Mathematics For Class IX Science - Unit 01 - SETS - Exercise 1.2

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Unit 01: SETS
EXERCISE 1.2

Q.1: Given that the sets A = {f, a, c, e} and B = {e, g, d, f } are subsets of the Universal Set U = {a, b, c, d, e, f, g}, list the elements of:
(1) A' (2) B' (3) A∩B (4) (AUB)' (5) A∩B' (6) A'∩B' (7) UUØ (8) U∩Ø

(1) A'
Solution:
A' = U - A
= {a,b,c,d,e,f,g} - {f,a,c,e} = {b,d,g}
∴ A' = {b,d,g} Ans.

(2) B'
Solution: B' = U - B'
= {a,b,c,d,e,f} - {e,g,d,f} = {a,b,c}
∴ B' = {a,b,c} Ans.

(3) A∩B
Solution:
A∩B = {f,a,c,e} ∩ {e,g,d,f}
∴ A∩B = {e,f} Ans.

(4) (AUB)'
Solution:
AUB = {f,a,c,e} U {e,g,d,f}
AUB = {a,c,d,e,f,g}
(AUB)' = U - (AUB) = {a,b,c,d,e,f,g} - {a,c,d,e,f,g}
∴ (AUB)' = {b} Ans.

(5) A∩B'
Solution:
B' = U - B = {a,b,c,d,e,f,g} - {e,g.d,f}
B' = {a,b,c}
A∩B' = {f,a,c,e} ∩ {a,b,c} ∴ A∩B' = {a,c } Ans.

(6) A'∩B'
Solution:
A' = U - A = { a,b,c,d,e,f,g) - {f,a,c,e}
A' {b,d,g}
B' = U - B = {a,b,c,d,e,f,g} - {e,g,d,f}
B' {a,b,c}
A'∩B' = {b, d, g} ∩ {a,b,c}
∴ A'∩B' = {b} Ans.

(7) UUØ
Solution:
UUØ = { a,b,c,d,e,f,g} U { } = { a,b,c,d,e,f,g }
∴ UUØ = { a,b,c,d,e,f,g } = U Ans.

(8) U∩Ø
Solution:
U∩Ø = {a,b,c,d,e,f,g} ∩ { }
∴ U∩Ø = { } Ans.

Q.2: Given the sets:
A = (x | x is positive even integer less than 10} and B = {x | x is positive odd integer less than 10} are subsets of the universal set U = {x | x is a positive integer less than 10} list the elements of:
(9) AUB' (10) A'∩B (11) A'∩B' (12) AΔB (13) A - B' (14) A'ΔB (15) (A'∩B)'

Solution:
In tabular form
A = {2,4,6,8}
B = {1,3,5,7,9}
U = {1,2,3,4,5,6,7,8,9}

(9) AUB'
Solution: A = {2,4,6,8}
B = {1,3,5,7,9}
U = {1,2,3,4,5,6,7,8,9}
B' = U-B = { 1,2,3,4,5,6,7,8,9} - (1,3,5,7,9}
B' = {2,4,6,8}
Now
AUB' =. {2.4,6,8} U {2,4,6,8}
∴ AUB' = {2,4,6,8} Ans.

(10) A'∩B
Solution:
A' = U - A = {1,2,3,4,5,6,7,8,9} - {2,4,6,8}
A' = (1,3,5,7,9)
A'∩B = (1,3,5,7,9) ∩ {1,3,5,7.9}
∴ A'∩B = { } Ans.

(11) A'∩B'
Solution:
A' = U - A = {1,2,3,4,5,6,7,8,9} - {2,4,6,8}
A' = {1,3,5,7,9}
B' = U - B = {1,2,3,4,5,6,7,8,9} - (1,3,5,7,9)
B' = (2,4,6,8)
A'∩B' = {1,3,5,7,9} ∩ {2,4,6.8}
∴ A'∩B' = { } Ans.

(12) AΔB
Solution:
As AΔB = AUB - A∩B
AUB = {2,4,6,8} U {1,3,5,7,9}
AUB = {1,2,3,4,5,6,7,8,9}
A∩B = {2,4,6,8} ∩ {1,3,5,7,9}
A∩B = { }
Now
AΔB = AUB - A∩B
AΔB = {1,2,3,4,5,6,7,8,9} - { }
∴ AΔB = {1,2,3,4,5,6,7,8,9} Ans.

(13) A - B'
Solution:
B' = U - B = { 1,2,3,4,5,6,7,8,9} - {1,3,5,7,9}
B' = {2,4,6,8}
A - B' = {2,4,6,8) - {2,4,6,8}
∴ A - B' = { } Ans.

(14) A'ΔB
Solution:
A' = U - A = {1,2,3,4,5,6,7,8,9} - {2,4,6,8}
A' = {1,3,5,7,9}
As A'ΔB = A'UB - A'∩B
A'UB = {1,3,5,7,9} U {1,3,5,7,9}
A'UB = {1,3,5,7,9}
A'∩B = {1,3,5,7,9} ∩ {1,3,5,7,9}
A'∩B = {1,3,5,7,9}
Now
A'ΔB = A'UB - A'∩B
A'ΔB = (1,3,5,7,9) - {1,3,5,7,9}
∴ A'ΔB = { }

(15) (A'∩B)'
Solution:
A' = U - A = { 1,2,3,4,5,6,7,8,9} - {2,4,6,8}
A' = {1,3,5,7,9}
As (A'∩B)' = U - (A'∩B)
A'∩B = { 1,3,5,7,9} ∩ {1,3,5,7,9}
A'∩B = {1,3,5,7,9}
Now
(A'∩B)' = U - (A'∩B)= {1,2,3,4,5,6,7,8,9} - {1,3,5,7,9}
∴ (A'∩B)' = {2,4,6,8} Ans.

(16) Draw the Venn diagrams for the sets in Questions # 9, 10, 11, 12, 13, 14 and 15.

(9) Venn diagrams AUB'

(10) Venn diagrams A'∩B

(11) Venn diagrams A'∩B'

(12) Venn diagrams AΔB

(13) Venn diagrams A - B'

(14) Venn diagrams A'ΔB

(15) Venn diagrams (A'∩B)'

If A = {1, 2, 3, 4} and B = {2, 4, 6, 8}, show that:
(17) AΔB = (AUB) - (A∩B)
(18) AΔB = (A - B) U (B - A)
(19) A - B = A - (A∩B)

(17) AΔB = (AUB) - (A∩B)
Solution:
Taking L.H.S:
AΔB = (1,2,3,4) Δ (2,4,6,8)
(By selecting uncommon members among both sets)
We get AΔB = {1,3,6,8} ....... (1)
Taking R.H.S:
AUB = (1,2,3,4) U (2,4,6,8)
AUB = (1,2,3,4,6,8)
A∩B = {1,2,3;4} ∩ {2,4,6,8)
A∩B = (2,4)
AUB - A∩B = (1,2,3,4,6,8) - {2,4}
We get AUB - A∩B = (1,3,6,8) ....... (2)
By (1) and (2)

AΔB = AUB - A∩B Hence Proved

(18) AΔB = (A - B) U (B - A)
Solution:
Taking L.H.S:
AΔB = {1,2,3,4} Δ {2,4,6,8}
AΔB = (1,3,6.8) ....... (1)
Taking R.H.S:
(A - B) U (B - A)
A - B = (1,2,3,4) - (2,4,6,8)
A - B = (1,3)
B - A = (2,4,6,8) - (1,2,3,4)
B - A = (6,8}
(A - B) U (B - A) = (1,3) - (6,8) = (1,3,6,8) ......... (2)
By (1) and (2)

AΔB = (A - B) U (B - A) Hence Proved. 

(19) A - B = A - (A∩B)
Solution:
Taking L.H.S:
A - B = (1,2,3,4) - (2,4,6,8)
A - B = (1,3) ....... (1)
Taking R.H.S:
A - (A∩B) = {1,2,3,4} - {(2,4,6,8) ∩ (2,4)}
A - (A∩B) = (1,2,3,4) - {2,4}
A - (A∩B) = {1,3} ....... (2)
By (1) and (2)

A - B = A - (A∩B Hence Proved.

Example No.1:
Let U = {1, 2, 3, 4, 5, 6, 7}, A = {1, 3, 5, 7} and B = {3, 4, 5, 6} Verify De Morgan's Laws.

Solution:
According to De Morgan's Laws
1. (AUB)' = A'∩B'
2. A'UB' = (A∩B)'

1. (AUB)' = A'∩B'
Taking L.H.S:
(AUB)' = U - AUB
AUB = {1, 3, 5, 7} U {3, 4, 5, 6}
AUB = {1,3,4,5,6,7}
U - AUB = {1, 2, 3, 4, 5, 6, 7} - {1,3,4,5,6,7}
∴ (AUB)' = {2} ..... (1)
Taking R.H.S:
A'∩B'
A' = U - A
A' = {1, 2, 3, 4, 5, 6, 7} - {1, 3, 5, 7}
A' = {2,4,6}
B' = U - B
B' = {1, 2, 3, 4, 5, 6, 7} - {3, 4, 5, 6}
B' = {1,2,7}
A'∩B' = {2,4,6} ∩ {1.2.7}
∴ A'∩B' = {2} .... (2)
By (1) and (2)

(AUB)' = A'∩B'

Hence De Morgan's Laws has proved

2. A'UB' = (A∩B)'
Taking L.H.S
A'UB'
A' = U - A
A' = {1, 2, 3, 4, 5, 6, 7} - {1, 3, 5, 7}
A' = {2,4,6}
B' = U - B
B' = {1, 2, 3, 4, 5, 6, 7} - {3, 4, 5, 6}
B' = {1,2,7}
A'UB' = {2,4,6} U {1,2,7}
∴ A'UB' = {1,2,4,6,7} ....... (1)
Taking R.H.S:
(A∩B)'
A∩B = {1, 3, 5, 7} ∩ {3, 4, 5, 6}
A∩B = {3,5}
(A∩B)' = U - (A∩B)
U - (A∩B) = {1, 2, 3, 4, 5, 6, 7} - {3.5}
∴ (A∩B)' = {1,2,4,6,7} ...... (2)
By (1) and (2)

A'UB' = (A∩B)'

Hence De Morgan's Law has proved

(20) If U = {1, 2, 3, . 20}, A = {1, 2, 4, 8, 10, 16, 20} and B = {2, 6, 8, 10, 14, 18}, verify De Morgan's Laws.

Solution:
According to De Morgan's Laws
1. (AUB)' = A'∩B'
2. A'UB' = (A∩B)'

1. (AUB)' = A'∩B'
Solution:
For L.H.S:
(AUB)' = U - AUB
AUB = {1,2,4,8,10,16,20} U {2,6,8,10,14,18)
AUB = {1,2,4,8,10,14,16,18,20)
U-(AUB)= (1,2,3 ....... 20} - { 1,2,4,8,10,14,16,18,20}
∴ (AUB)' = (5,6,7,9,11,12,13,14,15,17,19) ........ (1)
For R.H.S:
A'∩B'
A' = U - A = (1,2,3,4,5 ........ 20) - (1,2,4,8,10,16,20)
A' = {3,5,6,7,9,11,12,13,14,15,17,18,19}
B' = U - B = (1,2,3,4,5 ...... 20) - {2,4,6,8,10,14,18}
B' = {1,3,5,7,9,11,12,13,15,16,17,19}
A'∩B' = (3,5,6,7,9,11,12,13,14,15,17,18,19) ∩ (1,3,5,7,9,11,12,13,15,16,17,19)
A'∩B' = (5,6,7,9,11,12,13,14,15,17,19) ....... (2)
By (1) and (2)

(AUB)' = A'∩B'

Hence De Morgan's Laws has proved

2. A'UB' = (A∩B)'
Solution:
For L.H.S:
A'UB'
A' = U - A = (1,2,3,4,5 ........ 20) - (1,2,4,8,10,16,20)
A' = {3,5,6,7,9,11,12,13,14,15,17,18,19}
B' = U - B = (1,2,3,4,5 ...... 20) - {2,4,6,8,10,14,18}
B' = {1,3,5,7,9,11,12,13,15,16,17,19,20}
A'UB' = (3, 5, 6, 7, 9, 11, 12, 13, 14, 15, 17, 18, 19) U (1, 3, 5, 7, 9, 11, 12, 13, 15, 16, 17, 19,20}
∴ A'UB' = {1, 3, 5, 6, 7, 9, 11, 12, 13,1 4, 15, 16, 17, 18, 19, 20) ........ (1)
FOR R.H.S
(A∩B)'
A∩B = (1,2,4, 8, 10, 16, 20) ∩ {2, 4, 6, 8, 10, 14, 18}
A∩B = (2,4,8,10)
(A∩B)' = U- (A∩B)
U- (A∩B) = {1, 2, 3 ......... 20} - {2, 4, 8, 10}
∴ (A∩B)' = (1, 3, 5, 6, 7, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20) ....... (2)
By (1) and (2)

A'UB' = (A∩B)'

Hence De Morgan's Laws has proved

(21) Verify the commutative property of union and intersection for the following sets.
(a) A = {1, 2, 3, 4, 5}, B = {3, 5, 7, 9}
(b) A=(x|x ∈ Z+ and x ≤ 5) B = {x|x ∈ Z and l ≤ x ≤ 4}

(a) A = {1, 2, 3, 4, 5}, B = {3, 5, 7, 9}
Solution:
Commutative property of Union is

AUB = BUA

Taking L.H.S:
AUB = {1,2,3,4,5) U {3,5,7,9)
AUB = (1,2,3,4,5,7,9) ........ (1)
Taking R.H.S:
BUA = (3,5,7,9) U (1,2,3,4,5)
BUA = (1,2,3,4,5,7,9) .......... (2)
By (1) and (2)

AUB = BUA Hence Proved

Commutative property for intersection is

A∩B = B∩A

Taking L.H.S:
A∩B = {1,2,3,4,5} ∩ {3,5,7,9}
A∩B = (3,5) .......... (1)
Taking R.H.S:
B∩A = (3,5,7,9) ∩ (1,2,3,4,5)
B∩A = {3,5} ........ (2)
By (1) and (2)

A∩B = B∩A Hence Proved.

(b) A = (x|x ∈ Z+ and x ≤ 5) B = {x|x ∈ Z and l ≤ x ≤ 4}
Solution:
In tabular form:
A = {1,2,3,4,5}
B = {1,2,3,4}
Commutative property of union:

AUB = BUA

Taking L.H.S:
AUB = ( 1,2,3,4,5) U (1,2,3,4)
AUB = { 1,2,3,4,5) ....... (1)
Taking R.H.S:
BUA = (1,2,3,4) U (1,2,3,4,5)
BUA = (1,2,3,4,5) ........ (2)
By (1) and (2)

AUB = BUA Hence Proved.

Commutative property of intersection:

A∩B = B∩A

Taking L.H.S:
A∩B = {1,2,3,4,5) ∩ ( 1,2,3,4)
A∩B = {1,2,3,4) ....... (1)
Taking R.H.S:
A∩B = (1,2,3,4) ∩ ( 1,2,3,4,5)
A∩B = (1,2,3,4) .......... (2)
By (1) and (2)

A∩B = B∩A Hence proved

(22) Verify the following properties for the sets given below:
(i) Associative property of union and of intersection.
(ii) Distributive property of union over intersection.
(iii) Distributive property of intersection over union.
(a) A = {1, 2, 3, 4, 5}, B = {2, 4, 6, 8}, C = {4, 8, 10, 12,}
(b) A = {x|x ∈ Z+ and x ≤ 4}, B = {x|x ∈ Z and 0 < x < 5}, C = {1,2,3}

(a) A = (1,2,3,4,5), B = {2,4,6,8}; c = {4,8,10,12)
Solution:
(i) Associative property of union and of intersection:
Associative property of union

 AU(BUC) = (AUB) UC

Taking L.H.S:
AU(BUC)
BUC = (2,4,6,8) U (4,8,10,12)
BUC = (2,4,6,8,10,12)
AU(BUC) = (1,2,3,4,5) U (2,4,6,8,10,12)
AU(BUC) = (1,2,3,4,5,6,8,10,12) ........ (1)
Taking R.H.S:
(AUB) UC
AUB = (1,2,3,4,5) U (2,4,6,8)
AUB = (1,2,3,4,5,6,8)
(AUB)UC = (1,2,3,4,5,6,8) U (4,8,10,12)
(AUB)UC = (1,2,3,4,5,6,8,10,12) ......... (2)
By (1) and (2)

AU(BUC) = (AUB) UC Hence Proved

Associative property of intersection

A∩(B∩C) = (A∩B) ∩ C

Taking L.H.S:
A∩(B∩C)
B∩C = (2,4,6,8) ∩ (4,8,10,12)
B∩C = (4,8)
A∩(B∩C) = (1,2,3,4,5) ∩ (4,8)
A∩(B∩C) = (4) ........ (1)
Taking R.H.S:
(A∩B) ∩ C
A∩B = (1,2,3,4,5) ∩ (2,4,6,8)
A∩B = (2,4)
(A∩B)∩C = (2,4) ∩ (4,8,10,12)
(A∩B)∩C = (4) ......... (2)
By (1) and (2)

A∩(B∩C) = (A∩B) ∩ C Hence Proved

(ii) Distributive property of union over intersection:

AU(B∩C) = (AUB) ∩ (AUC)

Taking L.H.S:
AU(B∩C)
B∩C = (2,4,6,8) ∩ (4,8,10,12)
B∩C = (4,8)
AU(B∩C)= '{1,2,3,4,5} U {4,8}
AU(B∩C) = {1,2,3,4,5,8} ........ (1)
Taking R.H.S
(AUB) ∩ (AUC)
AUB = {1,2,3,4,5) U {2,4,6,8}
AUB = {1,2,3,4,5,6,8}
AUC = {1,2,3,4,5} U {4,8,10,12}
AUC = {1,2,3,4,5,8,10,12}
(AUB) ∩ (AUC) = {1,2,3,4,5,6,8} ∩ (1,2,3,4,5,8,10,12)
(AUB) ∩ (AUC) = {1,2,3,4,5,8) ....... (2)
By (1) and (2)

AU(B∩C) = (AUB) ∩ (AUC) Hence Proved

(iii) Distributive property of intersection over union:

A∩(BUC) = (A∩B) U (A∩C)

Taking L.H.S:
A∩(BUC)
BUC = {2,4,6,8} U {4,8,10,12}
BUC = {2,4,6,8,10,12}
A∩(BUC) = (1,2,3,4,5} ∩ {2,4,6,8,10,12}
A∩(BUC) = {2,4} ....... (1)
Taking R.H.S:
(A∩B) U (A∩C)
A∩B = {1,2,3,4,5} ∩ {2,4,6,8}
A∩B = {2,4)
A∩C = {1,2,3,4,5} ∩ {4,8,10,12}
A∩C = {4}
(A∩B) U (A∩C) = {2,4} U {4}
(A∩B) U (A∩C) = {2,4} ....... (2)
By (1) and (2)

A∩(BUC) = (A∩B) U (A∩C) Hence Proved.

(b) A = {x|x ∈ Z+ and x ≤ 4}, B = {x|x ∈ Z and 0 < x < 5}, C = {1,2,3}
Solution:
In Tabular Form:
A = {1,2,3,4}
B = {1,2,3,4}
C = {1,2,3}

(i) Associative property of union and of intersection:
Associative property of union

AU(BUC) = (AUB) UC

Taking L.H.S:
AU(BUC)
BUC = {1,2,3,4} U {1,2,3}
BUC = (1,2,3,4)
AU(BUC) = {1,2,3,4} U (1,2,3,4)
AU(BUC) = (1,2,3,4) ........ (1)
Taking R.H.S:
(AUB) UC
AUB = (1,2,3,4) U (1,2,3,4)
AUB = (1,2,3,4)
(AUB)UC = (1,2,3,4) U {1,2,3}
(AUB)UC = {1,2,3,4} ......... (2)
By (1) and (2)

AU(BUC) = (AUB) UC Hence Proved

Associative property of intersection

A∩(B∩C) = (A∩B) ∩ C

Taking L.H.S:
A∩(B∩C)
B∩C = {1,2,3,4} ∩ (1,2,3)
B∩C = (1,2,3)
A∩(B∩C) = (1,2,3,4) ∩ (1,2,3)
A∩(B∩C) = (1,2,3) ........ (1)
Taking R.H.S:
(A∩B) ∩ C
A∩B = (1,2,3,4) ∩ (1,2,3,4)
A∩B = (1,2,3,4)
(A∩B)∩C = (1,2,3,4) ∩ (1,2,3)
(A∩B)∩C = (1,2,3) ......... (2)
By (1) and (2)

A∩(B∩C) = (A∩B) ∩ C Hence Proved

(ii) Distributive property of union over intersection:

AU(B∩C) = (AUB) ∩ (AUC)

Taking L.H.S:
AU(B∩C)
B∩C = (1,2,3,4) ∩ (1,2,3)
B∩C = (1,2,3)
AU(B∩C)= '{1,2,3,4} U {1,2,3}
AU(B∩C) = {1,2,3,4} ........ (1)
Taking R.H.S
(AUB) ∩ (AUC)
AUB = {1,2,3,4) U {1,2,3,4}
AUB = {1,2,3,4}
AUC = {1,2,3,4} U {1,2,3}
AUC = {1,2,3,4}
(AUB) ∩ (AUC) = {1,2,3,4} ∩ (1,2,3,4)
(AUB) ∩ (AUC) = {1,2,3,4) ....... (2)
By (1) and (2)

AU(B∩C) = (AUB) ∩ (AUC) Hence Proved

(iii) Distributive property of intersection over union:

A∩(BUC) = (A∩B) U (A∩C)

Taking L.H.S:
A∩(BUC)
BUC = {1,2,3,4} U {1,2,3}
BUC = {1,2,3,4)
A∩(BUC) = (1,2,3,4} ∩ {1,2,3,4}
A∩(BUC) = {1,2,3,4} ....... (1)
Taking R.H.S:
(A∩B) U (A∩C)
A∩B = {1,2,3,4} ∩ {1,2,3,4}
A∩B = {1,2,3,4}
A∩C = {1,2,3,4} ∩ {1,2,3}
A∩C = {1,2,3}
(A∩B) U (A∩C) = {1,2,3,4} U {1,2,3}
(A∩B) U (A∩C) = {1,2,3,4} ....... (2)
By (1) and (2)

A∩(BUC) = (A∩B) U (A∩C) Hence Proved.