⇒ {2} ⊂ {2, 3, 4}
(iv) Since {b, a, e} and {a, b, c} have same elements namely a, b and c respectively
∴ {a, b, c} = {b, a, c}
(v) Since the sets {a, b, c} and {a, b, b, a, c} have same elements namely a, b and e respectively and
hence both sets are equal.
∴ {a, b, c} = {a, b, b, a, c}
(vi) Since, all members of {a, a, i} are the members of {a, j, u}
∴ {a, a, i} be a subset of {a, i, u} or {a, i, u) is a super set or contains (a, a, iI).
Thus, {a, i, u) ⊃ (a, a, i).
(vii) Since MUMBAI is not a separate capital of any country in Asia.
∴ MUMBAI ∉ {x : x is a capital city of countries in Asia}.
Question 4.
Examine whether the following statements are true or false:
(i) {a, b} ⊄ {b, c, a)
(ii) {1, 2, 3) ⊂ {1, 3, 5}
(iii) {a, e} ⊂ {a, b, c}
(iv) {a} ∈ (a, b, c}
(v) {a, e) ⊂ {x : x is a vowel in the English alphabet)
(vi) (x : x is an even natural less than 8) ⊂ (x : x is a natural number which divides 36}.
Solution:
(i) Clearly (a, b) be a subset of (b, c, a)
∴ {a, b} ⊂ {b, c, a}
Thus, given statement is false.
(ii) Clearly (1, 2, 3) be not a subset of (1, 3, 5}
∴ {1, 2, 3} ⊂ {1, 3, 5}
Thus, given statement is false.
(iii) True, since {a} be a subset of {a, b, c}.
(iv) False, clearly a, b, c ∈ {a, b, c} but {a} ∈ {a, b, c}
(v) True, clearly {a, e} be a subset of {a, e, i, o, u}
∴ {a, e} ⊂ {a, e, i, o, u}
(vi) Since even natural numbers less than 8 are 2, 4, 6.
∴ A = {x : x is a even natural number less than 8} – {2, 4, 6}
and B = {x 😡 isanatural numberwhichdivides 36} = {1, 2, 3, 4, 6, 9, 12, 18, 36}
Clearly {2, 4, 6)} ⊂ {1, 2, 3, 4, 69, 12, 18, 36}
Hence the given statement is true.
Question 5.
State whether each of the following statements is true or false:
(i) 2 ⊂ {1, 2, 3}
(ii) {2} ∈ {1, 2, 3}
(iii) {Φ} ∈ (1, 2, 3}
(iv) 0 ∈ Φ
(v) {1, 2} ⊂ {1, {2, 3}}
(vi) Φ ⊂ {Φ}.
Solution:
(i) False, since {2} ⊂ {1, 2, 3} while 2 ⊄ {1, 2, 3}
(ii) False, since subset of {1, 2, 3} are Φ, {1}, (1), {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}
∴ {2} ⊂ {1, 2, 3}
but (2) ∉ {1, 2, 3}
(iii) False, since empty set is a subset of every set
∴ Φ ⊂ {1, 2, 3}.
Thus Φ ∉ (1, 2, 3) since Φ is not a member of {1, 2, 3}.
(iv) False, 0 ∉ Φ as Φ = { } i.e. an empty set contains no clement.
(v) False, since the subsets of {1, {2, 3}} are Φ, {1}, {{2, 3}}, {1, (2, 3}}
Thus {1, 2} ⊄ {1, {2, 3}}
True, since empty set Φ be a subset of every set.
∴ Φ ⊂ {Φ}
Question 6.
Let A = {1, 2, {3, 4}, 5}. Which of the following statements are not true and why?
(i) {3, 4} ⊂ A
(ii){3, 4} ∈ A
(iii) {{3, 4}} ⊂ A
(iv) {1, 2, 5} ⊂ A
(v) {1, 2, 5} ∈ A
(vi) 1 ∈ A
(vii) 1 ⊂ A
(viii) Φ ∈ A
(ix) Φ ∈ A
(x) Φ ∈ A
(xi) {Φ} ⊂ A.
Solution:
Given A = {1, 2, {3, 4}, 5}
(i) Clearly {3, 4} be not a subset of A
∴ {3, 4} ⊄ A
∴ given statement is not true.
(ii) Clearly {3, 4} be a member of A
Thus {3, 4} ∈ A is true.
(iii) Clearly {{3, 4}} be a subset of A
Thus given statement is true.
(iv) Clearly {1, 2, 5} be a subset of A
∴ Given statement is true.
(v) Since 1, 2, 5 ∈ A but {1, 2, 5} A
As { 1, 2, 5) be a subset of A.
Thus, given statement is not true.
(vi) Clearly {1, 2, 3} be not a subset of A
Thus, given statement is not true.
(vii) Clearly 1 be a member of A
∴ 1 ∈ A
Thus given statement is true.
(viii) Clearly 1 ∈ A but {1} ∉ A as {1} ⊂ A.
∴ given statement is not true.
(ix) Clearly Φ ⊂ A but is not a member of set A.
Thus given statement is false.
(x) Since empty set be a subset of every set
∴ Φ ⊂ A is a true statement
(xi) Clearly Φ ⊂ A but {Φ} be a set containing element Φ
∴ {Φ} ⊄ A
Thus given statement is not true.
Question 7.
If ξ = {1, 2, 3, ……………., 10} and A = {|x| x is a prime factor of 66}. List the set A.
Solution:
Given = {1, 2, 3, …………………., 10}
Since prime factors of 66 are 2, 3, 11.
but A ⊂ ξ
∴ A = {2, 3}
Since 11 ∉ ξ .
Question 8.
Write down ati the subsets of the following sets:
(i) {a}
(ii) {a, b}
(iii) Φ
(iv) {Φ, 1).
Solution:
(i) Φ, {a}
(ii) Φ, {a}, {b}, {a, b}
(iii) Φ
(iv) Φ, Φ {1}, {Φ, 1}
Question 9.
Write the power set of A, where A = {- 1, 0, 2}.
Solution:
Given A = {- 1, 0, 2}
So power set of A = P(A) = set of all subsets of A
= {Φ, {- 1}, {0}, {2}, {- 1, 0}, {0, 2}, {- 1, 2}, {- 1, 0, 2}}
Question 10.
Write the number of proper subsets of A, where A = {- 3, – 1, 0, 1, 4}.
Solution:
Given A = {- 3, – 1, 0, 1, 4}
So cardinal number of set A = n(A)
= no. of distinct elements of A = 5
Thus the no. of subsets of A = 2n(A)
= 25 = 32
Hence the total no. of proper subsets of A = 32 – 1 = 31
[Since every set is a subset of itself so subset
A is not included in proper subset]
Question 11.
If ξ = {1, 2, 3, ……………., 40} and A = {x : x is a factor of 42}, then write n(A).
Solution:
Given ξ = {1, 2, 3, …………………., 40}
and A = {x : x is a factor of 42}
= {1, 2, 3, 6, 7, 14, 21, 42}
Since A ⊂ξ
∴ A = {1, 2, 3, 6, 7, 14, 21}
Thus n(A) = Cardinal number of A = 7
Question 12.
If ξ = {all digits in our number system} and A = {x : x is a multiple of 3}, then write n(A).
Solution:
Given ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} as A ⊂ ξ
and A = {0, 3, 6, 9)
Then n(A) = Cardinal number of A = 4.
Question 13.
Write the following sets as intervals:
(i){x : x ∈ R, – 3 < x ≤ 5}
(ii) {x : x ∈ R, – 5 < x< – 1}
(iii) {x : x ∈ R, 2 ≤ x ≤ 7}
(iv) {x : x ∈ R, 0 ≤ x < 3}
(v) {x : x ∈ R, x ≤ 5}
(vi) (x : x ∈ R, x < – 3}
(vii) {x : x ∈R x – 2}
Solution:
(i){x : x ∈ R, – 3 < x ≤ 5} = {- 3, 5]
(ii) {x : x ∈ R, – 5 < x < – 1} = {- 5, – 1}
since – 5 and – 1 both are excluded
(iii) {x : x ∈ R, + 2 ≤ x ≤ 7} = [2, 7]
(iv) {x : x ∈ R, 0 ≤ x < 3} = [0, 3)
(v) {x : x ∈ R, x ≤ 5} = {- ∞, 5)
(vi) {x : x ∈ R, x < – 3} = (- ∞, – 3)
(vii) {x : x ∈ R, x ≥ – 2} = [- 2, ∞)
Question 14.
Write the following intervals in set builder form:
(i) (- 2, 0]
(ii) (2, 7)
(iii) [- 5, – 2]
(iv) [- 9, 4)
(v) (3, ∞)
(vi) (- ∞, – 1]
(vii) (- ∞, 4).
Solution:
(i) (- 2, 0] = {x : x ∈ R, – 2 < x ≤ 0}
(ii) (2, 7) = {x : x ∈ R, 2 < x < 7}
(iii) [- 5, – 2] = [x : x ∈ R, – 5 ≤ x < – 2}
(iv) [- 9, 4) = {x : x ∈ R, – 9 ≤ x < 4}
(v) (3, – ∞) = {x : x ∈ R, x > 3}
(vi) (- ∞, – 1] = {x : x ∈ R, x ≤ – 1}
(vii) (- ∞, 4) = {x : x ∈ R, x < 4}
Question 15.
If ξ = {1, 2, 3, …………………., 12), A = {x |2x + 3 ≤ 18} and B = {x |x2 ≤ 40) ; write A and B in the roster form.
Solution:
Given ξ = {1, 2, 3, …………………. 12}
and A = {x |2x + 3 ≤ 18}
Since 2x + 3 ≤ 18
⇒ 2x ≤ 15,
⇒ x ≤ \(\frac{15}{2}\) = 7.5
∴ A = {1, 2, 3, 4, 5, 6, 7}
and B = {x |x2 ≤ 40} since x2 ≤ 40
Now 12 ≤ 40 ;
22 = 4 ≤ 40 ;
32 = 9 ≤ 40 ;
42 = 16 ≤ 40 ;
52 ≤ 40 ;
62 = 36 ≤ 40
Thus, B = {1, 2, 3, 4, 5, 6}
Question 16.
Let ξ = {0, 1, 2, 3 , 50}, A = {x : x = 6n, n ∈ W}, B = {x : x = 7n, n ∈ W} and C = {x : x is a factor of 72}, then
(i) write the sets A, B and C in roster form
(ii) state n(A), n(B) and n(C).
Solution:
Given ξ = {0, 1, 2, 3, ……………….. 50}
A = {x : x = 6n, n ∈ W}
since n ∈ W
∴ n = 0, 1, 2, 3, ………………………
when n = 0 ⇒ x = 0
when n = 1 ⇒ x = 6
when n = 2 ⇒ x = 12
when n = 3 ⇒ x = 18
when n = 4 ⇒ x =24
when n = 5 ⇒ x = 30
when n = 6 ⇒ x = 36
when n = 7 ⇒ x = 42
when n = 8 ⇒ x = 48
when n = 9 ⇒ x = 54
∴ A = {0, 6, 12, 18, 24, 30, 36, 42, 48}
B = {x : x = 7n, n ∈ W) = {0, 7, 14, 21, 28, 35, 42, 49}
and C = {x : x is a factor of 72) = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36)
as 72 ∉ ξ
(ii) Thus cardinal number of A = no. of distinct elements in A
= n(A) = 9
Cardinal no. of B = no. of distinct elements in B
n(B) = 8
and No. of distinct elements in C = n(C) = 11.
Question 17.
Given A = {x : x is a letter in the word ACCUMULATOR}
(i) Express A in roster form.
(ii) Find the cardinal number of the set of vowels in A.
(iii) Write down the power set of the set of vowels in A.
Solution:
(i) In roster form, A = {A, C, U, M, L,T, O, R}
(ii) Let B = {all vowels in A) = {A, U, O}
∴ n(B) = Cardinal no. of B = 3
(iii) Power set of set of vowels in A = {Φ, {A}, {U), {O}, {A, U), {A. O), {U, O), {A, U, O}}
Question 18.
List all the proper subsets of {0, 1, 2, 3}.
Solution:
Let A = {0, 1, 2, 3}
Thus, the required proper subsets of A is given by
Φ, (0), (1), {2}, {3), {O, I), {0, 2), {0, 3}, {1, 2}, {1, 3), {2, 3}, {0, 1, 2), {0, 1, 3), {0, 2, 3}, {1, 2, 3}
Question 19.
If the set A has five elements, how many subsets will A have ? If A has six elements, how many proper subsets will A have?
Solution:
Given set A has five elements
∴ Cardinal no. of A = n(A) = 5
Thus, No. of subsets of A = 2n(A)
= 25 = 32
Given A has six elements
∴ n(A) = 6
thus, no. of proper subsets of A = 2n(A) – 1
= 26 – 1 = 63
[Since the subset A of set A is excluded]
Question 20.
Two finite sets have m and k elements. If the number of subsets of the first set is 112 more than the number of subsets of the second set, then find the values of m and k.
Solution:
Let the two given sets be A and B
Then n(A) = m
and n(B) = k, n > k
Now, n(P(A)) = 2m ;
n(P(B)) = 2k
according to given condition ; we have
2m = 12 + 2k
⇒ 2m – 2k = 112
⇒2k (2m – k – 1) = 112 = 24 × 7
⇒ 2k = 24 and 2m – k – 1 = 7
⇒ k = 4 and 2m – k = 8 = 23
⇒ k = 4 and m – k = 3
⇒ k = 4 and m = 7
Hence, m = 7 and k = 4.