1.Answer: $B \cap C = \{4\}$. So, $A \times (B \cap C) = \{(1, 4), (2, 4), (3, 4)\}$.
2.Answer: $A$ contains first elements: $\{x, y, z\}$. $B$ contains second elements: $\{1, 2\}$. Both meet the cardinality criteria.
3.Answer: $A = \{-1, 0, 1\}$. Remaining 7 elements: $(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)$.
4.Proof: Let $x \in A, y \in B$. Since $(x,y) \in A \times B$ and $A \times B \subseteq C \times D$, $(x,y) \in C \times D$. Thus $x \in C \implies A \subseteq C$, and $y \in D \implies B \subseteq D$.
5.Answer: $x^2 - 3x = -2 \implies x^2 - 3x + 2 = 0 \implies x=1, 2$. Also, $y^2 - 5y = -6 \implies y^2 - 5y + 6 = 0 \implies y=2, 3$.
6.Answer: $A = \{1, 2\}$, $B = \{0, 1\}$. $A \cup B = \{0, 1, 2\}$, $A \cap B = \{1\}$. Product = $\{(0,1), (1,1), (2,1)\}$.
7.Answer: $n(B) = 24 / 3 = 8$. Subsets of $B$ = $2^8 = 256$.
8.Proof: $(x,y) \in A \times (B \cup C) \iff x \in A \text{ and } (y \in B \text{ or } y \in C) \iff (x \in A \text{ and } y \in B) \text{ or } (x \in A \text{ and } y \in C) \iff (x,y) \in (A \times B) \cup (A \times C)$.
9.Answer: $\{(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2)\}$.
10.Answer: $A = \{x, y, z\}$ and $B = \{a, b\}$.
11.Answer: $3y = 12 - x \implies y = \frac{12-x}{3}$. For $x, y \in \mathbb{N}$, pairs are $(3,3), (6,2), (9,1)$. Domain = $\{3, 6, 9\}$, Range = $\{1, 2, 3\}$.
12.Answer: $y = x - 3 \implies R = \{(11, 8), (13, 10)\}$. Inverse $R^{-1} = \{(8, 11), (10, 13)\}$.
13.Answer: $y = 3x$. $R = \{(1, 3), (2, 6), (3, 9), (4, 12)\}$. Domain = $\{1, 2, 3, 4\}$, Co-domain = $A$, Range = $\{3, 6, 9, 12\}$.
14.Answer: $n(A \times B) = 3 \times 2 = 6$. Total relations = $2^6 = 64$. Non-empty relations = $64 - 1 = 63$.
15.Answer: Prime numbers < 10 are 2, 3, 5, 7. $R = \{(2, 8), (3, 27), (5, 125), (7, 343)\}$.
16.Answer: $x^2 \le 25 \implies x \in \{-5, -4, -3, 0, 3, 4, 5\}$ (since $y^2$ must be a perfect square). Domain = $\{0, \pm 3, \pm 4, \pm 5\}$.
17.Answer: Range is symmetric to domain. Range = $\{0, \pm 3, \pm 4, \pm 5\}$.
18.Answer: $R_1 \cup R_2 = \{(1, 2), (2, 3), (2, 2), (3, 4)\}$. $R_1 \cap R_2 = \emptyset$.
19.Answer: No. $1/2 \not\le (1/2)^2 = 1/4$.
20.Answer: $|x| \le 3 - |y| \implies |x| \le 3$. So, $x \in \{-3, -2, -1, 0, 1, 2, 3\}$. This is the Domain.
21.Proof: Reflexive: $a-a=0$ (divisible by 2). Symmetric: $2|a-b \implies 2|-(a-b) \implies 2|b-a$. Transitive: $2|a-b$ and $2|b-c \implies 2|(a-b+b-c) \implies 2|a-c$. Thus, an equivalence relation.
22.Proof: Similar to Q21. $|a-a|=0$ (even). $|a-b|=|b-a|$ (even). If $|a-b|, |b-c|$ are even, $a,b,c$ have same parity, so $|a-c|$ is even.
23.Answer: $|1 - b|$ is even. $b$ must be odd. Hence, $\{1, 3, 5\}$.
24.Proof: Not reflexive: $1/2 \not\le (1/2)^3$. Not symmetric: $1 \le 2^3$ but $2 \not\le 1^3$. Not transitive: $3 \le (3/2)^3$ and $3/2 \le (6/5)^3$, but $3 \not\le (6/5)^3 = 216/125 \approx 1.728$.
25.Proof: $L_1 \parallel L_1$ (Reflexive). $L_1 \parallel L_2 \implies L_2 \parallel L_1$ (Symmetric). $L_1 \parallel L_2, L_2 \parallel L_3 \implies L_1 \parallel L_3$ (Transitive).
26.Proof: $|a-a| = 0$ (multiple of 4). $|a-b|$ is multiple of 4 $\implies |b-a|$ is multiple of 4. $a-b=4k_1, b-c=4k_2 \implies a-c=4(k_1+k_2)$, so transitive.
27.Answer: $[1] = \{x \in A : |x - 1| \text{ is a multiple of } 4\} = \{1, 5, 9\}$.
28.Proof: Reflexive: $a+b=b+a$. Symmetric: $a+d=b+c \implies c+b=d+a \implies (c,d) S (a,b)$. Transitive: $a+d=b+c$ and $c+f=d+e \implies a+d+c+f = b+c+d+e \implies a+f = b+e$.
29.Answer: Not reflexive ($1 \neq 1+1$). Not symmetric ($(1,2) \in R$ but $(2,1) \notin R$). Not transitive ($(1,2) \in R, (2,3) \in R \implies 3=1+1$ False, so $(1,3) \notin R$).
30.Answer: Let $A=\{1, 2, 3\}$. $R = \{(1, 1), (1, 2), (2, 1), (2, 2)\}$. Symmetric and transitive, but $(3,3) \notin R$, so not reflexive.
31.Proof: Not 1-1: $f(1)=f(-1)=1$. Not Onto: $y=-2$ has no real pre-image since $x^2 \ge 0$.
32.Proof: 1-1: $\frac{x_1-2}{x_1-3} = \frac{x_2-2}{x_2-3} \implies x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 2x_1 - 3x_2 + 6 \implies x_1 = x_2$. Onto: $y = \frac{x-2}{x-3} \implies x = \frac{3y-2}{y-1} \in \mathbb{R}-\{3\}$ for $y \neq 1$. Yes, bijective.
33.Proof: Not 1-1: $f(2) = 1$ and $f(5) = 1$. Not Onto: Range is $\{-1, 0, 1\}$, which is not equal to $\mathbb{R}$ (e.g., $y=2$ has no pre-image).
34.Answer: Injective: $x_1^3 = x_2^3 \implies x_1 = x_2$ in $\mathbb{Z}$. Not Surjective: $y=2$ implies $x = 2^{1/3} \notin \mathbb{Z}$.
35.Proof: One-one: Let $f(x_1) = f(x_2)$. If both odd, $x_1+1=x_2+1 \implies x_1=x_2$. If both even, $x_1-1=x_2-1 \implies x_1=x_2$. (Odd and even cannot map to same). Onto: Every even $y$ comes from odd $y-1$, every odd $y$ comes from even $y+1$.
36.Proof: $\sin x, \cos x$ are strictly monotonic in $[0, \pi/2]$, thus 1-1. But $(f+g)(0) = 0+1=1$ and $(f+g)(\pi/2) = 1+0=1$. Hence $f+g$ is many-one.
37.Answer: Total onto functions = $3^4 - {^3C_1}(2)^4 + {^3C_2}(1)^4 = 81 - 3(16) + 3 = 36$.
38.Answer: Number of bijections = $n! = 3! = 6$.
39.Proof: Not 1-1: $f(1)=f(2)=1$. Onto: For any $y \in \mathbb{N}$, $x = y+1 \ge 2$. If $y=1$, pre-image is 1 or 2. If $y>1$, pre-image is $y+1 > 2$. Range = $\mathbb{N}$.
40.Proof: 1-1: $3-4x_1 = 3-4x_2 \implies x_1 = x_2$. Onto: $y = 3-4x \implies x = \frac{3-y}{4} \in \mathbb{R}$. Yes, it is a bijection.
41.Answer: $(f \circ g)(x) = f\left(\frac{x}{x^2+1}\right) = 3\left(\frac{x}{x^2+1}\right)^2 - 5 = \frac{3x^2}{(x^2+1)^2} - 5$.
42.Proof: $f(f(x)) = \frac{4(\frac{4x+3}{6x-4})+3}{6(\frac{4x+3}{6x-4})-4} = \frac{16x+12+18x-12}{24x+18-24x+16} = \frac{34x}{34} = x$. Since $f \circ f = I$, $f^{-1} = f$.
43.Answer: $f(g(x)) = e^{\log_e x} = x$. $g(f(x)) = \log_e(e^x) = x \log_e e = x$. Yes, they are equal.
44.Answer: Let $y = \frac{2x-7}{4} \implies 4y = 2x - 7 \implies x = \frac{4y+7}{2}$. So, $f^{-1}(x) = \frac{4x+7}{2}$.
45.Proof: $y = (3x)^2 + 2(3x)(1) + 1 - 6 = (3x+1)^2 - 6 \implies (3x+1)^2 = y+6$. Since $x \ge 0$, $3x+1 = \sqrt{y+6} \implies x = \frac{\sqrt{y+6}-1}{3}$. Hence invertible.
46.Proof: $(f \circ f)(x) = \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1 + \frac{x^2}{1+x^2}}} = \frac{x}{\sqrt{1+2x^2}}$. By pattern, $(f \circ f \circ f)(x) = \frac{x}{\sqrt{1+3x^2}}$.
47.Answer: $f(-2) = |-2| = 2$. Then $g(2) = |2-1| = 1$. $(g \circ f)(-2) = 1$.
48.Answer: No. $f \circ g$ requires Range($g$) $\subseteq$ Domain($f$). Let $f(x) = \sqrt{x}$ (Domain $[0, \infty)$) and $g(x) = -x^2 - 1$. Range($g$) is $(-\infty, -1]$. $f \circ g$ is undefined in reals.
49.Answer: No. $f(2) = 1$ and $f(3) = 1$. It is a many-one function, therefore not bijective, so it has no inverse.
50.Answer: $g(7) = 7-7 = 0$. Then $f(0) = 0+7 = 7$.
51.Answer: Commutative: $\frac{ab}{4} = \frac{ba}{4}$. Associative: $(a*b)*c = (\frac{ab}{4})*c = \frac{abc}{16}$. $a*(b*c) = a*(\frac{bc}{4}) = \frac{abc}{16}$. Yes to both.
52.Answer: $a * e = a \implies \frac{ae}{4} = a \implies e = 4$.
53.Answer: $a * b = e \implies \frac{ab}{4} = 4 \implies ab = 16 \implies b = \frac{16}{a}$. Inverse is $16/a$.
54.Answer: $a * e = a \implies a + e - ae = a \implies e(1-a) = 0$. Since $a \neq 1$, $e = 0$.
55.Answer: $a * b = 0 \implies a + b - ab = 0 \implies b(1-a) = -a \implies b = \frac{a}{a-1}$. All elements are invertible.
56.Answer: For $e$ to be identity, $\text{HCF}(a, e) = a$ for all $a \in \mathbb{N}$. This requires $e$ to be a multiple of every natural number, which is impossible. No identity element exists.
57.Proof: For $a-b$, $a-e = a \implies e=0$. But $e-a = 0-a = -a \neq a$. Not commutative, no identity. For $a/b$, $a/e = a \implies e=1$. But $1/a \neq a$. No identity.
58.Answer: Table matrix: Row 1: 1,1,1,1,1. Row 2: 1,2,2,2,2. Row 3: 1,2,3,3,3. Row 4: 1,2,3,4,4. Row 5: 1,2,3,4,5.
59.Answer: Identity is 0. Inverse of 4 is $b$ such that $(4+b) \pmod 6 = 0$. Hence $b = 2$.
60.Proof: $(c,d) * (a,b) = (c+a, d+b)$. Since addition in $\mathbb{N}$ is commutative, $(c+a, d+b) = (a+c, b+d) = (a,b)*(c,d)$. Commutative.
