GSEB Class 11 Maths Solutions Chapter 1 Sets Exercise 1.5

Question 1. Let \( U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \), \( A = \{1, 2, 3, 4\} \), \( B = \{2, 4, 6, 8\} \), \( C = \{3, 4, 5, 6\} \) Find:
(i) \( A' \)
(ii) \( B' \)
(iii) \( (A \cup C)' \)
(iv) \( (A \cup B)' \)
(v) \( (A')' \)
(vi) \( (B - C)' \)
Answer:
(i) To find \( A' \), we subtract set A from the universal set U.
\( A' = U - A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{1, 2, 3, 4\} \)
\( = \{5, 6, 7, 8, 9\} \)
(ii) To find \( B' \), we remove set B from the universal set U.
\( B' = U - B = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2, 4, 6, 8\} \)
\( = \{1, 3, 5, 7, 9\} \)
(iii) To find \( (A \cup C)' \), first we combine sets A and C.
\( A \cup C = \{1, 2, 3, 4\} \cup \{3, 4, 5, 6\} = \{1, 2, 3, 4, 5, 6\} \)
Then we take the complement of this union from U.
\( (A \cup C)' = U - (A \cup C) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{1, 2, 3, 4, 5, 6\} \)
\( = \{7, 8, 9\} \)
(iv) To find \( (A \cup B)' \), first we combine sets A and B.
\( A \cup B = \{1, 2, 3, 4\} \cup \{2, 4, 6, 8\} = \{1, 2, 3, 4, 6, 8\} \)
Then we calculate the complement of this combined set from U.
\( (A \cup B)' = U - (A \cup B) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{1, 2, 3, 4, 6, 8\} \)
\( = \{5, 7, 9\} \)
(v) To find \( (A')' \), first we find the complement of A.
\( A' = U - A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{1, 2, 3, 4\} = \{5, 6, 7, 8, 9\} \)
Then we find the complement of \( A' \). According to the property \( (A')' = A \).
\( (A')' = U - A' = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{5, 6, 7, 8, 9\} \)
\( = \{1, 2, 3, 4\} \)
(vi) To find \( (B - C)' \), first we determine the difference between set B and set C.
\( B - C = \{2, 4, 6, 8\} - \{3, 4, 5, 6\} = \{2, 8\} \)
Then we find the complement of this difference from U.
\( (B - C)' = U - (B - C) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2, 8\} \)
\( = \{1, 3, 4, 5, 6, 7, 9\} \)
In simple words: For each part, we take elements from the universal set U that are not in the specified set or combination of sets. For complements like \( A' \), we remove A's elements from U. For unions like \( A \cup C \), we combine them first, then find the complement. For \( (A')' \), it always simplifies back to A itself.

Exam Tip: Remember De Morgan's Laws: \( (A \cup B)' = A' \cap B' \) and \( (A \cap B)' = A' \cup B' \). These can help verify your results for complements of unions and intersections.

Question 2. If \( U = \{a, b, c, d, e, f, g, h\} \), find the complements of the following sets:
\( A = \{a, b, c\} \)
\( B = \{d, e, f, g\} \)
\( C = \{a, c, e, g\} \)
\( D = \{f, g, h, a\} \)
Answer:
(i) To find the complement of A, we remove elements of A from U.
\( A' = U - A = \{a, b, c, d, e, f, g, h\} - \{a, b, c\} = \{d, e, f, g, h\} \)
(ii) To find the complement of B, we exclude elements of B from U.
\( B' = U - B = \{a, b, c, d, e, f, g, h\} - \{d, e, f, g\} = \{a, b, c, h\} \)
(iii) To find the complement of C, we subtract elements of C from U.
\( C' = U - C = \{a, b, c, d, e, f, g, h\} - \{a, c, e, g\} = \{b, d, f, h\} \)
(iv) To find the complement of D, we take out elements of D from U.
\( D' = U - D = \{a, b, c, d, e, f, g, h\} - \{f, g, h, a\} = \{b, c, d, e\} \)
In simple words: To find the complement of a set, simply list all the items in the universal set that are not present in the given set.

Exam Tip: When finding complements, always list all elements in the universal set first, then cross out the elements that are in the given set to avoid mistakes.

Question 3. Taking the set of natural numbers \( U = \{1, 2, 3, \ldots\} \) as the universal set, write down the complements of the following sets:
1. \( \{x: x \text{ is an even number}\} \)
2. \( \{x: x \text{ is an odd number}\} \)
3. \( \{x: x \text{ is a positive multiple of } 3\} \)
4. \( \{x: x \text{ is a prime number}\} \)
5. \( \{x: x \text{ is a natural number divisible by } 3 \text{ and } 5\} \)
6. \( \{x : x \text{ is a perfect square}\} \)
7. \( \{x: x \text{ is a perfect cube}\} \)
8. \( \{x: x + 5 = 8\} \)
9. \( \{x: 2x + 5 = 9\} \)
10. \( \{x: x \ge 7\} \)
11. \( \{x: x \in N \text{ and } 2x + 1 > 10\} \)
Answer:
1. The complement of "x is an even number" is "x is an odd natural number".
2. The complement of "x is an odd number" is "x is an even natural number".
3. The complement of "x is a positive multiple of 3" is "x is a natural number and x is not a multiple of 3".
4. The complement of "x is a prime number" is "x is a composite number or x equals 1".
5. The complement of "x is a natural number divisible by 3 and 5" is "x is a natural number and x is neither divisible by 3 nor by 5".
6. The complement of "x is a perfect square" is "x is a natural number, and x is not a perfect square".
7. The complement of "x is a perfect cube" is "x is a natural number and x is not a perfect cube".
8. For \( \{x: x + 5 = 8\} \), this means \( x = 3 \). So, the complement is "x is a natural number, and x is not 3".
9. For \( \{x: 2x + 5 = 9\} \), this means \( 2x = 4 \implies x = 2 \). So, the complement is "x is a natural number, and x is not 2".
10. For \( \{x: x \ge 7\} \), this means \( \{7, 8, 9, \ldots\} \). So, the complement is "x is a natural number, and x is less than 7".
11. For \( \{x: x \in N \text{ and } 2x + 1 > 10\} \), this means \( 2x > 9 \implies x > 4.5 \). So the set is \( \{5, 6, 7, \ldots\} \). Its complement is "x is a natural number and x is less than or equal to \( \frac{9}{2} \)" (which means \( x \le 4 \)).
In simple words: For each property, we describe the numbers that *do not* have that property, but are still natural numbers. For equations or inequalities, first solve them to find the set, then state its opposite.

Exam Tip: Remember that the universal set here is natural numbers. When finding complements, make sure your description of the complement set refers to natural numbers as well.

Question 4. If \( U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \), \( A = \{2, 4, 6, 8\} \), \( B = \{2, 3, 5, 7\} \). Verify that:
(i) \( (A \cup B)' = A' \cap B' \)
(ii) \( (A \cap B)' = A' \cup B' \)
Answer:
(i) Verification of \( (A \cup B)' = A' \cap B' \):
First, calculate the left-hand side (LHS): \( (A \cup B)' \)
\( A \cup B = \{2, 4, 6, 8\} \cup \{2, 3, 5, 7\} = \{2, 3, 4, 5, 6, 7, 8\} \)
\( (A \cup B)' = U - (A \cup B) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2, 3, 4, 5, 6, 7, 8\} = \{1, 9\} \)
Now, calculate the right-hand side (RHS): \( A' \cap B' \)
\( A' = U - A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2, 4, 6, 8\} = \{1, 3, 5, 7, 9\} \)
\( B' = U - B = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2, 3, 5, 7\} = \{1, 4, 6, 8, 9\} \)
\( A' \cap B' = \{1, 3, 5, 7, 9\} \cap \{1, 4, 6, 8, 9\} = \{1, 9\} \)
Since LHS = RHS, we have verified that \( (A \cup B)' = A' \cap B' \).
(ii) Verification of \( (A \cap B)' = A' \cup B' \):
First, calculate the left-hand side (LHS): \( (A \cap B)' \)
\( A \cap B = \{2, 4, 6, 8\} \cap \{2, 3, 5, 7\} = \{2\} \)
\( (A \cap B)' = U - (A \cap B) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2\} = \{1, 3, 4, 5, 6, 7, 8, 9\} \)
Now, calculate the right-hand side (RHS): \( A' \cup B' \)
\( A' = \{1, 3, 5, 7, 9\} \) (from part i)
\( B' = \{1, 4, 6, 8, 9\} \) (from part i)
\( A' \cup B' = \{1, 3, 5, 7, 9\} \cup \{1, 4, 6, 8, 9\} = \{1, 3, 4, 5, 6, 7, 8, 9\} \)
Since LHS = RHS, we have verified that \( (A \cap B)' = A' \cup B' \).
In simple words: We showed that De Morgan's Laws work. For the first law, the complement of A union B is the same as the complement of A intersected with the complement of B. For the second law, the complement of A intersection B is the same as the complement of A union the complement of B. We calculated both sides of each equation and found they were equal.

Exam Tip: Clearly show the steps for calculating both sides of the equation (LHS and RHS) separately before comparing them to prove set identities like De Morgan's Laws.

Question 5. Draw the Venn diagram for each of the following:
1. \( (A \cup B)' \)
2. \( A' \cap B' \)
3. \( (A \cap B)' \)
4. \( A' \cup B' \)
Answer:
1. The shaded area represents \( (A \cup B)' \), which includes all elements in the universal set U that are not in A or B.

2. The shaded area represents \( A' \cap B' \). According to De Morgan's Law, this is equivalent to \( (A \cup B)' \).

3. The shaded area represents \( (A \cap B)' \), which includes all elements in U except those that are in both A and B.

4. The shaded area represents \( A' \cup B' \). According to De Morgan's Law, this is equivalent to \( (A \cap B)' \).

In simple words: Venn diagrams show sets as circles inside a rectangle (the universal set). Shaded parts mean those elements are included. \( (A \cup B)' \) means everything outside both A and B. \( (A \cap B)' \) means everything except what is in both A and B. De Morgan's laws show that \( (A \cup B)' \) is the same as \( A' \cap B' \), and \( (A \cap B)' \) is the same as \( A' \cup B' \).

Exam Tip: Clearly label your circles (A, B) and the universal set (U) in your Venn diagrams. Ensure shading precisely matches the set operation being represented.

Question 6. Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A'?
Answer:
The universal set U contains all possible triangles in a plane. Set A includes all triangles that have at least one angle not equal to 60°. This means that set A includes all triangles that are not equilateral (because equilateral triangles have all three angles equal to 60°). Therefore, the complement of A, denoted as \( A' \), will be the set of triangles where all angles *are* 60°. This implies that \( A' \) is the set of all equilateral triangles.
\( A' = \text{set of equilateral triangles} \)
In simple words: If A is all triangles that are not equilateral, then A' must be all the triangles that *are* equilateral. Equilateral triangles have every angle at exactly 60 degrees.

Exam Tip: To find the complement of a descriptive set, think about the exact opposite property. If 'at least one' is in the original set, its complement will often use 'all' or 'none'.

Question 7. Fill in the blanks to make each of the following a true statement:
1. \( A \cup A' = \underline{\hspace{2cm}} \)
2. \( \phi' \cap A = \underline{\hspace{2cm}} \)
3. \( A \cap A' = \underline{\hspace{2cm}} \)
4. \( U' \cap A = \underline{\hspace{2cm}} \)
Answer:
Let U be the universal set.
1. \( A \cup A' = U \). The union of a set and its complement is always the universal set.
2. \( \phi' \cap A = U \cap A = A \). The complement of an empty set is the universal set, and the intersection of the universal set with any set A is A itself.
3. \( A \cap A' = \phi \). The intersection of a set and its complement is always an empty set.
4. \( U' \cap A = \phi \cap A = \phi \). The complement of the universal set is the empty set, and the intersection of the empty set with any set A is the empty set.
In simple words: For the first one, joining a set and everything outside it gives you the whole universe. For the second, everything outside nothing is the universe, and combining that with A just leaves A. For the third, a set and everything outside it have nothing in common. For the fourth, everything outside the universe is nothing, and combining nothing with A gives you nothing.

Exam Tip: Understand the basic properties of sets, such as the complement of the empty set (\( \phi' = U \)), the complement of the universal set (\( U' = \phi \)), and the union and intersection of a set with its complement.