NCERT Solutions for Class 11 Mathematics

• (i) Months of a year beginning with 'J': {January, June, July}. The collection is well-defined and identical for everyone. It is a set.

• (ii) Ten most talented writers of India: The criterion "most talented" is subjective, vague, and varies from person to person. It is not a set.

• (iii) Team of eleven best-cricket batsmen of the world: The definition of "best batsmen" is subjective and non-measurable. It is not a set.

• (iv) Collection of all boys in your class: The membership is clearly defined, identifiable, and verifiable. It is a set.

• (v) Collection of all natural numbers less than 100: {1, 2, 3, ..., 99} is precisely determined. It is a set.

• (vi) Collection of novels written by Munshi Prem Chand: The exact bibliographic list of novels by Prem Chand is definite and historical. It is a set.

• (vii) Collection of all even integers: {..., −4, −2, 0, 2, 4, ...} is mathematically well-defined. It is a set.

• (viii) Collection of questions in this Chapter: Finite, numbered, and fixed in the textbook. It is a set.

• (ix) Collection of most dangerous animals of the world: "Dangerous" is subjective and depends on individual perspectives. It is not a set.

• (i) 5 ∈ A  (5 is present in set A)
• (ii) 8 ∉ A  (8 is not in set A)
• (iii) 0 ∉ A  (0 is not an element of A)
• (iv) 4 ∈ A  (4 is an element of A)
• (v) 2 ∈ A  (2 is an element of A)
• (vi) 10 ∉ A  (10 is not in set A)

• (i) A: Integers $x$ satisfying $-3 \le x < 7$:
$$\mathbf{A = \{-3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}}$$
• (ii) B: Natural numbers less than 6 ($1, 2, 3, 4, 5$):
$$\mathbf{B = \{1, 2, 3, 4, 5\}}$$
• (iii) C: Two-digit numbers whose digits add up to 8 ($1+7, 2+6, 3+5, 4+4, 5+3, 6+2, 7+1, 8+0$):
$$\mathbf{C = \{17, 26, 35, 44, 53, 62, 71, 80\}}$$
• (iv) D: Prime divisors of 60. Prime factorisation of $60 = 2^2 \times 3 \times 5$:
$$\mathbf{D = \{2, 3, 5\}}$$
• (v) E: Unique letters in 'TRIGONOMETRY':
$$\mathbf{E = \{T, R, I, G, O, N, M, E, Y\}}$$
• (vi) F: Unique letters in 'BETTER':
$$\mathbf{F = \{B, E, T, R\}}$$

• (i) {3, 6, 9, 12}: Multiples of 3 up to $3 \times 4$:
$$\mathbf{\{x : x = 3n, \, n \in \mathbb{N} \text{ and } 1 \le n \le 4\}}$$
• (ii) {2, 4, 8, 16, 32}: Powers of 2 from $2^1$ to $2^5$:
$$\mathbf{\{x : x = 2^n, \, n \in \mathbb{N} \text{ and } 1 \le n \le 5\}}$$
• (iii) {5, 25, 125, 625}: Powers of 5 from $5^1$ to $5^4$:
$$\mathbf{\{x : x = 5^n, \, n \in \mathbb{N} \text{ and } 1 \le n \le 4\}}$$
• (iv) {2, 4, 6, ...}: Set of even natural numbers:
$$\mathbf{\{x : x \text{ is an even natural number}\} \quad \text{or} \quad \{x : x = 2n, \, n \in \mathbb{N}\}}$$
• (v) {1, 4, 9, ..., 100}: Squares of natural numbers up to $10^2$ ($1^2, 2^2, ..., 10^2$):
$$\mathbf{\{x : x = n^2, \, n \in \mathbb{N} \text{ and } 1 \le n \le 10\}}$$

• (i) A: Odd natural numbers:
$$\mathbf{A = \{1, 3, 5, 7, 9, ...\}}$$
• (ii) B: Integers strictly between $-0.5$ and $4.5$ ($0, 1, 2, 3, 4$):
$$\mathbf{B = \{0, 1, 2, 3, 4\}}$$
• (iii) C: Integers whose square is $\le 4$, i.e. $x \in \{-2, -1, 0, 1, 2\}$:
$$\mathbf{C = \{-2, -1, 0, 1, 2\}}$$
• (iv) D: Unique letters in 'LOYAL' (L appears twice, write once):
$$\mathbf{D = \{L, O, Y, A\}}$$
• (v) E: Calendar months with fewer than 31 days:
$$\mathbf{E = \{\text{February, April, June, September, November}\}}$$
• (vi) F: Consonants before 'k' (letters before 'k': a, b, c, d, e, f, g, h, i, j; vowels are a, e, i):
$$\mathbf{F = \{b, c, d, f, g, h, j\}}$$

• (i) {1, 2, 3, 6}: All natural divisors of 6. Matches with (c).
• (ii) {2, 3}: Prime numbers that divide 6. Matches with (a).
• (iii) {M, A, T, H, E, I, C, S}: Distinct letters in 'MATHEMATICS'. Matches with (d).
• (iv) {1, 3, 5, 7, 9}: Odd natural numbers less than 10. Matches with (b).

Summary of Matches:
(i) $\leftrightarrow$ (c),   (ii) $\leftrightarrow$ (a),   (iii) $\leftrightarrow$ (d),   (iv) $\leftrightarrow$ (b)

• (i) Odd natural numbers divisible by 2: By definition, no odd number is divisible by 2. The set contains no elements.
Hence, it is a null set.

• (ii) Set of even prime numbers: The number $2$ is an even prime number. So the set is $\{2\} \ne \phi$.
Hence, it is not a null set.

• (iii) {x : x ∈ N, x < 5 and x > 7}: No natural number can simultaneously be less than 5 and greater than 7.
Hence, it is a null set.

• (iv) Points common to two parallel lines: Parallel lines never intersect by definition; they have no points in common.
Hence, it is a null set.

• (i) Set of months of a year: Exactly 12 months. Definite count $\rightarrow$ Finite set.
• (ii) {1, 2, 3, ...}: Natural numbers continue indefinitely without bound $\rightarrow$ Infinite set.
• (iii) {1, 2, 3, ..., 99, 100}: Exactly 100 elements $\rightarrow$ Finite set.
• (iv) Positive integers greater than 100: {101, 102, 103, ...} has no upper bound $\rightarrow$ Infinite set.
• (v) Prime numbers less than 99: Primes {2, 3, 5, 7, ..., 97} is a definite count (25 primes) $\rightarrow$ Finite set.

• (i) Lines parallel to the x-axis: In the Cartesian plane, infinitely many horizontal lines $y = c$ (for every real $c$) can be drawn. $\rightarrow$ Infinite set.

• (ii) Letters in the English alphabet: Exactly 26 letters (A to Z). $\rightarrow$ Finite set.

• (iii) Numbers which are multiples of 5: {..., −15, −10, −5, 0, 5, 10, 15, ...} has no beginning or end. $\rightarrow$ Infinite set.

• (iv) Animals living on the earth: Although the number is extremely large, the count of living animals at any given instant is a finite whole number. $\rightarrow$ Finite set.

• (v) Circles passing through origin (0, 0): Circles with center $(h, k)$ and radius $r = \sqrt{h^2+k^2}$ can be drawn for any choice of center $(h, k) \in \mathbb{R}^2$. There are infinitely many such circles. $\rightarrow$ Infinite set.

• (i) A = {a, b, c, d}, B = {d, c, b, a}: Both sets contain exactly the same four elements. Yes, A = B.

• (ii) A = {4, 8, 12, 16}, B = {8, 4, 16, 18}: $12 \in A$ but $12 \notin B$, and $18 \in B$ but $18 \notin A$. No, A ≠ B.

• (iii) A = {2, 4, 6, 8, 10}, B = {x : x is positive even integer and x ≤ 10}: The positive even integers $\le 10$ are $\{2, 4, 6, 8, 10\}$. Exactly identical. Yes, A = B.

• (iv) A = {10, 20, 30, 40, ...}, B = {10, 15, 20, 25, ...}: $15 \in B$ but $15 \notin A$ (since 15 is not a multiple of 10). No, A ≠ B.

• (i) Pair A = {2, 3} and B:
Solve quadratic equation: $x^2 + 5x + 6 = 0$
$\implies (x + 2)(x + 3) = 0 \implies x = -2 \text{ or } x = -3$.
Thus, $B = \{-2, -3\}$.
Since $2 \in A$ but $2 \notin B$, and $-2 \in B$ but $-2 \notin A$, A ≠ B.

• (ii) Pair A and B:
Letters in 'FOLLOW': $A = \{F, O, L, W\}$.
Letters in 'WOLF': $B = \{W, O, L, F\}$.
Since both sets contain precisely the same four elements ${F, O, L, W}$, A = B.

Compare each set pair:

• $B = \{1, 2, 3, 4\}$ and $D = \{3, 1, 4, 2\}$: They contain exactly the same elements. Hence, B = D.
• $E = \{-1, 1\}$ and $G = \{1, -1\}$: Both contain elements $-1$ and $1$. Hence, E = G.
• $A = \{2, 4, 8, 12\}$: No other set has this exact combination ($C$ has 14 instead of 2).
• $F = \{0, a\}$ and $H = \{0, 1\}$: Elements differ ($a \ne 1$).

Conclusion: The only equal sets are B = D and E = G.

• (i) {2, 3, 4} ⊂ {1, 2, 3, 4, 5}  (all elements 2, 3, 4 belong to the RHS)
• (ii) {a, b, c} ⊄ {b, c, d}  (element $a$ does not belong to RHS)
• (iii) {Student of Class XI} ⊂ {Student of your school}  (every Class XI student is a school student)
• (iv) {Circle in plane} ⊄ {Circle with radius 1 unit}  (a circle can have radius 2, 3, etc., not in RHS)
• (v) {Triangle in plane} ⊄ {Rectangle in plane}  (triangles are 3-sided, rectangles are 4-sided)
• (vi) {Equilateral triangle in plane} ⊂ {Triangle in same plane}  (every equilateral triangle is a triangle)
• (vii) {Even natural numbers} ⊂ {Integers}  ({2, 4, 6, ...} is a subset of {..., -2, -1, 0, 1, 2, ...})

• (i) {a, b} ⊄ {b, c, a}: Both $a, b$ belong to ${b, c, a}$, so ${a, b} \subset \{b, c, a\}$. Hence, statement is False.

• (ii) {a, e} ⊂ {vowels}: English vowels are ${a, e, i, o, u}$. Since $a, e$ are vowels, the statement is True.

• (iii) {1, 2, 3} ⊂ {1, 3, 5}: The element $2 \in \{1, 2, 3\}$ but $2 \notin \{1, 3, 5\}$. Hence, statement is False.

• (iv) {a} ⊂ {a, b, c}: Element $a$ belongs to ${a, b, c}$. Hence, subset relation is True.

• (v) {a} ∈ {a, b, c}: The elements of ${a, b, c}$ are letters $a, b, c$, not the set ${a}$. Hence, statement is False ($a \in \{a, b, c\}$, but $\{a\} \notin \{a, b, c\}$).

• (vi) LHS: Even natural numbers less than 6 = ${2, 4}$.
RHS: Natural numbers dividing 36 = ${1, 2, 3, 4, 6, 9, 12, 18, 36}$.
Since $2 \in \text{RHS}$ and $4 \in \text{RHS}$, ${2, 4} \subset \text{RHS}$. Statement is True.

• (i) {3, 4} ⊂ A: Incorrect. ${3, 4}$ is an element of $A$, so ${3, 4} \in A$. For it to be a subset, its members 3 and 4 must belong to $A$, but $3 \notin A$ and $4 \notin A$.

• (ii) {3, 4} ∈ A: Correct. ${3, 4}$ is directly an element of $A$.

• (iii) {{3, 4}} ⊂ A: Correct. The element of this singleton set is ${3, 4}$, which is in $A$.

• (iv) 1 ∈ A: Correct. 1 is an element of $A$.

• (v) 1 ⊂ A: Incorrect. An element cannot be a subset; only a set enclosed in braces like ${1} \subset A$.

• (vi) {1, 2, 5} ⊂ A: Correct. Since $1, 2, 5 \in A$.

• (vii) {1, 2, 5} ∈ A: Incorrect. ${1, 2, 5}$ is a subset, not an element of $A$.

• (viii) {1, 2, 3} ⊂ A: Incorrect. $3$ is not an element of $A$ (it is inside ${3, 4}$).

• (ix) Φ ∈ A: Incorrect. $\phi$ is not an element listed inside $A$.

• (x) Φ ⊂ A: Correct. The empty set $\phi$ is a subset of every set.

• (xi) {Φ} ⊂ A: Incorrect. For ${\phi} \subset A$, we must have $\phi \in A$, which is false.

• (i) {a}: $n = 1 \implies 2^1 = 2$ subsets:
Φ, {a}

• (ii) {a, b}: $n = 2 \implies 2^2 = 4$ subsets:
Φ, {a}, {b}, {a, b}

• (iii) {1, 2, 3}: $n = 3 \implies 2^3 = 8$ subsets:
Φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}

• (iv) Φ: $n = 0 \implies 2^0 = 1$ subset:
Φ

Given $A = \phi$, the number of elements in $A$ is $n(A) = 0$.
The power set $P(A)$ is the collection of all subsets of $A$.
Number of elements in $P(A) = 2^{n(A)} = 2^0 = 1$.
Explicitly: $P(A) = \{\phi\}$.
Answer: P(A) has 1 element.

• (i) $-4 < x \le 6 \implies$ (−4, 6]  (Open at -4, closed at 6)
• (ii) $-12 < x < -10 \implies$ (−12, −10)  (Open at both ends)
• (iii) $0 \le x < 7 \implies$ [0, 7)  (Closed at 0, open at 7)
• (iv) $3 \le x \le 4 \implies$ [3, 4]  (Closed at both ends)

• (i) (−3, 0) = {x : x ∈ R, −3 < x < 0}
• (ii) [6, 12] = {x : x ∈ R, 6 ≤ x ≤ 12}
• (iii) (6, 12] = {x : x ∈ R, 6 < x ≤ 12}
• (iv) [−23, 5) = {x : x ∈ R, −23 ≤ x < 5}

A universal set $U$ must contain every element under consideration.
• (i) For right triangles: We can propose U = The set of all triangles in the plane (or the set of all polygons).
• (ii) For isosceles triangles: We can propose U = The set of all triangles in the plane (or the set of all 2D geometric shapes).

To be a universal set for $A, B, C$, the proposed set must contain all elements from $A \cup B \cup C$.
$A \cup B \cup C = \{0, 1, 2, 3, 4, 5, 6, 8\}$.
• (i) Missing 8 $\implies$ Cannot be universal set.
• (ii) $\phi$ contains no elements $\implies$ Cannot be universal set.
• (iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} contains every element ${0, 1, 2, 3, 4, 5, 6, 8}$. $\implies$ Valid Universal Set.
• (iv) Missing 0 $\implies$ Cannot be universal set.

Answer: (iii) is the correct option.

• (i) X ∪ Y: Elements in $X$ or $Y$:
$$\mathbf{X \cup Y = \{1, 2, 3, 5\}}$$
• (ii) A ∪ B:
$$\mathbf{A \cup B = \{a, b, c, e, i, o, u\}}$$
• (iii) A ∪ B:
$A = \{3, 6, 9, 12, ...\}$, $B = \{1, 2, 3, 4, 5\}$:
$$\mathbf{A \cup B = \{1, 2, 4, 5, 3, 6, 9, 12, ...\} = \{x : x = 1, 2, 4, 5 \text{ or a multiple of } 3\}}$$
• (iv) A ∪ B:
$A = \{2, 3, 4, 5, 6\}$, $B = \{7, 8, 9\}$:
$$\mathbf{A \cup B = \{2, 3, 4, 5, 6, 7, 8, 9\} = \{x : x \in \mathbb{N} \text{ and } 1 < x < 10\}}$$
• (v) A ∪ B:
$A \cup \phi = A$:
$$\mathbf{A \cup B = \{1, 2, 3\}}$$

• Check subset condition: Elements of $A$ are $a$ and $b$. Both $a \in B$ and $b \in B$.
Yes, A ⊂ B.

• Compute $A \cup B$:
$A \cup B = \{a, b\} \cup \{a, b, c\} = \{a, b, c\} = B$.
A ∪ B = {a, b, c} = B.

Given $A \subset B$, every element of $A$ already belongs to $B$.
Therefore, combining all elements of $A$ with $B$ adds no new elements beyond $B$.
$$A \cup B = B$$
Answer: A ∪ B = B.

• (i) A ∪ B: {1, 2, 3, 4} ∪ {3, 4, 5, 6} = {1, 2, 3, 4, 5, 6}
• (ii) A ∪ C: {1, 2, 3, 4} ∪ {5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8}
• (iii) B ∪ C: {3, 4, 5, 6} ∪ {5, 6, 7, 8} = {3, 4, 5, 6, 7, 8}
• (iv) B ∪ D: {3, 4, 5, 6} ∪ {7, 8, 9, 10} = {3, 4, 5, 6, 7, 8, 9, 10}
• (v) A ∪ B ∪ C: (A ∪ B) ∪ C = {1, 2, 3, 4, 5, 6} ∪ {5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8}
• (vi) A ∪ B ∪ D: (A ∪ B) ∪ D = {1, 2, 3, 4, 5, 6} ∪ {7, 8, 9, 10} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
• (vii) B ∪ C ∪ D: (B ∪ C) ∪ D = {3, 4, 5, 6, 7, 8} ∪ {7, 8, 9, 10} = {3, 4, 5, 6, 7, 8, 9, 10}

• (i) X ∩ Y: $X = \{1, 3, 5\}, Y = \{1, 2, 3\}$:
$$\mathbf{X \cap Y = \{1, 3\}}$$
• (ii) A ∩ B: $A = \{a, e, i, o, u\}, B = \{a, b, c\}$:
$$\mathbf{A \cap B = \{a\}}$$
• (iii) A ∩ B: $A = \{3, 6, 9, 12, ...\}, B = \{1, 2, 3, 4, 5\}$:
$$\mathbf{A \cap B = \{3\}}$$
• (iv) A ∩ B: $A = \{2, 3, 4, 5, 6\}, B = \{7, 8, 9\}$:
No elements in common:
$$\mathbf{A \cap B = \phi}$$
• (v) A ∩ B: $A = \{1, 2, 3\}, B = \phi$:
$$\mathbf{A \cap \phi = \phi}$$

• (i) A ∩ B: {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13} = {7, 9, 11}
• (ii) B ∩ C: {7, 9, 11, 13} ∩ {11, 13, 15} = {11, 13}
• (iii) A ∩ C ∩ D: (A ∩ C) ∩ D = {11} ∩ {15, 17} = Φ
• (iv) A ∩ C: {3, 5, 7, 9, 11} ∩ {11, 13, 15} = {11}
• (v) B ∩ D: {7, 9, 11, 13} ∩ {15, 17} = Φ
• (vi) A ∩ (B ∪ C): B ∪ C = {7, 9, 11, 13, 15}. Then A ∩ {7, 9, 11, 13, 15} = {7, 9, 11}
• (vii) A ∩ D: {3, 5, 7, 9, 11} ∩ {15, 17} = Φ
• (viii) A ∩ (B ∪ D): B ∪ D = {7, 9, 11, 13, 15, 17}. Then A ∩ {7, 9, 11, 13, 15, 17} = {7, 9, 11}
• (ix) (A ∩ B) ∩ (B ∪ C): {7, 9, 11} ∩ {7, 9, 11, 13, 15} = {7, 9, 11}
• (x) (A ∪ D) ∩ (B ∪ C): A ∪ D = {3, 5, 7, 9, 11, 15, 17}, B ∪ C = {7, 9, 11, 13, 15}.
Intersection = {7, 9, 11, 15}

• (i) A ∩ B: $\mathbb{N} \cap \{\text{even naturals}\} = \mathbf{B}$ = {x : x is an even natural number}

• (ii) A ∩ C: $\mathbb{N} \cap \{\text{odd naturals}\} = \mathbf{C}$ = {x : x is an odd natural number}

• (iii) A ∩ D: $\mathbb{N} \cap \{\text{prime numbers}\} = \mathbf{D}$ = {x : x is a prime number}

• (iv) B ∩ C: An even natural number is never odd $\implies$ Φ

• (v) B ∩ D: The only even prime number is 2 $\implies$ {2}

• (vi) C ∩ D: All prime numbers except the even prime 2 are odd $\implies$ {x : x is an odd prime number}

• (i) {1, 2, 3, 4} and {4, 5, 6}:
Intersection = $\{4\} \ne \phi$.
Not disjoint.

• (ii) {a, e, i, o, u} and {c, d, e, f}:
Both sets share the letter 'e': Intersection = $\{e\} \ne \phi$.
Not disjoint.

• (iii) {Even integers} and {Odd integers}:
No integer can be both even and odd simultaneously. Intersection = $\phi$.
Disjoint sets.

• (i) A − B: Remove 12 from A $\implies$ {3, 6, 9, 15, 18, 21}
• (ii) A − C: Remove 6, 12 from A $\implies$ {3, 9, 15, 18, 21}
• (iii) A − D: Remove 15 from A $\implies$ {3, 6, 9, 12, 18, 21}
• (iv) B − A: Remove 12 from B $\implies$ {4, 8, 16, 20}
• (v) C − A: Remove 6, 12 from C $\implies$ {2, 4, 8, 10, 14, 16}
• (vi) D − A: Remove 15 from D $\implies$ {5, 10, 20}
• (vii) B − C: Remove 4, 8, 16 from B $\implies$ {12, 20}
• (viii) B − D: Remove 20 from B $\implies$ {4, 8, 12, 16}
• (ix) C − B: Remove 4, 8, 12, 16 from C $\implies$ {2, 6, 10, 14}
• (x) D − B: Remove 20 from D $\implies$ {5, 10, 15}
• (xi) C − D: Remove 10 from C $\implies$ {2, 4, 6, 8, 12, 14, 16}
• (xii) D − C: Remove 10 from D $\implies$ {5, 15, 20}

• (i) X − Y: Elements in $X$ that are not in $Y$ (remove $b, d$ from $X$):
$$\mathbf{X - Y = \{a, c\}}$$
• (ii) Y − X: Elements in $Y$ that are not in $X$ (remove $b, d$ from $Y$):
$$\mathbf{Y - X = \{f, g\}}$$
• (iii) X ∩ Y: Common elements of $X$ and $Y$:
$$\mathbf{X \cap Y = \{b, d\}}$$

The real numbers $\mathbb{R}$ consist of all rational numbers $\mathbb{Q}$ and all irrational numbers.
By removing all rational numbers from the real number system, only irrational numbers remain.
$$\mathbb{R} - \mathbb{Q} = \mathbf{T} = \text{Set of irrational numbers}$$
Answer: R − Q is the set of irrational numbers.

• (i) {2, 3, 4, 5} and {3, 6}:
Intersection = $\{3\} \ne \phi$. The sets share the element 3.
False.

• (ii) {a, e, i, o, u} and {a, b, c, d}:
Intersection = $\{a\} \ne \phi$. The sets share the element 'a'.
False.

• (iii) {2, 6, 10, 14} and {3, 7, 11, 15}:
First set has even numbers, second has odd numbers. Intersection = $\phi$.
True.

• (iv) {2, 6, 10} and {3, 7, 11}:
No elements in common: $\{2, 6, 10\} \cap \{3, 7, 11\} = \phi$.
True.

• (i) A′: U − A = {1, ..., 9} − {1, 2, 3, 4} = {5, 6, 7, 8, 9}
• (ii) B′: U − B = {1, ..., 9} − {2, 4, 6, 8} = {1, 3, 5, 7, 9}
• (iii) (A ∪ C)′: A ∪ C = {1, 2, 3, 4, 5, 6}.
(A ∪ C)′ = U − {1, 2, 3, 4, 5, 6} = {7, 8, 9}
• (iv) (A ∪ B)′: A ∪ B = {1, 2, 3, 4, 6, 8}.
(A ∪ B)′ = U − {1, 2, 3, 4, 6, 8} = {5, 7, 9}
• (v) (A′)′: Law of double complementation: (A′)′ = A = {1, 2, 3, 4}
• (vi) (B − C)′: B − C = {2, 4, 6, 8} − {3, 4, 5, 6} = {2, 8}.
(B − C)′ = U − {2, 8} = {1, 3, 4, 5, 6, 7, 9}

• (i) A′: U − {a, b, c} = {d, e, f, g, h}
• (ii) B′: U − {d, e, f, g} = {a, b, c, h}
• (iii) C′: U − {a, c, e, g} = {b, d, f, h}
• (iv) D′: U − {f, g, h, a} = {b, c, d, e}

Universal set $U = \mathbb{N} = \{1, 2, 3, 4, ...\}$.

• • (i): {x : x is an odd natural number}
• • (ii): {x : x is an even natural number}
• • (iii): {x : x ∈ N and x is not a positive multiple of 3}
• • (iv): Primes complement in $\mathbb{N}$ includes 1 and composite numbers:
  {x : x ∈ N, x = 1 or x is a positive composite number}
• • (v): Divisible by 3 and 5 means divisible by 15. Complement:
  {x : x ∈ N and x is not divisible by 15} (or not divisible by 3 or not divisible by 5)
• • (vi): {x : x ∈ N and x is not a perfect square}
• • (vii): {x : x ∈ N and x is not a perfect cube}
• • (viii): $x + 5 = 8 \implies x = 3$. Complement:
  {x : x ∈ N and x ≠ 3} = $\mathbb{N} - \{3\}$
• • (ix): $2x + 5 = 9 \implies 2x = 4 \implies x = 2$. Complement:
  {x : x ∈ N and x ≠ 2} = $\mathbb{N} - \{2\}$
• • (x): Complement of $x \ge 7$:
  {x : x ∈ N and x < 7} = {1, 2, 3, 4, 5, 6}
• • (xi): $2x + 1 > 10 \implies 2x > 9 \implies x > 4.5$. Complement in $\mathbb{N}$:
  {x : x ∈ N and x ≤ 4} = {1, 2, 3, 4}

First compute $A'$ and $B'$: $$A' = U - A = \{1, 3, 5, 7, 9\}$$ $$B' = U - B = \{1, 4, 6, 8, 9\}$$ (i) Verification of (A ∪ B)′ = A′ ∩ B′:
• LHS: $A \cup B = \{2, 3, 4, 5, 6, 7, 8\}$.
$(A \cup B)' = U - \{2, 3, 4, 5, 6, 7, 8\} = \mathbf{\{1, 9\}}$.
• RHS: $A' \cap B' = \{1, 3, 5, 7, 9\} \cap \{1, 4, 6, 8, 9\} = \mathbf{\{1, 9\}}$.
Since $\text{LHS} = \text{RHS} = \{1, 9\}$, the identity is Verified.

(ii) Verification of (A ∩ B)′ = A′ ∪ B′:
• LHS: $A \cap B = \{2\}$.
$(A \cap B)' = U - \{2\} = \mathbf{\{1, 3, 4, 5, 6, 7, 8, 9\}}$.
• RHS: $A' \cup B' = \{1, 3, 5, 7, 9\} \cup \{1, 4, 6, 8, 9\} = \mathbf{\{1, 3, 4, 5, 6, 7, 8, 9\}}$.
Since $\text{LHS} = \text{RHS}$, the identity is Verified.

Below are official geometric Venn diagram representations showing shaded regions corresponding to each set identity:

• $U =$ Set of all triangles in a plane.
• $A =$ Set of triangles that have at least one angle $\ne 60^\circ$.
• Complement $A' = U - A$ is the set of triangles that do NOT have any angle different from $60^\circ$.
This means all three angles of the triangle must equal $60^\circ$ ($60^\circ + 60^\circ + 60^\circ = 180^\circ$).
A triangle whose all three internal angles are $60^\circ$ is an equilateral triangle.

Answer: A′ is the set of all equilateral triangles.

• (i) A ∪ A′: Union of a set with its complement covers the entire universal set $\implies$ U

• (ii) Φ′ ∩ A: Since $\phi' = U$, we have $U \cap A = \mathbf{A}$

• (iii) A ∩ A′: A set and its complement have no elements in common $\implies$ Φ

• (iv) U′ ∩ A: Since $U' = \phi$, we have $\phi \cap A = \mathbf{Φ}$

First write set A in roster form by solving $x^2 - 8x + 12 = 0$:
$$(x - 2)(x - 6) = 0 \implies x = 2 \text{ or } x = 6$$ So, $A = \{2, 6\}$.
The sets are: $$A = \{2, 6\}, \quad B = \{2, 4, 6\}, \quad C = \{2, 4, 6, 8, ...\}, \quad D = \{6\}$$ Now inspect subset relations:

• $D = \{6\}$: $6 \in A, 6 \in B, 6 \in C$.
  $\implies \mathbf{D \subset A, \, D \subset B, \, D \subset C}$
• $A = \{2, 6\}$: $2, 6 \in B$ and $2, 6 \in C$.
  $\implies \mathbf{A \subset B, \, A \subset C}$
• $B = \{2, 4, 6\}$: $2, 4, 6 \in C$.
  $\implies \mathbf{B \subset C}$

Summary: $D \subset A \subset B \subset C$.

• (i) False:
Let $A = \{1\}$ and $B = \{\{1\}, 2\}$.
Here $1 \in A$ and $A \in B$, but $1 \notin B$ (the element in $B$ is the set $\{1\}$, not the number $1$).

• (ii) False:
Let $A = \{1\}$, $B = \{1, 2\}$, and $C = \{\{1, 2\}, 3\}$.
Here $A \subset B$ and $B \in C$, but $A = \{1\} \notin C$ (the element in $C$ is $\{1, 2\}$).

• (iii) True (Proof):
Let $x \in A$.
Since $A \subset B$, $x \in B$.
Since $B \subset C$, $x \in C$.
Thus, every $x \in A \implies x \in C$. Hence, A ⊂ C.

• (iv) False:
Let $A = \{1, 2\}$, $B = \{0, 3\}$, $C = \{1, 2, 4\}$.
Here $A \not\subset B$ and $B \not\subset C$, but $A \subset C$.

• (v) False:
Let $A = \{1, 2, 3\}$, $B = \{2, 3, 4\}$. Here $A \not\subset B$.
Choose $x = 1 \in A$. Clearly $1 \notin B$.

• (vi) True (Proof by Contraposition):
Given $A \subset B$. Suppose $x \in A$.
Then by definition of subset, $x \in B$.
This contradicts the premise $x \notin B$.
Therefore, our assumption was false, and we must have x ∉ A.

To Prove: $B = C$, which is equivalent to proving $B \subset C$ and $C \subset B$.

Step 1: Express B using Distributive Law:
$$B = B \cap (A \cup B)$$ Given $A \cup B = A \cup C$, substitute it: $$B = B \cap (A \cup C)$$ Apply distributive law: $$B = (B \cap A) \cup (B \cap C) = (A \cap B) \cup (B \cap C)$$ Given $A \cap B = A \cap C$, substitute: $$B = (A \cap C) \cup (B \cap C) \quad \text{--- (1)}$$
Step 2: Express C using Distributive Law:
$$C = C \cap (A \cup C)$$ Since $A \cup C = A \cup B$: $$C = C \cap (A \cup B) = (C \cap A) \cup (C \cap B) = (A \cap C) \cup (B \cap C) \quad \text{--- (2)}$$
Step 3: Comparison:
From (1) and (2), the right-hand sides are identical: $$(A \cap C) \cup (B \cap C) = (A \cap C) \cup (B \cap C)$$ $$\implies \mathbf{B = C}$$ Hence proved.

To show equivalence, we show $(i) \implies (ii) \implies (iii) \implies (iv) \implies (i)$:

• (i) $\implies$ (ii): Let $A \subset B$. Then every element of $A$ is in $B$. Thus, there is no element in $A$ which is not in $B$. Hence, $A - B = \phi$.

• (ii) $\implies$ (iii): Let $A - B = \phi$. This means $A$ has no elements outside $B$, so $A \subset B$. Therefore, the union of all elements in $A$ and $B$ equals $B$, i.e., $A \cup B = B$.

• (iii) $\implies$ (iv): Let $A \cup B = B$. Then $A \subset B$. The common elements between $A$ and $B$ must be all of $A$. Hence, $A \cap B = A$.

• (iv) $\implies$ (i): Let $A \cap B = A$. This means every element of $A$ belongs to $B$. Hence, $A \subset B$.

Therefore, all four statements are completely equivalent.

Given: $A \subset B$.
To Prove: $(C - B) \subset (C - A)$.

Let $x \in (C - B)$.
$\implies x \in C \text{ and } x \notin B$.
Since $A \subset B$, if $x$ were in $A$, it would have to be in $B$.
Therefore, $x \notin B \implies x \notin A$.
So we have $x \in C \text{ and } x \notin A$.
$\implies x \in (C - A)$.

Since every element $x \in (C - B)$ also belongs to $(C - A)$: $$\mathbf{(C - B) \subset (C - A)}$$ Hence proved.

Given: $P(A) = P(B)$, where $P(X)$ denotes the power set of $X$.
To Prove: $A = B$.

Part 1: Prove A ⊂ B:
Let $x \in A$.
Then the singleton set ${x}$ is a subset of $A$, i.e., ${x} \subset A$.
By definition of power set, ${x} \in P(A)$.
Since $P(A) = P(B)$, ${x} \in P(B)$.
$\implies {x} \subset B \implies x \in B$.
Since every $x \in A$ belongs to $B$, we have A ⊂ B  --- (1)

Part 2: Prove B ⊂ A:
Let $y \in B \implies {y} \subset B \implies {y} \in P(B)$.
Since $P(B) = P(A)$, ${y} \in P(A)$.
$\implies {y} \subset A \implies y \in A$.
Thus, B ⊂ A  --- (2)

From (1) and (2), $A \subset B$ and $B \subset A \implies \mathbf{A = B}$.
Hence proved.

Answer: No, it is not true.

Counterexample:
Let $A = \{1\}$ and $B = \{2\}$.
• $P(A) = \{\phi, \{1\}\}$
• $P(B) = \{\phi, \{2\}\}$
• LHS: $P(A) \cup P(B) = \{\phi, \{1\}, \{2\}\}$

Now find RHS:
$A \cup B = \{1, 2\}$.
• RHS: $P(A \cup B) = \{\phi, \{1\}, \{2\}, \{1, 2\}\}$.

Observe that $\{1, 2\} \in P(A \cup B)$, but $\{1, 2\} \notin P(A) \cup P(B)$.
Hence, $\text{LHS} \ne \text{RHS}$. In general, $P(A) \cup P(B) \subset P(A \cup B)$, but equality does not hold.

Part 1: Prove A = (A ∩ B) ∪ (A − B):
RHS = $(A \cap B) \cup (A \cap B')$   [since $A - B = A \cap B'$]
Apply distributive property of $\cap$ over $\cup$:
RHS = $A \cap (B \cup B')$
Since $B \cup B' = U$ (universal set):
RHS = $A \cap U = A = \text{LHS}$.
Hence, A = (A ∩ B) ∪ (A − B).

Part 2: Prove A ∪ (B − A) = (A ∪ B):
LHS = $A \cup (B \cap A')$
Apply distributive property of $\cup$ over $\cap$:
LHS = $(A \cup B) \cap (A \cup A')$
Since $A \cup A' = U$:
LHS = $(A \cup B) \cap U = A \cup B = \text{RHS}$.
Hence, A ∪ (B − A) = A ∪ B.

(i) Show A ∪ (A ∩ B) = A (Absorption Law):
LHS = $(A \cap U) \cup (A \cap B)$   [since $A = A \cap U$]
$= A \cap (U \cup B)$   [by distributive law]
Since $U \cup B = U$:
$= A \cap U = A = \text{RHS}$.
Hence proved.

(ii) Show A ∩ (A ∪ B) = A:
LHS = $(A \cup \phi) \cap (A \cup B)$   [since $A = A \cup \phi$]
$= A \cup (\phi \cap B)$   [by distributive law]
Since $\phi \cap B = \phi$:
$= A \cup \phi = A = \text{RHS}$.
Hence proved.

To prove this, we construct a counterexample where $A \cap B = A \cap C$ but $B \ne C$.

Let: $$A = \{0, 1\}, \quad B = \{1, 2\}, \quad C = \{1, 3\}$$ Now calculate the intersections: $$A \cap B = \{0, 1\} \cap \{1, 2\} = \{1\}$$ $$A \cap C = \{0, 1\} \cap \{1, 3\} = \{1\}$$ Here, $A \cap B = A \cap C = \{1\}$.
However: $$B = \{1, 2\} \ne \{1, 3\} = C$$ Thus, A ∩ B = A ∩ C does NOT imply B = C.

Given: $A \cap X = B \cap X = \phi$ and $A \cup X = B \cup X$.
To Prove: $A = B$.

Start with $A$: $$A = A \cap (A \cup X)$$ Since $A \cup X = B \cup X$, substitute: $$A = A \cap (B \cup X)$$ Apply the Distributive Law of $\cap$ over $\cup$: $$A = (A \cap B) \cup (A \cap X)$$ Since given $A \cap X = \phi$: $$A = (A \cap B) \cup \phi = A \cap B \quad \text{--- (1)}$$
Now start with $B$: $$B = B \cap (B \cup X)$$ Since $B \cup X = A \cup X$: $$B = B \cap (A \cup X) = (B \cap A) \cup (B \cap X)$$ Since given $B \cap X = \phi$: $$B = (B \cap A) \cup \phi = B \cap A = A \cap B \quad \text{--- (2)}$$
Comparing (1) and (2): $$A = A \cap B \quad \text{and} \quad B = A \cap B \implies \mathbf{A = B}$$ Hence proved.

We need pairwise intersections to be non-empty, but all three sets cannot have any single element in common.

Let: $$A = \{1, 2\}$$ $$B = \{2, 3\}$$ $$C = \{1, 3\}$$ Now verify each condition:

• $A \cap B = \{1, 2\} \cap \{2, 3\} = \{2\} \ne \phi$
• $B \cap C = \{2, 3\} \cap \{1, 3\} = \{3\} \ne \phi$
• $A \cap C = \{1, 2\} \cap \{1, 3\} = \{1\} \ne \phi$

Now calculate the triple intersection: $$A \cap B \cap C = (A \cap B) \cap C = \{2\} \cap \{1, 3\} = \mathbf{\phi}$$ Hence, the sets $A = \{1, 2\}, B = \{2, 3\}, C = \{1, 3\}$ completely satisfy all required conditions.

Let $U$ be the universal set of all students surveyed, $T$ be the set of students taking tea, and $C$ be the set of students taking coffee.

Given:
• Total students surveyed $n(U) = 600$
• Students taking tea $n(T) = 150$
• Students taking coffee $n(C) = 225$
• Students taking both $n(T \cap C) = 100$

Step 1: Find students taking either tea or coffee $n(T \cup C)$:
$$n(T \cup C) = n(T) + n(C) - n(T \cap C)$$ $$n(T \cup C) = 150 + 225 - 100 = 375 - 100 = 275$$
Step 2: Find students taking neither tea nor coffee:
By De Morgan's Law, neither tea nor coffee is $(T \cup C)'$: $$n(T' \cap C') = n(T \cup C)' = n(U) - n(T \cup C)$$ $$n(T \cup C)' = 600 - 275 = \mathbf{325}$$
Answer: 325 students were taking neither tea nor coffee.

Let $H$ be the set of students knowing Hindi and $E$ be the set of students knowing English.

Given:
• Total students $n(H \cup E) = 100$ (or in group)
• Students who know Hindi $n(H) = 100$
• Students who know English $n(E) = 50$
• Students who know both $n(H \cap E) = 25$

To Find: Number of students who know Hindi only, which is $n(H - E)$.

Formula:
$$n(H - E) = n(H) - n(H \cap E)$$ $$n(H - E) = 100 - 25 = \mathbf{75}$$
Answer: 75 students know Hindi only.

Let $H, T, I$ denote the sets of people reading newspapers H, T, and I respectively.

Given data: $$n(U) = 60$$ $$n(H) = 25, \quad n(T) = 26, \quad n(I) = 26$$ $$n(H \cap T) = 11, \quad n(H \cap I) = 9, \quad n(T \cap I) = 8$$ $$n(H \cap T \cap I) = 3$$ (i) People who read at least one newspaper $n(H \cup T \cup I)$:
By the 3-Set Inclusion-Exclusion Principle: $$n(H \cup T \cup I) = n(H) + n(T) + n(I) - n(H \cap T) - n(H \cap I) - n(T \cap I) + n(H \cap T \cap I)$$ $$n(H \cup T \cup I) = 25 + 26 + 26 - 11 - 9 - 8 + 3$$ $$n(H \cup T \cup I) = 77 - 28 + 3 = 49 + 3 = \mathbf{52}$$
(ii) Number of people who read exactly one newspaper:
Break down each region:

• Read all 3 = $3$
• Read only H and T = $11 - 3 = 8$
• Read only H and I = $9 - 3 = 6$
• Read only T and I = $8 - 3 = 5$
• Read H only = $25 - (8 + 3 + 6) = 25 - 17 = 8$
• Read T only = $26 - (8 + 3 + 5) = 26 - 16 = 10$
• Read I only = $26 - (6 + 3 + 5) = 26 - 14 = 12$

Total reading exactly one newspaper: $$\text{Exactly one} = 8 + 10 + 12 = \mathbf{30}$$
Answers: (i) 52 people read at least one newspaper;   (ii) 30 people read exactly one newspaper.

Let $A, B, C$ represent the sets of people who liked products A, B, and C respectively.

Given:
• $n(A) = 21$
• $n(B) = 26$
• $n(C) = 29$
• $n(A \cap B) = 14$
• $n(C \cap A) = 12$
• $n(B \cap C) = 14$
• $n(A \cap B \cap C) = 8$

To Find: Number of people who liked product C only, denoted by $n(C \text{ only})$.

Step-by-step Region Breakdown:

• People who liked all three products = $n(A \cap B \cap C) = 8$
• People who liked C and A only = $n(C \cap A) - n(A \cap B \cap C) = 12 - 8 = 4$
• People who liked C and B only = $n(B \cap C) - n(A \cap B \cap C) = 14 - 8 = 6$

Now subtract these intersection portions from the total who liked C ($n(C) = 29$):
$$n(C \text{ only}) = n(C) - [\text{liked C and A only}] - [\text{liked C and B only}] - [\text{liked all three}]$$ $$n(C \text{ only}) = 29 - 4 - 6 - 8 = 29 - 18 = \mathbf{11}$$
Answer: 11 people liked product C only.