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) $y = 2 ^ { x }$

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

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 $( 0,1 )$

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 $C _ { 2 }$

6. Let $f ( x )$ be an odd function, and when $x \geq 0$, $f ( x ) = \mathrm { e } ^ { x } - 1$. Then when $x < 0$, $f ( x ) =$
A. $\mathrm { e } ^ { - x } - 1$
B. $\mathrm { e } ^ { - x } + 1$
C. $- \mathrm { e } ^ { - x } - 1$
D. $- \mathrm { e } ^ { - x } + 1$

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.

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)

12. Let the function $f ( x )$ and its derivative $f ^ { \prime } ( x )$ both have domain $\mathbf { R }$. Let $g ( x ) = f ^ { \prime } ( x )$. If $f \left( \frac { 3 } { 2 } - 2 x \right)$ and $g ( 2 + x )$ are both even functions, then
A. $f ( 0 ) = 0$
B. $g \left( - \frac { 1 } { 2 } \right) = 0$
C. $f ( - 1 ) = f ( 4 )$
D. $g ( - 1 ) = g ( 2 )$

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

If $y = (x - 1)^{2} + ax + \sin\left(x + \frac{\pi}{2}\right)$ is an even function, then $a =$ $\_\_\_\_$ .

Let $f(x)$ be an even function defined on $\mathbb{R}$ with period $2$. When $2 \leq x \leq 3$, $f(x) = 5 - 2x$. Then $f\left(-\frac{3}{4}\right) =$
A. $-\frac{1}{2}$
B. $-\frac{1}{4}$
C. $\frac{1}{4}$
D. $\frac{1}{2}$

Let $f(x)$ be an even function defined on $\mathbf{R}$ with period 2. When $2 \leq x \leq 3$, $f(x) = 5 - 2x$. Then $f\left(-\frac{3}{4}\right) =$
A. $-\frac{1}{2}$
B. $-\frac{1}{4}$
C. $\frac{1}{4}$
D. $\frac{1}{2}$

Justify that every function $h : I \rightarrow \mathbb{R}$ can be written uniquely in the form $h = p + i$ with $p : I \rightarrow \mathbb{R}$ an even function and $i : I \rightarrow \mathbb{R}$ an odd function.

Let $f : \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow \mathbb{R}$ be given by $$f(x) = (\log(\sec x + \tan x))^3$$ Then
(A) $f(x)$ is an odd function
(B) $f(x)$ is a one-one function
(C) $f(x)$ is an onto function
(D) $f(x)$ is an even function

Let $f ( x ) = \sin \left( \frac { \pi } { 6 } \sin \left( \frac { \pi } { 2 } \sin x \right) \right)$ for all $x \in \mathbb { R }$ and $g ( x ) = \frac { \pi } { 2 } \sin x$ for all $x \in \mathbb { R }$. Let $( f \circ g ) ( x )$ denote $f ( g ( x ) )$ and $( g \circ f ) ( x )$ denote $g ( f ( x ) )$. Then which of the following is (are) true?
(A) Range of $f$ is $\left[ - \frac { 1 } { 2 } , \frac { 1 } { 2 } \right]$
(B) Range of $f \circ g$ is $\left[ - \frac { 1 } { 2 } , \frac { 1 } { 2 } \right]$
(C) $\lim _ { x \rightarrow 0 } \frac { f ( x ) } { g ( x ) } = \frac { \pi } { 6 }$
(D) There is an $x \in \mathbb { R }$ such that $( g \circ f ) ( x ) = 1$

Let $f ( x ) = \frac { 1 } { 7 - \sin 5 x }$ be a function defined on $\mathbf { R }$. Then the range of the function $f ( x )$ is equal to:
(1) $\left[ \frac { 1 } { 7 } , \frac { 1 } { 6 } \right]$
(2) $\left[ \frac { 1 } { 8 } , \frac { 1 } { 5 } \right]$
(3) $\left[ \frac { 1 } { 7 } , \frac { 1 } { 5 } \right]$
(4) $\left[ \frac { 1 } { 8 } , \frac { 1 } { 6 } \right]$

Let $[ x ]$ denote the greatest integer less than or equal to $x$. Then the domain of $f ( x ) = \sec ^ { - 1 } ( 2 [ x ] + 1 )$ is :
(1) $( - \infty , - 1 ] \cup [ 0 , \infty )$
(2) $( - \infty , - 1 ] \cup [ 1 , \infty )$
(3) $( - \infty , \infty )$
(4) $( - \infty , \infty ) - \{ 0 \}$

The graph of the function $f ( x ) = x ^ { 2 } - 2 x + 3$ is translated $a$ units to the right and $b$ units downward to obtain the graph of the function $g ( x ) = x ^ { 2 } - 8 x + 14$.
Accordingly, what is the value of the expression $| \mathbf { a } | + | \mathbf { b } |$?
A) 4
B) 5
C) 6
D) 7
E) 8

The graph of the function $f$ is given below.
Given that $\mathbf { g } ( \mathbf { x } ) = \mathbf { 3 } - \mathbf { f } ( \mathbf { x } - \mathbf { 2 } )$, what is the sum $\mathbf { g } ( - \mathbf { 2 } ) + \mathbf { g } ( \mathbf { 5 } )$?
A) - 3
B) - 1
C) 1
D) 2
E) 3

A function f defined on the set R of real numbers

• For every $x \in [ -10,10 ]$, $f ( x ) = | x |$
• For every $x \in R$, $f ( x ) = f ( x + 20 )$

satisfies these properties. Accordingly, what is the value of $f ( 117 )$?
A) 3
B) 4
C) 6
D) 7
E) 9

The function f on the set of real numbers is defined for every real number x as
$$f ( x ) = \left\{ \begin{array} { c c } x + 2 , & x < 0 \\ x , & x \geq 0 \end{array} \right.$$
Accordingly, what is the value of the sum $\sum _ { k = - 3 } ^ { 4 } f ( k )$?
A) 8
B) 10
C) 12
D) 14
E) 16

In the rectangular coordinate plane, the graph of a function f defined on the interval $[ 0,2 ]$ is given below.
Accordingly, I. $( f \circ f ) ( x ) = 2$ II. $( f \circ f ) ( x ) = 1$ III. $( f \circ f ) ( x ) = 0$ Which of these equalities are satisfied for exactly two different values of x?
A) Only I
B) Only II
C) Only III
D) I and II
E) II and III