Recall that $\arcsin(t)$ (also known as $\sin^{-1}(t)$) is a function with domain $[-1,1]$ and range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Consider the function $f(x) := \arcsin(\sin(x))$ and answer the following questions as a series of four letters (T for True and F for False) in order.
(a) The function $f(x)$ is well defined for all real numbers $x$.
(b) The function $f(x)$ is continuous wherever it is defined.
(c) The function $f(x)$ is differentiable wherever it is continuous.

Recall the function $\arctan(x)$, also denoted as $\tan^{-1}(x)$. Complete the sentence: $$\arctan(20202019) + \arctan(20202021) \quad\underline{\hspace{2cm}}\quad 2\arctan(20202020),$$ because in the relevant region, the graph of $y = \arctan(x)$ $\_\_\_\_$.
Fill in the first blank with one of the following: is less than / is equal to / is greater than. Fill in the second blank with a single correct reason consisting of one of the following phrases: is bounded / is continuous / has positive first derivative / has negative first derivative / has positive second derivative / has negative second derivative / has an inflection point.

For a constant $a$, define the function $f ( x )$ as $$f ( x ) = \lim _ { n \rightarrow \infty } \frac { ( a - 2 ) x ^ { 2 n + 1 } + 2 x } { 3 x ^ { 2 n } + 1 }$$ What is the sum of all values of $a$ such that $( f \circ f ) ( 1 ) = \frac { 5 } { 4 }$? [4 points]
(1) $\frac { 11 } { 2 }$
(2) $\frac { 13 } { 2 }$
(3) $\frac { 15 } { 2 }$
(4) $\frac { 17 } { 2 }$
(5) $\frac { 19 } { 2 }$

15. If the graph of function $y = f ( x )$ can be obtained by rotating the graph of function $y = \lg ( x + 1 )$ counterclockwise by $\frac { \pi } { 2 }$ around the origin O, then $f ( x ) =$
A. $10 ^ { - x } - 1$.
B. $10 ^ { x } - 1$.
C. $1 - 10 ^ { - x }$.
D. $1 - 10 ^ { x }$.

14. Let $g(x)$ be a function defined on $\mathbb{R}$ with period 1. If the function $f(x) = x + g(x)$ has range $[-2, 5]$ on the interval $[0, 1]$, then the range of $f(x)$ on the interval $[0, 3]$ is $\_\_\_\_$ II. Multiple Choice Questions (Total: 20 points) This section contains 4 questions. Each question has exactly one correct answer. Candidates should shade the corresponding box on the answer sheet. Each correct answer is worth 5 points; otherwise, zero points are awarded.

2. Which of the following functions is an odd function?
A. $y = \sqrt { x }$
B. $y = | \sin x |$
C. $y = \cos x$
D. $y = e ^ { x } - e ^ { - x }$

Which of the following functions is an even function?\n(A) $y = x ^ { 2 } \sin x$\n(B) $y = x ^ { 2 } \cos x$\n(C) $y = | \ln x |$\n(D) $y = 2 ^ { x }$

4. Among the following functions, which one is both an even function and has a zero point?
(A) $y = \ln x$
(B) $y = x ^ { 2 } + 1$
(C) $y = \sin x$
(D) $y = \cos x$

Let the function $\left\{ a _ { \mathrm { n } } \right\} =$ , then $f(-2) + f(0) =$
(A) $3$
(B) $6$
(C) $9$
(D) $12$

6. The sign function is defined as $\operatorname{sgn} x = \begin{cases} 1, & x > 0, \\ 0, & x = 0, \\ -1, & x < 0. \end{cases}$ Let $f(x)$ be an increasing function on $\mathbf{R}$, and $g(x) = f(x) - f(ax)$ where $a > 1$. Then
A. $\operatorname{sgn}[g(x)] = \operatorname{sgn} x$
B. $\operatorname{sgn}[g(x)] = -\operatorname{sgn} x$
C. $\operatorname{sgn}[g(x)] = \operatorname{sgn}[f(x)]$
D. $\operatorname{sgn}[g(x)] = -\operatorname{sgn}[f(x)]$

8. Let the function $f ( x ) = \ln ( 1 + x ) - \ln ( 1 - x )$. Then $f ( x )$ is
A. an odd function and increasing on $( 0,1 )$
B. an odd function and decreasing on $( 0,1 )$
C. an even function and increasing on $( 0,1 )$
D. an even function and decreasing on $( 0,1 )$

14. If the function $f ( x ) = \left\{ \begin{array} { l } - x + 6 , x \leq 2 , \\ 3 + \log _ { a } x , x > 2 , \end{array} ( a > 0 \right.$ and $a \neq 1 )$ has range $[ 4 , + \infty )$, then the range of the real number $a$ is $\_\_\_\_$.

15. Given functions $f ( x ) = 2 ^ { x } , g ( x ) = \hat { x } ^ { 2 } + a _ { 2 }$ (where $a \in R$). For unequal real numbers $x _ { 1 } , x _ { 2 }$, let $m = \frac { f \left( x _ { 1 } \right) - f \left( x _ { 2 } \right) } { x _ { 1 } - x _ { 2 } } , n = \frac { g \left( x _ { 1 } \right) - g \left( x _ { 2 } \right) } { x _ { 1 } - x _ { 2 } }$. Consider the following propositions:
(1) For any unequal real numbers $x _ { 1 } , x _ { 2 }$, we have $m > 0$; (2) For any $a$ and any unequal real numbers $x _ { 1 } , x _ { 2 }$, we have $n > 0$;
(3) For any $a$, there exist unequal real numbers $x _ { 1 } , x _ { 2 }$ such that $m = n$; (4) For any $a$, there exist unequal real numbers $x _ { 1 } , x _ { 2 }$ such that $m = - n$. The true propositions are \_\_\_\_ (write the numbers of all true propositions).
III. Solution Questions:

17. (11 points)
A student uses the ``five-point method'' to sketch the graph of $f(x) = A\sin(\omega x + \varphi)$ (where $\omega > 0$, $|\varphi| < \frac{\pi}{2}$) over one period and creates a table with partial data filled in as follows:
\begin{tabular}{ | c | c | c | c | c | c | } \hline $\omega x + \varphi$ & 0 & $\frac{\pi}{2}$ & $\pi$ & $\frac{3\pi}{2}$ & $2\pi$ \hline $x$ & & $\frac Thus $\overrightarrow { P B } \cdot \overrightarrow { D E } = 0$, that is, $P B \perp D E$. Since $E F \perp P B$ and $D E \cap E F = E$, we have $P B \perp$ plane $D E F$. Because $\overrightarrow { P C } = ( 0,1 , - 1 ) , \overrightarrow { D E } \cdot \overrightarrow { P C } = 0$, then $D E \perp P C$, so $D E \perp$ plane $P B C$. From $D E \perp$ plane $P B C$ and $P B \perp$ plane $D E F$, we know that all four faces of tetrahedron $B D E F$ are right triangles, that is, tetrahedron $B D E F$ is an orthocentric tetrahedron, with right angles on its four faces being $\angle D E B , \angle D E F , \angle E F B , \angle D F B$ respectively.
[Figure]
Solution diagram 1 for Question 19
[Figure]
Solution diagram 2 for Question 19
(II) Since $P D \perp$ plane $A B C D$, we have $\overrightarrow { D P } = ( 0,0,1 )$ is a normal vector to plane $A B C D$; From (I), $P B \perp$ plane $D E F$, so $\overrightarrow { B P } = ( - \lambda , - 1,1 )$ is a normal vector to plane $D E F$. If the dihedral angle between plane $D E F$ and plane $A B C D$ is $\frac { \pi } { 3 }$, then $\cos \frac { \pi } { 3 } = \left| \frac { \overrightarrow { B P } \cdot \overrightarrow { D P } } { | \overrightarrow { B P } | \cdot | \overrightarrow { D P } | } \right| = \left| \frac { 1 } { \sqrt { \lambda ^ { 2 } + 2 } } \right| = \frac { 1 } { 2 }$, solving gives $\lambda = \sqrt { 2 }$. Therefore $\frac { D C } { B C } = \frac { 1 } { \lambda } = \frac { \sqrt { 2 } } { 2 }$. Thus when the dihedral angle between plane $D E F$ and plane $A B C D$ is $\frac { \pi } { 3 }$, we have $\frac { D C } { B C } = \frac { \sqrt { 2 } } { 2 }$.

Given curves $C _ { 1 }: y = \cos x$ and $C _ { 2 }: y = \sin \left( 2 x + \frac { \pi } { 6 } \right)$, which of the following conclusions is correct?
A. Stretch the horizontal coordinates of points on $C _ { 1 }$ to twice the original length while keeping the vertical coordinates unchanged, then shift the resulting curve to the right by $\frac { \pi } { 6 }$ units to obtain curve $C _ { 2 }$
B. Stretch the horizontal coordinates of points on $C _ { 1 }$ to twice the original length while keeping the vertical coordinates unchanged, then shift the resulting curve to the left by $\frac { \pi } { 12 }$ units to obtain curve $C _ { 2 }$
C. Compress the horizontal coordinates of points on $C _ { 1 }$ to half the original length while keeping the vertical coordinates unchanged, then shift the resulting curve to the right by $\frac { \pi } { 6 }$ units to obtain curve $C _ { 2 }$
D. Compress the horizontal coordinates of points on $C _ { 1 }$ to half the original length while keeping the vertical coordinates unchanged, then shift the resulting curve to the left by $\frac { \pi } { 12 }$ units to obtain curve $C _ { 2 }$

It is known that the function $f(x)$ is an odd function defined on $\mathbb{R}$. When $x \in (-\infty, 0)$, $f(x) = 2x^3 + x^2$. Then $f(2) = $ \_\_\_\_

Let $f ( x )$ be an odd function with domain $( - \infty , + \infty )$, satisfying $f ( 1 - x ) = f ( 1 + x )$ and $f ( 1 ) = 2$. Then $f ( 1 ) + f ( 2 ) + f ( 3 ) + \cdots + f ( 50 ) =$
A. $- 50$
B. $0$
C. $2$
D. $50$

11. Let $f ( x )$ be an even function with domain $\mathbf { R }$ that is monotonically decreasing on $( 0 , + \infty )$ . Then
A. $f \left( \log _ { 3 } \frac { 1 } { 4 } \right) > f \left( 2 ^ { - \frac { 3 } { 2 } } \right) > f \left( 2 ^ { - \frac { 2 } { 3 } } \right)$
B. $f \left( \log _ { 3 } \frac { 1 } { 4 } \right) > f \left( 2 ^ { - \frac { 2 } { 3 } } \right) > f \left( 2 ^ { - \frac { 3 } { 2 } } \right)$
C. $f \left( 2 ^ { - \frac { 3 } { 2 } } \right) > f \left( 2 ^ { - \frac { 2 } { 3 } } \right) > f \left( \log _ { 3 } \frac { 1 } { 4 } \right)$
D. $f \left( 2 ^ { - \frac { 2 } { 3 } } \right) > f \left( 2 ^ { - \frac { 3 } { 2 } } \right) > f \left( \log _ { 3 } \frac { 1 } { 4 } \right)$

12. Let the domain of function $f ( x )$ be $\mathbf { R }$, satisfying $f ( x + 1 ) = 2 f ( x )$, and when $x \in ( 0,1 ]$, $f ( x ) = x ( x - 1 )$. If for all $x \in ( - \infty , m ]$, we have $f ( x ) \geq - \frac { 8 } { 9 }$, then the range of $m$ is
A. $\left( - \infty , \frac { 9 } { 4 } \right]$
B. $\left( - \infty , \frac { 7 } { 3 } \right]$
C. $\left( - \infty , \frac { 5 } { 2 } \right]$
D. $\left( - \infty , \frac { 8 } { 3 } \right]$

II. Fill-in-the-Blank Questions: 4 questions in total, 5 points each, 20 points total.

12. Let $f ( x )$ be an even function with domain $\mathbf { R }$ , and monotonically decreasing on $( 0 , + \infty )$ . Then
A. $f \left( \log _ { 3 } \frac { 1 } { 4 } \right) > f \left( 2 ^ { - \frac { 3 } { 2 } } \right) > f \left( 2 ^ { - \frac { 2 } { 3 } } \right)$
B. $f \left( \log _ { 3 } \frac { 1 } { 4 } \right) > f \left( 2 ^ { - \frac { 2 } { 3 } } \right) > f \left( 2 ^ { - \frac { 3 } { 2 } } \right)$
C. $f \left( 2 ^ { - \frac { 3 } { 2 } } \right) > f \left( 2 ^ { - \frac { 2 } { 3 } } \right) > f \left( \log _ { 3 } \frac { 1 } { 4 } \right)$
D. $f \left( 2 ^ { - \frac { 2 } { 3 } } \right) > f \left( 2 ^ { - \frac { 3 } { 2 } } \right) > f \left( \log _ { 3 } \frac { 1 } { 4 } \right)$ II. Fill-in-the-Blank Questions: This section has 4 questions, each worth 5 points, for a total of 20 points.

Let the function $f ( x ) = \ln | 2 x + 1 | - \ln | 2 x - 1 |$ . Then $f ( x )$
A． is an even function and monotonically increasing on $\left( \frac { 1 } { 2 } , + \infty \right)$
B． is an odd function and monotonically decreasing on $\left( - \frac { 1 } { 2 } , \frac { 1 } { 2 } \right)$
C． is an even function and monotonically increasing on $\left( - \infty , - \frac { 1 } { 2 } \right)$
D． is an odd function and monotonically decreasing on $\left( - \infty , - \frac { 1 } { 2 } \right)$

8. Given that the domain of function $f ( x )$ is $\mathbf { R }$, $f ( x + 2 )$ is an even function, and $f ( 2 x + 1 )$ is an odd function, then ( )
A. $f \left( - \frac { 1 } { 2 } \right) = 0$
B. $f ( - 1 ) = 0$
C. $f ( 2 ) = 0$
D. $f ( 4 ) = 0$
【Answer】B 【Solution】 【Analysis】Derive that $f ( x )$ is a periodic function with period 4. From the given conditions, deduce that $f ( 1 ) = 0$, and combine with the given conditions to reach the conclusion.
【Detailed Solution】Since $f ( x + 2 )$ is an even function, we have $f ( 2 + x ) = f ( 2 - x )$, which gives $f ( x + 3 ) = f ( 1 - x )$. Since $f ( 2 x + 1 )$ is an odd function, we have $f ( 1 - 2 x ) = - f ( 2 x + 1 )$, so $f ( 1 - x ) = - f ( x + 1 )$. Therefore, $f ( x + 3 ) = - f ( x + 1 ) = f ( x - 1 )$, that is, $f ( x ) = f ( x + 4 )$. Thus, $f ( x )$ is a periodic function with period 4. Since $F ( x ) = f ( 2 x + 1 )$ is an odd function, we have $F ( 0 ) = f ( 1 ) = 0$. Therefore, $f ( - 1 ) = - f ( 1 ) = 0$, while the other three options are unknown. Therefore, the answer is: B.
II. Multiple Choice Questions: This section contains 4 questions, each worth 5 points, for a total of 20 points. For each question, there may be multiple correct options. Full marks (5 points) are awarded for selecting all correct options, 2 points for partially correct selections, and 0 points for any incorrect selection.

12. Let $f ( x )$ be an odd function with domain $\mathbb{R}$, and $f ( 1 + x ) = f ( - x )$. If $f \left( - \frac { 1 } { 3 } \right) = \frac { 1 } { 3 }$, then $f \left( \frac { 5 } { 3 } \right) =$
A. $- \frac { 5 } { 3 }$
B. $- \frac { 1 } { 3 }$
C. $\frac { 1 } { 3 }$
D. $\frac { 5 } { 3 }$

II. Fill-in-the-Blank Questions: This section contains 4 questions, each worth 5 points, for a total of 20 points.

12. Let the domain of function $f(x)$ be $\mathbb{R}$. $f(x+1)$ is an odd function, $f(x+2)$ is an even function. When $x \in [1,2]$, $f(x) = ax^2 + b$. If $f(0) + f(3) = 6$, then $f\left(\frac{4}{2}\right) =$
A. $-\frac{9}{4}$
B. $-\frac{3}{2}$
C. $\frac{7}{4}$
D. $\frac{5}{2}$

II. Fill in the Blank Questions (4 questions in total, 5 points each, 20 points total)

Given that the domain of function $f ( x )$ is $\mathbb { R }$, $f ( x + 2 )$ is an even function, and $f ( 2 x + 1 )$ is an odd function, then
A. $f \left( - \frac { 1 } { 2 } \right) = 0$
B. $f ( - 1 ) = 0$
C. $f ( 2 ) = 0$
D. $f ( 4 ) = 0$