Domain and Range Questions in English

(A) The relation $R$ is defined on the set of natural numbers $N = \{1, 2, 3, \dots\}$ by the equation $x + 2y = 8$.
We can rewrite this as $x = 8 - 2y$.
Since $x \in N$,we must have $x > 0$,which implies $8 - 2y > 0$,so $2y < 8$,or $y < 4$.
Since $y \in N$,the possible values for $y$ are $1, 2, 3$.
Now,we find the corresponding values of $x$ for each $y$:
If $y = 1$,$x = 8 - 2(1) = 6$.
If $y = 2$,$x = 8 - 2(2) = 4$.
If $y = 3$,$x = 8 - 2(3) = 2$.
Thus,the relation $R$ is the set of ordered pairs: $R = \{(6, 1), (4, 2), (2, 3)\}$.
The domain of $R$ is the set of all first elements of the ordered pairs in $R$.
Therefore,the domain is $\{2, 4, 6\}$.

(C) The relation $R$ is defined as $x^2 + y^2 \le 4$,where $x, y \in Z$.
The domain is the set of all values of $x$ for which there exists at least one $y \in Z$ such that $x^2 + y^2 \le 4$.
If $x = 0$,then $0^2 + y^2 \le 4 \implies y^2 \le 4 \implies y \in \{-2, -1, 0, 1, 2\}$.
If $x = 1$,then $1^2 + y^2 \le 4 \implies y^2 \le 3 \implies y \in \{-1, 0, 1\}$.
If $x = -1$,then $(-1)^2 + y^2 \le 4 \implies y^2 \le 3 \implies y \in \{-1, 0, 1\}$.
If $x = 2$,then $2^2 + y^2 \le 4 \implies 4 + y^2 \le 4 \implies y^2 \le 0 \implies y = 0$.
If $x = -2$,then $(-2)^2 + y^2 \le 4 \implies 4 + y^2 \le 4 \implies y^2 \le 0 \implies y = 0$.
If $|x| > 2$,then $x^2 > 4$,which means no integer $y$ exists such that $x^2 + y^2 \le 4$.
Thus,the set of all possible values for the domain is $\{-2, -1, 0, 1, 2\}$.

(A) First,factorize the numerator and denominator:
$f(x) = \frac{e^x \ln x \cdot 5^{(x^2 + 2)}(x - 2)(x - 5)}{(2x - 3)(x - 4)}$
Note that the function is defined for $x > 0$ due to $\ln x$ and $x \neq \frac{3}{2}, x \neq 4$.
In the interval $\left( \frac{3}{2}, 4 \right)$,the function is continuous.
Evaluating the limits at the boundaries:
$\lim_{x \to (3/2)^+} f(x) = \frac{(+) \cdot \ln(1.5) \cdot (+) \cdot (1.5 - 2)(1.5 - 5)}{(+) \cdot (0^+)} = \frac{(+) \cdot (+) \cdot (+) \cdot (-) \cdot (-)}{(+) \cdot (0^+)} = \frac{+}{0^+} = \infty$
Wait,let us re-evaluate the sign:
At $x \to (3/2)^+$,the denominator $(2x-3)$ is $(0^+)$ and $(x-4)$ is negative.
Numerator: $e^x$ is $(+)$,$\ln x$ is $(+)$,$5^{(x^2+2)}$ is $(+)$,$(x-2)$ is negative,$(x-5)$ is negative.
So,$\frac{(+) \cdot (+) \cdot (+) \cdot (-) \cdot (-)}{(+) \cdot (-)} = \frac{+}{-} = -\infty$.
At $x \to 4^-$,the denominator $(2x-3)$ is $(+)$,$(x-4)$ is $(0^-)$.
Numerator: $e^x$ is $(+)$,$\ln x$ is $(+)$,$5^{(x^2+2)}$ is $(+)$,$(x-2)$ is $(+)$,$(x-5)$ is negative.
So,$\frac{(+) \cdot (+) \cdot (+) \cdot (+) \cdot (-)}{(+) \cdot (0^-)} = \frac{-}{-} = \infty$.
Since the function is continuous on $\left( \frac{3}{2}, 4 \right)$ and goes from $-\infty$ to $\infty$,the range is $(-\infty, \infty)$.

(C) For the function $f(x) = \frac{\cot^{-1} x}{\sqrt{x^2 - [x^2]}}$ to be defined,the denominator must be non-zero and the expression inside the square root must be positive.
$1$. The expression inside the square root is $x^2 - [x^2]$,which is equal to the fractional part of $x^2$,denoted as $\{x^2\}$.
$2$. The condition for the square root to be defined and non-zero is $\{x^2\} > 0$.
$3$. We know that for any real number $y$,$0 \le \{y\} < 1$. Thus,$\{x^2\} > 0$ implies $\{x^2\} \neq 0$.
$4$. The fractional part $\{x^2\} = 0$ if and only if $x^2$ is an integer,i.e.,$x^2 = n$ for some non-negative integer $n \in \{0, 1, 2, \dots\}$.
$5$. This means $x = \pm \sqrt{n}$ where $n \in \{0, 1, 2, \dots\}$.
$6$. Therefore,the domain is all real numbers except where $x^2$ is an integer,which is $R - \{\pm \sqrt{n} : n \in I^+ \cup \{0\}\}$.

(A) Let $\tan x = t$. Since $x \in [0, \frac{\pi}{2})$,$t \in [0, \infty)$.
$f(x) = \frac{t^n}{1 + t + t^2 + \dots + t^{2n}}$.
For $t=0$,$f(x) = 0$.
For $t > 0$,divide numerator and denominator by $t^n$:
$f(x) = \frac{1}{t^n + t^{n-1} + \dots + 1 + \dots + \frac{1}{t^{n-1}} + \frac{1}{t^n}}$.
Using the $AM$-$GM$ inequality,$t^k + \frac{1}{t^k} \ge 2$ for $t > 0$.
Thus,the denominator is $\ge 2 + 2 + \dots + 2 + 1 = 2n + 1$.
Therefore,$f(x) \le \frac{1}{2n+1}$.
The equality holds when $t=1$,i.e.,$x = \frac{\pi}{4}$.
Since $f(0) = 0$ and the maximum value is $\frac{1}{2n+1}$,the range is $[0, \frac{1}{2n+1}]$.
Since $n \in N$,$2n+1 \ge 3$,so $\frac{1}{2n+1} < 1$. The only integer in the range $[0, \frac{1}{2n+1}]$ is $0$. Thus,it contains exactly one integral point.

(C) For the function $f(x) = (1 + \frac{1}{x})^x$ to be defined,the base must be positive,i.e.,$1 + \frac{1}{x} > 0$.
Solving the inequality: $\frac{x+1}{x} > 0$.
Using the sign scheme method,we identify the critical points at $x = -1$ and $x = 0$.
The expression $\frac{x+1}{x}$ is positive in the intervals $(-\infty, -1)$ and $(0, \infty)$.
Thus,the domain of the function is $(-\infty, -1) \cup (0, \infty)$.

(A) Let $y = (1 + \frac{1}{x})^x$.
Taking the natural logarithm on both sides,we get $\ln(y) = x \ln(1 + \frac{1}{x})$.
As $x \to \infty$,$\ln(y) = x(\frac{1}{x} - \frac{1}{2x^2} + \dots) = 1 - \frac{1}{2x} + \dots \to 1$,so $y \to e$.
As $x \to -\infty$,$\ln(y) = x(\frac{1}{x} - \frac{1}{2x^2} + \dots) = 1 - \frac{1}{2x} + \dots \to 1$,so $y \to e$.
As $x \to 0^+$,$\ln(y) = x \ln(1 + \frac{1}{x}) \to 0 \cdot \infty$. Using $L$'Hopital's rule,$\lim_{x \to 0^+} \frac{\ln(1 + 1/x)}{1/x} = \lim_{t \to \infty} \frac{\ln(1+t)}{t} = 0$,so $y \to e^0 = 1$.
As $x \to -1^+$,$f(x) \to 0^0$ (undefined/approaching $0$).
By analyzing the derivative $f'(x) = f(x) [\ln(1 + \frac{1}{x}) - \frac{1}{x+1}]$,it can be shown that the function is strictly increasing on $(-\infty, -1)$ and $(0, \infty)$.
The range of the function is $(1, e)$.

(A) For $f(x) = \sqrt{\ln(2\lambda \cos x + 5)}$ to be defined for all $x \in R$, the expression inside the square root must be non-negative: $\ln(2\lambda \cos x + 5) \geq 0$.
This implies $2\lambda \cos x + 5 \geq e^0 = 1$, which simplifies to $2\lambda \cos x \geq -4$, or $\lambda \cos x \geq -2$.
Since this must hold for all $x \in R$, we consider the range of $\cos x$, which is $[-1, 1]$.
If $\lambda > 0$, the minimum value of $\lambda \cos x$ is $-\lambda$. Thus, $-\lambda \geq -2 \implies \lambda \leq 2$. So, $\lambda \in \{1, 2\}$.
If $\lambda < 0$, the minimum value of $\lambda \cos x$ is $\lambda$ (since $\cos x$ can be $1$). Thus, $\lambda \geq -2$. So, $\lambda \in \{-1, -2\}$.
If $\lambda = 0$, the expression becomes $\ln(5) \geq 0$, which is true.
Thus, the possible integral values for $\lambda$ are $\{-2, -1, 0, 1, 2\}$.
The total number of such values is $5$.

(C) Let $f(x) = \frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 2}$ and $g(x) = 3x^3$.
To find the number of real roots,we analyze the intersection of the graphs of $f(x)$ and $g(x)$.
The function $f(x)$ has vertical asymptotes at $x = 0$,$x = 1$,and $x = 2$.
By plotting the graph of $f(x)$ and the cubic function $g(x) = 3x^3$,we observe the points of intersection.
As shown in the graph,the curves intersect at $4$ distinct points.
Therefore,the equation has $4$ real roots,so $k = 4$.

(A) Given the function $f(x) = \frac{\sqrt{1 - x^2}}{1 + |x|}$.
Since the domain of $\sqrt{1 - x^2}$ is $x \in [-1, 1]$,we can substitute $x = \sin \theta$ where $\theta \in [-\pi/2, \pi/2]$.
Then $|x| = |\sin \theta|$ and $\sqrt{1 - x^2} = \sqrt{1 - \sin^2 \theta} = \sqrt{\cos^2 \theta} = |\cos \theta|$.
So,$f(x) = \frac{|\cos \theta|}{1 + |\sin \theta|}$.
Let $t = |\sin \theta|$,where $t \in [0, 1]$. Then $|\cos \theta| = \sqrt{1 - t^2}$.
Thus,$y = \frac{\sqrt{1 - t^2}}{1 + t} = \frac{\sqrt{(1 - t)(1 + t)}}{1 + t} = \sqrt{\frac{1 - t}{1 + t}}$.
As $t$ increases from $0$ to $1$,the expression $\frac{1 - t}{1 + t}$ decreases from $1$ to $0$.
Therefore,the range of $y = \sqrt{\frac{1 - t}{1 + t}}$ is $[0, 1]$.

(A) For the function $f(x) = \sqrt{\frac{4 - x^2}{[x] + 2}}$ to be defined,the expression inside the square root must be non-negative and the denominator must not be zero.
Case $1$: $\frac{4 - x^2}{[x] + 2} \geq 0$ and $[x] + 2 > 0$.
This implies $4 - x^2 \geq 0$ and $[x] > -2$.
$x^2 \leq 4 \Rightarrow x \in [-2, 2]$.
$[x] > -2 \Rightarrow x \geq -1$.
Intersection: $x \in [-1, 2]$.
Case $2$: $\frac{4 - x^2}{[x] + 2} \geq 0$ and $[x] + 2 < 0$.
This implies $4 - x^2 \leq 0$ and $[x] < -2$.
$x^2 \geq 4 \Rightarrow x \in (-\infty, -2] \cup [2, \infty)$.
$[x] < -2 \Rightarrow x < -2$.
Intersection: $x \in (-\infty, -2)$.
Combining both cases,the domain is $x \in (-\infty, -2) \cup [-1, 2]$.

(A) For the function $f(x)$ to be defined,the expression inside the square root must be non-negative and the argument of the logarithm must be positive.
$1$. Argument of logarithm must be positive:
$\frac{10x - 4}{4 - x^2} > 0 \Rightarrow \frac{2(5x - 2)}{(2 - x)(2 + x)} > 0 \Rightarrow \frac{5x - 2}{(x - 2)(x + 2)} < 0$
Using the wavy curve method,the solution for this inequality is $x \in \left( -\infty, -2 \right) \cup \left( 0.4, 2 \right)$.
$2$. Expression inside square root must be non-negative:
$\log_2\left(\frac{10x - 4}{4 - x^2}\right) - 1 \geq 0 \Rightarrow \log_2\left(\frac{10x - 4}{4 - x^2}\right) \geq 1$
$\Rightarrow \frac{10x - 4}{4 - x^2} \geq 2^1 \Rightarrow \frac{10x - 4}{4 - x^2} - 2 \geq 0$
$\Rightarrow \frac{10x - 4 - 2(4 - x^2)}{4 - x^2} \geq 0 \Rightarrow \frac{10x - 4 - 8 + 2x^2}{4 - x^2} \geq 0$
$\Rightarrow \frac{2x^2 + 10x - 12}{4 - x^2} \geq 0 \Rightarrow \frac{2(x^2 + 5x - 6)}{-(x^2 - 4)} \geq 0$
$\Rightarrow \frac{(x + 6)(x - 1)}{(x - 2)(x + 2)} \leq 0$
Using the wavy curve method for $\frac{(x + 6)(x - 1)}{(x - 2)(x + 2)} \leq 0$,the intervals are $\left[ -6, -2 \right) \cup \left[ 1, 2 \right)$.
Since this interval is a subset of the domain condition from step $1$,the final domain is $\left[ -6, -2 \right) \cup \left[ 1, 2 \right)$.

(C) Given $f(x) = \cos(\pi(|x| + 2[x]))$.
Since $\cos(2n\pi + \theta) = \cos(\theta)$ for any integer $n$,and $2[x]$ is an integer,we have:
$f(x) = \cos(2\pi[x] + \pi|x|) = \cos(\pi|x|)$.
Now,check for even/odd property:
$f(-x) = \cos(\pi|-x|) = \cos(\pi|x|) = f(x)$.
Since $f(-x) = f(x)$,the function is an even function.
Regarding periodicity,$f(x) = \cos(\pi|x|)$ is periodic with period $T = 2$.
Regarding the range,since $\cos(\theta)$ oscillates between $-1$ and $1$,the range of $f(x) = \cos(\pi|x|)$ is $[-1, 1]$.
Regarding $f(x) = |f(x)|$,this is only true when $f(x) \geq 0$,which is not true for all $x$ (e.g.,at $x = 1$,$f(1) = \cos(\pi) = -1$).
Thus,the correct statement is that the range of $f(x)$ is $[-1, 1]$.

(C) Given the interval $-\frac{\pi}{4} < x < \frac{\pi}{4}$.
Since $\pi \approx 3.14$,we have $-\frac{3.14}{4} < x < \frac{3.14}{4}$,which means $-0.785 < x < 0.785$.
The greatest integer function $[x]$ takes values based on the range of $x$:
For $-0.785 < x < 0$,$[x] = -1$.
For $0 \le x < 0.785$,$[x] = 0$.
Thus,the possible values for $[x]$ are $\{-1, 0\}$.
Now,substitute these values into $f(x) = \cos[x]$:
If $[x] = -1$,$f(x) = \cos(-1) = \cos 1$.
If $[x] = 0$,$f(x) = \cos 0 = 1$.
Therefore,the range of the function is $\{\cos 1, 1\}$.

(B) The function $f(x) = \log|5\{x\} - 2x|$ is defined when $5\{x\} - 2x \neq 0$.
We need to find the values of $x$ for which $5\{x\} - 2x = 0$.
Since $x = [x] + \{x\}$,we have $\{x\} = x - [x]$.
Substituting this into the equation: $5(x - [x]) - 2x = 0$.
$5x - 5[x] - 2x = 0 \Rightarrow 3x = 5[x] \Rightarrow [x] = \frac{3x}{5}$.
Since $[x]$ is an integer,$\frac{3x}{5}$ must be an integer. Let $\frac{3x}{5} = k$,where $k \in Z$. Then $x = \frac{5k}{3}$.
Also,we know that $k \le x < k+1$.
Substituting $x = \frac{5k}{3}$: $k \le \frac{5k}{3} < k+1$.
$3k \le 5k < 3k + 3$.
From $3k \le 5k$,we get $2k \ge 0 \Rightarrow k \ge 0$.
From $5k < 3k + 3$,we get $2k < 3 \Rightarrow k < 1.5$.
Since $k$ is an integer,$k$ can be $0$ or $1$.
If $k = 0$,$x = \frac{5(0)}{3} = 0$.
If $k = 1$,$x = \frac{5(1)}{3} = \frac{5}{3}$.
Thus,the set $A = \{0, \frac{5}{3}\}$.
The number of elements $n(A) = 2$.

(C) To find the range of $f(x) = \frac{(x + 1)^4}{x^4 + 1}$,we observe the following:
$1$. Since $(x+1)^4 \geq 0$ and $x^4 + 1 > 0$ for all $x \in R$,the minimum value of $f(x)$ is $0$ (at $x = -1$).
$2$. To find the maximum value,we use the power mean inequality or Jensen's inequality. For the convex function $g(t) = t^4$,by Jensen's inequality:
$\frac{g(x) + g(1)}{2} \geq g\left(\frac{x+1}{2}\right)$
$\frac{x^4 + 1}{2} \geq \left(\frac{x+1}{2}\right)^4 = \frac{(x+1)^4}{16}$
$3$. Rearranging the inequality:
$\frac{(x+1)^4}{x^4 + 1} \leq \frac{16}{2} = 8$
$4$. Thus,the range of the function is $[0, 8]$.

(C) The domain of $f(x)$ is determined by the conditions: $(x + 4)(1 - x) \ge 0$ and $x > 0$.
This implies $x \in [-4, 1]$ and $x > 0$,so the domain is $x \in (0, 1]$.
We have $f(x) = \sqrt{-x^2 - 3x + 4} + \log_{1/2} x$.
Since $\sqrt{-x^2 - 3x + 4}$ is a decreasing function on $(0, 1]$ and $\log_{1/2} x$ is also a decreasing function on $(0, 1]$,the sum $f(x)$ is a strictly decreasing function.
As $x \to 0^+$,$f(x) \to \infty$.
At $x = 1$,$f(1) = \sqrt{(1 + 4)(1 - 1)} - \log_2 1 = 0 - 0 = 0$.
Since the function is strictly decreasing on $(0, 1]$,the range of $f(x)$ is $[0, \infty)$.
The least integer in the range $[0, \infty)$ is $0$.

(B) $f(x) = \sum_{r=1}^n r + \sum_{r=1}^n [\cos(\frac{x}{r})]$.
Since $r$ is an integer,$[r + \cos(\frac{x}{r})] = r + [\cos(\frac{x}{r})]$.
$f(x) = \frac{n(n+1)}{2} + \sum_{r=1}^n [\cos(\frac{x}{r})]$.
For $x \in [0, \pi]$,$0 \le \frac{x}{r} \le \pi$. Thus,$\cos(\frac{x}{r}) \in [-1, 1]$.
If $x=0$,then $\cos(\frac{x}{r}) = 1$ for all $r$,so $[\cos(\frac{x}{r})] = 1$. Thus,$f(0) = \frac{n(n+1)}{2} + n = \frac{n^2+3n}{2}$.
If $x \in (0, \pi]$,then $\cos(\frac{x}{r}) < 1$,so $[\cos(\frac{x}{r})] \le 0$. Specifically,for $x \in (0, \pi]$,$\cos(\frac{x}{r}) \in [-1, 1)$.
If $x = \frac{\pi}{2}$,$\cos(\frac{\pi}{2r}) \ge 0$ for all $r \ge 1$,so $[\cos(\frac{\pi}{2r})] = 0$. Thus,$f(\frac{\pi}{2}) = \frac{n(n+1)}{2}$.
If $x = \pi$,$\cos(\frac{\pi}{r}) = -1$ for $r=1$ and $\cos(\frac{\pi}{r}) \in (-1, 1)$ for $r > 1$. Thus,$[\cos(\pi)] = -1$ and $[\cos(\frac{\pi}{r})] = 0$ for $r > 1$. Thus,$f(\pi) = \frac{n(n+1)}{2} - 1 = \frac{n^2+n-2}{2}$.
Therefore,the range is $\{\frac{n^2+n-2}{2}, \frac{n^2+n}{2}, \frac{n^2+3n}{2}\}$.

(C) We know that for any real $x$,the range of $\sin x$ is $[-1, 1]$.
So,$-1 \leq \sin x \leq 1$.
Multiplying by $-3$,we get $-3 \leq -3 \sin x \leq 3$.
Adding $2$ to all parts,we get $2 - 3 \leq 2 - 3 \sin x \leq 2 + 3$,which simplifies to $-1 \leq 2 - 3 \sin x \leq 5$.
Now,we consider the function $f(x) = \frac{1}{2 - 3 \sin x}$.
Since $2 - 3 \sin x$ takes values in $[-1, 5]$ excluding $0$ (where the function is undefined),we analyze the intervals.
When $2 - 3 \sin x \in [-1, 0)$,the value of $\frac{1}{2 - 3 \sin x}$ lies in $(-\infty, -1]$.
When $2 - 3 \sin x \in (0, 5]$,the value of $\frac{1}{2 - 3 \sin x}$ lies in $[1/5, \infty)$.
Thus,the range is $(-\infty, -1] \cup [1/5, \infty)$.

(D) Given $f(x) = 3 - x^2$ where $1 \le x \le 4$.
For the function $f(2x)$,we substitute $2x$ in place of $x$:
$f(2x) = 3 - (2x)^2 = 3 - 4x^2$.
The condition $1 \le x \le 4$ for $f(x)$ implies that for $f(2x)$,the argument $2x$ must satisfy $1 \le 2x \le 4$,which simplifies to $\frac{1}{2} \le x \le 2$.
For the logarithmic function $\log_e(f(2x))$ to be defined,the argument must be strictly positive:
$f(2x) > 0 \implies 3 - 4x^2 > 0$.
$4x^2 < 3 \implies x^2 < \frac{3}{4} \implies |x| < \frac{\sqrt{3}}{2}$.
Since $x$ must be positive from the domain constraint,we have $0 \le x < \frac{\sqrt{3}}{2}$.
Combining the conditions $\frac{1}{2} \le x \le 2$ and $x < \frac{\sqrt{3}}{2}$,we get the intersection:
$\frac{1}{2} \le x < \frac{\sqrt{3}}{2}$.
Thus,the domain is $[\frac{1}{2}, \frac{\sqrt{3}}{2})$.

(D) For the function $f(x) = \sqrt{\ln(m\sin x + 4)}$ to be defined for all $x \in R$,the expression inside the square root must be non-negative,i.e.,$\ln(m\sin x + 4) \ge 0$.
This implies $m\sin x + 4 \ge e^0$,which simplifies to $m\sin x + 4 \ge 1$.
Thus,$m\sin x \ge -3$.
Since the domain is $R$,this inequality must hold for all $x \in R$. The range of $\sin x$ is $[-1, 1]$.
If $m > 0$,the minimum value of $m\sin x$ is $-m$,so $-m \ge -3 \Rightarrow m \le 3$.
If $m < 0$,the minimum value of $m\sin x$ is $m$,so $m \ge -3$.
If $m = 0$,$0 \ge -3$,which is true.
Combining these,$m \in [-3, 3]$.
The integral values of $m$ are $\{-3, -2, -1, 0, 1, 2, 3\}$.
Counting these,there are $7$ possible integral values.

(C) Given $f(x) = \frac{2x^4-14x^2-8x+49}{x^4-7x^2-4x+23}$.
We can rewrite the numerator as:
$2(x^4-7x^2-4x+23) + 3 = 2x^4-14x^2-8x+46+3 = 2x^4-14x^2-8x+49$.
So,$f(x) = \frac{2(x^4-7x^2-4x+23) + 3}{x^4-7x^2-4x+23} = 2 + \frac{3}{x^4-7x^2-4x+23}$.
Let $h(x) = x^4-7x^2-4x+23$.
We can express $h(x)$ as:
$h(x) = (x^2-4)^2 + (x-2)^2 + 3$.
Since $(x^2-4)^2 \geq 0$ and $(x-2)^2 \geq 0$,the minimum value of $h(x)$ is $3$ (at $x=2$).
Thus,the range of $h(x)$ is $[3, \infty)$.
Now,$f(x) = 2 + \frac{3}{h(x)}$.
As $h(x) \in [3, \infty)$,$\frac{3}{h(x)} \in (0, 1]$.
Therefore,$f(x) \in (2+0, 2+1] = (2, 3]$.
Comparing $(2, 3]$ with $(a, b]$,we get $a=2$ and $b=3$.
Thus,$a+b = 2+3 = 5$.

(C) For the function $f(x)$ to be defined,the expression inside the square root must be strictly positive:
$(x + 1)(e^x - 1)(x - 4)(x + 5)(x - 6) > 0$
Identify the critical points by setting each factor to zero:
$x + 1 = 0 \implies x = -1$
$e^x - 1 = 0 \implies x = 0$
$x - 4 = 0 \implies x = 4$
$x + 5 = 0 \implies x = -5$
$x - 6 = 0 \implies x = 6$
The critical points are $\{-5, -1, 0, 4, 6\}$.
Arrange these points on the number line in increasing order: $-5, -1, 0, 4, 6$.
Using the wavy curve method (sign scheme):
For $x > 6$,the expression is positive.
For $4 < x < 6$,the expression is negative.
For $0 < x < 4$,the expression is positive.
For $-1 < x < 0$,the expression is negative.
For $-5 < x < -1$,the expression is positive.
For $x < -5$,the expression is negative.
Since we require the expression to be strictly greater than zero,the domain is the union of intervals where the expression is positive:
$x \in (-5, -1) \cup (0, 4) \cup (6, \infty)$.

(D) The given function is $y = \frac{|x-x^2|}{x^2-x}$.
First,we simplify the expression inside the absolute value: $x-x^2 = -x(x-1)$.
Thus,$|x-x^2| = |-x(x-1)| = |x| \cdot |x-1|$.
The denominator is $x^2-x = x(x-1)$.
So,$y = \frac{|x| \cdot |x-1|}{x(x-1)}$.
The function is defined for $x \neq 0$ and $x \neq 1$.
Case $1$: If $x > 1$,then $x > 0$ and $x-1 > 0$. Thus,$|x| = x$ and $|x-1| = x-1$.
$y = \frac{x(x-1)}{x(x-1)} = 1$.
Case $2$: If $0 < x < 1$,then $x > 0$ and $x-1 < 0$. Thus,$|x| = x$ and $|x-1| = -(x-1)$.
$y = \frac{x \cdot -(x-1)}{x(x-1)} = -1$.
Case $3$: If $x < 0$,then $x < 0$ and $x-1 < 0$. Thus,$|x| = -x$ and $|x-1| = -(x-1)$.
$y = \frac{-x \cdot -(x-1)}{x(x-1)} = \frac{x(x-1)}{x(x-1)} = 1$.
Combining these,the function is $y = 1$ for $x < 0$ or $x > 1$,and $y = -1$ for $0 < x < 1$. The points $x=0$ and $x=1$ are excluded from the domain.

(D) Let $f_n(x)$ be the function with $n$ logarithms.
For $n=1$,$f_1(x) = \log_{10}(x)$. The domain is $x > 0$,which is $(0, \infty) = (10^0, \infty)$.
For $n=2$,$f_2(x) = \log_{10}(\log_{10}(x))$. We require $\log_{10}(x) > 0$,which implies $x > 10^0 = 1$. Thus,the domain is $(1, \infty) = (10^1, \infty)$ is incorrect; let us re-evaluate.
Actually,for $n=1$,$\log(x) > 0 \implies x > 10^0 = 1$. Wait,the domain of $\log(x)$ is $x > 0$.
Let $L_1 = \log(x) > 0 \implies x > 10^0 = 1$.
Let $L_2 = \log(\log(x)) > 0 \implies \log(x) > 10^0 = 1 \implies x > 10^1 = 10$.
Let $L_3 = \log(\log(\log(x))) > 0 \implies \log(\log(x)) > 10^0 = 1 \implies \log(x) > 10^1 = 10 \implies x > 10^{10}$.
Following this pattern,for $n$ logarithms,the condition is $x > 10^{10^{...^{10}}}$ ($n-1$ times).
However,if the question implies the base is $10$ and we are looking for the domain of the composition,the standard result for $n$ nested logs is $x > 10^{10^{...}}$ ($n-1$ times).
Given the options,if $n=1$,domain is $(0, \infty)$. If $n=2$,domain is $(1, \infty)$. If $n=3$,domain is $(10, \infty)$.
This matches $(10^{n-2}, \infty)$ for $n \ge 2$.

(C) Let $f(x) = 2^{(x^2 - 3)^3+27}$.
To find the least value of $f(x)$,we need to find the least value of the exponent $g(x) = (x^2 - 3)^3 + 27$.
Since the function $h(u) = u^3$ is a strictly increasing function,the range of $(x^2 - 3)^3$ depends on the range of $x^2 - 3$.
We know that $x^2 \ge 0$ for all real $x$,so $x^2 - 3 \ge -3$.
Let $u = x^2 - 3$. Then $u \in [-3, \infty)$.
The function $g(u) = u^3 + 27$ is strictly increasing for $u \in [-3, \infty)$.
The minimum value of $g(u)$ occurs at the smallest value of $u$,which is $u = -3$.
Thus,$g_{min} = (-3)^3 + 27 = -27 + 27 = 0$.
Therefore,the least value of $f(x) = 2^{g(x)}$ is $2^0 = 1$.

(D) Given the function $f(x) = \sin x - \sqrt{3} \cos x + 1$.
We can rewrite this expression by multiplying and dividing by $2$:
$f(x) = 2 \left( \frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x \right) + 1$
Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we get:
$f(x) = 2 \sin \left( x - \frac{\pi}{3} \right) + 1$
We know that the range of $\sin \theta$ is $[-1, 1]$.
Therefore,the range of $2 \sin \left( x - \frac{\pi}{3} \right)$ is $[-2, 2]$.
Adding $1$ to the entire range,we get the range of $f(x)$ as $[-2 + 1, 2 + 1] = [-1, 3]$.
Since the function is onto,the codomain $S$ must be equal to the range of the function.
Thus,$S = [-1, 3]$.
Therefore,option $D$ is correct.

(D) For the function $f(x) = [\sin x] \cos \left( \frac{\pi}{[x - 1]} \right)$ to be defined,the denominator inside the cosine function must not be zero.
Thus,$[x - 1] \neq 0$.
By the definition of the Greatest Integer Function,$[x - 1] = 0$ when $0 \leq x - 1 < 1$.
Adding $1$ to all parts of the inequality,we get $1 \leq x < 2$.
Therefore,the function is undefined for $x \in [1, 2)$.
The domain of the function is all real numbers except the interval $[1, 2)$,which is written as $R - [1, 2)$.

(C) We know that the expression $\cos x + \sin x$ can be written as $\sqrt{2} \left( \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x \right) = \sqrt{2} \sin(x + \frac{\pi}{4})$.
Since the range of $\sin(x + \frac{\pi}{4})$ is $[-1, 1]$,the range of $\cos x + \sin x$ is $[-\sqrt{2}, \sqrt{2}]$.
Given $\sqrt{2} \approx 1.414$,the range of $\cos x + \sin x$ is approximately $[-1.414, 1.414]$.
The function $f(x) = [\cos x + \sin x]$ represents the Greatest Integer Function $(G.I.F.)$ of values in the interval $[-1.414, 1.414]$.
The integers contained in this interval are $\{-1, 0, 1\}$.
Therefore,the range of $f(x)$ is $\{-1, 0, 1\}$.

(B) Given the function $f(x) = \frac{x}{1 + |x|}$ for $x \in R$.
Case $1$: If $x \ge 0$,then $|x| = x$.
$f(x) = \frac{x}{1 + x} = \frac{x + 1 - 1}{1 + x} = 1 - \frac{1}{1 + x}$.
As $x$ increases from $0$ to $\infty$,$1 + x$ increases from $1$ to $\infty$,so $\frac{1}{1 + x}$ decreases from $1$ to $0$.
Thus,$f(x)$ increases from $0$ to $1$. So,the range for $x \ge 0$ is $[0, 1)$.
Case $2$: If $x < 0$,then $|x| = -x$.
$f(x) = \frac{x}{1 - x} = \frac{-(1 - x) + 1}{1 - x} = -1 + \frac{1}{1 - x}$.
As $x$ decreases from $0$ to $-\infty$,$1 - x$ increases from $1$ to $\infty$,so $\frac{1}{1 - x}$ decreases from $1$ to $0$.
Thus,$f(x)$ decreases from $0$ to $-1$. So,the range for $x < 0$ is $(-1, 0)$.
Combining both cases,the range of the function is $(-1, 0) \cup [0, 1) = (-1, 1)$.

(A) Let $y = \frac{x}{x^2 + 1}$.
Rearranging the equation,we get $y(x^2 + 1) = x$,which implies $yx^2 - x + y = 0$.
Since $x$ is a real number,the discriminant $D$ of this quadratic equation in $x$ must be greater than or equal to $0$.
$D = (-1)^2 - 4(y)(y) \ge 0$.
$1 - 4y^2 \ge 0$.
$4y^2 \le 1$,which means $y^2 \le \frac{1}{4}$.
Taking the square root,we get $|y| \le \frac{1}{2}$,which implies $y \in [ - \frac{1}{2}, \frac{1}{2} ]$.
Thus,the range of $f$ is $[ - \frac{1}{2}, \frac{1}{2} ]$.

(A) Let $y = \frac{x^2}{1 - x^2}$.
We want to find the range of $f(x)$.
$y(1 - x^2) = x^2$
$y - yx^2 = x^2$
$y = x^2(1 + y)$
$x^2 = \frac{y}{1 + y}$.
Since $x^2 \ge 0$,we must have $\frac{y}{1 + y} \ge 0$.
Using the sign scheme for the inequality,we get $y \in (-\infty, -1) \cup [0, \infty)$.
Also,$x^2 = \frac{y}{1 + y} \neq 1$ (since $x \neq \pm 1$),so $\frac{y}{1 + y} \neq 1 \implies y \neq y + 1$,which is always true.
Thus,the range is $R - [-1, 0)$.
For the function to be surjective,the codomain $A$ must be equal to the range.
Therefore,$A = R - [-1, 0)$.

(C) The function is $f(x) = \frac{1}{4 - x^2} + \log(x^3 - x)$.
For the function to be defined,two conditions must be satisfied simultaneously:
$1$. The denominator of the rational part must not be zero: $4 - x^2 \neq 0 \implies x^2 \neq 4 \implies x \neq \pm 2$.
$2$. The argument of the logarithm must be strictly positive: $x^3 - x > 0$.
Factoring the inequality: $x(x^2 - 1) > 0 \implies x(x - 1)(x + 1) > 0$.
Using the sign scheme (Wavy Curve Method),the solution for $x(x - 1)(x + 1) > 0$ is $x \in (-1, 0) \cup (1, \infty)$.
Now,we must exclude the points where the denominator is zero ($x = 2$ and $x = -2$).
Since $x = -2$ is not in the interval $(-1, 0) \cup (1, \infty)$,we only need to exclude $x = 2$.
Thus,the domain is $x \in (-1, 0) \cup (1, 2) \cup (2, \infty)$.

(D) The function is defined as $f(x) = \frac{x[x]}{1+x^2}$ for $x \in (1, 3)$.
We split the domain based on the greatest integer function $[x]$:
Case $1$: $x \in (1, 2)$,then $[x] = 1$. So,$f(x) = \frac{x}{1+x^2}$.
For $x \in (1, 2)$,$f'(x) = \frac{(1+x^2)(1) - x(2x)}{(1+x^2)^2} = \frac{1-x^2}{(1+x^2)^2} < 0$. Thus,$f(x)$ is strictly decreasing.
As $x \to 1^+$,$f(x) \to \frac{1}{1+1^2} = \frac{1}{2}$. As $x \to 2^-$,$f(x) \to \frac{2}{1+2^2} = \frac{2}{5}$. So,$f(x) \in (\frac{2}{5}, \frac{1}{2})$.
Case $2$: $x \in [2, 3)$,then $[x] = 2$. So,$f(x) = \frac{2x}{1+x^2}$.
For $x \in [2, 3)$,$f'(x) = \frac{(1+x^2)(2) - 2x(2x)}{(1+x^2)^2} = \frac{2-2x^2}{(1+x^2)^2} < 0$. Thus,$f(x)$ is strictly decreasing.
At $x = 2$,$f(2) = \frac{2(2)}{1+2^2} = \frac{4}{5}$. As $x \to 3^-$,$f(x) \to \frac{2(3)}{1+3^2} = \frac{6}{10} = \frac{3}{5}$. So,$f(x) \in (\frac{3}{5}, \frac{4}{5}]$.
Combining both cases,the range is $(\frac{2}{5}, \frac{1}{2}) \cup (\frac{3}{5}, \frac{4}{5}]$.

(B) The domain of the natural logarithm function $\log x$ is the set of all positive real numbers,i.e.,$x \in (0, \infty)$.
For $x \leq 0$,the expression $\log x$ is not defined in the set of real numbers.
For $x > 0$,let $y = e^{\log x}$.
Taking the natural logarithm on both sides,we get $\log y = \log(e^{\log x})$.
Using the property $\log(a^b) = b \log a$,we have $\log y = (\log x) \cdot \log e$.
Since $\log e = 1$,we get $\log y = \log x$.
By the one-to-one property of the logarithmic function,$y = x$.
Therefore,the equation $x = e^{\log x}$ is true only for $x > 0$.

(A) The relation $R$ is defined on the set of integers $Z$ as $R = \{(a, b) : a, b \in Z, a - b \in Z\}$.
Since the difference between any two integers $a$ and $b$ is always an integer,every pair $(a, b)$ where $a, b \in Z$ satisfies the condition $a - b \in Z$.
Therefore,$R = Z \times Z$.
The domain of $R$ is the set of all first elements of the ordered pairs in $R$,which is $Z$.
The range of $R$ is the set of all second elements of the ordered pairs in $R$,which is $Z$.
Thus,the domain is $Z$ and the range is $Z$.

(N/A) The completed table is as follows:

$x$	$-4$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y = f(x) = x^2$	$16$	$9$	$4$	$1$	$0$	$1$	$4$	$9$	$16$

Domain of $f = R$ (the set of all real numbers).
Range of $f = [0, \infty)$ (the set of all non-negative real numbers).

(N/A) The completed table is given below:

$x$	$-2$	$-1.5$	$-1$	$-0.5$	$0.25$	$0.5$	$1$	$1.5$	$2$
$y = \frac{1}{x}$	$-0.5$	$-0.67$	$-1$	$-2$	$4$	$2$	$1$	$0.67$	$0.5$

The domain of the function $f(x) = \frac{1}{x}$ is $R - \{0\}$ because the function is undefined at $x = 0$.
The range of the function is $R - \{0\}$ because for any $y \in R - \{0\}$,there exists $x = \frac{1}{y}$ such that $f(x) = y$.

(N/A) The given function is $f(x) = -|x|$,where $x \in \mathbb{R}$.
We know that the absolute value function is defined as:
$|x| = \begin{cases} x, & \text{if } x \ge 0 \\ -x, & \text{if } x < 0 \end{cases}$
Therefore,the function $f(x) = -|x|$ is defined as:
$f(x) = \begin{cases} -x, & \text{if } x \ge 0 \\ x, & \text{if } x < 0 \end{cases}$
Since $f(x)$ is defined for all real numbers $x$,the domain of $f$ is $\mathbb{R}$ (the set of all real numbers).
For any $x \in \mathbb{R}$,$|x| \ge 0$. Multiplying by $-1$,we get $-|x| \le 0$. Thus,$f(x) \le 0$ for all $x$.
Therefore,the range of $f$ is $(-\infty, 0]$.

(A) The function is defined when $9 - x^{2} \ge 0$.
This implies $x^{2} \le 9$,which means $-3 \le x \le 3$.
Thus,the domain of $f(x)$ is $[-3, 3]$.
Since $f(x) = \sqrt{9 - x^{2}}$,the minimum value occurs at $x = \pm 3$,giving $f(x) = 0$.
The maximum value occurs at $x = 0$,giving $f(x) = \sqrt{9} = 3$.
Therefore,the range of $f(x)$ is $[0, 3]$.

(D) Given the function $f(x) = 2 - 3x$ for $x > 0$.
Since $x > 0$,we multiply both sides by $3$:
$3x > 0$
Now,multiply by $-1$ (which reverses the inequality sign):
$-3x < 0$
Add $2$ to both sides:
$2 - 3x < 2 - 0$
$f(x) < 2$
Thus,the range of the function $f$ is the set of all real numbers less than $2$.
Therefore,the range of $f$ is $(-\infty, 2)$.

(N/A) Given the function $f(x) = x^{2} + 2$,where $x \in \mathbb{R}$.
Since $x$ is a real number,the square of any real number is always non-negative:
$x^{2} \geq 0$
Adding $2$ to both sides of the inequality:
$x^{2} + 2 \geq 0 + 2$
$x^{2} + 2 \geq 2$
Since $f(x) = x^{2} + 2$,we have:
$f(x) \geq 2$
Therefore,the range of the function $f$ is the set of all real numbers greater than or equal to $2$.
In interval notation,the range is $[2, \infty)$.

(N/A) Given the function $f(x) = x$,where $x \in \mathbb{R}$.
By definition,the range of a function is the set of all possible output values (images) for the given domain.
Since $x$ can be any real number,the output $f(x)$ will also be any real number.
Therefore,the range of $f$ is the set of all real numbers,denoted by $\mathbb{R}$.

(A) The function $f(x) = \frac{x^{2}+3x+5}{x^{2}-5x+4}$ is defined for all real values of $x$ except where the denominator is zero.
Set the denominator equal to zero: $x^{2}-5x+4 = 0$.
Factor the quadratic expression: $(x-4)(x-1) = 0$.
This gives $x = 4$ or $x = 1$.
Therefore,the function is undefined at $x = 1$ and $x = 4$.
The domain of the function is the set of all real numbers except $1$ and $4$,which is written as $R - \{1, 4\}$.

(A) The given function is $f(x) = \frac{x^{2}+2x+1}{x^{2}-8x+12}$.
To find the domain,we set the denominator equal to zero to find the values of $x$ for which the function is undefined:
$x^{2}-8x+12 = 0$
$(x-6)(x-2) = 0$
This gives $x = 6$ or $x = 2$.
The function $f(x)$ is defined for all real numbers except at $x = 2$ and $x = 6$.
Therefore,the domain of $f$ is $R - \{2, 6\}$.

(A) The given real function is $f(x) = \sqrt{x-1}$.
For the function to be defined in the set of real numbers,the expression inside the square root must be non-negative:
$x - 1 \geq 0 \Rightarrow x \geq 1$.
Therefore,the domain of $f$ is the set of all real numbers greater than or equal to $1$,i.e.,$[1, \infty)$.
Since $\sqrt{x-1} \geq 0$ for all $x \geq 1$,the output values of the function are always non-negative.
As $x$ takes all values in $[1, \infty)$,$\sqrt{x-1}$ takes all values in $[0, \infty)$.
Thus,the range of $f$ is $[0, \infty)$.

(A) The given real function is $f(x) = |x - 1|$.
Since the absolute value function $|x - 1|$ is defined for all real numbers $x$,the domain of $f$ is the set of all real numbers,$R$.
For any real number $x$,the absolute value $|x - 1|$ is always greater than or equal to $0$.
Thus,the range of $f$ is the set of all non-negative real numbers,which is $[0, \infty)$.

(A) Let $y = f(x) = \frac{x^2}{1+x^2}$.
Since $x^2 \ge 0$ for all $x \in R$,the numerator is non-negative and the denominator $1+x^2 \ge 1$.
Thus,$y \ge 0$.
Now,$y(1+x^2) = x^2 \implies y + yx^2 = x^2 \implies y = x^2(1-y) \implies x^2 = \frac{y}{1-y}$.
Since $x^2 \ge 0$,we must have $\frac{y}{1-y} \ge 0$.
This inequality holds when $0 \le y < 1$.
Therefore,the range of $f$ is $[0, 1)$.

(A) Given $A = \{9, 10, 11, 12, 13\}$ and $f(n) = \text{highest prime factor of } n$.
We calculate $f(n)$ for each $n \in A$:
$f(9) = \text{highest prime factor of } 3^2 = 3$.
$f(10) = \text{highest prime factor of } 2 \times 5 = 5$.
$f(11) = \text{highest prime factor of } 11 = 11$.
$f(12) = \text{highest prime factor of } 2^2 \times 3 = 3$.
$f(13) = \text{highest prime factor of } 13 = 13$.
The range of $f$ is the set of all output values: $\{f(9), f(10), f(11), f(12), f(13)\} = \{3, 5, 11, 3, 13\}$.
Removing duplicates,the range is $\{3, 5, 11, 13\}$.

(B) For the function $f(x) = \frac{\operatorname{cosec}^{-1} x}{\sqrt{x - [x]}}$ to be defined,two conditions must be satisfied:
$1$. The domain of $\operatorname{cosec}^{-1} x$ is $(-\infty, -1] \cup [1, \infty)$.
$2$. The denominator $\sqrt{x - [x]}$ must be non-zero and real. Since $x - [x] = \{x\}$ (the fractional part of $x$),we require $\{x\} > 0$,which means $x$ cannot be an integer.
Combining these conditions,$x$ must be in $(-\infty, -1] \cup [1, \infty)$ and $x \notin \mathbb{Z}$.
This corresponds to all non-integers in the set $(-\infty, -1] \cup [1, \infty)$,which is equivalent to all non-integers except the interval $(-1, 1)$.