MATH 350 - DISCRETE
MATHEMATICS - Operations on sets
Question Bank - Set 5
Liberty University
Question 1
Question
Let A={2,4,6,8,10},B={3,6,9,12}, and C={5,10,15}. Find (A∪B)∩C.
Solution
Step 1: First, find A∪B, the union of sets Aand B.
A∪B={2,3,4,6,8,9,10,12}
Step 2: Next, find the intersection of A∪Band C.
(A∪B)∩C={10}
Therefore, (A∪B)∩C={10}.
Question 2
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find A∪Band
A∩B.
Solution
Step 1: To find A∪B, we need to determine all the elements that are in either
set Aor set B, or both.
A∪B={x∈Z|1≤x≤10 or 5 ≤x≤15}
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Step 2: To find A∩B, we need to determine all the elements that are in
both set Aand set B.
A∩B={x∈Z|1≤x≤10 and 5 ≤x≤15}
A∩B={5,6,7,8,9,10}
Question 3
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find the
intersection and union of sets Aand B.
Solution
Step 1: To find the intersection of sets Aand B, we need to determine the
elements that are common to both sets.
A∩B={x∈Z|5≤x≤10}
Step 2: To find the union of sets Aand B, we need to combine all elements
from both sets, removing duplicates.
A∪B={x∈Z|1≤x≤15}
Question 4
Question
Let A={x∈N|1≤x≤10}and B={x∈N|5≤x≤15}. Find A∪B,
A∩B, and A\B.
Solution
Step 1: To find A∪B, we need to determine all elements that are in either set
Aor set B.A={1,2,3,4,5,6,7,8,9,10}B={5,6,7,8,9,10,11,12,13,14,15}
Therefore, A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Step 2: To find A∩B, we need to determine all elements that are common
to both set Aand set B.A∩B={5,6,7,8,9,10}.
Step 3: To find A\B, we need to determine all elements that are in set A
but not in set B.A\B={1,2,3,4}.
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Question 5
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|4≤x≤12}. Find the set
resulting from the operation A∪B.
Solution
Step 1: List the elements in sets Aand B.
A={2,3,4,5,6,7,8,9,10}
B={4,5,6,7,8,9,10,11,12}
Step 2: Combine the elements from both sets without including any dupli-
cates.
A∪B={2,3,4,5,6,7,8,9,10,11,12}
Therefore, A∪B={2,3,4,5,6,7,8,9,10,11,12}.
Question 6
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find the
intersection A∩Band the union A∪B.
Solution
Step 1: To find the intersection A∩B, we look for elements that are common
to both sets Aand B.
A∩B={x∈Z|xis in both Aand B}
={x∈Z|5≤x≤10}
={5,6,7,8,9,10}.
Step 2: To find the union A∪B, we combine all elements from sets Aand
B.A∪B={x∈Z|xis in Aor B}
={x∈Z|2≤x≤15}
={2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 7
Question
Let A={x| −3≤x≤3}and B={x| −2≤x≤2}. Find the set
(A∩B)∪(A∩Bc).
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Solution
Step 1: Find A∩B, the intersection of sets Aand B.
A∩B={x| −3≤x≤2}
Step 2: Find A∩Bc, the intersection of set Aand the complement of set B.
A∩Bc={x|2< x ≤3}
Step 3: Find (A∩B)∪(A∩Bc), the union of the two previously found sets.
(A∩B)∪(A∩Bc) = {x| −3≤x≤3}
Question 8
Question
Let A={x∈R|x2−4x+ 3 <0}and B={x∈R|x2−2x−3>0}. Find
A∩B.
Solution
Step 1: Find the values of xthat satisfy the inequality x2−4x+ 3 <0.
The quadratic x2−4x+ 3 factors as (x−1)(x−3).
To find where x2−4x+ 3 is less than 0, we want the signs of the factors
to alternate. This occurs between the roots at x= 1 and x= 3.
So, A={x∈R|1<x<3}.
Step 2: Find the values of xthat satisfy the inequality x2−2x−3>0.
The quadratic x2−2x−3 factors as (x−3)(x+ 1).
To find where x2−2x−3 is greater than 0, we want the signs of the factors
to be positive. This occurs for x < −1 or x > 3.
So, B={x∈R|x < −1 or x > 3}.
Step 3: Find the intersection A∩B.
A∩Brepresents the set of elements that are in both Aand B. So,
A∩B={x∈R|1<x<3 and x < −1 or x > 3}.
Simplifying the condition, we find A∩B={x∈R|1<x<3 and x > 3},
which is an empty set since no real number satisfies both inequalities
simultaneously.
Therefore, A∩B=∅.
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Question 9
Question
Let A={x∈Z|0≤x≤10}and B={y∈N|5≤y≤15}. Determine the
set A∪B.
Solution
To determine the set A∪B, we need to find all elements that are in either Aor
Bor both.
Step 1: Determine the elements in A.A={0,1,2,3,4,5,6,7,8,9,10}.
Step 2: Determine the elements in B.B={5,6,7,8,9,10,11,12,13,14,15}.
Step 3: Determine A∪B.A∪B={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 10
Question
Let A={x∈Z|2≤x≤8}and B={x∈Z|5≤x≤10}. Find the elements
of the set A∩Band A∪B.
Solution
Step 1: Find A∩B(intersection of sets Aand B).
A∩B={x∈Z|2≤x≤8}∩{x∈Z|5≤x≤10}
A∩B={x∈Z|5≤x≤8}={5,6,7,8}
Step 2: Find A∪B(union of sets Aand B).
A∪B={x∈Z|2≤x≤8}∪{x∈Z|5≤x≤10}
A∪B={x∈Z|2≤x≤10}={2,3,4,5,6,7,8,9,10}
Therefore, the elements of A∩Bare 5, 6, 7, and 8, and the elements of A∪B
are 2, 3, 4, 5, 6, 7, 8, 9, and 10.
Question 11
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Find the set (A∪B)c∩(A∩B).
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Solution
Step 1: First, we find A∪B:
A∪B={1,2,3,4,5}∪{3,4,5,6,7}={1,2,3,4,5,6,7}
Step 2: Next, we find (A∪B)c:
(A∪B)c={1,2,3,4,5,6,7}c=∅
Step 3: Then, we find A∩B:
A∩B={1,2,3,4,5}∩{3,4,5,6,7}={3,4,5}
Step 4: Finally, we find (A∪B)c∩(A∩B):
∅∩{3,4,5}=∅
Therefore, (A∪B)c∩(A∩B) = ∅.
Question 12
Question
Let A={1,2,3,4},B={3,4,5,6}, and C={2,3,6,7}. Determine the set
(A∪B)∩C.
Solution
Step 1: Find A∪B. Step 2: Find (A∪B)∩C.
Step 1: Find A∪B. Since A={1,2,3,4}and B={3,4,5,6}, we have
A∪B={1,2,3,4}∪{3,4,5,6}={1,2,3,4,5,6}.
Step 2: Find (A∪B)∩C. Now, we need to find the intersection of A∪B
and C. Since A∪B={1,2,3,4,5,6}and C={2,3,6,7}, we have
(A∪B)∩C={1,2,3,4,5,6}∩{2,3,6,7}.
To find the intersection, we look for elements that are common to both sets.
In this case, the common elements are 2, 3, and 6. Therefore,
(A∪B)∩C={2,3,6}.
So, (A∪B)∩C={2,3,6}.
Question 13
Question
Let A={x∈Z|1≤x≤10},B={x∈Z|5≤x≤15}, and C={x∈Z|
1≤x≤5}. Compute (A∪B)∩C.
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Solution
Step 1: First, let’s find A∪B, the union of sets Aand B.
A∪B={x∈Z|1≤x≤15}
Step 2: Next, let’s find (A∪B)∩C, the intersection of A∪Band C.
(A∪B)∩C={x∈Z|1≤x≤5}
Therefore, (A∪B)∩C={1,2,3,4,5}.
Question 14
Question
Let Aand Bbe sets such that |A|= 8, |B|= 10, and |A∪B|= 14. If |A∩B|= 4,
find |A∩B′|, where B′denotes the complement of set B.
Solution
Step 1: Recall that for any two sets Xand Y, the principle of inclusion-exclusion
states that
|X∪Y|=|X|+|Y|−|X∩Y|.
Step 2: Let’s use the principle of inclusion-exclusion to find |A∪B|. We
have:
|A∪B|=|A|+|B|−|A∩B|.
Substitute the given values:
14 = 8 + 10 −4.
Step 3: Now, let’s find |A∩B′|.
Since |A∪B|=|A|+|B|−|A∩B|, we can rearrange to find |A∩B|:
|A∩B|=|A|+|B|−|A∪B|= 8 + 10 −14 = 4.
Step 4: We know that |A∩B|=|A|−|A∩B′|. Substitute the known values
to find |A∩B′|:
4=8− |A∩B′|.
Step 5: Solving for |A∩B′|, we get
|A∩B′|= 8 −4 = 4.
Therefore, |A∩B′|= 4.
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Question 15
Question
Let A={1,2,3,4,5},B={3,4,5,6,7}, and C={5,6,7,8,9}. Determine
(A∪B)∩(B∪C).
Solution
Step 1: Find A∪B.
A∪B={1,2,3,4,5}∪{3,4,5,6,7}
={1,2,3,4,5,6,7}
Step 2: Find B∪C.
B∪C={3,4,5,6,7}∪{5,6,7,8,9}
={3,4,5,6,7,8,9}
Step 3: Find (A∪B)∩(B∪C).
(A∪B)∩(B∪C) = {1,2,3,4,5,6,7}∩{3,4,5,6,7,8,9}
={3,4,5,6,7}
Therefore, (A∪B)∩(B∪C) = {3,4,5,6,7}.
Question 16
Question
Let A={1,2,3},B={2,3,4}, and C={3,4,5}. Find (A∪B)∩C.
Solution
Step 1: First, find the union of sets Aand B.
Step 1: A∪B={1,2,3}∪{2,3,4}
A∪B={1,2,3,4}
Step 2: Next, find the intersection of the result from Step 1 and set C.
Step 2: (A∪B)∩C={1,2,3,4}∩{3,4,5}
(A∪B)∩C={3,4}
Therefore, (A∪B)∩C={3,4}.
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Question 17
Question
Let A={x∈R:x2−7x+ 10 >0}and B={x∈R:x2−6x+ 5 >0}. Find
the set A∩B.
Solution
Step 1: First, find the values of xthat satisfy the inequalities for sets Aand
Bseparately. For set A: We need to solve the inequality x2−7x+ 10 >0.
Factorizing the quadratic gives (x−5)(x−2) >0. From this, we can see that
x∈(2,5). Therefore, A={x∈R: 2 <x<5}.
Step 2: Now, find the values of xfor set B. For set B: We need to solve the
inequality x2−6x+ 5 >0. Factorizing the quadratic gives (x−5)(x−1) >0.
From this, we can see that x∈(−∞,1) ∪(5,+∞). Therefore, B={x∈R:
x < 1 or x > 5}.
Step 3: Finally, find the intersection of sets Aand B. To find A∩B, we
need to find the values of xthat satisfy both inequalities. By construction, set
Arequires 2 <x<5 and set Brequires x < 1 or x > 5. Thus, the only overlap
between the two sets is (2,1). Therefore, A∩B={x∈R: 2 < x < 5} ∩ {x∈
R:x < 1 or x > 5}=∅.
Question 18
Question
Let A,B, and Cbe sets such that A∪B=A∪Cand A∩B=A∩C. Prove
or disprove that B=C.
Solution
To prove or disprove that B=Cbased on the given information, we will consider
the definitions of union and intersection of sets.
Step 1: Recall the definition of set equality. Two sets Xand Yare equal,
denoted X=Y, if and only if every element of Xis an element of Yand vice
versa.
Step 2: Prove the statement. We are given that A∪B=A∪Cand
A∩B=A∩C. From the definition of union, A∪Bincludes all elements in
Aand B(at least once), and A∪Cincludes all elements in Aand C(at least
once). Similarly, from the definition of intersection, A∩Bincludes all elements
that are common to both Aand B, and A∩Cincludes all elements that are
common to both Aand C.
Step 3: Analyze the sets Band C. Based on the given information: -
A∪B=A∪Cimplies that any element in B(which is not in A) must also be
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in Cto have the same union with A. - A∩B=A∩Cimplies that any element
in B(which is in A) must also be in Cto have the same intersection with A.
Step 4: Conclusion. Since Band Cmust contain the same elements outside
of Aand the same elements inside of Ain order to satisfy the given conditions,
we can conclude that B=C.
Question 19
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find
(A∪B)∩(A∩B).
Solution
Step 1: First, let’s find A∪B.
A∪B={x∈Z|1≤x≤10}∪{x∈Z|5≤x≤15}
In this case, A∪Bwill be the set of all integers that are in either set Aor
set B(or both).
Step 2: By examining the two sets, we can see that A∪Bwill be from the
smallest value in Ato the largest value in B, which is from 1 to 15. Therefore,
A∪B={x∈Z|1≤x≤15}
Step 3: Next, let’s find A∩B.
A∩B={x∈Z|1≤x≤10}∩{x∈Z|5≤x≤15}
A∩Bis the set of all integers that are in both set Aand set B.
Step 4: By comparing the two sets, we see that the set of common values
between Aand Bis from 5 to 10. Therefore,
A∩B={x∈Z|5≤x≤10}
Step 5: Finally, let’s find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {x∈Z|1≤x≤15}∩{x∈Z|5≤x≤10}
Step 6: By comparing the two sets, we can see that the common values
between (A∪B) and (A∩B) are from 5 to 10. Therefore,
(A∪B)∩(A∩B) = {x∈Z|5≤x≤10}
Question 20
Question
Let A={x∈R|0<x<2}and B={x∈R|1<x<3}. Find A∩B.
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Solution
Step 1: To find A∩B, we need to determine the elements that are common to
both sets Aand B.
Step 2: Set Acontains all real numbers xsuch that 0 <x<2, and set B
contains all real numbers xsuch that 1 <x<3.
Step 3: To find A∩B, we need to find the values of xthat satisfy both
0<x<2 and 1 <x<3.
Step 4: The intersection of the two sets can be found by determining the
range in which the inequalities of both sets overlap.
Step 5: We see that the overlapping range is 1 <x<2, which means
A∩B={x∈R|1<x<2}.
Question 21
Question
Let A={x∈N: 1 ≤x≤10}and B={x∈N: 5 ≤x≤15}. Determine the
set (A∩B)∪(A∩B∁).
Solution
Step 1: Find A∩B.
A∩B={x∈N: 1 ≤x≤10}∩{x∈N: 5 ≤x≤15}
={5,6,7,8,9,10}
Step 2: Find B∁, the complement of set B.
B∁={x∈N:x < 5 or x > 15}
={1,2,3,4,16,17,18,19, ...}
Step 3: Find A∩B∁.
A∩B∁={x∈N: 1 ≤x≤10}∩{x∈N:x < 5 or x > 15}
={1,2,3,4}
Step 4: Find (A∩B)∪(A∩B∁).
(A∩B)∪(A∩B∁) = {5,6,7,8,9,10}∪{1,2,3,4}
={1,2,3,4,5,6,7,8,9,10}
Therefore, (A∩B)∪(A∩B∁) = {1,2,3,4,5,6,7,8,9,10}.
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Question 22
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Determine A∩B,
A∪B, and A\B.
Solution
Step 1: To find A∩B(the intersection of sets Aand B), we look for elements
that are in both sets.
A∩B={x∈Z|5≤x≤10}.
Step 2: Next, to find A∪B(the union of sets Aand B), we combine all
elements in both sets without repetitions.
A∪B={x∈Z|1≤x≤15}.
Step 3: Finally, to find A\B(the set difference of set Awith respect to set
B), we look for elements that are in set Abut not in set B.
A\B={x∈Z|1≤x≤4}.
Question 23
Question
Let A={x∈R| −2≤x≤3}and B={x∈R|x2<4}. Find the set A∪B.
Solution
Step 1: Let’s first determine the elements of set Aand set B. - Set A: The
set Aconsists of all real numbers xsuch that −2≤x≤3. Therefore, A=
{−2,−1,0,1,2,3}. - Set B: The set Bconsists of all real numbers xsuch that
x2<4. This means xcan be in the range −2<x<2. So, B={−2,−1,0,1}.
Step 2: Now, let’s find the union of sets Aand B, denoted by A∪B. The
union of two sets Aand Bis the set of all elements that are in A, in B, or in
both Aand B.
Step 3: Combining the elements of sets Aand B, we have A∪B={−2,−1,0,1,2,3}.
Therefore, the set A∪Bis {−2,−1,0,1,2,3}.
Question 24
Question
Let A,B, and Cbe sets such that A∪B=A∩C. Prove or disprove: B⊆C.
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Solution
To prove or disprove B⊆C, we need to show whether every element in set B
is also in set C.
Step 1: Let’s start by assuming that x∈B.
Step 2: Since A∪B=A∩C, we know that x∈A∪Bif and only if
x∈A∩C.
Step 3: If x∈A∪B, then either x∈Aor x∈B.
Step 4: Since x∈B(by our assumption), we have two cases to consider:
Case 1: If x∈A, then x∈A∩C, which implies x∈C.
Case 2: If x∈B, then x∈A∩C, which implies x∈C.
Step 5: In either case, we have shown that if x∈B, then x∈C. Therefore,
B⊆C.
Step 6: Hence, we have proved that if A∪B=A∩C, then B⊆C.
Question 25
Question
Let Aand Bbe sets such that |A|= 10 and |B|= 15. If |A∪B|= 20 and
|A∩B|= 5, find |A∩B′|where B′denotes the complement of set B.
Solution
Step 1: We know that the union of two sets can be calculated using the formula
|A∪B|=|A|+|B|−|A∩B|. Therefore, we can find |B|using the known values.
|A∪B|=|A|+|B|−|A∩B|=⇒20 = 10 + 15 −5 = 20 =⇒ |B|= 15
Step 2: To find |A∩B′|, we can use the formula |A∩B′|=|A|−|A∩B|.
Since we already know |A∩B|= 5, we can calculate |A∩B′|.
|A∩B′|=|A|−|A∩B|= 10 −5=5
Therefore, |A∩B′|= 5.
Question 26
Question
Let A={x∈R|x2−4x+ 3 = 0}and B={x∈R|x2−3x−10 = 0}. Find
the set A∩B.
13
Solution
Step 1: Find the solutions to the equations x2−4x+3 = 0 and x2−3x−10 = 0.
For x2−4x+ 3 = 0: Factoring gives (x−1)(x−3) = 0, so the solutions
are x= 1 and x= 3.
For x2−3x−10 = 0: Factoring gives (x−5)(x+ 2) = 0, so the solutions
are x= 5 and x=−2.
Step 2: Find A∩B, the intersection of sets Aand B. This is the set of all
elements that are in both Aand B.
A∩B={x∈R|x∈Aand x∈B}
={1,3}∩{5,−2}
={}
Therefore, A∩B=∅.
Question 27
Question
Let A={1,2,3,4}and B={3,4,5,6}. Define the set Cas the set of all
elements which are in either Aor B, but not both. Find the set C.
Solution
Step 1: First, let’s find the elements that are in Aor B, but not both. Step
2: The elements in A∪Bare {1,2,3,4,5,6}. Step 3: The elements in A∩B
are {3,4}. Step 4: The elements that are in either Aor B, but not both, are
{1,2,5,6}. Step 5: Therefore, the set Cis C={1,2,5,6}.
Question 28
Question
Let Aand Bbe sets such that |A∩B|= 8, |A\B|= 12, and |B\A|= 15. Find
|A∪B|.
Solution
Step 1: Recall the formula for the cardinality of the union of two sets:
|A∪B|=|A|+|B|−|A∩B|.
Step 2: We are given |A∩B|= 8, |A\B|= 12, and |B\A|= 15. Let’s denote
|A∩B|=x,|A\B|=y, and |B\A|=zfor simplicity.
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Step 3: We know that |A|=|A∩B|+|A\B|=x+yand |B|=|A∩B|+
|B\A|=x+z.
Step 4: Substituting x= 8, y= 12, and z= 15, we have |A|= 8 + 12 = 20
and |B|= 8 + 15 = 23.
Step 5: Now, we can find |A∪B|using the formula:
|A∪B|=|A|+|B|−|A∩B|= 20 + 23 −8 = 43 −8 = 35.
Therefore, |A∪B|= 35.
Question 29
Question
Let A={x∈R| −3≤x≤3}and B={x∈R| −2≤x≤4}. Find the set
(A∩B)∪(A∩Bc).
Solution
Step 1: Find A∩B.
A∩B={x∈R| −3≤x≤3}∩{x∈R| −2≤x≤4}
={x∈R| −2≤x≤3}
Step 2: Find A∩Bc.
A∩Bc={x∈R| −3≤x≤3}∩{x∈R|x < −2 or x > 4}
={x∈R| −3≤x≤ −2}∪{x∈R|3<x<4}
Step 3: Find (A∩B)∪(A∩Bc).
(A∩B)∪(A∩Bc) = {x∈R| −2≤x≤3}∪{x∈R| −3≤x≤ −2}∪{x∈R|3<x<4}
={x∈R| −3≤x≤4}
Therefore, (A∩B)∪(A∩Bc) = {x∈R| −3≤x≤4}.
Question 30
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Find each of the following sets:
a) A∪B
b) A∩B
c) A\B
d) B\A
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Solution
a) To find the union of sets Aand B(A∪B), we need to combine all unique
elements from both sets.
Step 1: List out all elements in Aand B:
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: Combine all unique elements from Aand B:
A∪B={1,2,3,4,5,6,7}
b) To find the intersection of sets Aand B(A∩B), we need to identify
elements that are common to both sets.
Step 1: List out all elements in Aand B:
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: Identify elements that are common to both sets:
A∩B={3,4,5}
c) To find Aminus B(A\B), we need to list out elements that are in A
but not in B.
Step 1: List out all elements in Aand B:
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: List out elements that are in Abut not in B:
A\B={1,2}
d) To find Bminus A(B\A), we need to list out elements that are in B
but not in A.
Step 1: List out all elements in Aand B:
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: List out elements that are in Bbut not in A:
B\A={6,7}
16
Question 5
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|4≤x≤12}. Find the set
resulting from the operation A∪B.
Solution
Step 1: List the elements in sets Aand B.
A={2,3,4,5,6,7,8,9,10}
B={4,5,6,7,8,9,10,11,12}
Step 2: Combine the elements from both sets without including any dupli-
cates.
A∪B={2,3,4,5,6,7,8,9,10,11,12}
Therefore, A∪B={2,3,4,5,6,7,8,9,10,11,12}.
Question 6
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find the
intersection A∩Band the union A∪B.
Solution
Step 1: To find the intersection A∩B, we look for elements that are common
to both sets Aand B.
A∩B={x∈Z|xis in both Aand B}
={x∈Z|5≤x≤10}
={5,6,7,8,9,10}.
Step 2: To find the union A∪B, we combine all elements from sets Aand
B.A∪B={x∈Z|xis in Aor B}
={x∈Z|2≤x≤15}
={2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 7
Question
Let A={x| −3≤x≤3}and B={x| −2≤x≤2}. Find the set
(A∩B)∪(A∩Bc).
3
Solution
Step 1: Find A∩B, the intersection of sets Aand B.
A∩B={x| −3≤x≤2}
Step 2: Find A∩Bc, the intersection of set Aand the complement of set B.
A∩Bc={x|2< x ≤3}
Step 3: Find (A∩B)∪(A∩Bc), the union of the two previously found sets.
(A∩B)∪(A∩Bc) = {x| −3≤x≤3}
Question 8
Question
Let A={x∈R|x2−4x+ 3 <0}and B={x∈R|x2−2x−3>0}. Find
A∩B.
Solution
Step 1: Find the values of xthat satisfy the inequality x2−4x+ 3 <0.
The quadratic x2−4x+ 3 factors as (x−1)(x−3).
To find where x2−4x+ 3 is less than 0, we want the signs of the factors
to alternate. This occurs between the roots at x= 1 and x= 3.
So, A={x∈R|1<x<3}.
Step 2: Find the values of xthat satisfy the inequality x2−2x−3>0.
The quadratic x2−2x−3 factors as (x−3)(x+ 1).
To find where x2−2x−3 is greater than 0, we want the signs of the factors
to be positive. This occurs for x < −1 or x > 3.
So, B={x∈R|x < −1 or x > 3}.
Step 3: Find the intersection A∩B.
A∩Brepresents the set of elements that are in both Aand B. So,
A∩B={x∈R|1<x<3 and x < −1 or x > 3}.
Simplifying the condition, we find A∩B={x∈R|1<x<3 and x > 3},
which is an empty set since no real number satisfies both inequalities
simultaneously.
Therefore, A∩B=∅.
4
Question 9
Question
Let A={x∈Z|0≤x≤10}and B={y∈N|5≤y≤15}. Determine the
set A∪B.
Solution
To determine the set A∪B, we need to find all elements that are in either Aor
Bor both.
Step 1: Determine the elements in A.A={0,1,2,3,4,5,6,7,8,9,10}.
Step 2: Determine the elements in B.B={5,6,7,8,9,10,11,12,13,14,15}.
Step 3: Determine A∪B.A∪B={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 10
Question
Let A={x∈Z|2≤x≤8}and B={x∈Z|5≤x≤10}. Find the elements
of the set A∩Band A∪B.
Solution
Step 1: Find A∩B(intersection of sets Aand B).
A∩B={x∈Z|2≤x≤8}∩{x∈Z|5≤x≤10}
A∩B={x∈Z|5≤x≤8}={5,6,7,8}
Step 2: Find A∪B(union of sets Aand B).
A∪B={x∈Z|2≤x≤8}∪{x∈Z|5≤x≤10}
A∪B={x∈Z|2≤x≤10}={2,3,4,5,6,7,8,9,10}
Therefore, the elements of A∩Bare 5, 6, 7, and 8, and the elements of A∪B
are 2, 3, 4, 5, 6, 7, 8, 9, and 10.
Question 11
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Find the set (A∪B)c∩(A∩B).
5
Solution
Step 1: First, we find A∪B:
A∪B={1,2,3,4,5}∪{3,4,5,6,7}={1,2,3,4,5,6,7}
Step 2: Next, we find (A∪B)c:
(A∪B)c={1,2,3,4,5,6,7}c=∅
Step 3: Then, we find A∩B:
A∩B={1,2,3,4,5}∩{3,4,5,6,7}={3,4,5}
Step 4: Finally, we find (A∪B)c∩(A∩B):
∅∩{3,4,5}=∅
Therefore, (A∪B)c∩(A∩B) = ∅.
Question 12
Question
Let A={1,2,3,4},B={3,4,5,6}, and C={2,3,6,7}. Determine the set
(A∪B)∩C.
Solution
Step 1: Find A∪B. Step 2: Find (A∪B)∩C.
Step 1: Find A∪B. Since A={1,2,3,4}and B={3,4,5,6}, we have
A∪B={1,2,3,4}∪{3,4,5,6}={1,2,3,4,5,6}.
Step 2: Find (A∪B)∩C. Now, we need to find the intersection of A∪B
and C. Since A∪B={1,2,3,4,5,6}and C={2,3,6,7}, we have
(A∪B)∩C={1,2,3,4,5,6}∩{2,3,6,7}.
To find the intersection, we look for elements that are common to both sets.
In this case, the common elements are 2, 3, and 6. Therefore,
(A∪B)∩C={2,3,6}.
So, (A∪B)∩C={2,3,6}.
Question 13
Question
Let A={x∈Z|1≤x≤10},B={x∈Z|5≤x≤15}, and C={x∈Z|
1≤x≤5}. Compute (A∪B)∩C.
6
Solution
Step 1: First, let’s find A∪B, the union of sets Aand B.
A∪B={x∈Z|1≤x≤15}
Step 2: Next, let’s find (A∪B)∩C, the intersection of A∪Band C.
(A∪B)∩C={x∈Z|1≤x≤5}
Therefore, (A∪B)∩C={1,2,3,4,5}.
Question 14
Question
Let Aand Bbe sets such that |A|= 8, |B|= 10, and |A∪B|= 14. If |A∩B|= 4,
find |A∩B′|, where B′denotes the complement of set B.
Solution
Step 1: Recall that for any two sets Xand Y, the principle of inclusion-exclusion
states that
|X∪Y|=|X|+|Y|−|X∩Y|.
Step 2: Let’s use the principle of inclusion-exclusion to find |A∪B|. We
have:
|A∪B|=|A|+|B|−|A∩B|.
Substitute the given values:
14 = 8 + 10 −4.
Step 3: Now, let’s find |A∩B′|.
Since |A∪B|=|A|+|B|−|A∩B|, we can rearrange to find |A∩B|:
|A∩B|=|A|+|B|−|A∪B|= 8 + 10 −14 = 4.
Step 4: We know that |A∩B|=|A|−|A∩B′|. Substitute the known values
to find |A∩B′|:
4=8− |A∩B′|.
Step 5: Solving for |A∩B′|, we get
|A∩B′|= 8 −4 = 4.
Therefore, |A∩B′|= 4.
7
Question 15
Question
Let A={1,2,3,4,5},B={3,4,5,6,7}, and C={5,6,7,8,9}. Determine
(A∪B)∩(B∪C).
Solution
Step 1: Find A∪B.
A∪B={1,2,3,4,5}∪{3,4,5,6,7}
={1,2,3,4,5,6,7}
Step 2: Find B∪C.
B∪C={3,4,5,6,7}∪{5,6,7,8,9}
={3,4,5,6,7,8,9}
Step 3: Find (A∪B)∩(B∪C).
(A∪B)∩(B∪C) = {1,2,3,4,5,6,7}∩{3,4,5,6,7,8,9}
={3,4,5,6,7}
Therefore, (A∪B)∩(B∪C) = {3,4,5,6,7}.
Question 16
Question
Let A={1,2,3},B={2,3,4}, and C={3,4,5}. Find (A∪B)∩C.
Solution
Step 1: First, find the union of sets Aand B.
Step 1: A∪B={1,2,3}∪{2,3,4}
A∪B={1,2,3,4}
Step 2: Next, find the intersection of the result from Step 1 and set C.
Step 2: (A∪B)∩C={1,2,3,4}∩{3,4,5}
(A∪B)∩C={3,4}
Therefore, (A∪B)∩C={3,4}.
8
Question 17
Question
Let A={x∈R:x2−7x+ 10 >0}and B={x∈R:x2−6x+ 5 >0}. Find
the set A∩B.
Solution
Step 1: First, find the values of xthat satisfy the inequalities for sets Aand
Bseparately. For set A: We need to solve the inequality x2−7x+ 10 >0.
Factorizing the quadratic gives (x−5)(x−2) >0. From this, we can see that
x∈(2,5). Therefore, A={x∈R: 2 <x<5}.
Step 2: Now, find the values of xfor set B. For set B: We need to solve the
inequality x2−6x+ 5 >0. Factorizing the quadratic gives (x−5)(x−1) >0.
From this, we can see that x∈(−∞,1) ∪(5,+∞). Therefore, B={x∈R:
x < 1 or x > 5}.
Step 3: Finally, find the intersection of sets Aand B. To find A∩B, we
need to find the values of xthat satisfy both inequalities. By construction, set
Arequires 2 <x<5 and set Brequires x < 1 or x > 5. Thus, the only overlap
between the two sets is (2,1). Therefore, A∩B={x∈R: 2 < x < 5} ∩ {x∈
R:x < 1 or x > 5}=∅.
Question 18
Question
Let A,B, and Cbe sets such that A∪B=A∪Cand A∩B=A∩C. Prove
or disprove that B=C.
Solution
To prove or disprove that B=Cbased on the given information, we will consider
the definitions of union and intersection of sets.
Step 1: Recall the definition of set equality. Two sets Xand Yare equal,
denoted X=Y, if and only if every element of Xis an element of Yand vice
versa.
Step 2: Prove the statement. We are given that A∪B=A∪Cand
A∩B=A∩C. From the definition of union, A∪Bincludes all elements in
Aand B(at least once), and A∪Cincludes all elements in Aand C(at least
once). Similarly, from the definition of intersection, A∩Bincludes all elements
that are common to both Aand B, and A∩Cincludes all elements that are
common to both Aand C.
Step 3: Analyze the sets Band C. Based on the given information: -
A∪B=A∪Cimplies that any element in B(which is not in A) must also be
9
in Cto have the same union with A. - A∩B=A∩Cimplies that any element
in B(which is in A) must also be in Cto have the same intersection with A.
Step 4: Conclusion. Since Band Cmust contain the same elements outside
of Aand the same elements inside of Ain order to satisfy the given conditions,
we can conclude that B=C.
Question 19
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find
(A∪B)∩(A∩B).
Solution
Step 1: First, let’s find A∪B.
A∪B={x∈Z|1≤x≤10}∪{x∈Z|5≤x≤15}
In this case, A∪Bwill be the set of all integers that are in either set Aor
set B(or both).
Step 2: By examining the two sets, we can see that A∪Bwill be from the
smallest value in Ato the largest value in B, which is from 1 to 15. Therefore,
A∪B={x∈Z|1≤x≤15}
Step 3: Next, let’s find A∩B.
A∩B={x∈Z|1≤x≤10}∩{x∈Z|5≤x≤15}
A∩Bis the set of all integers that are in both set Aand set B.
Step 4: By comparing the two sets, we see that the set of common values
between Aand Bis from 5 to 10. Therefore,
A∩B={x∈Z|5≤x≤10}
Step 5: Finally, let’s find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {x∈Z|1≤x≤15}∩{x∈Z|5≤x≤10}
Step 6: By comparing the two sets, we can see that the common values
between (A∪B) and (A∩B) are from 5 to 10. Therefore,
(A∪B)∩(A∩B) = {x∈Z|5≤x≤10}
Question 20
Question
Let A={x∈R|0<x<2}and B={x∈R|1<x<3}. Find A∩B.
10
Solution
Step 1: To find A∩B, we need to determine the elements that are common to
both sets Aand B.
Step 2: Set Acontains all real numbers xsuch that 0 <x<2, and set B
contains all real numbers xsuch that 1 <x<3.
Step 3: To find A∩B, we need to find the values of xthat satisfy both
0<x<2 and 1 <x<3.
Step 4: The intersection of the two sets can be found by determining the
range in which the inequalities of both sets overlap.
Step 5: We see that the overlapping range is 1 <x<2, which means
A∩B={x∈R|1<x<2}.
Question 21
Question
Let A={x∈N: 1 ≤x≤10}and B={x∈N: 5 ≤x≤15}. Determine the
set (A∩B)∪(A∩B∁).
Solution
Step 1: Find A∩B.
A∩B={x∈N: 1 ≤x≤10}∩{x∈N: 5 ≤x≤15}
={5,6,7,8,9,10}
Step 2: Find B∁, the complement of set B.
B∁={x∈N:x < 5 or x > 15}
={1,2,3,4,16,17,18,19, ...}
Step 3: Find A∩B∁.
A∩B∁={x∈N: 1 ≤x≤10}∩{x∈N:x < 5 or x > 15}
={1,2,3,4}
Step 4: Find (A∩B)∪(A∩B∁).
(A∩B)∪(A∩B∁) = {5,6,7,8,9,10}∪{1,2,3,4}
={1,2,3,4,5,6,7,8,9,10}
Therefore, (A∩B)∪(A∩B∁) = {1,2,3,4,5,6,7,8,9,10}.
11
Question 22
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Determine A∩B,
A∪B, and A\B.
Solution
Step 1: To find A∩B(the intersection of sets Aand B), we look for elements
that are in both sets.
A∩B={x∈Z|5≤x≤10}.
Step 2: Next, to find A∪B(the union of sets Aand B), we combine all
elements in both sets without repetitions.
A∪B={x∈Z|1≤x≤15}.
Step 3: Finally, to find A\B(the set difference of set Awith respect to set
B), we look for elements that are in set Abut not in set B.
A\B={x∈Z|1≤x≤4}.
Question 23
Question
Let A={x∈R| −2≤x≤3}and B={x∈R|x2<4}. Find the set A∪B.
Solution
Step 1: Let’s first determine the elements of set Aand set B. - Set A: The
set Aconsists of all real numbers xsuch that −2≤x≤3. Therefore, A=
{−2,−1,0,1,2,3}. - Set B: The set Bconsists of all real numbers xsuch that
x2<4. This means xcan be in the range −2<x<2. So, B={−2,−1,0,1}.
Step 2: Now, let’s find the union of sets Aand B, denoted by A∪B. The
union of two sets Aand Bis the set of all elements that are in A, in B, or in
both Aand B.
Step 3: Combining the elements of sets Aand B, we have A∪B={−2,−1,0,1,2,3}.
Therefore, the set A∪Bis {−2,−1,0,1,2,3}.
Question 24
Question
Let A,B, and Cbe sets such that A∪B=A∩C. Prove or disprove: B⊆C.
12
Solution
To prove or disprove B⊆C, we need to show whether every element in set B
is also in set C.
Step 1: Let’s start by assuming that x∈B.
Step 2: Since A∪B=A∩C, we know that x∈A∪Bif and only if
x∈A∩C.
Step 3: If x∈A∪B, then either x∈Aor x∈B.
Step 4: Since x∈B(by our assumption), we have two cases to consider:
Case 1: If x∈A, then x∈A∩C, which implies x∈C.
Case 2: If x∈B, then x∈A∩C, which implies x∈C.
Step 5: In either case, we have shown that if x∈B, then x∈C. Therefore,
B⊆C.
Step 6: Hence, we have proved that if A∪B=A∩C, then B⊆C.
Question 25
Question
Let Aand Bbe sets such that |A|= 10 and |B|= 15. If |A∪B|= 20 and
|A∩B|= 5, find |A∩B′|where B′denotes the complement of set B.
Solution
Step 1: We know that the union of two sets can be calculated using the formula
|A∪B|=|A|+|B|−|A∩B|. Therefore, we can find |B|using the known values.
|A∪B|=|A|+|B|−|A∩B|=⇒20 = 10 + 15 −5 = 20 =⇒ |B|= 15
Step 2: To find |A∩B′|, we can use the formula |A∩B′|=|A|−|A∩B|.
Since we already know |A∩B|= 5, we can calculate |A∩B′|.
|A∩B′|=|A|−|A∩B|= 10 −5=5
Therefore, |A∩B′|= 5.
Question 26
Question
Let A={x∈R|x2−4x+ 3 = 0}and B={x∈R|x2−3x−10 = 0}. Find
the set A∩B.
13
Solution
Step 1: Find the solutions to the equations x2−4x+3 = 0 and x2−3x−10 = 0.
For x2−4x+ 3 = 0: Factoring gives (x−1)(x−3) = 0, so the solutions
are x= 1 and x= 3.
For x2−3x−10 = 0: Factoring gives (x−5)(x+ 2) = 0, so the solutions
are x= 5 and x=−2.
Step 2: Find A∩B, the intersection of sets Aand B. This is the set of all
elements that are in both Aand B.
A∩B={x∈R|x∈Aand x∈B}
={1,3}∩{5,−2}
={}
Therefore, A∩B=∅.
Question 27
Question
Let A={1,2,3,4}and B={3,4,5,6}. Define the set Cas the set of all
elements which are in either Aor B, but not both. Find the set C.
Solution
Step 1: First, let’s find the elements that are in Aor B, but not both. Step
2: The elements in A∪Bare {1,2,3,4,5,6}. Step 3: The elements in A∩B
are {3,4}. Step 4: The elements that are in either Aor B, but not both, are
{1,2,5,6}. Step 5: Therefore, the set Cis C={1,2,5,6}.
Question 28
Question
Let Aand Bbe sets such that |A∩B|= 8, |A\B|= 12, and |B\A|= 15. Find
|A∪B|.
Solution
Step 1: Recall the formula for the cardinality of the union of two sets:
|A∪B|=|A|+|B|−|A∩B|.
Step 2: We are given |A∩B|= 8, |A\B|= 12, and |B\A|= 15. Let’s denote
|A∩B|=x,|A\B|=y, and |B\A|=zfor simplicity.
14
Step 3: We know that |A|=|A∩B|+|A\B|=x+yand |B|=|A∩B|+
|B\A|=x+z.
Step 4: Substituting x= 8, y= 12, and z= 15, we have |A|= 8 + 12 = 20
and |B|= 8 + 15 = 23.
Step 5: Now, we can find |A∪B|using the formula:
|A∪B|=|A|+|B|−|A∩B|= 20 + 23 −8 = 43 −8 = 35.
Therefore, |A∪B|= 35.
Question 29
Question
Let A={x∈R| −3≤x≤3}and B={x∈R| −2≤x≤4}. Find the set
(A∩B)∪(A∩Bc).
Solution
Step 1: Find A∩B.
A∩B={x∈R| −3≤x≤3}∩{x∈R| −2≤x≤4}
={x∈R| −2≤x≤3}
Step 2: Find A∩Bc.
A∩Bc={x∈R| −3≤x≤3}∩{x∈R|x < −2 or x > 4}
={x∈R| −3≤x≤ −2}∪{x∈R|3<x<4}
Step 3: Find (A∩B)∪(A∩Bc).
(A∩B)∪(A∩Bc) = {x∈R| −2≤x≤3}∪{x∈R| −3≤x≤ −2}∪{x∈R|3<x<4}
={x∈R| −3≤x≤4}
Therefore, (A∩B)∪(A∩Bc) = {x∈R| −3≤x≤4}.
Question 30
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Find each of the following sets:
a) A∪B
b) A∩B
c) A\B
d) B\A
15
Solution
a) To find the union of sets Aand B(A∪B), we need to combine all unique
elements from both sets.
Step 1: List out all elements in Aand B:
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: Combine all unique elements from Aand B:
A∪B={1,2,3,4,5,6,7}
b) To find the intersection of sets Aand B(A∩B), we need to identify
elements that are common to both sets.
Step 1: List out all elements in Aand B:
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: Identify elements that are common to both sets:
A∩B={3,4,5}
c) To find Aminus B(A\B), we need to list out elements that are in A
but not in B.
Step 1: List out all elements in Aand B:
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: List out elements that are in Abut not in B:
A\B={1,2}
d) To find Bminus A(B\A), we need to list out elements that are in B
but not in A.
Step 1: List out all elements in Aand B:
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: List out elements that are in Bbut not in A:
B\A={6,7}
16
Question 5
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|4≤x≤12}. Find the set
resulting from the operation A∪B.
Solution
Step 1: List the elements in sets Aand B.
A={2,3,4,5,6,7,8,9,10}
B={4,5,6,7,8,9,10,11,12}
Step 2: Combine the elements from both sets without including any dupli-
cates.
A∪B={2,3,4,5,6,7,8,9,10,11,12}
Therefore, A∪B={2,3,4,5,6,7,8,9,10,11,12}.
Question 6
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find the
intersection A∩Band the union A∪B.
Solution
Step 1: To find the intersection A∩B, we look for elements that are common
to both sets Aand B.
A∩B={x∈Z|xis in both Aand B}
={x∈Z|5≤x≤10}
={5,6,7,8,9,10}.
Step 2: To find the union A∪B, we combine all elements from sets Aand
B.A∪B={x∈Z|xis in Aor B}
={x∈Z|2≤x≤15}
={2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 7
Question
Let A={x| −3≤x≤3}and B={x| −2≤x≤2}. Find the set
(A∩B)∪(A∩Bc).
3
Solution
Step 1: Find A∩B, the intersection of sets Aand B.
A∩B={x| −3≤x≤2}
Step 2: Find A∩Bc, the intersection of set Aand the complement of set B.
A∩Bc={x|2< x ≤3}
Step 3: Find (A∩B)∪(A∩Bc), the union of the two previously found sets.
(A∩B)∪(A∩Bc) = {x| −3≤x≤3}
Question 8
Question
Let A={x∈R|x2−4x+ 3 <0}and B={x∈R|x2−2x−3>0}. Find
A∩B.
Solution
Step 1: Find the values of xthat satisfy the inequality x2−4x+ 3 <0.
The quadratic x2−4x+ 3 factors as (x−1)(x−3).
To find where x2−4x+ 3 is less than 0, we want the signs of the factors
to alternate. This occurs between the roots at x= 1 and x= 3.
So, A={x∈R|1<x<3}.
Step 2: Find the values of xthat satisfy the inequality x2−2x−3>0.
The quadratic x2−2x−3 factors as (x−3)(x+ 1).
To find where x2−2x−3 is greater than 0, we want the signs of the factors
to be positive. This occurs for x < −1 or x > 3.
So, B={x∈R|x < −1 or x > 3}.
Step 3: Find the intersection A∩B.
A∩Brepresents the set of elements that are in both Aand B. So,
A∩B={x∈R|1<x<3 and x < −1 or x > 3}.
Simplifying the condition, we find A∩B={x∈R|1<x<3 and x > 3},
which is an empty set since no real number satisfies both inequalities
simultaneously.
Therefore, A∩B=∅.
4
Question 9
Question
Let A={x∈Z|0≤x≤10}and B={y∈N|5≤y≤15}. Determine the
set A∪B.
Solution
To determine the set A∪B, we need to find all elements that are in either Aor
Bor both.
Step 1: Determine the elements in A.A={0,1,2,3,4,5,6,7,8,9,10}.
Step 2: Determine the elements in B.B={5,6,7,8,9,10,11,12,13,14,15}.
Step 3: Determine A∪B.A∪B={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 10
Question
Let A={x∈Z|2≤x≤8}and B={x∈Z|5≤x≤10}. Find the elements
of the set A∩Band A∪B.
Solution
Step 1: Find A∩B(intersection of sets Aand B).
A∩B={x∈Z|2≤x≤8}∩{x∈Z|5≤x≤10}
A∩B={x∈Z|5≤x≤8}={5,6,7,8}
Step 2: Find A∪B(union of sets Aand B).
A∪B={x∈Z|2≤x≤8}∪{x∈Z|5≤x≤10}
A∪B={x∈Z|2≤x≤10}={2,3,4,5,6,7,8,9,10}
Therefore, the elements of A∩Bare 5, 6, 7, and 8, and the elements of A∪B
are 2, 3, 4, 5, 6, 7, 8, 9, and 10.
Question 11
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Find the set (A∪B)c∩(A∩B).
5
Solution
Step 1: First, we find A∪B:
A∪B={1,2,3,4,5}∪{3,4,5,6,7}={1,2,3,4,5,6,7}
Step 2: Next, we find (A∪B)c:
(A∪B)c={1,2,3,4,5,6,7}c=∅
Step 3: Then, we find A∩B:
A∩B={1,2,3,4,5}∩{3,4,5,6,7}={3,4,5}
Step 4: Finally, we find (A∪B)c∩(A∩B):
∅∩{3,4,5}=∅
Therefore, (A∪B)c∩(A∩B) = ∅.
Question 12
Question
Let A={1,2,3,4},B={3,4,5,6}, and C={2,3,6,7}. Determine the set
(A∪B)∩C.
Solution
Step 1: Find A∪B. Step 2: Find (A∪B)∩C.
Step 1: Find A∪B. Since A={1,2,3,4}and B={3,4,5,6}, we have
A∪B={1,2,3,4}∪{3,4,5,6}={1,2,3,4,5,6}.
Step 2: Find (A∪B)∩C. Now, we need to find the intersection of A∪B
and C. Since A∪B={1,2,3,4,5,6}and C={2,3,6,7}, we have
(A∪B)∩C={1,2,3,4,5,6}∩{2,3,6,7}.
To find the intersection, we look for elements that are common to both sets.
In this case, the common elements are 2, 3, and 6. Therefore,
(A∪B)∩C={2,3,6}.
So, (A∪B)∩C={2,3,6}.
Question 13
Question
Let A={x∈Z|1≤x≤10},B={x∈Z|5≤x≤15}, and C={x∈Z|
1≤x≤5}. Compute (A∪B)∩C.
6
Solution
Step 1: First, let’s find A∪B, the union of sets Aand B.
A∪B={x∈Z|1≤x≤15}
Step 2: Next, let’s find (A∪B)∩C, the intersection of A∪Band C.
(A∪B)∩C={x∈Z|1≤x≤5}
Therefore, (A∪B)∩C={1,2,3,4,5}.
Question 14
Question
Let Aand Bbe sets such that |A|= 8, |B|= 10, and |A∪B|= 14. If |A∩B|= 4,
find |A∩B′|, where B′denotes the complement of set B.
Solution
Step 1: Recall that for any two sets Xand Y, the principle of inclusion-exclusion
states that
|X∪Y|=|X|+|Y|−|X∩Y|.
Step 2: Let’s use the principle of inclusion-exclusion to find |A∪B|. We
have:
|A∪B|=|A|+|B|−|A∩B|.
Substitute the given values:
14 = 8 + 10 −4.
Step 3: Now, let’s find |A∩B′|.
Since |A∪B|=|A|+|B|−|A∩B|, we can rearrange to find |A∩B|:
|A∩B|=|A|+|B|−|A∪B|= 8 + 10 −14 = 4.
Step 4: We know that |A∩B|=|A|−|A∩B′|. Substitute the known values
to find |A∩B′|:
4=8− |A∩B′|.
Step 5: Solving for |A∩B′|, we get
|A∩B′|= 8 −4 = 4.
Therefore, |A∩B′|= 4.
7
Question 15
Question
Let A={1,2,3,4,5},B={3,4,5,6,7}, and C={5,6,7,8,9}. Determine
(A∪B)∩(B∪C).
Solution
Step 1: Find A∪B.
A∪B={1,2,3,4,5}∪{3,4,5,6,7}
={1,2,3,4,5,6,7}
Step 2: Find B∪C.
B∪C={3,4,5,6,7}∪{5,6,7,8,9}
={3,4,5,6,7,8,9}
Step 3: Find (A∪B)∩(B∪C).
(A∪B)∩(B∪C) = {1,2,3,4,5,6,7}∩{3,4,5,6,7,8,9}
={3,4,5,6,7}
Therefore, (A∪B)∩(B∪C) = {3,4,5,6,7}.
Question 16
Question
Let A={1,2,3},B={2,3,4}, and C={3,4,5}. Find (A∪B)∩C.
Solution
Step 1: First, find the union of sets Aand B.
Step 1: A∪B={1,2,3}∪{2,3,4}
A∪B={1,2,3,4}
Step 2: Next, find the intersection of the result from Step 1 and set C.
Step 2: (A∪B)∩C={1,2,3,4}∩{3,4,5}
(A∪B)∩C={3,4}
Therefore, (A∪B)∩C={3,4}.
8
Question 17
Question
Let A={x∈R:x2−7x+ 10 >0}and B={x∈R:x2−6x+ 5 >0}. Find
the set A∩B.
Solution
Step 1: First, find the values of xthat satisfy the inequalities for sets Aand
Bseparately. For set A: We need to solve the inequality x2−7x+ 10 >0.
Factorizing the quadratic gives (x−5)(x−2) >0. From this, we can see that
x∈(2,5). Therefore, A={x∈R: 2 <x<5}.
Step 2: Now, find the values of xfor set B. For set B: We need to solve the
inequality x2−6x+ 5 >0. Factorizing the quadratic gives (x−5)(x−1) >0.
From this, we can see that x∈(−∞,1) ∪(5,+∞). Therefore, B={x∈R:
x < 1 or x > 5}.
Step 3: Finally, find the intersection of sets Aand B. To find A∩B, we
need to find the values of xthat satisfy both inequalities. By construction, set
Arequires 2 <x<5 and set Brequires x < 1 or x > 5. Thus, the only overlap
between the two sets is (2,1). Therefore, A∩B={x∈R: 2 < x < 5} ∩ {x∈
R:x < 1 or x > 5}=∅.
Question 18
Question
Let A,B, and Cbe sets such that A∪B=A∪Cand A∩B=A∩C. Prove
or disprove that B=C.
Solution
To prove or disprove that B=Cbased on the given information, we will consider
the definitions of union and intersection of sets.
Step 1: Recall the definition of set equality. Two sets Xand Yare equal,
denoted X=Y, if and only if every element of Xis an element of Yand vice
versa.
Step 2: Prove the statement. We are given that A∪B=A∪Cand
A∩B=A∩C. From the definition of union, A∪Bincludes all elements in
Aand B(at least once), and A∪Cincludes all elements in Aand C(at least
once). Similarly, from the definition of intersection, A∩Bincludes all elements
that are common to both Aand B, and A∩Cincludes all elements that are
common to both Aand C.
Step 3: Analyze the sets Band C. Based on the given information: -
A∪B=A∪Cimplies that any element in B(which is not in A) must also be
9
in Cto have the same union with A. - A∩B=A∩Cimplies that any element
in B(which is in A) must also be in Cto have the same intersection with A.
Step 4: Conclusion. Since Band Cmust contain the same elements outside
of Aand the same elements inside of Ain order to satisfy the given conditions,
we can conclude that B=C.
Question 19
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find
(A∪B)∩(A∩B).
Solution
Step 1: First, let’s find A∪B.
A∪B={x∈Z|1≤x≤10}∪{x∈Z|5≤x≤15}
In this case, A∪Bwill be the set of all integers that are in either set Aor
set B(or both).
Step 2: By examining the two sets, we can see that A∪Bwill be from the
smallest value in Ato the largest value in B, which is from 1 to 15. Therefore,
A∪B={x∈Z|1≤x≤15}
Step 3: Next, let’s find A∩B.
A∩B={x∈Z|1≤x≤10}∩{x∈Z|5≤x≤15}
A∩Bis the set of all integers that are in both set Aand set B.
Step 4: By comparing the two sets, we see that the set of common values
between Aand Bis from 5 to 10. Therefore,
A∩B={x∈Z|5≤x≤10}
Step 5: Finally, let’s find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {x∈Z|1≤x≤15}∩{x∈Z|5≤x≤10}
Step 6: By comparing the two sets, we can see that the common values
between (A∪B) and (A∩B) are from 5 to 10. Therefore,
(A∪B)∩(A∩B) = {x∈Z|5≤x≤10}
Question 20
Question
Let A={x∈R|0<x<2}and B={x∈R|1<x<3}. Find A∩B.
10
Solution
Step 1: To find A∩B, we need to determine the elements that are common to
both sets Aand B.
Step 2: Set Acontains all real numbers xsuch that 0 <x<2, and set B
contains all real numbers xsuch that 1 <x<3.
Step 3: To find A∩B, we need to find the values of xthat satisfy both
0<x<2 and 1 <x<3.
Step 4: The intersection of the two sets can be found by determining the
range in which the inequalities of both sets overlap.
Step 5: We see that the overlapping range is 1 <x<2, which means
A∩B={x∈R|1<x<2}.
Question 21
Question
Let A={x∈N: 1 ≤x≤10}and B={x∈N: 5 ≤x≤15}. Determine the
set (A∩B)∪(A∩B∁).
Solution
Step 1: Find A∩B.
A∩B={x∈N: 1 ≤x≤10}∩{x∈N: 5 ≤x≤15}
={5,6,7,8,9,10}
Step 2: Find B∁, the complement of set B.
B∁={x∈N:x < 5 or x > 15}
={1,2,3,4,16,17,18,19, ...}
Step 3: Find A∩B∁.
A∩B∁={x∈N: 1 ≤x≤10}∩{x∈N:x < 5 or x > 15}
={1,2,3,4}
Step 4: Find (A∩B)∪(A∩B∁).
(A∩B)∪(A∩B∁) = {5,6,7,8,9,10}∪{1,2,3,4}
={1,2,3,4,5,6,7,8,9,10}
Therefore, (A∩B)∪(A∩B∁) = {1,2,3,4,5,6,7,8,9,10}.
11
Question 22
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Determine A∩B,
A∪B, and A\B.
Solution
Step 1: To find A∩B(the intersection of sets Aand B), we look for elements
that are in both sets.
A∩B={x∈Z|5≤x≤10}.
Step 2: Next, to find A∪B(the union of sets Aand B), we combine all
elements in both sets without repetitions.
A∪B={x∈Z|1≤x≤15}.
Step 3: Finally, to find A\B(the set difference of set Awith respect to set
B), we look for elements that are in set Abut not in set B.
A\B={x∈Z|1≤x≤4}.
Question 23
Question
Let A={x∈R| −2≤x≤3}and B={x∈R|x2<4}. Find the set A∪B.
Solution
Step 1: Let’s first determine the elements of set Aand set B. - Set A: The
set Aconsists of all real numbers xsuch that −2≤x≤3. Therefore, A=
{−2,−1,0,1,2,3}. - Set B: The set Bconsists of all real numbers xsuch that
x2<4. This means xcan be in the range −2<x<2. So, B={−2,−1,0,1}.
Step 2: Now, let’s find the union of sets Aand B, denoted by A∪B. The
union of two sets Aand Bis the set of all elements that are in A, in B, or in
both Aand B.
Step 3: Combining the elements of sets Aand B, we have A∪B={−2,−1,0,1,2,3}.
Therefore, the set A∪Bis {−2,−1,0,1,2,3}.
Question 24
Question
Let A,B, and Cbe sets such that A∪B=A∩C. Prove or disprove: B⊆C.
12
Solution
To prove or disprove B⊆C, we need to show whether every element in set B
is also in set C.
Step 1: Let’s start by assuming that x∈B.
Step 2: Since A∪B=A∩C, we know that x∈A∪Bif and only if
x∈A∩C.
Step 3: If x∈A∪B, then either x∈Aor x∈B.
Step 4: Since x∈B(by our assumption), we have two cases to consider:
Case 1: If x∈A, then x∈A∩C, which implies x∈C.
Case 2: If x∈B, then x∈A∩C, which implies x∈C.
Step 5: In either case, we have shown that if x∈B, then x∈C. Therefore,
B⊆C.
Step 6: Hence, we have proved that if A∪B=A∩C, then B⊆C.
Question 25
Question
Let Aand Bbe sets such that |A|= 10 and |B|= 15. If |A∪B|= 20 and
|A∩B|= 5, find |A∩B′|where B′denotes the complement of set B.
Solution
Step 1: We know that the union of two sets can be calculated using the formula
|A∪B|=|A|+|B|−|A∩B|. Therefore, we can find |B|using the known values.
|A∪B|=|A|+|B|−|A∩B|=⇒20 = 10 + 15 −5 = 20 =⇒ |B|= 15
Step 2: To find |A∩B′|, we can use the formula |A∩B′|=|A|−|A∩B|.
Since we already know |A∩B|= 5, we can calculate |A∩B′|.
|A∩B′|=|A|−|A∩B|= 10 −5=5
Therefore, |A∩B′|= 5.
Question 26
Question
Let A={x∈R|x2−4x+ 3 = 0}and B={x∈R|x2−3x−10 = 0}. Find
the set A∩B.
13
Solution
Step 1: Find the solutions to the equations x2−4x+3 = 0 and x2−3x−10 = 0.
For x2−4x+ 3 = 0: Factoring gives (x−1)(x−3) = 0, so the solutions
are x= 1 and x= 3.
For x2−3x−10 = 0: Factoring gives (x−5)(x+ 2) = 0, so the solutions
are x= 5 and x=−2.
Step 2: Find A∩B, the intersection of sets Aand B. This is the set of all
elements that are in both Aand B.
A∩B={x∈R|x∈Aand x∈B}
={1,3}∩{5,−2}
={}
Therefore, A∩B=∅.
Question 27
Question
Let A={1,2,3,4}and B={3,4,5,6}. Define the set Cas the set of all
elements which are in either Aor B, but not both. Find the set C.
Solution
Step 1: First, let’s find the elements that are in Aor B, but not both. Step
2: The elements in A∪Bare {1,2,3,4,5,6}. Step 3: The elements in A∩B
are {3,4}. Step 4: The elements that are in either Aor B, but not both, are
{1,2,5,6}. Step 5: Therefore, the set Cis C={1,2,5,6}.
Question 28
Question
Let Aand Bbe sets such that |A∩B|= 8, |A\B|= 12, and |B\A|= 15. Find
|A∪B|.
Solution
Step 1: Recall the formula for the cardinality of the union of two sets:
|A∪B|=|A|+|B|−|A∩B|.
Step 2: We are given |A∩B|= 8, |A\B|= 12, and |B\A|= 15. Let’s denote
|A∩B|=x,|A\B|=y, and |B\A|=zfor simplicity.
14
Step 3: We know that |A|=|A∩B|+|A\B|=x+yand |B|=|A∩B|+
|B\A|=x+z.
Step 4: Substituting x= 8, y= 12, and z= 15, we have |A|= 8 + 12 = 20
and |B|= 8 + 15 = 23.
Step 5: Now, we can find |A∪B|using the formula:
|A∪B|=|A|+|B|−|A∩B|= 20 + 23 −8 = 43 −8 = 35.
Therefore, |A∪B|= 35.
Question 29
Question
Let A={x∈R| −3≤x≤3}and B={x∈R| −2≤x≤4}. Find the set
(A∩B)∪(A∩Bc).
Solution
Step 1: Find A∩B.
A∩B={x∈R| −3≤x≤3}∩{x∈R| −2≤x≤4}
={x∈R| −2≤x≤3}
Step 2: Find A∩Bc.
A∩Bc={x∈R| −3≤x≤3}∩{x∈R|x < −2 or x > 4}
={x∈R| −3≤x≤ −2}∪{x∈R|3<x<4}
Step 3: Find (A∩B)∪(A∩Bc).
(A∩B)∪(A∩Bc) = {x∈R| −2≤x≤3}∪{x∈R| −3≤x≤ −2}∪{x∈R|3<x<4}
={x∈R| −3≤x≤4}
Therefore, (A∩B)∪(A∩Bc) = {x∈R| −3≤x≤4}.
Question 30
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Find each of the following sets:
a) A∪B
b) A∩B
c) A\B
d) B\A
15
Solution
a) To find the union of sets Aand B(A∪B), we need to combine all unique
elements from both sets.
Step 1: List out all elements in Aand B:
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: Combine all unique elements from Aand B:
A∪B={1,2,3,4,5,6,7}
b) To find the intersection of sets Aand B(A∩B), we need to identify
elements that are common to both sets.
Step 1: List out all elements in Aand B:
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: Identify elements that are common to both sets:
A∩B={3,4,5}
c) To find Aminus B(A\B), we need to list out elements that are in A
but not in B.
Step 1: List out all elements in Aand B:
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: List out elements that are in Abut not in B:
A\B={1,2}
d) To find Bminus A(B\A), we need to list out elements that are in B
but not in A.
Step 1: List out all elements in Aand B:
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: List out elements that are in Bbut not in A:
B\A={6,7}
16
Question 5
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|4≤x≤12}. Find the set
resulting from the operation A∪B.
Solution
Step 1: List the elements in sets Aand B.
A={2,3,4,5,6,7,8,9,10}
B={4,5,6,7,8,9,10,11,12}
Step 2: Combine the elements from both sets without including any dupli-
cates.
A∪B={2,3,4,5,6,7,8,9,10,11,12}
Therefore, A∪B={2,3,4,5,6,7,8,9,10,11,12}.
Question 6
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find the
intersection A∩Band the union A∪B.
Solution
Step 1: To find the intersection A∩B, we look for elements that are common
to both sets Aand B.
A∩B={x∈Z|xis in both Aand B}
={x∈Z|5≤x≤10}
={5,6,7,8,9,10}.
Step 2: To find the union A∪B, we combine all elements from sets Aand
B.A∪B={x∈Z|xis in Aor B}
={x∈Z|2≤x≤15}
={2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 7
Question
Let A={x| −3≤x≤3}and B={x| −2≤x≤2}. Find the set
(A∩B)∪(A∩Bc).
3
Solution
Step 1: Find A∩B, the intersection of sets Aand B.
A∩B={x| −3≤x≤2}
Step 2: Find A∩Bc, the intersection of set Aand the complement of set B.
A∩Bc={x|2< x ≤3}
Step 3: Find (A∩B)∪(A∩Bc), the union of the two previously found sets.
(A∩B)∪(A∩Bc) = {x| −3≤x≤3}
Question 8
Question
Let A={x∈R|x2−4x+ 3 <0}and B={x∈R|x2−2x−3>0}. Find
A∩B.
Solution
Step 1: Find the values of xthat satisfy the inequality x2−4x+ 3 <0.
The quadratic x2−4x+ 3 factors as (x−1)(x−3).
To find where x2−4x+ 3 is less than 0, we want the signs of the factors
to alternate. This occurs between the roots at x= 1 and x= 3.
So, A={x∈R|1<x<3}.
Step 2: Find the values of xthat satisfy the inequality x2−2x−3>0.
The quadratic x2−2x−3 factors as (x−3)(x+ 1).
To find where x2−2x−3 is greater than 0, we want the signs of the factors
to be positive. This occurs for x < −1 or x > 3.
So, B={x∈R|x < −1 or x > 3}.
Step 3: Find the intersection A∩B.
A∩Brepresents the set of elements that are in both Aand B. So,
A∩B={x∈R|1<x<3 and x < −1 or x > 3}.
Simplifying the condition, we find A∩B={x∈R|1<x<3 and x > 3},
which is an empty set since no real number satisfies both inequalities
simultaneously.
Therefore, A∩B=∅.
4
Question 9
Question
Let A={x∈Z|0≤x≤10}and B={y∈N|5≤y≤15}. Determine the
set A∪B.
Solution
To determine the set A∪B, we need to find all elements that are in either Aor
Bor both.
Step 1: Determine the elements in A.A={0,1,2,3,4,5,6,7,8,9,10}.
Step 2: Determine the elements in B.B={5,6,7,8,9,10,11,12,13,14,15}.
Step 3: Determine A∪B.A∪B={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 10
Question
Let A={x∈Z|2≤x≤8}and B={x∈Z|5≤x≤10}. Find the elements
of the set A∩Band A∪B.
Solution
Step 1: Find A∩B(intersection of sets Aand B).
A∩B={x∈Z|2≤x≤8}∩{x∈Z|5≤x≤10}
A∩B={x∈Z|5≤x≤8}={5,6,7,8}
Step 2: Find A∪B(union of sets Aand B).
A∪B={x∈Z|2≤x≤8}∪{x∈Z|5≤x≤10}
A∪B={x∈Z|2≤x≤10}={2,3,4,5,6,7,8,9,10}
Therefore, the elements of A∩Bare 5, 6, 7, and 8, and the elements of A∪B
are 2, 3, 4, 5, 6, 7, 8, 9, and 10.
Question 11
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Find the set (A∪B)c∩(A∩B).
5
Solution
Step 1: First, we find A∪B:
A∪B={1,2,3,4,5}∪{3,4,5,6,7}={1,2,3,4,5,6,7}
Step 2: Next, we find (A∪B)c:
(A∪B)c={1,2,3,4,5,6,7}c=∅
Step 3: Then, we find A∩B:
A∩B={1,2,3,4,5}∩{3,4,5,6,7}={3,4,5}
Step 4: Finally, we find (A∪B)c∩(A∩B):
∅∩{3,4,5}=∅
Therefore, (A∪B)c∩(A∩B) = ∅.
Question 12
Question
Let A={1,2,3,4},B={3,4,5,6}, and C={2,3,6,7}. Determine the set
(A∪B)∩C.
Solution
Step 1: Find A∪B. Step 2: Find (A∪B)∩C.
Step 1: Find A∪B. Since A={1,2,3,4}and B={3,4,5,6}, we have
A∪B={1,2,3,4}∪{3,4,5,6}={1,2,3,4,5,6}.
Step 2: Find (A∪B)∩C. Now, we need to find the intersection of A∪B
and C. Since A∪B={1,2,3,4,5,6}and C={2,3,6,7}, we have
(A∪B)∩C={1,2,3,4,5,6}∩{2,3,6,7}.
To find the intersection, we look for elements that are common to both sets.
In this case, the common elements are 2, 3, and 6. Therefore,
(A∪B)∩C={2,3,6}.
So, (A∪B)∩C={2,3,6}.
Question 13
Question
Let A={x∈Z|1≤x≤10},B={x∈Z|5≤x≤15}, and C={x∈Z|
1≤x≤5}. Compute (A∪B)∩C.
6
Solution
Step 1: First, let’s find A∪B, the union of sets Aand B.
A∪B={x∈Z|1≤x≤15}
Step 2: Next, let’s find (A∪B)∩C, the intersection of A∪Band C.
(A∪B)∩C={x∈Z|1≤x≤5}
Therefore, (A∪B)∩C={1,2,3,4,5}.
Question 14
Question
Let Aand Bbe sets such that |A|= 8, |B|= 10, and |A∪B|= 14. If |A∩B|= 4,
find |A∩B′|, where B′denotes the complement of set B.
Solution
Step 1: Recall that for any two sets Xand Y, the principle of inclusion-exclusion
states that
|X∪Y|=|X|+|Y|−|X∩Y|.
Step 2: Let’s use the principle of inclusion-exclusion to find |A∪B|. We
have:
|A∪B|=|A|+|B|−|A∩B|.
Substitute the given values:
14 = 8 + 10 −4.
Step 3: Now, let’s find |A∩B′|.
Since |A∪B|=|A|+|B|−|A∩B|, we can rearrange to find |A∩B|:
|A∩B|=|A|+|B|−|A∪B|= 8 + 10 −14 = 4.
Step 4: We know that |A∩B|=|A|−|A∩B′|. Substitute the known values
to find |A∩B′|:
4=8− |A∩B′|.
Step 5: Solving for |A∩B′|, we get
|A∩B′|= 8 −4 = 4.
Therefore, |A∩B′|= 4.
7
Question 15
Question
Let A={1,2,3,4,5},B={3,4,5,6,7}, and C={5,6,7,8,9}. Determine
(A∪B)∩(B∪C).
Solution
Step 1: Find A∪B.
A∪B={1,2,3,4,5}∪{3,4,5,6,7}
={1,2,3,4,5,6,7}
Step 2: Find B∪C.
B∪C={3,4,5,6,7}∪{5,6,7,8,9}
={3,4,5,6,7,8,9}
Step 3: Find (A∪B)∩(B∪C).
(A∪B)∩(B∪C) = {1,2,3,4,5,6,7}∩{3,4,5,6,7,8,9}
={3,4,5,6,7}
Therefore, (A∪B)∩(B∪C) = {3,4,5,6,7}.
Question 16
Question
Let A={1,2,3},B={2,3,4}, and C={3,4,5}. Find (A∪B)∩C.
Solution
Step 1: First, find the union of sets Aand B.
Step 1: A∪B={1,2,3}∪{2,3,4}
A∪B={1,2,3,4}
Step 2: Next, find the intersection of the result from Step 1 and set C.
Step 2: (A∪B)∩C={1,2,3,4}∩{3,4,5}
(A∪B)∩C={3,4}
Therefore, (A∪B)∩C={3,4}.
8
Question 17
Question
Let A={x∈R:x2−7x+ 10 >0}and B={x∈R:x2−6x+ 5 >0}. Find
the set A∩B.
Solution
Step 1: First, find the values of xthat satisfy the inequalities for sets Aand
Bseparately. For set A: We need to solve the inequality x2−7x+ 10 >0.
Factorizing the quadratic gives (x−5)(x−2) >0. From this, we can see that
x∈(2,5). Therefore, A={x∈R: 2 <x<5}.
Step 2: Now, find the values of xfor set B. For set B: We need to solve the
inequality x2−6x+ 5 >0. Factorizing the quadratic gives (x−5)(x−1) >0.
From this, we can see that x∈(−∞,1) ∪(5,+∞). Therefore, B={x∈R:
x < 1 or x > 5}.
Step 3: Finally, find the intersection of sets Aand B. To find A∩B, we
need to find the values of xthat satisfy both inequalities. By construction, set
Arequires 2 <x<5 and set Brequires x < 1 or x > 5. Thus, the only overlap
between the two sets is (2,1). Therefore, A∩B={x∈R: 2 < x < 5} ∩ {x∈
R:x < 1 or x > 5}=∅.
Question 18
Question
Let A,B, and Cbe sets such that A∪B=A∪Cand A∩B=A∩C. Prove
or disprove that B=C.
Solution
To prove or disprove that B=Cbased on the given information, we will consider
the definitions of union and intersection of sets.
Step 1: Recall the definition of set equality. Two sets Xand Yare equal,
denoted X=Y, if and only if every element of Xis an element of Yand vice
versa.
Step 2: Prove the statement. We are given that A∪B=A∪Cand
A∩B=A∩C. From the definition of union, A∪Bincludes all elements in
Aand B(at least once), and A∪Cincludes all elements in Aand C(at least
once). Similarly, from the definition of intersection, A∩Bincludes all elements
that are common to both Aand B, and A∩Cincludes all elements that are
common to both Aand C.
Step 3: Analyze the sets Band C. Based on the given information: -
A∪B=A∪Cimplies that any element in B(which is not in A) must also be
9
in Cto have the same union with A. - A∩B=A∩Cimplies that any element
in B(which is in A) must also be in Cto have the same intersection with A.
Step 4: Conclusion. Since Band Cmust contain the same elements outside
of Aand the same elements inside of Ain order to satisfy the given conditions,
we can conclude that B=C.
Question 19
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find
(A∪B)∩(A∩B).
Solution
Step 1: First, let’s find A∪B.
A∪B={x∈Z|1≤x≤10}∪{x∈Z|5≤x≤15}
In this case, A∪Bwill be the set of all integers that are in either set Aor
set B(or both).
Step 2: By examining the two sets, we can see that A∪Bwill be from the
smallest value in Ato the largest value in B, which is from 1 to 15. Therefore,
A∪B={x∈Z|1≤x≤15}
Step 3: Next, let’s find A∩B.
A∩B={x∈Z|1≤x≤10}∩{x∈Z|5≤x≤15}
A∩Bis the set of all integers that are in both set Aand set B.
Step 4: By comparing the two sets, we see that the set of common values
between Aand Bis from 5 to 10. Therefore,
A∩B={x∈Z|5≤x≤10}
Step 5: Finally, let’s find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {x∈Z|1≤x≤15}∩{x∈Z|5≤x≤10}
Step 6: By comparing the two sets, we can see that the common values
between (A∪B) and (A∩B) are from 5 to 10. Therefore,
(A∪B)∩(A∩B) = {x∈Z|5≤x≤10}
Question 20
Question
Let A={x∈R|0<x<2}and B={x∈R|1<x<3}. Find A∩B.
10
Solution
Step 1: To find A∩B, we need to determine the elements that are common to
both sets Aand B.
Step 2: Set Acontains all real numbers xsuch that 0 <x<2, and set B
contains all real numbers xsuch that 1 <x<3.
Step 3: To find A∩B, we need to find the values of xthat satisfy both
0<x<2 and 1 <x<3.
Step 4: The intersection of the two sets can be found by determining the
range in which the inequalities of both sets overlap.
Step 5: We see that the overlapping range is 1 <x<2, which means
A∩B={x∈R|1<x<2}.
Question 21
Question
Let A={x∈N: 1 ≤x≤10}and B={x∈N: 5 ≤x≤15}. Determine the
set (A∩B)∪(A∩B∁).
Solution
Step 1: Find A∩B.
A∩B={x∈N: 1 ≤x≤10}∩{x∈N: 5 ≤x≤15}
={5,6,7,8,9,10}
Step 2: Find B∁, the complement of set B.
B∁={x∈N:x < 5 or x > 15}
={1,2,3,4,16,17,18,19, ...}
Step 3: Find A∩B∁.
A∩B∁={x∈N: 1 ≤x≤10}∩{x∈N:x < 5 or x > 15}
={1,2,3,4}
Step 4: Find (A∩B)∪(A∩B∁).
(A∩B)∪(A∩B∁) = {5,6,7,8,9,10}∪{1,2,3,4}
={1,2,3,4,5,6,7,8,9,10}
Therefore, (A∩B)∪(A∩B∁) = {1,2,3,4,5,6,7,8,9,10}.
11
Question 22
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Determine A∩B,
A∪B, and A\B.
Solution
Step 1: To find A∩B(the intersection of sets Aand B), we look for elements
that are in both sets.
A∩B={x∈Z|5≤x≤10}.
Step 2: Next, to find A∪B(the union of sets Aand B), we combine all
elements in both sets without repetitions.
A∪B={x∈Z|1≤x≤15}.
Step 3: Finally, to find A\B(the set difference of set Awith respect to set
B), we look for elements that are in set Abut not in set B.
A\B={x∈Z|1≤x≤4}.
Question 23
Question
Let A={x∈R| −2≤x≤3}and B={x∈R|x2<4}. Find the set A∪B.
Solution
Step 1: Let’s first determine the elements of set Aand set B. - Set A: The
set Aconsists of all real numbers xsuch that −2≤x≤3. Therefore, A=
{−2,−1,0,1,2,3}. - Set B: The set Bconsists of all real numbers xsuch that
x2<4. This means xcan be in the range −2<x<2. So, B={−2,−1,0,1}.
Step 2: Now, let’s find the union of sets Aand B, denoted by A∪B. The
union of two sets Aand Bis the set of all elements that are in A, in B, or in
both Aand B.
Step 3: Combining the elements of sets Aand B, we have A∪B={−2,−1,0,1,2,3}.
Therefore, the set A∪Bis {−2,−1,0,1,2,3}.
Question 24
Question
Let A,B, and Cbe sets such that A∪B=A∩C. Prove or disprove: B⊆C.
12
Solution
To prove or disprove B⊆C, we need to show whether every element in set B
is also in set C.
Step 1: Let’s start by assuming that x∈B.
Step 2: Since A∪B=A∩C, we know that x∈A∪Bif and only if
x∈A∩C.
Step 3: If x∈A∪B, then either x∈Aor x∈B.
Step 4: Since x∈B(by our assumption), we have two cases to consider:
Case 1: If x∈A, then x∈A∩C, which implies x∈C.
Case 2: If x∈B, then x∈A∩C, which implies x∈C.
Step 5: In either case, we have shown that if x∈B, then x∈C. Therefore,
B⊆C.
Step 6: Hence, we have proved that if A∪B=A∩C, then B⊆C.
Question 25
Question
Let Aand Bbe sets such that |A|= 10 and |B|= 15. If |A∪B|= 20 and
|A∩B|= 5, find |A∩B′|where B′denotes the complement of set B.
Solution
Step 1: We know that the union of two sets can be calculated using the formula
|A∪B|=|A|+|B|−|A∩B|. Therefore, we can find |B|using the known values.
|A∪B|=|A|+|B|−|A∩B|=⇒20 = 10 + 15 −5 = 20 =⇒ |B|= 15
Step 2: To find |A∩B′|, we can use the formula |A∩B′|=|A|−|A∩B|.
Since we already know |A∩B|= 5, we can calculate |A∩B′|.
|A∩B′|=|A|−|A∩B|= 10 −5=5
Therefore, |A∩B′|= 5.
Question 26
Question
Let A={x∈R|x2−4x+ 3 = 0}and B={x∈R|x2−3x−10 = 0}. Find
the set A∩B.
13
Solution
Step 1: Find the solutions to the equations x2−4x+3 = 0 and x2−3x−10 = 0.
For x2−4x+ 3 = 0: Factoring gives (x−1)(x−3) = 0, so the solutions
are x= 1 and x= 3.
For x2−3x−10 = 0: Factoring gives (x−5)(x+ 2) = 0, so the solutions
are x= 5 and x=−2.
Step 2: Find A∩B, the intersection of sets Aand B. This is the set of all
elements that are in both Aand B.
A∩B={x∈R|x∈Aand x∈B}
={1,3}∩{5,−2}
={}
Therefore, A∩B=∅.
Question 27
Question
Let A={1,2,3,4}and B={3,4,5,6}. Define the set Cas the set of all
elements which are in either Aor B, but not both. Find the set C.
Solution
Step 1: First, let’s find the elements that are in Aor B, but not both. Step
2: The elements in A∪Bare {1,2,3,4,5,6}. Step 3: The elements in A∩B
are {3,4}. Step 4: The elements that are in either Aor B, but not both, are
{1,2,5,6}. Step 5: Therefore, the set Cis C={1,2,5,6}.
Question 28
Question
Let Aand Bbe sets such that |A∩B|= 8, |A\B|= 12, and |B\A|= 15. Find
|A∪B|.
Solution
Step 1: Recall the formula for the cardinality of the union of two sets:
|A∪B|=|A|+|B|−|A∩B|.
Step 2: We are given |A∩B|= 8, |A\B|= 12, and |B\A|= 15. Let’s denote
|A∩B|=x,|A\B|=y, and |B\A|=zfor simplicity.
14
Step 3: We know that |A|=|A∩B|+|A\B|=x+yand |B|=|A∩B|+
|B\A|=x+z.
Step 4: Substituting x= 8, y= 12, and z= 15, we have |A|= 8 + 12 = 20
and |B|= 8 + 15 = 23.
Step 5: Now, we can find |A∪B|using the formula:
|A∪B|=|A|+|B|−|A∩B|= 20 + 23 −8 = 43 −8 = 35.
Therefore, |A∪B|= 35.
Question 29
Question
Let A={x∈R| −3≤x≤3}and B={x∈R| −2≤x≤4}. Find the set
(A∩B)∪(A∩Bc).
Solution
Step 1: Find A∩B.
A∩B={x∈R| −3≤x≤3}∩{x∈R| −2≤x≤4}
={x∈R| −2≤x≤3}
Step 2: Find A∩Bc.
A∩Bc={x∈R| −3≤x≤3}∩{x∈R|x < −2 or x > 4}
={x∈R| −3≤x≤ −2}∪{x∈R|3<x<4}
Step 3: Find (A∩B)∪(A∩Bc).
(A∩B)∪(A∩Bc) = {x∈R| −2≤x≤3}∪{x∈R| −3≤x≤ −2}∪{x∈R|3<x<4}
={x∈R| −3≤x≤4}
Therefore, (A∩B)∪(A∩Bc) = {x∈R| −3≤x≤4}.
Question 30
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Find each of the following sets:
a) A∪B
b) A∩B
c) A\B
d) B\A
15
Solution
a) To find the union of sets Aand B(A∪B), we need to combine all unique
elements from both sets.
Step 1: List out all elements in Aand B:
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: Combine all unique elements from Aand B:
A∪B={1,2,3,4,5,6,7}
b) To find the intersection of sets Aand B(A∩B), we need to identify
elements that are common to both sets.
Step 1: List out all elements in Aand B:
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: Identify elements that are common to both sets:
A∩B={3,4,5}
c) To find Aminus B(A\B), we need to list out elements that are in A
but not in B.
Step 1: List out all elements in Aand B:
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: List out elements that are in Abut not in B:
A\B={1,2}
d) To find Bminus A(B\A), we need to list out elements that are in B
but not in A.
Step 1: List out all elements in Aand B:
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: List out elements that are in Bbut not in A:
B\A={6,7}
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