1. Decide, among the following sets, which sets are subsets of one and another:
A={x : x ∈ R and x satisfy $\displaystyle \small x^{2}$– 8x + 12 = 0}, B ={ 2, 4, 6 }, C ={ 2, 4, 6, 8, . . . }, D ={ 6 }.
A={2,6}, B={2,4,6}, C={2,4,6,8,...}, D={6}
∴ A⊂B, A⊂C
B⊂C
D⊂A, D⊂B, D⊂C
2. In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
(i) If x ∈ A and A ∈ B , then x ∈ B
False
Let A={1,2} and B={1,{1,2},3}
2∈A and {1,2}∈{1,{1,2},3}
∴ A∈B
But 2∉{1,{1,2},3} i.e. 2∉B
(ii) If A ⊂ B and B ∈ C , then A ∈ C
False
Let A={1,2}, B={1,2,3,4}, C={{1,2,3,4}, 5,6}
{1,2}⊂{1,2,3,4} i.e. A⊂B
{1,2,3,4}∈{{1,2,3,4},5,6} i.e. B∈C
But {1,2}∉{{1,2,3,4},5,6} i.e. A∉C
(iii) If A ⊂ B and B ⊂ C , then A ⊂ C
True
Let A={1,2}, B={1,2,3,4}, C={1,2,3,4,5,6}
{1,2}⊂{1,2,3,4} i.e. A⊂B
{1,2,3,4}⊂{1,2,3,4,5,6} i.e. B⊂C
{1,2}⊂{1,2,3,4,5,6} i.e. A⊂C
(iv) If A ⊄ B and B ⊄ C , then A ⊄ C
False
Let A={1,2}, B={3,4,5}, C={1,2,7,8}
{1,2}⊄{3,4,5} i.e. A⊄B
{3,4,5}⊄{1,2,7,8} i.e. B⊄C
But {1,2}⊂{1,2,7,8} i.e. A⊂C
∴ if A⊄B and B⊄C, need not imply that A⊄C
(v) If x ∈ A and A ⊄ B , then x ∈ B
False
Let A={1,2}, B={3,4,5}
2∈{1,2} i.e. x∈A
{1,2}⊄{3,4,5} i.e. A⊄B
2∉{3,4,5} i.e. x∉B
(vi) If A ⊂ B and x ∉ B , then x ∉ A
True
Let A={1,2}, B={1,2,3,4}
{1,2}⊂{1,2,3,4} i.e. A⊂B
5∉{1,2,3,4} i.e. x∉B
5∉{1,2} i.e. x∉A
3. Let A, B, and C be the sets such that A∪B = A∪C and A∩B = A∩C. Show that B=C.
Let A, B, and C be the sets such that A∪B=A∪C and A∩B=A∩C
Let x∈B
⇒ x∈(A∪B)
⇒ x∈(A∪C) [∵(A∪B)=(A∪C)]
⇒ x∈A or x∈C
⇒ x∈A ⇒ x∈(A∩B)
⇒ x∈(A∩C) [∵(A∩B)=(A∩C)]
⇒ x∈C
⇒ x∈B and x∈C ⇒ B⊂C ...(i)
Let y∈C
⇒ y∈(A∪C)
⇒ y∈(A∪B) [∵(A∪B)=(A∪C)]
⇒ y∈A or y∈B
⇒ y∈A ⇒ y∈(A∩C)
⇒ y∈(A∩B) [∵(A∩B)=(A∩C)]
⇒ y∈B
⇒ y∈C and y∈B⇒ C⊂B ...(ii)
∴ From (i) and (ii), since B⊂C and C⊂B ⇒ B=C
4. Show that the following four conditions are equivalent :
(i) A⊂B (ii) A–B=φ (iii) A∪B=B (iv) A∩B=A
Conditions (i) and (ii) are equivalent
Let A-B=ϕ
There exists no x such that x∈A but x∉B
All elements of A are in B i.e. for every x∈A, x∈B ⇒A⊂B
Conditions (ii) and (iii) are equivalent
Let A-B=ϕ
There exists no x such that x∈A but x∉B
All elements of A are in B i.e. for every x∈A, x∈B ⇒A⊂B ⇒ A∪B=B
Conditions (iii) and (iv) are equivalent
A∪B=B ⇒All elements of A are in B ⇒All elements of A are common in A and B ⇒A∩B=A
5. Show that if A⊂B, then C–B⊂C–A.
x∈ (C-B) ⇒ x∈C but x∉B
⇒x∈C and x∉A [∵A⊂B]
⇒ x∈(C-A)
∴ C-B⊂C-A
6. Assume that P(A)=P(B). Show that A=B
Let x be an element of set A. Then there exist a subset of A, say X such that X={x}
Now, X⊂A ⇒ X∈P(A)
⇒ X∈P(B) [∵P(A)=P(B)]
⇒X⊂B ⇒ x∈B
Thus x∈A and x∈B
∴ A⊂B ...(i)
Let y be an element of set B. Then there exist a subset of B, say Y such that Y={y}
Now, Y⊂B ⇒ Y∈P(B)
⇒ Y∈P(A) [∵P(A)=P(B)]
⇒Y⊂A ⇒ y∈A
Thus y∈B and y∈A
∴B⊂A ...(ii)
From (i) and (ii), since A⊂B and B⊂A ⇒ A=B
7. Is it true that for any sets A and B, P(A)∪P(B)=P(A∪B)? Justify your answer.
False
Let A={1}, B={3}
A∪B={1,3}
P(A)={ϕ, {1}}
P(B)= {ϕ, {3}}
P(A)∪P(B) = {ϕ,{1},{3}}
P(A∪B)={ϕ, {1},{3},{1,3}}
∴ P(A)∪P(B) ≠ P(A∪B)
8. Show that for any sets A and B, A=(A∩B)∪(A–B) and A∪(B–A)=(A∪B)
A=(A∩B)∪(A–B)
(A∩B)∪(A-B) = (A∩B)∪(A∩B') [∵A-B=A∩B']
= A∩(B∪B')
=A∩U [∵(B∪B')=U]
=A
A∪(B–A)=(A∪B)
A∪(B-A) = A∪(B∩A') [∵(B-A)=B∩A']
=(A∪B)∩(A∪A')
=(A∪B)∩U
=(A∪B)
9. Using properties of sets, show that
(i) A ∪ ( A ∩ B ) = A
A∪(A∩B) = (A∪A)∩(A∪B) [Distributive law]
=A∩(A∪B) [∵A∪A = A]
=A [∵A⊂(A∪B)]
(ii) A ∩ ( A ∪ B ) = A.
A∩(A∪B)=(A∩A)∪(A∩B) [Distributive law]
=A∪(A∩B) [∵A∩A = A]
=A [∵A∩B⊂A]
10. Show that A∩B=A∩C need not imply B=C.
Let A={a,b}, B={a,c}, C={a,d}
A∩B={a}
A∩C={a}
∴ A∩B = A∩C
But still B ≠ C
11. Let A and B be sets. If A∩X=B∩X=φ and A∪X=B∪X for some set X, show that A=B.
(Hints A=A∩(A∪X), B=B∩(B∪X) and use Distributive law )
A=A∩(A∪X) = A∩(B∪X) [∵ A∪X=B∪X]
A=(A∩B)∪(A∩X)
A=(A∩B)∪ϕ [∵(A∩X)=ϕ]
A= (A∩B) ⇒ A⊂B ...(i)
B=B∩(B∪X) = B∩(A∪X) [∵ A∪X=B∪X]
B=(B∩A)∪(B∩X)
B=(B∩A)∪ϕ [∵(B∩X)=ϕ]
B= (B∩A) ⇒ B⊂A ...(ii)
From (i) and (ii), since A⊂B and B⊂A ⇒ A=B
12. Find sets A, B, C such that A∩B, B∩C and A∩C are non-empty sets and A∩B∩C=φ.
Let A={a,b}, B={b,c}, C={a,c}
A∩B={b} ≠ ϕ
B∩C={c} ≠ ϕ
A∩C={a} ≠ ϕ
(A∩B)∩C = {b}∩{a,c} = ϕ
13. In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?
Let C and T be the set of students taking coffee and tea respectively
It is given that,
Total number of students = 600
n(C)=225, n(T)=150, n(T∩C)=100
n(T∪C) = n(T) + n(C) - n(T∩C)
n(T∪C)= 150 + 225 - 100 = 275
Number of students who were taking neither tea nor coffee = 600 - n(T∪C) = 600 -275 = 325
14. In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?
Let H and E be set of students knowing Hindi and English respectively.
It is given that,
n(H) = 100, n(E) = 50 , n(H∩E) = 25
n(H∪E) = n(H) + n(E) - n(H∩E)
n(H∪E) = 100 + 50 - 25 = 125
15. In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find:
It is given that,
n(H) = 25, n(T) = 26, n(I) = 26, n(H∩I) = 9, n(H∩T) = 11, n(T∩I) = 8, n(H∩T∩I) = 3
(i) the number of people who read at least one of the newspapers.
n(H∪T∪I) = n(H) + n(T) + n(I) - n(H∩I) - n(H∩T) - n(T∩I) + n(H∩T∩I)
n(H∪T∪I) = 25+26+26-9-11-8+3 = 52
(ii) the number of people who read exactly one newspaper.
n(H∪T∪I) - (8+3+6+5) = 52 - 22 = 30
16. In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.
It is given that,
n(A) = 21, n(B) = 26, n(C) = 29, n(A∩B) =14, n(A∩C) =12, n(B∩C) =14, n(A∩B∩C)=8
People who liked C only = 29-(4+8+6) = 11
A={x : x ∈ R and x satisfy $\displaystyle \small x^{2}$– 8x + 12 = 0}, B ={ 2, 4, 6 }, C ={ 2, 4, 6, 8, . . . }, D ={ 6 }.
A={2,6}, B={2,4,6}, C={2,4,6,8,...}, D={6}
∴ A⊂B, A⊂C
B⊂C
D⊂A, D⊂B, D⊂C
2. In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
(i) If x ∈ A and A ∈ B , then x ∈ B
False
Let A={1,2} and B={1,{1,2},3}
2∈A and {1,2}∈{1,{1,2},3}
∴ A∈B
But 2∉{1,{1,2},3} i.e. 2∉B
(ii) If A ⊂ B and B ∈ C , then A ∈ C
False
Let A={1,2}, B={1,2,3,4}, C={{1,2,3,4}, 5,6}
{1,2}⊂{1,2,3,4} i.e. A⊂B
{1,2,3,4}∈{{1,2,3,4},5,6} i.e. B∈C
But {1,2}∉{{1,2,3,4},5,6} i.e. A∉C
(iii) If A ⊂ B and B ⊂ C , then A ⊂ C
True
Let A={1,2}, B={1,2,3,4}, C={1,2,3,4,5,6}
{1,2}⊂{1,2,3,4} i.e. A⊂B
{1,2,3,4}⊂{1,2,3,4,5,6} i.e. B⊂C
{1,2}⊂{1,2,3,4,5,6} i.e. A⊂C
(iv) If A ⊄ B and B ⊄ C , then A ⊄ C
False
Let A={1,2}, B={3,4,5}, C={1,2,7,8}
{1,2}⊄{3,4,5} i.e. A⊄B
{3,4,5}⊄{1,2,7,8} i.e. B⊄C
But {1,2}⊂{1,2,7,8} i.e. A⊂C
∴ if A⊄B and B⊄C, need not imply that A⊄C
(v) If x ∈ A and A ⊄ B , then x ∈ B
False
Let A={1,2}, B={3,4,5}
2∈{1,2} i.e. x∈A
{1,2}⊄{3,4,5} i.e. A⊄B
2∉{3,4,5} i.e. x∉B
(vi) If A ⊂ B and x ∉ B , then x ∉ A
True
Let A={1,2}, B={1,2,3,4}
{1,2}⊂{1,2,3,4} i.e. A⊂B
5∉{1,2,3,4} i.e. x∉B
5∉{1,2} i.e. x∉A
3. Let A, B, and C be the sets such that A∪B = A∪C and A∩B = A∩C. Show that B=C.
Let A, B, and C be the sets such that A∪B=A∪C and A∩B=A∩C
Let x∈B
⇒ x∈(A∪B)
⇒ x∈(A∪C) [∵(A∪B)=(A∪C)]
⇒ x∈A or x∈C
⇒ x∈A ⇒ x∈(A∩B)
⇒ x∈(A∩C) [∵(A∩B)=(A∩C)]
⇒ x∈C
⇒ x∈B and x∈C ⇒ B⊂C ...(i)
Let y∈C
⇒ y∈(A∪C)
⇒ y∈(A∪B) [∵(A∪B)=(A∪C)]
⇒ y∈A or y∈B
⇒ y∈A ⇒ y∈(A∩C)
⇒ y∈(A∩B) [∵(A∩B)=(A∩C)]
⇒ y∈B
⇒ y∈C and y∈B⇒ C⊂B ...(ii)
∴ From (i) and (ii), since B⊂C and C⊂B ⇒ B=C
4. Show that the following four conditions are equivalent :
(i) A⊂B (ii) A–B=φ (iii) A∪B=B (iv) A∩B=A
Conditions (i) and (ii) are equivalent
Let A-B=ϕ
There exists no x such that x∈A but x∉B
All elements of A are in B i.e. for every x∈A, x∈B ⇒A⊂B
Conditions (ii) and (iii) are equivalent
Let A-B=ϕ
There exists no x such that x∈A but x∉B
All elements of A are in B i.e. for every x∈A, x∈B ⇒A⊂B ⇒ A∪B=B
Conditions (iii) and (iv) are equivalent
A∪B=B ⇒All elements of A are in B ⇒All elements of A are common in A and B ⇒A∩B=A
5. Show that if A⊂B, then C–B⊂C–A.
x∈ (C-B) ⇒ x∈C but x∉B
⇒x∈C and x∉A [∵A⊂B]
⇒ x∈(C-A)
∴ C-B⊂C-A
6. Assume that P(A)=P(B). Show that A=B
Let x be an element of set A. Then there exist a subset of A, say X such that X={x}
Now, X⊂A ⇒ X∈P(A)
⇒ X∈P(B) [∵P(A)=P(B)]
⇒X⊂B ⇒ x∈B
Thus x∈A and x∈B
∴ A⊂B ...(i)
Let y be an element of set B. Then there exist a subset of B, say Y such that Y={y}
Now, Y⊂B ⇒ Y∈P(B)
⇒ Y∈P(A) [∵P(A)=P(B)]
⇒Y⊂A ⇒ y∈A
Thus y∈B and y∈A
∴B⊂A ...(ii)
From (i) and (ii), since A⊂B and B⊂A ⇒ A=B
7. Is it true that for any sets A and B, P(A)∪P(B)=P(A∪B)? Justify your answer.
False
Let A={1}, B={3}
A∪B={1,3}
P(A)={ϕ, {1}}
P(B)= {ϕ, {3}}
P(A)∪P(B) = {ϕ,{1},{3}}
P(A∪B)={ϕ, {1},{3},{1,3}}
∴ P(A)∪P(B) ≠ P(A∪B)
8. Show that for any sets A and B, A=(A∩B)∪(A–B) and A∪(B–A)=(A∪B)
A=(A∩B)∪(A–B)
(A∩B)∪(A-B) = (A∩B)∪(A∩B') [∵A-B=A∩B']
= A∩(B∪B')
=A∩U [∵(B∪B')=U]
=A
A∪(B–A)=(A∪B)
A∪(B-A) = A∪(B∩A') [∵(B-A)=B∩A']
=(A∪B)∩(A∪A')
=(A∪B)∩U
=(A∪B)
9. Using properties of sets, show that
(i) A ∪ ( A ∩ B ) = A
A∪(A∩B) = (A∪A)∩(A∪B) [Distributive law]
=A∩(A∪B) [∵A∪A = A]
=A [∵A⊂(A∪B)]
(ii) A ∩ ( A ∪ B ) = A.
A∩(A∪B)=(A∩A)∪(A∩B) [Distributive law]
=A∪(A∩B) [∵A∩A = A]
=A [∵A∩B⊂A]
10. Show that A∩B=A∩C need not imply B=C.
Let A={a,b}, B={a,c}, C={a,d}
A∩B={a}
A∩C={a}
∴ A∩B = A∩C
But still B ≠ C
11. Let A and B be sets. If A∩X=B∩X=φ and A∪X=B∪X for some set X, show that A=B.
(Hints A=A∩(A∪X), B=B∩(B∪X) and use Distributive law )
A=A∩(A∪X) = A∩(B∪X) [∵ A∪X=B∪X]
A=(A∩B)∪(A∩X)
A=(A∩B)∪ϕ [∵(A∩X)=ϕ]
A= (A∩B) ⇒ A⊂B ...(i)
B=B∩(B∪X) = B∩(A∪X) [∵ A∪X=B∪X]
B=(B∩A)∪(B∩X)
B=(B∩A)∪ϕ [∵(B∩X)=ϕ]
B= (B∩A) ⇒ B⊂A ...(ii)
From (i) and (ii), since A⊂B and B⊂A ⇒ A=B
12. Find sets A, B, C such that A∩B, B∩C and A∩C are non-empty sets and A∩B∩C=φ.
Let A={a,b}, B={b,c}, C={a,c}
A∩B={b} ≠ ϕ
B∩C={c} ≠ ϕ
A∩C={a} ≠ ϕ
(A∩B)∩C = {b}∩{a,c} = ϕ
13. In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?
Let C and T be the set of students taking coffee and tea respectively
It is given that,
Total number of students = 600
n(C)=225, n(T)=150, n(T∩C)=100
n(T∪C) = n(T) + n(C) - n(T∩C)
n(T∪C)= 150 + 225 - 100 = 275
Number of students who were taking neither tea nor coffee = 600 - n(T∪C) = 600 -275 = 325
14. In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?
Let H and E be set of students knowing Hindi and English respectively.
It is given that,
n(H) = 100, n(E) = 50 , n(H∩E) = 25
n(H∪E) = n(H) + n(E) - n(H∩E)
n(H∪E) = 100 + 50 - 25 = 125
15. In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find:
It is given that,
n(H) = 25, n(T) = 26, n(I) = 26, n(H∩I) = 9, n(H∩T) = 11, n(T∩I) = 8, n(H∩T∩I) = 3
(i) the number of people who read at least one of the newspapers.
n(H∪T∪I) = n(H) + n(T) + n(I) - n(H∩I) - n(H∩T) - n(T∩I) + n(H∩T∩I)
n(H∪T∪I) = 25+26+26-9-11-8+3 = 52
(ii) the number of people who read exactly one newspaper.
n(H∪T∪I) - (8+3+6+5) = 52 - 22 = 30
16. In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.
It is given that,
n(A) = 21, n(B) = 26, n(C) = 29, n(A∩B) =14, n(A∩C) =12, n(B∩C) =14, n(A∩B∩C)=8
People who liked C only = 29-(4+8+6) = 11
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