Domain and Range Questions in English

(D) Let $f(x) = \frac{x^2 - 1}{x^2 + 1}$.
We can rewrite the function as:
$f(x) = \frac{x^2 + 1 - 2}{x^2 + 1} = 1 - \frac{2}{x^2 + 1}$.
Since $x^2 \ge 0$ for all real numbers $x$,we have $x^2 + 1 \ge 1$.
This implies that $0 < \frac{1}{x^2 + 1} \le 1$.
Multiplying by $2$,we get $0 < \frac{2}{x^2 + 1} \le 2$.
Now,multiplying by $-1$ reverses the inequality:
$-2 \le -\frac{2}{x^2 + 1} < 0$.
Adding $1$ to all parts:
$1 - 2 \le 1 - \frac{2}{x^2 + 1} < 1 + 0$.
$-1 \le f(x) < 1$.
Thus,the minimum value of $f(x)$ is $-1$,which is attained when $x = 0$.

(A) The function is given by $f(x) = \sin x - \sqrt{3} \cos x + 1$.
We know that the expression $a \sin x + b \cos x$ lies in the interval $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 1$ and $b = -\sqrt{3}$.
Thus,the range of $\sin x - \sqrt{3} \cos x$ is $[-\sqrt{1^2 + (-\sqrt{3})^2}, \sqrt{1^2 + (-\sqrt{3})^2}] = [-\sqrt{1+3}, \sqrt{1+3}] = [-2, 2]$.
Adding $1$ to the entire inequality,we get:
$-2 + 1 \le \sin x - \sqrt{3} \cos x + 1 \le 2 + 1$.
$-1 \le f(x) \le 3$.
Since the function $f$ is onto,the codomain $S$ must be equal to the range of the function.
Therefore,$S = [-1, 3]$.

(B) The function is defined as $f(x) = \frac{|x - 3|}{x - 3}$.
For the function to be defined,the denominator must not be zero,so $x - 3 \neq 0$,which implies $x \neq 3$.
Thus,the domain is $R - \{3\}$.
If $x > 3$,then $|x - 3| = x - 3$,so $f(x) = \frac{x - 3}{x - 3} = 1$.
If $x < 3$,then $|x - 3| = -(x - 3)$,so $f(x) = \frac{-(x - 3)}{x - 3} = -1$.
Therefore,the range is $\{-1, 1\}$.
The correct option is $B$.

(D) The greatest integer function is defined as $f(x) = [x]$,where $[x]$ denotes the greatest integer less than or equal to $x$.
By definition,for any real number $x$,the value of $[x]$ is always an integer.
Therefore,the range of the greatest integer function is the set of all integers,denoted by $\mathbb{Z}$.

(B) For the function $f(x) = \frac{\sin^{-1}(3 - x)}{\ln(|x| - 2)}$ to be defined,the following conditions must be met:
$1$. The argument of $\sin^{-1}$ must be in $[-1, 1]$:
$-1 \le 3 - x \le 1$
$-4 \le -x \le -2$
$2 \le x \le 4$
So,the domain of the numerator is $[2, 4]$.
$2$. The argument of $\ln$ must be positive,and the denominator cannot be zero:
$|x| - 2 > 0 \implies |x| > 2 \implies x \in (-\infty, -2) \cup (2, \infty)$.
Also,$\ln(|x| - 2) \neq 0 \implies |x| - 2 \neq 1 \implies |x| \neq 3 \implies x \neq 3, x \neq -3$.
Combining these,the domain of the denominator is $(-\infty, -3) \cup (-3, -2) \cup (2, 3) \cup (3, \infty)$.
$3$. The domain of $f(x)$ is the intersection of these sets:
$[2, 4] \cap ((-\infty, -3) \cup (-3, -2) \cup (2, 3) \cup (3, \infty)) = (2, 3) \cup (3, 4]$.
Thus,the correct option is $B$.

(C) The function $f(x) = \log |x^2 - 9|$ is defined when $|x^2 - 9| > 0$.
Since the absolute value $|x^2 - 9|$ is always non-negative,the condition $|x^2 - 9| > 0$ is satisfied for all $x$ except where $x^2 - 9 = 0$.
Solving $x^2 - 9 = 0$ gives $x^2 = 9$,which implies $x = 3$ or $x = -3$.
Therefore,the function is undefined at $x = 3$ and $x = -3$.
Thus,the domain of the function is $R - \{-3, 3\}$.

(C) The function $f(x) = \log |\log x|$ is defined if the argument of the outer logarithm is strictly positive,i.e.,$|\log x| > 0$,and the argument of the inner logarithm is strictly positive,i.e.,$x > 0$.
For $|\log x| > 0$,we must have $\log x \neq 0$,which implies $x \neq 1$.
Combining the conditions $x > 0$ and $x \neq 1$,we get the domain as $x \in (0, 1) \cup (1, \infty)$.

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

(B) The function $f(x)$ is defined if the following conditions are met:
$1$. The domain of $\sec^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$,which can be written as $R - (-1, 1)$.
$2$. The denominator $\sqrt{x - [x]}$ must be defined and non-zero. The expression inside the square root must be strictly positive: $x - [x] > 0$.
$3$. Since $x - [x] = \{x\}$ (the fractional part of $x$),the condition $x - [x] > 0$ implies that $x$ cannot be an integer. Thus,$x \in R - Z$.
$4$. Combining these two conditions,the function is defined for $x \in (R - (-1, 1)) \cap (R - Z)$.
$5$. This intersection is equivalent to $R - ((-1, 1) \cup \{n \mid n \in Z\})$.

(B) Given the function $f(x) = x^2 - 6x + 7$.
We can rewrite the function by completing the square:
$f(x) = (x^2 - 6x + 9) - 9 + 7$
$f(x) = (x - 3)^2 - 2$
Since $(x - 3)^2 \ge 0$ for all real $x$,the minimum value of $(x - 3)^2$ is $0$.
Therefore,the minimum value of $f(x)$ is $0 - 2 = -2$.
As $x \to \infty$ or $x \to -\infty$,$f(x) \to \infty$.
Thus,the range of the function is $[-2, \infty)$.

(B) For the function $f(x) = \sqrt{\log \frac{1}{|\sin x|}}$ to be defined,the expression inside the square root must be non-negative:
$\log \frac{1}{|\sin x|} \ge 0$
Since $\log 1 = 0$,this implies $\frac{1}{|\sin x|} \ge 1$,which means $|\sin x| \le 1$.
Also,the argument of the logarithm must be strictly positive: $\frac{1}{|\sin x|} > 0$,which is always true for $|\sin x| \neq 0$.
Additionally,the denominator cannot be zero: $|\sin x| \neq 0$.
$|\sin x| \neq 0 \implies \sin x \neq 0 \implies x \neq n\pi$ for any integer $n \in I$.
Thus,the domain is $R - \{ n\pi : n \in I \}$.

(C) For the function $f(x) = \log (\sqrt {x - 4} + \sqrt {6 - x} )$ to be defined,the expression inside the logarithm must be positive,and the expressions inside the square roots must be non-negative.
$1$. For $\sqrt{x-4}$ to be defined,$x - 4 \ge 0$,which implies $x \ge 4$.
$2$. For $\sqrt{6-x}$ to be defined,$6 - x \ge 0$,which implies $x \le 6$.
$3$. The sum $\sqrt{x-4} + \sqrt{6-x}$ is always non-negative. Since the logarithm is defined for positive values,we check if the sum can be zero. The sum is zero only if $x-4=0$ and $6-x=0$,which is impossible ($x=4$ and $x=6$ simultaneously). Thus,the sum is always positive for $x \in [4, 6]$.
Combining these conditions,the domain is $x \in [4, 6]$.

(B) Given the function $f(x) = [\log_{10}(\frac{5x - x^2}{4})]^{1/2}$.
For $f(x)$ to be defined,the expression inside the square root must be non-negative,and the argument of the logarithm must be positive.
First,we require $\log_{10}(\frac{5x - x^2}{4}) \ge 0$.
This implies $\frac{5x - x^2}{4} \ge 10^0$,which simplifies to $\frac{5x - x^2}{4} \ge 1$.
Multiplying by $4$,we get $5x - x^2 \ge 4$,or $x^2 - 5x + 4 \le 0$.
Factoring the quadratic,we have $(x - 1)(x - 4) \le 0$.
This inequality holds for $x \in [1, 4]$.
Since for all $x \in [1, 4]$,the term $\frac{5x - x^2}{4}$ is positive (specifically,it ranges from $1$ to $1.5625$),the logarithm is always defined.
Thus,the domain is $[1, 4]$.

(C) For the function $f(x) = \log_{3+x}(x^2 - 1)$ to be defined,the following conditions must be satisfied:
$1$. The argument of the logarithm must be positive: $x^2 - 1 > 0 \implies x^2 > 1 \implies x \in (-\infty, -1) \cup (1, \infty)$.
$2$. The base of the logarithm must be positive and not equal to $1$: $3 + x > 0 \implies x > -3$ and $3 + x \neq 1 \implies x \neq -2$.
Combining these conditions:
$x > -3$ and $x \neq -2$ and $(x < -1$ or $x > 1)$.
Intersection of these sets:
$(-3, -2) \cup (-2, -1) \cup (1, \infty)$.

(B) For the function $f(x) = \sqrt{\sin 2x}$ to be defined,the expression under the square root must be non-negative:
$\sin 2x \geq 0$.
The sine function is non-negative in the first and second quadrants,i.e.,in the intervals $[2n\pi, 2n\pi + \pi]$.
Therefore,$2n\pi \leq 2x \leq 2n\pi + \pi$.
Dividing by $2$,we get $n\pi \leq x \leq n\pi + \frac{\pi}{2}$.
Thus,the domain is $[n\pi, n\pi + \frac{\pi}{2}]$.

(D) 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: $4 - x^2 \neq 0$ $\Rightarrow x^2 \neq 4$ $\Rightarrow x \neq \pm 2$.
$2$. The argument of the logarithm must be positive: $x^3 - x > 0$.
Factoring the inequality: $x(x^2 - 1) > 0 \Rightarrow x(x - 1)(x + 1) > 0$.
Using the wavy curve method (sign scheme) on the number line with critical points $-1, 0, 1$,the expression is positive in the intervals $(-1, 0) \cup (1, \infty)$.
Combining the conditions: $x \in (-1, 0) \cup (1, \infty)$ and $x \neq \pm 2$.
Since $2$ lies in the interval $(1, \infty)$,we must exclude it.
Thus,the domain is $D = (-1, 0) \cup (1, 2) \cup (2, \infty)$.

(B) For the function $f(x) = \sqrt{2 - 2x - x^2}$ to be defined,the expression under the square root must be non-negative:
$2 - 2x - x^2 \ge 0$
Multiply by $-1$ and reverse the inequality sign:
$x^2 + 2x - 2 \le 0$
Complete the square for the quadratic expression:
$(x^2 + 2x + 1) - 1 - 2 \le 0$
$(x + 1)^2 - 3 \le 0$
$(x + 1)^2 \le 3$
Taking the square root on both sides:
$-\sqrt{3} \le x + 1 \le \sqrt{3}$
Subtract $1$ from all parts:
$-1 - \sqrt{3} \le x \le -1 + \sqrt{3}$
Thus,the domain is $[-1 - \sqrt{3}, -1 + \sqrt{3}]$.

(B) For the function $f(x) = \frac{x - 3}{(x - 1)\sqrt{x^2 - 4}}$ to be defined,the expression inside the square root must be strictly positive and the denominator must not be zero.
$1$. The condition for the square root is $x^2 - 4 > 0$,which implies $x^2 > 4$,so $x \in (-\infty, -2) \cup (2, \infty)$.
$2$. The condition for the denominator is $(x - 1)\sqrt{x^2 - 4} \neq 0$. This means $x - 1 \neq 0$ (so $x \neq 1$) and $\sqrt{x^2 - 4} \neq 0$ (so $x \neq \pm 2$).
$3$. Combining these conditions,we need $x \in (-\infty, -2) \cup (2, \infty)$ and $x \neq 1$. Since $1$ is not in the interval $(-\infty, -2) \cup (2, \infty)$,the domain remains $(-\infty, -2) \cup (2, \infty)$.

(B) For the function $\sqrt{\log \left\{ \frac{5x - x^2}{6} \right\}}$ to be defined,the expression inside the square root must be non-negative: $\log \left\{ \frac{5x - x^2}{6} \right\} \ge 0$.
This implies $\frac{5x - x^2}{6} \ge 10^0$,which simplifies to $\frac{5x - x^2}{6} \ge 1$.
Multiplying by $6$,we get $5x - x^2 \ge 6$,which rearranges to $x^2 - 5x + 6 \le 0$.
Factoring the quadratic,we get $(x - 2)(x - 3) \le 0$.
This inequality holds for $x \in [2, 3]$.
Thus,the domain is $[2, 3]$.

(C) For the function to be defined,the expressions inside the square roots must satisfy specific conditions:
$(i)$ For $\sqrt{2 - x}$ to be defined,$2 - x \ge 0$,which implies $x \le 2$.
(ii) For $\frac{1}{\sqrt{9 - x^2}}$ to be defined,the denominator must be non-zero and the value inside the square root must be positive: $9 - x^2 > 0$.
This implies $x^2 < 9$,which means $-3 < x < 3$.
To find the domain,we take the intersection of both conditions: $x \le 2$ $AND$ $-3 < x < 3$.
The intersection of $(-\infty, 2]$ and $(-3, 3)$ is $(-3, 2]$.
Therefore,the domain is $(-3, 2]$.

(D) For the function $f(x) = \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x}$ to be defined:
$1$. The expression under the square root must be non-negative:
$1 + x \ge 0 \implies x \ge -1$
$1 - x \ge 0 \implies x \le 1$
Combining these,we get $x \in [-1, 1]$.
$2$. The denominator cannot be zero:
$x \neq 0$
Combining both conditions,the domain is $[-1, 1] \setminus \{0\}$.

(D) The function $f(x) = \sqrt{x - x^2} + \sqrt{4 + x} + \sqrt{4 - x}$ is defined if all the expressions under the square roots are non-negative.
$1$. For $\sqrt{4 + x}$,we require $4 + x \ge 0$,which implies $x \ge -4$.
$2$. For $\sqrt{4 - x}$,we require $4 - x \ge 0$,which implies $x \le 4$.
$3$. For $\sqrt{x - x^2}$,we require $x - x^2 \ge 0$,which is $x(1 - x) \ge 0$. This inequality holds when $0 \le x \le 1$.
To find the domain,we take the intersection of these intervals:
$x \in [-4, \infty) \cap (- \infty, 4] \cap [0, 1] = [0, 1]$.
Thus,the domain of the function is $[0, 1]$.

(A) Given the domain is $-1 \le x \le 1$.
Since $f(x) = \frac{1}{1 + e^x}$ is a strictly decreasing function,the range is $[f(1), f(-1)]$.
Calculating the values:
$f(1) = \frac{1}{1 + e^1} = \frac{1}{1 + e}$
$f(-1) = \frac{1}{1 + e^{-1}} = \frac{1}{1 + \frac{1}{e}} = \frac{e}{e + 1}$
Thus,the range is $\left[ \frac{1}{1+e}, \frac{e}{1+e} \right]$.

(C) The function $f(x) = \sqrt{\log({x^2} - 6x + 6)}$ is defined when $\log({x^2} - 6x + 6) \ge 0$.
This implies ${x^2} - 6x + 6 \ge 10^0$,which simplifies to ${x^2} - 6x + 6 \ge 1$.
Rearranging the inequality,we get ${x^2} - 6x + 5 \ge 0$.
Factoring the quadratic expression,we have $(x - 5)(x - 1) \ge 0$.
Solving this inequality,we find that $x \le 1$ or $x \ge 5$.
Thus,the domain of the function is $(-\infty, 1] \cup [5, \infty)$.

(D) For the function $f(x) = \sqrt{1 - \frac{1}{x}}$ to be defined,the expression inside the square root must be non-negative:
$1 - \frac{1}{x} \ge 0$
$\frac{x - 1}{x} \ge 0$
Using the wavy curve method (sign scheme) for the critical points $x = 0$ and $x = 1$:
The expression is positive in the intervals $(-\infty, 0)$ and $[1, \infty)$.
Note that $x = 0$ is excluded because the denominator cannot be zero.
Thus,the domain is $(-\infty, 0) \cup [1, \infty)$.

(D) The given function is $f(x) = \frac{x^2 - 3x + 2}{x^2 + x - 6}$.
Factorizing the numerator: $x^2 - 3x + 2 = (x - 2)(x - 1)$.
Factorizing the denominator: $x^2 + x - 6 = (x + 3)(x - 2)$.
Thus,$f(x) = \frac{(x - 2)(x - 1)}{(x + 3)(x - 2)}$.
The function is defined for all real values of $x$ except where the denominator is zero.
Setting the denominator to zero: $(x + 3)(x - 2) = 0$,which gives $x = -3$ or $x = 2$.
Therefore,the domain of the function is the set of all real numbers except $2$ and $-3$.
Domain = $\{x : x \in R, x \neq 2, x \neq -3\}$.

(A) The function is defined when $x^2 - 1 > 0$.
This implies $x^2 > 1$.
Taking the square root on both sides,we get $|x| > 1$.
This inequality holds when $x < -1$ or $x > 1$.
Therefore,the domain is $( - \infty, -1) \cup (1, \infty)$.

(A) For the function $y = \frac{1}{\sqrt{|x| - x}}$ 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 the inequality becomes $-x > x$,which simplifies to $0 > 2x$,or $x < 0$.
Thus,the domain of the function is $( - \infty, 0)$.

(D) For the function $f(x) = \sqrt{x^2 - 1} + \sqrt{x^2 + 1}$ to be defined,both square root terms must be non-negative.
$1$. For $\sqrt{x^2 - 1}$,we require $x^2 - 1 \ge 0$,which implies $x^2 \ge 1$. This holds when $x \le -1$ or $x \ge 1$.
$2$. For $\sqrt{x^2 + 1}$,we require $x^2 + 1 \ge 0$,which is true for all real numbers $x \in \mathbb{R}$.
Taking the intersection of these two conditions,the domain is $x \in (-\infty, -1] \cup [1, \infty)$,which can also be written as $(-\infty, \infty) - (-1, 1)$.

(A) For the function $f(x) = \exp (\sqrt {5x - 3 - 2{x^2}} )$ to be defined,the expression under the square root must be non-negative:
$5x - 3 - 2{x^2} \ge 0$
Multiply by $-1$ and reverse the inequality sign:
$2{x^2} - 5x + 3 \le 0$
Factor the quadratic expression:
$2{x^2} - 2x - 3x + 3 \le 0$
$2x(x - 1) - 3(x - 1) \le 0$
$(2x - 3)(x - 1) \le 0$
$2(x - 1)(x - 1.5) \le 0$
Using the wavy curve method,the expression is $\le 0$ for $x \in [1, 1.5]$.
Thus,the domain is $\left[ {1, \frac{3}{2}} \right]$.

(A) Given the function $f(x) = \sec \left( \frac{\pi}{4} \cos^2 x \right)$.
We know that for any real $x$,the range of $\cos^2 x$ is $[0, 1]$.
Therefore,the argument of the secant function,$\theta = \frac{\pi}{4} \cos^2 x$,lies in the interval $[0, \frac{\pi}{4}]$.
Since the secant function is strictly increasing on the interval $[0, \frac{\pi}{4}]$,we evaluate the function at the boundaries:
At $\cos^2 x = 0$,$f(x) = \sec(0) = 1$.
At $\cos^2 x = 1$,$f(x) = \sec \left( \frac{\pi}{4} \right) = \sqrt{2}$.
Thus,the range of $f(x)$ is $[1, \sqrt{2}]$.

(C) Let $y = \frac{x^2 + x + 2}{x^2 + x + 1}$.
We can rewrite the function as $y = \frac{(x^2 + x + 1) + 1}{x^2 + x + 1} = 1 + \frac{1}{x^2 + x + 1}$.
Let $g(x) = x^2 + x + 1$. The minimum value of $g(x)$ occurs at $x = -1/2$,which is $g(-1/2) = (-1/2)^2 - 1/2 + 1 = 1/4 - 1/2 + 1 = 3/4$.
Since $x^2 + x + 1$ is a parabola opening upwards,its range is $[3/4, \infty)$.
Therefore,the range of $\frac{1}{x^2 + x + 1}$ is $(0, 1 / (3/4)] = (0, 4/3]$.
Adding $1$ to this range,we get the range of $f(x)$ as $(1 + 0, 1 + 4/3] = (1, 7/3]$.

(B) The function is defined as $f(x) = [x] - x$,where $[x]$ denotes the greatest integer function.
We know that for any real number $x$,$x = [x] + \{x\}$,where $\{x\}$ is the fractional part of $x$.
Therefore,$[x] - x = -\{x\}$.
The fractional part function $\{x\}$ takes values in the interval $[0, 1)$.
Thus,$-\{x\}$ takes values in the interval $(-1, 0]$.
Hence,the range of the function $f(x) = [x] - x$ is $(-1, 0]$.

(B) Given the function $f(x) = \frac{x + 2}{|x + 2|}$.
We know that the absolute value function $|x + 2|$ is defined as:
$|x + 2| = \begin{cases} x + 2, & \text{if } x + 2 > 0 \text{ (i.e., } x > -2) \\ -(x + 2), & \text{if } x + 2 < 0 \text{ (i.e., } x < -2) \end{cases}$
For $x > -2$,$f(x) = \frac{x + 2}{x + 2} = 1$.
For $x < -2$,$f(x) = \frac{x + 2}{-(x + 2)} = -1$.
Note that the function is undefined at $x = -2$ because the denominator becomes zero.
Therefore,the range of the function $f(x)$ is $\{-1, 1\}$.

(B) Let $y = \frac{x^2}{x^2 + 1}$.
Since $x^2 \ge 0$ for all $x \in R$,the numerator is non-negative and the denominator is always greater than the numerator.
As $x \to \pm \infty$,$f(x) \to 1$,but $f(x)$ never reaches $1$ because $x^2 < x^2 + 1$.
At $x = 0$,$f(0) = \frac{0}{0 + 1} = 0$.
Thus,the range of the function is $[0, 1)$.

(B) Let $f(x) = \frac{1}{2 - \sin 3x}$.
We know that for any real $x$,the range of $\sin 3x$ is $[-1, 1]$.
So,$-1 \le \sin 3x \le 1$.
Multiplying by $-1$,we get $-1 \le -\sin 3x \le 1$.
Adding $2$ to all parts,we get $2 - 1 \le 2 - \sin 3x \le 2 + 1$,which simplifies to $1 \le 2 - \sin 3x \le 3$.
Taking the reciprocal,the inequality signs reverse: $\frac{1}{1} \ge \frac{1}{2 - \sin 3x} \ge \frac{1}{3}$.
Thus,$\frac{1}{3} \le f(x) \le 1$.
Therefore,the range of the function is $[\frac{1}{3}, 1]$.

(B) We know the fundamental trigonometric identity: $\sin^2(\theta) + \cos^2(\theta) = 1$ for any real number $\theta$.
In the given function,let $\theta = x^4$.
Since $x^4$ is a real number for all $x \in \mathbb{R}$,we have $f(x) = \sin^2(x^4) + \cos^2(x^4) = 1$.
Since the function value is always $1$ regardless of the input $x$,the range of the function is the singleton set $\{1\}$.

(B) We know that for any real number $x$,the range of $\sin x$ is $[-1, 1]$.
Thus,$-1 \le \sin x \le 1$.
Multiplying by $-7$,we get $-7 \le -7 \sin x \le 7$ (note the inequality signs flip).
Adding $9$ to all parts,we get $9 - 7 \le 9 - 7 \sin x \le 9 + 7$.
This simplifies to $2 \le f(x) \le 16$.
Therefore,the range of the function is $[2, 16]$.

(B) Let $y = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}$.
$y(x^2 + 2x - 7) = x^2 + 34x - 71$
$x^2(y - 1) + x(2y - 34) - 7y + 71 = 0$.
For $x$ to be real,the discriminant $D \ge 0$:
$D = (2y - 34)^2 - 4(y - 1)(-7y + 71) \ge 0$
$4(y - 17)^2 - 4(-7y^2 + 71y + 7y - 71) \ge 0$
$(y^2 - 34y + 289) - (-7y^2 + 78y - 71) \ge 0$
$8y^2 - 112y + 360 \ge 0$
$y^2 - 14y + 45 \ge 0$
$(y - 5)(y - 9) \ge 0$.
Thus,$y \le 5$ or $y \ge 9$.
The range is $( - \infty, 5] \cup [9, \infty)$.

(C) Given $f(x) = \cos^2 x + \sin^4 x$.
We can rewrite this as $f(x) = \cos^2 x + \sin^2 x \cdot \sin^2 x$.
Since $\sin^2 x = 1 - \cos^2 x$,we have $f(x) = \cos^2 x + \sin^2 x(1 - \cos^2 x)$.
$f(x) = \cos^2 x + \sin^2 x - \sin^2 x \cos^2 x$.
Since $\cos^2 x + \sin^2 x = 1$,we get $f(x) = 1 - \sin^2 x \cos^2 x$.
Multiplying and dividing by $4$,we get $f(x) = 1 - \frac{1}{4}(4 \sin^2 x \cos^2 x) = 1 - \frac{1}{4} \sin^2(2x)$.
Since $0 \le \sin^2(2x) \le 1$,we have $1 - \frac{1}{4}(1) \le f(x) \le 1 - \frac{1}{4}(0)$.
Thus,$\frac{3}{4} \le f(x) \le 1$.
Therefore,$f(R) \in \left[ \frac{3}{4}, 1 \right]$.

(A) The function $f(x) = \log_e(x - [x])$ is defined when the argument of the logarithm is strictly positive.
So,we require $x - [x] > 0$.
We know that the fractional part function is defined as $\{x\} = x - [x]$.
The range of the fractional part function is $[0, 1)$.
For $\log_e(\{x\})$ to be defined,we need $\{x\} > 0$.
Since $\{x\} = 0$ when $x$ is an integer $(x \in Z)$,the function is undefined for all integers.
For all non-integer values $(x \in R - Z)$,$\{x\} > 0$.
Therefore,the domain of the function is $R - Z$.

(C) Let $y = \frac{x^2}{x^2 + 1}$.
Since $x^2 \ge 0$ for all real $x$,the numerator is always non-negative and the denominator is always greater than the numerator.
As $x \to \pm \infty$,$y \to 1$,but $y$ never reaches $1$ because $x^2 < x^2 + 1$.
At $x = 0$,$y = \frac{0}{0+1} = 0$.
Since $x^2 \ge 0$ and $x^2 + 1 > 0$,the value of $y$ is always non-negative.
Thus,the range of the function is $[0, 1)$.

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

(D) Given the equation $2^x + 2^y = 2$.
We can rewrite this as $2^y = 2 - 2^x$.
For $y$ to be a real number,the value of $2^y$ must be strictly greater than $0$.
Therefore,$2 - 2^x > 0$.
This implies $2 > 2^x$.
Since the base $2 > 1$,the inequality holds when the exponents satisfy $1 > x$.
Thus,the domain of $x$ is $( - \infty , 1)$.

(D) Given $f(x) = \frac{x^2 - 1}{x^2 + 1}$.
We can rewrite the function as $f(x) = \frac{x^2 + 1 - 2}{x^2 + 1} = 1 - \frac{2}{x^2 + 1}$.
Since $x^2 \ge 0$ for all real $x$,we have $x^2 + 1 \ge 1$,which implies $0 < \frac{2}{x^2 + 1} \le 2$.
Subtracting this inequality from $1$,we get $1 - 2 \le 1 - \frac{2}{x^2 + 1} < 1 - 0$.
Thus,$-1 \le f(x) < 1$.
The minimum value is attained when $x^2 = 0$,i.e.,$x = 0$.
$f(0) = \frac{0^2 - 1}{0^2 + 1} = -1$.
Therefore,the minimum value of $f$ is $-1$.

(B) Let $y = \frac{x}{x^2 + 4}$.
$yx^2 - x + 4y = 0$.
Since $x \in \mathbb{R}$,the discriminant $D \ge 0$.
$D = (-1)^2 - 4(y)(4y) \ge 0$.
$1 - 16y^2 \ge 0$.
$16y^2 \le 1$.
$y^2 \le \frac{1}{16}$.
Therefore,$\frac{-1}{4} \le y \le \frac{1}{4}$.

(B) The function $f(x) = 4^{-x^2} + \cos^{-1}\left( \frac{x}{2} - 1 \right) + \log(\cos x)$ is defined if all its components are defined.
$1$. For $\cos^{-1}\left( \frac{x}{2} - 1 \right)$ to be defined,we must have $-1 \leq \frac{x}{2} - 1 \leq 1$.
Adding $1$ to all parts: $0 \leq \frac{x}{2} \leq 2$.
Multiplying by $2$: $0 \leq x \leq 4$.
$2$. For $\log(\cos x)$ to be defined,we must have $\cos x > 0$.
In the interval $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$,$\cos x > 0$ is always true.
$3$. Combining these conditions with the given interval $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$:
We need $x \in [0, 4]$ $AND$ $x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$.
Since $\frac{\pi}{2} \approx 1.57$,the intersection is $[0, \frac{\pi}{2})$.
Thus,the largest interval is $\left[ 0, \frac{\pi}{2} \right)$.

(A) The function is defined as $f(x) = \frac{1}{\sqrt{x-[x]}}$.
Domain of $f$:
We know that $0 \leq x-[x] < 1$ for all $x \in \mathbb{R}$.
Also,$x-[x] = 0$ when $x \in \mathbb{Z}$.
Since the denominator cannot be zero,$x-[x] > 0$,which implies $x \notin \mathbb{Z}$.
Thus,the domain is $\mathbb{R} - \mathbb{Z}$.
Range of $f$:
For $x \in \mathbb{R} - \mathbb{Z}$,we have $0 < x-[x] < 1$.
Taking the square root,$0 < \sqrt{x-[x]} < 1$.
Taking the reciprocal,$\frac{1}{\sqrt{x-[x]}} > 1$.
Therefore,$f(x) \in (1, \infty)$.
Hence,the range of $f$ is $(1, \infty)$.

(A) For $y$ to be a real value,the expression under the square root must be non-negative:
$\frac{(x + 1)(x - 3)}{(x - 2)} \ge 0$
We use the sign scheme (Wavy Curve Method) for the expression $f(x) = \frac{(x + 1)(x - 3)}{(x - 2)}$.
The critical points are $x = -1, 2, 3$.
Testing the intervals:
$1$. For $x \in (-1, 2)$,let $x = 0$: $\frac{(1)(-3)}{(-2)} = \frac{3}{2} > 0$ (Valid).
$2$. For $x \in [3, \infty)$,let $x = 4$: $\frac{(5)(1)}{(2)} = \frac{5}{2} > 0$ (Valid).
$3$. At $x = -1$,$y = 0$ (Valid).
$4$. At $x = 3$,$y = 0$ (Valid).
$5$. At $x = 2$,the expression is undefined.
Combining these,the solution is $x \in [-1, 2) \cup [3, \infty)$.