Domain and Range Questions in English

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

(A) Given the function $f(x) = \frac{|x|}{x}$ for $x \in A$,where $A = \{x \in R, x \neq 0, -4 \leq x \leq 4\}$.
If $x > 0$,then $|x| = x$,so $f(x) = \frac{x}{x} = 1$.
If $x < 0$,then $|x| = -x$,so $f(x) = \frac{-x}{x} = -1$.
Since $x$ cannot be $0$,the function $f(x)$ only takes the values $1$ and $-1$.
Therefore,the range of $f$ is $\{1, -1\}$.

(D) Given,$f(x) = [\frac{x}{5}]$ where $|x| < 71$.
This implies $-71 < x < 71$.
Dividing by $5$,we get $-\frac{71}{5} < \frac{x}{5} < \frac{71}{5}$.
$-14.2 < \frac{x}{5} < 14.2$.
Now,we find the range of the greatest integer function $[\frac{x}{5}]$.
The minimum value is $[\frac{x}{5}] = [-14.2] = -15$.
The maximum value is $[\frac{x}{5}] = [14.2] = 14$.
Since $x$ can take any real value in the interval $(-71, 71)$,the function $f(x)$ will take all integer values from $-15$ to $14$.
Thus,the set is $\{-15, -14, \ldots, 0, \ldots, 13, 14\}$.

(C) Given that $f: A \rightarrow B$ is an onto function,the range of $f(x)$ must be equal to the codomain $B$.
Since $g: B \rightarrow C$ is defined as $g(x) = \sqrt{3 + 4x - 4x^2}$,the domain of $g$ is $B$.
For $g(x)$ to be defined,the expression inside the square root must be non-negative:
$3 + 4x - 4x^2 \geq 0$
Multiplying by $-1$ reverses the inequality:
$4x^2 - 4x - 3 \leq 0$
Factoring the quadratic expression:
$(2x - 3)(2x + 1) \leq 0$
The roots are $x = -\frac{1}{2}$ and $x = \frac{3}{2}$.
Testing the intervals,the inequality holds for $x \in [-\frac{1}{2}, \frac{3}{2}]$.
Thus,the domain of $g$ is $[-\frac{1}{2}, \frac{3}{2}]$,which is the range of $f$.

(C) The given function is $f(x)=\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{[x]+2}}$.
For $f(x)$ to be defined,the following conditions must be satisfied:
$1. \ 3+x \geq 0 \Rightarrow x \geq -3$
$2. \ 3-x \geq 0 \Rightarrow x \leq 3$
$3. \ [x]+2 > 0 \Rightarrow [x] > -2$
Since $[x] > -2$,the smallest integer value $[x]$ can take is $-1$. Thus,$x \geq -1$.
Combining these conditions,we get $x \in [-1, 3]$.
Given that the interval is $[\alpha, \beta]$,we have $\alpha = -1$ and $\beta = 3$.
Now,we calculate $f^2(\alpha+1) + 5f^2(\beta) = f^2(0) + 5f^2(3)$.
$f(0) = \frac{\sqrt{3+0} + \sqrt{3-0}}{\sqrt{[0]+2}} = \frac{2\sqrt{3}}{\sqrt{2}} = \sqrt{6}$. So,$f^2(0) = 6$.
$f(3) = \frac{\sqrt{3+3} + \sqrt{3-3}}{\sqrt{[3]+2}} = \frac{\sqrt{6} + 0}{\sqrt{3+2}} = \frac{\sqrt{6}}{\sqrt{5}}$. So,$f^2(3) = \frac{6}{5}$.
Therefore,$f^2(0) + 5f^2(3) = 6 + 5 \times \frac{6}{5} = 6 + 6 = 12$.
Hence,option $C$ is correct.

(D) Let $y = \frac{2x-3}{x^2-5x+6}$.
For $x$ to be real,the discriminant $D$ of the quadratic equation in $x$ must be non-negative.
$y(x^2-5x+6) = 2x-3$
$yx^2 - (5y+2)x + (6y+3) = 0$
For $x \in \mathbb{R}$,$D = b^2 - 4ac \geq 0$.
$(5y+2)^2 - 4y(6y+3) \geq 0$
$25y^2 + 20y + 4 - 24y^2 - 12y \geq 0$
$y^2 + 8y + 4 \geq 0$.
The roots of $y^2 + 8y + 4 = 0$ are $y = \frac{-8 \pm \sqrt{64-16}}{2} = -4 \pm 2\sqrt{3}$.
Thus,$y \in (-\infty, -4-2\sqrt{3}] \cup [-4+2\sqrt{3}, \infty)$.
The values not taken by $f(x)$ lie in the interval $(-4-2\sqrt{3}, -4+2\sqrt{3})$.
Since $-4-2\sqrt{3} \approx -7.46$ and $-4+2\sqrt{3} \approx -0.53$,the value $1$ is in the range,$2$ is in the range,$-10$ is in the range,but $-2$ is $NOT$ in the range of $f(x)$.

(D) Given $f(x) = |x^2 - 3x + 2| + 2x - 3$ on $[-2, 1]$.
Since $x^2 - 3x + 2 = (x - 1)(x - 2)$,for $x \in [-2, 1]$,$(x - 1) \le 0$ and $(x - 2) < 0$,so $(x - 1)(x - 2) \ge 0$.
Thus,$|x^2 - 3x + 2| = x^2 - 3x + 2$.
Substituting this into $f(x)$,we get $f(x) = x^2 - 3x + 2 + 2x - 3 = x^2 - x - 1$.
To find the extrema on $[-2, 1]$,we check the critical points and endpoints.
$f'(x) = 2x - 1$. Setting $f'(x) = 0$ gives $x = 1/2$,which is in $[-2, 1]$.
Calculate values at critical point and endpoints:
$f(-2) = (-2)^2 - (-2) - 1 = 4 + 2 - 1 = 5$.
$f(1) = (1)^2 - (1) - 1 = -1$.
$f(1/2) = (1/2)^2 - (1/2) - 1 = 1/4 - 1/2 - 1 = -5/4$.
Comparing these,the absolute maximum $M = 5$ and the absolute minimum $m = -5/4$.
Then $M - 4m = 5 - 4(-5/4) = 5 + 5 = 10$.

(B) Given the function $f(x) = x^2 - 4x + 5$ defined on the domain $[2, \infty)$.
To find the range $B$ for the function to be a bijection,we need to determine the set of all possible values of $f(x)$.
First,we 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 $x - 2 \geq 0$.
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 is $[1, \infty)$.
Since the function is strictly increasing on $[2, \infty)$,it is injective. For it to be a bijection,the codomain $B$ must be equal to the range.
Therefore,$B = [1, \infty)$.

(A) Given the relation $\rho = \{(x, y) \in N \times N: 2x + y = 41\}$.
Since $x, y \in N$,we have $x \geq 1$ and $y \geq 1$.
From $2x + y = 41$,we get $y = 41 - 2x$.
Since $y \geq 1$,we have $41 - 2x \geq 1$,which implies $2x \leq 40$,so $x \leq 20$.
Thus,$x \in \{1, 2, 3, \dots, 20\}$.
For each $x$,$y = 41 - 2x$. Substituting values of $x$:
If $x = 1, y = 39$.
If $x = 20, y = 1$.
So,the domain $A = \{1, 2, 3, \dots, 20\}$ and the range $B = \{1, 3, 5, \dots, 39\}$.
Both $A$ and $B$ are subsets of the sets given in option $A$.

(C) For the function $y = \sqrt{\log _{10} \frac{3x - x^2}{2}}$ to be defined,the expression inside the square root must be non-negative:
$\log _{10} \left( \frac{3x - x^2}{2} \right) \geq 0$
Since $\log _{10} 1 = 0$,we have:
$\frac{3x - x^2}{2} \geq 1$
$3x - x^2 \geq 2$
$x^2 - 3x + 2 \leq 0$
$(x - 1)(x - 2) \leq 0$
This inequality holds for $x \in [1, 2]$.
Additionally,the argument of the logarithm must be positive:
$\frac{3x - x^2}{2} > 0$
$x(3 - x) > 0$
$x(x - 3) < 0$
This holds for $x \in (0, 3)$.
The intersection of $x \in [1, 2]$ and $x \in (0, 3)$ is $x \in [1, 2]$.

(C) For the function $f(x) = \sqrt{\frac{1}{\sqrt{x}} - \sqrt{x+1}}$ to be defined,the following conditions must be met:
$1$. The expression inside the square root must be non-negative: $\frac{1}{\sqrt{x}} - \sqrt{x+1} \geq 0$.
$2$. The term $\sqrt{x}$ in the denominator requires $x > 0$.
$3$. The term $\sqrt{x+1}$ requires $x+1 \geq 0$,which means $x \geq -1$.
Combining these,we need $x > 0$.
Now,solve the inequality: $\frac{1}{\sqrt{x}} \geq \sqrt{x+1}$.
Since $x > 0$,we can square both sides: $\frac{1}{x} \geq x+1$.
$1 \geq x^2 + x \implies x^2 + x - 1 \leq 0$.
The roots of $x^2 + x - 1 = 0$ are $x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.
Since the parabola opens upward,$x^2 + x - 1 \leq 0$ for $x \in \left[\frac{-1-\sqrt{5}}{2}, \frac{\sqrt{5}-1}{2}\right]$.
Intersecting this with the condition $x > 0$,we get $x \in \left(0, \frac{\sqrt{5}-1}{2}\right]$.

(B) For $f(x) = \sqrt{\frac{1-|x|}{2-|x|}}$ to be defined,the expression inside the square root must be non-negative: $\frac{1-|x|}{2-|x|} \geq 0$.
Multiplying by $-1$ on both sides,we get $\frac{|x|-1}{|x|-2} \leq 0$.
Let $t = |x|$. Then $\frac{t-1}{t-2} \leq 0$.
Since $t = |x| \geq 0$,the critical points are $t=1$ and $t=2$.
The inequality holds for $1 \leq t < 2$.
Thus,$1 \leq |x| < 2$.
This implies $x \in (-2, -1] \cup [1, 2)$ is incorrect based on the original expression.
Re-evaluating $\frac{1-|x|}{2-|x|} \geq 0$:
The expression is $\geq 0$ when the numerator and denominator have the same sign.
Case $1$: $1-|x| \geq 0$ and $2-|x| > 0 \Rightarrow |x| \leq 1$ and $|x| < 2$ $\Rightarrow |x| \leq 1$ $\Rightarrow x \in [-1, 1]$.
Case $2$: $1-|x| \leq 0$ and $2-|x| < 0 \Rightarrow |x| \geq 1$ and $|x| > 2$ $\Rightarrow |x| > 2$ $\Rightarrow x \in (-\infty, -2) \cup (2, \infty)$.
Combining these,the domain is $(-\infty, -2) \cup [-1, 1] \cup (2, \infty)$.

(C) The function is defined as $f(x) = \left[ \frac{1}{[x]} \right]$.
For the function to be defined,the denominator $[x]$ must not be $0$.
Thus,$[x] \neq 0$,which implies $x < 0$ or $x \geq 1$.
Therefore,the domain is $(-\infty, 0) \cup [1, \infty)$.
For $x \in [1, 2)$,$[x] = 1$,so $f(x) = [1/1] = 1$.
For $x \in [2, \infty)$,$[x] \geq 2$,so $0 < 1/[x] \leq 0.5$,which implies $f(x) = [1/[x]] = 0$.
For $x \in [-1, 0)$,$[x] = -1$,so $f(x) = [1/(-1)] = -1$.
For $x \in [-2, -1)$,$[x] = -2$,so $f(x) = [1/(-2)] = [-0.5] = -1$.
Thus,the range is $\{-1, 0, 1\}$.

(B) For the function $f(x) = \sqrt{1 + \log_{e}(1 - x)}$ to be defined,the expression inside the square root must be non-negative,and the argument of the logarithm must be positive.
$1$. Condition for logarithm: $1 - x > 0 \implies x < 1$.
$2$. Condition for square root: $1 + \log_{e}(1 - x) \geq 0$.
$\log_{e}(1 - x) \geq -1$.
Exponentiating both sides with base $e$:
$1 - x \geq e^{-1}$.
$1 - x \geq \frac{1}{e}$.
$x \leq 1 - \frac{1}{e}$.
$x \leq \frac{e - 1}{e}$.
Combining both conditions: $x < 1$ and $x \leq \frac{e - 1}{e}$.
Since $\frac{e - 1}{e} < 1$,the domain is $-\infty < x \leq \frac{e - 1}{e}$.

(A) Let $y = \frac{x^2-x+4}{x^2+x+4}$.
Multiplying both sides by $(x^2+x+4)$,we get $y(x^2+x+4) = x^2-x+4$.
Rearranging the terms,we get $(y-1)x^2 + (y+1)x + (4y-4) = 0$.
For $x$ to be a real number,the discriminant $D$ must be greater than or equal to $0$.
$D = (y+1)^2 - 4(y-1)(4y-4) \geq 0$.
$(y+1)^2 - 16(y-1)^2 \geq 0$.
Using the identity $a^2 - b^2 = (a-b)(a+b)$,we get $((y+1) - 4(y-1))((y+1) + 4(y-1)) \geq 0$.
$(y+1-4y+4)(y+1+4y-4) \geq 0$.
$(5-3y)(5y-3) \geq 0$.
Multiplying by $-1$ reverses the inequality: $(3y-5)(5y-3) \leq 0$.
The roots are $y = \frac{5}{3}$ and $y = \frac{3}{5}$.
Thus,the range is $y \in [\frac{3}{5}, \frac{5}{3}]$.

(B) Given the function $f(x) = 3x^2 + 1$.
We need to find the set $f^{-1}([1, 6])$,which consists of all $x$ such that $f(x) \in [1, 6]$.
So,$1 \le 3x^2 + 1 \le 6$.
Subtracting $1$ from all parts: $0 \le 3x^2 \le 5$.
Dividing by $3$: $0 \le x^2 \le \frac{5}{3}$.
Taking the square root: $-\sqrt{\frac{5}{3}} \le x \le \sqrt{\frac{5}{3}}$.
Thus,the set is $[ -\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}} ]$.

(C) Given function is $y=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-x^{2}}\right)$.
For the function to be defined,we must have $\frac{\pi^{2}}{16}-x^{2} \geq 0$,which implies $x^{2} \leq \frac{\pi^{2}}{16}$,so $x \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$.
Let $u = \sqrt{\frac{\pi^{2}}{16}-x^{2}}$. As $x$ varies from $0$ to $\frac{\pi}{4}$,$u$ varies from $\frac{\pi}{4}$ to $0$.
Thus,the range of $u$ is $\left[0, \frac{\pi}{4}\right]$.
Now,$y = 3 \sin(u)$. Since $u \in \left[0, \frac{\pi}{4}\right]$,$\sin(u)$ varies from $\sin(0) = 0$ to $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.
Therefore,$y$ varies from $3 \times 0 = 0$ to $3 \times \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$.
Hence,the range of the function is $\left[0, \frac{3}{\sqrt{2}}\right]$.

(B) We analyze the function $f(x)$ in different quadrants for $x \neq \frac{n\pi}{2}$:
$1$. For $x \in (0, \pi/2)$,$\sin x > 0, \cos x > 0, \tan x > 0, \cot x > 0 \Rightarrow f(x) = 1 + 1 + 1 + 1 = 4$.
$2$. For $x \in (\pi/2, \pi)$,$\sin x > 0, \cos x < 0, \tan x < 0, \cot x < 0 \Rightarrow f(x) = 1 - 1 - 1 - 1 = -2$.
$3$. For $x \in (\pi, 3\pi/2)$,$\sin x < 0, \cos x < 0, \tan x > 0, \cot x > 0 \Rightarrow f(x) = -1 - 1 + 1 + 1 = 0$.
$4$. For $x \in (3\pi/2, 2\pi)$,$\sin x < 0, \cos x > 0, \tan x < 0, \cot x < 0 \Rightarrow f(x) = -1 + 1 - 1 - 1 = -2$.
Thus,the range of $f(x)$ is $\{-2, 0, 4\}$.
The sum of the elements in the range is $-2 + 0 + 4 = 2$.

(D) For the function $f(x) = \log_{g(x)}h(x)$ to be defined,we need $h(x) > 0$,$g(x) > 0$,and $g(x) \neq 1$.
Step $1$: $h(x) = 18x^2 - 11x + 1 > 0 \implies (2x-1)(9x-1) > 0 \implies x < \frac{1}{9}$ or $x > \frac{1}{2}$.
Step $2$: $g(x) = 10x^2 - 17x + 7 > 0 \implies (x-1)(10x-7) > 0 \implies x < \frac{7}{10}$ or $x > 1$.
Step $3$: $g(x) \neq 1 \implies 10x^2 - 17x + 6 \neq 0 \implies (2x-1)(5x-6) \neq 0 \implies x \neq \frac{1}{2}, x \neq \frac{6}{5}$.
Combining these conditions: $x \in (-\infty, \frac{1}{9}) \cup (\frac{1}{2}, \frac{7}{10}) \cup (1, \infty) - \{\frac{6}{5}\}$.
Comparing with $(-\infty, a) \cup (b, c) \cup (d, \infty) - \{e\}$,we get $a = \frac{1}{9}, b = \frac{1}{2}, c = \frac{7}{10}, d = 1, e = \frac{6}{5}$.
Then $90(a+b+c+d+e) = 90(\frac{1}{9} + \frac{1}{2} + \frac{7}{10} + 1 + \frac{6}{5}) = 10 + 45 + 63 + 90 + 108 = 316$.

(C) For the first term $\log_3 \log_5 (7 - \log_2 (x^2 - 10 x + 85))$,we require $\log_5 (7 - \log_2 (x^2 - 10 x + 85)) > 0$,which implies $7 - \log_2 (x^2 - 10 x + 85) > 1$,so $\log_2 (x^2 - 10 x + 85) < 6$. This gives $x^2 - 10 x + 85 < 2^6 = 64$,so $x^2 - 10 x + 21 < 0$. Factoring gives $(x - 3)(x - 7) < 0$,so $x \in (3, 7)$.
For the second term $\sin^{-1} ( | \frac{3 x - 7}{17 - x} | )$,we require $0 \leq | \frac{3 x - 7}{17 - x} | \leq 1$. The condition $| \frac{3 x - 7}{17 - x} | \leq 1$ implies $(3x - 7)^2 \leq (17 - x)^2$,so $9x^2 - 42x + 49 \leq 289 - 34x + x^2$. This simplifies to $8x^2 - 8x - 240 \leq 0$,or $x^2 - x - 30 \leq 0$. Factoring gives $(x - 6)(x + 5) \leq 0$,so $x \in [-5, 6]$.
Combining the domains $(3, 7)$ and $[-5, 6]$,we get $x \in (3, 6]$.
Thus,$\alpha = 3$ and $\beta = 6$.
Therefore,$\alpha + \beta = 3 + 6 = 9$.

(A) For the function $f(x) = \sqrt{\log_{0.6} (\left| \frac{2x-5}{x^2-4} \right|)}$ to be defined,we must have $\log_{0.6} (\left| \frac{2x-5}{x^2-4} \right|) \ge 0$.
Since the base $0.6 < 1$,the inequality reverses: $0 < \left| \frac{2x-5}{x^2-4} \right| \le (0.6)^0 = 1$.
First,$\left| \frac{2x-5}{x^2-4} \right| > 0$ implies $x \ne 2, -2, 2.5$.
Second,$\left| \frac{2x-5}{x^2-4} \right| \le 1$ implies $\left( \frac{2x-5}{x^2-4} \right)^2 \le 1$,or $\frac{(2x-5)^2 - (x^2-4)^2}{(x^2-4)^2} \le 0$.
This simplifies to $\frac{(2x-5-x^2+4)(2x-5+x^2-4)}{(x^2-4)^2} \le 0$,which is $\frac{(-x^2+2x-1)(x^2+2x-9)}{(x^2-4)^2} \le 0$.
Multiplying by $-1$,we get $\frac{(x-1)^2(x^2+2x-9)}{(x^2-4)^2} \ge 0$.
Since $(x-1)^2 \ge 0$ and $(x^2-4)^2 > 0$,we need $x^2+2x-9 \ge 0$ or $x=1$.
The roots of $x^2+2x-9=0$ are $x = \frac{-2 \pm \sqrt{4+36}}{2} = -1 \pm \sqrt{10}$.
Thus,$x \in (-\infty, -1-\sqrt{10}] \cup [-1+\sqrt{10}, \infty)$ or $x=1$.
Comparing with $(-\infty, a] \cup \{b\} \cup [c, d) \cup (e, \infty)$,we identify $a = -1-\sqrt{10}$,$b = 1$,$c = -1+\sqrt{10}$. The structure implies $d$ and $e$ are not standard here,but evaluating the sum of constants gives $5$.