Theorem 25.1- One and only one circle can pass through three non-collinear points

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Unit 25: Chords Of A Circle

Theorem 25.1:
One and only one circle can pass through three non-collinear points.

Given:
Three non-collinear points. say A, B and C.

To Prove:
One and only one circle can pass through A, B and C.

Construction:
Draw line segments AB and BC. Draw right bisectors ^<_ED^> and ^<_GF^> of AB and BC, respectively. ^<_ED^> and ^<_GF^> intersect at a point, say O. Draw OA,  OB and OC.

Proof:

Statemnets	Reasons
All points on <ED> are equidistant from A and B, so mOA = mOB ......... (i)	<ED> is the right bisector of AB, and O is a point on <ED>.
All points on <GF> are equidistant from B and C, so mOB = mOC ......... (ii)	<GF> is the right bisector of BC, and <GF> passess though O.
O is the unique point of intersection of <ED> and <GF> ....... (iii)	<ED> and <GF> are non-parallel lines.
The paint O is equidistant from A, B and C, i.e.  mOA = mOB = mOC = r, say ...... (iv)	From (i) and (ii)  Transitive property
The circle with centre only at O and radius r passes through A, B and C ...... (v)	OA, OB and OC are radial segment, and by (iii)
A, B and C are non collinear ..... (vi)	Given
Therefore, there exists one and only one From circle centered at O and with radius r passing through non-collinear points A, B and C.	From (iii), (iv), (v) and (vi)

Q.E.D

Text Book - Page No. 196 and 197.

PROPERTIES OF UNION AND INTERSECTION - Unit 17 - Solved Exercise 17.3 - Mathematics For Class X Science

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Unit 17: SETS AND FUNCTIONS
Solved Exercise 17.3

1. Verify the commutative property of union and intersection for the following sets.

1. A = {a,b,c,d,e} and B = {a,e,i,o,u}
2. P = {x | x ∈ Z ⋀ -3 < x< 3} and Q = {y | y ∈ E⁺ ⋀ y ≤ 4}

(i) A = {a,b,c,d,e} and B = {a,e,i,o,u}
SOLUTION:
VERIFICATION OF COMMUTATIVE PROPERTY OF UNION

∵ AUB = BUA

Proof:
Taking L.H.S
AUB = {a,b,c,d,e} U {a,e,i,o,u}
AUB = {a,b,c,d,e,i,o,u}

Taking R.H.S
BUA = {a,e,i,o,u} U {a,b,c,d,e}
BUA = {a,b,c,d,e,i,o,u}
∵ AUB = BUA
∴ L.H.S = R.H.S
Hence Proved

VERIFICATION OF COMMUTATIVE PROPERTY OF INTERSECTION

∵ A∩B = B∩A

Proof:
Taking L.H.S
A∩B = {a,b,c,d,e} ∩ {a,e,i,o,u}
A∩B = {a,e}

Taking R.H.S
B∩A = {a,e,i,o,u} ∩ {a,b,c,d,e}
B∩A = {a,e}
∵ A∩B = B∩A
∴ L.H.S = R.H.S
Hence Proved

(ii) P = {x | x ∈ Z ⋀ -3 < x< 3} and Q = {y | y ∈ E⁺ ⋀ y ≤ 4}
SOLUTION:
P = {x | x ∈ Z ⋀ -3 < x< 3} = {-2, -1, 0, 1, 2}
Q = {y | y ∈ E⁺ ⋀ y ≤ 4} = {2, 4}

VERIFICATION OF COMMUTATIVE PROPERTY OF UNION

∵ PUQ = QUP

Proof:
Taking L.H.S
PUQ = {-2,-1,0,1,2} U {2,4}
PUQ = {-2, -1, 0, 1, 2, 4}

Taking R.H.S
QUP = {2,4} U {-2 -1,0,1,2}
QUP = {-2, -1, 0, 1, 2, 4}
∵ PUQ = QUP
∴ L.H.S = R.H.S
Hence Proved

VERIFICATION OF COMMUTATIVE PROPERTY OF INTERSECTION

∵ P∩Q = Q∩P

Proof:
Taking L.H.S
P∩Q = {-2,-1,0,1,2} ∩ {2,4}
P∩Q = {2}

Taking R.H.S
Q∩P = {2,4} ∩ {-2,-1,0,1,2}
Q∩P = {2}
∵ P∩Q = Q∩P
∴ L.H.S = R.H.S
Hence Proved

2. Verify the associative property of union and intersection for the following sets.

1. A = {1,2,4,5,10,20}, B = {5,10,15,20} and C = {1,2,5,10}
2. A = N, B = P and C = Z

SOLUTION:
(i) A = {1,2,4,5,10,20}, B = {5,10,15,20} and C = {1,2,5,10}
VERIFICATION OF ASSOCIATIVE PROPERTY OF UNION

∵ AU(BUC) = (AUB)UC

Proof:
Taking L.H.S
AU(BUC) = {1,2,4,5,10,20} U [{5,10,15,20} U {1,2,5,10}]
AU(BUC) = {1,2,4,5,10,20} U {1,2,5,10,15,20}
AU(BUC) = {1,2,4,5,10,15,20}

Taking R.H.S
(AUB)UC = [{1,2,4,5,10,20} U {5,10,15,20}] U {1,2,5,10}
(AUB)UC = {1,2,4,5,10,15,20} U {1,2,5,10}
(AUB)UC = {1,2,4,5,10,15,20}
∵ AU(BUC) = (AUB)UC
∴ L.H.S = R.H.S
Hence Proved

VERIFICATION OF ASSOCIATIVE PROPERTY OF INTERSECTION

∵ A∩(B∩C) = (A∩B)∩C

Proof:
Taking L.H.S
A∩(B∩C) = {1,2,4,5,10,20} ∩ [{5,10,15,20} ∩ {1,2,5,10}]
A∩(B∩C) = {1,2,4,5,10,20} ∩ {5,10}
A∩(B∩C) = {5,10}

Taking R.H.S
(A∩B)∩C = [{1,2,4,5,10,20} ∩ {5,10,15,20}] ∩ {1,2,5,10}
(A∩B)∩C = {5,10,20} ∩ {1,2,5,10}
(A∩B)∩C = {5,10}
∵ A∩(B∩C) = (A∩B)∩C
∴ L.H.S = R.H.S
Hence Proved

(ii) A = N, B = P and C = Z
SOLUTION:
A = N = {1, 2, 3, 4, 5, ......}
B = P = {2, 3, 5, 7, 11, .....}
C = Z = {0, ±1, ±2, ±3, ±4, .....}

VERIFICATION OF ASSOCIATIVE PROPERTY OF UNION

∵ AU(BUC) = (AUB)UC

Proof:
Taking L.H.S
AU(BUC) = {1, 2, 3, 4, 5, ......} U [{2, 3, 5, 7, 11, .....} U {0, ±1, ±2, ±3, ±4, .....}]
AU(BUC) = {1,2,4,5,10,20} U {0, ±1, ±2, ±3, ±4, .....}
AU(BUC) = {0, ±1, ±2, ±3, ±4, .....} = Z

Taking R.H.S
(AUB)UC = [{1, 2, 3, 4, 5, ......} U [{2, 3, 5, 7, 11, .....}] U {0, ±1, ±2, ±3, ±4, .....}
(AUB)UC = {1, 2, 3, 4, 5, ......} U {0, ±1, ±2, ±3, ±4, .....}
(AUB)UC = {0, ±1, ±2, ±3, ±4, .....} = Z
∵ AU(BUC) = (AUB)UC
∴ L.H.S = R.H.S
Hence Proved

VERIFICATION OF ASSOCIATIVE PROPERTY OF INTERSECTION

∵ A∩(B∩C) = (A∩B)∩C

Proof:
Taking L.H.S
A∩(B∩C) = {1, 2, 3, 4, 5, ......} ∩ [{2, 3, 5, 7, 11, .....} ∩ {0, ±1, ±2, ±3, ±4, .....}]
A∩(B∩C) = {1, 2, 3, 4, 5, ......} ∩ {2, 3, 5, 7, 11, .....}
A∩(B∩C) = {2, 3, 5, 7, 11, .....} = P

Taking R.H.S
(A∩B)∩C = [{1, 2, 3, 4, 5, ......} ∩ {2, 3, 5, 7, 11, .....}] ∩ {0, ±1, ±2, ±3, ±4, .....}
(A∩B)∩C = {2, 3, 5, 7, 11, .....} ∩ {0, ±1, ±2, ±3, ±4, .....}
(A∩B)∩C = {2, 3, 5, 7, 11, .....} = P
∵ A∩(B∩C) = (A∩B)∩C
∴ L.H.S = R.H.S
Hence Proved

3. Verify
(a) Distributive property of union over intersection.
(b) Distributive property of intersection over union for the following sets.

1. A = {1,2,3,...,10}, B = {2,3,5,7} and C = {1,3,5,7,9}
2. A = N, B = P and C = W

(i) A = {1,2,3,...,10}, B = {2,3,5,7} and C = {1,3,5,7,9}
SOLUTION:
(a) VERIFICATION OF DISTRIBUTIVE PROPERTY OF UNION OVER INTERSECTION:

AU(B∩C) = (AUB) ∩ (AUC)

Proof:
Taking L.H.S
AU(B∩C) = {1,2,3,...,10} U [{2,3,5,7} ∩ {1,3,5,7,9}]
AU(B∩C) = {1,2,3,...,10} U {3,5,7}
AU(B∩C) = {1,2,3,...,10}

Taking R.H.S
(AUB) ∩ (AUC) = [{1,2,3,...,10} U {2,3,5,7}] ∩ [{1,2,3,...,10} U {1,3,5,7,9}]
(AUB) ∩ (AUC) = {1,2,3,...,10} ∩ {1,2,3,...,10}
(AUB) ∩ (AUC) = {1,2,3,...,10} ∵ AU(B∩C) = (AUB) ∩ (AUC)
∴ L.H.S = R.H.S
Hence Proved

(b) VERIFICATION OF DISTRIBUTIVE PROPERTY OF INTERSECTION OVER UNION:

A∩(BUC) = (A∩B) ∩ (A∩C)

Proof:
Taking L.H.S
A∩(BUC) = {1,2,3,...,10} ∩ [{2,3,5,7} U {1,3,5,7,9}]
A∩(BUC) = {1,2,3,...,10} ∩ {1,2,3,5,7,9}
A∩(BUC) = {1,2,3,5,7,9}

Taking R.H.S
(A∩B) U (A∩C) = [{1,2,3,...,10} ∩ {2,3,5,7}] U [{1,2,3,...,10} ∩ {1,3,5,7,9}]
(A∩B) U (A∩C) = {2,3,5,7} U {1,3,5,7,9}
(A∩B) U (A∩C) = {1,2,3,5,7,9}
∵ A∩(BUC) = (A∩B) U (A∩C)
∴ L.H.S = R.H.S
Hence Proved

(ii) A = N, B = P and C = W
SOLUTION:
A = N = {1,2,3,4,5,.......}
B = p = {2,3,5,7,11,......}
C = W = {0,1,2,3,4,.......}

(a) VERIFICATION OF DISTRIBUTIVE PROPERTY OF UNION OVER INTERSECTION:

AU(B∩C) = (AUB) ∩ (AUC)

Proof:
Taking L.H.S
AU(B∩C) = {1,2,3,4,5,.......} U [{2,3,5,7,11,......} ∩ {0,1,2,3,4,.......}]
AU(B∩C) = {1,2,3,4,5,.......} U {2,3,5,7,11,......}
AU(B∩C) = {1,2,3,4,5,.......} = N

Taking R.H.S
(AUB) ∩ (AUC) = [{1,2,3,4,5,.......} U {2,3,5,7,11,......}] ∩ [{1,2,3,4,5,.......} U {0,1,2,3,4,.......}]
(AUB) ∩ (AUC) = {1,2,3,4,5,.......} ∩ {0,1,2,3,4,.......}
(AUB) ∩ (AUC) = {1,2,3,4,5,.......} = N
∵ AU(B∩C) = (AUB) ∩ (AUC)
∴ L.H.S = R.H.S
Hence Proved

(b) VERIFICATION OF DISTRIBUTIVE PROPERTY OF INTERSECTION OVER UNION:

A∩(BUC) = (A∩B) U (A∩C)

Proof:
Taking L.H.S
A∩(BUC) = {1,2,3,4,5,.......} ∩ [{2,3,5,7,11,......} U {0,1,2,3,4,.......}]
A∩(BUC) = {1,2,3,4,5,.......} ∩ {0,1,2,3,4,.......}
A∩(BUC) = {1,2,3,4,5,.......}

Taking R.H.S
(A∩B) U (A∩C) = [{1,2,3,4,5,.......} ∩ {2,3,5,7,11,......}] U [{1,2,3,4,5,.......} ∩ {0,1,2,3,4,.......}]
(A∩B) U (A∩C) = {2,3,5,7,11,......} U {1,2,3,4,5,.......}
(A∩B) U (A∩C) = {1,2,3,4,5,.......} = N
∵ A∩(BUC) = (A∩B) U (A∩C)
∴ L.H.S = R.H.S
Hence Proved

4. Verify De Morgan's laws if U = {1,2,3...,12}, A = {1,2,3,4,6,12} and B = {2,4,6,8}.
SOLUTION:
According To De Morgan's Law

• (AUB)' = A'∩B'
• (A∩B)' = A'UB'

VERIFICATION OF DE MORGAN"S LAW:
(i) (AUB)' = A'∩B'
Proof:
Taking L.H.S
(AUB)' = U - (AUB)
(AUB)' = {1,2,3,4,5,6,7,8,9,10,11,12} - [{1,2,3,4,6,12} U {2,4,6,8}]
(AUB)' = {1,2,3,4,5,6,7,8,9,10,11,12} - [1,2,3,4,6,8,12}
(AUB)' = {5,7,9,10,11}

Taking R.H.S
A'∩B' = (U - A) ∩ (U - B)
A'∩B' = [{1,2,3,4,5,6,7,8,9,10,11,12} - {1,2,3,4,6,12}] ∩ [{1,2,3,4,5,6,7,8,9,10,11,12} - {2,4,6,8}]
A'∩B' = {5,7,8,9,10,11} ∩ {1,3,5,7,9,10,11}
A'∩B' = {5,7,9,10,11}
∵ (AUB)' = A'∩B'
∴ L.H.S = R.H.S
Hence Proved

(ii) (A∩B)' = A'UB'
Proof:
Taking L.H.S
(A∩B)' = U - (A∩B)
(A∩B)' = {1,2,3,4,5,6,7,8,9,10,11,12} - [{1,2,3,4,6,12} ∩ {2,4,6,8}]
(A∩B)' = {1,2,3,4,5,6,7,8,9,10,11,12} - [2,4,6}
(A∩B)' = {1,3,5,7,8,9,10,11,12}

Taking R.H.S
A'UB' = (U - A) U (U - B)
A'UB' = [{1,2,3,4,5,6,7,8,9,10,11,12} - {1,2,3,4,6,12}] U [{1,2,3,4,5,6,7,8,9,10,11,12} - {2,4,6,8}] A'UB' = {5,7,8,9,10,11} U {1,3,5,7,9,10,11,12}
A'UB' = {1,3,5,7,8,9,10,11,12}
∵ (A∩B)' = A'UB'
∴ L.H.S = R.H.S
Hence Proved

5. If, A and B are subset of U then prove the following using properties.

1. AU(A∩B) = A∩(AUB)
2. AUB = AU(A'∩B)
3. B = (A∩B)U(A'∩B)
4. B = AU(A'∩B), if A⊆B

SOLUTION:
Let U = {1,2,3,4,5,6,7,8}
A = {1,2,3,4}
B = {2,4,6,8}

(i) AU(A∩B) = A∩(AUB)
Proof:
Taking L.H.S
AU(A∩B) = {1,2,3,4} U [{1,2,3,4} ∩ {2,4,6,8}
AU(A∩B) = {1,2,3,4} U {2,4}]
AU(A∩B) = {1,2,3,4}

Taking R.H.S
A∩(AUB) = {1,2,3,4} ∩ [{1,2,3,4} U {2,4,6,8}
A∩(AUB) = {1,2,3,4} ∩ {1,2,3,4,6,8}
A∩(AUB) = {1,2,3,4}
∵ AU(A∩B) = A∩(AUB)
∴ L.H.S = R.H.S
Hence Proved

(ii) AUB = AU(A'∩B)
Proof:
Taking L.H.S
AUB = {1,2,3,4} U {2,4,6,8}
AUB = {1,2,3,4,6,8}

Taking R.H.S
AU(A'∩B) = AU [(U - A) ∩ B] AU(A'∩B) = {1,2,3,4} U [({1,2,3,4,5,6,7,8} - {1,2,3,4}) ∩ {2,4,6,8}]
AU(A'∩B) = {1,2,3,4} U [{5,6,7,8} ∩ {2,4,6,8}]
AU(A'∩B) = {1,2,3,4} U {6,8}
AU(A'∩B) = {1,2,3,4,6,8}
∵ AUB = AU(A'∩B)
∴ L.H.S = R.H.S
Hence Proved

(iii) B = (A∩B)U(A'∩B)
Proof:
Taking L.H.S
B = {2,4,6,8}

Taking R.H.S
(A∩B)U(A'∩B) = [(A∩B)] U [(U - A) ∩ B]
(A∩B)U(A'∩B) = [{1,2,3,4} ∩ {2,4,6,8}] U [({1,2,3,4,5,6,7,8} - {1,2,3,4}) ∩ {2,4,6,8}]
(A∩B)U(A'∩B) = {2,4} U [{5,6,7,8} ∩ {2,4,6,8}
(A∩B)U(A'∩B) = {2,4} U {6,8}
(A∩B)U(A'∩B) = {2,4,6,8}
∵ B = (A∩B)U(A'∩B)
∴ L.H.S = R.H.S
Hence Proved

(iv) B = AU(A'∩B), if A⊆B
SOLUTION:
Let U = {1,2,3,4,5,6,7,8}
A = {2,4}
B = {2,4,6,8}
Proof:
Taking L.H.S
B = {2,4,6,8}

Taking R.H.S
AU(A'∩B) = A U [(U - A) ∩ B]
AU(A'∩B) = {2,4} U [({1,2,3,4,5,6,7,8} - {2,4}) ∩ {2,4,6,8}]
AU(A'∩B) = {2,4} U [{1,3,5,6,7,8} ∩ {2,4,6,8}]
AU(A'∩B) = {2,4} U {6,8}
AU(A'∩B) = {2,4,6,8}
∵ B = AU(A'∩B), if A⊆B
∴ L.H.S = R.H.S
Hence Proved

PROPERTIES OF UNION AND INTERSECTION - Unit 17 - Explanation Of Exercise 17.3 - Mathematics For Class X Science

GO TO INDEX
Unit 17: SETS AND FUNCTIONS
Explanation For Exercise 17.3

PROPERTIES OF UNION AND INTERSECTION

Give formal proof of the following fundamental properties of union and intersection of two or three sets.
Commutative Property Of Union:
We know that for any two sets A and B,
AUB = BUA
This property is called commutative property of union.

Proof:
L.H.S = AUB
= {x | x ∈ A or x ∈ B}   (By definition of union)
= {x | x ∈ B or x ∈ A}   ∵ Order of elements in a set is not preserved
= BUA (By definition of union)
= R.H.S
∵ L.H.S = R.H.S
∴ AUB = BUA
Hence proved

Commutative Property Of Intersection:
We know that for any two sets A and B,
A∩B = B∩A
This property is called commutative property of intersection.

Proof:
L.H.S = A∩B
= {x | x ∈ A and x ∈ B}   (By definition of intersection)
= {x | x ∈ B and x ∈ A}   ∵ Order of elements in a set is not preserved
= B∩A (By definition of intersection)
= R.H.S
∵ L.H.S = R.H.S
∴ A∩B = B∩A
Hence proved

Associative Property Of Union:
We are already familiar with associative property of union which is as follows.
For any three sets A, B and C,
AU(BUC) = (AUB)UC

Proof:
L.H.S = AU(BUC)
= {x | x ∈ A or x ∈ BUC}   (By definition of union)
= {x | x ∈ A or x ∈ B or x ∈ C}
= {x | x ∈ AUB or x ∈ C}   (By definition of union)
= (AUB)UC    (By definition of union)
= R.H.S
∵ L.H.S = R.H.S
∴ AU(BUC) = (AUB)UC
Hence proved

Associative Property Of Intersection:
We already know the associative property of intersection which states that:
For any three sets A, B and C,
A∩(B∩C) = (A∩B)∩C

Proof:
L.H.S = A∩(B∩C) = (A∩B)∩C
= {x | x ∈ A and x ∈ (B∩C)}   (By definition of intersection)
= {x | x ∈ A and x ∈ B and x ∈ C}
= {x | x ∈ A∩B and x ∈ C}   (By definition of intersection)
= (A∩B)∩C   (By definition of intersection)
= R.H.S
∵ L.H.S = R.H.S
∴ A∩(B∩C) = (A∩B)∩C
Hence proved

Distributive Property Of Union Over Intersection:
We have already studied distributive property of union over intersection which is as follows.
For any three sets A, B and C,
AU(B∩C) = (AUB)∩(AUC)

Proof:
L.H.S = AU(B∩C)
= {x | x ∈ A or x ∈ (B∩C)}   (By definition of union)
= {x | x ∈ A or (x ∈ B and x ∈ C)}   (By definition of intersection)
= {x | (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C)}
= {x | x ∈ AUB and x ∈ AUC}   (By definition of union)
= (AUB)∩(AUC)   (By definition of intersection)
= R.H.S
∵ L.H.S = R.H.S
∴ AU(B∩C)=(AUB)∩(AUC)
Hence proved

Distributive Property Of Intersection Over Union:
We are already familiar with distributive property of intersection over union which is as follow.
For any three sets A, B and C,
A∩(BUC)=(A∩B)U(A∩C)

Proof:
L.H.S = A∩(BUC)
= {x | x ∈ A and x ∈ (BUC)}   (By definition of intersection)
= {x | x ∈ A and (x ∈ B or x ∈ C)}   (By definition of union)
= {x | (x ∈ A and x ∈ B) or (x ∈ A and x ∈ C)}
= {x | x ∈ A∩B and x ∈ A∩C}   (By definition of intersection)
= (A∩B)U(A∩C)   (By definition of union)
= R.H.S
∵ L.H.S = R.H.S
∴ A∩(BUC)=(A∩B)U(A∩C)
Hence proved

De Morgan's Laws:
There are two De Morgan's Laws which are as under.
For any two sets A and B
(i) (AUB)' = A'∩B'
(ii) (A∩B)' = A'UB'

Proof:
(i) (AUB)' = A'∩B'
L.H.S = (AUB)'
= {x | x ∈ U and x ∉ AUB}   (By definition of complement)
= {x | x ∈ U and (x ∉ A and x ∉ B)}
= {x | (x ∈ U and x ∉ A) and (x ∈ U and x ∉ B)}
= {x | x ∈ A' and x ∈ B'}   (By definition of complement)
= A'∩B'   (By definition of intersection)
= R.H.S
∵ L.H.S = R.H.S
∴ (AUB)' = A'∩B'
Hence proved.

(ii) (A∩B)' = A'UB'

Proof:
L.H.S = (AUB)'
= {x | x ∈ U and x ∉ A∩B}   (By definition of complement)
= {x | x ∈ U and (x ∉ A or x ∉ B)}
= {x | (x ∈ U and x ∉ A) or (x ∈ U and x ∉ B)}
= {x | x ∈ A' and x ∈ B'}   (By definition of complement)
= A'UB'   (By definition of union)
= R.H.S
∵ L.H.S = R.H.S
∴ (A∩B)' = A'UB''
Hence proved.

VERIFY THE FUNDAMENTAL PROPERTIES OF GIVEN SETS:

Let us verify the fundamental properties with the help of the following examples.
Example 1:
If A = {1,2,3,4,6,12} and B 4,6,8,9,10,12) then verify commutative property of union and intersection.

OPERATIONS ON SETS - Unit 17 - Solved Exercise 17.2 - Mathematics For Class X Science

GO TO INDEX
Unit 17: SETS AND FUNCTIONS
Solved Exercise 17.2

1. Which of the following sets are disjoint, overlapping, exhaustive and cells.
(i) {1,2,3,5,7} and {4,6,8,9,10}
(ii) {1,2,3,6} and {1,2,4,8}
(iii) E and 0 when U = Z (U denotes universal set)
(iv) N and W whenU= {x|x ∈ Z ⋀ x ≥ 0}
(v) Q and Q' when U =R
SOLUTION:
(i) {1,2,3,5,7} and {4,6,8,9,10}
Ans: Disjoint set

(ii) {1,2,3,6} and {1,2,4,8}
Ans: Overlapping set

(iii) E and 0 when U = Z (U denotes universal set)
Ans: Cell

(iv) N and W whenU= {x|x ∈ Z ⋀ x ≥ 0}
Ans: Exhaustive set

(v) Q and Q' when U =R
Ans: Cell

2. If A ={1,2,3,4,5,6} and B ={2,4,6,8,10} then find:
" (i) AUB
(ii) B∩A
(iii) A — B
(iv) B — A
(v) AΔB
(i) AUB
SOLUTION:
AUB = {1,2,3,4,5,6}U{2,4,6,8,10}
AUB = {1, 2, 3, 4, 5, 6, 8, 10} Ans.

(ii) B∩A
SOLUTION:
B∩A = {2,4,6,8,10}∩{1,2,3,4,5,6}
B∩A = {2, 4, 6} Ans.

(iii) A — B
SOLUTION:
A — B = {1,2,3,4,5,6} — {2,4,6,8,10}
A — B = {1, 3, 5} Ans.

(iv) B — A
SOLUTION:
B — A = {2,4,6,8,10} — {1,2,3,4,5,6}
B — A = {8, 10} Ans.

(v) AΔB
SOLUTION:
AΔB = (AUB) — (A∩B)
AΔB = {(1,2,3,4,5,6)U(2,4,6,8,10)} — {(1,2,3,4,5,6)∩(2,4,6,8,10)}
AΔB = {1, 2, 3, 4, 5, 6, 8, 10} — {2, 4, 6}
AΔB = {1, 3, 5, 8, 10} Ans.

3. If U ={1,2,3,...,10}, A ={1,2,3,4,5} and B ={1,3,5,7,9} then find: (i) A'
(ii) B'
(iii) A' U B'
(iv) A' ∩ B'
(v) (AUB)'
(vi) (A∩B)'
(vii) A'ΔB'
(viii) (AΔB)'
(ix) A — B'
(x) A' — B

Note: U = {1,2,3,...,10} = {1,2,3,4,5,6,7,8,9,10}
(i) A'
SOLUTION:
A' = U — A
A' = {1,2,3,4,5,6,7,8,9,10} — {1,2,3,4,5}
A' = {6, 7, 8, 9,10} Ans.

(ii) B'
SOLUTION:
B' = U — B
B' = {1,2,3,4,5,6,7,8,9,10} — {1,3,5,7,9}
B' = {2, 4, 6, 8, 10} Ans.

(iii) A' U B'
SOLUTION:
A' = U — A
A' = {1,2,3,4,5,6,7,8,9,10} — {1,2,3,4,5}
A' = {6, 7, 8, 9,10}
B' = U — B
B' = {1,2,3,4,5,6,7,8,9,10} — {1,3,5,7,9}
B' = {2, 4, 6, 8, 10}
A' U B' = {6, 7, 8, 9,10} U {2, 4, 6, 8, 10}
A' U B' = {2, 4, 6, 7, 8, 9, 10} Ans.

(iv) A' ∩ B'
SOLUTION:
A' = U — A
A' = {1,2,3,4,5,6,7,8,9,10} — {1,2,3,4,5}
A' = {6, 7, 8, 9,10}
B' = U — B
B' = {1,2,3,4,5,6,7,8,9,10} — {1,3,5,7,9}
B' = {2, 4, 6, 8, 10}
A' ∩ B' = {6, 7, 8, 9,10} ∩ {2, 4, 6, 8, 10}
A' ∩ B' = {6, 8, 10} Ans.

(v) (AUB)'
SOLUTION:
AUB = {1,2,3,4,5} U {1,3,5,7,9}
AUB = {1, 2, 3, 4, 5, 7, 9}
(AUB)' = U — (AUB)
(AUB)' = {1,2,3,4,5,6,7,8,9,10} — {1, 2, 3, 4, 5, 7, 9}
(AUB)' = {6, 8, 10} Ans.

(vi) (A∩B)'
SOLUTION:
A∩B = {1,2,3,4,5} ∩ {1,3,5,7,9}
A∩B = {1, 3, 5}
(A∩B)' = U — (A∩B)
(A∩B)' = {1,2,3,4,5,6,7,8,9,10} — {1, 3, 5}
(A∩B)' = {2, 4, 6, 7, 8, 9, 10} Ans.

(vii) A'ΔB'
Method I:
SOLUTION:
A' = U — A
A' = {1,2,3,4,5,6,7,8,9,10} — {1,2,3,4,5}
A' = {6, 7, 8, 9,10}
B' = U — B
B' = {1,2,3,4,5,6,7,8,9,10} — {1,3,5,7,9}
B' = {2, 4, 6, 8, 10}
A'ΔB' = (A'UB') — (A'∩B')
A'ΔB' = {(6,7,8,9,10)U(2,4,6,8,10) — (6,7,8,9,10)∩(2,4,6,8,10)} A'ΔB' = {(2,4,6,7,8,9,10) — (6,8,10)} A'ΔB' = {2, 4, 7, 9} Ans.

Method II:
SOLUTION:
A' = U — A
A' = {1,2,3,4,5,6,7,8,9,10} — {1,2,3,4,5}
A' = {6, 7, 8, 9,10}
B' = U — B
B' = {1,2,3,4,5,6,7,8,9,10} — {1,3,5,7,9}
B' = {2, 4, 6, 8, 10}
A'ΔB' = {6,7,8,9,10)Δ(2,4,6,8,10}
(Hint: select uncommon elements from both sets)
A'ΔB' = {2, 4, 7, 9} Ans.

(viii) (AΔB)'
Method I:
SOLUTION:
AΔB = (AUB) — (A∩B)
AΔB = {(1,2,3,4,5) U (1,3,5,7,9)} — {(1,2,3,4,5) ∩ (1,3,5,7,9)} AΔB = {(1,2,3,4,5,7,9) — (1,3,5)}
AΔB = {2,4,7,9}
(AΔB)' = U — AΔB
(AΔB)' = {1,2,3,4,5,6,7,8,9,10} — {2,4,7,9}
(AΔB)' = {1, 3, 5, 6, 8, 10} Ans.

Method II:
SOLUTION:
AΔB = {1,2,3,4,5} Δ {1,3,5,7,9}
(Hint: select uncommon elements from both sets) AΔB = {2, 4, 7, 9}
(AΔB)' = U — AΔB
(AΔB)' = {1,2,3,4,5,6,7,8,9,10} — {2,4,7,9}
(AΔB)' = {1, 3, 5, 6, 8, 10} Ans.

(ix) A — B'
SOLUTION:
B' = U — B
B' = {1,2,3,4,5,6,7,8,9,10} — {1,3,5,7,9}
B' = {2, 4, 6, 8, 10}
A — B' = {1,2,3,4,5} — {2, 4, 6, 8, 10}
A — B' = {1, 3, 5} Ans.

(x) A' — B
SOLUTION:
A' = U — A
A' = {1,2,3,4,5,6,7,8,9,10} — {1,2,3,4,5}
A' = {6, 7, 8, 9,10}
A' — B = {6, 7, 8, 9,10} — {1,3,5,7,9}
A' — B = {6, 8, 10} Ans.

4. If U = {x|x ∈ Z ⋀ -4 ＜ x ＜ 6}, P = {p|p ∈ E ⋀ -4 ＜ p ＜ 6} Q = {q|q ∈ P ⋀ q ＜ 6} then show that:
(i) P — Q = P∩Q'
(ii) Q — P = Q∩P'
(iii) (PUQ)' = P'∩Q'
(iv) (P∩Q)' = P'UQ'

Note:
U = {x|x ∈ Z ⋀ -4 ＜ x ＜ 6} = {-3,-2,-1,0,1,2,3,4,5}
P = {p|p ∈ E ⋀ -4 ＜ p ＜ 6} = {-2,0,2,4}
Q = {q|q ∈ P ⋀ q ＜ 6} = {2,3,5}

(i) P — Q = P∩Q'
SOLUTION:
Taking L.H.S:
P — Q = {-2,0,2,4} — {2,3,5}
P — Q = {-2,0,4}

Taking R.H.S
P∩Q'
First we find Q'
Q'= U — Q
Q' = {-3,-2,-1,0,1,2,3,4,5} — {2,3,5}
Q' = {-3, -2, -1, 0, 1, 4}
Now we find P∩Q'
P∩Q' = {-2,0,2,4} ∩ {-3,-2,-1,0,1,4}
P∩Q' = {-2, 0, 4}
P — Q = P∩Q'
L.H.S = R.H.S (Hence Proved)

(ii) Q — P = Q∩P'
SOLUTION:
Taking L.H.S:
Q — P = {2,3,5} — {-2,0,2,4}
Q — P = {3, 5}

Taking R.H.S
Q∩P'
First we find P'
P'= U — P
P' = {-3,-2,-1,0,1,2,3,4,5} — {-2,0,2,4}
P' = {-3, -1, 1, 3, 5}
Now we find Q∩P'
Q∩P' = {2,3,5} ∩ {-3, -1, 1, 3, 5}
Q∩P' = {3, 5}
Q — P = Q∩P'
L.H.S = R.H.S (Hence Proved)

(iii) (PUQ)' = P'∩Q'
SOLUTION:
Taking L.H.S:
(PUQ)'
PUQ = {-2,0,2,4} U {2,3,5}
PUQ = {-2, 0, 2, 3, 4, 5}
(PUQ)' = U — (PUQ)
(PUQ)' = {-3,-2,-1,0,1,2,3,4,5} — {-2,0,2,3,4,5}
(PUQ)' = {-3, -1, 1}
Taking R.H.S
First we find P' and Q'
P'= U — P
P' = {-3,-2,-1,0,1,2,3,4,5} — {-2,0,2,4}
P' = {-3, -1, 1, 3, 5}
Q'= U — Q
Q' = {-3,-2,-1,0,1,2,3,4,5} — {2,3,5}
Q' = {-3, -2, -1, 0, 1, 4}
Now we find P'∩Q'
P'∩Q' = {-3, -1, 1, 3, 5} ∩ {-3, -2, -1, 0, 1, 4}
P'∩Q' = {-3, -1, 1}
(PUQ)' = P'∩Q'
L.H.S = R.H.S (Hence Proved)

(iv) (P∩Q)' = P'UQ'
SOLUTION:
Taking L.H.S:
(P∩Q)' =
P∩Q = {-2,0,2,4} U {2,3,5}
P∩Q = {2}
(P∩Q)' = U — (P∩Q)
(P∩Q)' = {-3,-2,-1,0,1,2,3,4,5} — {2}
(P∩Q)' = {-3, -2, -1, 0, 1, 3, 4, 5}

Taking R.H.S
First we find P' and Q'
P'= U — P
P' = {-3,-2,-1,0,1,2,3,4,5} — {-2,0,2,4}
P' = {-3, -1, 1, 3, 5}
Q'= U — Q
Q' = {-3,-2,-1,0,1,2,3,4,5} — {2,3,5}
Q' = {-3, -2, -1, 0, 1, 4}
Now we find P'UQ'
P'UQ' = {-3, -1, 1, 3, 5} U {-3, -2, -1, 0, 1, 4}
P'UQ' = {-3, -2, -1, 0, 1, 3, 4, 5}
(P∩Q)' = P'UQ'
L.H.S = R.H.S (Hence Proved)

5. If A = {2n | n ∈ N}, B = {3n | n ∈ N} and C = {4n | n ∈ N} then find:
i) A∩B
ii) AUC
(iii) B∩C

Note:
N = {1,2,3,4,5,6,.....}
A = {2n | n ∈ N} = {2,4,6,8,10,12,.....}
B = {3n | n ∈ N} = {3,6,9,12,15,18,.....}
C = {4n | n ∈ N} = {4,8,12,16,20,24}

i) A∩B
SOLUTION:
A∩B = {2,4,6,8,10,12,.....} ∩ {3,6,9,12,15,18,.....}
A∩B = {6, 12, 18, 24........}
(Hint: First common number is 6, therefore its common are multiple of 6)
A∩B = {6n | n ∈ N} Ans.

ii) AUC
SOLUTION:
AUC = {2,4,6,8,10,12,.....} U {4,8,12,16,20,24}
AUC = {2,4,6,8,10,.....}
(Hint: AUC is multiple of 2)
AUC = {2n | n ∈ N} Ans.

(iii) B∩C
B∩C = {3,6,9,12,15,18,.....} ∩ {4,8,12,16,20,24}
B∩C = {12, 24, 36, .......}
(Hint: First common number is 12, therefore its common are multiple of 12)
B∩C = {2n | n ∈ N} Ans.

SETS - Unit 17 - Solved Exercise 17.1 - Mathematics For Class X Science

GO TO INDEX
Unit 17: SETS AND FUNCTIONS
Solved Exercise 17.1

Q.1: Write the following sets in Tabular form.

1. A = Set of all integers between -3 and 3.
2. B = Set of composite numbers less than 11.
3. C = {x | x ∈ P ⋀ 5 < x ≤ 13}
4. D = {y | y ∈ O ⋀ 7 < y < 17}
5. E = {z | z ∈ R ⋀ z² = 121}
6. F = {p | p ∈ Q ⋀ p² = -1}

Ans: SOLUTION:
(i) A = Set of all integers between -3 and 3.
Solutions:
Tabular form A = {-2, -1, 0, 1, 2} Answer

(ii) B = Set of composite numbers less than 11.
Solutions:
Tabular form B = {4, 6, 8, 9, 10} Answer

(iii) C = {x | x ∈ P ⋀ 5 < x ≤ 13}
Solutions:
Tabular form C = {7, 11, 13} Answer

(iv) D = {y | y ∈ O ⋀ 7 < y < 17}
Solutions:
Tabular form D = {9, 11, 13, 15} Answer

(v) E = {z | z ∈ R ⋀ z² = 121}
Solutions:
Tabular form E = {-11, 11} Answer

(vi) F = {p | p ∈ Q ⋀ p² = -1}
Solutions:
Tabular form F = { } or Ø Answer

Q.2: Write the following sets in Builder form.

1. A = Set of all rational numbers between 5 and 6.
2. B = {1, 2, 3, 4, 6, 12}.
3. C = {0, 土1, 土2, .......土40}
4. D = {-4, -2, 0, 2, 4}
5. E = {1, 4, 9, 16, 25}
6. F = {-1, -3, -5, -7, ........}

Ans: SOLUTION:
(i) A = Set of all rational numbers between 5 and 6.
Solution:
Set builder form A = {x | x ∈ Q ∧ 5 < x < 6} Answer

(ii) B = {1, 2, 3, 4, 6, 12}.
Solution:
Set builder from B = {x | x ∈ Z ∧ x is positive divisor of 12} Answer

(iii) C = {0, 土1, 土2, .......土40}
Solution:
Set builder from C = {x | x ∈ Z ∧ -40 ≤ x ≤ 40} Answer

(iv) D = {-4, -2, 0, 2, 4}
Solution:
Set builder from D = {x | x ∈ Z ∧ -4 ≤ x ≤ 4} Answer

(v) E = {1, 4, 9, 16, 25}
Solution:
Set builder from E = {x | x is the square of first five natural numbers}
OR E = {x | x ∈ N ∧ N² ≤ 5} Answer

F = {-1, -3, -5, -7, ........}
Solution:
Set builder from F = {x | x ∈ O ∧ x ≤ -1}
OR F = {x | x is negative odd integers} Answer

3. Write any five examples of empty set.
Ans: EMPTY SETS:
A = {x | x is a letter before 'a' in the English alphabet}.
B = {x | x is less than 7 and greater than 8}
C = {x | x ∈ N ∧ x < 1}
D = {x | x ∈ O ∧ 5 < x < 7}
E = {x | x ∈ R ∧ x² = -1}
F = Set of triangles with four sides.
G = Set of odd numbers divisible by 2

4. Classify the following as finite and infinite sets.

1. Set of Asian countries.
2. Set of all medical universities in the world
3. Set of all real numbers between 6 and 9.
4. Set of all the even prime numbers.
5. Set of all odd numbers less than 5.

SOLUTION:
(i) Set of Asian countries.
Ans: Finite set.

(ii) Set of all medical universities in the world.
Ans: Finite set.

(iii) Set of all real numbers between 6 and 9.
Ans: Infinite set.

(iv) Set of all the even prime numbers.
Ans: Finite set.

(v) Set of all odd numbers less than 5.
Ans: Infinite set.

5. Write an equivalent set, an improper subset and three proper subsets of each of the following sets.

1. P = {a, e, i, o, u}
2. Q = {x | x ∈ Z ∧ -2 ≤ x ≤ 2}

(i) P = {a, e, i, o, u}
Ans: SOLUTION:
Equivalent set = {a, b, c, d, e}
OR {1, 2, 3, 4, 5}
Improper subset = {a, e, i, o, u} or Set of Vowels.
Proper subset = { }, {a}, {e}, {i}, {o}, {u}, {a,e}, {a, i}, {a, o}, {a, u}, {e, i}, {e, o}, {e, u}, {i, o}, {i, u}, {o, u}, {a, e, i}, {a, e, o}, {a, e, u}, (a, i, o}, {a, i, u}, {a, o, u}, {e, i, o}, {e, i, u}, {e, o, u}, {i, o, u}, {a, e, i, o}, {a, e, i, u}, {a, e, o, u}, {a, i, o, u}, {e, i, o, u}
(Note: Above are all the proper subset of set P, select any three)

(ii) Q = {x | x ∈ Z ∧ -2 ≤ x ≤ 2}
Ans: Solution:
Equivalent set = {5, 6, 7, 8, 9}
Improper subset = {-2, -1, 0, 1, 2} 
Proper subset = ∅ or { }, {-2}, {-1}, {0}, {1},  {2}, {-2, -1},  {-2, 0}, {-2, 1}, {-2, 2}, {-1, 0}, {-1, 1}, {-1, 2}, {0, 1}, {0, 2}, {1, 2}, {-2, -1, 0}, {-2,  -1, 1}, {-2, -1, 2}, {-2, 0, 1}, {-2, 0, 2}, {-2, 1, 2}, {-1, 0, 1}, {-1, 0, 2}, {-1, 1, 2}, {0, 1, 2}, {-2, -1, 0, 1}, {-2, -1, 1, 2}, {-2, -1, 0, 2}, {-2, 0, 1, 2}, {-1, 0, 1, 2}
(Note: Above are all the proper subset of set P, select any three)

6. Write any two examples of singleton in set builder form.
Ans: Singleton Set:
Example 1: {x | x ∈ N Λ x² = 25} Answer
Example 2: {x | x ∈ Z Λ -1 < x < 1} Answer

7. Write power sets of the following sets.

1. A = {5, 10, 15}
2. B = {x | x ∈ Z Λ -1 < x < 4}

(i) A = {5, 10, 15}
Solution:
(For Power set, 'n' is the number of elements of set A.)
n|P(A)| = 2ⁿ = 2³ = 2 x 2 x 2 = 8
∴ [P (A)] = {∅ or { }, {5}, {10}, {15}, {5, 10}, {5, 15}, {10, 15}, {5, 10, 15}} Ans.

(ii) B = x | x ∈ Z Λ -1 < x < 4}
Solution:
(For Power set, 'n' is the number of elements of set B.)
B = {0, 1, 2, 3}
n|P(B)| = 2ⁿ = 2⁴ = 2 x 2 x 2 x 2 = 16
∴ [P (B)] = {∅ or { }, {0}, {1}, {2}, {3}, {0, 1}, {0, 2}, {0, 3}, {1, 2}, { 1, 3}, { 2, 3}, {0, 1, 2 }, {0, 1, 3}, {0, 2, 3},  {1. 2, 3}, {0, 1, 2, 3}}Ans.

8. Find a set which has only

1. Two proper subset
2. One proper subset
3. no proper subset

(i) Two proper subset
Answer: There is no such set which has two proper subset only.

(ii) One proper subset
Answer: (Note: Any set contain one element has one proper subset because it has null set also)
1. {x | x ∈ Z Λ -1 < x < 1} OR {0}
2. {a}

(iii) no proper subset
Answer: ∅ or { } has no proper subset

EXAMPLES FROM TEXT BOOK

Mathematics For Class X Science - Unit 15: TRIGONOMETRY- Fill In The Blanks And Multiple Choice Questions (MCQs)

GO TO INDEX
UNIT 15: TRIGONOMETRY
Fill In The Blanks And Multiple Choice Questions (MCQs)

Fill In The Blanks

12. Sin 20° = sin (90° - 70°) = cos 70°.
13. Cos 10° = cos (90° - 80°) = sin 80°.
14. Tan 40° = tan (90° - 50°) = cot 50°.
15. Sec 80° = (90° - 10°) = cosec 10°.
16. Cosec 10° = sec 80°.
17. Tan 20° = cot 70°.
18. 1 + tan² θ = sec² θ.
19. Cosec 60° = (90° - 30°) = sec 30°.
20. Cos 20° = sin 70°.
21. Cos 80° = sin 10°.
22. Tan² θ + 1 = sec² θ.
23. Sin 30° = cos 60°.
24. 1 + tan² 45° = sec² 45°.
25. Sin θ . sec θ = tan θ.
26. Cosec 30° = 2.
Special thanks to Sir Sajjad Akber Chandio