1.Answer: Not an equivalence relation. It is reflexive ($1+x^2>0$) and symmetric, but not transitive. Counterexample: Let $x = 1, y = -1/2, z = -4$. $1+xy = 1/2 > 0$ and $1+yz = 3 > 0$, but $1+xz = -3 \not> 0$.
2.Answer: Number of partitions of a 3-element set (Bell Number $B_3$). Partitions: 3 blocks (1), 2 blocks (3), 1 block (1). Total = 5.
3.Answer: $R$ lacks $(3,2), (3,1), (2,1)$ to be symmetric. To make it equivalence with minimal removal, we must remove pairs destroying symmetry/transitivity. Minimum to remove is 2 pairs: $(1,2), (2,3)$ or similar to isolate elements. Actually, removing $(1,2), (2,3), (1,3)$ yields identity relation. Minimum removal = 3 ordered pairs.
4.Proof: Reflexive: $A \subseteq A$. Anti-Symmetric: $A \subseteq B, B \subseteq A \implies A=B$. Transitive: $A \subseteq B, B \subseteq C \implies A \subseteq C$. It is a partial order. Not symmetric because $A \subseteq B \not\implies B \subseteq A$, hence not equivalence.
5.Proof: Rewrite condition: $\frac{b+c}{bc} = \frac{a+d}{ad} \implies \frac{1}{c} + \frac{1}{b} = \frac{1}{d} + \frac{1}{a} \implies \frac{1}{b} - \frac{1}{a} = \frac{1}{d} - \frac{1}{c}$. This shows $f(x,y) = \frac{1}{y} - \frac{1}{x}$. Since it's equating functional values, it's an equivalence relation.
6.Answer: Factor: $(a-b)(a-6b) = 0 \implies a=b$ or $a=6b$. Reflexive (since $a=a$ satisfies). Not symmetric ($6 R 1$ holds since $6=6(1)$, but $1 R 6$ false). Not transitive ($36 R 6, 6 R 1$, but $36 \neq 1$ and $36 \neq 6(1)$).
7.Proof: Reflexive: $(a,a) \in R_1, (a,a) \in R_2 \implies (a,a) \in R_1 \cap R_2$. Symmetric: $(a,b) \in R_1 \cap R_2 \implies (b,a) \in R_1, R_2 \implies (b,a) \in R_1 \cap R_2$. Transitive similarly holds.
8.Answer: No. Let $A = \{1, 2, 3\}$. $R_1 = \{(1,1), (2,2), (3,3), (1,2), (2,1)\}$ and $R_2 = \{(1,1), (2,2), (3,3), (2,3), (3,2)\}$. $R_1 \cup R_2$ has $(1,2)$ and $(2,3)$ but not $(1,3)$, so it is not transitive.
9.Answer: Not 1-1: $f(2) = f(-2) = 3/5$. Not Onto: $y = 1 - \frac{2}{x^2+1}$. Since $x^2 \ge 0$, $\frac{2}{x^2+1} \in (0, 2]$. Hence $y \in [-1, 1)$. Range is $[-1, 1) \neq \mathbb{R}$.
10.Proof: $f'(x) = 1 + \cos x \ge 0$. It is strictly increasing except at isolated points where $f'=0$, hence injective. As $x \to \pm \infty$, $f(x) \to \pm \infty$, continuous, hence surjective. Bijective.
11.Answer: Not 1-1 because $f(-1) = e^1 = e$ and $f(1) = e^1 = e$. However, it is Onto because range of $|x|$ is $[0, \infty)$, so $e^{|x|}$ yields $[1, \infty)$. Not bijective.
12.Answer: 1-1: $\frac{x_1-1}{x_1-2} = \frac{x_2-1}{x_2-2} \implies x_1 = x_2$. Onto: $y = \frac{x-1}{x-2} \implies x = \frac{2y-1}{y-1} \in A$ for $y \neq 1$. Bijective. $f^{-1}(x) = \frac{2x-1}{x-1}$.
13.Answer: 1-1: $\frac{x_1}{1+x_1} = \frac{x_2}{1+x_2} \implies x_1 = x_2$. Not onto: Range is $[0, 1)$, since $\frac{x}{1+x} < 1$ for all $x \in [0, \infty)$. Co-domain is $[0, \infty)$.
14.Answer: Not 1-1: $f(1) = 0$ and $f(3) = 0$. Onto: For any $y \in \mathbb{Z}$, let $n = 2y$ (even), then $f(2y) = 2y/2 = y$. Thus, onto.
15.Answer: Formula: $3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5 = 243 - 3(32) + 3(1) = 243 - 96 + 3 = 150$.
16.Answer: For square root: $\log_{0.5}(x^2 - 5x + 7) \ge 0 \implies x^2 - 5x + 7 \le 1 \implies x^2 - 5x + 6 \le 0 \implies (x-2)(x-3) \le 0$. Domain = $[2, 3]$. Note: $x^2-5x+7 > 0$ is naturally satisfied here.
17.Answer: $f(-5/3) = [-1.66] = -2 \implies g(-2) = 2$. Next, $g(-5/3) = 5/3 = 1.66 \implies f(1.66) = 1$. Difference: $2 - 1 = 1$.
18.Answer: $(f \circ g)(x) = 2(x^2+2)-1 = 2x^2+3$. $(g \circ f)(x) = (2x-1)^2+2 = 4x^2-4x+3$. Equate: $2x^2+3 = 4x^2-4x+3 \implies 2x^2-4x = 0 \implies x=0, 2$.
19.Answer: $x - [x] > 0 \implies \{x\} > 0$. Fractional part is always $\ge 0$, and $>0$ when $x$ is not an integer. Domain: $\mathbb{R} - \mathbb{Z}$.
20.Answer: $f(x) = \frac{e^{2x}-1}{e^{2x}+1} = 1 - \frac{2}{e^{2x}+1}$. As $x \to \infty, f(x) \to 1$. As $x \to -\infty, f(x) \to -1$. Since $e^{2x}>0$, Range = $(-1, 1)$.
21.Answer: $f(cx+d) = a(cx+d)+b = acx + ad + b$. $g(ax+b) = c(ax+b)+d = acx + bc + d$. Equating constants: $ad+b = bc+d \implies d(a-1) = b(c-1)$.
22.Answer: By induction: $f_2(x) = \frac{x}{\sqrt{1+2x^2}}$, $f_3(x) = \frac{x}{\sqrt{1+3x^2}}$. Hence $f_n(x) = \frac{x}{\sqrt{1+nx^2}}$.
23.Proof: Set $x=y=0 \implies f(0) = f(0)+f(0) \implies f(0)=0$. Set $y=-x \implies f(0) = f(x)+f(-x) \implies 0 = f(x)+f(-x) \implies f(-x) = -f(x)$.
24.Answer: $y = \log_a(x + \sqrt{x^2+1}) \implies a^y = x + \sqrt{x^2+1}$. Also $-y = \log_a(\sqrt{x^2+1} - x) \implies a^{-y} = \sqrt{x^2+1} - x$. Subtracting gives $a^y - a^{-y} = 2x \implies f^{-1}(y) = \frac{a^y - a^{-y}}{2}$.
25.Answer: Not 1-1. For $x \in [1,2]$, $f(x) = (x-1) - (x-2) = 1$. The function is constant on $[1,2]$, meaning multiple inputs yield $1$. Not bijective.
26.Answer: $\sin^{-1}$ requires $-1 \le \frac{2x-3}{5} \le 1 \implies -5 \le 2x-3 \le 5 \implies -1 \le x \le 4$. $\log$ requires $4-x^2 > 0 \implies x \in (-2, 2)$. Intersection: $[-1, 4] \cap (-2, 2) = [-1, 2)$.
27.Answer: Reflexivity fixes $n$ diagonal pairs. For remaining $n(n-1)/2 = 6$ symmetric pairs, each can either be in relation or not. $2^6 = 64$.
28.Answer: $pq = 60$. Factors: $(1,60), (2,30), (3,20), (4,15), (5,12), (6,10)$. Max sum $p+q = 1+60 = 61$. Min sum $p+q = 6+10 = 16$.
29.Answer: $f(x) = \frac{(x-1)(x-2)}{(x+3)(x-2)}$. Domain is $\mathbb{R} - \{-3, 2\}$. Reduced $f(x) = \frac{x-1}{x+3}$. As $x \to \pm \infty, y \to 1$. Hole at $x=2 \implies y = \frac{2-1}{2+3} = 1/5$. Range is $\mathbb{R} - \{1, 1/5\}$.
30.Answer: Eq 1: $2f(x)+3f(-x)=15-4x$. Substitute $-x$: Eq 2: $2f(-x)+3f(x)=15+4x \implies f(-x) = \frac{15+4x-3f(x)}{2}$. Sub into Eq 1: $2f(x) + 3\left(\frac{15+4x-3f(x)}{2}\right) = 15-4x$. Multiply by 2: $4f(x) + 45 + 12x - 9f(x) = 30 - 8x \implies -5f(x) = -15 - 20x \implies f(x) = 4x + 3$.
31.Answer: Not reflexive ($|-2| \not\le -2$). Not symmetric ($|-1| \le 2$ but $|2| \not\le -1$). Transitive ($|x| \le y \implies y \ge 0 \implies |y|=y$. $|y| \le z \implies |x| \le |y| \le z \implies |x| \le z$).
32.Answer: No. $f(0) = 0$, $f(1) = 0$, $f(-1) = 0$. Many-to-one function.
33.Answer: Domain of $g$ is $\mathbb{R}$. Range of $g$ is $[-1, \infty)$. Domain of $f$ is $\mathbb{R}$. So $f \circ g$ is defined for all $\mathbb{R}$. Domain = $\mathbb{R}$.
34.Answer: Commutative: $a*b = a+b-ab = b+a-ba = b*a$. Associative: $(a*b)*c = (a+b-ab)+c-(a+b-ab)c = a+b+c-ab-ac-bc+abc$. $a*(b*c)$ gives same. Yes to both.
35.Answer: $a * e = a \implies \frac{ae}{3} = a \implies e = 3$.
36.Answer: $x * x^{-1} = 3 \implies \frac{x \cdot x^{-1}}{3} = 3 \implies x^{-1} = \frac{9}{x}$.
37.Proof: Commutative is obvious. $(a*b)*c = (a+b+ab)+c+(a+b+ab)c = a+b+c+ab+bc+ac+abc$. Same for $a*(b*c)$. Note $a*b = (a+1)(b+1)-1$, hence closed in $\mathbb{R}-\{-1\}$.
38.Answer: $a * e = a \implies a+e+ae = a \implies e(1+a) = 0 \implies e = 0$.
39.Answer: $2 * x = 0 \implies 2+x+2x = 0 \implies 3x = -2 \implies x = -2/3$.
40.Proof: A binary operation is a function from $A \times A \to A$. Domain has $n \times n = n^2$ elements. Co-domain has $n$ elements. Total functions = $n^{n^2}$.
41.Answer: Number of commutative operations is $n^{\frac{n(n+1)}{2}}$. For $n=3$, it is $3^{3(4)/2} = 3^6 = 729$.
42.Proof: $(f \circ g)(1) = f(g(1)) = f(3) = 1$. $(f \circ g)(2) = f(g(2)) = f(1) = 2$. $(f \circ g)(3) = f(g(3)) = f(2) = 3$. Hence $f \circ g = \{(1,1), (2,2), (3,3)\} = I_A$.
43.Answer: Let $y = 10^{x+1} \implies \log_{10} y = x+1 \implies x = \log_{10} y - 1$. So, $f^{-1}(x) = \log_{10} x - 1$.
44.Answer: For $x \ge 1$, $x(x-1)$ is strictly increasing from $0 \to \infty$. Thus $f(x)$ is strictly increasing $1 \to \infty$, hence bijective. $y = 2^{x^2-x} \implies x^2-x-\log_2 y = 0 \implies x = \frac{1 + \sqrt{1+4\log_2 y}}{2}$ (taking positive root for $x \ge 1$).
45.Proof: For $x \ge 0, f(x) = \frac{x}{1+x} = 1 - \frac{1}{1+x}$ (increasing). For $x < 0, f(x) = \frac{x}{1-x} = -1 + \frac{1}{1-x}$ (increasing). Continuous at 0. Hence strictly increasing on $\mathbb{R}$.
46.Answer: $f(x) = x^3 \implies \max(x, x^3) = x^3 \implies x^3 \ge x \implies x(x-1)(x+1) \ge 0 \implies x \in [-1, 0] \cup [1, \infty)$.
47.Answer: This is equivalent to choosing 3 distinct numbers from $\{1, 2, 3, 4, 5\}$ and arranging them in ascending order (which can be done in 1 way). Total = $\binom{5}{3} = 10$.
48.Answer: Reflexive: $x-x = 0 \in \mathbb{Z}$. Symmetric: $x-y \in \mathbb{Z} \implies -(x-y) \in \mathbb{Z} \implies y-x \in \mathbb{Z}$. Transitive: $(x-y) \in \mathbb{Z}, (y-z) \in \mathbb{Z} \implies (x-y)+(y-z) = x-z \in \mathbb{Z}$. Yes, it is.
49.Answer: $y = \frac{2^{2x}-1}{2^{2x}+1} \implies y \cdot 2^{2x} + y = 2^{2x} - 1 \implies 2^{2x}(1-y) = 1+y \implies 2^{2x} = \frac{1+y}{1-y} \implies x = \frac{1}{2} \log_2 \left( \frac{1+y}{1-y} \right)$.
50.Proof: Let $z \in C$. Since $g$ is onto, $\exists y \in B$ s.t. $g(y) = z$. Since $f$ is onto, $\exists x \in A$ s.t. $f(x) = y$. Then $g(f(x)) = g(y) = z$. Hence $g \circ f$ is onto.
51.Answer: $f(-\sqrt{\pi/2}) = (-\sqrt{\pi/2})^2 = \pi/2$. Then $g(\pi/2) = \sin(\pi/2) = 1$.
52.Answer: $f(x+y) - \frac{(x+y)^2}{2} = f(x) - \frac{x^2}{2} + f(y) - \frac{y^2}{2}$. Let $g(x) = f(x) - \frac{x^2}{2} \implies g(x+y) = g(x)+g(y) \implies g(x) = cx$. $f(x) = cx + \frac{x^2}{2}$. $f(1) = c+1/2 = 1 \implies c=1/2$. $f(x) = \frac{x^2+x}{2}$.
53.Proof: $f(x)+f(y) = \log\left(\frac{1+x}{1-x}\right) + \log\left(\frac{1+y}{1-y}\right) = \log\left(\frac{(1+x)(1+y)}{(1-x)(1-y)}\right) = \log\left(\frac{1+xy+x+y}{1+xy-(x+y)}\right)$. Divide num/den by $1+xy$: $\log\left(\frac{1+\frac{x+y}{1+xy}}{1-\frac{x+y}{1+xy}}\right) = f\left(\frac{x+y}{1+xy}\right)$.
54.Answer: $f'(x) = 1 - \frac{1}{x^2}$. $f'(x) = 0$ at $x=1$. Decreasing on $(0, 1)$, increasing on $(1, \infty)$. Therefore, many-one (not injective). E.g., $f(2) = f(1/2) = 2.5$.
55.Answer: Equivalence classes form a partition. Partitions of $\{1,2,3\}$ into 2 blocks: $\{\{1,2\}, \{3\}\}, \{\{1,3\}, \{2\}\}, \{\{2,3\}, \{1\}\}$. There are 3 such relations (Stirling Number of Second Kind $S(3,2)$).
56.Answer: $(x,y) \in R \circ R \implies \exists z$ s.t. $(x,z) \in R$ and $(z,y) \in R \implies z=2x, y=2z \implies y = 4x$. $R \circ R = \{(x,y): y=4x\}$.
57.Answer: $f(f(x)) = \sqrt{a^2 - (\sqrt{a^2-x^2})^2} = \sqrt{a^2 - (a^2-x^2)} = \sqrt{x^2} = |x|$. Domain: $a^2-x^2 \ge 0 \implies x \in [-|a|, |a|]$.
58.Answer: Standard result: $f(x) = \pm x^n + 1$. $f(3) = 28 \implies \pm 3^n + 1 = 28 \implies 3^n = 27 \implies n=3$ (positive). $f(x) = x^3 + 1$. $f(4) = 4^3 + 1 = 65$.
59.Answer: Suppose periodic with period $T>0$. $\cos((x+T)^2) = \cos(x^2)$ for all $x$. Differentiate: $-2(x+T)\sin((x+T)^2) = -2x\sin(x^2)$. Put $x=0 \implies -2T\sin(T^2) = 0 \implies T^2 = n\pi$. Put $x=\sqrt{\pi} \implies$ fails for constant $T$. Not periodic.
60.Answer: Number of onto functions from set of 5 to set of 3: $3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5 = 243 - 3(32) + 3 = 150$.
