Q1. Let U = { 1, 2, 3, 4, 5, 6, 7, 8, 9 },
A = { 1, 2, 3, 4},
B = { 2, 4, 6, 8 } and
C = { 3, 4, 5, 6 }. Find
(i) A′ (ii) B′ (iii) (A ∪ C)′ (iv) (A ∪ B)′ (v) (A′)′ (vi) (B – C)′

Solution

\( \begin{aligned} U &= \{1,2,3,4,5,6,7,8,9\} \\ A &= \{1,2,3,4\} \\ B &= \{2,4,6,8\} \\ C &= \{3,4,5,6\} \end{aligned} \)

Since complements are taken with respect to the universal set \(U\), we determine each result accordingly.

\( \begin{aligned} \text{(i)}\quad A' &= U - A = \{5,6,7,8,9\} \end{aligned} \)

\( \begin{aligned} \text{(ii)}\quad B' &= U - B = \{1,3,5,7,9\} \end{aligned} \)

First, \(A \cup C = \{1,2,3,4,5,6\}\).

\( \begin{aligned} \text{(iii)}\quad (A \cup C)' &= U - (A \cup C) = \{7,8,9\} \end{aligned} \)

Next, \(A \cup B = \{1,2,3,4,6,8\}\).

\( \begin{aligned} \text{(iv)}\quad (A \cup B)' &= U - (A \cup B) = \{5,7,9\} \end{aligned} \)

\( \begin{aligned} \text{(v)}\quad (A')' &= A = \{1,2,3,4\} \end{aligned} \)

Now, \(B - C = \{2,8\}\).

\( \begin{aligned} \text{(vi)}\quad (B - C)' &= U - (B - C) = \{1,3,4,5,6,7,9\} \end{aligned} \)

Q2. If U = { a, b, c, d, e, f, g, h}, find the complements of the following sets :
(i) A = {a, b, c}
(ii) B = {d, e, f, g}
(iii) C = {a, c, e, g}
(iv) D = { f, g, h, a}

Solution

\( \begin{aligned} U &= \{a,b,c,d,e,f,g,h\} \end{aligned} \)

The complement of a set is obtained by removing all its elements from the universal set \(U\).

\( \begin{aligned} \text{(i)}\quad A &= \{a,b,c\} \\ A' &= \{d,e,f,g,h\} \end{aligned} \)

\( \begin{aligned} \text{(ii)}\quad B &= \{d,e,f,g\} \\ B' &= \{a,b,c,h\} \end{aligned} \)

\( \begin{aligned} \text{(iii)}\quad C &= \{a,c,e,g\} \\ C' &= \{b,d,f,h\} \end{aligned} \)

\( \begin{aligned} \text{(iv)}\quad D &= \{f,g,h,a\} \\ D' &= \{b,c,d,e\} \end{aligned} \)

Q3. Taking the set of natural numbers as the universal set, write down the complements of the following sets:
(i) {x : x is an even natural number}
(ii) { x : x is an odd natural number }
(iii) {x : x is a positive multiple of 3}
(iv) { x : x is a prime number }
(v) {x : x is a natural number divisible by 3 and 5}
(vi) { x : x is a perfect square }
(vii) { x : x is a perfect cube}
(viii) { x : x + 5 = 8 }
(ix) { x : 2x + 5 = 9}
(x) { x : x ≥ 7 }
(xi) { x : x ∈ N and 2x + 1 > 10 }

Solution

\( U=\{1,2,3,4,5,\ldots\} \)

Here, the universal set is the set of all natural numbers. The complement of each set consists of all natural numbers that do not satisfy the given condition.

\( \begin{aligned} \text{(i)}\quad A &= \{x : x \text{ is an even natural number}\} \\ A' &= \{x : x \text{ is an odd natural number}\} \end{aligned} \)

\( \begin{aligned} \text{(ii)}\quad B &= \{x : x \text{ is an odd natural number}\} \\ B' &= \{x : x \text{ is an even natural number}\} \end{aligned} \)

\( \begin{aligned} \text{(iii)}\quad C &= \{x : x \text{ is a positive multiple of } 3\} \\ C' &= \{x : x \text{ is not a positive multiple of } 3\} \end{aligned} \)

\( \begin{aligned} \text{(iv)}\quad D &= \{x : x \text{ is a prime number}\} \\ D' &= \{x : x \text{ is a non-prime natural number}\} \end{aligned} \)

\( \begin{aligned} \text{(v)}\quad E &= \{x : x \text{ is divisible by both } 3 \text{ and } 5\} \\ E' &= \{x : x \text{ is not divisible by both } 3 \text{ and } 5\} \end{aligned} \)

\( \begin{aligned} \text{(vi)}\quad F &= \{x : x \text{ is a perfect square}\} \\ F' &= \{x : x \text{ is not a perfect square}\} \end{aligned} \)

\( \begin{aligned} \text{(vii)}\quad G &= \{x : x \text{ is a perfect cube}\} \\ G' &= \{x : x \text{ is not a perfect cube}\} \end{aligned} \)

\( \begin{aligned} \text{(viii)}\quad H &= \{x : x+5=8\} \\ H' &= \{x : x+5 \neq 8\} \end{aligned} \)

\( \begin{aligned} \text{(ix)}\quad I &= \{x : 2x+5=9\} \\ I' &= \{x : 2x+5 \neq 9\} \end{aligned} \)

\( \begin{aligned} \text{(x)}\quad J &= \{x : x \ge 7\} \\ J' &= \{x : 1 \le x < 7,\; x \in \mathbb{N}\} \end{aligned} \)

\( \begin{aligned} \text{(xi)}\quad K &= \{x : x \in \mathbb{N} \text{ and } 2x+1>10\} \\ K' &= \{x : x \in \mathbb{N} \text{ and } 2x+1 \le 10\} \end{aligned} \)

Q4. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8} and B = { 2, 3, 5, 7}. Verify that
(i) (A ∪ B)′ = A′ ∩ B′
(ii) (A ∩ B)′ = A′ ∪ B′

Solution

\( \begin{aligned} U &= \{1,2,3,4,5,6,7,8,9\} \\ A &= \{2,4,6,8\} \\ B &= \{2,3,5,7\} \end{aligned} \)

We verify each identity by evaluating both sides using the given universal set.

\( \begin{aligned} \text{(i)}\quad (A \cup B)' &= A' \cap B' \end{aligned} \)

\( \begin{aligned} A \cup B &= \{2,3,4,5,6,7,8\} \\ (A \cup B)' &= U - (A \cup B) = \{1,9\} \end{aligned} \)

\( \begin{aligned} A' &= U - A = \{1,3,5,7,9\} \\ B' &= U - B = \{1,4,6,8,9\} \end{aligned} \)

\( \begin{aligned} A' \cap B' &= \{1,9\} \end{aligned} \)

Thus, \( (A \cup B)' = A' \cap B' \) and the identity is verified.

\( \begin{aligned} \text{(ii)}\quad (A \cap B)' &= A' \cup B' \end{aligned} \)

\( \begin{aligned} A \cap B &= \{2\} \\ (A \cap B)' &= U - (A \cap B) = \{1,3,4,5,6,7,8,9\} \end{aligned} \)

\( \begin{aligned} A' &= \{1,3,5,7,9\} \\ B' &= \{1,4,6,8,9\} \end{aligned} \)

\( \begin{aligned} A' \cup B' &= \{1,3,4,5,6,7,8,9\} \end{aligned} \)

Hence, \( (A \cap B)' = A' \cup B' \) and the second identity is also verified.

Q5. Draw appropriate Venn diagram for each of the following :
(i) (A ∪ B)′,
(ii) A′ ∩ B′,
(iii) (A ∩ B)′,
(iv) A′ ∪ B′

Solution

Q6. 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′?

Solution

\( U=\{x : x \text{ is a triangle in a plane}\} \)

Here, the universal set \(U\) consists of all possible triangles in a plane.

\( A=\{x : x \text{ is a triangle with at least one angle different from } 60^\circ\} \)

The complement of \(A\) consists of all triangles that do not satisfy the given condition.

\( A'=\{x : x \text{ is a triangle in which all angles are } 60^\circ\} \)

Thus, \(A'\) is the set of all equilateral triangles.

Q7. Fill in the blanks to make each of the following a true statement :
(i) A ∪ A′ = . . .
(ii) φ′ ∩ A = . . .
(iii) A ∩ A′ = . . .
(iv) U′ ∩ A = . . .

Solution

Let \(U\) denote the universal set and \(A\) be any subset of \(U\). We evaluate each expression using basic properties of sets and their complements.

\( \begin{aligned} \text{(i)}\quad A \cup A' &= U \end{aligned} \)

The union of a set with its complement contains every element of the universal set.

\( \begin{aligned} \text{(ii)}\quad \varnothing' \cap A &= U \cap A = A \end{aligned} \)

Since the complement of the empty set is the universal set, its intersection with \(A\) gives back the set \(A\).

\( \begin{aligned} \text{(iii)}\quad A \cap A' &= \varnothing \end{aligned} \)

A set and its complement have no common elements, so their intersection is the empty set.

\( \begin{aligned} \text{(iv)}\quad U' \cap A &= \varnothing \cap A = \varnothing \end{aligned} \)

The complement of the universal set is the empty set, and its intersection with any set is again the empty set.