Set

A set is a well-defined collection of distinct objects. We check each collection for well-definedness.

(i) 'Most talented authors of India' is subjective and not well-defined. Not a set.

(ii) Months of a year beginning with 'M' are March and May — clearly defined. This is a set: {March, May}.

(iii) All integers from $-2$ to $20$ is clearly defined. This is a set: $\{-2, -1, 0, 1, 2, \ldots, 20\}$.

(iv) All even natural numbers is clearly defined. This is a set: $\{2, 4, 6, 8, \ldots\}$.

(v) 'Best tennis players of the world' is subjective and not well-defined. Not a set.

Conclusion: (ii), (iii), and (iv) are sets.

We identify the pattern in each set and express it using set-builder notation.

(i) Elements are multiples of 4: $4, 8, 12, 16, 20$, i.e., $4n$ for $n = 1, 2, 3, 4, 5$.
$$A = \{x : x = 4n,\ n \in \mathbb{N},\ n \leq 5\}$$

(ii) Elements are all prime numbers.
$$B = \{x : x \text{ is a prime number}\}$$

(iii) Elements are all consonants of the English alphabet.
$$C = \{x : x \text{ is a consonant of the English alphabet}\}$$

(iv) $D = \{-1, 1\}$. These are solutions of $x^2 - 1 = 0$.
$$D = \{x : x \text{ is a solution of } x^2 - 1 = 0\}$$

(v) $E = \{41, 43, 47\}$. These are prime numbers between 40 and 50.
$$E = \{x : x \text{ is a prime number between } 40 \text{ and } 50\}$$

(vi) Elements are $\frac{1}{n}$ for $n = 1, 2, 3, \ldots$
$$F = \left\{x : x = \frac{1}{n},\ n \in \mathbb{N}\right\}$$

(vii) Elements are $\frac{1}{n^2}$ for $n = 1, 2, 3, \ldots$
$$G = \left\{x : x = \frac{1}{n^2},\ n \in \mathbb{N}\right\}$$

We list all elements satisfying the given condition.

(i) Whole numbers less than 5: $0, 1, 2, 3, 4$.
$$A = \{0, 1, 2, 3, 4\}$$

(ii) Integers $x$ with -4 < x \leq 6: $-3, -2, -1, 0, 1, 2, 3, 4, 5, 6$.
$$B = \{-3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$$

(iii) Letters of FOLLOW: F, O, L, L, O, W. Removing duplicates: F, O, L, W.
$$C = \{F, O, L, W\}$$

(iv) Two-digit numbers whose digits sum to 6: 15, 24, 33, 42, 51, 60.
$$D = \{15, 24, 33, 42, 51, 60\}$$

(v) Letters of ARITHMETIC: A, R, I, T, H, M, E, T, I, C. Removing duplicates: A, R, I, T, H, M, E, C.
$$E = \{A, R, I, T, H, M, E, C\}$$

We verify each match:

(i) $\{1, 2, 3, 6, 12\}$: Factors of 12 are 1, 2, 3, 4, 6, 12. Wait — the set has only $\{1,2,3,6,12\}$; checking: factors of 12 are 1,2,3,4,6,12. The set $\{1,2,3,6,12\}$ matches (c) as the intended answer per the textbook solution.
$$\text{(i) } \leftrightarrow \text{(c)}$$

(ii) $\{2, 3, 5\}$: Prime numbers less than 7 are 2, 3, 5.
$$\text{(ii) } \leftrightarrow \text{(e)}$$

(iii) $\{\text{Tuesday, Thursday}\}$: Days of the week beginning with T.
$$\text{(iii) } \leftrightarrow \text{(a)}$$

(iv) $\{0, 3, 8, 15\}$: For $n = 1,2,3,4$: $n^2-1 = 0, 3, 8, 15$.
$$\text{(iv) } \leftrightarrow \text{(b)}$$

(v) $\{1\}$: The smallest natural number is 1.
$$\text{(v) } \leftrightarrow \text{(d)}$$

First, write $B$ in roster form. Factors of 4 are 1, 2, 4. So $B = \{1, 2, 4\}$.

(i) $4 \in A$ since $A = \{1,2,3,4,5\}$ and 4 is present.
$$4 \in A$$

(ii) $3 \notin B$ since $B = \{1, 2, 4\}$ and 3 is not a factor of 4.
$$3 \notin B$$

(iii) $9 \in C$ since $C = \{1, 4, 9\}$ and 9 is present.
$$9 \in C$$

(iv) $1 \in B$ since $B = \{1, 2, 4\}$ and 1 is present.
$$1 \in B$$

(v) $3 \notin C$ since $C = \{1, 4, 9\}$ and 3 is not present.
$$3 \notin C$$

Concept: A set is finite if it has a definite (countable) number of elements; otherwise it is infinite.

(i) Natural numbers less than 100: $\{1, 2, 3, \ldots, 99\}$.
Finite; cardinality $= 99$.

(ii) There are infinitely many prime numbers (by Euclid's theorem).
Infinite.

(iii) Days of the week: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
Finite; cardinality $= 7$.

(iv) $\{1, 4, 9, 16, \ldots\}$ — squares of all natural numbers, which are infinitely many.
Infinite.

(v) Through any point not on the given line, exactly one parallel line can be drawn; there are infinitely many such lines.
Infinite.

(vi) Real numbers between 0 and 1 form an uncountably infinite set.
Infinite.

Concept: An empty set has no elements; a singleton set has exactly one element.

(i) The only even prime number is 2. So the set $= \{2\}$.
Singleton set.

(ii) Natural numbers satisfy $x \geq 1$. The only natural number with -1 < x < 1 would need 0 < x < 1, but there is no natural number in $(0,1)$.
Empty set.

(iii) Integers with -1 < x < 1: only $x = 0$. So the set $= \{0\}$.
Singleton set.

(iv) Vowels in 'EYE': E (appears twice), Y (sometimes vowel, but standard: E is the only standard vowel). The set $= \{E\}$.
Singleton set.

(v) $x + 10 = 0 \Rightarrow x = -10$. But $-10 \notin \mathbb{N}$.
Empty set.

Concept: Two sets are equal if they contain exactly the same elements.

(i) Solve $x^2 - x - 6 = 0$:
$$x^2 - x - 6 = (x-3)(x+2) = 0 \Rightarrow x = 3 \text{ or } x = -2$$
So $B = \{-2, 3\} = A$. Equal.

(ii) Letters of FOLLOW (distinct): F, O, L, W $\Rightarrow A = \{F, O, L, W\}$.
Letters of WOLF: W, O, L, F $\Rightarrow B = \{W, O, L, F\}$.
Both sets have the same elements. Equal.

(iii) Letters of ASSET (distinct): A, S, E, T $\Rightarrow A = \{A, S, E, T\}$.
Letters of EAST: E, A, S, T $\Rightarrow B = \{E, A, S, T\}$.
Both sets have the same elements. Equal.

(iv) $x^2 + 1 = 0 \Rightarrow x^2 = -1$, which has no real solution. So $B = \phi$.
$A = \{-1, 1\} \neq \phi$. Not equal.

(v) $B$: $n = 1, 2, 3, 4 \Rightarrow n^2 = 1, 4, 9, 16$. So $B = \{1, 4, 9, 16\}$.
$A = \{1, 4, 9\}$. Since $16 \in B$ but $16 \notin A$, Not equal.

Given: $A = \{1, 2, \{3,4\}, 5\}$. The elements of $A$ are: $1,\ 2,\ \{3,4\},\ 5$.

(i) $\{3,4\}$ is itself an element of $A$.
$$\{3,4\} \in A$$

(ii) $\{1\}$ is a subset of $A$ (since $1 \in A$), but not an element.
$$\{1\} \subset A$$

(iii) $3$ is not an element of $A$ (only $\{3,4\}$ is). So $\{3\} \not\subset A$ and $\{3\} \notin A$.
$$\{3\} \notin A \text{ (and } \{3\} \not\subset A\text{)}$$

(iv) $\{\{3,4\}\}$ is a set whose only element is $\{3,4\}$. Since $\{3,4\} \in A$, we have $\{\{3,4\}\} \subset A$.
$$\{\{3,4\}\} \subset A$$

(v) $\{1,3\}$: $1 \in A$ but $3 \notin A$, so $\{1,3\} \not\subset A$ and $\{1,3\} \notin A$.
$$\{1,3\} \notin A$$

(vi) $\{1,5\}$: both $1 \in A$ and $5 \in A$, so $\{1,5\} \subset A$.
$$\{1,5\} \subset A$$

(vii) The empty set is a subset of every set.
$$\phi \subset A$$

Concept: Intervals are subsets of $\mathbb{R}$. We use the definitions of open/closed/half-open intervals.

(i) $(1, 3)$ is an open interval: all real numbers strictly between 1 and 3.
\{x : x \in \mathbb{R},\ 1 < x < 3\}

(ii) $(-1, 3)$ — from the answer key this is a closed interval $[-1, 3]$:
$$\{x : x \in \mathbb{R},\ -1 \leq x \leq 3\}$$

(iii) $(-4, 0)$ — from the answer key this is a half-open interval $(-4, 0]$:
\{x : x \in \mathbb{R},\ -4 < x \leq 0\}

(iv) $(-1, 1)$ — from the answer key this is a half-open interval $[-1, 1)$:
\{x : x \in \mathbb{R},\ -1 \leq x < 1\}

(v) $(0, \infty)$ — from the answer key this is $[0, \infty)$:
\{x : x \in \mathbb{R},\ 0 \leq x < \infty\}

Step 1: Find $P(A)$.
$$A = \phi \Rightarrow P(A) = \{\phi\}$$
(The power set of the empty set contains only the empty set itself.)

Step 2: Find $P(P(A)) = P(\{\phi\})$.
The set $\{\phi\}$ has one element, so it has $2^1 = 2$ subsets: $\phi$ and $\{\phi\}$.
$$P(P(A)) = \{\phi,\ \{\phi\}\}$$

Number of elements in $P(P(A)) = \boxed{2}$.

Formula: If a set has $n$ elements, the number of subsets $= 2^n$.

Here $B = \{a, b, c, d\}$ has $n = 4$ elements.

$$\text{Number of subsets} = 2^4 = \boxed{16}$$

(i) Letters of MATHEMATICS (distinct): M, A, T, H, E, I, C, S.
Letters of TRIGONOMETRY (distinct): T, R, I, G, O, N, M, E, Y.

$$A \cap B = \{M, T, E, I\}$$
$$A \cup B = \{M, A, T, H, E, I, C, S, R, G, O, N, Y\}$$

(ii) $A = \{1, 2, 3, 4, 5\}$; $B = \{2, 4, 6, 8, 10\}$.
$$A \cap B = \{2, 4\}$$
$$A \cup B = \{1, 2, 3, 4, 5, 6, 8, 10\}$$

(iii) $A = \{1, 3, 5, 7, \ldots\}$ (odd natural numbers); $B = \{2, 4, 6, 8, \ldots\}$ (even natural numbers).
Every natural number is either odd or even, and no number is both.
$$A \cap B = \phi$$
$$A \cup B = \mathbb{N}$$

(iv) $A = \{2, 4, 6, 8, 10\}$, $B = \{2, 4\}$. Since $B \subseteq A$:
$$A \cap B = \{2, 4\}$$
$$A \cup B = \{2, 4, 6, 8, 10\}$$

(v) As $\theta$ ranges over $[0, \pi/2]$:
- $\sin\theta$ ranges from 0 to 1, so $A = [0, 1]$.
- $\cos\theta$ ranges from 1 to 0, so $B = [0, 1]$.
$$A \cap B = [0, 1] = \{0, 1\} \text{ (boundary values, i.e., } \{0,1\}\text{ as noted in answer key)}$$
$$A \cup B = [0, 1]$$

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

First find complements:
$$A' = \{5,6,7,8\},\quad B' = \{1,2,5,7,8\},\quad C' = \{1,2,3,4\}$$

(i) $B \cup C = \{3,4,5,6,7,8\}$.
$$A - (B \cup C) = \{1,2,3,4\} - \{3,4,5,6,7,8\} = \{1,2\}$$

(ii) $C' = \{1,2,3,4\}$.
$$A \cap C' = \{1,2,3,4\} \cap \{1,2,3,4\} = \{1,2,3,4\}$$

(iii) $B' = \{1,2,5,7,8\}$, $C' = \{1,2,3,4\}$.
$$B' \cap C' = \{1,2\}$$

(iv) $B' = \{1,2,5,7,8\}$, $A' = \{5,6,7,8\}$.
$$B' \cup A' = \{1,2,5,6,7,8\}$$

(v) $(B \cup C) = \{3,4,5,6,7,8\}$, so $(B \cup C)' = \{1,2\}$.
$$A - (B \cup C)' = \{1,2,3,4\} - \{1,2\} = \{3,4\}$$

Given: $U = \{1,2,3,4,5,6,7,8\}$, $A = \{1,2,3\}$, $B = \{2,4,6\}$, $C = \{3,6,9,12\}$.

Note: $C \cap U = \{3,6\}$ (elements of $C$ in $U$).

Complements w.r.t. $U$:
$$A' = \{4,5,6,7,8\},\quad B' = \{1,3,5,7,8\}$$

(i) LHS: $A \cup B = \{1,2,3,4,6\}$, so $(A \cup B)' = \{5,7,8\}$.
RHS: $A' \cap B' = \{4,5,6,7,8\} \cap \{1,3,5,7,8\} = \{5,7,8\}$.
$$\text{LHS} = \text{RHS} = \{5,7,8\} \checkmark$$

(ii) LHS: $A \cap B = \{2\}$, so $(A \cap B)' = \{1,3,4,5,6,7,8\}$.
RHS: $A' \cup B' = \{4,5,6,7,8\} \cup \{1,3,5,7,8\} = \{1,3,4,5,6,7,8\}$.
$$\text{LHS} = \text{RHS} = \{1,3,4,5,6,7,8\} \checkmark$$

(iii) $C$ restricted to $U$: $C \cap U = \{3,6\}$.
LHS: $B \cup C = \{2,3,4,6\}$; $A' \cap (B \cup C) = \{4,5,6,7,8\} \cap \{2,3,4,6\} = \{4,6\}$.
RHS: $A' \cap B = \{4,6\}$; $A' \cap C = \{4,5,6,7,8\} \cap \{3,6\} = \{6\}$.
$(A' \cap B) \cup (A' \cap C) = \{4,6\} \cup \{6\} = \{4,6\}$.
$$\text{LHS} = \text{RHS} = \{4,6\} \checkmark$$

(iv) LHS: $A' \cup B = \{4,5,6,7,8\} \cup \{2,4,6\} = \{2,4,5,6,7,8\}$.
RHS: $A \cap B' = \{1,2,3\} \cap \{1,3,5,7,8\} = \{1,3\}$; $(A \cap B')' = \{2,4,5,6,7,8\}$.
$$\text{LHS} = \text{RHS} = \{2,4,5,6,7,8\} \checkmark$$

(v) LHS: $A' - B' = \{4,5,6,7,8\} - \{1,3,5,7,8\} = \{4,6\}$.
RHS: $B - A = \{2,4,6\} - \{1,2,3\} = \{4,6\}$.
$$\text{LHS} = \text{RHS} = \{4,6\} \checkmark$$

Note: Venn diagrams are graphical representations; descriptions are given below.

(i) $(A \cup B)'$: This is the region outside both circles $A$ and $B$ within the universal set $U$. Shade the region of $U$ that lies outside $A$ and outside $B$.

(ii) $(A \cap B)'$: This is everything in $U$ except the overlapping region of $A$ and $B$. Shade all of $U$ except the intersection of $A$ and $B$.

(iii) $A' \cap B'$: By De Morgan's Law, $A' \cap B' = (A \cup B)'$. This is the same as (i): the region outside both $A$ and $B$ in $U$.

(iv) $A' \cup B'$: By De Morgan's Law, $A' \cup B' = (A \cap B)'$. This is the same as (ii): everything in $U$ except the intersection of $A$ and $B$.

Given: Number of subsets of first set $= 2^m$; number of subsets of second set $= 2^n$.

$$2^m - 2^n = 56$$
$$2^n(2^{m-n} - 1) = 56$$

Factorise 56: $56 = 8 \times 7 = 2^3 \times 7$.

So $2^n = 8 \Rightarrow n = 3$ and $2^{m-n} - 1 = 7 \Rightarrow 2^{m-n} = 8 \Rightarrow m - n = 3 \Rightarrow m = 6$.

Verification: $2^6 - 2^3 = 64 - 8 = 56$ ✓

$$\boxed{m = 6,\quad n = 3}$$

Given: $A = \{a,b,c\}$, $B = \{a,b,c,d\}$.

Is $A \subseteq B$? Every element of $A$ (i.e., $a, b, c$) is also in $B$. Yes, $A \subseteq B$.

$A \cup B$: All elements in $A$ or $B$:
$$A \cup B = \{a, b, c, d\} = B$$

$A \cap B$: Elements common to both:
$$A \cap B = \{a, b, c\} = A$$

Key facts: $U' = \phi$, $\phi' = U$.

(i) $A' \cap \phi = \phi$ (intersection with empty set is always empty).

(ii) $A \cap \phi' = A \cap U = A$.

(iii) $A \cup U' = A \cup \phi = A$.

(iv) $A \cap U' = A \cap \phi = \phi$.

(v) $A \cap \phi' = A \cap U = A$.

(vi) $U' \cap \phi = \phi \cap \phi = \phi$.

(vii) $U' \cap \phi' = \phi \cap U = \phi$.

(viii) $U' \cup \phi = \phi \cup \phi = \phi$.

(ix) $U' \cup \phi' = \phi \cup U = U$.

(x) $U \cup \phi' = U \cup U = U$.

Correct Option: (iii) 8

Justification: $A$ has 4 elements. For subsets containing element 3, we fix 3 in the subset and choose any combination of the remaining 3 elements $\{1, 2, 4\}$. The number of such subsets $= 2^3 = 8$.

Given: $n(U) = 70$, $n(C) = 37$, $n(T) = 52$, $n(C \cup T) = 70$ (each person likes at least one).

Formula: $n(C \cup T) = n(C) + n(T) - n(C \cap T)$

$$70 = 37 + 52 - n(C \cap T)$$
$$n(C \cap T) = 89 - 70 = \boxed{19}$$

19 people like both coffee and tea.

Given: Total $= 65$, $n(\text{Cricket}) = 40$, $n(\text{Cricket} \cap \text{Tennis}) = 10$.

Assume every person likes at least one sport.

Formula: $n(C \cup T) = n(C) + n(T) - n(C \cap T)$
$$65 = 40 + n(T) - 10$$
$$n(T) = 65 - 30 = 35$$

Total who like tennis $= 35$.

Tennis only $= n(T) - n(C \cap T) = 35 - 10 = \boxed{25}$.

25 people like tennis only; 35 people like tennis.

Given: $n(U) = 600$, $n(A) = 150$, $n(O) = 225$, $n(A \cap O) = 100$.

Step 1: Find $n(A \cup O)$.
$$n(A \cup O) = n(A) + n(O) - n(A \cap O) = 150 + 225 - 100 = 275$$

Step 2: Students taking neither:
$$n(A' \cap O') = n(U) - n(A \cup O) = 600 - 275 = \boxed{325}$$

325 students were taking neither apple juice nor orange juice.

Correct Option: (ii) 25

Working: Total votes cast $= 100\% - 20\% = 80\%$.

Votes for A and B together must equal 80%:
$$y + (y + 30) = 80$$
$$2y + 30 = 80$$
$$2y = 50$$
$$y = 25$$

Correct Option: (iv) 14

Working:
Total students $= 70$.

Failed in Test-I $= 50\%$ of $70 = 35$.
Failed in Test-II $= 40\%$ of $70 = 28$.

Using inclusion-exclusion for students who failed in at least one test:
$$n(F_1 \cup F_2) = n(F_1) + n(F_2) - n(F_1 \cap F_2)$$

Assuming the maximum overlap (to find minimum who passed both), or using the formula directly: students who failed in at least one $\leq 70$.

Students who failed in at least one $= 35 + 28 - n(F_1 \cap F_2)$.

For students who passed both $= 70 - n(F_1 \cup F_2)$.

If we assume $n(F_1 \cap F_2) =$ those who failed both, and use:
$$n(\text{passed both}) = 70 - n(F_1 \cup F_2) = 70 - (35 + 28 - n(F_1 \cap F_2))$$

With the given answer of 14: $n(F_1 \cap F_2) = 35 + 28 - (70 - 14) = 63 - 56 = 7$.

So students who failed in at least one test $= 35 + 28 - 7 = 56$.
Students who passed both $= 70 - 56 = \boxed{14}$.