Set Theory Questions in English

(D) relation from $X$ to $Y$ is a subset of the Cartesian product $X \times Y$.
For $R_1$: If $x=1, y=3 \in Y$; $x=2, y=4 \notin Y$; $x=3, y=5 \in Y$; $x=4, y=6 \notin Y$; $x=5, y=7 \in Y$. Since $y$ must be in $Y$ for all $x \in X$,$R_1$ is not a relation from $X$ to $Y$ because it contains pairs where $y \notin Y$.
For $R_2$: All pairs $(x, y)$ satisfy $x \in X$ and $y \in Y$. Thus,$R_2 \subseteq X \times Y$,so it is a relation.
For $R_3$: All pairs $(x, y)$ satisfy $x \in X$ and $y \in Y$. Thus,$R_3 \subseteq X \times Y$,so it is a relation.
Therefore,both $(B)$ and $(C)$ are relations from $X$ to $Y$.

(C) Given that $n(A) = 2$ and $n(B) = 3$.
The number of elements in the Cartesian product $A \times B$ is given by $n(A \times B) = n(A) \times n(B) = 2 \times 3 = 6$.
$A$ relation from set $A$ to set $B$ is defined as any subset of the Cartesian product $A \times B$.
The total number of subsets of a set with $m$ elements is $2^m$.
Therefore,the total number of relations from $A$ to $B$ is $2^{n(A \times B)} = 2^6 = 64$.

(B) The relation $R$ is defined on the set of natural numbers $N$ as $\{(a, b) : a - b = 3\}$ or $\{(a, b) : b - a = 3\}$.
Given the options,we look for pairs where the difference is $3$.
If $b = n$,then $a = n + 3$.
For $n = 1, a = 4 \implies (4, 1)$.
For $n = 2, a = 5 \implies (5, 2)$.
For $n = 3, a = 6 \implies (6, 3)$.
Thus,the relation is $R = \{(4, 1), (5, 2), (6, 3), .....\}$.

(B) Given that ${N_a} = \{ an : n \in N \}$.
This represents the set of all multiples of $a$.
Therefore,${N_3} = \{ 3, 6, 9, 12, 15, 18, 21, 24, \dots \}$ and ${N_4} = \{ 4, 8, 12, 16, 20, 24, \dots \}$.
The intersection ${N_3} \cap {N_4}$ consists of elements that are common to both sets,which are the common multiples of $3$ and $4$.
The least common multiple of $3$ and $4$ is $12$.
Thus,${N_3} \cap {N_4} = \{ 12, 24, 36, \dots \} = {N_{12}}$.
In general,${N_a} \cap {N_b} = {N_{\text{lcm}(a, b)}}$.

(B) The number of elements in the union of two sets is given by the formula: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.
Given $n(A) = 3$ and $n(B) = 6$.
To find the minimum number of elements in $A \cup B$,we need to maximize the number of elements in the intersection $n(A \cap B)$.
The maximum possible value for $n(A \cap B)$ is the number of elements in the smaller set,which is $3$ (since $A \subseteq B$ is possible).
Substituting these values: $n(A \cup B) = 3 + 6 - 3 = 6$.
Therefore,the minimum number of elements in $A \cup B$ is $6$.

(D) The set $A$ represents a circle with center $(0, 0)$ and radius $r = 5$. The equation is $x^2 + y^2 = 25$.
The set $B$ represents an ellipse with the equation $x^2 + 9y^2 = 144$. Dividing by $144$,we get $\frac{x^2}{144} + \frac{9y^2}{144} = 1$,which simplifies to $\frac{x^2}{12^2} + \frac{y^2}{4^2} = 1$.
This is an ellipse with semi-major axis $a = 12$ and semi-minor axis $b = 4$.
Since the circle has a radius of $5$,and the ellipse has a minor axis of $4$ and a major axis of $12$,the circle intersects the ellipse at four distinct points because the circle's radius is greater than the semi-minor axis but less than the semi-major axis of the ellipse.
Therefore,$A \cap B$ contains four points.

(B) The set difference $A - B$ is defined as the set of all elements that are in $A$ but not in $B$.
Mathematically,$A - B = \{x : x \in A \text{ and } x \notin B\}$.
Since $x \notin B$ is equivalent to $x \in \bar B$ (where $\bar B$ is the complement of $B$),we can write:
$A - B = A \cap \bar B$.
Therefore,the correct option is $B$.

(A) First,we solve the quadratic equation for set $A$: $x^2 - 5x + 6 = 0$.
Factoring the quadratic,we get $(x - 2)(x - 3) = 0$,so $x = 2$ or $x = 3$. Thus,$A = \{2, 3\}$.
Next,we find the intersection of sets $B$ and $C$: $B \cap C = \{2, 4\} \cap \{4, 5\} = \{4\}$.
Finally,we calculate the Cartesian product $A \times (B \cap C) = \{2, 3\} \times \{4\}$.
This results in the set of ordered pairs $\{(2, 4), (3, 4)\}$.

(C) Let the number of newspapers be $x$.
Each student reads $5$ newspapers,so the total number of readings by all $300$ students is $300 \times 5 = 1500$.
Each newspaper is read by $60$ students,so the total number of readings is also $x \times 60$.
Equating the two,we get $60x = 1500$.
Solving for $x$,we get $x = 1500 / 60 = 25$.
Therefore,the number of newspapers is $25$.

(C) The Cartesian product of two sets $A$ and $B$ is defined as $A \times B = \{(a, b) : a \in A, b \in B\}$.
For the intersection $(A \times B) \cap (B \times A)$,an element $(x, y)$ must satisfy $(x, y) \in A \times B$ and $(x, y) \in B \times A$.
This implies $x \in A, y \in B$ and $x \in B, y \in A$.
Therefore,$x \in (A \cap B)$ and $y \in (A \cap B)$.
Given $A = \{1, 2, 3, 4, 5\}$ and $B = \{2, 3, 6, 7\}$,the intersection is $A \cap B = \{2, 3\}$.
The number of elements in $A \cap B$ is $n(A \cap B) = 2$.
The number of elements in $(A \times B) \cap (B \times A)$ is given by $n(A \cap B) \times n(A \cap B) = 2 \times 2 = 4$.

(B) The function $f(x) = \log(x^2)$ is defined for all $x \in \mathbb{R} \setminus \{0\}$ because $x^2 > 0$ for all non-zero real numbers.
Using the logarithmic property $\log(a^n) = n \log a$,we must ensure the domain remains consistent.
Since $x^2 = |x|^2$,we can write $\log(x^2) = \log(|x|^2) = 2 \log |x|$.
The function $2 \log |x|$ is also defined for all $x \in \mathbb{R} \setminus \{0\}$.
Therefore,the equivalent function is $2 \log |x|$.

(B) Given the function $f(x) = \frac{x - |x|}{|x|}$.
To find $f(-1)$,substitute $x = -1$ into the function.
$f(-1) = \frac{-1 - |-1|}{|-1|}$.
Since $|-1| = 1$,we have:
$f(-1) = \frac{-1 - 1}{1} = \frac{-2}{1} = -2$.

(D) Given $f(x) = 4x^3 + 3x^2 + 3x + 4$.
To find $x^3 f\left( \frac{1}{x} \right)$,first substitute $\frac{1}{x}$ for $x$ in the function $f(x)$:
$f\left( \frac{1}{x} \right) = 4\left( \frac{1}{x} \right)^3 + 3\left( \frac{1}{x} \right)^2 + 3\left( \frac{1}{x} \right) + 4$
$f\left( \frac{1}{x} \right) = \frac{4}{x^3} + \frac{3}{x^2} + \frac{3}{x} + 4$
Now,multiply this expression by $x^3$:
$x^3 f\left( \frac{1}{x} \right) = x^3 \left( \frac{4}{x^3} + \frac{3}{x^2} + \frac{3}{x} + 4 \right)$
$x^3 f\left( \frac{1}{x} \right) = 4 + 3x + 3x^2 + 4x^3$
Rearranging the terms,we get:
$x^3 f\left( \frac{1}{x} \right) = 4x^3 + 3x^2 + 3x + 4 = f(x)$
Therefore,the correct option is $D$.

(B) Given the function $f(x) = 2x + |x|$.
First,calculate $f(2x)$:
$f(2x) = 2(2x) + |2x| = 4x + 2|x|$.
Next,calculate $f(-x)$:
$f(-x) = 2(-x) + |-x| = -2x + |x|$.
Now,substitute these into the expression $f(2x) + f(-x) - f(x)$:
$= (4x + 2|x|) + (-2x + |x|) - (2x + |x|)$.
$= 4x + 2|x| - 2x + |x| - 2x - |x|$.
Combine the like terms:
$= (4x - 2x - 2x) + (2|x| + |x| - |x|)$.
$= 0x + 2|x| = 2|x|$.

(D) Given $f(x) = \cos([\pi^2]x) + \cos([- \pi^2]x)$.
Since $\pi^2 \approx 9.86$,we have $[\pi^2] = 9$.
Also,$[-\pi^2] = [-9.86] = -10$.
Thus,$f(x) = \cos(9x) + \cos(-10x) = \cos(9x) + \cos(10x)$.
Using the sum-to-product formula $\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$:
$f(x) = 2\cos\left(\frac{19x}{2}\right)\cos\left(\frac{x}{2}\right)$.
Now,evaluate $f\left(\frac{\pi}{2}\right)$:
$f\left(\frac{\pi}{2}\right) = 2\cos\left(\frac{19\pi}{4}\right)\cos\left(\frac{\pi}{4}\right)$.
Since $\cos\left(\frac{19\pi}{4}\right) = \cos\left(4\pi + \frac{3\pi}{4}\right) = \cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}$ and $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$:
$f\left(\frac{\pi}{2}\right) = 2 \times \left(-\frac{1}{\sqrt{2}}\right) \times \left(\frac{1}{\sqrt{2}}\right) = 2 \times \left(-\frac{1}{2}\right) = -1$.

(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 $R - \{3\}$.
Now,consider the two cases for the absolute value:
Case $1$: If $x > 3$,then $|x - 3| = x - 3$. Therefore,$f(x) = \frac{x - 3}{x - 3} = 1$.
Case $2$: If $x < 3$,then $|x - 3| = -(x - 3)$. Therefore,$f(x) = \frac{-(x - 3)}{x - 3} = -1$.
Thus,the range of the function is the set $\{1, -1\}$.

(C) The function $f(x) = \log |x^2 - 9|$ is defined only when the argument of the logarithm is strictly positive.
That is,$|x^2 - 9| > 0$.
Since the absolute value $|x^2 - 9|$ is always non-negative,the condition $|x^2 - 9| > 0$ is satisfied for all real numbers $x$ except where $|x^2 - 9| = 0$.
Setting $x^2 - 9 = 0$,we get $x^2 = 9$,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 all real numbers except $\{-3, 3\}$,which is written as $R - \{-3, 3\}$.

(C) The function $f(x) = \log |\log x|$ is defined if the argument of the logarithm is strictly positive.
$1$. The inner logarithm $\log x$ is defined only when $x > 0$.
$2$. The outer logarithm $\log |\log x|$ is defined when $|\log x| > 0$.
$3$. The condition $|\log x| > 0$ implies $\log x \neq 0$,which means $x \neq 1$.
$4$. Combining both conditions,we require $x > 0$ and $x \neq 1$.
Therefore,the domain is $x \in (0, 1) \cup (1, \infty)$.

(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: $x^2 + 3x + 2 \neq 0$.
Factoring the denominator: $(x + 1)(x + 2) \neq 0$,which implies $x \neq -1$ and $x \neq -2$.
Combining these conditions,we require $x > -3$ and $x \notin \{-1, -2\}$.
Thus,the domain is $(-3, \infty) - \{-1, -2\}$.

(B) Given the function $f(x) = x^2 - 6x + 7$.
To find the range,we complete 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 the expression is $0 - 2 = -2$.
As $x \to \infty$ or $x \to -\infty$,$f(x) \to \infty$.
Therefore,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 $\log_a x \ge 0$ implies $x \ge 1$ (for base $10$ or $e$),we have:
$\frac{1}{|\sin x|} \ge 1$
$1 \ge |\sin x|$
This inequality is always true for all $x$ where $\sin x$ is defined,provided $|\sin x| \neq 0$.
Since $\sin x = 0$ at $x = n\pi$ for $n \in I$,the term $\frac{1}{|\sin x|}$ becomes undefined at these points.
Therefore,the domain is all real numbers except where $\sin x = 0$.
Domain $= R - \{ n\pi, n \in I \}$.

(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 $\sqrt {x - 4} + \sqrt {6 - x}$ must be strictly greater than $0$. Since both square root terms are non-negative,their sum is always $\ge 0$. The sum is $0$ only if $x-4=0$ and $6-x=0$ simultaneously,which is impossible ($x=4$ and $x=6$). Thus,the sum is always positive for $x \in [4, 6]$.
Combining these conditions,we get $4 \le x \le 6$.
Therefore,the domain is $[4, 6]$.

(B) For the function $f(x) = {\left[ {{{\log }_{10}}\left( {\frac{{5x - {x^2}}}{4}} \right)} \right]^{1/2}}$ to be defined,the expression inside the square root must be non-negative:
${\log _{10}}\left( {\frac{{5x - {x^2}}}{4}} \right) \ge 0$
Since the base of the logarithm is $10 > 1$,the inequality remains unchanged when we exponentiate:
$\frac{{5x - {x^2}}}{4} \ge {10^0}$
$\frac{{5x - {x^2}}}{4} \ge 1$
$5x - {x^2} \ge 4$
${x^2} - 5x + 4 \le 0$
Factoring the quadratic expression:
$(x - 1)(x - 4) \le 0$
This inequality holds when $x$ lies between the roots $1$ and $4$ inclusive.
Therefore,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: $x^2 - 1 > 0$,which implies $x^2 > 1$,so $x < -1$ or $x > 1$.
$2$. The base of the logarithm must be positive and not equal to $1$: $3 + x > 0$ and $3 + x \neq 1$.
From $3 + x > 0$,we get $x > -3$.
From $3 + x \neq 1$,we get $x \neq -2$.
Combining all these conditions: $x > -3$,$x \neq -2$,and ($x < -1$ or $x > 1$).
Intersection of these sets gives the domain: $D_f = (-3, -2) \cup (-2, -1) \cup (1, \infty)$.

(B) For the function $f(x) = \sqrt{\sin 2x}$ to be defined,the expression inside the square root must be non-negative.
Therefore,$\sin 2x \ge 0$.
We know that $\sin \theta \ge 0$ when $\theta$ lies in the first or second quadrant,i.e.,$2n\pi \le \theta \le 2n\pi + \pi$.
Substituting $\theta = 2x$,we get $2n\pi \le 2x \le 2n\pi + \pi$.
Dividing the entire inequality by $2$,we get $n\pi \le x \le n\pi + \frac{\pi}{2}$.
Thus,the domain is $[n\pi, n\pi + \frac{\pi}{2}]$.
Hence,option $(b)$ is correct.

(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: $x^3 - x > 0$.
Factorizing the expression: $x(x^2 - 1) > 0 \Rightarrow x(x - 1)(x + 1) > 0$.
Using the wavy curve method (sign scheme) for the critical points $x = -1, 0, 1$,the expression is positive in the intervals $(-1, 0) \cup (1, \infty)$.
Combining these conditions,we must exclude $x = 2$ from the interval $(1, \infty)$.
Thus,the domain is $(-1, 0) \cup (1, 2) \cup (2, \infty)$.

(B) For the function $f(x) = \sqrt{2 - 2x - x^2}$ to be defined,the expression under the square root must be non-negative.
Thus,$2 - 2x - x^2 \ge 0$.
Multiplying by $-1$,we get $x^2 + 2x - 2 \le 0$.
To solve this inequality,we complete the square:
$x^2 + 2x + 1 - 1 - 2 \le 0$
$(x + 1)^2 - 3 \le 0$
$(x + 1)^2 \le 3$
Taking the square root on both sides,we get:
$-\sqrt{3} \le x + 1 \le \sqrt{3}$
Subtracting $1$ from all parts:
$-1 - \sqrt{3} \le x \le -1 + \sqrt{3}$.
Therefore,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 greater than zero,and the denominator must not be zero.
$1$. The condition for the square root is $x^2 - 4 > 0$,which implies $x^2 > 4$. This gives $x > 2$ or $x < -2$.
$2$. The condition for the denominator is $(x - 1)\sqrt{x^2 - 4} \neq 0$. This implies $x - 1 \neq 0$ (so $x \neq 1$) and $\sqrt{x^2 - 4} \neq 0$ (so $x^2 - 4 \neq 0$,which means $x \neq \pm 2$).
$3$. 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 is $( - \infty, - 2) \cup (2, \infty)$.

(B) For the function $f(x) = \sqrt{\log \left( \frac{5x - x^2}{6} \right)}$ to be defined,the expression inside the square root must be non-negative:
$\log \left( \frac{5x - x^2}{6} \right) \ge 0$
Since the base of the logarithm is $10$ (implied),this inequality is equivalent to:
$\frac{5x - x^2}{6} \ge 10^0$
$\frac{5x - x^2}{6} \ge 1$
$5x - x^2 \ge 6$
$x^2 - 5x + 6 \le 0$
Factoring the quadratic expression,we get:
$(x - 2)(x - 3) \le 0$
The inequality holds when $x$ lies between the roots of the quadratic equation,inclusive:
$2 \le x \le 3$
Thus,the domain is $[2, 3]$.

(C) For the function $f(x) = \sqrt{2 - x} - \frac{1}{\sqrt{9 - x^2}}$ to be defined:
$1$. The expression under the square root in the numerator must be non-negative: $2 - x \ge 0 \Rightarrow x \le 2$.
$2$. The expression under the square root in the denominator must be strictly positive: $9 - x^2 > 0 \Rightarrow x^2 < 9 \Rightarrow |x| < 3$,which means $-3 < x < 3$.
$3$. Combining these two conditions: $x \le 2$ $AND$ $-3 < x < 3$.
$4$. The intersection of these intervals 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 $[-1, 1] - \{0\}$.

(D) For the function $f(x) = \sqrt{x - x^2} + \sqrt{4 + x} + \sqrt{4 - x}$ to be defined,the expressions under the square roots must be 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:
$x \in [-4, \infty) \cap (- \infty, 4] \cap [0, 1]$.
The intersection of these sets is $[0, 1]$.
Therefore,the domain of the function is $[0, 1]$.

(A) The domain of a function $f(x)$ is the set of all possible input values $(x)$ for which the function is defined.
For the function $f(x) = \frac{1}{1 + e^x}$,the denominator must not be zero.
Since $e^x > 0$ for all real values of $x$,it follows that $1 + e^x > 1$.
Therefore,$1 + e^x$ is never zero for any real number $x$.
Thus,the function is defined for all real numbers.
The domain is $(-\infty, \infty)$.

(C) The function $f(x) = \sqrt{\log(x^2 - 6x + 6)}$ is defined when the expression inside the square root is non-negative,i.e.,$\log(x^2 - 6x + 6) \ge 0$.
Since $\log(y) \ge 0$ implies $y \ge 1$ (assuming base $10$ or $e$),we have:
$x^2 - 6x + 6 \ge 1$
$x^2 - 6x + 5 \ge 0$
Factoring the quadratic expression:
$(x - 5)(x - 1) \ge 0$
To solve this inequality,we find the critical points $x = 1$ and $x = 5$. Testing the intervals $(-\infty, 1]$,$(1, 5)$,and $[5, \infty)$,we find that the expression is non-negative in the intervals $(-\infty, 1]$ and $[5, \infty)$.
Additionally,we must ensure the argument of the logarithm is positive: $x^2 - 6x + 6 > 0$. Since $x^2 - 6x + 6 \ge 1$ is already satisfied,the condition $x^2 - 6x + 6 > 0$ is automatically satisfied.
Therefore,the domain is $( - \infty, 1 ] \cup [ 5, \infty )$.

(D) For the function $f(x) = \sqrt{1 - \frac{1}{x}}$ to be defined,the expression under the square root must be non-negative:
$1 - \frac{1}{x} \ge 0$
$\frac{x - 1}{x} \ge 0$
To solve this inequality,we find the critical points where the numerator and denominator are zero,which are $x = 1$ and $x = 0$.
Using the sign scheme (Wavy Curve Method):
For $x > 1$,$\frac{x-1}{x} > 0$ (positive).
For $0 < x < 1$,$\frac{x-1}{x} < 0$ (negative).
For $x < 0$,$\frac{x-1}{x} > 0$ (positive).
Since we need the expression to be $\ge 0$,the solution is $x \in (-\infty, 0) \cup [1, \infty)$.
Note: $x = 0$ is excluded because the denominator cannot be zero.

(A) The function is defined as $f(x) = \frac{1}{\sqrt{x^2 - 1}}$.
For $f(x)$ to be defined,the expression inside the square root must be strictly greater than zero because it is in the denominator.
Therefore,$x^2 - 1 > 0$.
This implies $x^2 > 1$.
Solving the inequality $x^2 > 1$,we get $|x| > 1$.
This means $x > 1$ or $x < -1$.
In interval notation,this is expressed as $x \in ( - \infty, -1) \cup (1, \infty)$.

(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 by definition of the absolute value function:
If $x \ge 0$,then $|x| = x$,which implies $x > x$,which is impossible.
If $x < 0$,then $|x| = -x$,which implies $-x > x$,or $0 > 2x$,which simplifies to $x < 0$.
Therefore,the function is defined for all $x < 0$.
The domain is $( - \infty, 0)$.

(D) The function is given by $f(x) = \sqrt{x^2 - 1} + \sqrt{x^2 + 1}$.
For the function to be defined in the set of real numbers,the expressions under the square roots must be non-negative.
For the first term $\sqrt{x^2 - 1}$,we require $x^2 - 1 \ge 0$,which implies $x^2 \ge 1$. This inequality holds when $x \in (-\infty, -1] \cup [1, \infty)$.
For the second term $\sqrt{x^2 + 1}$,we require $x^2 + 1 \ge 0$. Since $x^2 \ge 0$ for all real $x$,$x^2 + 1 \ge 1$ is always true for all $x \in \mathbb{R}$.
The domain of the function $f(x)$ is the intersection of the domains of the two terms.
Thus,the domain is $(-\infty, -1] \cup [1, \infty)$,which can be written as $(-\infty, \infty) - (-1, 1)$.

(D) 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.
Thus,$5x - 3 - 2{x^2} \ge 0$.
Multiplying by $-1$,we get $2{x^2} - 5x + 3 \le 0$.
Factoring the quadratic expression: $2{x^2} - 2x - 3x + 3 \le 0 \implies 2x(x - 1) - 3(x - 1) \le 0 \implies (2x - 3)(x - 1) \le 0$.
This can be written as $2(x - 3/2)(x - 1) \le 0$.
Using the wavy curve method (sign scheme),the expression is $\le 0$ for $x \in [1, 3/2]$.
Therefore,the domain is $[1, 3/2]$.

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

(C) Let $y = \frac{x^2 + x + 2}{x^2 + x + 1}$.
We can rewrite the function as $y = \frac{(x^2 + x + 1) + 1}{x^2 + x + 1} = 1 + \frac{1}{x^2 + x + 1}$.
To find the range,we need to find the range of the expression $g(x) = x^2 + x + 1$.
Completing the square for $g(x)$,we get $g(x) = (x + 1/2)^2 + 3/4$.
The minimum value of $g(x)$ is $3/4$ (at $x = -1/2$) and the maximum value is $\infty$.
Thus,the range of $g(x)$ is $[3/4, \infty)$.
Now,the range of $\frac{1}{g(x)}$ is $(0, 1/(3/4)] = (0, 4/3]$.
Adding $1$ to this range,the range of $f(x)$ is $(1, 1 + 4/3] = (1, 7/3]$.

(D) Given the function $f(x) = a\cos(bx + c) + d$.
We know that the range of the cosine function $\cos(\theta)$ is $[-1, 1]$.
Therefore,$-1 \le \cos(bx + c) \le 1$.
Multiplying by $a$ (assuming $a > 0$),we get $-a \le a\cos(bx + c) \le a$.
Adding $d$ to all parts,we get $d - a \le a\cos(bx + c) + d \le d + a$.
Thus,the range of $f(x)$ is $[d - a, d + a]$.

(D) Given the function $f(x) = \cos x - \sin x$.
We know that any expression of the form $a \cos x + b \sin x$ can be written as $R \cos(x + \alpha)$ or $R \sin(x + \beta)$,where $R = \sqrt{a^2 + b^2}$.
Here,$a = 1$ and $b = -1$.
So,$R = \sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$.
Thus,$f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x \right) = \sqrt{2} \cos(x + \frac{\pi}{4})$.
Since the range of $\cos(\theta)$ is $[-1, 1]$,the range of $\sqrt{2} \cos(x + \frac{\pi}{4})$ is $[-\sqrt{2}, \sqrt{2}]$.
Therefore,the correct option is $D$.

(A) 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 $x \to \pm \infty$,$y \to 1$,but $y$ never reaches $1$ because $x^2 < x^2 + 1$.
At $x = 0$,$y = 0$.
Thus,the range of the function is $[0, 1)$.

(B) Given function is $f(x) = \cos 2x - \sin 2x$.
We can rewrite this as $f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos 2x - \frac{1}{\sqrt{2}} \sin 2x \right)$.
Using the identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$,we get $f(x) = \sqrt{2} \cos(2x + \frac{\pi}{4})$.
Since the range of $\cos \theta$ is $[-1, 1]$,the range of $f(x)$ is $[-\sqrt{2}, \sqrt{2}]$.
Since $\sqrt{2} \approx 1.414$,the interval $[-\sqrt{2}, \sqrt{2}]$ is approximately $[-1.414, 1.414]$.
The set $[-1, 1]$ is a subset of $[-1.414, 1.414]$.
Therefore,the range of $f(x)$ contains the set $[-1, 1]$.

(B) Given the function $f(x) = \frac{1}{2 - \sin 3x}$.
We know that the range of $\sin 3x$ is $[-1, 1]$.
Let $u = \sin 3x$,then $u \in [-1, 1]$.
The function becomes $f(u) = \frac{1}{2 - u}$.
To find the range,we evaluate the function at the boundaries of $u$:
When $u = -1$,$f(-1) = \frac{1}{2 - (-1)} = \frac{1}{3}$.
When $u = 1$,$f(1) = \frac{1}{2 - 1} = 1$.
Since the function is continuous and monotonic within this interval,the range of the function is $[\frac{1}{3}, 1]$.

(B) We know that the range of the sine function $\sin x$ is $[-1, 1]$.
This means $-1 \le \sin x \le 1$.
Multiplying by $-7$,we get $7 \ge -7 \sin x \ge -7$,which can be written as $-7 \le -7 \sin x \le 7$.
Adding $9$ to all parts of the inequality,we get $9 - 7 \le 9 - 7 \sin x \le 9 + 7$.
Thus,$2 \le 9 - 7 \sin x \le 16$.
Therefore,the range of the function $f(x) = 9 - 7 \sin x$ is $[2, 16]$.

(B) Let $y = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}$.
Rearranging the terms,we get $y(x^2 + 2x - 7) = x^2 + 34x - 71$.
$x^2(y - 1) + x(2y - 34) - 7y + 71 = 0$.
For $x$ to be a real number,the discriminant $D = B^2 - 4AC \ge 0$.
$D = (2y - 34)^2 - 4(y - 1)(-7y + 71) \ge 0$.
$4(y - 17)^2 - 4(-7y^2 + 71y + 7y - 71) \ge 0$.
$(y^2 - 34y + 289) - (-7y^2 + 78y - 71) \ge 0$.
$y^2 - 34y + 289 + 7y^2 - 78y + 71 \ge 0$.
$8y^2 - 112y + 360 \ge 0$.
Dividing by $8$,we get $y^2 - 14y + 45 \ge 0$.
$(y - 5)(y - 9) \ge 0$.
Thus,the range is $y \in ( - \infty, 5] \cup [9, \infty)$.

(C) We know that the identity ${\sin ^{ - 1}} \theta + {\cos ^{ - 1}} \theta = \frac{\pi }{2}$ is valid for all $\theta \in [-1, 1]$.
In this problem,$\theta = \sqrt x$.
For $\sqrt x$ to be defined,we must have $x \ge 0$.
For the inverse trigonometric functions to be defined,we must have $-1 \le \sqrt x \le 1$.
Since $\sqrt x$ is always non-negative,the condition $-1 \le \sqrt x \le 1$ simplifies to $0 \le \sqrt x \le 1$.
Squaring the inequality,we get $0^2 \le x \le 1^2$,which results in $0 \le x \le 1$.
Therefore,the interval for which the equation holds is $x \in [0, 1]$.

(C) Given $f(x) = \cos^2 x + \sin^4 x$.
We can rewrite the expression as:
$f(x) = \cos^2 x + \sin^2 x \cdot \sin^2 x$
$f(x) = \cos^2 x + (1 - \cos^2 x) \sin^2 x$
Alternatively,using $\sin^2 x = 1 - \cos^2 x$:
$f(x) = \cos^2 x + (1 - \cos^2 x)^2$
$f(x) = \cos^2 x + 1 - 2\cos^2 x + \cos^4 x$
$f(x) = \cos^4 x - \cos^2 x + 1$
Let $t = \cos^2 x$,where $t \in [0, 1]$.
Then $g(t) = t^2 - t + 1$.
This is a parabola opening upwards with vertex at $t = -(-1)/(2 \cdot 1) = 1/2$.
Since $1/2 \in [0, 1]$,the minimum value is $g(1/2) = (1/4) - (1/2) + 1 = 3/4$.
The maximum value occurs at the boundaries $t=0$ or $t=1$:
$g(0) = 1$ and $g(1) = 1 - 1 + 1 = 1$.
Thus,the range is $[3/4, 1]$.