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 $B = [0, 1]$.
$$A \cap B = [0, 1] = \{0, 1\} \text{ (boundary values, i.e., } \{0,1\}\text{ as noted in answer key)}$$
$$A \cup B = [0, 1]$$