Domain and Range Questions in English

(C) For $f(x) = \sqrt{x}$,the domain is $[0, \infty)$.
For $g(x) = \sqrt{1-x}$,the domain is $(-\infty, 1]$.
The domain of $f+g, f-g,$ and $g-f$ is the intersection of the domains of $f$ and $g$,which is $[0, 1]$.
For $f/g$,we require $g(x) \neq 0$,so $1-x \neq 0 \implies x \neq 1$. The domain is $[0, 1)$.
For $g/f$,we require $f(x) \neq 0$,so $x \neq 0$. The domain is $(0, 1]$.
The common domain for all these functions is the intersection of $[0, 1], [0, 1),$ and $(0, 1]$,which is $(0, 1)$.

(D) Let the argument of the logarithm be $g(x) = 3 + \cos(\frac{3\pi}{4} + x) + \cos(\frac{\pi}{4} + x) + \cos(\frac{\pi}{4} - x) - \cos(\frac{3\pi}{4} - x)$.
Using the sum-to-product formulas $\cos(A+B) + \cos(A-B) = 2\cos A \cos B$ and $\cos(A-B) - \cos(A+B) = 2\sin A \sin B$:
$g(x) = 3 + [\cos(\frac{3\pi}{4} + x) - \cos(\frac{3\pi}{4} - x)] + [\cos(\frac{\pi}{4} + x) + \cos(\frac{\pi}{4} - x)]$
$g(x) = 3 - 2\sin(\frac{3\pi}{4})\sin(x) + 2\cos(\frac{\pi}{4})\cos(x)$
Since $\sin(\frac{3\pi}{4}) = \frac{1}{\sqrt{2}}$ and $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$:
$g(x) = 3 - 2(\frac{1}{\sqrt{2}})\sin(x) + 2(\frac{1}{\sqrt{2}})\cos(x) = 3 + \sqrt{2}(\cos x - \sin x)$.
We know that $-\sqrt{2} \leq \cos x - \sin x \leq \sqrt{2}$.
Thus,$3 + \sqrt{2}(-\sqrt{2}) \leq g(x) \leq 3 + \sqrt{2}(\sqrt{2})$,
which simplifies to $3 - 2 \leq g(x) \leq 3 + 2$,or $1 \leq g(x) \leq 5$.
Since $f(x) = \log_{\sqrt{5}}(g(x))$,the range is $[\log_{\sqrt{5}}(1), \log_{\sqrt{5}}(5)]$.
Since $\log_{\sqrt{5}}(1) = 0$ and $\log_{\sqrt{5}}(5) = \frac{\log_5(5)}{\log_5(\sqrt{5})} = \frac{1}{1/2} = 2$,the range is $[0, 2]$.

(C) For the function $f(x) = \sqrt{\frac{|[x]|-2}{|[x]|-3}}$ to be defined,the expression inside the square root must be non-negative and the denominator must be non-zero.
$\frac{|[x]|-2}{|[x]|-3} \geq 0$ and $|[x]|-3 \neq 0$.
Let $Y = [x]$. We solve $\frac{|Y|-2}{|Y|-3} \geq 0$.
The critical points are $|Y| = 2$ and $|Y| = 3$,which means $Y \in \{-3, -2, 2, 3\}$.
Testing intervals for $|Y|$:
$1$. If $|Y| < 2$,then $\frac{-}{-} > 0$ (True). This corresponds to $-2 < Y < 2$,so $[x] \in \{-1, 0, 1\}$,which implies $x \in [-1, 2)$.
$2$. If $2 \leq |Y| < 3$,then $\frac{+}{-} < 0$ (False).
$3$. If $|Y| > 3$,then $\frac{+}{+} > 0$ (True). This corresponds to $Y > 3$ or $Y < -3$,so $[x] \geq 4$ or $[x] \leq -4$,which implies $x \in [4, \infty)$ or $x \in (-\infty, -3)$.
Combining these,the domain is $(-\infty, -3) \cup [-1, 2) \cup [4, \infty)$.
Comparing with $(-\infty, a) \cup [b, c) \cup [4, \infty)$,we get $a = -3$,$b = -1$,and $c = 2$.
Thus,$a+b+c = -3 + (-1) + 2 = -2$.

(D) For the function $f(x)$ to be defined,the following conditions must be satisfied:
$1$. The argument of $\cos^{-1}$ must be in $[-1, 1]$: $-1 \leq \frac{x^{2}-5x+6}{x^{2}-9} \leq 1$.
Solving $\frac{x^{2}-5x+6}{x^{2}-9} - 1 \leq 0 \implies \frac{(x-2)(x-3)}{(x-3)(x+3)} - 1 \leq 0 \implies \frac{x-2}{x+3} - 1 \leq 0 \implies \frac{-5}{x+3} \leq 0 \implies x+3 > 0 \implies x > -3$.
Solving $\frac{x^{2}-5x+6}{x^{2}-9} + 1 \geq 0 \implies \frac{(x-2)(x-3)}{(x-3)(x+3)} + 1 \geq 0 \implies \frac{x-2}{x+3} + 1 \geq 0 \implies \frac{2x+1}{x+3} \geq 0 \implies x \in (-\infty, -3) \cup [-\frac{1}{2}, \infty)$.
Taking the intersection,we get $x \in [-\frac{1}{2}, \infty)$ (excluding $x=3$ where the expression is undefined).
$2$. The argument of $\log_{e}$ must be positive: $x^{2}-3x+2 > 0 \implies (x-1)(x-2) > 0 \implies x \in (-\infty, 1) \cup (2, \infty)$.
$3$. The denominator cannot be zero: $\log_{e}(x^{2}-3x+2) \neq 0 \implies x^{2}-3x+2 \neq 1 \implies x^{2}-3x+1 \neq 0 \implies x \neq \frac{3 \pm \sqrt{5}}{2}$.
Combining all conditions: $x \in [-\frac{1}{2}, 1) \cup (2, \infty) - \{\frac{3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}\}$.

(C) For the function $f(x) = \sin^{-1}[2x^2 - 3] + \log_2(\log_{1/2}(x^2 - 5x + 5))$ to be defined:
$1.$ For $\sin^{-1}[2x^2 - 3]$,the argument must satisfy $-1 \leq [2x^2 - 3] \leq 1$. Since the range of $\sin^{-1}$ is $[-1, 1]$,we require $-1 \leq [2x^2 - 3] < 1$ (because if $[2x^2 - 3] = 1$,then $2x^2 - 3 \geq 1$,which is allowed,but the log term imposes stricter bounds).
Actually,for $\sin^{-1}(y)$,$y \in [-1, 1]$. Thus,$-1 \leq [2x^2 - 3] \leq 1$. This implies $-1 \leq 2x^2 - 3 < 2$,so $2 \leq 2x^2 < 5$,which gives $1 \leq x^2 < 2.5$. Thus $x \in (-\sqrt{2.5}, -1] \cup [1, \sqrt{2.5})$.
$2.$ For $\log_2(\log_{1/2}(x^2 - 5x + 5))$,we need $\log_{1/2}(x^2 - 5x + 5) > 0$. Since the base $1/2 < 1$,this implies $0 < x^2 - 5x + 5 < (1/2)^0 = 1$.
$3.$ Solving $x^2 - 5x + 5 > 0$: Roots are $\frac{5 \pm \sqrt{25 - 20}}{2} = \frac{5 \pm \sqrt{5}}{2}$. So $x \in (-\infty, \frac{5-\sqrt{5}}{2}) \cup (\frac{5+\sqrt{5}}{2}, \infty)$.
$4.$ Solving $x^2 - 5x + 5 < 1$: $x^2 - 5x + 4 < 0 \Rightarrow (x-1)(x-4) < 0 \Rightarrow x \in (1, 4)$.
Intersection of all conditions:
From $1$: $x \in [1, \sqrt{2.5}) \cup (-\sqrt{2.5}, -1]$.
From $2, 3, 4$: $x \in (1, \frac{5-\sqrt{5}}{2})$.
Combining these,the domain is $(1, \frac{5-\sqrt{5}}{2})$.

(D) Given,$f(x) = \frac{16x^2 - 96x + 153}{x - 3}$.
Let $f(x) = y$.
Then,$y = \frac{16x^2 - 96x + 153}{x - 3}$.
$y(x - 3) = 16x^2 - 96x + 153$.
$16x^2 - (96 + y)x + (153 + 3y) = 0$.
Since $x$ is a real number,the discriminant $D \geq 0$.
$D = (96 + y)^2 - 4(16)(153 + 3y) \geq 0$.
$9216 + 192y + y^2 - 64(153 + 3y) \geq 0$.
$9216 + 192y + y^2 - 9792 - 192y \geq 0$.
$y^2 - 576 \geq 0$.
$y^2 \geq 576$.
This implies $y \in (-\infty, -24] \cup [24, \infty)$.
Thus,the least positive value of $f(x)$ is $24$.

(B) Given $f(x) = \sqrt{|x|} - \log(1 + |x|)$.
Since $f(x)$ is an even function,we consider $g(t) = \sqrt{t} - \log(1 + t)$ for $t \geq 0$,where $t = |x|$.
To check the behavior of $g(t)$,we find its derivative:
$g'(t) = \frac{1}{2\sqrt{t}} - \frac{1}{1 + t} = \frac{1 + t - 2\sqrt{t}}{2\sqrt{t}(1 + t)} = \frac{(\sqrt{t} - 1)^2}{2\sqrt{t}(1 + t)}$.
Since $g'(t) > 0$ for all $t > 0$ $(t \neq 1)$,the function $g(t)$ is strictly increasing for $t \geq 0$.
As $t \to \infty$,$g(t) = \sqrt{t}(1 - \frac{\log(1 + t)}{\sqrt{t}}) \to \infty$. Thus,$f(x)$ is not bounded above,so assertion $I$ is false.
At $t = 0$,$g(0) = 0$. Since $g(t)$ is increasing for $t \geq 0$,the minimum value of $g(t)$ is $g(0) = 0$. Thus,$f(x) \geq 0$ for all $x$. So,assertion $II$ is true.
Therefore,$I$ is false and $II$ is true.

(C) Given the polynomial $P(x) = 4x^3 - 3x$ for $x \in \left(-\frac{1}{2}, \frac{1}{2}\right)$.
To find the range,we analyze the derivative $P'(x) = 12x^2 - 3 = 3(4x^2 - 1) = 3(2x - 1)(2x + 1)$.
For $x \in \left(-\frac{1}{2}, \frac{1}{2}\right)$,we have $4x^2 < 1$,which implies $4x^2 - 1 < 0$.
Thus,$P'(x) < 0$ for all $x$ in the given interval,meaning $P(x)$ is a strictly decreasing function.
Since $P(x)$ is continuous and strictly decreasing,the range is $(P(1/2), P(-1/2))$.
Calculating the values at the boundaries:
$P(1/2) = 4(1/8) - 3(1/2) = 1/2 - 3/2 = -1$.
$P(-1/2) = 4(-1/8) - 3(-1/2) = -1/2 + 3/2 = 1$.
Since the interval is open,the range is $(-1, 1)$.

(A) Given $f(x) = \sqrt{2-x-x^2}$ and $g(x) = \cos x$.
For $f(x)$ to be defined,$2-x-x^2 \geq 0 \Rightarrow x^2+x-2 \leq 0 \Rightarrow (x+2)(x-1) \leq 0 \Rightarrow x \in [-2, 1]$.
For $f(g(x))$,we need $g(x) \in [-2, 1]$. Since $-1 \leq \cos x \leq 1$,this is always true for all $x \in \mathbb{R}$. Thus,$\text{Domain of } f(g(x)) = \mathbb{R}$.
For $f((g(x))^2)$,we need $(g(x))^2 \in [-2, 1]$. Since $0 \leq \cos^2 x \leq 1$,this is always true for all $x \in \mathbb{R}$. Thus,$\text{Domain of } f((g(x))^2) = \mathbb{R}$.
Therefore,$\text{Domain of } f((g(x))^2) = \text{Domain of } f(g(x))$,so statement $I$ is true.
For $g(f(x))$,we need $f(x)$ to be defined,so $x \in [-2, 1]$. Thus,$\text{Domain of } g(f(x)) = [-2, 1]$.
Since $\text{Domain of } f(g(x)) = \mathbb{R}$ and $\text{Domain of } g(f(x)) = [-2, 1]$,statements $II$ and $III$ are false.
For $g((f(x))^3)$,we need $f(x)$ to be defined,so $x \in [-2, 1]$. Thus,$\text{Domain of } g((f(x))^3) = [-2, 1]$. This is not equal to $\mathbb{R}$,so statement $IV$ is false.
Hence,only statement $I$ is true.

(B) The function $f(x) = \sqrt{4 - \sqrt{2x + 5}}$ is defined if the expressions under the square roots are non-negative.
First,for the inner square root: $2x + 5 \geq 0 \implies x \geq -\frac{5}{2}$.
Second,for the outer square root: $4 - \sqrt{2x + 5} \geq 0 \implies 4 \geq \sqrt{2x + 5}$.
Squaring both sides (since both are non-negative): $16 \geq 2x + 5 \implies 11 \geq 2x \implies x \leq \frac{11}{2}$.
Combining these conditions,the domain of $f(x)$ is $x \in \left[ -\frac{5}{2}, \frac{11}{2} \right]$.
The mid-point of the domain is calculated as $\frac{-\frac{5}{2} + \frac{11}{2}}{2} = \frac{\frac{6}{2}}{2} = \frac{3}{2}$.

(C) The function $f(x) = \log_{\{x\}}[x] + \log_{[x]}\{x\}$ is defined under the following conditions:
$1$. For $\log_{\{x\}}[x]$ to be defined:
Base $\{x\} > 0$ and $\{x\} \neq 1$. Since $\{x\} = x - [x]$,$\{x\} \in [0, 1)$. Thus,$\{x\} \in (0, 1)$.
Argument $[x] > 0$,which implies $x \geq 1$. Since $\{x\} \neq 0$,$x$ cannot be an integer.
$2$. For $\log_{[x]}\{x\}$ to be defined:
Base $[x] > 0$ and $[x] \neq 1$. This implies $[x] \geq 2$,so $x \geq 2$.
Argument $\{x\} > 0$,which implies $x$ is not an integer.
Combining these conditions,we require $x \geq 2$ and $x \notin \mathbb{Z}$.
Among the given options,the interval $(2, 3)$ satisfies $x > 2$ and $x$ is not an integer.
Therefore,the correct option is $(c)$.

(D) Let $y = f(x) = (\sin x)^{\sin x}$ for $x \in (0, \pi)$.
Taking the natural logarithm on both sides,we get $\ln y = \sin x \ln(\sin x)$.
Let $u = \sin x$. Since $x \in (0, \pi)$,the range of $u$ is $(0, 1]$.
We define $g(u) = u^{\ln u}$ for $u \in (0, 1]$.
To find the range,we differentiate $g(u)$ with respect to $u$:
$g'(u) = u^u (1 + \ln u)$.
Setting $g'(u) = 0$,we get $1 + \ln u = 0$,which implies $\ln u = -1$,so $u = e^{-1} = \frac{1}{e}$.
At $u = \frac{1}{e}$,$g(\frac{1}{e}) = (e^{-1})^{e^{-1}} = e^{-1/e}$.
As $u \to 0^+$,$g(u) = u^u \to 1$ (using the limit $\lim_{u \to 0^+} u^u = 1$).
At $u = 1$,$g(1) = 1^1 = 1$.
Since $e^{-1/e} \approx 0.692$,the minimum value is $e^{-1/e}$ and the maximum value is $1$.
Thus,the range is $[e^{-1/e}, 1]$.

(A) We know that $-\sqrt{2} \leq \sin x - \cos x \leq \sqrt{2}$.
Multiplying by $\sqrt{2}$,we get $-2 \leq \sqrt{2}(\sin x - \cos x) \leq 2$.
Let $k = \sqrt{2}(\sin x - \cos x)$,so $-2 \leq k \leq 2$.
The function is $f(x) = \log_{\sqrt{m}}(k + m - 2)$.
Given the range of $f$ is $[0, 2]$,we have $0 \leq \log_{\sqrt{m}}(k + m - 2) \leq 2$.
This implies $(\sqrt{m})^0 \leq k + m - 2 \leq (\sqrt{m})^2$,which simplifies to $1 \leq k + m - 2 \leq m$.
Solving for $k$,we get $3 - m \leq k \leq 2$.
Comparing this with $-2 \leq k \leq 2$,we equate the lower bounds: $3 - m = -2$.
Thus,$m = 5$.

(B) For the function $f(x) = \frac{\log_{(x+1)}(x-2)}{x^2 - 2x - 3}$ to be defined:
$1$. The argument of the logarithm must be positive: $x - 2 > 0 \Rightarrow x > 2$.
$2$. The base of the logarithm must be positive and not equal to $1$: $x + 1 > 0 \Rightarrow x > -1$ and $x + 1 \neq 1 \Rightarrow x \neq 0$.
$3$. The denominator must not be zero: $x^2 - 2x - 3 \neq 0$.
Factoring the denominator: $(x - 3)(x + 1) \neq 0$,which implies $x \neq 3$ and $x \neq -1$.
Combining all conditions: $x > 2$ and $x \neq 3$.
Thus,the domain is $(2, \infty) - \{3\}$.

(A) Let $y = \sqrt{3-x} + \sqrt{2+x}$. The domain is determined by $3-x \ge 0$ and $2+x \ge 0$,which gives $-2 \le x \le 3$.
Squaring both sides,we get $y^2 = (3-x) + (2+x) + 2\sqrt{(3-x)(2+x)} = 5 + 2\sqrt{6+x-x^2}$.
To find the range,we analyze the expression $g(x) = 6+x-x^2$. This is a downward-opening parabola with vertex at $x = -\frac{b}{2a} = -\frac{1}{2(-1)} = \frac{1}{2}$.
The maximum value of $g(x)$ is $g(1/2) = 6 + 1/2 - 1/4 = 6 + 1/4 = 25/4$.
The minimum value of $g(x)$ occurs at the endpoints of the domain,$x = -2$ or $x = 3$. $g(-2) = 6-2-4 = 0$ and $g(3) = 6+3-9 = 0$.
Thus,$0 \le g(x) \le 25/4$.
Substituting this into $y^2 = 5 + 2\sqrt{g(x)}$,we get $5 + 2\sqrt{0} \le y^2 \le 5 + 2\sqrt{25/4}$.
$5 \le y^2 \le 5 + 2(5/2) = 10$.
Since $y \ge 0$,the range is $[\sqrt{5}, \sqrt{10}]$.

(D) Given $f(x) = \frac{[x]}{1+x^2}$ for $x \in (2, 6)$.
Since $[x]$ is the greatest integer function,we analyze the intervals:
For $x \in [2, 3)$,$[x] = 2$,so $f(x) = \frac{2}{1+x^2}$. As $x$ increases from $2$ to $3$,$f(x)$ decreases from $\frac{2}{1+2^2} = \frac{2}{5}$ to $\frac{2}{1+3^2} = \frac{2}{10} = \frac{1}{5}$. Range: $(\frac{1}{5}, \frac{2}{5}]$.
For $x \in [3, 4)$,$[x] = 3$,so $f(x) = \frac{3}{1+x^2}$. Range: $(\frac{3}{17}, \frac{3}{10}]$.
For $x \in [4, 5)$,$[x] = 4$,so $f(x) = \frac{4}{1+x^2}$. Range: $(\frac{4}{26}, \frac{4}{17}] = (\frac{2}{13}, \frac{4}{17}]$.
For $x \in [5, 6)$,$[x] = 5$,so $f(x) = \frac{5}{1+x^2}$. Range: $(\frac{5}{37}, \frac{5}{26}]$.
The union of these ranges is $(\frac{5}{37}, \frac{2}{5}]$. Note that the function is strictly decreasing in each interval $[n, n+1)$,and the values at the right endpoints are excluded. The set of values is the union of these intervals,which simplifies to $(\frac{5}{37}, \frac{2}{5}]$.

(A) Let $y = \frac{x^2+2x+1}{x^2-8x+12}$.
By cross-multiplying,we get:
$y(x^2-8x+12) = x^2+2x+1$
$yx^2 - 8xy + 12y = x^2 + 2x + 1$
$x^2(y-1) - x(8y+2) + (12y-1) = 0$.
Case $1$: If $y \neq 1$,for $x$ to be real,the discriminant $D \geq 0$.
$D = (-(8y+2))^2 - 4(y-1)(12y-1) \geq 0$
$(64y^2 + 32y + 4) - 4(12y^2 - 13y + 1) \geq 0$
$64y^2 + 32y + 4 - 48y^2 + 52y - 4 \geq 0$
$16y^2 + 84y \geq 0$
$4y(4y + 21) \geq 0$.
This gives $y \in (-\infty, -\frac{21}{4}] \cup [0, \infty)$.
Since $y \neq 1$ was assumed,we check if $y=1$ is possible.
Case $2$: If $y = 1$,then $x^2+2x+1 = x^2-8x+12$,which simplifies to $10x = 11$,or $x = \frac{11}{10}$.
Since $x = \frac{11}{10}$ is in the domain $R - \{2, 6\}$,$y=1$ is included in the range.
Thus,the range is $(-\infty, -\frac{21}{4}] \cup [0, \infty)$.

(A) The function is given by $f(x) = \frac{1}{\sqrt{\lceil x \rceil - x}}$.
For the function to be defined,the expression inside the square root must be strictly positive: $\lceil x \rceil - x > 0$,which implies $\lceil x \rceil > x$.
This inequality holds for all $x \notin \mathbb{Z}$. If $x \in \mathbb{Z}$,then $\lceil x \rceil = x$,so $\lceil x \rceil - x = 0$,making the denominator zero.
Thus,the domain $A = \mathbb{R} - \mathbb{Z}$.
For $x \notin \mathbb{Z}$,we know that $\lceil x \rceil = \lfloor x \rfloor + 1$. Therefore,$\lceil x \rceil - x = \lfloor x \rfloor + 1 - x = 1 - (x - \lfloor x \rfloor) = 1 - \{x\}$,where $\{x\}$ is the fractional part of $x$.
Since $x \notin \mathbb{Z}$,$0 < \{x\} < 1$,which implies $0 < 1 - \{x\} < 1$.
Then,$0 < \sqrt{1 - \{x\}} < 1$,and consequently $f(x) = \frac{1}{\sqrt{1 - \{x\}}} > 1$.
Thus,the range $B = (1, \infty)$.
Now,$A \cap B = (\mathbb{R} - \mathbb{Z}) \cap (1, \infty) = (1, \infty) - \mathbb{Z}$. Since the intersection with $(1, \infty)$ only includes positive integers,$(1, \infty) - \mathbb{Z} = (1, \infty) - \mathbb{N}$. Thus,$(S1)$ is true.
$A \cup B = (\mathbb{R} - \mathbb{Z}) \cup (1, \infty) = \mathbb{R} - \{0, -1, -2, \dots\}$. This is not equal to $(1, \infty)$. Thus,$(S2)$ is false.
Therefore,only $(S1)$ is true.

(A) For the function to be defined,both parts must be valid.
$1$. For $\log_e\left(\frac{6x^2+5x+1}{2x-1}\right)$,we need $\frac{6x^2+5x+1}{2x-1} > 0$.
$\frac{(3x+1)(2x+1)}{2x-1} > 0$.
Using the wavy curve method,the critical points are $x = -1/2, -1/3, 1/2$.
The inequality holds for $x \in (-1/2, -1/3) \cup (1/2, \infty) \dots (A)$.
$2$. For $\cos^{-1}\left(\frac{2x^2-3x+4}{3x-5}\right)$,we need $-1 \le \frac{2x^2-3x+4}{3x-5} \le 1$ and $3x-5 \neq 0$.
Solving $\frac{2x^2-3x+4}{3x-5} \le 1 \implies \frac{2x^2-6x+9}{3x-5} \le 0$. Since $2x^2-6x+9$ has $D < 0$,it is always positive. Thus,$3x-5 < 0 \implies x < 5/3 \dots (B)$.
Solving $\frac{2x^2-3x+4}{3x-5} \ge -1 \implies \frac{2x^2-1}{3x-5} \ge 0$.
Critical points are $x = -1/\sqrt{2}, 1/\sqrt{2}, 5/3$.
The inequality holds for $x \in [-1/\sqrt{2}, 1/\sqrt{2}] \cup (5/3, \infty) \dots (C)$.
Intersection $A \cap B \cap C = (-1/2, -1/3) \cup (1/2, 1/\sqrt{2}]$.
Here $\alpha = -1/2, \beta = -1/3, \gamma = 1/2, \delta = 1/\sqrt{2}$.
$18(\alpha^2+\beta^2+\gamma^2+\delta^2) = 18(1/4 + 1/9 + 1/4 + 1/2) = 18(1/2 + 1/9 + 1/2) = 18(1 + 1/9) = 18 + 2 = 20$.

(C) For the function $f(x) = \frac{1}{\sqrt{[x]^2 - 3[x] - 10}}$ to be defined,the expression inside the square root must be strictly positive:
$[x]^2 - 3[x] - 10 > 0$
Let $t = [x]$. Then $t^2 - 3t - 10 > 0$.
Factoring the quadratic: $(t - 5)(t + 2) > 0$.
This inequality holds when $t < -2$ or $t > 5$.
Since $t = [x]$,we have $[x] < -2$ or $[x] > 5$.
If $[x] < -2$,then $[x] \leq -3$,which implies $x < -2$.
If $[x] > 5$,then $[x] \geq 6$,which implies $x \geq 6$.
Thus,the domain is $(-\infty, -2) \cup [6, \infty)$.

(B) For the function $f(x)$ to be defined,we require the argument of $\sin^{-1}$ to be in $[-1, 1]$ and the base of the logarithm to be positive and not equal to $1$.
First,$\frac{6+2 \log_3 x}{-5x} > 0$ and $x > 0, x \neq \frac{1}{3}$. Since $x > 0$,we need $6+2 \log_3 x < 0$,so $\log_3 x < -3$,which means $x < 3^{-3} = \frac{1}{27}$. Thus,$x \in (0, \frac{1}{27})$.
Next,$-1 \leq \log_{3x} \left(\frac{6+2 \log_3 x}{-5x}\right) \leq 1$. Since $x < \frac{1}{27}$,$3x < \frac{1}{9} < 1$,so the inequality reverses when removing the logarithm: $(3x)^1 \leq \frac{6+2 \log_3 x}{-5x} \leq (3x)^{-1}$.
Solving $15x^2 + 6 + 2 \log_3 x \geq 0$ and $6 + 2 \log_3 x + \frac{5}{3} \geq 0$ leads to $x \in [3^{-23/6}, \frac{1}{27})$.
Thus,the domain $D = [3^{-23/6}, \frac{1}{27})$.
Since $3^{-23/6} < x < \frac{1}{27}$,we have $[x] = 0$,so $g(x) = x$. The range is $(3^{-23/6}, \frac{1}{27}) = (\alpha, \beta)$.
Then $\alpha = 3^{-23/6}$ and $\beta = \frac{1}{27}$.
Calculating $\alpha^2 + \frac{5}{\beta} = (3^{-23/6})^2 + 5(27) = 3^{-23/3} + 135$. Since $3^{-23/3}$ is very small,the value is approximately $135$.

(B) The domain of $f(x)$ is the intersection of the domains of its three components.
$(i)$ For $\log_e(4x^2 + 11x + 6)$,we require $4x^2 + 11x + 6 > 0$.
Factoring gives $(4x + 3)(x + 2) > 0$,so $x \in (-\infty, -2) \cup (-\frac{3}{4}, \infty)$.
(ii) For $\sin^{-1}(4x + 3)$,we require $-1 \le 4x + 3 \le 1$.
$-4 \le 4x \le -2$,which implies $x \in [-1, -\frac{1}{2}]$.
(iii) For $\cos^{-1}\left(\frac{10x + 6}{3}\right)$,we require $-1 \le \frac{10x + 6}{3} \le 1$.
$-3 \le 10x + 6 \le 3$,so $-9 \le 10x \le -3$,which implies $x \in [-\frac{9}{10}, -\frac{3}{10}]$.
Taking the intersection of all three intervals:
$x \in ((-\infty, -2) \cup (-\frac{3}{4}, \infty)) \cap [-1, -\frac{1}{2}] \cap [-\frac{9}{10}, -\frac{3}{10}]$.
Intersection of $[-1, -\frac{1}{2}]$ and $[-\frac{9}{10}, -\frac{3}{10}]$ is $[-\frac{9}{10}, -\frac{1}{2}]$.
Now,intersecting this with $(-\infty, -2) \cup (-\frac{3}{4}, \infty)$:
Since $-\frac{9}{10} < -\frac{3}{4}$,the intersection is $(-\frac{3}{4}, -\frac{1}{2}]$.
Thus,$\alpha = -\frac{3}{4}$ and $\beta = -\frac{1}{2}$.
Then $|\alpha + \beta| = |-\frac{3}{4} - \frac{1}{2}| = |-\frac{5}{4}| = \frac{5}{4}$.
Finally,$36|\alpha + \beta| = 36 \times \frac{5}{4} = 9 \times 5 = 45$.

(C) For the function $f(x) = \cos^{-1}\left(\frac{2-|x|}{4}\right) + (\log_e(3-x))^{-1}$,we need to satisfy the domain conditions for both parts.
$1$. For $\cos^{-1}\left(\frac{2-|x|}{4}\right)$,the argument must be in $[-1, 1]$:
$-1 \leq \frac{2-|x|}{4} \leq 1$
$-4 \leq 2-|x| \leq 4$
$-6 \leq -|x| \leq 2$
Since $|x| \geq 0$,we have $-|x| \leq 2$ always true. Thus,$|x| \leq 6$,which implies $x \in [-6, 6]$.
$2$. For $(\log_e(3-x))^{-1}$,we need $\log_e(3-x) \neq 0$ and $3-x > 0$:
$3-x > 0 \Rightarrow x < 3$.
$\log_e(3-x) \neq 0 \Rightarrow 3-x \neq 1 \Rightarrow x \neq 2$.
Combining these conditions:
$x \in [-6, 6] \cap (-\infty, 3) \setminus \{2\} = [-6, 3) \setminus \{2\}$.
Comparing this with $[-\alpha, \beta) \setminus \{\gamma\}$,we get $\alpha = 6$,$\beta = 3$,and $\gamma = 2$.
Therefore,$\alpha + \beta + \gamma = 6 + 3 + 2 = 11$.

(B) For the function to be defined,we need:
$1$) $\frac{2x+3}{4x^2+x-3} > 0$
$2$) $-1 \leq \frac{2x-1}{x+2} \leq 1$
Step $1$: Solve $\frac{2x+3}{(4x-3)(x+1)} > 0$.
The critical points are $x = -3/2, -1, 3/4$. Using the wavy curve method,the solution is $x \in (-3/2, -1) \cup (3/4, \infty)$.
Step $2$: Solve $-1 \leq \frac{2x-1}{x+2} \leq 1$.
Part $A$: $\frac{2x-1}{x+2} + 1 \geq 0 \implies \frac{3x+1}{x+2} \geq 0$. Solution: $x \in (-\infty, -2) \cup [-1/3, \infty)$.
Part $B$: $\frac{2x-1}{x+2} - 1 \leq 0 \implies \frac{x-3}{x+2} \leq 0$. Solution: $x \in (-2, 3]$.
Intersection of Part $A$ and Part $B$: $x \in [-1/3, 3]$.
Step $3$: Find the intersection of Step $1$ and Step $2$.
$((-3/2, -1) \cup (3/4, \infty)) \cap [-1/3, 3] = (3/4, 3]$.
Thus,$\alpha = 3/4$ and $\beta = 3$.
Step $4$: Calculate $5\beta - 4\alpha$.
$5(3) - 4(3/4) = 15 - 3 = 12$.

(C) The function is $f(x) = \frac{\sqrt{x^2-25}}{4-x^2} + \log_{10}(x^2+2x-15)$.
For the square root term,$x^2-25 \geq 0 \Rightarrow x \in (-\infty, -5] \cup [5, \infty)$.
For the denominator,$4-x^2 \neq 0 \Rightarrow x \neq \pm 2$.
For the logarithm,$x^2+2x-15 > 0$ $\Rightarrow (x+5)(x-3) > 0$ $\Rightarrow x \in (-\infty, -5) \cup (3, \infty)$.
Taking the intersection of all conditions:
$x \in ((-\infty, -5] \cup [5, \infty)) \cap (-\infty, -5) \cup (3, \infty) \cap \mathbb{R} \setminus \{-2, 2\}$.
This simplifies to $x \in (-\infty, -5) \cup [5, \infty)$.
Comparing with $(-\infty, \alpha) \cup [\beta, \infty)$,we get $\alpha = -5$ and $\beta = 5$.
Therefore,$\alpha^2 + \beta^3 = (-5)^2 + 5^3 = 25 + 125 = 150$.

(A) For $\sin^{-1}(u)$,we need $-1 \leq u \leq 1$. So,$-1 \leq \frac{3x-22}{2x-19} \leq 1$. Solving this inequality gives $x \in (5, 8.2]$.
For $\log_e(v)$,we need $v > 0$. So,$\frac{3x^2-8x+5}{x^2-3x-10} > 0$,which simplifies to $\frac{(3x-5)(x-1)}{(x-5)(x+2)} > 0$. The solution is $x \in (-\infty, -2) \cup (1, 5/3) \cup (5, \infty)$.
Taking the intersection of both conditions,the domain is $(5, 8.2]$,which is $(5, 41/5]$.
Here,$\alpha = 5$ and $\beta = 41/5$.
Thus,$3\alpha + 10\beta = 3(5) + 10(41/5) = 15 + 82 = 97$.

(D) We know that for any real $x$,the range of $\sin 5x$ is $[-1, 1]$.
Therefore,$-\sin 5x$ also lies in the interval $[-1, 1]$.
Adding $7$ to all parts,we get $7 - \sin 5x \in [7 - 1, 7 + 1]$,which simplifies to $7 - \sin 5x \in [6, 8]$.
Taking the reciprocal,the range of $f(x) = \frac{1}{7 - \sin 5x}$ is $\left[\frac{1}{8}, \frac{1}{6}\right]$.

(A) We know that the range of $g(x) = \sin 3x + \cos 3x$ is $[-\sqrt{2}, \sqrt{2}]$.
Thus,the denominator $2 + \sin 3x + \cos 3x$ ranges from $[2 - \sqrt{2}, 2 + \sqrt{2}]$.
Therefore,the range of $f(x) = \frac{1}{2 + \sin 3x + \cos 3x}$ is $[a, b] = \left[\frac{1}{2 + \sqrt{2}}, \frac{1}{2 - \sqrt{2}}\right]$.
Rationalizing the endpoints: $a = \frac{2 - \sqrt{2}}{4 - 2} = \frac{2 - \sqrt{2}}{2} = 1 - \frac{1}{\sqrt{2}}$ and $b = \frac{2 + \sqrt{2}}{4 - 2} = \frac{2 + \sqrt{2}}{2} = 1 + \frac{1}{\sqrt{2}}$.
Given $\alpha = \frac{a + b}{2}$ and $\beta = \sqrt{ab}$,we need to find $\frac{\alpha}{\beta} = \frac{a + b}{2\sqrt{ab}}$.
$a + b = \frac{2 - \sqrt{2} + 2 + \sqrt{2}}{2} = 2$.
$ab = \left(\frac{2 - \sqrt{2}}{2}\right) \left(\frac{2 + \sqrt{2}}{2}\right) = \frac{4 - 2}{4} = \frac{2}{4} = \frac{1}{2}$.
$\frac{\alpha}{\beta} = \frac{2}{2\sqrt{1/2}} = \frac{1}{1/\sqrt{2}} = \sqrt{2}$.

(C) Given $f(x) = \frac{(x-1)(x-5)}{(x-2)(x-3)}$.
$(A)$ For $-1 < x < 1$,$f(x) > 0$ and $f(x) < 1$. Thus,$f(x)$ satisfies $(p, r, s)$.
$(B)$ For $1 < x < 2$,$f(x) < 0$ and $f(x) < 1$. Thus,$f(x)$ satisfies $(q, s)$.
$(C)$ For $3 < x < 5$,$f(x) < 0$ and $f(x) < 1$. Thus,$f(x)$ satisfies $(q, s)$.
$(D)$ For $x > 5$,$f(x) > 0$ and $f(x) < 1$. Thus,$f(x)$ satisfies $(p, r, s)$.

(A) Since $f(x)$ is onto,$A$ is the range of $f(x)$.
$f'(x) = 6x^2 - 30x + 36 = 6(x-2)(x-3)$.
Critical points are $x=2$ and $x=3$.
Evaluating at boundaries and critical points:
$f(0) = 7$
$f(2) = 2(8) - 15(4) + 36(2) + 7 = 16 - 60 + 72 + 7 = 35$
$f(3) = 2(27) - 15(9) + 36(3) + 7 = 54 - 135 + 108 + 7 = 34$.
Since $f(x)$ is continuous on $[0,3]$,the range $A = [7, 35]$.
For $g(x) = \frac{x^{2025}}{x^{2025}+1}$,as $x \in [0, \infty)$,$x^{2025} \in [0, \infty)$.
Thus,$g(x) = 1 - \frac{1}{x^{2025}+1}$.
At $x=0$,$g(0) = 0$. As $x \to \infty$,$g(x) \to 1$.
So,the range $B = [0, 1)$.
$S = \{x \in \mathbb{Z} : x \in [7, 35] \cup [0, 1)\} = \{0, 7, 8, 9, \dots, 35\}$.
The number of elements in $S$ is $1 + (35 - 7 + 1) = 1 + 29 = 30$.

(A) For $f_1(x) = \log _5(18 x-x^2-77)$,the domain requires $18 x-x^2-77 > 0$.
$x^2-18 x+77 < 0 \implies (x-7)(x-11) < 0$.
Thus,$x \in (7, 11)$,so $\alpha = 7$ and $\beta = 11$.
For $f_2(x) = \log _{(x-1)}\left(\frac{2 x^2+3 x-2}{x^2-3 x-4}\right)$,the domain requires:
$1) \ x-1 > 0 \implies x > 1$
$2) \ x-1 \neq 1 \implies x \neq 2$
$3) \ \frac{2 x^2+3 x-2}{x^2-3 x-4} > 0 \implies \frac{(2 x-1)(x+2)}{(x-4)(x+1)} > 0$.
Using the sign scheme for $\frac{(2 x-1)(x+2)}{(x-4)(x+1)} > 0$,the solution is $x \in (-\infty, -2) \cup (-1, 1/2) \cup (4, \infty)$.
Taking the intersection with $x > 1$ and $x \neq 2$,we get $x \in (4, \infty)$.
Thus,$\gamma = 4$ and $\delta = \infty$.
Finally,$\alpha^2+\beta^2+\gamma^2 = 7^2+11^2+4^2 = 49+121+16 = 186$.

(A) For the function to be defined,the expressions inside the square roots in the denominator must be strictly positive.
$1) \ x + |x| > 0$
If $x \leq 0$,then $x + |x| = 0$,which makes the denominator zero. Thus,we must have $x > 0$.
So,$x \in (0, \infty)$.
$2) \ 10 + 3x - x^2 > 0$
Multiplying by $-1$,we get $x^2 - 3x - 10 < 0$.
$(x - 5)(x + 2) < 0$.
This inequality holds for $x \in (-2, 5)$.
Taking the intersection of $x \in (0, \infty)$ and $x \in (-2, 5)$,we get the domain $(a, b) = (0, 5)$.
Therefore,$a = 0$ and $b = 5$.
Calculating the required value: $(1+a)^2 + b^2 = (1+0)^2 + 5^2 = 1 + 25 = 26$.

(B) The given function is $f(x) = \log_e\left(\frac{2x-3}{5+4x}\right) + \sin^{-1}\left(\frac{4+3x}{2-x}\right)$.
For the function to be defined,we need:
$1) \frac{2x-3}{5+4x} > 0$
$2) -1 \leq \frac{4+3x}{2-x} \leq 1$
Solving condition $(1)$:
$\frac{2x-3}{5+4x} > 0 \Rightarrow x \in \left(-\infty, -\frac{5}{4}\right) \cup \left(\frac{3}{2}, \infty\right)$.
Solving condition $(2)$:
$-1 \leq \frac{4+3x}{2-x} \leq 1$
$\Rightarrow \frac{4+3x}{2-x} + 1 \geq 0$ and $\frac{4+3x}{2-x} - 1 \leq 0$
$\Rightarrow \frac{4+3x+2-x}{2-x} \geq 0$ and $\frac{4+3x-2+x}{2-x} \leq 0$
$\Rightarrow \frac{6+2x}{2-x} \geq 0$ and $\frac{2+4x}{2-x} \leq 0$
$\Rightarrow \frac{x+3}{x-2} \leq 0$ and $\frac{x+1/2}{x-2} \geq 0$
From $\frac{x+3}{x-2} \leq 0$,we get $x \in [-3, 2)$.
From $\frac{x+1/2}{x-2} \geq 0$,we get $x \in (-\infty, -1/2] \cup (2, \infty)$.
Intersection of these gives $x \in [-3, -1/2]$.
Now,taking the intersection with condition $(1)$:
$x \in ([-3, -1/2]) \cap ((-\infty, -5/4) \cup (3/2, \infty)) = [-3, -5/4)$.
Thus,$\alpha = -3$ and $\beta = -5/4$.
Therefore,$\alpha^2 + 4\beta = (-3)^2 + 4(-5/4) = 9 - 5 = 4$.

(A) For the function $f(x) = \log_7(1 - \log_4(x^2 - 9x + 18))$ to be defined,we must have $1 - \log_4(x^2 - 9x + 18) > 0$ and $x^2 - 9x + 18 > 0$.
Step $1$: Solve $x^2 - 9x + 18 > 0$.
$(x - 3)(x - 6) > 0 \implies x \in (-\infty, 3) \cup (6, \infty)$.
Step $2$: Solve $1 - \log_4(x^2 - 9x + 18) > 0$.
$\log_4(x^2 - 9x + 18) < 1 \implies x^2 - 9x + 18 < 4^1$.
$x^2 - 9x + 14 < 0 \implies (x - 2)(x - 7) < 0 \implies x \in (2, 7)$.
Step $3$: Find the intersection of the two conditions.
$x \in ((-\infty, 3) \cup (6, \infty)) \cap (2, 7) = (2, 3) \cup (6, 7)$.
Thus,$(\alpha, \beta) = (2, 3)$ and $(\gamma, \delta) = (6, 7)$.
Therefore,$\alpha + \beta + \gamma + \delta = 2 + 3 + 6 + 7 = 18$.

(A) For $f(x)$,we require $\log_3 \log_7(8 - \log_2(x^2 + 4x + 5)) > 0$.
This implies $\log_7(8 - \log_2(x^2 + 4x + 5)) > 1$,so $8 - \log_2(x^2 + 4x + 5) > 7$.
Thus,$\log_2(x^2 + 4x + 5) < 1$,which means $x^2 + 4x + 5 < 2$,or $x^2 + 4x + 3 < 0$.
Factoring gives $(x + 3)(x + 1) < 0$,so $x \in (-3, -1)$. Thus,$\alpha = -3$ and $\beta = -1$.
For $g(x)$,we require $-1 \leq \frac{7x + 10}{x - 2} \leq 1$.
Solving $\frac{7x + 10}{x - 2} \leq 1$ $\Rightarrow \frac{7x + 10 - x + 2}{x - 2} \leq 0$ $\Rightarrow \frac{6x + 12}{x - 2} \leq 0$ $\Rightarrow \frac{x + 2}{x - 2} \leq 0$,so $x \in [-2, 2)$.
Solving $\frac{7x + 10}{x - 2} \geq -1$ $\Rightarrow \frac{7x + 10 + x - 2}{x - 2} \geq 0$ $\Rightarrow \frac{8x + 8}{x - 2} \geq 0$ $\Rightarrow \frac{x + 1}{x - 2} \geq 0$,so $x \in (-\infty, -1] \cup (2, \infty)$.
The intersection is $x \in [-2, -1]$. Thus,$\gamma = -2$ and $\delta = -1$.
Finally,$\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (-3)^2 + (-1)^2 + (-2)^2 + (-1)^2 = 9 + 1 + 4 + 1 = 15$.

(D) Let $y = \frac{5-x}{x^2-3x+2}$.
$y(x^2-3x+2) = 5-x$
$yx^2 - 3xy + 2y = 5-x$
$yx^2 + (1-3y)x + (2y-5) = 0$.
If $y=0$,then $x=5$,which is a valid value.
If $y \neq 0$,for $x$ to be real,the discriminant $D \geq 0$.
$D = (1-3y)^2 - 4(y)(2y-5) \geq 0$
$1 + 9y^2 - 6y - 8y^2 + 20y \geq 0$
$y^2 + 14y + 1 \geq 0$.
Solving $y^2 + 14y + 1 = 0$ using the quadratic formula:
$y = \frac{-14 \pm \sqrt{196-4}}{2} = \frac{-14 \pm \sqrt{192}}{2} = \frac{-14 \pm 8\sqrt{3}}{2} = -7 \pm 4\sqrt{3}$.
Thus,$y \in (-\infty, -7-4\sqrt{3}] \cup [-7+4\sqrt{3}, \infty)$.
Here,$\alpha = -7-4\sqrt{3}$ and $\beta = -7+4\sqrt{3}$.
$\alpha^2 + \beta^2 = (-7-4\sqrt{3})^2 + (-7+4\sqrt{3})^2$
$= (49 + 48 + 56\sqrt{3}) + (49 + 48 - 56\sqrt{3})$
$= 97 + 97 = 194$.

(B) For $f(x) = \cos^{-1}\left(\frac{4x+5}{3x-7}\right)$,we require $-1 \leq \frac{4x+5}{3x-7} \leq 1$.
Case $1$: $\frac{4x+5}{3x-7} \geq -1$ $\Rightarrow \frac{4x+5+3x-7}{3x-7} \geq 0$ $\Rightarrow \frac{7x-2}{3x-7} \geq 0$. This gives $x \in (-\infty, 2/7] \cup (7/3, \infty)$.
Case $2$: $\frac{4x+5}{3x-7} \leq 1$ $\Rightarrow \frac{4x+5-3x+7}{3x-7} \leq 0$ $\Rightarrow \frac{x+12}{3x-7} \leq 0$. This gives $x \in [-12, 7/3)$.
Taking the intersection,the domain of $f(x)$ is $[-12, 2/7]$. Thus,$\alpha = -12$ and $\beta = 2/7$.
For $g(x) = \log_2(2-6\log_{27}(2x+5))$,we require $2-6\log_{27}(2x+5) > 0$ and $2x+5 > 0$.
$6\log_{27}(2x+5) < 2$ $\Rightarrow \log_{27}(2x+5) < 1/3$ $\Rightarrow 2x+5 < (27)^{1/3} = 3$ $\Rightarrow x < -1$.
Also,$2x+5 > 0 \Rightarrow x > -5/2$.
So,the domain of $g(x)$ is $(-5/2, -1)$. Thus,$\gamma = -5/2$ and $\delta = -1$.
Now,$|7(\alpha+\beta) + 4(\gamma+\delta)| = |7(-12 + 2/7) + 4(-5/2 - 1)| = |-84 + 2 - 10 - 4| = |-96| = 96$.

(B) Given the equation $[x]^2-5[x]+6=0$. Let $[x] = y$.
Then $y^2-5y+6=0$.
Factoring the quadratic equation,we get $(y-2)(y-3)=0$.
So,$[x]=2$ or $[x]=3$.
If $[x]=2$,then $2 \le x < 3$.
If $[x]=3$,then $3 \le x < 4$.
Combining these intervals,we get $x \in [2, 4)$.

(D) Given,$f(x) = \frac{\log_2(x+3)}{x^2+3x+2}$.
For the function to be defined,the denominator must not be zero:
$x^2 + 3x + 2 \neq 0$
$(x+1)(x+2) \neq 0$
$x \neq -1$ and $x \neq -2$.
Also,the argument of the logarithm must be positive:
$x + 3 > 0$
$x > -3$.
Combining these conditions,the domain is $x \in (-3, \infty) - \{-1, -2\}$.

(D) Given the equation $2^x + 2^y = 2$.
Rearranging for $2^y$,we get $2^y = 2 - 2^x$.
Taking the logarithm base $2$ on both sides,$y = \log_2(2 - 2^x)$.
For the function to be defined,the argument of the logarithm must be strictly positive: $2 - 2^x > 0$.
This implies $2^x < 2$.
Since $2 = 2^1$,we have $2^x < 2^1$.
Since the base $2 > 1$,the inequality holds for $x < 1$.
Thus,the domain is $(-\infty, 1)$,which is written as $-\infty < x < 1$.

(D) Given the equation $2^x + 2^y = 2$.
Rearranging for $2^y$,we get $2^y = 2 - 2^x$.
Since $2^y > 0$ for all real $y$,the expression $2 - 2^x$ must be strictly greater than $0$.
$2 - 2^x > 0$
$2^x < 2$
$2^x < 2^1$
Since the base $2 > 1$,the inequality holds for the exponents:
$x < 1$.
Thus,the domain of the function is $(-\infty, 1)$,which is written as $-\infty < x < 1$.

(D) Given equation: $[x]^2-5[x]+6=0$
Let $[x] = a$.
Then the equation becomes $a^2-5a+6=0$.
Factoring the quadratic equation: $(a-2)(a-3)=0$.
This gives $a=2$ or $a=3$.
Substituting back $[x]=a$,we get $[x]=2$ or $[x]=3$.
For $[x]=2$,the range of $x$ is $x \in [2,3)$.
For $[x]=3$,the range of $x$ is $x \in [3,4)$.
Combining these intervals,we get $x \in [2,3) \cup [3,4) = [2,4)$.

(D) Given the equation $2^x+2^y=2$.
We can rewrite this as $2^y = 2 - 2^x$.
For $y$ to be defined,the argument of the logarithm must be positive: $2 - 2^x > 0$.
This implies $2^x < 2$.
Since the base $2 > 1$,the inequality holds when $x < 1$.
Thus,the domain is $(-\infty, 1)$ or $-\infty < x < 1$.

(D) Given the equation $2^x + 2^y = 2$.
Rearranging for $2^y$,we get $2^y = 2 - 2^x$.
Since $2^y > 0$ for all real $y$,the expression $2 - 2^x$ must be strictly greater than $0$.
$2 - 2^x > 0 \Rightarrow 2^x < 2$.
Since $2 = 2^1$,we have $2^x < 2^1$.
Since the base $2 > 1$,the inequality holds for $x < 1$.
Thus,the domain is $(-\infty, 1)$,which is written as $-\infty < x < 1$.

(A) Given the equation: $2[2x - 5] - 1 = 7$ \\
Using the property $[x + n] = [x] + n$ for integer $n$,we have $[2x - 5] = [2x] - 5$ \\
Substituting this into the equation: $2([2x] - 5) - 1 = 7$ \\
$2[2x] - 10 - 1 = 7$ \\
$2[2x] - 11 = 7$ \\
$2[2x] = 18$ \\
$[2x] = 9$ \\
By the definition of the greatest integer function,$[y] = n \Rightarrow n \leq y < n + 1$ \\
So,$9 \leq 2x < 10$ \\
Dividing by $2$,we get $\frac{9}{2} \leq x < 5$ \\
Thus,$x \in \left[\frac{9}{2}, 5\right)$

(D) The function $f(x) = e^{|x| \sin x}$ is defined for all real values of $x$.
Hence,the domain of $f(x)$ is $\mathbb{R}$.
Since $|x| \sin x$ can take any real value from $-\infty$ to $\infty$,the expression $e^{|x| \sin x}$ will always be strictly greater than $0$.
Therefore,the range of $f(x)$ is $(0, \infty)$.

(B) For the function $f(x) = \frac{1}{4-x^2} + \log_{10}(x^3-x)$ to be defined:
$1$. The denominator 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 positive: $x^3 - x > 0$.
Factorizing the expression: $x(x^2 - 1) > 0 \implies x(x-1)(x+1) > 0$.
Using the wavy curve method (sign scheme),the inequality $x(x-1)(x+1) > 0$ holds for $x \in (-1, 0) \cup (1, \infty)$.
Combining the conditions $x \neq \pm 2$ and $x \in (-1, 0) \cup (1, \infty)$,we exclude $x = 2$ from the interval $(1, \infty)$.
Thus,the domain is $x \in (-1, 0) \cup (1, 2) \cup (2, \infty)$.

(C) The function is defined as $f(x) = \frac{1}{\sqrt{x+|x|}}$.
For the function to be defined,the expression inside the square root must be strictly greater than zero:
$x + |x| > 0$.
Case $1$: If $x > 0$,then $|x| = x$. Substituting this,we get $x + x = 2x > 0$,which implies $x > 0$.
Case $2$: If $x < 0$,then $|x| = -x$. Substituting this,we get $x + (-x) = 0$. Since $0$ is not greater than $0$,this case yields no solution.
Case $3$: If $x = 0$,then $x + |x| = 0 + 0 = 0$. Since the denominator cannot be zero,$x = 0$ is not in the domain.
Therefore,the domain of the function is $(0, \infty)$.

(B) The function $f(x) = \sqrt{x-1} + \sqrt{6-x}$ is defined if and only if the expressions under the square roots are non-negative.
For $\sqrt{x-1}$ to be defined,$x - 1 \geq 0$,which implies $x \geq 1$.
For $\sqrt{6-x}$ to be defined,$6 - x \geq 0$,which implies $x \leq 6$.
Combining these two conditions,we get $1 \leq x \leq 6$.
Therefore,the domain of $f(x)$ is $[1, 6]$.

(B) For the function $f(x) = \log _{10}(x^2-5x+6)$ to be defined,the argument of the logarithm must be strictly positive.
$x^2-5x+6 > 0$
Factorizing the quadratic expression:
$(x-2)(x-3) > 0$
Using the wavy curve method or sign scheme,the expression is positive when $x$ lies outside the roots $2$ and $3$.
Thus,$x < 2$ or $x > 3$.
Therefore,the domain is $x \in (-\infty, 2) \cup (3, \infty)$.