MATH 350 - DISCRETE
MATHEMATICS - Operations on sets
Question Bank - Set 4
Liberty University
Question 1
Question
Let A={x∈R|x2<4}and B={x∈R|x > −2}. Determine A∩B.
Solution
Step 1: Find A.
A={x∈R|x2<4}
={x∈R| −2<x<2}
= (−2,2)
Step 2: Find B.
B={x∈R|x > −2}
= (−2,∞)
Step 3: Find A∩B.
A∩B= (−2,2) ∩(−2,∞)
= (−2,2)
Therefore, A∩B= (−2,2).
Question 2
Question
Let A={x∈Q|x2<2}and B={x∈Q|x2>2}. Determine A∩B.
Solution
Step 1: To find A∩B, we first need to determine A∩B.
A∩B={x∈Q|x2<2}∩{x∈Q|x2>2}
={x∈Q|x2<2 and x2>2}
Step 2: Since x2<2 and x2>2 cannot hold true simultaneously for any
rational number x, we have A∩B=∅(the empty set).
Step 3: The complement of the empty set is the universal set, so A∩B=Q.
Therefore, A∩B=Q.
Question 3
Question
Let A={x∈Z|x2−4x−5 = 0}and B={x∈Z|x2−x−6 = 0}. Find
A∪Band A∩B.
Solution
Step 1: Find the solutions to the equations defining sets Aand B.
For set A: The equation x2−4x−5 = 0 can be factored as (x−5)(x+1) = 0.
So, the solutions are x= 5 and x=−1. Therefore, A={5,−1}.
For set B: The equation x2−x−6 = 0 can be factored as (x−3)(x+2) = 0.
So, the solutions are x= 3 and x=−2. Therefore, B={3,−2}.
Step 2: Determine A∪B(the union of sets Aand B). A∪Bis the set of
all elements that are in Aor in B(or both). Therefore, A∪B={5,−1,3,−2}.
Step 3: Determine A∩B(the intersection of sets Aand B). A∩Bis the set
of all elements that are in both Aand B. Since A={5,−1}and B={3,−2},
A∩B=∅(the empty set) because there are no elements common to both sets.
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 take the union of sets Aand B, which means
combining all the elements in both sets without repetition.
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
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Step 2: To find A∩B, we need to find the intersection of sets Aand B,
which means finding the elements that are common to both sets.
A∩B={5,6,7,8,9,10}
Step 3: To find A\B, we need to determine the set of elements that are in
Abut not in B.
A\B={1,2,3,4}
Question 5
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find A∩B.
Solution
To find A∩B, we need to determine the elements that are common to both sets
Aand B.
Step 1: Write out the elements of set Aand B.
A={2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
Step 2: Identify the common elements between sets Aand B.
A∩B={5,6,7,8,9,10}
Step 3: Write the final answer.
A∩B={5,6,7,8,9,10}
Question 6
Question
Let A={x∈R|x2<5}and B={y∈R|y2≥3}. Determine A∪B.
Solution
To find A∪B, we need to find the set of all elements that are in either set Aor
set B.
Step 1: Find all elements in set A. Set A={x∈R|x2<5}can be
rewritten as A={x∈R| −√5<x<√5}. Therefore, set Acontains all real
numbers between −√5 and √5.
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Step 2: Find all elements in set B. Set B={y∈R|y2≥3}can be
rewritten as B={y∈R|y≤ −√3 or y≥√3}. Therefore, set Bcontains all
real numbers less than or equal to −√3 or greater than or equal to √3.
Step 3: Find A∪B.The union of sets Aand B, denoted as A∪B, contains
all the elements that are in either set A, set B, or both.
Therefore, A∪B={x∈R|x≤ −√3 or −√5<x<√5 or x≥√3}.
Hence, A∪B= (−∞,−√3] ∪(−√5,√5) ∪[√3,∞).
Question 7
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Define the set Cas C=A∪Band
Das D=A∩B. Find the Cartesian product C×D.
Solution
To find the Cartesian product C×D, we first need to determine the sets Cand
D.
Step 1: Calculate C=A∪B.A={1,2,3,4,5}B={3,4,5,6,7}C=
A∪B={1,2,3,4,5,6,7}
Step 2: Calculate D=A∩B.A={1,2,3,4,5}B={3,4,5,6,7}D=
A∩B={3,4,5}
Step 3: Find C×D.C={1,2,3,4,5,6,7}D={3,4,5}C×D=
{(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,3),(3,4),(3,5),(4,3),(4,4),(4,5),(5,3),(5,4),(5,5),(6,3),(6,4),(6,5),(7,3),(7,4),(7,5)}
Therefore, C×Dis {(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,3),(3,4),(3,5),
(4,3),(4,4),(4,5),(5,3),(5,4),(5,5),(6,3),(6,4),(6,5),(7,3),(7,4),(7,5)}.
Question 8
Question
Let A={x∈R| −3≤x≤5}and B={x∈R|0< x < 6}. Compute the
following set:
A∪B−A∩B
Solution
We first find A∩Band then compute A∪B−(A∩B).
Step 1: Find A∩BTo find A∩B, we need to identify the elements that
are in both Aand B.
A∩B={x∈R| −3≤x≤5 and 0 <x<6}
Intersecting the intervals [−3,5] and (0,6), we see that A∩B= (0,5].
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Step 2: Compute A∪B−A∩BTo compute A∪B−A∩B, we first find
A∪B:
A∪B={x∈R| −3≤x≤5 or 0 <x<6}
Now, we subtract the intersection A∩B= (0,5] from the union A∪B:
A∪B−(A∩B) = {x∈R|(−3≤x≤5 or 0 <x<6) and x /∈(0,5]}
Simplifying the above expression, we get:
A∪B−(A∩B) = (−3,0] ∪(5,6)
Therefore, A∪B−A∩B= (−3,0] ∪(5,6).
Question 9
Question
Let A={x∈R|x2−2x−8<0}and B={x∈R|x2−2x−8≥0}. Find
A∩B.
Solution
Step 1: We begin by finding the solutions to the inequality x2−2x−8<0. This
can be done by determining the roots of the quadratic equation x2−2x−8=0
and analyzing the sign of the expression in each interval.
The roots of x2−2x−8 = 0 can be found using the quadratic formula:
x=−(−2) ±p(−2)2−4(1)(−8)
2(1) =2±√4 + 32
2=2±√36
2=2±6
2
So, x=−2 or x= 4.
Step 2: Now, we analyze the sign of x2−2x−8 in the intervals (−∞,−2),
(−2,4), and (4,∞). Choosing test points x=−3, x= 0, and x= 5 respectively:
x=−3: (−3)2−2(−3) −8 = 9 + 6 −8=7⇒positive
x= 0: 02−2(0) −8 = −8⇒negative
x= 5: 52−2(5) −8 = 25 −10 −8=7⇒positive
Step 3: Based on the sign analysis, x2−2x−8<0 for x∈(−2,4). Therefore,
A= (−2,4).
Step 4: Next, we find the solutions to the inequality x2−2x−8≥0 by
analyzing the sign of x2−2x−8 in the intervals (−∞,−2), (−2,4), and (4,∞)
using the same test points.
Step 5: Based on the sign analysis, x2−2x−8≥0 for x∈(−∞,−2]∪[4,∞).
Therefore, B= (−∞,−2] ∪[4,∞).
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Step 6: Finally, we find the intersection of Aand Bby taking A∩B. Since
A= (−2,4) and B= (−∞,−2] ∪[4,∞), we have:
A∩B= (−2,4) ∩((−∞,−2] ∪[4,∞)) = (−2,4) ∩[4,∞) = ∅
Therefore, A∩B=∅.
Question 10
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Determine the
set (A∩B)∪(A∩Bc).
Solution
Step 1: Find A∩B.
A∩Bconsists of all elements that are in both sets Aand B. So, we need
to find the intersection of the two sets.
A∩B={x∈Z|2≤x≤10}∩{x∈Z|5≤x≤15}
To find the intersection, we take the elements that are common in both sets,
which is the set {5,6,7,8,9,10}.
Step 2: Find A∩Bc.
A∩Bcconsists of all elements that are in set Aand not in set B. So, we
need to find the complement of set Band then find the intersection.
Bc={x∈Z|x < 5 or x > 15}
Therefore, A∩Bc={x∈Z|2≤x≤4} ∪ {x∈Z|11 ≤x≤15}, which
simplifies to {2,3,4,11,12,13,14,15}.
Step 3: Find (A∩B)∪(A∩Bc).
Now, we combine the two intersection sets we found in Step 1 and Step 2.
(A∩B)∪(A∩Bc) = {5,6,7,8,9,10}∪{2,3,4,11,12,13,14,15}
This gives us the set {2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 11
Question
Let A={x∈Z|2≤x≤7}and B={x∈Z|4≤x≤9}. Find A∪B,A∩B,
and A\B.
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Solution
Let’s first list the elements of set Aand set B.
Step 1: Listing the elements of set A
A={2,3,4,5,6,7}
Step 2: Listing the elements of set B
B={4,5,6,7,8,9}
Step 3: Finding the union A∪B
A∪B={2,3,4,5,6,7,8,9}
Step 4: Finding the intersection A∩B
A∩B={4,5,6,7}
Step 5: Finding the set difference A\B
A\B={2,3}
Therefore, A∪B={2,3,4,5,6,7,8,9},A∩B={4,5,6,7}, and A\B=
{2,3}.
Question 12
Question
Let A={x∈R|1≤x < 5}and B={x∈R|3≤x < 7}. Find A∪Band
A∩B.
Solution
Step 1: To find A∪B(the union of sets Aand B), we need to find all elements
that are in either set A, set B, or both.
A∪B={x∈R|1≤x < 5 or 3 ≤x < 7}
Step 2: Simplify the condition: A∪B={x∈R|1≤x < 7}.
So, A∪B={x∈R|1≤x < 7}.
Step 3: To find A∩B(the intersection of sets Aand B), we need to find all
elements that are common to both sets Aand B.
A∩B={x∈R|3≤x < 5}
Step 4: Simplify the condition: A∩B={x∈R|3≤x < 5}.
Therefore, A∩B={x∈R|3≤x < 5}.
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Question 13
Question
Let Aand Bbe two sets such that |A|= 7, |B|= 8, and |A∩B|= 5. If
|A∪B|= 10, find |A∩B′|where B′is the complement of set B.
Solution
Step 1: First, observe that |A∪B|=|A|+|B|−|A∩B|. We can use this
formula to determine |A∩B′|.
Step 2: Substitute the given values into the formula:
|A∪B|=|A|+|B|−|A∩B|=⇒10 = 7 + 8 −5
Step 3: Simplify the equation:
10 = 10
Step 4: Since the equation holds true, we can conclude that the given values
satisfy the condition.
Step 5: We know that |A∪B|=|A|+|B|−|A∩B|. We can rearrange this
formula to solve for |A∩B′|:
|A∩B′|=|A|+|B|−|A∪B|
Step 6: Substitute the given values into the formula:
|A∩B′|= 7 + 8 −10 = 5
Step 7: Therefore, |A∩B′|= 5.
Question 14
Question
Let A={x∈Z:x2−2x−8=0}and B={y∈Z:y2−y−6=0}. Determine
the intersection of sets Aand B.
Solution
Step 1: First, we find the elements in set Aby solving x2−2x−8 = 0. The
solutions to this quadratic equation are x=−2 and x= 4. So, set A={−2,4}.
Step 2: Next, we find the elements in set Bby solving y2−y−6 = 0. The
solutions to this quadratic equation are y=−2 and y= 3. So, set B={−2,3}.
Step 3: Now, we find the intersection of sets Aand Bto get the common
elements. The intersection of Aand Bis given by A∩B={−2}.
Therefore, the intersection of sets Aand Bis {−2}.
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Question 15
Question
Let A={x|xis a prime number less than 10}and B={y|yis a perfect square less than 20}.
Find the set A∪B.
Solution
To find A∪B, we need to determine the set that contains all elements that are
in A, in B, or in both Aand B.
Step 1: Determine the elements in set A. The prime numbers less than 10
are 2, 3, 5, and 7. Thus, A={2,3,5,7}.
Step 2: Determine the elements in set B. The perfect squares less than 20
are 1, 4, 9, 16. Thus, B={1,4,9,16}.
Step 3: Find A∪B. The union of sets Aand Bincludes all elements that
are in A, in B, or in both Aand B.T heref ore,A∪B={1,2,3,4,5,7,9,16}.
Question 16
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Determine the set A∩(A∪B).
Solution
Step 1: First, we need to find A∪B, which is the union of sets Aand B. Step
2: A∪B={1,2,3,4,5,6,7}. Step 3: Next, we find the intersection of Awith
A∪B. Step 4: A∩(A∪B) = {1,2,3,4,5}.
Question 17
Question
Let A,B, and Cbe sets such that |A∩B|= 20, |B∩C|= 30, |A∩C|= 15,
|A∩B∩C|= 10, |A∪B∪C|= 60. Determine |A∩B∩C′|where C′denotes
the complement of set C.
Solution
Step 1: Apply the principle of inclusion-exclusion to find |A∩B∩C|.
|A∪B∪C|=|A|+|B|+|C|−|A∩B|−|B∩C|−|A∩C|+|A∩B∩C|
60 = |A|+|B|+|C| − 20 −30 −15 + 10
60 = |A|+|B|+|C| − 55
|A|+|B|+|C|= 115
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Step 2: Use the values calculated to find |A∩B∩C′|.
|A∩B∩C′|=|A∩B|−|A∩B∩C|
= 20 −10
= 10
Therefore, |A∩B∩C′|= 10.
Question 18
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|6≤x≤15}. Find the set
A∩B.
Solution
Step 1: Write down the elements in set Aand set B.
A={1,2,3,4,5,6,7,8,9,10}
B={6,7,8,9,10,11,12,13,14,15}
Step 2: Identify the common elements in sets Aand B.
A∩B={6,7,8,9,10}
Therefore, A∩B={6,7,8,9,10}.
Question 19
Question
Let A={x∈Z|2≤x≤7}and B={x∈Z|3≤x≤9}. Find the set A∪B.
Solution
Step 1: To find A∪B, we need to consider all elements that are either in set A,
set B, or in both sets.
Step 2: Set Aincludes the integers from 2 to 7, while set Bincludes the
integers from 3 to 9. To find A∪B, we need to list out all the unique elements
in the union of set Aand set B.
Step 3: The elements of set Aare: {2,3,4,5,6,7}, and the elements of set
Bare: {3,4,5,6,7,8,9}.
Step 4: The union of sets Aand B, denoted A∪B, is the set containing all
the distinct elements from both sets.
Therefore, A∪B={2,3,4,5,6,7,8,9}.
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Question 20
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Determine the
set of all possible values of xin the set A−B.
Solution
Step 1: To find A−B, we need to determine the values that are in Abut not in
B. Step 2: Acontains integers from 1 to 10 inclusive, and Bcontains integers
from 5 to 15 inclusive. Step 3: So, A−Bwill include the values in Athat are
not in B. Step 4: The values in Athat are not in Bare 1, 2, 3, and 4. Step 5:
Therefore, the set of all possible values of xin A−Bis {1,2,3,4}.
Question 21
Question
Let A={n∈Z|1≤n≤10}and B={n∈Z|nis a prime number}. Find
(A∩B)∪(A∩Bc).
Solution
Step 1: First, let’s determine A∩B, the intersection of sets Aand B. Since
Acontains the integers from 1 to 10, and Bcontains only prime numbers, we
have A∩B={2,3,5,7}.
Step 2: Next, let’s find A∩Bc, the intersection of Aand the complement of
B. The complement of Bconsists of all non-prime numbers between 1 and 10.
So, Bc={1,4,6,8,9,10}. Therefore, A∩Bc={1,4,6,8,9,10}∩{n∈Z|1≤
n≤10}={1,4,6,8,9,10}.
Step 3: Finally, we find (A∩B)∪(A∩Bc), the union of A∩Band A∩Bc.
(A∩B)∪(A∩Bc) = {2,3,5,7}∪{1,4,6,8,9,10}={1,2,3,4,5,6,7,8,9,10}.
Therefore, (A∩B)∪(A∩Bc) = {1,2,3,4,5,6,7,8,9,10}.
Question 22
Question
Let A={n∈Z|2≤n≤10}and B={n∈Z|7≤n≤15}. Find the set
A∪B.
Solution
To find A∪B, we need to determine all the elements that are in A, in B, or in
both sets.
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Step 1: Find elements in set AWe know that A={n∈Z|2≤n≤10}.
Therefore, Acontains the integers 2, 3, 4, 5, 6, 7, 8, 9, and 10.
Step 2: Find elements in set BWe know that B={n∈Z|7≤n≤15}.
Therefore, Bcontains the integers 7, 8, 9, 10, 11, 12, 13, 14, and 15.
Step 3: Determine A∪BThe union A∪Bincludes all the elements that
are in set A, set B, or both. So, A∪B={2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 23
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|x∈Aand x∈B}
Step 2: The elements in set Aare {1,2,3,4,5,6,7,8,9,10}, and the elements
in set Bare {5,6,7,8,9,10,11,12,13,14,15}.
Step 3: Therefore, the intersection of sets Aand Bis
A∩B={5,6,7,8,9,10}
Step 4: To find the union of sets Aand B, we need to combine all elements
from both sets without duplicates.
A∪B={x|x∈Aor x∈B}
Step 5: The elements in the union of sets Aand Bare
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Question 24
Question
Let Aand Bbe sets such that |A|= 10, |B|= 15. If |A∪B|= 20, find |A∩B|.
Solution
Step 1: In order to find |A∩B|, we can use the principle of inclusion-exclusion:
Step 2: The principle of inclusion-exclusion states that
|A∪B|=|A|+|B|−|A∩B|.
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Step 3: Substituting the given values, we have
20 = 10 + 15 − |A∩B|.
Step 4: Simplifying the equation gives
20 = 25 − |A∩B|.
Step 5: Solving for |A∩B|yields
|A∩B|= 25 −20 = 5.
Step 6: Therefore, |A∩B|= 5.
Question 25
Question
Let A={x∈Z:−4≤x≤5}and B={x∈Z:−2≤x≤8}. Determine the
set A∩(A∪B).
Solution
Step 1: First, we find A∪B.
A∪B={x∈Z:−4≤x≤8}={−4,−3,−2,−1,0,1,2,3,4,5,6,7,8}
Step 2: Next, we find A∩(A∪B).
A∩(A∪B) = {x∈Z:x∈Aand x∈(A∪B)}
A∩(A∪B) = {x∈Z:−4≤x≤5 and −4≤x≤8}
A∩(A∪B) = {x∈Z:−4≤x≤5}={−4,−3,−2,−1,0,1,2,3,4,5}
Therefore, the set A∩(A∪B) is {−4,−3,−2,−1,0,1,2,3,4,5}.
Question 26
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|4≤x≤12}. Find the set C
defined as C=A∩B′, where B′is the complement of set B.
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Solution
Given: A={x∈Z|1≤x≤10}
B={x∈Z|4≤x≤12}
First, let’s find B′, the complement of set B:
B′={x∈Z|x /∈B}
Since Bincludes integers from 4 to 12, B′will include all integers except
those. Therefore:
B′={x∈Z|x < 4 or x > 12}
Hence,
B′={x∈Z|x < 4}∪{x∈Z|x > 12}={x∈Z|x < 4}
Now, let’s find C=A∩B′:
C={x∈Z|x∈Aand x∈B′}
C={x∈Z|1≤x≤10 and x < 4}
Since there are no integers that satisfy both conditions (1 ≤x≤10 and x < 4),
the intersection A∩B′is an empty set.
Therefore, the set Cis:
C=∅
Question 27
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Define the set Cas C=A∪Band
the set Das D=A∩B. Find |C∩D|.
Solution
Step 1: First, find C=A∪Bby taking the union of sets Aand B.
C={1,2,3,4,5}∪{3,4,5,6,7}={1,2,3,4,5,6,7}
Step 2: Next, find D=A∩Bby taking the intersection of sets Aand B.
D={1,2,3,4,5}∩{3,4,5,6,7}={3,4,5}
Step 3: Calculate C∩Dby finding the intersection of sets Cand D.
C∩D={1,2,3,4,5,6,7}∩{3,4,5}={3,4,5}
Step 4: Finally, find the cardinality of the set C∩Dto get |C∩D|.
|C∩D|=|{3,4,5}| = 3
Therefore, |C∩D|= 3.
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Question 28
Question
Let Aand Bbe two sets such that |A∩B|= 10, |A\B|= 15, and |B\A|= 20.
If |A∪B|= 50, find |A∩Bc|.
Solution
Step 1: We can use the principle of inclusion-exclusion to find |A∪B|.
|A∪B|=|A|+|B|−|A∩B|
=|A\B|+|B\A|+|A∩B|+|A∩B|
= 15 + 20 + 10 + 10
= 55.
Step 2: Since |A∪B|= 50, there must be 5 elements counted twice. These
5 elements are in both A\Band B\A.
Step 3: Now, we need to find |A∩Bc|=|A\B|.
|A∩Bc|=|A\B|
=|A|−|A∩B|
=|A| − 10.
Step 4: To find |A|, we use the formula for the union of two sets:
|A∪B|=|A|+|B|−|A∩B|
50 = |A|+|B| − 10
50 = |A|+|B| − 10
Step 5: We also know that |A\B|= 15 and |B\A|= 20. This implies that
Ahas 15 + 10 = 25 elements and Bhas 20 + 10 = 30 elements.
Step 6: Substituting the values back into the equation:
50 = 25 + 30 −10
50 = 45
|A|= 45.
Step 7: Finally, substitute |A|= 45 into |A∩Bc|=|A| − 10 to get:
|A∩Bc|= 45 −10
= 35.
Therefore, |A∩Bc|= 35.
15
Question 29
Question
Let A,B, and Cbe sets such that A∪B=A∪Cand A∩B=A∩C. Prove
that B=C.
Solution
Let’s prove this by first showing that B⊆Cand C⊆B.
Step 1: Showing B⊆CSince A∪B=A∪C, it follows that B⊆A∪C.
Thus, b∈B=⇒b∈Aor b∈C. But since A∩B=A∩C, we have b∈Aif
and only if b∈C. Therefore, b∈B=⇒b∈C, which implies B⊆C.
Step 2: Showing C⊆BBy symmetry, we can also show that C⊆B
using a similar argument. Since A∪B=A∪C, we have C⊆A∪B. Similarly,
a∈A=⇒a∈Bor a∈C. Since A∩B=A∩C, we have a∈Aif and only if
a∈B. Therefore, a∈C=⇒a∈B, which implies C⊆B.
Step 3: Conclusion From Step 1 and Step 2, we have shown that B⊆C
and C⊆B. Therefore, B=C.
Question 30
Question
Let A={x∈Z|2≤x≤8}and B={x∈Z|5≤x≤12}. Find the set
A∪B.
Solution
Step 1: First, list out the elements in set Aand set B.
A={2,3,4,5,6,7,8}
B={5,6,7,8,9,10,11,12}
Step 2: To find A∪B, we need to combine all unique elements from sets A
and B.
A∪B={2,3,4,5,6,7,8,9,10,11,12}
Therefore, the set A∪Bconsists of all integers from 2 to 12.
16
Solution
Step 1: To find A∩B, we first need to determine A∩B.
A∩B={x∈Q|x2<2}∩{x∈Q|x2>2}
={x∈Q|x2<2 and x2>2}
Step 2: Since x2<2 and x2>2 cannot hold true simultaneously for any
rational number x, we have A∩B=∅(the empty set).
Step 3: The complement of the empty set is the universal set, so A∩B=Q.
Therefore, A∩B=Q.
Question 3
Question
Let A={x∈Z|x2−4x−5 = 0}and B={x∈Z|x2−x−6 = 0}. Find
A∪Band A∩B.
Solution
Step 1: Find the solutions to the equations defining sets Aand B.
For set A: The equation x2−4x−5 = 0 can be factored as (x−5)(x+1) = 0.
So, the solutions are x= 5 and x=−1. Therefore, A={5,−1}.
For set B: The equation x2−x−6 = 0 can be factored as (x−3)(x+2) = 0.
So, the solutions are x= 3 and x=−2. Therefore, B={3,−2}.
Step 2: Determine A∪B(the union of sets Aand B). A∪Bis the set of
all elements that are in Aor in B(or both). Therefore, A∪B={5,−1,3,−2}.
Step 3: Determine A∩B(the intersection of sets Aand B). A∩Bis the set
of all elements that are in both Aand B. Since A={5,−1}and B={3,−2},
A∩B=∅(the empty set) because there are no elements common to both sets.
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 take the union of sets Aand B, which means
combining all the elements in both sets without repetition.
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
2
Step 2: To find A∩B, we need to find the intersection of sets Aand B,
which means finding the elements that are common to both sets.
A∩B={5,6,7,8,9,10}
Step 3: To find A\B, we need to determine the set of elements that are in
Abut not in B.
A\B={1,2,3,4}
Question 5
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find A∩B.
Solution
To find A∩B, we need to determine the elements that are common to both sets
Aand B.
Step 1: Write out the elements of set Aand B.
A={2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
Step 2: Identify the common elements between sets Aand B.
A∩B={5,6,7,8,9,10}
Step 3: Write the final answer.
A∩B={5,6,7,8,9,10}
Question 6
Question
Let A={x∈R|x2<5}and B={y∈R|y2≥3}. Determine A∪B.
Solution
To find A∪B, we need to find the set of all elements that are in either set Aor
set B.
Step 1: Find all elements in set A. Set A={x∈R|x2<5}can be
rewritten as A={x∈R| −√5<x<√5}. Therefore, set Acontains all real
numbers between −√5 and √5.
3
Step 2: Find all elements in set B. Set B={y∈R|y2≥3}can be
rewritten as B={y∈R|y≤ −√3 or y≥√3}. Therefore, set Bcontains all
real numbers less than or equal to −√3 or greater than or equal to √3.
Step 3: Find A∪B.The union of sets Aand B, denoted as A∪B, contains
all the elements that are in either set A, set B, or both.
Therefore, A∪B={x∈R|x≤ −√3 or −√5<x<√5 or x≥√3}.
Hence, A∪B= (−∞,−√3] ∪(−√5,√5) ∪[√3,∞).
Question 7
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Define the set Cas C=A∪Band
Das D=A∩B. Find the Cartesian product C×D.
Solution
To find the Cartesian product C×D, we first need to determine the sets Cand
D.
Step 1: Calculate C=A∪B.A={1,2,3,4,5}B={3,4,5,6,7}C=
A∪B={1,2,3,4,5,6,7}
Step 2: Calculate D=A∩B.A={1,2,3,4,5}B={3,4,5,6,7}D=
A∩B={3,4,5}
Step 3: Find C×D.C={1,2,3,4,5,6,7}D={3,4,5}C×D=
{(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,3),(3,4),(3,5),(4,3),(4,4),(4,5),(5,3),(5,4),(5,5),(6,3),(6,4),(6,5),(7,3),(7,4),(7,5)}
Therefore, C×Dis {(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,3),(3,4),(3,5),
(4,3),(4,4),(4,5),(5,3),(5,4),(5,5),(6,3),(6,4),(6,5),(7,3),(7,4),(7,5)}.
Question 8
Question
Let A={x∈R| −3≤x≤5}and B={x∈R|0< x < 6}. Compute the
following set:
A∪B−A∩B
Solution
We first find A∩Band then compute A∪B−(A∩B).
Step 1: Find A∩BTo find A∩B, we need to identify the elements that
are in both Aand B.
A∩B={x∈R| −3≤x≤5 and 0 <x<6}
Intersecting the intervals [−3,5] and (0,6), we see that A∩B= (0,5].
4
Step 2: Compute A∪B−A∩BTo compute A∪B−A∩B, we first find
A∪B:
A∪B={x∈R| −3≤x≤5 or 0 <x<6}
Now, we subtract the intersection A∩B= (0,5] from the union A∪B:
A∪B−(A∩B) = {x∈R|(−3≤x≤5 or 0 <x<6) and x /∈(0,5]}
Simplifying the above expression, we get:
A∪B−(A∩B) = (−3,0] ∪(5,6)
Therefore, A∪B−A∩B= (−3,0] ∪(5,6).
Question 9
Question
Let A={x∈R|x2−2x−8<0}and B={x∈R|x2−2x−8≥0}. Find
A∩B.
Solution
Step 1: We begin by finding the solutions to the inequality x2−2x−8<0. This
can be done by determining the roots of the quadratic equation x2−2x−8=0
and analyzing the sign of the expression in each interval.
The roots of x2−2x−8 = 0 can be found using the quadratic formula:
x=−(−2) ±p(−2)2−4(1)(−8)
2(1) =2±√4 + 32
2=2±√36
2=2±6
2
So, x=−2 or x= 4.
Step 2: Now, we analyze the sign of x2−2x−8 in the intervals (−∞,−2),
(−2,4), and (4,∞). Choosing test points x=−3, x= 0, and x= 5 respectively:
x=−3: (−3)2−2(−3) −8 = 9 + 6 −8=7⇒positive
x= 0: 02−2(0) −8 = −8⇒negative
x= 5: 52−2(5) −8 = 25 −10 −8=7⇒positive
Step 3: Based on the sign analysis, x2−2x−8<0 for x∈(−2,4). Therefore,
A= (−2,4).
Step 4: Next, we find the solutions to the inequality x2−2x−8≥0 by
analyzing the sign of x2−2x−8 in the intervals (−∞,−2), (−2,4), and (4,∞)
using the same test points.
Step 5: Based on the sign analysis, x2−2x−8≥0 for x∈(−∞,−2]∪[4,∞).
Therefore, B= (−∞,−2] ∪[4,∞).
5
Step 6: Finally, we find the intersection of Aand Bby taking A∩B. Since
A= (−2,4) and B= (−∞,−2] ∪[4,∞), we have:
A∩B= (−2,4) ∩((−∞,−2] ∪[4,∞)) = (−2,4) ∩[4,∞) = ∅
Therefore, A∩B=∅.
Question 10
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Determine the
set (A∩B)∪(A∩Bc).
Solution
Step 1: Find A∩B.
A∩Bconsists of all elements that are in both sets Aand B. So, we need
to find the intersection of the two sets.
A∩B={x∈Z|2≤x≤10}∩{x∈Z|5≤x≤15}
To find the intersection, we take the elements that are common in both sets,
which is the set {5,6,7,8,9,10}.
Step 2: Find A∩Bc.
A∩Bcconsists of all elements that are in set Aand not in set B. So, we
need to find the complement of set Band then find the intersection.
Bc={x∈Z|x < 5 or x > 15}
Therefore, A∩Bc={x∈Z|2≤x≤4} ∪ {x∈Z|11 ≤x≤15}, which
simplifies to {2,3,4,11,12,13,14,15}.
Step 3: Find (A∩B)∪(A∩Bc).
Now, we combine the two intersection sets we found in Step 1 and Step 2.
(A∩B)∪(A∩Bc) = {5,6,7,8,9,10}∪{2,3,4,11,12,13,14,15}
This gives us the set {2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 11
Question
Let A={x∈Z|2≤x≤7}and B={x∈Z|4≤x≤9}. Find A∪B,A∩B,
and A\B.
6
Solution
Let’s first list the elements of set Aand set B.
Step 1: Listing the elements of set A
A={2,3,4,5,6,7}
Step 2: Listing the elements of set B
B={4,5,6,7,8,9}
Step 3: Finding the union A∪B
A∪B={2,3,4,5,6,7,8,9}
Step 4: Finding the intersection A∩B
A∩B={4,5,6,7}
Step 5: Finding the set difference A\B
A\B={2,3}
Therefore, A∪B={2,3,4,5,6,7,8,9},A∩B={4,5,6,7}, and A\B=
{2,3}.
Question 12
Question
Let A={x∈R|1≤x < 5}and B={x∈R|3≤x < 7}. Find A∪Band
A∩B.
Solution
Step 1: To find A∪B(the union of sets Aand B), we need to find all elements
that are in either set A, set B, or both.
A∪B={x∈R|1≤x < 5 or 3 ≤x < 7}
Step 2: Simplify the condition: A∪B={x∈R|1≤x < 7}.
So, A∪B={x∈R|1≤x < 7}.
Step 3: To find A∩B(the intersection of sets Aand B), we need to find all
elements that are common to both sets Aand B.
A∩B={x∈R|3≤x < 5}
Step 4: Simplify the condition: A∩B={x∈R|3≤x < 5}.
Therefore, A∩B={x∈R|3≤x < 5}.
7
Question 13
Question
Let Aand Bbe two sets such that |A|= 7, |B|= 8, and |A∩B|= 5. If
|A∪B|= 10, find |A∩B′|where B′is the complement of set B.
Solution
Step 1: First, observe that |A∪B|=|A|+|B|−|A∩B|. We can use this
formula to determine |A∩B′|.
Step 2: Substitute the given values into the formula:
|A∪B|=|A|+|B|−|A∩B|=⇒10 = 7 + 8 −5
Step 3: Simplify the equation:
10 = 10
Step 4: Since the equation holds true, we can conclude that the given values
satisfy the condition.
Step 5: We know that |A∪B|=|A|+|B|− |A∩B|. We can rearrange this
formula to solve for |A∩B′|:
|A∩B′|=|A|+|B|−|A∪B|
Step 6: Substitute the given values into the formula:
|A∩B′|= 7 + 8 −10 = 5
Step 7: Therefore, |A∩B′|= 5.
Question 14
Question
Let A={x∈Z:x2−2x−8=0}and B={y∈Z:y2−y−6=0}. Determine
the intersection of sets Aand B.
Solution
Step 1: First, we find the elements in set Aby solving x2−2x−8 = 0. The
solutions to this quadratic equation are x=−2 and x= 4. So, set A={−2,4}.
Step 2: Next, we find the elements in set Bby solving y2−y−6 = 0. The
solutions to this quadratic equation are y=−2 and y= 3. So, set B={−2,3}.
Step 3: Now, we find the intersection of sets Aand Bto get the common
elements. The intersection of Aand Bis given by A∩B={−2}.
Therefore, the intersection of sets Aand Bis {−2}.
8
Question 15
Question
Let A={x|xis a prime number less than 10}and B={y|yis a perfect square less than 20}.
Find the set A∪B.
Solution
To find A∪B, we need to determine the set that contains all elements that are
in A, in B, or in both Aand B.
Step 1: Determine the elements in set A. The prime numbers less than 10
are 2, 3, 5, and 7. Thus, A={2,3,5,7}.
Step 2: Determine the elements in set B. The perfect squares less than 20
are 1, 4, 9, 16. Thus, B={1,4,9,16}.
Step 3: Find A∪B. The union of sets Aand Bincludes all elements that
are in A, in B, or in both Aand B.T heref ore,A∪B={1,2,3,4,5,7,9,16}.
Question 16
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Determine the set A∩(A∪B).
Solution
Step 1: First, we need to find A∪B, which is the union of sets Aand B. Step
2: A∪B={1,2,3,4,5,6,7}. Step 3: Next, we find the intersection of Awith
A∪B. Step 4: A∩(A∪B) = {1,2,3,4,5}.
Question 17
Question
Let A,B, and Cbe sets such that |A∩B|= 20, |B∩C|= 30, |A∩C|= 15,
|A∩B∩C|= 10, |A∪B∪C|= 60. Determine |A∩B∩C′|where C′denotes
the complement of set C.
Solution
Step 1: Apply the principle of inclusion-exclusion to find |A∩B∩C|.
|A∪B∪C|=|A|+|B|+|C|−|A∩B|−|B∩C|−|A∩C|+|A∩B∩C|
60 = |A|+|B|+|C| − 20 −30 −15 + 10
60 = |A|+|B|+|C| − 55
|A|+|B|+|C|= 115
9
Step 2: Use the values calculated to find |A∩B∩C′|.
|A∩B∩C′|=|A∩B|−|A∩B∩C|
= 20 −10
= 10
Therefore, |A∩B∩C′|= 10.
Question 18
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|6≤x≤15}. Find the set
A∩B.
Solution
Step 1: Write down the elements in set Aand set B.
A={1,2,3,4,5,6,7,8,9,10}
B={6,7,8,9,10,11,12,13,14,15}
Step 2: Identify the common elements in sets Aand B.
A∩B={6,7,8,9,10}
Therefore, A∩B={6,7,8,9,10}.
Question 19
Question
Let A={x∈Z|2≤x≤7}and B={x∈Z|3≤x≤9}. Find the set A∪B.
Solution
Step 1: To find A∪B, we need to consider all elements that are either in set A,
set B, or in both sets.
Step 2: Set Aincludes the integers from 2 to 7, while set Bincludes the
integers from 3 to 9. To find A∪B, we need to list out all the unique elements
in the union of set Aand set B.
Step 3: The elements of set Aare: {2,3,4,5,6,7}, and the elements of set
Bare: {3,4,5,6,7,8,9}.
Step 4: The union of sets Aand B, denoted A∪B, is the set containing all
the distinct elements from both sets.
Therefore, A∪B={2,3,4,5,6,7,8,9}.
10
Question 20
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Determine the
set of all possible values of xin the set A−B.
Solution
Step 1: To find A−B, we need to determine the values that are in Abut not in
B. Step 2: Acontains integers from 1 to 10 inclusive, and Bcontains integers
from 5 to 15 inclusive. Step 3: So, A−Bwill include the values in Athat are
not in B. Step 4: The values in Athat are not in Bare 1, 2, 3, and 4. Step 5:
Therefore, the set of all possible values of xin A−Bis {1,2,3,4}.
Question 21
Question
Let A={n∈Z|1≤n≤10}and B={n∈Z|nis a prime number}. Find
(A∩B)∪(A∩Bc).
Solution
Step 1: First, let’s determine A∩B, the intersection of sets Aand B. Since
Acontains the integers from 1 to 10, and Bcontains only prime numbers, we
have A∩B={2,3,5,7}.
Step 2: Next, let’s find A∩Bc, the intersection of Aand the complement of
B. The complement of Bconsists of all non-prime numbers between 1 and 10.
So, Bc={1,4,6,8,9,10}. Therefore, A∩Bc={1,4,6,8,9,10}∩{n∈Z|1≤
n≤10}={1,4,6,8,9,10}.
Step 3: Finally, we find (A∩B)∪(A∩Bc), the union of A∩Band A∩Bc.
(A∩B)∪(A∩Bc) = {2,3,5,7}∪{1,4,6,8,9,10}={1,2,3,4,5,6,7,8,9,10}.
Therefore, (A∩B)∪(A∩Bc) = {1,2,3,4,5,6,7,8,9,10}.
Question 22
Question
Let A={n∈Z|2≤n≤10}and B={n∈Z|7≤n≤15}. Find the set
A∪B.
Solution
To find A∪B, we need to determine all the elements that are in A, in B, or in
both sets.
11
Step 1: Find elements in set AWe know that A={n∈Z|2≤n≤10}.
Therefore, Acontains the integers 2, 3, 4, 5, 6, 7, 8, 9, and 10.
Step 2: Find elements in set BWe know that B={n∈Z|7≤n≤15}.
Therefore, Bcontains the integers 7, 8, 9, 10, 11, 12, 13, 14, and 15.
Step 3: Determine A∪BThe union A∪Bincludes all the elements that
are in set A, set B, or both. So, A∪B={2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 23
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|x∈Aand x∈B}
Step 2: The elements in set Aare {1,2,3,4,5,6,7,8,9,10}, and the elements
in set Bare {5,6,7,8,9,10,11,12,13,14,15}.
Step 3: Therefore, the intersection of sets Aand Bis
A∩B={5,6,7,8,9,10}
Step 4: To find the union of sets Aand B, we need to combine all elements
from both sets without duplicates.
A∪B={x|x∈Aor x∈B}
Step 5: The elements in the union of sets Aand Bare
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Question 24
Question
Let Aand Bbe sets such that |A|= 10, |B|= 15. If |A∪B|= 20, find |A∩B|.
Solution
Step 1: In order to find |A∩B|, we can use the principle of inclusion-exclusion:
Step 2: The principle of inclusion-exclusion states that
|A∪B|=|A|+|B|−|A∩B|.
12
Step 3: Substituting the given values, we have
20 = 10 + 15 − |A∩B|.
Step 4: Simplifying the equation gives
20 = 25 − |A∩B|.
Step 5: Solving for |A∩B|yields
|A∩B|= 25 −20 = 5.
Step 6: Therefore, |A∩B|= 5.
Question 25
Question
Let A={x∈Z:−4≤x≤5}and B={x∈Z:−2≤x≤8}. Determine the
set A∩(A∪B).
Solution
Step 1: First, we find A∪B.
A∪B={x∈Z:−4≤x≤8}={−4,−3,−2,−1,0,1,2,3,4,5,6,7,8}
Step 2: Next, we find A∩(A∪B).
A∩(A∪B) = {x∈Z:x∈Aand x∈(A∪B)}
A∩(A∪B) = {x∈Z:−4≤x≤5 and −4≤x≤8}
A∩(A∪B) = {x∈Z:−4≤x≤5}={−4,−3,−2,−1,0,1,2,3,4,5}
Therefore, the set A∩(A∪B) is {−4,−3,−2,−1,0,1,2,3,4,5}.
Question 26
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|4≤x≤12}. Find the set C
defined as C=A∩B′, where B′is the complement of set B.
13
Solution
Given: A={x∈Z|1≤x≤10}
B={x∈Z|4≤x≤12}
First, let’s find B′, the complement of set B:
B′={x∈Z|x /∈B}
Since Bincludes integers from 4 to 12, B′will include all integers except
those. Therefore:
B′={x∈Z|x < 4 or x > 12}
Hence,
B′={x∈Z|x < 4}∪{x∈Z|x > 12}={x∈Z|x < 4}
Now, let’s find C=A∩B′:
C={x∈Z|x∈Aand x∈B′}
C={x∈Z|1≤x≤10 and x < 4}
Since there are no integers that satisfy both conditions (1 ≤x≤10 and x < 4),
the intersection A∩B′is an empty set.
Therefore, the set Cis:
C=∅
Question 27
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Define the set Cas C=A∪Band
the set Das D=A∩B. Find |C∩D|.
Solution
Step 1: First, find C=A∪Bby taking the union of sets Aand B.
C={1,2,3,4,5}∪{3,4,5,6,7}={1,2,3,4,5,6,7}
Step 2: Next, find D=A∩Bby taking the intersection of sets Aand B.
D={1,2,3,4,5}∩{3,4,5,6,7}={3,4,5}
Step 3: Calculate C∩Dby finding the intersection of sets Cand D.
C∩D={1,2,3,4,5,6,7}∩{3,4,5}={3,4,5}
Step 4: Finally, find the cardinality of the set C∩Dto get |C∩D|.
|C∩D|=|{3,4,5}| = 3
Therefore, |C∩D|= 3.
14
Question 28
Question
Let Aand Bbe two sets such that |A∩B|= 10, |A\B|= 15, and |B\A|= 20.
If |A∪B|= 50, find |A∩Bc|.
Solution
Step 1: We can use the principle of inclusion-exclusion to find |A∪B|.
|A∪B|=|A|+|B|−|A∩B|
=|A\B|+|B\A|+|A∩B|+|A∩B|
= 15 + 20 + 10 + 10
= 55.
Step 2: Since |A∪B|= 50, there must be 5 elements counted twice. These
5 elements are in both A\Band B\A.
Step 3: Now, we need to find |A∩Bc|=|A\B|.
|A∩Bc|=|A\B|
=|A|−|A∩B|
=|A| − 10.
Step 4: To find |A|, we use the formula for the union of two sets:
|A∪B|=|A|+|B|−|A∩B|
50 = |A|+|B| − 10
50 = |A|+|B| − 10
Step 5: We also know that |A\B|= 15 and |B\A|= 20. This implies that
Ahas 15 + 10 = 25 elements and Bhas 20 + 10 = 30 elements.
Step 6: Substituting the values back into the equation:
50 = 25 + 30 −10
50 = 45
|A|= 45.
Step 7: Finally, substitute |A|= 45 into |A∩Bc|=|A| − 10 to get:
|A∩Bc|= 45 −10
= 35.
Therefore, |A∩Bc|= 35.
15
Question 29
Question
Let A,B, and Cbe sets such that A∪B=A∪Cand A∩B=A∩C. Prove
that B=C.
Solution
Let’s prove this by first showing that B⊆Cand C⊆B.
Step 1: Showing B⊆CSince A∪B=A∪C, it follows that B⊆A∪C.
Thus, b∈B=⇒b∈Aor b∈C. But since A∩B=A∩C, we have b∈Aif
and only if b∈C. Therefore, b∈B=⇒b∈C, which implies B⊆C.
Step 2: Showing C⊆BBy symmetry, we can also show that C⊆B
using a similar argument. Since A∪B=A∪C, we have C⊆A∪B. Similarly,
a∈A=⇒a∈Bor a∈C. Since A∩B=A∩C, we have a∈Aif and only if
a∈B. Therefore, a∈C=⇒a∈B, which implies C⊆B.
Step 3: Conclusion From Step 1 and Step 2, we have shown that B⊆C
and C⊆B. Therefore, B=C.
Question 30
Question
Let A={x∈Z|2≤x≤8}and B={x∈Z|5≤x≤12}. Find the set
A∪B.
Solution
Step 1: First, list out the elements in set Aand set B.
A={2,3,4,5,6,7,8}
B={5,6,7,8,9,10,11,12}
Step 2: To find A∪B, we need to combine all unique elements from sets A
and B.
A∪B={2,3,4,5,6,7,8,9,10,11,12}
Therefore, the set A∪Bconsists of all integers from 2 to 12.
16
Solution
Step 1: To find A∩B, we first need to determine A∩B.
A∩B={x∈Q|x2<2}∩{x∈Q|x2>2}
={x∈Q|x2<2 and x2>2}
Step 2: Since x2<2 and x2>2 cannot hold true simultaneously for any
rational number x, we have A∩B=∅(the empty set).
Step 3: The complement of the empty set is the universal set, so A∩B=Q.
Therefore, A∩B=Q.
Question 3
Question
Let A={x∈Z|x2−4x−5 = 0}and B={x∈Z|x2−x−6 = 0}. Find
A∪Band A∩B.
Solution
Step 1: Find the solutions to the equations defining sets Aand B.
For set A: The equation x2−4x−5 = 0 can be factored as (x−5)(x+1) = 0.
So, the solutions are x= 5 and x=−1. Therefore, A={5,−1}.
For set B: The equation x2−x−6 = 0 can be factored as (x−3)(x+2) = 0.
So, the solutions are x= 3 and x=−2. Therefore, B={3,−2}.
Step 2: Determine A∪B(the union of sets Aand B). A∪Bis the set of
all elements that are in Aor in B(or both). Therefore, A∪B={5,−1,3,−2}.
Step 3: Determine A∩B(the intersection of sets Aand B). A∩Bis the set
of all elements that are in both Aand B. Since A={5,−1}and B={3,−2},
A∩B=∅(the empty set) because there are no elements common to both sets.
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 take the union of sets Aand B, which means
combining all the elements in both sets without repetition.
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
2
Step 2: To find A∩B, we need to find the intersection of sets Aand B,
which means finding the elements that are common to both sets.
A∩B={5,6,7,8,9,10}
Step 3: To find A\B, we need to determine the set of elements that are in
Abut not in B.
A\B={1,2,3,4}
Question 5
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find A∩B.
Solution
To find A∩B, we need to determine the elements that are common to both sets
Aand B.
Step 1: Write out the elements of set Aand B.
A={2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
Step 2: Identify the common elements between sets Aand B.
A∩B={5,6,7,8,9,10}
Step 3: Write the final answer.
A∩B={5,6,7,8,9,10}
Question 6
Question
Let A={x∈R|x2<5}and B={y∈R|y2≥3}. Determine A∪B.
Solution
To find A∪B, we need to find the set of all elements that are in either set Aor
set B.
Step 1: Find all elements in set A. Set A={x∈R|x2<5}can be
rewritten as A={x∈R| −√5<x<√5}. Therefore, set Acontains all real
numbers between −√5 and √5.
3
Step 2: Find all elements in set B. Set B={y∈R|y2≥3}can be
rewritten as B={y∈R|y≤ −√3 or y≥√3}. Therefore, set Bcontains all
real numbers less than or equal to −√3 or greater than or equal to √3.
Step 3: Find A∪B.The union of sets Aand B, denoted as A∪B, contains
all the elements that are in either set A, set B, or both.
Therefore, A∪B={x∈R|x≤ −√3 or −√5<x<√5 or x≥√3}.
Hence, A∪B= (−∞,−√3] ∪(−√5,√5) ∪[√3,∞).
Question 7
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Define the set Cas C=A∪Band
Das D=A∩B. Find the Cartesian product C×D.
Solution
To find the Cartesian product C×D, we first need to determine the sets Cand
D.
Step 1: Calculate C=A∪B.A={1,2,3,4,5}B={3,4,5,6,7}C=
A∪B={1,2,3,4,5,6,7}
Step 2: Calculate D=A∩B.A={1,2,3,4,5}B={3,4,5,6,7}D=
A∩B={3,4,5}
Step 3: Find C×D.C={1,2,3,4,5,6,7}D={3,4,5}C×D=
{(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,3),(3,4),(3,5),(4,3),(4,4),(4,5),(5,3),(5,4),(5,5),(6,3),(6,4),(6,5),(7,3),(7,4),(7,5)}
Therefore, C×Dis {(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,3),(3,4),(3,5),
(4,3),(4,4),(4,5),(5,3),(5,4),(5,5),(6,3),(6,4),(6,5),(7,3),(7,4),(7,5)}.
Question 8
Question
Let A={x∈R| −3≤x≤5}and B={x∈R|0< x < 6}. Compute the
following set:
A∪B−A∩B
Solution
We first find A∩Band then compute A∪B−(A∩B).
Step 1: Find A∩BTo find A∩B, we need to identify the elements that
are in both Aand B.
A∩B={x∈R| −3≤x≤5 and 0 <x<6}
Intersecting the intervals [−3,5] and (0,6), we see that A∩B= (0,5].
4
Step 2: Compute A∪B−A∩BTo compute A∪B−A∩B, we first find
A∪B:
A∪B={x∈R| −3≤x≤5 or 0 <x<6}
Now, we subtract the intersection A∩B= (0,5] from the union A∪B:
A∪B−(A∩B) = {x∈R|(−3≤x≤5 or 0 <x<6) and x /∈(0,5]}
Simplifying the above expression, we get:
A∪B−(A∩B) = (−3,0] ∪(5,6)
Therefore, A∪B−A∩B= (−3,0] ∪(5,6).
Question 9
Question
Let A={x∈R|x2−2x−8<0}and B={x∈R|x2−2x−8≥0}. Find
A∩B.
Solution
Step 1: We begin by finding the solutions to the inequality x2−2x−8<0. This
can be done by determining the roots of the quadratic equation x2−2x−8=0
and analyzing the sign of the expression in each interval.
The roots of x2−2x−8 = 0 can be found using the quadratic formula:
x=−(−2) ±p(−2)2−4(1)(−8)
2(1) =2±√4 + 32
2=2±√36
2=2±6
2
So, x=−2 or x= 4.
Step 2: Now, we analyze the sign of x2−2x−8 in the intervals (−∞,−2),
(−2,4), and (4,∞). Choosing test points x=−3, x= 0, and x= 5 respectively:
x=−3: (−3)2−2(−3) −8 = 9 + 6 −8=7⇒positive
x= 0: 02−2(0) −8 = −8⇒negative
x= 5: 52−2(5) −8 = 25 −10 −8=7⇒positive
Step 3: Based on the sign analysis, x2−2x−8<0 for x∈(−2,4). Therefore,
A= (−2,4).
Step 4: Next, we find the solutions to the inequality x2−2x−8≥0 by
analyzing the sign of x2−2x−8 in the intervals (−∞,−2), (−2,4), and (4,∞)
using the same test points.
Step 5: Based on the sign analysis, x2−2x−8≥0 for x∈(−∞,−2]∪[4,∞).
Therefore, B= (−∞,−2] ∪[4,∞).
5
Step 6: Finally, we find the intersection of Aand Bby taking A∩B. Since
A= (−2,4) and B= (−∞,−2] ∪[4,∞), we have:
A∩B= (−2,4) ∩((−∞,−2] ∪[4,∞)) = (−2,4) ∩[4,∞) = ∅
Therefore, A∩B=∅.
Question 10
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Determine the
set (A∩B)∪(A∩Bc).
Solution
Step 1: Find A∩B.
A∩Bconsists of all elements that are in both sets Aand B. So, we need
to find the intersection of the two sets.
A∩B={x∈Z|2≤x≤10}∩{x∈Z|5≤x≤15}
To find the intersection, we take the elements that are common in both sets,
which is the set {5,6,7,8,9,10}.
Step 2: Find A∩Bc.
A∩Bcconsists of all elements that are in set Aand not in set B. So, we
need to find the complement of set Band then find the intersection.
Bc={x∈Z|x < 5 or x > 15}
Therefore, A∩Bc={x∈Z|2≤x≤4} ∪ {x∈Z|11 ≤x≤15}, which
simplifies to {2,3,4,11,12,13,14,15}.
Step 3: Find (A∩B)∪(A∩Bc).
Now, we combine the two intersection sets we found in Step 1 and Step 2.
(A∩B)∪(A∩Bc) = {5,6,7,8,9,10}∪{2,3,4,11,12,13,14,15}
This gives us the set {2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 11
Question
Let A={x∈Z|2≤x≤7}and B={x∈Z|4≤x≤9}. Find A∪B,A∩B,
and A\B.
6
Solution
Let’s first list the elements of set Aand set B.
Step 1: Listing the elements of set A
A={2,3,4,5,6,7}
Step 2: Listing the elements of set B
B={4,5,6,7,8,9}
Step 3: Finding the union A∪B
A∪B={2,3,4,5,6,7,8,9}
Step 4: Finding the intersection A∩B
A∩B={4,5,6,7}
Step 5: Finding the set difference A\B
A\B={2,3}
Therefore, A∪B={2,3,4,5,6,7,8,9},A∩B={4,5,6,7}, and A\B=
{2,3}.
Question 12
Question
Let A={x∈R|1≤x < 5}and B={x∈R|3≤x < 7}. Find A∪Band
A∩B.
Solution
Step 1: To find A∪B(the union of sets Aand B), we need to find all elements
that are in either set A, set B, or both.
A∪B={x∈R|1≤x < 5 or 3 ≤x < 7}
Step 2: Simplify the condition: A∪B={x∈R|1≤x < 7}.
So, A∪B={x∈R|1≤x < 7}.
Step 3: To find A∩B(the intersection of sets Aand B), we need to find all
elements that are common to both sets Aand B.
A∩B={x∈R|3≤x < 5}
Step 4: Simplify the condition: A∩B={x∈R|3≤x < 5}.
Therefore, A∩B={x∈R|3≤x < 5}.
7
Question 13
Question
Let Aand Bbe two sets such that |A|= 7, |B|= 8, and |A∩B|= 5. If
|A∪B|= 10, find |A∩B′|where B′is the complement of set B.
Solution
Step 1: First, observe that |A∪B|=|A|+|B|−|A∩B|. We can use this
formula to determine |A∩B′|.
Step 2: Substitute the given values into the formula:
|A∪B|=|A|+|B|−|A∩B|=⇒10 = 7 + 8 −5
Step 3: Simplify the equation:
10 = 10
Step 4: Since the equation holds true, we can conclude that the given values
satisfy the condition.
Step 5: We know that |A∪B|=|A|+|B|− |A∩B|. We can rearrange this
formula to solve for |A∩B′|:
|A∩B′|=|A|+|B|−|A∪B|
Step 6: Substitute the given values into the formula:
|A∩B′|= 7 + 8 −10 = 5
Step 7: Therefore, |A∩B′|= 5.
Question 14
Question
Let A={x∈Z:x2−2x−8=0}and B={y∈Z:y2−y−6=0}. Determine
the intersection of sets Aand B.
Solution
Step 1: First, we find the elements in set Aby solving x2−2x−8 = 0. The
solutions to this quadratic equation are x=−2 and x= 4. So, set A={−2,4}.
Step 2: Next, we find the elements in set Bby solving y2−y−6 = 0. The
solutions to this quadratic equation are y=−2 and y= 3. So, set B={−2,3}.
Step 3: Now, we find the intersection of sets Aand Bto get the common
elements. The intersection of Aand Bis given by A∩B={−2}.
Therefore, the intersection of sets Aand Bis {−2}.
8
Question 15
Question
Let A={x|xis a prime number less than 10}and B={y|yis a perfect square less than 20}.
Find the set A∪B.
Solution
To find A∪B, we need to determine the set that contains all elements that are
in A, in B, or in both Aand B.
Step 1: Determine the elements in set A. The prime numbers less than 10
are 2, 3, 5, and 7. Thus, A={2,3,5,7}.
Step 2: Determine the elements in set B. The perfect squares less than 20
are 1, 4, 9, 16. Thus, B={1,4,9,16}.
Step 3: Find A∪B. The union of sets Aand Bincludes all elements that
are in A, in B, or in both Aand B.T heref ore,A∪B={1,2,3,4,5,7,9,16}.
Question 16
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Determine the set A∩(A∪B).
Solution
Step 1: First, we need to find A∪B, which is the union of sets Aand B. Step
2: A∪B={1,2,3,4,5,6,7}. Step 3: Next, we find the intersection of Awith
A∪B. Step 4: A∩(A∪B) = {1,2,3,4,5}.
Question 17
Question
Let A,B, and Cbe sets such that |A∩B|= 20, |B∩C|= 30, |A∩C|= 15,
|A∩B∩C|= 10, |A∪B∪C|= 60. Determine |A∩B∩C′|where C′denotes
the complement of set C.
Solution
Step 1: Apply the principle of inclusion-exclusion to find |A∩B∩C|.
|A∪B∪C|=|A|+|B|+|C|−|A∩B|−|B∩C|−|A∩C|+|A∩B∩C|
60 = |A|+|B|+|C| − 20 −30 −15 + 10
60 = |A|+|B|+|C| − 55
|A|+|B|+|C|= 115
9
Step 2: Use the values calculated to find |A∩B∩C′|.
|A∩B∩C′|=|A∩B|−|A∩B∩C|
= 20 −10
= 10
Therefore, |A∩B∩C′|= 10.
Question 18
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|6≤x≤15}. Find the set
A∩B.
Solution
Step 1: Write down the elements in set Aand set B.
A={1,2,3,4,5,6,7,8,9,10}
B={6,7,8,9,10,11,12,13,14,15}
Step 2: Identify the common elements in sets Aand B.
A∩B={6,7,8,9,10}
Therefore, A∩B={6,7,8,9,10}.
Question 19
Question
Let A={x∈Z|2≤x≤7}and B={x∈Z|3≤x≤9}. Find the set A∪B.
Solution
Step 1: To find A∪B, we need to consider all elements that are either in set A,
set B, or in both sets.
Step 2: Set Aincludes the integers from 2 to 7, while set Bincludes the
integers from 3 to 9. To find A∪B, we need to list out all the unique elements
in the union of set Aand set B.
Step 3: The elements of set Aare: {2,3,4,5,6,7}, and the elements of set
Bare: {3,4,5,6,7,8,9}.
Step 4: The union of sets Aand B, denoted A∪B, is the set containing all
the distinct elements from both sets.
Therefore, A∪B={2,3,4,5,6,7,8,9}.
10
Question 20
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Determine the
set of all possible values of xin the set A−B.
Solution
Step 1: To find A−B, we need to determine the values that are in Abut not in
B. Step 2: Acontains integers from 1 to 10 inclusive, and Bcontains integers
from 5 to 15 inclusive. Step 3: So, A−Bwill include the values in Athat are
not in B. Step 4: The values in Athat are not in Bare 1, 2, 3, and 4. Step 5:
Therefore, the set of all possible values of xin A−Bis {1,2,3,4}.
Question 21
Question
Let A={n∈Z|1≤n≤10}and B={n∈Z|nis a prime number}. Find
(A∩B)∪(A∩Bc).
Solution
Step 1: First, let’s determine A∩B, the intersection of sets Aand B. Since
Acontains the integers from 1 to 10, and Bcontains only prime numbers, we
have A∩B={2,3,5,7}.
Step 2: Next, let’s find A∩Bc, the intersection of Aand the complement of
B. The complement of Bconsists of all non-prime numbers between 1 and 10.
So, Bc={1,4,6,8,9,10}. Therefore, A∩Bc={1,4,6,8,9,10}∩{n∈Z|1≤
n≤10}={1,4,6,8,9,10}.
Step 3: Finally, we find (A∩B)∪(A∩Bc), the union of A∩Band A∩Bc.
(A∩B)∪(A∩Bc) = {2,3,5,7}∪{1,4,6,8,9,10}={1,2,3,4,5,6,7,8,9,10}.
Therefore, (A∩B)∪(A∩Bc) = {1,2,3,4,5,6,7,8,9,10}.
Question 22
Question
Let A={n∈Z|2≤n≤10}and B={n∈Z|7≤n≤15}. Find the set
A∪B.
Solution
To find A∪B, we need to determine all the elements that are in A, in B, or in
both sets.
11
Step 1: Find elements in set AWe know that A={n∈Z|2≤n≤10}.
Therefore, Acontains the integers 2, 3, 4, 5, 6, 7, 8, 9, and 10.
Step 2: Find elements in set BWe know that B={n∈Z|7≤n≤15}.
Therefore, Bcontains the integers 7, 8, 9, 10, 11, 12, 13, 14, and 15.
Step 3: Determine A∪BThe union A∪Bincludes all the elements that
are in set A, set B, or both. So, A∪B={2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 23
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|x∈Aand x∈B}
Step 2: The elements in set Aare {1,2,3,4,5,6,7,8,9,10}, and the elements
in set Bare {5,6,7,8,9,10,11,12,13,14,15}.
Step 3: Therefore, the intersection of sets Aand Bis
A∩B={5,6,7,8,9,10}
Step 4: To find the union of sets Aand B, we need to combine all elements
from both sets without duplicates.
A∪B={x|x∈Aor x∈B}
Step 5: The elements in the union of sets Aand Bare
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Question 24
Question
Let Aand Bbe sets such that |A|= 10, |B|= 15. If |A∪B|= 20, find |A∩B|.
Solution
Step 1: In order to find |A∩B|, we can use the principle of inclusion-exclusion:
Step 2: The principle of inclusion-exclusion states that
|A∪B|=|A|+|B|−|A∩B|.
12
Step 3: Substituting the given values, we have
20 = 10 + 15 − |A∩B|.
Step 4: Simplifying the equation gives
20 = 25 − |A∩B|.
Step 5: Solving for |A∩B|yields
|A∩B|= 25 −20 = 5.
Step 6: Therefore, |A∩B|= 5.
Question 25
Question
Let A={x∈Z:−4≤x≤5}and B={x∈Z:−2≤x≤8}. Determine the
set A∩(A∪B).
Solution
Step 1: First, we find A∪B.
A∪B={x∈Z:−4≤x≤8}={−4,−3,−2,−1,0,1,2,3,4,5,6,7,8}
Step 2: Next, we find A∩(A∪B).
A∩(A∪B) = {x∈Z:x∈Aand x∈(A∪B)}
A∩(A∪B) = {x∈Z:−4≤x≤5 and −4≤x≤8}
A∩(A∪B) = {x∈Z:−4≤x≤5}={−4,−3,−2,−1,0,1,2,3,4,5}
Therefore, the set A∩(A∪B) is {−4,−3,−2,−1,0,1,2,3,4,5}.
Question 26
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|4≤x≤12}. Find the set C
defined as C=A∩B′, where B′is the complement of set B.
13
Solution
Given: A={x∈Z|1≤x≤10}
B={x∈Z|4≤x≤12}
First, let’s find B′, the complement of set B:
B′={x∈Z|x /∈B}
Since Bincludes integers from 4 to 12, B′will include all integers except
those. Therefore:
B′={x∈Z|x < 4 or x > 12}
Hence,
B′={x∈Z|x < 4}∪{x∈Z|x > 12}={x∈Z|x < 4}
Now, let’s find C=A∩B′:
C={x∈Z|x∈Aand x∈B′}
C={x∈Z|1≤x≤10 and x < 4}
Since there are no integers that satisfy both conditions (1 ≤x≤10 and x < 4),
the intersection A∩B′is an empty set.
Therefore, the set Cis:
C=∅
Question 27
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Define the set Cas C=A∪Band
the set Das D=A∩B. Find |C∩D|.
Solution
Step 1: First, find C=A∪Bby taking the union of sets Aand B.
C={1,2,3,4,5}∪{3,4,5,6,7}={1,2,3,4,5,6,7}
Step 2: Next, find D=A∩Bby taking the intersection of sets Aand B.
D={1,2,3,4,5}∩{3,4,5,6,7}={3,4,5}
Step 3: Calculate C∩Dby finding the intersection of sets Cand D.
C∩D={1,2,3,4,5,6,7}∩{3,4,5}={3,4,5}
Step 4: Finally, find the cardinality of the set C∩Dto get |C∩D|.
|C∩D|=|{3,4,5}| = 3
Therefore, |C∩D|= 3.
14
Question 28
Question
Let Aand Bbe two sets such that |A∩B|= 10, |A\B|= 15, and |B\A|= 20.
If |A∪B|= 50, find |A∩Bc|.
Solution
Step 1: We can use the principle of inclusion-exclusion to find |A∪B|.
|A∪B|=|A|+|B|−|A∩B|
=|A\B|+|B\A|+|A∩B|+|A∩B|
= 15 + 20 + 10 + 10
= 55.
Step 2: Since |A∪B|= 50, there must be 5 elements counted twice. These
5 elements are in both A\Band B\A.
Step 3: Now, we need to find |A∩Bc|=|A\B|.
|A∩Bc|=|A\B|
=|A|−|A∩B|
=|A| − 10.
Step 4: To find |A|, we use the formula for the union of two sets:
|A∪B|=|A|+|B|−|A∩B|
50 = |A|+|B| − 10
50 = |A|+|B| − 10
Step 5: We also know that |A\B|= 15 and |B\A|= 20. This implies that
Ahas 15 + 10 = 25 elements and Bhas 20 + 10 = 30 elements.
Step 6: Substituting the values back into the equation:
50 = 25 + 30 −10
50 = 45
|A|= 45.
Step 7: Finally, substitute |A|= 45 into |A∩Bc|=|A| − 10 to get:
|A∩Bc|= 45 −10
= 35.
Therefore, |A∩Bc|= 35.
15
Question 29
Question
Let A,B, and Cbe sets such that A∪B=A∪Cand A∩B=A∩C. Prove
that B=C.
Solution
Let’s prove this by first showing that B⊆Cand C⊆B.
Step 1: Showing B⊆CSince A∪B=A∪C, it follows that B⊆A∪C.
Thus, b∈B=⇒b∈Aor b∈C. But since A∩B=A∩C, we have b∈Aif
and only if b∈C. Therefore, b∈B=⇒b∈C, which implies B⊆C.
Step 2: Showing C⊆BBy symmetry, we can also show that C⊆B
using a similar argument. Since A∪B=A∪C, we have C⊆A∪B. Similarly,
a∈A=⇒a∈Bor a∈C. Since A∩B=A∩C, we have a∈Aif and only if
a∈B. Therefore, a∈C=⇒a∈B, which implies C⊆B.
Step 3: Conclusion From Step 1 and Step 2, we have shown that B⊆C
and C⊆B. Therefore, B=C.
Question 30
Question
Let A={x∈Z|2≤x≤8}and B={x∈Z|5≤x≤12}. Find the set
A∪B.
Solution
Step 1: First, list out the elements in set Aand set B.
A={2,3,4,5,6,7,8}
B={5,6,7,8,9,10,11,12}
Step 2: To find A∪B, we need to combine all unique elements from sets A
and B.
A∪B={2,3,4,5,6,7,8,9,10,11,12}
Therefore, the set A∪Bconsists of all integers from 2 to 12.
16
Solution
Step 1: To find A∩B, we first need to determine A∩B.
A∩B={x∈Q|x2<2}∩{x∈Q|x2>2}
={x∈Q|x2<2 and x2>2}
Step 2: Since x2<2 and x2>2 cannot hold true simultaneously for any
rational number x, we have A∩B=∅(the empty set).
Step 3: The complement of the empty set is the universal set, so A∩B=Q.
Therefore, A∩B=Q.
Question 3
Question
Let A={x∈Z|x2−4x−5 = 0}and B={x∈Z|x2−x−6 = 0}. Find
A∪Band A∩B.
Solution
Step 1: Find the solutions to the equations defining sets Aand B.
For set A: The equation x2−4x−5 = 0 can be factored as (x−5)(x+1) = 0.
So, the solutions are x= 5 and x=−1. Therefore, A={5,−1}.
For set B: The equation x2−x−6 = 0 can be factored as (x−3)(x+2) = 0.
So, the solutions are x= 3 and x=−2. Therefore, B={3,−2}.
Step 2: Determine A∪B(the union of sets Aand B). A∪Bis the set of
all elements that are in Aor in B(or both). Therefore, A∪B={5,−1,3,−2}.
Step 3: Determine A∩B(the intersection of sets Aand B). A∩Bis the set
of all elements that are in both Aand B. Since A={5,−1}and B={3,−2},
A∩B=∅(the empty set) because there are no elements common to both sets.
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 take the union of sets Aand B, which means
combining all the elements in both sets without repetition.
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
2
Step 2: To find A∩B, we need to find the intersection of sets Aand B,
which means finding the elements that are common to both sets.
A∩B={5,6,7,8,9,10}
Step 3: To find A\B, we need to determine the set of elements that are in
Abut not in B.
A\B={1,2,3,4}
Question 5
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find A∩B.
Solution
To find A∩B, we need to determine the elements that are common to both sets
Aand B.
Step 1: Write out the elements of set Aand B.
A={2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
Step 2: Identify the common elements between sets Aand B.
A∩B={5,6,7,8,9,10}
Step 3: Write the final answer.
A∩B={5,6,7,8,9,10}
Question 6
Question
Let A={x∈R|x2<5}and B={y∈R|y2≥3}. Determine A∪B.
Solution
To find A∪B, we need to find the set of all elements that are in either set Aor
set B.
Step 1: Find all elements in set A. Set A={x∈R|x2<5}can be
rewritten as A={x∈R| −√5<x<√5}. Therefore, set Acontains all real
numbers between −√5 and √5.
3
Step 2: Find all elements in set B. Set B={y∈R|y2≥3}can be
rewritten as B={y∈R|y≤ −√3 or y≥√3}. Therefore, set Bcontains all
real numbers less than or equal to −√3 or greater than or equal to √3.
Step 3: Find A∪B.The union of sets Aand B, denoted as A∪B, contains
all the elements that are in either set A, set B, or both.
Therefore, A∪B={x∈R|x≤ −√3 or −√5<x<√5 or x≥√3}.
Hence, A∪B= (−∞,−√3] ∪(−√5,√5) ∪[√3,∞).
Question 7
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Define the set Cas C=A∪Band
Das D=A∩B. Find the Cartesian product C×D.
Solution
To find the Cartesian product C×D, we first need to determine the sets Cand
D.
Step 1: Calculate C=A∪B.A={1,2,3,4,5}B={3,4,5,6,7}C=
A∪B={1,2,3,4,5,6,7}
Step 2: Calculate D=A∩B.A={1,2,3,4,5}B={3,4,5,6,7}D=
A∩B={3,4,5}
Step 3: Find C×D.C={1,2,3,4,5,6,7}D={3,4,5}C×D=
{(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,3),(3,4),(3,5),(4,3),(4,4),(4,5),(5,3),(5,4),(5,5),(6,3),(6,4),(6,5),(7,3),(7,4),(7,5)}
Therefore, C×Dis {(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,3),(3,4),(3,5),
(4,3),(4,4),(4,5),(5,3),(5,4),(5,5),(6,3),(6,4),(6,5),(7,3),(7,4),(7,5)}.
Question 8
Question
Let A={x∈R| −3≤x≤5}and B={x∈R|0< x < 6}. Compute the
following set:
A∪B−A∩B
Solution
We first find A∩Band then compute A∪B−(A∩B).
Step 1: Find A∩BTo find A∩B, we need to identify the elements that
are in both Aand B.
A∩B={x∈R| −3≤x≤5 and 0 <x<6}
Intersecting the intervals [−3,5] and (0,6), we see that A∩B= (0,5].
4
Step 2: Compute A∪B−A∩BTo compute A∪B−A∩B, we first find
A∪B:
A∪B={x∈R| −3≤x≤5 or 0 <x<6}
Now, we subtract the intersection A∩B= (0,5] from the union A∪B:
A∪B−(A∩B) = {x∈R|(−3≤x≤5 or 0 <x<6) and x /∈(0,5]}
Simplifying the above expression, we get:
A∪B−(A∩B) = (−3,0] ∪(5,6)
Therefore, A∪B−A∩B= (−3,0] ∪(5,6).
Question 9
Question
Let A={x∈R|x2−2x−8<0}and B={x∈R|x2−2x−8≥0}. Find
A∩B.
Solution
Step 1: We begin by finding the solutions to the inequality x2−2x−8<0. This
can be done by determining the roots of the quadratic equation x2−2x−8=0
and analyzing the sign of the expression in each interval.
The roots of x2−2x−8 = 0 can be found using the quadratic formula:
x=−(−2) ±p(−2)2−4(1)(−8)
2(1) =2±√4 + 32
2=2±√36
2=2±6
2
So, x=−2 or x= 4.
Step 2: Now, we analyze the sign of x2−2x−8 in the intervals (−∞,−2),
(−2,4), and (4,∞). Choosing test points x=−3, x= 0, and x= 5 respectively:
x=−3: (−3)2−2(−3) −8 = 9 + 6 −8=7⇒positive
x= 0: 02−2(0) −8 = −8⇒negative
x= 5: 52−2(5) −8 = 25 −10 −8=7⇒positive
Step 3: Based on the sign analysis, x2−2x−8<0 for x∈(−2,4). Therefore,
A= (−2,4).
Step 4: Next, we find the solutions to the inequality x2−2x−8≥0 by
analyzing the sign of x2−2x−8 in the intervals (−∞,−2), (−2,4), and (4,∞)
using the same test points.
Step 5: Based on the sign analysis, x2−2x−8≥0 for x∈(−∞,−2]∪[4,∞).
Therefore, B= (−∞,−2] ∪[4,∞).
5
Step 6: Finally, we find the intersection of Aand Bby taking A∩B. Since
A= (−2,4) and B= (−∞,−2] ∪[4,∞), we have:
A∩B= (−2,4) ∩((−∞,−2] ∪[4,∞)) = (−2,4) ∩[4,∞) = ∅
Therefore, A∩B=∅.
Question 10
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Determine the
set (A∩B)∪(A∩Bc).
Solution
Step 1: Find A∩B.
A∩Bconsists of all elements that are in both sets Aand B. So, we need
to find the intersection of the two sets.
A∩B={x∈Z|2≤x≤10}∩{x∈Z|5≤x≤15}
To find the intersection, we take the elements that are common in both sets,
which is the set {5,6,7,8,9,10}.
Step 2: Find A∩Bc.
A∩Bcconsists of all elements that are in set Aand not in set B. So, we
need to find the complement of set Band then find the intersection.
Bc={x∈Z|x < 5 or x > 15}
Therefore, A∩Bc={x∈Z|2≤x≤4} ∪ {x∈Z|11 ≤x≤15}, which
simplifies to {2,3,4,11,12,13,14,15}.
Step 3: Find (A∩B)∪(A∩Bc).
Now, we combine the two intersection sets we found in Step 1 and Step 2.
(A∩B)∪(A∩Bc) = {5,6,7,8,9,10}∪{2,3,4,11,12,13,14,15}
This gives us the set {2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 11
Question
Let A={x∈Z|2≤x≤7}and B={x∈Z|4≤x≤9}. Find A∪B,A∩B,
and A\B.
6
Solution
Let’s first list the elements of set Aand set B.
Step 1: Listing the elements of set A
A={2,3,4,5,6,7}
Step 2: Listing the elements of set B
B={4,5,6,7,8,9}
Step 3: Finding the union A∪B
A∪B={2,3,4,5,6,7,8,9}
Step 4: Finding the intersection A∩B
A∩B={4,5,6,7}
Step 5: Finding the set difference A\B
A\B={2,3}
Therefore, A∪B={2,3,4,5,6,7,8,9},A∩B={4,5,6,7}, and A\B=
{2,3}.
Question 12
Question
Let A={x∈R|1≤x < 5}and B={x∈R|3≤x < 7}. Find A∪Band
A∩B.
Solution
Step 1: To find A∪B(the union of sets Aand B), we need to find all elements
that are in either set A, set B, or both.
A∪B={x∈R|1≤x < 5 or 3 ≤x < 7}
Step 2: Simplify the condition: A∪B={x∈R|1≤x < 7}.
So, A∪B={x∈R|1≤x < 7}.
Step 3: To find A∩B(the intersection of sets Aand B), we need to find all
elements that are common to both sets Aand B.
A∩B={x∈R|3≤x < 5}
Step 4: Simplify the condition: A∩B={x∈R|3≤x < 5}.
Therefore, A∩B={x∈R|3≤x < 5}.
7
Question 13
Question
Let Aand Bbe two sets such that |A|= 7, |B|= 8, and |A∩B|= 5. If
|A∪B|= 10, find |A∩B′|where B′is the complement of set B.
Solution
Step 1: First, observe that |A∪B|=|A|+|B|−|A∩B|. We can use this
formula to determine |A∩B′|.
Step 2: Substitute the given values into the formula:
|A∪B|=|A|+|B|−|A∩B|=⇒10 = 7 + 8 −5
Step 3: Simplify the equation:
10 = 10
Step 4: Since the equation holds true, we can conclude that the given values
satisfy the condition.
Step 5: We know that |A∪B|=|A|+|B|− |A∩B|. We can rearrange this
formula to solve for |A∩B′|:
|A∩B′|=|A|+|B|−|A∪B|
Step 6: Substitute the given values into the formula:
|A∩B′|= 7 + 8 −10 = 5
Step 7: Therefore, |A∩B′|= 5.
Question 14
Question
Let A={x∈Z:x2−2x−8=0}and B={y∈Z:y2−y−6=0}. Determine
the intersection of sets Aand B.
Solution
Step 1: First, we find the elements in set Aby solving x2−2x−8 = 0. The
solutions to this quadratic equation are x=−2 and x= 4. So, set A={−2,4}.
Step 2: Next, we find the elements in set Bby solving y2−y−6 = 0. The
solutions to this quadratic equation are y=−2 and y= 3. So, set B={−2,3}.
Step 3: Now, we find the intersection of sets Aand Bto get the common
elements. The intersection of Aand Bis given by A∩B={−2}.
Therefore, the intersection of sets Aand Bis {−2}.
8
Question 15
Question
Let A={x|xis a prime number less than 10}and B={y|yis a perfect square less than 20}.
Find the set A∪B.
Solution
To find A∪B, we need to determine the set that contains all elements that are
in A, in B, or in both Aand B.
Step 1: Determine the elements in set A. The prime numbers less than 10
are 2, 3, 5, and 7. Thus, A={2,3,5,7}.
Step 2: Determine the elements in set B. The perfect squares less than 20
are 1, 4, 9, 16. Thus, B={1,4,9,16}.
Step 3: Find A∪B. The union of sets Aand Bincludes all elements that
are in A, in B, or in both Aand B.T heref ore,A∪B={1,2,3,4,5,7,9,16}.
Question 16
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Determine the set A∩(A∪B).
Solution
Step 1: First, we need to find A∪B, which is the union of sets Aand B. Step
2: A∪B={1,2,3,4,5,6,7}. Step 3: Next, we find the intersection of Awith
A∪B. Step 4: A∩(A∪B) = {1,2,3,4,5}.
Question 17
Question
Let A,B, and Cbe sets such that |A∩B|= 20, |B∩C|= 30, |A∩C|= 15,
|A∩B∩C|= 10, |A∪B∪C|= 60. Determine |A∩B∩C′|where C′denotes
the complement of set C.
Solution
Step 1: Apply the principle of inclusion-exclusion to find |A∩B∩C|.
|A∪B∪C|=|A|+|B|+|C|−|A∩B|−|B∩C|−|A∩C|+|A∩B∩C|
60 = |A|+|B|+|C| − 20 −30 −15 + 10
60 = |A|+|B|+|C| − 55
|A|+|B|+|C|= 115
9
Step 2: Use the values calculated to find |A∩B∩C′|.
|A∩B∩C′|=|A∩B|−|A∩B∩C|
= 20 −10
= 10
Therefore, |A∩B∩C′|= 10.
Question 18
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|6≤x≤15}. Find the set
A∩B.
Solution
Step 1: Write down the elements in set Aand set B.
A={1,2,3,4,5,6,7,8,9,10}
B={6,7,8,9,10,11,12,13,14,15}
Step 2: Identify the common elements in sets Aand B.
A∩B={6,7,8,9,10}
Therefore, A∩B={6,7,8,9,10}.
Question 19
Question
Let A={x∈Z|2≤x≤7}and B={x∈Z|3≤x≤9}. Find the set A∪B.
Solution
Step 1: To find A∪B, we need to consider all elements that are either in set A,
set B, or in both sets.
Step 2: Set Aincludes the integers from 2 to 7, while set Bincludes the
integers from 3 to 9. To find A∪B, we need to list out all the unique elements
in the union of set Aand set B.
Step 3: The elements of set Aare: {2,3,4,5,6,7}, and the elements of set
Bare: {3,4,5,6,7,8,9}.
Step 4: The union of sets Aand B, denoted A∪B, is the set containing all
the distinct elements from both sets.
Therefore, A∪B={2,3,4,5,6,7,8,9}.
10
Question 20
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Determine the
set of all possible values of xin the set A−B.
Solution
Step 1: To find A−B, we need to determine the values that are in Abut not in
B. Step 2: Acontains integers from 1 to 10 inclusive, and Bcontains integers
from 5 to 15 inclusive. Step 3: So, A−Bwill include the values in Athat are
not in B. Step 4: The values in Athat are not in Bare 1, 2, 3, and 4. Step 5:
Therefore, the set of all possible values of xin A−Bis {1,2,3,4}.
Question 21
Question
Let A={n∈Z|1≤n≤10}and B={n∈Z|nis a prime number}. Find
(A∩B)∪(A∩Bc).
Solution
Step 1: First, let’s determine A∩B, the intersection of sets Aand B. Since
Acontains the integers from 1 to 10, and Bcontains only prime numbers, we
have A∩B={2,3,5,7}.
Step 2: Next, let’s find A∩Bc, the intersection of Aand the complement of
B. The complement of Bconsists of all non-prime numbers between 1 and 10.
So, Bc={1,4,6,8,9,10}. Therefore, A∩Bc={1,4,6,8,9,10}∩{n∈Z|1≤
n≤10}={1,4,6,8,9,10}.
Step 3: Finally, we find (A∩B)∪(A∩Bc), the union of A∩Band A∩Bc.
(A∩B)∪(A∩Bc) = {2,3,5,7}∪{1,4,6,8,9,10}={1,2,3,4,5,6,7,8,9,10}.
Therefore, (A∩B)∪(A∩Bc) = {1,2,3,4,5,6,7,8,9,10}.
Question 22
Question
Let A={n∈Z|2≤n≤10}and B={n∈Z|7≤n≤15}. Find the set
A∪B.
Solution
To find A∪B, we need to determine all the elements that are in A, in B, or in
both sets.
11
Step 1: Find elements in set AWe know that A={n∈Z|2≤n≤10}.
Therefore, Acontains the integers 2, 3, 4, 5, 6, 7, 8, 9, and 10.
Step 2: Find elements in set BWe know that B={n∈Z|7≤n≤15}.
Therefore, Bcontains the integers 7, 8, 9, 10, 11, 12, 13, 14, and 15.
Step 3: Determine A∪BThe union A∪Bincludes all the elements that
are in set A, set B, or both. So, A∪B={2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 23
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|x∈Aand x∈B}
Step 2: The elements in set Aare {1,2,3,4,5,6,7,8,9,10}, and the elements
in set Bare {5,6,7,8,9,10,11,12,13,14,15}.
Step 3: Therefore, the intersection of sets Aand Bis
A∩B={5,6,7,8,9,10}
Step 4: To find the union of sets Aand B, we need to combine all elements
from both sets without duplicates.
A∪B={x|x∈Aor x∈B}
Step 5: The elements in the union of sets Aand Bare
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Question 24
Question
Let Aand Bbe sets such that |A|= 10, |B|= 15. If |A∪B|= 20, find |A∩B|.
Solution
Step 1: In order to find |A∩B|, we can use the principle of inclusion-exclusion:
Step 2: The principle of inclusion-exclusion states that
|A∪B|=|A|+|B|−|A∩B|.
12
Step 3: Substituting the given values, we have
20 = 10 + 15 − |A∩B|.
Step 4: Simplifying the equation gives
20 = 25 − |A∩B|.
Step 5: Solving for |A∩B|yields
|A∩B|= 25 −20 = 5.
Step 6: Therefore, |A∩B|= 5.
Question 25
Question
Let A={x∈Z:−4≤x≤5}and B={x∈Z:−2≤x≤8}. Determine the
set A∩(A∪B).
Solution
Step 1: First, we find A∪B.
A∪B={x∈Z:−4≤x≤8}={−4,−3,−2,−1,0,1,2,3,4,5,6,7,8}
Step 2: Next, we find A∩(A∪B).
A∩(A∪B) = {x∈Z:x∈Aand x∈(A∪B)}
A∩(A∪B) = {x∈Z:−4≤x≤5 and −4≤x≤8}
A∩(A∪B) = {x∈Z:−4≤x≤5}={−4,−3,−2,−1,0,1,2,3,4,5}
Therefore, the set A∩(A∪B) is {−4,−3,−2,−1,0,1,2,3,4,5}.
Question 26
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|4≤x≤12}. Find the set C
defined as C=A∩B′, where B′is the complement of set B.
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Solution
Given: A={x∈Z|1≤x≤10}
B={x∈Z|4≤x≤12}
First, let’s find B′, the complement of set B:
B′={x∈Z|x /∈B}
Since Bincludes integers from 4 to 12, B′will include all integers except
those. Therefore:
B′={x∈Z|x < 4 or x > 12}
Hence,
B′={x∈Z|x < 4}∪{x∈Z|x > 12}={x∈Z|x < 4}
Now, let’s find C=A∩B′:
C={x∈Z|x∈Aand x∈B′}
C={x∈Z|1≤x≤10 and x < 4}
Since there are no integers that satisfy both conditions (1 ≤x≤10 and x < 4),
the intersection A∩B′is an empty set.
Therefore, the set Cis:
C=∅
Question 27
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Define the set Cas C=A∪Band
the set Das D=A∩B. Find |C∩D|.
Solution
Step 1: First, find C=A∪Bby taking the union of sets Aand B.
C={1,2,3,4,5}∪{3,4,5,6,7}={1,2,3,4,5,6,7}
Step 2: Next, find D=A∩Bby taking the intersection of sets Aand B.
D={1,2,3,4,5}∩{3,4,5,6,7}={3,4,5}
Step 3: Calculate C∩Dby finding the intersection of sets Cand D.
C∩D={1,2,3,4,5,6,7}∩{3,4,5}={3,4,5}
Step 4: Finally, find the cardinality of the set C∩Dto get |C∩D|.
|C∩D|=|{3,4,5}| = 3
Therefore, |C∩D|= 3.
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Question 28
Question
Let Aand Bbe two sets such that |A∩B|= 10, |A\B|= 15, and |B\A|= 20.
If |A∪B|= 50, find |A∩Bc|.
Solution
Step 1: We can use the principle of inclusion-exclusion to find |A∪B|.
|A∪B|=|A|+|B|−|A∩B|
=|A\B|+|B\A|+|A∩B|+|A∩B|
= 15 + 20 + 10 + 10
= 55.
Step 2: Since |A∪B|= 50, there must be 5 elements counted twice. These
5 elements are in both A\Band B\A.
Step 3: Now, we need to find |A∩Bc|=|A\B|.
|A∩Bc|=|A\B|
=|A|−|A∩B|
=|A| − 10.
Step 4: To find |A|, we use the formula for the union of two sets:
|A∪B|=|A|+|B|−|A∩B|
50 = |A|+|B| − 10
50 = |A|+|B| − 10
Step 5: We also know that |A\B|= 15 and |B\A|= 20. This implies that
Ahas 15 + 10 = 25 elements and Bhas 20 + 10 = 30 elements.
Step 6: Substituting the values back into the equation:
50 = 25 + 30 −10
50 = 45
|A|= 45.
Step 7: Finally, substitute |A|= 45 into |A∩Bc|=|A| − 10 to get:
|A∩Bc|= 45 −10
= 35.
Therefore, |A∩Bc|= 35.
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Question 29
Question
Let A,B, and Cbe sets such that A∪B=A∪Cand A∩B=A∩C. Prove
that B=C.
Solution
Let’s prove this by first showing that B⊆Cand C⊆B.
Step 1: Showing B⊆CSince A∪B=A∪C, it follows that B⊆A∪C.
Thus, b∈B=⇒b∈Aor b∈C. But since A∩B=A∩C, we have b∈Aif
and only if b∈C. Therefore, b∈B=⇒b∈C, which implies B⊆C.
Step 2: Showing C⊆BBy symmetry, we can also show that C⊆B
using a similar argument. Since A∪B=A∪C, we have C⊆A∪B. Similarly,
a∈A=⇒a∈Bor a∈C. Since A∩B=A∩C, we have a∈Aif and only if
a∈B. Therefore, a∈C=⇒a∈B, which implies C⊆B.
Step 3: Conclusion From Step 1 and Step 2, we have shown that B⊆C
and C⊆B. Therefore, B=C.
Question 30
Question
Let A={x∈Z|2≤x≤8}and B={x∈Z|5≤x≤12}. Find the set
A∪B.
Solution
Step 1: First, list out the elements in set Aand set B.
A={2,3,4,5,6,7,8}
B={5,6,7,8,9,10,11,12}
Step 2: To find A∪B, we need to combine all unique elements from sets A
and B.
A∪B={2,3,4,5,6,7,8,9,10,11,12}
Therefore, the set A∪Bconsists of all integers from 2 to 12.
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