Domain and Range Questions in English

(B) The function $f(x) = \frac{x+2}{9-x^{2}}$ is defined for all real values of $x$ except where the denominator is zero.
Setting the denominator to zero: $9 - x^{2} = 0$.
This implies $x^{2} = 9$,so $x = \pm 3$.
Thus,the function is undefined at $x = 3$ and $x = -3$.
Therefore,the domain of the function is the set of all real numbers except $\{-3, 3\}$,which is written as $R - \{-3, 3\}$.

(C) For the function $f(y) = \frac{\cos^{-1}(y-5)}{\sqrt{25-y^2}}$ to be defined,the following conditions must be satisfied:
$1$. The argument of $\cos^{-1}$ must be in the interval $[-1, 1]$,so $-1 \leq y-5 \leq 1$. Adding $5$ to all sides gives $4 \leq y \leq 6$.
$2$. The denominator must be non-zero and the expression inside the square root must be positive,so $25-y^2 > 0$,which implies $y^2 < 25$,or $-5 < y < 5$.
$3$. Taking the intersection of the two conditions: $y \in [4, 6]$ and $y \in (-5, 5)$.
$4$. The intersection is $4 \leq y < 5$,which can be written as the interval $[4, 5)$.
Therefore,the domain is $[4, 5)$.

(C) For $f(x) = \sqrt{x}$ to be defined in the set of real numbers,the expression under the square root must be non-negative.
Therefore,$x \geq 0$.
The domain is the set of all non-negative real numbers,which is represented as $[0, \infty)$ or $R^{+} \cup \{0\}$.
Thus,the correct option is $C$.

(D) Given the relation $R = \{(x, y) : y = x + \frac{6}{x}, x, y \in N, x < 6\}$.
We test each value of $x \in \{1, 2, 3, 4, 5\}$ to see if $y$ is a natural number $(N)$:
For $x = 1, y = 1 + \frac{6}{1} = 7 \in N$.
For $x = 2, y = 2 + \frac{6}{2} = 2 + 3 = 5 \in N$.
For $x = 3, y = 3 + \frac{6}{3} = 3 + 2 = 5 \in N$.
For $x = 4, y = 4 + \frac{6}{4} = 4 + 1.5 = 5.5 \notin N$.
For $x = 5, y = 5 + \frac{6}{5} = 5 + 1.2 = 6.2 \notin N$.
Thus,the relation $R = \{(1, 7), (2, 5), (3, 5)\}$.
The domain is the set of first elements: $\{1, 2, 3\}$.
The range is the set of second elements: $\{5, 7\}$.

(B) For the function $f(x) = \sqrt{\frac{x-2}{3-x}}$ to be defined,the expression under the square root must be non-negative: $\frac{x-2}{3-x} \geq 0$.
This implies that the numerator and denominator must have opposite signs or the numerator must be zero.
Specifically,$x-2 \geq 0$ and $3-x > 0$.
From $x-2 \geq 0$,we get $x \geq 2$.
From $3-x > 0$,we get $x < 3$.
Combining these two conditions,we get $2 \leq x < 3$.
Thus,the domain of $f(x)$ is $[2, 3)$.

(C) Let $y = \frac{x^2+x+2}{x^2+x+1}$.
$y(x^2+x+1) = x^2+x+2$
$(y-1)x^2 + (y-1)x + (y-2) = 0$.
For $x \in R$,the discriminant $D \geq 0$.
$D = (y-1)^2 - 4(y-1)(y-2) \geq 0$.
$(y-1)[(y-1) - 4(y-2)] \geq 0$.
$(y-1)(-3y+7) \geq 0$.
$(y-1)(3y-7) \leq 0$.
Thus,$1 < y \leq \frac{7}{3}$.
Since $x^2+x+1 = (x+1/2)^2 + 3/4 > 0$,$y$ can never be $1$ because the numerator $x^2+x+2$ is always greater than the denominator $x^2+x+1$.
Therefore,the range is $\left(1, \frac{7}{3}\right]$.

(B) Let $y = \frac{x^2}{x^2+1}$.
$y(x^2+1) = x^2$
$yx^2 + y = x^2$
$x^2(y-1) = -y$
$x^2 = \frac{y}{1-y}$.
For $x$ to be a real number,$x^2 \geq 0$.
Therefore,$\frac{y}{1-y} \geq 0$.
This implies $y(1-y) \geq 0$ and $y \neq 1$.
Solving the inequality $y(y-1) \leq 0$,we get $0 \leq y < 1$.
Thus,the range of the function is $[0, 1)$.

(C) Given the set $A = \{10, 11, 12, 14, 26\}$.
We find the prime factorization for each element in $A$:
$10 = 2 \times 5$,so the highest prime factor is $5$.
$11 = 11$,so the highest prime factor is $11$.
$12 = 2^2 \times 3$,so the highest prime factor is $3$.
$14 = 2 \times 7$,so the highest prime factor is $7$.
$26 = 2 \times 13$,so the highest prime factor is $13$.
Thus,the range of $f$ is the set of all these highest prime factors: $\{3, 5, 7, 11, 13\}$.
Therefore,the correct option is $C$.

(A) Given the function $f(x) = 3 + 2^x + 4^x$.
Since $2^x > 0$ for all $x \in \mathbb{R}$,it follows that $4^x = (2^x)^2 > 0$.
Let $t = 2^x$,where $t \in (0, \infty)$.
Then the function becomes $g(t) = 3 + t + t^2$.
Since $t > 0$,the minimum value of $t + t^2$ approaches $0$ as $t \to 0^+$.
As $t \to \infty$,$t + t^2 \to \infty$.
Therefore,the range of $t + t^2$ for $t > 0$ is $(0, \infty)$.
Adding $3$ to this range,the range of $f(x)$ is $(3 + 0, 3 + \infty) = (3, \infty)$.

(D) Let $y = \frac{x-3}{5-x}$.
$y(5-x) = x-3$
$5y - xy = x - 3$
$5y + 3 = x + xy$
$5y + 3 = x(1+y)$
$x = \frac{5y+3}{1+y}$.
For $x$ to be defined,the denominator $1+y \neq 0$,which means $y \neq -1$.
Thus,the range of the function is $R - \{-1\}$.
Therefore,the correct option is $D$.

(C) Given the function $f(x) = \frac{x^2-4}{x-2}$.
Since the domain is $R-\{2\}$,we can simplify the expression for $x \neq 2$:
$f(x) = \frac{(x-2)(x+2)}{x-2} = x+2$.
As $x$ can take any real value except $2$,the value of $x+2$ can take any real value except $2+2 = 4$.
Therefore,the range of the function is $R-\{4\}$.

(A) For the function $f(x)$ to be surjective,the codomain $A$ must be equal to the range of $f(x)$.
Let $y = \frac{x^2}{1-x^2}$.
Rearranging for $x^2$,we get $y(1-x^2) = x^2$,which implies $y - yx^2 = x^2$,so $y = x^2(1+y)$.
Thus,$x^2 = \frac{y}{1+y}$.
Since $x^2 \geq 0$ and $x \neq \pm 1$,we must have $\frac{y}{1+y} \geq 0$ and $x^2 \neq 1$ (which implies $\frac{y}{1+y} \neq 1$,so $y \neq 1+y$,which is always true for any finite $y$).
The inequality $\frac{y}{1+y} \geq 0$ holds when $y \in (-\infty, -1) \cup [0, \infty)$.
Therefore,the range of $f(x)$ is $(-\infty, -1) \cup [0, \infty)$,which can be written as $R - [-1, 0)$.
Hence,$A = R - [-1, 0)$.

(C) Given the equation: $[x]^{2} - 5[x] + 6 = 0$.
Let $t = [x]$. Then the equation becomes $t^{2} - 5t + 6 = 0$.
Factoring the quadratic equation: $(t - 3)(t - 2) = 0$.
This gives $t = 2$ or $t = 3$.
Thus,$[x] = 2$ or $[x] = 3$.
By the definition of the greatest integer function:
If $[x] = 2$,then $x \in [2, 3)$.
If $[x] = 3$,then $x \in [3, 4)$.
Combining these intervals,we get $x \in [2, 3) \cup [3, 4) = [2, 4)$.
Therefore,the correct option is $C$.

(C) 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 $[x] = 3$.
By the definition of the greatest integer function:
If $[x] = 2$,then $x \in [2, 3)$.
If $[x] = 3$,then $x \in [3, 4)$.
Combining these intervals,we get $x \in [2, 3) \cup [3, 4) = [2, 4)$.
Thus,the correct option is $C$.

(C) 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 $x \in (2, \infty)$,we have $x > 2$.
Subtracting $2$ from both sides,we get $x - 2 > 0$.
Squaring both sides,we get $(x - 2)^2 > 0$.
Adding $1$ to both sides,we get $(x - 2)^2 + 1 > 1$.
Therefore,$f(x) > 1$.
Thus,the range of the function $f$ is $(1, \infty)$.

(B) The function $f(x) = \sqrt{\cos x}$ is defined only when $\cos x \geq 0$.
In the interval $[0, 2\pi]$,$\cos x$ is non-negative in the first quadrant $[0, \frac{\pi}{2}]$ and the fourth quadrant $[\frac{3\pi}{2}, 2\pi]$.
Thus,the domain for $n=0$ is $[0, \frac{\pi}{2}] \cup [\frac{3\pi}{2}, 2\pi]$.

(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:
$f(x) = (x^{2} - 4x + 4) + 1$
$f(x) = (x - 2)^{2} + 1$
Since the domain is $x \in [2, \infty)$,we have $x - 2 \geq 0$.
Therefore,$(x - 2)^{2} \geq 0$.
Adding $1$ to both sides,we get $(x - 2)^{2} + 1 \geq 1$.
Thus,$f(x) \geq 1$.
The range of the function $f$ is $[1, \infty)$.

(B) For the function $f(x) = \frac{1}{\sqrt{(x - 2)(x - 5)}}$ to be defined,the expression inside the square root must be strictly greater than zero.
$\Rightarrow (x - 2)(x - 5) > 0$
The critical points are $x = 2$ and $x = 5$.
Testing the intervals $(-\infty, 2)$,$(2, 5)$,and $(5, \infty)$:
For $x < 2$,$(x - 2)(x - 5) > 0$.
For $2 < x < 5$,$(x - 2)(x - 5) < 0$.
For $x > 5$,$(x - 2)(x - 5) > 0$.
Thus,the domain is $x \in (-\infty, 2) \cup (5, \infty)$.

(D) For the function $f(x) = \frac{1}{\log_{10}(1-x)} + \sqrt{x+2}$ to be defined:
$1$. The term inside the square root must be non-negative: $x + 2 \geq 0 \implies x \geq -2$.
$2$. The argument of the logarithm must be positive: $1 - x > 0 \implies x < 1$.
$3$. The denominator cannot be zero: $\log_{10}(1-x) \neq 0 \implies 1 - x \neq 10^0 \implies 1 - x \neq 1 \implies x \neq 0$.
Combining these conditions: $x \geq -2$,$x < 1$,and $x \neq 0$.
Thus,the domain is $x \in [-2, 0) \cup (0, 1)$.

(D) Given function,$f(x) = \frac{x}{1-|x|}$.
For the function to be defined,the denominator must not be zero.
$1 - |x| \neq 0$
$|x| \neq 1$
$x \neq 1$ and $x \neq -1$.
Therefore,the domain is all real numbers except $1$ and $-1$,which is written as $R - \{-1, 1\}$.

(A) For the function $f(x) = \frac{1}{\sqrt{[x]^2 - [x] - 6}}$ to be defined,the expression inside the square root must be strictly positive:
$[x]^2 - [x] - 6 > 0$
Factoring the quadratic expression:
$([x] - 3)([x] + 2) > 0$
This inequality holds when:
$[x] > 3$ or $[x] < -2$
If $[x] > 3$,then the smallest integer value for $[x]$ is $4$,which implies $x \geq 4$,or $x \in [4, \infty)$.
If $[x] < -2$,then the largest integer value for $[x]$ is $-3$,which implies $x < -2$,or $x \in (-\infty, -2)$.
Combining these,the domain is $x \in (-\infty, -2) \cup [4, \infty)$.

(C) For the function $f(x) = \sqrt{x^{2}-7x+12}$ to be defined,the expression inside the square root must be non-negative:
$x^{2}-7x+12 \geq 0$
Factorizing the quadratic expression:
$(x-4)(x-3) \geq 0$
Using the sign scheme method for the inequality,the expression is non-negative when $x \leq 3$ or $x \geq 4$.
Thus,the domain is $x \in (-\infty, 3] \cup [4, \infty)$.

(D) Given the function $f(x) = \sin [x]$ for $-\frac{\pi}{4} < x < \frac{\pi}{4}$.
We know that $\frac{\pi}{4} \approx 0.785$ and $-\frac{\pi}{4} \approx -0.785$.
The interval for $x$ is $(-0.785, 0.785)$.
For $x \in [0, 0.785)$,the greatest integer value $[x] = 0$. Thus,$f(x) = \sin(0) = 0$.
For $x \in (-0.785, 0)$,the greatest integer value $[x] = -1$. Thus,$f(x) = \sin(-1) = -\sin(1)$.
Therefore,the range of the function is $\{0, -\sin 1\}$.

(D) Given the function $f: R \rightarrow R$ defined by $f(x) = \frac{1}{x}$.
For a function to be defined on the domain $R$,every element in $R$ must have a corresponding image in the codomain.
At $x = 0$,the expression $f(0) = \frac{1}{0}$ is undefined in the set of real numbers $R$.
Since $0 \in R$ and $f(0)$ does not exist,the function $f$ is not well-defined on the domain $R$.
Therefore,$f$ is not defined.

(B) Given the function $f(x) = \sqrt{9 - x^2}$.
For the function to be defined,the expression inside the square root must be non-negative:
$9 - x^2 \geq 0 \Rightarrow x^2 \leq 9 \Rightarrow -3 \leq x \leq 3$.
Thus,the domain of the function is $x \in [-3, 3]$.
When $x = 0$,$f(0) = \sqrt{9 - 0} = 3$.
When $x = \pm 3$,$f(\pm 3) = \sqrt{9 - 9} = 0$.
Since the square root function always yields non-negative values,the minimum value is $0$ and the maximum value is $3$.
Therefore,the range of the function is $[0, 3]$.

(A) The domain of $(f + g)$ is the intersection of the domains of $f(x)$ and $g(x)$.
For $f(x) = \frac{1}{2} - \tan^{-1}\left(\frac{\pi x}{2}\right)$,the given domain is $-1 < x < 1$.
For $g(x) = \sqrt{3 + 4x - 4x^2}$ to be defined,we must have $3 + 4x - 4x^2 \geq 0$.
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: $\left(-\frac{1}{2}, 1\right)$.

(C) Let $f(x) = ax^2 + bx + c$. The given equation is $f(x) + f(1/x) = f(x)f(1/x)$,which can be rewritten as $f(x)f(1/x) - f(x) - f(1/x) + 1 = 1$,or $(f(x) - 1)(f(1/x) - 1) = 1$.
Let $g(x) = f(x) - 1$. Then $g(x)g(1/x) = 1$.
Since $f(x)$ is a quadratic polynomial,$g(x)$ is also a quadratic polynomial. Let $g(x) = ax^2 + bx + c'$.
Then $(ax^2 + bx + c')(a/x^2 + b/x + c') = 1$.
Comparing coefficients,we find $g(x) = x^2$ or $g(x) = 1/x^2$ (not a polynomial) or $g(x) = -x^2$ or $g(x) = 1$.
Given $f(-1) = 0$,we have $g(-1) = f(-1) - 1 = -1$.
If $g(x) = x^2$,$g(-1) = 1 \neq -1$.
If $g(x) = -x^2$,$g(-1) = -1$.
Thus $f(x) - 1 = -x^2$,which means $f(x) = 1 - x^2$.
The range of $f(x) = 1 - x^2$ is $(-\infty, 1]$.

(B) The given function is $f(x) = \cos^{-1}\left(\frac{x-3}{2}\right) - \log_{10}(4-x)$.
For the logarithmic function $\log_{10}(4-x)$ to be defined,the argument must be strictly positive:
$4 - x > 0 \implies x < 4$.
For the inverse trigonometric function $\cos^{-1}\left(\frac{x-3}{2}\right)$ to be defined,the argument must lie in the interval $[-1, 1]$:
$-1 \leq \frac{x-3}{2} \leq 1$.
Multiplying by $2$,we get $-2 \leq x - 3 \leq 2$.
Adding $3$ to all parts,we get $1 \leq x \leq 5$.
The domain of $f(x)$ is the intersection of the two conditions:
$x < 4$ and $1 \leq x \leq 5$.
Therefore,the domain is $x \in [1, 4)$.

(D) For $f(x)$ to be a real-valued function,the expression inside the square root must be non-negative: $\frac{[x]-1}{[x]^2-[x]-6} \ge 0$.
Let $n = [x]$. Then $\frac{n-1}{n^2-n-6} \ge 0$,which simplifies to $\frac{n-1}{(n-3)(n+2)} \ge 0$.
Using the wavy curve method,the inequality holds for $n \in (-2, 1] \cup (3, \infty)$.
Since $n$ is an integer $(n = [x])$,the possible values for $n$ are $\{-1, 0, 1, 4, 5, 6, \dots\}$.
If $n = -1$,then $-1 \le x < 0$.
If $n = 0$,then $0 \le x < 1$.
If $n = 1$,then $1 \le x < 2$.
Combining these,we get $[-1, 2)$.
If $n \ge 4$,then $x \ge 4$.
Thus,the domain is $[-1, 2) \cup [4, \infty)$.

(C) For the function $f(x) = \log(x^2 - 1) + x \operatorname{coth}^{-1} x$ to be defined:
$1$. The argument of the logarithm must be positive: $x^2 - 1 > 0$,which implies $x^2 > 1$,so $x \in (-\infty, -1) \cup (1, \infty)$.
$2$. The inverse hyperbolic cotangent function $\operatorname{coth}^{-1} x$ is defined for $|x| > 1$,which means $x \in (-\infty, -1) \cup (1, \infty)$.
Combining both conditions,the domain is $x \in (-\infty, -1) \cup (1, \infty)$,which can be written as $R - [-1, 1]$.

(A) For the function $f(x) = \sqrt{\frac{|x|-2}{|x|-3}}$ to be well-defined,the expression inside the square root must be non-negative and the denominator must not be zero.
So,$\frac{|x|-2}{|x|-3} \geq 0$ and $|x|-3 \neq 0$.
Let $t = |x|$,where $t \geq 0$. The inequality becomes $\frac{t-2}{t-3} \geq 0$.
The critical points are $t=2$ and $t=3$.
Testing intervals for $t$:
$1$) If $0 \leq t < 2$,$\frac{t-2}{t-3} > 0$ (e.g.,$t=1$,$\frac{-1}{-2} = 0.5 > 0$).
$2$) If $2 \leq t < 3$,$\frac{t-2}{t-3} \leq 0$ (e.g.,$t=2.5$,$\frac{0.5}{-0.5} = -1 < 0$).
$3$) If $t > 3$,$\frac{t-2}{t-3} > 0$ (e.g.,$t=4$,$\frac{2}{1} = 2 > 0$).
Thus,the solution for $t$ is $0 \leq t \leq 2$ or $t > 3$.
Substituting back $|x| = t$,we get $0 \leq |x| \leq 2$ or $|x| > 3$.
$|x| \leq 2 \implies x \in [-2, 2]$.
$|x| > 3 \implies x \in (-\infty, -3) \cup (3, \infty)$.
Combining these,the domain is $(-\infty, -3) \cup [-2, 2] \cup (3, \infty)$.

(C) For the function $f(x) = \sqrt{\log_e\left(\frac{1}{(x-2)^2}\right)} + \sin^{-1}(x^2-2)$ to be defined:
$1$. The term inside the square root must be non-negative: $\log_e\left(\frac{1}{(x-2)^2}\right) \ge 0$.
This implies $\frac{1}{(x-2)^2} \ge e^0 = 1$,so $(x-2)^2 \le 1$,which means $|x-2| \le 1$,giving $1 \le x \le 3$. Since $x \neq 2$,the domain is $[1, 2) \cup (2, 3]$.
$2$. The term inside $\sin^{-1}$ must be in $[-1, 1]$: $-1 \le x^2-2 \le 1$.
This implies $1 \le x^2 \le 3$,so $x \in [-\sqrt{3}, -1] \cup [1, \sqrt{3}]$.
$3$. Taking the intersection of both conditions: $[1, 2) \cup (2, 3] \cap ([-\sqrt{3}, -1] \cup [1, \sqrt{3}])$.
This results in $[1, \sqrt{3}] \setminus \{2\}$. Since $2$ is not in $[1, \sqrt{3}]$,the domain is $[1, \sqrt{3}]$.

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

(B) Given,$f(x) = \frac{1}{\sqrt{\log_{0.5}(2x - 3)}} + \sqrt{4 - 9x^2}$.
For $f(x)$ to be defined,the following conditions must hold:
$1. \log_{0.5}(2x - 3) > 0$ $\Rightarrow 2x - 3 < (0.5)^0$ $\Rightarrow 2x - 3 < 1$ $\Rightarrow 2x < 4$ $\Rightarrow x < 2$.
$2. 2x - 3 > 0$ $\Rightarrow 2x > 3$ $\Rightarrow x > \frac{3}{2}$.
$3. 4 - 9x^2 \geq 0$ $\Rightarrow 9x^2 \leq 4$ $\Rightarrow x^2 \leq \frac{4}{9}$ $\Rightarrow -\frac{2}{3} \leq x \leq \frac{2}{3}$.
Combining these conditions: $(x < 2) \cap (x > \frac{3}{2}) \cap (-\frac{2}{3} \leq x \leq \frac{2}{3})$.
Since there is no value of $x$ that satisfies $x > \frac{3}{2}$ and $x \leq \frac{2}{3}$ simultaneously,the domain is the null set.

(D) For the function $f(x) = \sqrt{2+x} + \sqrt{3-x}$ to be defined,the expressions under the square roots must be non-negative. \\ $2+x \geq 0 \Rightarrow x \geq -2$ \\ $3-x \geq 0 \Rightarrow x \leq 3$ \\ Combining these two conditions,we get $-2 \leq x \leq 3$. \\ Therefore,the domain is $x \in [-2, 3]$.

(D) For the function $f(x) = \log_2 \log_3 \log_5(x^2 - 5x + 11)$ to be defined,we require:
$1$. $\log_5(x^2 - 5x + 11) > 0$ $\Rightarrow x^2 - 5x + 11 > 5^0 = 1$ $\Rightarrow x^2 - 5x + 10 > 0$. Since the discriminant $D = (-5)^2 - 4(1)(10) = 25 - 40 = -15 < 0$ and the coefficient of $x^2$ is positive,this is true for all $x \in \mathbb{R}$.
$2$. $\log_3 \log_5(x^2 - 5x + 11) > 0$ $\Rightarrow \log_5(x^2 - 5x + 11) > 3^0 = 1$ $\Rightarrow x^2 - 5x + 11 > 5^1 = 5$ $\Rightarrow x^2 - 5x + 6 > 0$.
Factoring the quadratic: $(x - 2)(x - 3) > 0$.
This inequality holds when $x \in (-\infty, 2) \cup (3, \infty)$.

(A) For the function $f(x) = \sqrt{9 - \sqrt{x^2 - 144}}$ to be defined,the expressions under the square roots must be non-negative:
$1$. $x^2 - 144 \geq 0$ $\Rightarrow x^2 \geq 144$ $\Rightarrow |x| \geq 12$.
$2$. $9 - \sqrt{x^2 - 144} \geq 0 \Rightarrow 9 \geq \sqrt{x^2 - 144}$.
Squaring both sides,we get $81 \geq x^2 - 144$ $\Rightarrow x^2 \leq 225$ $\Rightarrow |x| \leq 15$.
Combining both conditions,we have $12 \leq |x| \leq 15$.
This implies $x \in [-15, -12] \cup [12, 15]$.

(B) Given the equation $2^x+2^y=2$.
Since $2^y > 0$ for all real $y$,we must have $2-2^x > 0$.
This implies $2^x < 2$.
Taking the logarithm base $2$ on both sides,we get $x < 1$.
Therefore,the domain of the function is $x \in (-\infty, 1)$.

(B) Given $f(x)=\sqrt{2-x^2}$ and $g(x)=\ln (1-x)$.
The domain of $(f+g)(x)$ is the intersection of the domains of $f(x)$ and $g(x)$.
For $f(x)=\sqrt{2-x^2}$,we require $2-x^2 \geq 0$,which implies $x^2 \leq 2$. Thus,$x \in [-\sqrt{2}, \sqrt{2}]$. So,$D_1 = [-\sqrt{2}, \sqrt{2}]$.
For $g(x)=\ln (1-x)$,we require $1-x > 0$,which implies $x < 1$. So,$D_2 = (-\infty, 1)$.
The domain of $(f+g)(x)$ is $D_1 \cap D_2 = [-\sqrt{2}, \sqrt{2}] \cap (-\infty, 1) = [-\sqrt{2}, 1)$.

(B) For the expression to be a real number,the following conditions must be satisfied:
$1$. The expression inside the square root must be non-negative: $|x|^2 - 2|x| - 8 \geq 0$.
Let $|x| = t$,then $t^2 - 2t - 8 \geq 0 \Rightarrow (t-4)(t+2) \geq 0$.
Since $t = |x| \geq 0$,we have $t \geq 4$,which means $|x| \geq 4$,so $x \in (-\infty, -4] \cup [4, \infty)$.
$2$. The argument of the logarithm must be positive: $2 - x - x^2 > 0 \Rightarrow x^2 + x - 2 < 0$.
$(x+2)(x-1) < 0 \Rightarrow x \in (-2, 1)$.
$3$. The denominator must not be zero: $\log(2 - x - x^2) \neq 0$ $\Rightarrow 2 - x - x^2 \neq 1$ $\Rightarrow x^2 + x - 1 \neq 0$.
$x \neq \frac{-1 \pm \sqrt{5}}{2}$.
Combining the conditions: $x \in ((-\infty, -4] \cup [4, \infty)) \cap (-2, 1)$.
Since there is no intersection between these sets,the solution is the empty set $\phi$.

(A) For $f(x) = \sin \log \left( \frac{\sqrt{4-x^2}}{1-x} \right)$ to be defined,the argument of the logarithm must be strictly positive.
First,the square root $\sqrt{4-x^2}$ is defined when $4-x^2 \geq 0$,which implies $x \in [-2, 2]$.
Second,the expression inside the logarithm must be positive: $\frac{\sqrt{4-x^2}}{1-x} > 0$.
Since $\sqrt{4-x^2} \geq 0$,we must have $1-x > 0$ (i.e.,$x < 1$) and $\sqrt{4-x^2} \neq 0$.
Thus,$x < 1$ and $x \neq \pm 2$.
Combining $x \in [-2, 2]$ with $x < 1$ and $x \neq \pm 2$,we get $x \in (-2, 1)$.

(D) For the function $f(x) = \frac{\log_2(x+3)}{\sqrt{x^2+3x+2}}$ to be defined:
$1$. The argument of the logarithm must be positive: $x+3 > 0 \implies x > -3$.
$2$. The expression inside the square root in the denominator must be strictly positive: $x^2+3x+2 > 0$.
Factoring the quadratic: $(x+2)(x+1) > 0$.
This inequality holds when $x \in (-\infty, -2) \cup (-1, \infty)$.
Combining the conditions $x > -3$ and $x \in (-\infty, -2) \cup (-1, \infty)$,we get the intersection:
$x \in (-3, -2) \cup (-1, \infty)$.

(A) For the function $f(x) = \frac{\sqrt{2-x} + \sqrt{1+x}}{\sqrt{x+3}}$ to be defined,the expressions under the square roots must be non-negative,and the denominator must not be zero.
$1$. For $\sqrt{2-x}$,we require $2-x \geq 0$,which implies $x \leq 2$.
$2$. For $\sqrt{1+x}$,we require $1+x \geq 0$,which implies $x \geq -1$.
$3$. For $\sqrt{x+3}$ in the denominator,we require $x+3 > 0$,which implies $x > -3$.
Combining these conditions: $x \leq 2$,$x \geq -1$,and $x > -3$.
The intersection of these intervals is $[-1, 2]$.

(A) Given the function $f(x) = \frac{\sqrt{\log _{0.5}(x-3)}}{\sqrt{x-1}}$.
For the square root in the numerator to be defined,we must have $\log _{0.5}(x-3) \geq 0$.
Since the base $0.5 < 1$,the inequality reverses: $x-3 \leq (0.5)^0$,which gives $x-3 \leq 1$,so $x \leq 4$.
Also,for the logarithm to be defined,$x-3 > 0$,which implies $x > 3$.
For the denominator to be defined and non-zero,$x-1 > 0$,which implies $x > 1$.
Taking the intersection of all conditions: $(x \leq 4) \cap (x > 3) \cap (x > 1)$,we get $3 < x \leq 4$.
Thus,the domain is $(3, 4]$.

(B) The domain of $f/g$ is the set of all $x$ such that $f(x)$ is defined,$g(x)$ is defined,and $g(x) \neq 0$.
For $f(x) = \sqrt{\frac{x+1}{x+3}}$ to be defined,we need $\frac{x+1}{x+3} \geq 0$. This holds for $x \in (-\infty, -3) \cup [-1, \infty)$.
For $g(x) = \sqrt{\frac{2-x}{x+3}}$ to be defined and $g(x) \neq 0$,we need $\frac{2-x}{x+3} > 0$. Multiplying by $-1$,we get $\frac{x-2}{x+3} < 0$,which holds for $x \in (-3, 2)$.
The domain of $f/g$ is the intersection of these two sets: $((-\infty, -3) \cup [-1, \infty)) \cap (-3, 2) = [-1, 2)$.

(D) Given,$f(x) = \sqrt{\frac{2-|x|}{3-|x|}}$.
For $f(x)$ to be defined,we must have $\frac{2-|x|}{3-|x|} \geq 0$.
Let $t = |x|$,where $t \geq 0$. The inequality becomes $\frac{2-t}{3-t} \geq 0$,which is equivalent to $\frac{t-2}{t-3} \geq 0$.
Using the wavy curve method for $t \geq 0$,the solution for $t$ is $t \in [0, 2] \cup (3, \infty)$.
Now,substitute $t = |x|$ back:
Case $I$: $0 \leq |x| \leq 2 \Rightarrow x \in [-2, 2]$.
Case $II$: $|x| > 3 \Rightarrow x \in (-\infty, -3) \cup (3, \infty)$.
Combining these,the domain is $x \in (-\infty, -3) \cup [-2, 2] \cup (3, \infty)$.

(C) The function is $f(x) = \frac{-5}{4x^2+1} + \sqrt{x^2-4}$.
For the function to be defined,the denominator $4x^2+1$ must not be zero,and the expression inside the square root must be non-negative.
Since $4x^2+1 \geq 1$ for all $x \in R$,the denominator is never zero.
For the square root,we require $x^2 - 4 \geq 0$.
$x^2 \geq 4$
$|x| \geq 2$
This implies $x \in (-\infty, -2] \cup [2, \infty)$.
Thus,the domain is $(-\infty, -2] \cup [2, \infty)$.

(B) For the function $f(x) = \sqrt{\frac{1-|x|}{2-|x|}}$ to be defined,the expression inside the square root must be non-negative and the denominator must not be zero.
$1$. $\frac{1-|x|}{2-|x|} \geq 0$
$2$. $2-|x| \neq 0 \implies |x| \neq 2 \implies x \neq \pm 2$
Let $t = |x|$,where $t \geq 0$. The inequality becomes $\frac{1-t}{2-t} \geq 0$.
Multiplying by $-1$ in numerator and denominator: $\frac{t-1}{t-2} \geq 0$.
Using the wavy curve method for $t \geq 0$,the solution for $t$ is $t \in [0, 1] \cup (2, \infty)$.
Now,substitute $t = |x|$ back:
Case $I$: $0 \leq |x| \leq 1 \implies x \in [-1, 1]$.
Case $II$: $|x| > 2 \implies x \in (-\infty, -2) \cup (2, \infty)$.
Combining these,the domain is $x \in (-\infty, -2) \cup [-1, 1] \cup (2, \infty)$.

(B) Given the function $f(x) = \frac{1}{\sqrt{|x|-x}}$.
For $f(x)$ to be defined,the expression inside the square root must be strictly positive:
$|x| - x > 0$
$|x| > x$
We know that for $x \geq 0$,$|x| = x$,so $x > x$ is false.
For $x < 0$,$|x| = -x$,so $-x > x$,which implies $-2x > 0$,or $x < 0$.
Thus,the function is defined for all $x \in (-\infty, 0)$.
Therefore,the domain of $f(x)$ is $(-\infty, 0)$.

(B) For the function $f(x) = ([x]^2 - [x] - 2)^{-1/2}$ 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 when $[x] < -1$ or $[x] > 2$.
Case $1$: If $[x] < -1$,then $x < -1$.
Case $2$: If $[x] > 2$,then $[x] \geq 3$,which implies $x \geq 3$.
Combining these,the domain is $(-\infty, -1) \cup [3, \infty)$.
This can be written as $R - [-1, 3)$.