Domain and Range Questions in English

(C) For the domain $A$,we require $|x| - x^2 > 0$.
Since $|x|^2 = x^2$,this is $|x| - |x|^2 > 0$,which implies $|x|(1 - |x|) > 0$.
This holds when $0 < |x| < 1$,so $x \in (-1, 0) \cup (0, 1)$. Thus,$A = (-1, 0) \cup (0, 1)$.
For the range $B$,let $y = \frac{1}{\sqrt{|x| - x^2}}$.
As $x \to 0$,$|x| - x^2 \to 0^+$,so $y \to \infty$.
As $|x| \to 1$,$|x| - x^2 \to 0^+$,so $y \to \infty$.
The maximum value of $|x| - x^2$ occurs at $|x| = 1/2$,giving $1/2 - 1/4 = 1/4$.
The minimum value of the denominator is $0$ (exclusive) and the maximum value is $1/4$.
Thus,$\sqrt{|x| - x^2} \in (0, 1/2]$.
Therefore,$y \in [2, \infty)$,so $B = [2, \infty)$.
Finally,$A \cup B = ((-1, 0) \cup (0, 1)) \cup [2, \infty)$.

(D) For the function $f(x)$ to be defined,both parts must be defined.
First,consider $\cos^{-1} \left( \frac{1 + x^2}{2 x} \right)$. This is defined if $-1 \leq \frac{1 + x^2}{2 x} \leq 1$.
This implies $\left| \frac{1 + x^2}{2 x} \right| \leq 1$,which means $|1 + x^2| \leq |2x|$.
Since $1 + x^2 \geq 2|x|$ is only true when $|x| = 1$,we have $x = 1$ or $x = -1$.
Now,check the first part $\sqrt{\cos(\sin x)}$.
For $x = 1$,$\cos(\sin 1) > 0$ since $\sin 1 \approx 0.84$ radians,which is in the first quadrant.
For $x = -1$,$\cos(\sin(-1)) = \cos(-\sin 1) = \cos(\sin 1) > 0$.
Since both conditions are satisfied only at $x = 1$ and $x = -1$,the domain is $\{-1, 1\}$.

(C) Given function: $f(x) = \sqrt[3]{\frac{x-2}{2x^2-7x+5}} + \log(x^2-x-2)$.
For the domain,the cube root is defined for all real values where the denominator is non-zero,and the logarithm is defined for positive arguments.
$1$. For the denominator of the cube root: $2x^2-7x+5 \neq 0$ $\Rightarrow (2x-5)(x-1) \neq 0$ $\Rightarrow x \neq 1, x \neq \frac{5}{2}$.
$2$. For the logarithm: $x^2-x-2 > 0 \Rightarrow (x-2)(x+1) > 0$. This inequality holds for $x \in (-\infty, -1) \cup (2, \infty)$.
Combining these conditions: We need $x \in (-\infty, -1) \cup (2, \infty)$ such that $x \neq 1$ and $x \neq \frac{5}{2}$.
Since $1$ is not in the interval $(-\infty, -1) \cup (2, \infty)$,we only need to exclude $\frac{5}{2}$.
Thus,the domain is $(-\infty, -1) \cup (2, \frac{5}{2}) \cup (\frac{5}{2}, \infty)$.

(A) Given $f(x) = \frac{1}{\sqrt{|x - |x||}}$.
For the function to be defined,the expression inside the square root must be strictly positive: $|x - |x|| > 0$.
If $x \ge 0$,then $|x| = x$,so $|x - x| = 0$,which is not $> 0$.
If $x < 0$,then $|x| = -x$,so $|x - (-x)| = |2x| = -2x$. Since $x < 0$,$-2x > 0$.
Thus,the domain $A = (-\infty, 0)$.
For $x \in (-\infty, 0)$,$f(x) = \frac{1}{\sqrt{-2x}}$. As $x$ ranges from $(-\infty, 0)$,$-2x$ ranges from $(0, \infty)$,and $\sqrt{-2x}$ ranges from $(0, \infty)$.
Therefore,the range $B = (0, \infty)$.
Finally,$A \cap B = (-\infty, 0) \cap (0, \infty) = \phi$.

(D) The function $f(x)=\sin ^{-1}(x^2-1)-3 \log _3(3^x-2)$ is defined when both terms are defined.
For $\sin ^{-1}(x^2-1)$,we require $-1 \leq x^2-1 \leq 1$,which implies $0 \leq x^2 \leq 2$,so $x \in [-\sqrt{2}, \sqrt{2}]$.
For $\log _3(3^x-2)$,we require $3^x-2 > 0$,which implies $3^x > 2$,so $x > \log _3 2$.
The domain of $f(x)$ is the intersection: $[-\sqrt{2}, \sqrt{2}] \cap (\log _3 2, \infty) = (\log _3 2, \sqrt{2}]$.
The function is not defined for $x \in (-\infty, \log _3 2] \cup (\sqrt{2}, \infty)$.
Comparing this with $(-\infty, a] \cup (b, \infty)$,we get $a = \log _3 2$ and $b = \sqrt{2}$.
Thus,$3^a + b^2 = 3^{\log _3 2} + (\sqrt{2})^2 = 2 + 2 = 4$.

(D) For the function $f(x)$ to be defined,the following conditions must be satisfied:
$1$. The expression inside the square root in the numerator must be non-negative: $\log_{10}\left(\frac{x}{x-2}\right) \geq 0$.
This implies $\frac{x}{x-2} \geq 10^0$,so $\frac{x}{x-2} \geq 1$.
$\frac{x}{x-2} - 1 \geq 0$ $\Rightarrow \frac{x - (x-2)}{x-2} \geq 0$ $\Rightarrow \frac{2}{x-2} > 0$.
This holds when $x-2 > 0$,i.e.,$x > 2$.
$2$. The expression inside the square root in the denominator must be strictly positive: $[x]^2 - 5[x] + 6 > 0$.
$([x]-2)([x]-3) > 0$.
This implies $[x] < 2$ or $[x] > 3$.
If $[x] < 2$,then $x < 2$. If $[x] > 3$,then $x \geq 4$.
$3$. Combining the conditions $x > 2$ and ($x < 2$ or $x \geq 4$),we get $x \in [4, \infty)$.

(C) The function is defined as $f(x) = \sqrt{\frac{|x|-x}{x-[x]}}$.
For the numerator $\sqrt{|x|-x}$ to be defined,we need $|x|-x \geq 0$,which implies $|x| \geq x$. This is true for all $x \in R$.
For the denominator $\sqrt{x-[x]}$ to be defined and non-zero,we need $x-[x] > 0$.
We know that $x-[x] = \{x\}$,where $\{x\}$ is the fractional part of $x$.
The condition $\{x\} > 0$ holds for all $x \notin Z$ (all real numbers except integers).
If $x \in Z$,then $\{x\} = 0$,which makes the denominator zero and the function undefined.
Therefore,the domain of the function is $R-Z$.

(A) For the combination ${ }^{n} C_{r}$ to be defined,we must have $n \geq r \geq 0$ and $n, r \in \mathbb{Z}_{\geq 0}$.
Here,$n = 16-x$ and $r = 2x-1$.
$1$) $r \geq 0 \implies 2x-1 \geq 0 \implies x \geq \frac{1}{2}$.
$2$) $n \geq r \implies 16-x \geq 2x-1 \implies 17 \geq 3x \implies x \leq \frac{17}{3} \approx 5.66$.
$3$) $n \geq 0 \implies 16-x \geq 0 \implies x \leq 16$.
Combining these,we get $\frac{1}{2} \leq x \leq 5.66$.
Since $n$ and $r$ must be non-negative integers,$x$ must be an integer such that $2x-1$ is a non-negative integer and $16-x$ is an integer $\geq 2x-1$.
The possible integer values for $x$ in the interval $[0.5, 5.66]$ are $x \in \{1, 2, 3, 4, 5\}$.

(A) Given: $f(x) = \frac{\sqrt{6x^2+5x-6}}{\sqrt{4-x}-\sqrt{x+4}}$
For $f(x)$ to be defined:
$(1)$ The numerator must be real: $6x^2+5x-6 \geq 0$
$(3x-2)(2x+3) \geq 0 \Rightarrow x \in (-\infty, -\frac{3}{2}] \cup [\frac{2}{3}, \infty)$
$(2)$ The denominator must be non-zero: $\sqrt{4-x} - \sqrt{x+4} \neq 0$
$4-x \neq x+4$ $\Rightarrow 2x \neq 0$ $\Rightarrow x \neq 0$
$(3)$ The square root terms must be defined:
$4-x \geq 0 \Rightarrow x \leq 4$
$x+4 \geq 0 \Rightarrow x \geq -4$
Combining these conditions: $x \in [-4, 4] \cap ((-\infty, -\frac{3}{2}] \cup [\frac{2}{3}, \infty)) \cap \{x \neq 0\}$
Since $0$ is not in the interval $(-\infty, -\frac{3}{2}] \cup [\frac{2}{3}, \infty)$,the condition $x \neq 0$ is automatically satisfied.
Thus,the domain is $[-4, -\frac{3}{2}] \cup [\frac{2}{3}, 4]$.

(A) For the function $f(x) = \sqrt{\frac{2x^2 - 7x + 5}{3x^2 - 5x - 2}}$ to be defined,the expression inside the square root must be non-negative:
$\frac{2x^2 - 7x + 5}{3x^2 - 5x - 2} \geq 0$
Factorizing the numerator and denominator:
Numerator: $2x^2 - 7x + 5 = 2x^2 - 2x - 5x + 5 = 2x(x - 1) - 5(x - 1) = (2x - 5)(x - 1)$
Denominator: $3x^2 - 5x - 2 = 3x^2 - 6x + x - 2 = 3x(x - 2) + 1(x - 2) = (3x + 1)(x - 2)$
So,the inequality is $\frac{(2x - 5)(x - 1)}{(3x + 1)(x - 2)} \geq 0$
The critical points are $x = -\frac{1}{3}, 1, 2, \frac{5}{2}$.
Using the wavy curve method (sign scheme) on the number line:
- For $x > \frac{5}{2}$,the expression is positive.
- For $2 < x < \frac{5}{2}$,the expression is negative.
- For $1 < x < 2$,the expression is positive.
- For $-\frac{1}{3} < x < 1$,the expression is negative.
- For $x < -\frac{1}{3}$,the expression is positive.
Including the points where the numerator is zero $(x = 1, \frac{5}{2})$ and excluding points where the denominator is zero $(x = -\frac{1}{3}, 2)$:
The domain is $\left(-\infty, -\frac{1}{3}\right) \cup [1, 2) \cup \left[\frac{5}{2}, \infty\right)$.
Thus,option $A$ is correct.

(A) The function is defined as $f(x) = \sqrt{\frac{[x]-x}{x-[x]}}$.
For any $x \in R$,let $x = [x] + \{x\}$,where $0 \leq \{x\} < 1$.
Then $x - [x] = \{x\}$.
If $x \notin Z$,then $\{x\} \neq 0$,so we can write the expression as:
$f(x) = \sqrt{\frac{-\{x\}}{\{x\}}} = \sqrt{-1} = i$.
Since $i$ is not a real number,$f(x)$ is not real for any $x \notin Z$.
If $x \in Z$,then $[x] = x$,which makes the denominator $x - [x] = 0$.
Division by zero is undefined,so $f(x)$ is not defined for $x \in Z$.
Therefore,there are no real values of $x$ for which $f(x)$ is real.
The set of such values is the empty set,denoted by $\phi$.

(C) Given $a^x + a^y = a$.
Since $f: A \rightarrow A$,the domain and range are both $A$.
For $y$ to be defined,we must have $a^y > 0$,which implies $a - a^x > 0$,so $a^x < a$.
Since $a > 1$,taking $\log_a$ on both sides gives $x < 1$.
Thus,the domain is $x \in (-\infty, 1)$.
Since the range must also be $A$,we have $y = \log_a(a - a^x)$.
As $x \rightarrow -\infty$,$a^x \rightarrow 0$,so $y \rightarrow \log_a(a) = 1$.
As $x \rightarrow 1^-$,$a^x \rightarrow a^-$,so $a - a^x \rightarrow 0^+$,which means $y \rightarrow -\infty$.
Thus,the range is $(-\infty, 1)$.
Therefore,$A = (-\infty, 1)$.

(C) Given,$f(x) = \frac{1}{2} - \tan \left(\frac{\pi x}{2}\right)$ with domain $(-1, 1)$ and $g(x) = \sqrt{3 + 4x - 4x^2}$.
To find the domain of $(f + g)$,we need the intersection of the domains of $f(x)$ and $g(x)$.
For $g(x)$ to be defined,$3 + 4x - 4x^2 \geq 0$.
Multiplying by $-1$,we get $4x^2 - 4x - 3 \leq 0$.
Factoring the quadratic,$(2x + 1)(2x - 3) \leq 0$.
This inequality holds for $x \in \left[-\frac{1}{2}, \frac{3}{2}\right]$.
The domain of $(f + g)$ is the intersection of $(-1, 1)$ and $\left[-\frac{1}{2}, \frac{3}{2}\right]$.
Intersection $= (-1, 1) \cap \left[-\frac{1}{2}, \frac{3}{2}\right] = \left[-\frac{1}{2}, 1\right)$.

(B) Given,$f(x) = \frac{1}{\sqrt{|x|-x}}$.
For the function $f(x)$ to be defined,the expression inside the square root must be strictly positive:
$|x| - x > 0$
$|x| > x$
This inequality holds true for all negative real numbers,i.e.,$x < 0$.
Therefore,the domain is $(-\infty, 0)$.

(D) The function is $f(x) = \sec^{-1}(3x - 4) + \tanh^{-1}\left(\frac{x + 3}{5}\right)$.
For $\sec^{-1}(3x - 4)$ to be defined,we must have $|3x - 4| \geq 1$.
This implies $3x - 4 \leq -1$ or $3x - 4 \geq 1$.
$3x \leq 3 \Rightarrow x \leq 1$ and $3x \geq 5 \Rightarrow x \geq \frac{5}{3}$.
So,the domain of the first part is $x \in (-\infty, 1] \cup [\frac{5}{3}, \infty)$.
For $\tanh^{-1}\left(\frac{x + 3}{5}\right)$ to be defined,we must have $-1 < \frac{x + 3}{5} < 1$.
$-5 < x + 3 < 5$.
$-8 < x < 2$.
Taking the intersection of $x \in (-\infty, 1] \cup [\frac{5}{3}, \infty)$ and $x \in (-8, 2)$:
$x \in (-8, 1] \cup [\frac{5}{3}, 2)$.

(B) For the function $f(x) = \sqrt{\log_{10}\left(\frac{5x - x^2}{4}\right)}$ to be defined,the expression inside the square root must be non-negative: $\log_{10}\left(\frac{5x - x^2}{4}\right) \geq 0$.
This implies $\frac{5x - x^2}{4} \geq 10^0$,which simplifies to $\frac{5x - x^2}{4} \geq 1$.
Multiplying by $4$,we get $5x - x^2 \geq 4$,or $x^2 - 5x + 4 \leq 0$.
Factoring the quadratic,we have $(x - 1)(x - 4) \leq 0$.
This inequality holds when $x \in [1, 4]$.
Thus,the domain of $f(x)$ is $[1, 4]$.

(B) For $f(x)=\frac{x-[x]}{\sqrt{|x|-x}}$ to be defined,the denominator must be non-zero and the expression inside the square root must be positive:
$|x|-x > 0 \Rightarrow |x| > x$.
This inequality holds for all $x < 0$. Thus,$A = (-\infty, 0)$.
For $g(x)=\frac{x-[x]}{\sqrt{|x|+x}}$ to be defined,the denominator must be non-zero and the expression inside the square root must be positive:
$|x|+x > 0 \Rightarrow |x| > -x$.
This inequality holds for all $x > 0$. Thus,$B = (0, \infty)$.
Since $A = (-\infty, 0)$ and $B = (0, \infty)$,their intersection is empty:
$A \cap B = \phi$.

(C) Given the function $f:[-3,2] \rightarrow [0, \sqrt[3]{x}]$ defined as $f(n) = \begin{cases} 2+\sqrt[3]{n}, & -3 \leq n \leq -1 \\ n^{2/3}, & -1 < n \leq 2 \end{cases}$.
For the function to be onto,the range of $f(n)$ must be equal to the codomain $[0, \sqrt[3]{x}]$.
First,evaluate the function at the boundaries and critical points:
For $-3 \leq n \leq -1$,$f(n)$ increases from $f(-3) = 2 + \sqrt[3]{-3} \approx 0.55$ to $f(-1) = 2 + (-1)^{1/3} = 2 - 1 = 1$.
For $-1 < n \leq 2$,$f(n) = n^{2/3}$. At $n = -1$,$f(-1) = (-1)^{2/3} = 1$. At $n = 0$,$f(0) = 0$. At $n = 2$,$f(2) = 2^{2/3} = \sqrt[3]{4}$.
Since the function is onto,the maximum value of the function must be the upper bound of the codomain.
The maximum value is $\max(f(-1), f(2)) = \max(1, \sqrt[3]{4}) = \sqrt[3]{4}$.
Thus,$\sqrt[3]{x} = \sqrt[3]{4}$,which implies $x = 4$.

(C) For the function $f(x) = \sqrt{\frac{x-[x]}{\log(x^2-x)}}$ to be defined,the following conditions must be met:
$1$. The expression inside the square root must be non-negative: $\frac{x-[x]}{\log(x^2-x)} \geq 0$.
$2$. The denominator must not be zero: $\log(x^2-x) \neq 0 \Rightarrow x^2-x \neq 1$.
$3$. The argument of the logarithm must be positive: $x^2-x > 0$.
Since $x-[x] = \{x\} \geq 0$ for all $x \in R$,the condition $\frac{x-[x]}{\log(x^2-x)} \geq 0$ simplifies to $\log(x^2-x) > 0$.
This implies $x^2-x > 1$,or $x^2-x-1 > 0$.
Solving $x^2-x-1 = 0$ gives $x = \frac{1 \pm \sqrt{5}}{2}$.
The inequality $x^2-x-1 > 0$ holds for $x \in \left(-\infty, \frac{1-\sqrt{5}}{2}\right) \cup \left(\frac{1+\sqrt{5}}{2}, \infty\right)$.
Also,we must ensure $x^2-x > 0$,which is satisfied in the intervals above.
Thus,the domain is $R \setminus \left[\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right]$.

(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.
$1$. Denominator condition: $[x] + 2 \neq 0 \Rightarrow [x] \neq -2$. This implies $x \notin [-2, -1)$.
$2$. Inequality condition: $\frac{4-x^2}{[x]+2} \geq 0$.
Case $I$: If $[x] + 2 > 0$,then $[x] > -2$,which means $x \geq -1$.
Then $4 - x^2 \geq 0$ $\Rightarrow x^2 \leq 4$ $\Rightarrow x \in [-2, 2]$.
Combining $x \geq -1$ and $x \in [-2, 2]$,we get $x \in [-1, 2]$.
Case $II$: If $[x] + 2 < 0$,then $[x] < -2$,which means $x < -2$.
Then $4 - x^2 \leq 0$ $\Rightarrow x^2 \geq 4$ $\Rightarrow x \in (-\infty, -2] \cup [2, \infty)$.
Combining $x < -2$ and $x \in (-\infty, -2] \cup [2, \infty)$,we get $x \in (-\infty, -2)$.
Combining both cases,the domain is $(-\infty, -2) \cup [-1, 2]$.

(D) The function $f(x) = \sin^{-1}\left[\log_4\left(\frac{x}{4}\right)\right] + \sqrt{17x - x^2 - 16}$ is defined if both parts are defined.
First,for $\sin^{-1}(u)$,we must have $u \in [-1, 1]$:
$-1 \leq \log_4\left(\frac{x}{4}\right) \leq 1$
$4^{-1} \leq \frac{x}{4} \leq 4^1$
$\frac{1}{4} \leq \frac{x}{4} \leq 4$
$1 \leq x \leq 16$
Second,for the square root $\sqrt{17x - x^2 - 16}$,we must have $17x - x^2 - 16 \geq 0$:
$x^2 - 17x + 16 \leq 0$
$(x - 16)(x - 1) \leq 0$
$1 \leq x \leq 16$
Combining both conditions,the domain is $[1, 16]$.

(A) For the function $f(x) = \frac{1}{\sqrt{[x]^2 - [x] - 2}}$ to be defined,the expression inside the square root must be strictly positive:
$[x]^2 - [x] - 2 > 0$
Factorizing the quadratic expression:
$([x] - 2)([x] + 1) > 0$
This inequality holds if $[x] > 2$ or $[x] < -1$.
If $[x] > 2$,then the smallest integer value is $3$,so $x \geq 3$.
If $[x] < -1$,then the largest integer value is $-2$,so $x < -1$.
Thus,the domain is $x \in (-\infty, -1) \cup [3, \infty)$.

(A) For the function $f(x) = \sqrt{\log_a(x - [x])}$ to be defined,the expression inside the square root must be non-negative and the logarithm must be defined.
$1$. The term $(x - [x])$ represents the fractional part of $x$,denoted as $\{x\}$. Since $0 \leq \{x\} < 1$,and for the logarithm $\log_a(\{x\})$ to be defined,we must have $\{x\} > 0$,which implies $x \notin Z$.
$2$. For the square root to be defined,$\log_a(\{x\}) \geq 0$.
$3$. If $a > 1$,then $\{x\} \geq a^0 = 1$. Since $\{x\} < 1$,this is impossible.
$4$. If $0 < a < 1$,then $\{x\} \leq a^0 = 1$. Since $\{x\} < 1$ is always true for $x \notin Z$,this condition holds.
$5$. Thus,$x \in R - Z$ and $a \in (0, 1)$.

(B) The function $f(x) = \log \left[(2.5)^{3-x^2} - (0.4)^{x+9}\right]$ is defined if the argument of the logarithm is strictly positive:
$(2.5)^{3-x^2} - (0.4)^{x+9} > 0$
$\Rightarrow (2.5)^{3-x^2} > (0.4)^{x+9}$
Since $0.4 = \frac{4}{10} = \frac{2}{5} = (2.5)^{-1}$,we can write:
$(2.5)^{3-x^2} > (2.5)^{-(x+9)}$
Since the base $2.5 > 1$,the inequality holds for the exponents:
$3 - x^2 > -x - 9$
$x^2 - x - 12 < 0$
$(x - 4)(x + 3) < 0$
This inequality holds for $x \in (-3, 4)$.

(B) Given that the function $f(x)$ is defined on the interval $[-1, 1]$.
For the function $g(x) = f(5x + 4)$ to be defined,the argument $(5x + 4)$ must lie within the domain of $f$,which is $[-1, 1]$.
Therefore,we have the inequality:
$-1 \leq 5x + 4 \leq 1$
Subtracting $4$ from all parts of the inequality:
$-1 - 4 \leq 5x \leq 1 - 4$
$-5 \leq 5x \leq -3$
Dividing all parts by $5$:
$-1 \leq x \leq -\frac{3}{5}$
Thus,the function $g(x)$ is defined on the interval $[-1, -\frac{3}{5}]$.

(D) Let $y = f(x) = \frac{2+x}{2-x}$.
$y(2-x) = 2+x$
$2y - xy = 2 + x$
$2y - 2 = x + xy$
$2(y-1) = x(1+y)$
$x = \frac{2(y-1)}{y+1}$.
For $x$ to be a real number,the denominator $y+1 \neq 0$,which implies $y \neq -1$.
Thus,the range of $f(x)$ is $R - \{-1\}$.

(B) For $f(x)$ to be a real-valued function,the domain requires $x+1 \ge 0$,so $x \ge -1$.
As $x$ varies from $-1$ to $\infty$,the term $\sqrt{x+1}$ varies from $0$ to $\infty$.
Consequently,the denominator $\sqrt{x+1}+4$ varies from $4$ to $\infty$.
Thus,the argument of the tangent function,$\theta = \frac{\pi}{\sqrt{x+1}+4}$,varies from $0$ to $\frac{\pi}{4}$.
Since the tangent function is strictly increasing in the interval $[0, \frac{\pi}{4}]$,the range of $f(x) = \tan(\theta)$ is $[\tan(0), \tan(\frac{\pi}{4})]$.
This simplifies to $[0, 1]$.
Since none of the options match $[0, 1]$,and assuming the question implies the range of the function as defined,the closest logical interval based on standard function analysis is $(0, 1]$ if $x > -1$.

(C) Let $y = \frac{x^2+x+a}{x^2-x+a}$.
Rearranging the terms,we get $y(x^2-x+a) = x^2+x+a$.
$yx^2 - yx + ay = x^2 + x + a$.
$(y-1)x^2 - (y+1)x + a(y-1) = 0$.
For $f$ to be a surjection,this quadratic equation in $x$ must have real roots for every $y$ in the range.
Thus,the discriminant $D \geq 0$.
$D = (-(y+1))^2 - 4(y-1)(a(y-1)) \geq 0$.
$(y+1)^2 - 4a(y-1)^2 \geq 0$.
$y^2 + 2y + 1 - 4a(y^2 - 2y + 1) \geq 0$.
$(1-4a)y^2 + (2+8a)y + (1-4a) \geq 0$.
For this to hold for all $y$,the coefficient of $y^2$ must be positive,i.e.,$1-4a > 0 \Rightarrow a < 1/4$.
Also,the discriminant of this quadratic in $y$ must be $\leq 0$.
$(2+8a)^2 - 4(1-4a)^2 \leq 0$.
$4(1+4a)^2 - 4(1-4a)^2 \leq 0$.
$(1+4a-1+4a)(1+4a+1-4a) \leq 0$.
$(8a)(2) \leq 0$ $\Rightarrow 16a \leq 0$ $\Rightarrow a \leq 0$.
Combining $a < 1/4$ and $a \leq 0$,we get $a \in (-\infty, 0]$.

(A) Let $g(x) = 5 + 4x - x^2$.
We can rewrite this as $g(x) = -(x^2 - 4x - 5) = -(x^2 - 4x + 4 - 9) = 9 - (x - 2)^2$.
Since $(x - 2)^2 \geq 0$,the maximum value of $g(x)$ is $9$ (at $x = 2$).
Also,for the logarithm to be defined,we require $g(x) > 0$,so $9 - (x - 2)^2 > 0$,which implies $(x - 2)^2 < 9$,or $-3 < x - 2 < 3$,meaning $-1 < x < 5$.
As $x$ varies in $(-1, 5)$,$g(x)$ takes all values in the interval $(0, 9]$.
Since $f(x) = \log_3(g(x))$,the range of $f(x)$ is $(\log_3(0^+), \log_3(9)]$.
As $g(x) \to 0^+$,$\log_3(g(x)) \to -\infty$.
As $g(x) \to 9$,$\log_3(g(x)) \to \log_3(9) = 2$.
Therefore,the range is $(-\infty, 2]$.

(B) Given the function $f(x) = x^2 - 4x + 5$ defined on the domain $[2, \infty)$.
We can rewrite the function by completing the square:
$f(x) = (x^2 - 4x + 4) + 1 = (x - 2)^2 + 1$.
Since the domain is $x \in [2, \infty)$,the minimum value of $(x - 2)^2$ is $0$ (at $x = 2$).
As $x \rightarrow \infty$,$(x - 2)^2 \rightarrow \infty$.
Therefore,the range of $f(x)$ is $[0 + 1, \infty) = [1, \infty)$.

(B) Let $g(x) = x^4 - 2x^2 + 3 = (x^2 - 1)^2 + 2$.
Since $(x^2 - 1)^2 \geq 0$,the minimum value of $g(x)$ is $2$ (at $x^2 = 1$) and the maximum value is $\infty$.
Thus,the range of $g(x)$ is $[2, \infty)$.
Now,$f(x) = \log_{0.5}(g(x)) = \log_{1/2}(g(x)) = -\log_2(g(x))$.
Since $g(x) \in [2, \infty)$,we have $\log_2(g(x)) \in [\log_2 2, \log_2 \infty) = [1, \infty)$.
Multiplying by $-1$,we get $-\log_2(g(x)) \in (-\infty, -1]$.
Therefore,the range of $f(x)$ is $(-\infty, -1]$.

(B) Let $g(x) = -x^2-6x-5$. This is a downward-opening parabola.
The maximum value of $g(x)$ is given by $-\frac{D}{4a}$,where $D = b^2-4ac = (-6)^2 - 4(-1)(-5) = 36 - 20 = 16$.
The maximum value is $-\frac{16}{4(-1)} = 4$.
Thus,the range of $g(x)$ is $(-\infty, 4]$.
Since the function is $f(x) = -\sqrt{g(x)}$,the expression inside the square root must be non-negative,so $g(x) \in [0, 4]$.
Taking the square root,$\sqrt{g(x)} \in [0, 2]$.
Multiplying by $-1$,we get $f(x) \in [-2, 0]$.

(B) Given the function $f(x) = \frac{1}{x - |x|}$.
For $x \geq 0$,we have $|x| = x$,so the denominator $x - |x| = 0$. Thus,the function is undefined for $x \geq 0$.
For $x < 0$,we have $|x| = -x$,so the denominator becomes $x - (-x) = 2x$.
Therefore,$f(x) = \frac{1}{2x}$ for $x < 0$.
As $x$ ranges from $(-\infty, 0)$,the value of $2x$ ranges from $(-\infty, 0)$.
Consequently,the value of $\frac{1}{2x}$ ranges from $(-\infty, 0)$.
Thus,the range of the function is $(-\infty, 0)$.

(B) To find the range,we analyze the function in three intervals:
$1$. For $x < -1$,$f(x) = 2x - 3$. As $x \to -1^-$,$f(x) \to 2(-1) - 3 = -5$. Since $x < -1$,$f(x) < -5$. So,the range for this part is $(-\infty, -5)$.
$2$. For $-1 \leq x \leq 1$,$f(x) = 1 - x^2$. The minimum value is at $x = -1$ or $x = 1$,which is $1 - (1)^2 = 0$. The maximum value is at $x = 0$,which is $1 - 0 = 1$. So,the range for this part is $[0, 1]$.
$3$. For $x > 1$,$f(x) = 3x^2 + 2$. As $x \to 1^+$,$f(x) \to 3(1)^2 + 2 = 5$. Since $x > 1$,$f(x) > 5$. So,the range for this part is $(5, \infty)$.
Combining these,the total range is $(-\infty, -5) \cup [0, 1] \cup (5, \infty)$.

(B) Given $f(x) = \frac{1}{\sqrt{[x]^2+[x]-2}}$.
For $f(x)$ to be defined,we must have $[x]^2+[x]-2 > 0$.
Let $[x] = t$. Then $t^2+t-2 > 0$,which factors as $(t+2)(t-1) > 0$.
This implies $t < -2$ or $t > 1$.
Since $t = [x]$ is an integer,$[x] \in \{\dots, -4, -3\} \cup \{2, 3, 4, \dots\}$.
Case $1$: If $[x] \geq 2$,then $[x]^2+[x]-2$ takes values $2^2+2-2 = 4, 3^2+3-2 = 10, 4^2+4-2 = 18, \dots$.
The function values are $\frac{1}{\sqrt{4}}, \frac{1}{\sqrt{10}}, \dots$,i.e.,$(0, \frac{1}{2}]$.
Case $2$: If $[x] \leq -3$,then $[x]^2+[x]-2$ takes values $(-3)^2-3-2 = 4, (-4)^2-4-2 = 10, \dots$.
The function values are $\frac{1}{\sqrt{4}}, \frac{1}{\sqrt{10}}, \dots$,i.e.,$(0, \frac{1}{2}]$.
Combining both cases,the range is $(0, \frac{1}{2}]$.

(B) Given the function $f(x) = \sqrt{\frac{1-x^2}{1+x^2}}$.
For the domain $D$,we require $\frac{1-x^2}{1+x^2} \geq 0$. Since $1+x^2 > 0$ for all real $x$,we need $1-x^2 \geq 0$,which implies $x^2 \leq 1$,so $x \in [-1, 1]$. Thus,$D = [-1, 1]$.
For the range $G$,let $y = \sqrt{\frac{1-x^2}{1+x^2}}$. Since $x \in [-1, 1]$,$1-x^2$ ranges from $0$ to $1$ and $1+x^2$ ranges from $1$ to $2$. Thus,$\frac{1-x^2}{1+x^2}$ ranges from $0$ to $1$. Taking the square root,$y \in [0, 1]$. Thus,$G = [0, 1]$.
Finally,$D \cap G = [-1, 1] \cap [0, 1] = [0, 1]$.

(B) Given the function $f(x) = |x-2| + |x-3|$.
We analyze the function in three intervals: $x \leq 2$,$2 < x < 3$,and $x \geq 3$.
For $x \leq 2$,$f(x) = -(x-2) - (x-3) = -x + 2 - x + 3 = -2x + 5$. Since $x \leq 2$,$-2x \geq -4$,so $f(x) \geq -4 + 5 = 1$.
For $2 < x < 3$,$f(x) = (x-2) - (x-3) = x - 2 - x + 3 = 1$.
For $x \geq 3$,$f(x) = (x-2) + (x-3) = 2x - 5$. Since $x \geq 3$,$2x \geq 6$,so $f(x) \geq 6 - 5 = 1$.
Combining these,the minimum value of the function is $1$ and it takes all values greater than or equal to $1$.
Thus,the range is $[1, \infty)$.
Therefore,option $(B)$ is correct.

(C) The expression $x-[x]$ represents the fractional part of $x$,denoted as $\{x\}$.
Since $x$ is a real number,the fractional part $\{x\}$ lies in the interval $[0, 1)$.
However,the denominator $\sqrt{x-[x]}$ cannot be zero,so $x-[x] \neq 0$.
Therefore,$x-[x] \in (0, 1)$.
As $x-[x]$ approaches $0$ from the right,$\frac{1}{\sqrt{x-[x]}}$ approaches $\infty$.
As $x-[x]$ approaches $1$ from the left,$\frac{1}{\sqrt{x-[x]}}$ approaches $1$.
Thus,the range of $f(x)$ is $(1, \infty)$.

(B) Given function: $y(x) = \cos x - 3$.
Since the domain of $\cos x$ is the set of all real numbers $R$,the domain of $y(x)$ is $R$.
We know that for any $x \in R$,$-1 \leq \cos x \leq 1$.
Subtracting $3$ from all parts of the inequality:
$-1 - 3 \leq \cos x - 3 \leq 1 - 3$
$-4 \leq y(x) \leq -2$.
Thus,the range of the function is $[-4, -2]$.
Therefore,the domain is $R$ and the range is $[-4, -2]$.

(B) The function is defined as $y(x) = f(x) = \begin{cases} \frac{5x}{(x-3)(x+3)}, & x \neq -1 \\ -1, & x = -1 \end{cases}$.
For $f(x)$ to be a well-defined function $f: A \rightarrow B$,$A$ must be the domain of the function.
The expression $\frac{5x}{(x-3)(x+3)}$ is defined for all real numbers except where the denominator is zero.
Setting the denominator to zero: $(x-3)(x+3) = 0 \implies x = 3$ or $x = -3$.
At $x = -1$,the function is explicitly defined as $y(-1) = -1$,which is a valid real number.
Therefore,the domain $A$ includes all real numbers except $3$ and $-3$.
Thus,$A = R \setminus \{-3, 3\}$.

(A) Given the function $f(x) = -3x - 3$.
To find the domain,we set $f(x)$ equal to each element in the range:
$(i)$ For $f(x) = 3$: $3 = -3x - 3$ $\Rightarrow 6 = -3x$ $\Rightarrow x = -2$.
(ii) For $f(x) = -6$: $-6 = -3x - 3$ $\Rightarrow -3 = -3x$ $\Rightarrow x = 1$.
(iii) For $f(x) = -9$: $-9 = -3x - 3$ $\Rightarrow -6 = -3x$ $\Rightarrow x = 2$.
(iv) For $f(x) = -18$: $-18 = -3x - 3$ $\Rightarrow -15 = -3x$ $\Rightarrow x = 5$.
Thus,the domain of $f$ is $\{-2, 1, 2, 5\}$.
Comparing this with the given options,$-1$ is not in the domain of $f$.

(C) The expression $\frac{2x-1}{x^3+4x^2+3x}$ is defined for all real numbers $x$ except where the denominator is zero.
Set the denominator equal to zero:
$x^3+4x^2+3x = 0$
$x(x^2+4x+3) = 0$
$x(x+1)(x+3) = 0$
Thus,the expression is undefined at $x = 0$,$x = -1$,and $x = -3$.
Therefore,the set is $R - \{0, -1, -3\}$.

(B) Given,$f(x) = \frac{1}{2 - \cos 3x}$.
We know that for any $x \in R$,$-1 \leq \cos 3x \leq 1$.
Multiplying by $-1$,we get $-1 \leq -\cos 3x \leq 1$.
Adding $2$ to all sides,we get $2 - 1 \leq 2 - \cos 3x \leq 2 + 1$,which simplifies to $1 \leq 2 - \cos 3x \leq 3$.
Taking the reciprocal,the inequality sign reverses: $\frac{1}{3} \leq \frac{1}{2 - \cos 3x} \leq \frac{1}{1}$.
Thus,$\frac{1}{3} \leq f(x) \leq 1$.
Therefore,the range of $f$ is $[1/3, 1]$.

(B) Given $f(x) = [2x] - 2[x]$ for all $x \in R$.
Case $1$: If $x$ is an integer,let $x = n$ where $n \in Z$.
Then $f(n) = [2n] - 2[n] = 2n - 2n = 0$.
Case $2$: If $x$ is not an integer,let $x = n + f$ where $n \in Z$ and $0 < f < 1$.
Then $f(x) = [2(n + f)] - 2[n + f] = [2n + 2f] - 2n = 2n + [2f] - 2n = [2f]$.
Since $0 < f < 1$,we have $0 < 2f < 2$.
If $0 < f < 0.5$,then $[2f] = 0$.
If $0.5 \leq f < 1$,then $[2f] = 1$.
Thus,the range of $f(x)$ is $\{0, 1\}$.

(C) Given $f(x) = \cos^2 x + \sin^4 x$.
Using $\cos^2 x = 1 - \sin^2 x$,we get:
$f(x) = 1 - \sin^2 x + \sin^4 x = 1 - \sin^2 x(1 - \sin^2 x) = 1 - \sin^2 x \cos^2 x$.
Multiplying and dividing by $4$,we get:
$f(x) = 1 - \frac{4 \sin^2 x \cos^2 x}{4} = 1 - \frac{(2 \sin x \cos x)^2}{4} = 1 - \frac{\sin^2 2x}{4}$.
Since $0 \leq \sin^2 2x \leq 1$,we have:
$0 \leq \frac{\sin^2 2x}{4} \leq \frac{1}{4}$.
Multiplying by $-1$ and adding $1$:
$1 - 0 \geq 1 - \frac{\sin^2 2x}{4} \geq 1 - \frac{1}{4}$.
$1 \geq f(x) \geq \frac{3}{4}$.
Thus,the range $f(R) = \left[\frac{3}{4}, 1\right]$.