1. The solution set of the inequality $\frac { 2 - x } { x + 4 } > 0$ is $\_\_\_\_$ $( - 4,2 )$.
Analysis: This examines the method for solving fractional inequalities. $\frac { 2 - x } { x + 4 } > 0$ is equivalent to $( x - 2 ) ( x + 4 ) < 0$, so $- 4 < x < 2$

2. The solution set of the inequality $\frac { 2 - x } { x + 4 } > 0$ is $\_\_\_\_$.

15. The maximum value of the objective function $z = x + y$ subject to the linear constraints $\left\{ \begin{array} { l } 2 x + y \leq 3 , \\ x + 2 y \leq 3 , \\ x \geq 0 , \\ y \geq 0 \end{array} \right.$ is
A. 1
B. $\frac { 3 } { 2 }$
C. 2
D. 3
16. ``$x = 2 k \pi + \frac { \pi } { 4 } ( k \in \mathbf { Z } )$'' is a condition for ``$\tan x = 1$'' that is
A. sufficient but not necessary
B. necessary but not sufficient
C. necessary and sufficient
D. neither sufficient nor necessary

22. (Total Score: 16 points) Subproblem 1: 3 points, Subproblem 2: 5 points, Subproblem 3: 8 points.
If real numbers $x , y , m$ satisfy $| x - m | < | y - m |$ , then $x$ is said to be closer to $m$ than $y$ is.
(1) If $x ^ { 2 } - 1$ is closer to 3 than to 0, find the range of $x$;
(2) For any two distinct positive numbers $a , b$ , prove that $a ^ { 2 } b + a b ^ { 2 }$ is closer to $2 a b \sqrt { a b }$ than $a ^ { 3 } + b ^ { 3 }$ is;
(3) Given that the domain of function $f ( x )$ is $D = \{ x \mid x \neq k \pi , k \in \mathbf { Z } , x \in \mathbf { R } \}$ . For any $x \in D$ , $f ( x )$ equals whichever of $1 + \sin x$ and $1 - \sin x$ is closer to 0. Write the analytical expression for $f ( x )$ and indicate its parity, minimum positive period, minimum value, and monotonicity (proofs of conclusions are not required).

1. If the universal set $U = \mathbb{R}$, set $A = \{ x \mid x \geq 1 \}$, then $C_U A =$ $\_\_\_\_$

6. The solution to the inequality $\frac{1}{x} < 1$ is $\_\_\_\_$

16. If $a, b \in \mathbb{R}$ and $ab > 0$, then which of the following inequalities always holds?
(A) $a^2 + b^2 > 2ab$
(B) $a + b \geq 2\sqrt{ab}$
(C) $\frac{1}{a} + \frac{1}{b} > \frac{2}{\sqrt{ab}}$
(D) $\frac{b}{a} + \frac{a}{b} \geq 2$

If set $A = \{ x \mid -5 < x < 2 \}$, $B = \{ x \mid -3 < x < 3 \}$, then $A \cap B =$

1. Let set $A = \{ \mathrm { x } / ( \mathrm { x } + 1 ) ( \mathrm { x } - 2 ) < 0 \}$, set $B = \{ \mathrm { x } / 1 < \mathrm { x } < 3 \}$, then $A \cup B =$
A. $\{ X / - 1 < X < 3 \}$ B. $\{ X / - 1 < X < 1 \}$ C. $\{ X / 1 < X < 2 \}$ D. $\{ X / 2 < X < 3 \}$

1. Given sets $\mathrm { P } = \left\{ x \mid x ^ { 2 } - 2 x \geq 3 \right\} , \mathrm { Q } = \{ x \mid 2 < x < 4 \}$ , then $\mathrm { P } \cap \mathrm { Q } =$ $\_\_\_\_$

1. Given sets $P = \left\{ x \mid x ^ { 2 } - 2 x \geq 0 \right\} , Q = \{ x \mid 1 < x \leq 2 \}$ , then $\left( \complement _ { \mathbb{R} } P \right) \cap Q =$
A. $[0,1)$
B. $( 0,2 ]$
C. $( 1,2 )$
D. $[ 1,2 ]$

2. If $x , y$ satisfy $\left\{ \begin{array} { l } x - y \leqslant 0 , \\ x + y \leqslant 1 , \\ x \geqslant 0 , \end{array} \right.$ then the maximum value of $z = x + 2 y$ is
A. $0$
B. $1$
C. $\frac { 3 } { 2 }$
D. $2$

2. Let variables $x , y$ satisfy the constraint conditions $x - 2y \geq 0$, $x \leq 2$, $y \geq 0$. Then the maximum value of the objective function $z = 3x + y$ is
(A) 7
(B) 8
(C) 9
(D) 14

Variables $x, y$ satisfy the constraints $\left\{\begin{array}{c}x + 2 \geq 0, \\ x - y + 3 \geq 0, \\ 2x + y - 3 \leq 0,\end{array}\right.$ then the maximum value of the objective function $Z = x + 6y$ is
(A) 3
(B) 4
(C) 18
(D) 40

3. Let $x \in R$. Then ``$x > 1$'' is ``$x ^ { 3 } > 1$'' a
A. sufficient but not necessary condition
B. necessary but not sufficient condition
C. necessary and sufficient condition
D. neither sufficient nor necessary condition

4. If variables $x , y$ satisfy the constraint conditions $\left\{ \begin{array} { c } x + y \geq 1 , \\ y - x \leq 1 , \text { then } z = 2 x - y \text { has a minimum value of } \\ x \leq 1 , \end{array} \right.$
A. $-1$
B. 0
C. 1
D. 2

4. Let $a , b$ be positive real numbers. Then ``$a > b > 1$'' is a \_\_\_\_ condition for ``$\log _ { 2 } a > \log _ { 2 } b > 0$''
(A) necessary and sufficient condition
(B) sufficient but not necessary condition
(C) necessary but not sufficient condition
(D) neither sufficient nor necessary condition

4. Let $x \in \mathbb{R}$. Then ``$1 < x < 2$'' is ``$|x - 2| < 1$'' a
(A) sufficient but not necessary condition
(B) necessary but not sufficient condition
(C) necessary and sufficient condition
(D) neither sufficient nor necessary condition

Let $x \in \mathbb{R}$. Then ``$|x - 2| < 1$'' is ``$x^2 + x - 2 > 0$'' a
(A) sufficient but not necessary condition
(B) necessary but not sufficient condition
(C) necessary and sufficient condition
(D) neither sufficient nor necessary condition

5. Given that $\mathrm { x } , \mathrm { y }$ satisfy the constraints $\left\{ \begin{array} { c } x - y \geq 0 \\ x + y - 4 \leq 0 \\ y \geq 1 \end{array} \right.$, then the maximum value of $\mathrm { z } = - 2 \mathrm { x } + \mathrm { y }$ is
(A) $- 1$
(B) $- 2$
(C) $- 5$
(D) $1$

5. If variables $x$ and $y$ satisfy the constraint conditions $\left\{ \begin{array} { l } x + 2 y \geq 0 , \\ x - y \leq 0 , \\ x - 2 y + 2 \geq 0 , \end{array} \right.$ then the minimum value of $z = 2 x - y$ equals
A. $- \frac { 5 } { 2 }$
B. $- 2$
C. $- \frac { 3 } { 2 }$
D. 2

7. If real numbers $\mathrm { a } , \mathrm { b }$ satisfy $\frac { 1 } { a } + \frac { 2 } { b } = \sqrt { a b }$, then the minimum value of $ab$ is
A. $\sqrt { 2 }$
B. 2
C. $2 \sqrt { 2 }$
D. 4

9. Let real numbers $x , y$ satisfy $\left\{ \begin{array} { c } 2 x + y \leq 10 , \\ x + 2 y \leq 14 , \\ x + y \geq 6 , \end{array} \right.$ Then the maximum value of $x y$ is
(A) $\frac { 25 } { 2 }$
(B) $\frac { 49 } { 2 }$
(C) 12
(D) 16

If the system of inequalities $\left\{ \begin{array} { c } x + y - 2 \leq 0 \\ x + 2 y - 2 \geq 0 \\ x - y + 2 m \geq 0 \end{array} \right.$ represents a triangular region with area equal to $\frac { 4 } { 3 }$, then the value of $m$ is
(A) $-3$
(B) $1$
(C) $\frac { 4 } { 3 }$
(D) $3$

12. Let variables $\mathrm { x } , \mathrm { y }$ satisfy the constraints $\left\{ \begin{array} { c } x + y \leq 4 \\ x - y \leq 2 \\ 3 x - y \geq 0 \end{array} \right.$. Then the maximum value of $3 x + y$ is $\_\_\_\_$.