1.Answer: $A \times B = \{(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)\}$
2.Answer: $P \times P \times P = \{(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2)\}$
3.Answer: $x/3 + 1 = 5/3 \implies x/3 = 2/3 \implies x = 2$. And $y - 2/3 = 1/3 \implies y = 1$.
4.Answer: $A$ contains first elements: $\{p, m\}$. $B$ contains second elements: $\{q, r\}$.
5.Answer: $n(A \times B) = 2 \times 2 = 4$. Number of subsets = $2^4 = 16$.
6.Answer: $n(B \times A) = n(B) \times n(A) = 5 \times 4 = 20$.
7.Answer: True. Since there are no elements to pair, the Cartesian product is empty.
8.Proof: Let $(x,y) \in A \times C \implies x \in A, y \in C$. Since $A \subseteq B$, $x \in B$. Hence $(x,y) \in B \times C$.
10.Answer: $n(B) = 15 / 3 = 5$.
11.Answer: $R = \{(1, 2), (2, 3), (3, 4)\}$. Note: $x=4$ gives $y=5 \notin A$.
12.Answer: Domain = $\{1, 2, 3\}$. Range = $\{2, 3, 4\}$.
13.Answer: $A \times A = \{(a, a), (a, b), (b, a), (b, b)\}$.
14.Answer: Total elements in $A \times B = 3 \times 2 = 6$. Total relations = $2^6 = 64$.
15.Answer: If $y=1, x=6$. If $y=2, x=4$. If $y=3, x=2$. $R = \{(6, 1), (4, 2), (2, 3)\}$.
16.Answer: $x^2 \le 9 \implies x \in \{1, 2, 3\}$. Domain = $\{1, 2, 3\}$.
17.Answer: Yes, because $2, 3, 4$ are all natural numbers.
18.Answer: $R^{-1} = \{(2, 1), (4, 3), (6, 5)\}$.
19.Answer: The range of $R^{-1}$ is equal to the domain of $R$, which is $\{a, b, c\}$.
20.Answer: For any integer $a$, $a-b$ is an integer for all integers $b$. Domain = $\mathbb{Z}$, Range = $\mathbb{Z}$.
21.Answer: Yes, because $(1,1), (2,2), (3,3)$ are all in $R$.
22.Answer: No. $(1,2) \in R$ but $(2,1) \notin R$.
23.Answer: No. $2 \le 3$ so $(2,3) \in R$, but $3 \not\le 2$ so $(3,2) \notin R$.
24.Answer: Yes. Reflexive ($a=a$), Symmetric ($a=b \implies b=a$), Transitive ($a=b, b=c \implies a=c$).
25.Answer: If $(a,b) \in R$ and $(b,a) \in R \implies a=b$. Example: Subset relation $A \subseteq B$.
26.Answer: No. If $L_1 \perp L_2$ and $L_2 \perp L_3$, then $L_1 \parallel L_3$, not perpendicular.
27.Answer: $R = \{(1, 2), (2, 1)\}$. It is symmetric, but $(1,1) \notin R$ (not reflexive) and $(1,2), (2,1) \in R \implies (1,1) \notin R$ (not transitive).
28.Answer: $(1,2) \in R$ and $(2,1) \in R$, but $(1,1) \notin R$. So, not transitive.
29.Answer: They partition the set $X$ into mutually disjoint subsets whose union is $X$.
30.Answer: $[1] = \{\dots, -5, -2, 1, 4, 7, \dots\}$ (Elements leaving remainder 1 when divided by 3).
31.Answer: Yes. Every element in the domain $\{2, 3, 4\}$ has a unique image.
32.Proof: One-one: $2x_1 = 2x_2 \implies x_1 = x_2$. Not Onto: $y=3 \implies x=1.5 \notin \mathbb{N}$.
33.Answer: No. $f(2) = 4$ and $f(-2) = 4$. Different inputs yield the same output.
34.Proof: $x_1^3 = x_2^3 \implies x_1 = x_2$ (one-one). Let $y \in \mathbb{R}$, $x = y^{1/3} \in \mathbb{R}$ (onto). Bijective.
35.Answer: One-one: Yes. Onto: No, because $7 \in B$ has no pre-image.
36.Answer: $f(x) = x^2$ mapping from $\mathbb{R} \to [0, \infty)$.
37.Answer: Total functions = $n(B)^{n(A)} = 2^3 = 8$.
38.Answer: ${^4P_3} = 4 \times 3 \times 2 = 24$.
39.Proof: Not one-one: $f(1.2) = 1$ and $f(1.5) = 1$. Not onto: Range is $\mathbb{Z} \neq \mathbb{R}$.
40.Answer: Not injective. $f(1) = 2$ and $f(-1) = 2$.
41.Answer: $f(g(x)) = \sin(x^2)$. $g(f(x)) = (\sin x)^2 = \sin^2 x$.
42.Answer: $g(2) = 2^2+1 = 5$. Then $f(5) = 2(5)+3 = 13$.
43.Answer: $f\left(\frac{x-1}{x+1}\right) = \frac{\frac{x-1}{x+1} - 1}{\frac{x-1}{x+1} + 1} = \frac{x-1-x-1}{x-1+x+1} = \frac{-2}{2x} = -\frac{1}{x}$.
44.Answer: The function must be a bijection (both one-one and onto).
45.Answer: $y = 3x - 4 \implies 3x = y + 4 \implies x = \frac{y+4}{3}$. Thus, $f^{-1}(x) = \frac{x+4}{3}$.
46.Answer: $y = \frac{2x+3}{x-1} \implies xy-y = 2x+3 \implies x(y-2) = y+3 \implies x = \frac{y+3}{y-2}$. $f^{-1}(x) = \frac{x+3}{x-2}$.
47.Proof: $f(f^{-1}(x)) = (x^{1/3})^3 = x$. Valid inverse.
48.Proof: $g(f(x_1)) = g(f(x_2)) \implies f(x_1) = f(x_2)$ (since $g$ is 1-1) $\implies x_1 = x_2$ (since $f$ is 1-1).
49.Answer: $g \circ f = \{(1, 3), (2, 1), (3, 2)\}$.
50.Answer: False. The reversal rule states $(f \circ g)^{-1} = g^{-1} \circ f^{-1}$.
51.Answer: No. $2 * 3 = -1$ but $3 * 2 = 1$.
52.Answer: $(a*b)*c = (ab/2)*c = abc/4$. $a*(b*c) = a*(bc/2) = abc/4$. Yes, associative.
53.Answer: $a * e = a \implies a + e - 5 = a \implies e = 5$. Identity is $5$.
54.Answer: $a * a^{-1} = e \implies a + a^{-1} - 5 = 5 \implies a^{-1} = 10 - a$.
55.Answer: $\text{LCM}(20, 16) = 80$.
56.Answer: Yes, because $\text{HCF}(a, b) = \text{HCF}(b, a)$.
57.Answer: Table with headers $\{1,2,3\}$. Rows: (1: 1,2,3), (2: 2,2,3), (3: 3,3,3).
58.Answer: $4+5 = 9$. Remainder of $9/6$ is $3$. Answer $= 3$.
59.Answer: $3 \times 4 = 12$. Remainder of $12/5$ is $2$. Answer $= 2$.
60.Answer: Yes. $a * b = 2^{ab} = 2^{ba} = b * a$.
