Sets

NCERT Class 11 Maths • Exercise 1.5

1. Finding Complements ($A’ = U – A$)

Given: $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$
$A = \{1, 2, 3, 4\}, B = \{2, 4, 6, 8\}, C = \{3, 4, 5, 6\}$

Q1: Find the following complements.

(i) $A’$

Remove elements of A from U.
$A’ = \{5, 6, 7, 8, 9\}$

(ii) $B’$

Remove $\{2, 4, 6, 8\}$ from U.
$B’ = \{1, 3, 5, 7, 9\}$

(iii) $(A \cup C)’$

First, find $A \cup C = \{1, 2, 3, 4, 5, 6\}$
$(A \cup C)’ = \{7, 8, 9\}$

(iv) $(A \cup B)’$

$A \cup B = \{1, 2, 3, 4, 6, 8\}$
$(A \cup B)’ = \{5, 7, 9\}$

(v) $(A’)’$

Complement of a complement is the set itself.
$(A’)’ = A = \{1, 2, 3, 4\}$

(vi) $(B – C)’$

$B – C = \{2, 8\}$ (Remove 4, 6)
$(B – C)’ = \{1, 3, 4, 5, 6, 7, 9\}$

2. Complements (Alphabets)

Given: $U = \{a, b, c, d, e, f, g, h\}$

Q2: Find complements of the following.

(i) $A = \{a, b, c\}$

$A’ = \{d, e, f, g, h\}$

(ii) $B = \{d, e, f, g\}$

$B’ = \{a, b, c, h\}$

(iii) $C = \{a, c, e, g\}$

$C’ = \{b, d, f, h\}$

(iv) $D = \{f, g, h, a\}$

$D’ = \{b, c, d, e\}$

3. Universal Set: Natural Numbers ($N$)

Q3: Write down the complements.

(i) $\{x : x \text{ is an even natural number}\}$

Complement: $\{x : x \text{ is an odd natural number}\}$

(ii) $\{x : x \text{ is an odd natural number}\}$

Complement: $\{x : x \text{ is an even natural number}\}$

(iii) $\{x : x \text{ is a positive multiple of 3}\}$

Complement: $\{x : x \in N \text{ and } x \text{ is not a multiple of 3}\}$

(iv) $\{x : x \text{ is a prime number}\}$

Complement: $\{x : x \text{ is a positive composite number and } x = 1\}$

(v) $\{x : x \text{ is divisible by 3 and 5}\}$

Complement: $\{x : x \in N \text{ and } x \text{ is not divisible by 15}\}$

(vi) $\{x : x \text{ is a perfect square}\}$

Complement: $\{x : x \in N \text{ and } x \text{ is not a perfect square}\}$

(vii) $\{x : x \text{ is a perfect cube}\}$

Complement: $\{x : x \in N \text{ and } x \text{ is not a perfect cube}\}$

(viii) $\{x : x + 5 = 8\}$

Given set = $\{3\}$.
Complement: $\{x : x \in N \text{ and } x \neq 3\}$

(ix) $\{x : 2x + 5 = 9\}$

$2x = 4 \Rightarrow x = 2$. Given set = $\{2\}$.
Complement: $\{x : x \in N \text{ and } x \neq 2\}$

(x) $\{x : x \ge 7\}$

Complement: $\{x : x \in N \text{ and } x < 7\} = \{1, 2, 3, 4, 5, 6\}$

(xi) $\{x : x \in N \text{ and } 2x + 1 > 10\}$

$2x > 9 \Rightarrow x > 4.5$. So set is $\{5, 6, 7, …\}$.
Complement: $\{x : x \in N \text{ and } x \le 4.5\} = \{1, 2, 3, 4\}$

4. Verification of De Morgan’s Laws

Given: $U=\{1..9\}, A=\{2,4,6,8\}, B=\{2,3,5,7\}$

Q4: Verify (i) $(A \cup B)’ = A’ \cap B’$ and (ii) $(A \cap B)’ = A’ \cup B’$

Prerequisites:
$A’ = \{1, 3, 5, 7, 9\}$
$B’ = \{1, 4, 6, 8, 9\}$

(i) Verify $(A \cup B)’ = A’ \cap B’$

LHS: $A \cup B = \{2, 3, 4, 5, 6, 7, 8\}$
$(A \cup B)’ = \{1, 9\}$

RHS: $A’ \cap B’ = \{1, 3, 5, 7, 9\} \cap \{1, 4, 6, 8, 9\}$
$= \{1, 9\}$
LHS = RHS (Verified)

(ii) Verify $(A \cap B)’ = A’ \cup B’$

LHS: $A \cap B = \{2\}$
$(A \cap B)’ = \{1, 3, 4, 5, 6, 7, 8, 9\}$

RHS: $A’ \cup B’ = \{1, 3, 5, 7, 9\} \cup \{1, 4, 6, 8, 9\}$
$= \{1, 3, 4, 5, 6, 7, 8, 9\}$
LHS = RHS (Verified)

5. Venn Diagrams

Q5: Draw appropriate Venn diagrams.

(i) $(A \cup B)’$

This represents the region outside both circles A and B.

[Image of Venn diagram for complement of A union B]

(ii) $A’ \cap B’$

By De Morgan’s Law, $A’ \cap B’ = (A \cup B)’$. The diagram is the same as (i).

[Image of Venn diagram for A complement intersection B complement]

(iii) $(A \cap B)’$

This represents everything except the common intersection area.

[Image of Venn diagram for complement of A intersection B]

(iv) $A’ \cup B’$

By De Morgan’s Law, $A’ \cup B’ = (A \cap B)’$. The diagram is the same as (iii).

6. Set of Triangles

Q6: Find $A’$ given properties of triangles.

Given:
$U$ = Set of all triangles.
$A$ = Set of triangles with at least one angle different from $60^\circ$.

Logic:
The complement of “at least one different from $60^\circ$” is “none different from $60^\circ$”.
This means all three angles are equal to $60^\circ$.

Answer:
$A’ = $ Set of all Equilateral Triangles.

7. Fill in the Blanks

Q7: Complete the statements.

(i) $A \cup A’ = \dots$
Everything in A plus everything not in A makes everything.
Answer: $U$

(ii) $\phi’ \cap A = \dots$
$\phi’ = U$. So, $U \cap A = A$.
Answer: $A$

(iii) $A \cap A’ = \dots$
No element can be in A and not in A simultaneously.
Answer: $\phi$

(iv) $U’ \cap A = \dots$
$U’ = \phi$. So, $\phi \cap A = \phi$.
Answer: $\phi$