20. (12 marks) (I) Let the daily production quantities of products $A$ and $B$ be $x$ and $y$ respectively, with corresponding profit $z$. Then we have
$$\left\{ \begin{array} { l } 2 x + 1.5 y \leq W \\ x + 1.5 y \leq 12 \\ 2 x - y \geq 0 \\ x \geq 0 , \quad y \geq 0 \end{array} \right.$$
The objective function is $z = 1000x + 1200y$.
[Figure]
Solution diagram 1 for Question 20
[Figure]
Solution diagram 2 for Question 20
[Figure]
Solution diagram 3 for Question 20
When $W = 12$, the planar region represented by (1) is shown in Figure 1, with three vertices $A ( 0,0 ) , B ( 2.4,4.8 ) , C ( 6,0 )$ respectively. Rewriting $z = 1000 x + 1200 y$ as $y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$, when $x = 2.4 , y = 4.8$, the line $l : y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$ has maximum $y$-intercept, maximum profit $Z = z _ { \max } = 2.4 \times 1000 + 4.8 \times 1200 = 8160$. When $W = 15$, the planar region represented by (1) is shown in Figure 2, with three vertices $A ( 0,0 ) , B ( 3,6 ) , C ( 7.5,0 )$ respectively. Rewriting $z = 1000 x + 1200 y$ as $y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$, when $x = 3 , y = 6$, the line $l : y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$ has maximum $y$-intercept, maximum profit $Z = z _ { \max } = 3 \times 1000 + 6 \times 1200 = 10200$. When $W = 18$, the planar region represented by (1) is shown in Figure 3, with four vertices $A ( 0,0 ) , B ( 3,6 ) , C ( 6,4 ) , D ( 9,0 )$ respectively. Rewriting $z = 1000 x + 1200 y$ as $y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$, when $x = 6 , y = 4$, the line $l : y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$ has maximum $y$-intercept, maximum profit $Z = z _ { \max } = 6 \times 1000 + 4 \times 1200 = 10800$. Thus the distribution of maximum profit $Z$ is

$Z$	8160	10200	10800
$P$	0.3	0.5	0.2

Therefore, $E ( Z ) = 8160 \times 0.3 + 10200 \times 0.5 + 10800 \times 0.2 = 9708$. (II) From (I), the probability that daily maximum profit exceeds 10000 yuan is $p _ { 1 } = P ( Z > 10000 ) = 0.5 + 0.2 = 0.7$. By the binomial distribution, the probability that at least one day out of 3 days has maximum profit exceeding 10000 yuan is $p = 1 - \left( 1 - p _ { 1 } \right) ^ { 3 } = 1 - 0.3 ^ { 3 } = 0.973$.

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

3. Execute the flowchart shown in Figure 1. If the input is $n = 3$, then the output is
A. $\frac { 6 } { 7 }$
B. $\frac { 3 } { 7 }$
C. $\frac { 8 } { 9 }$
D. $\frac { 4 } { 9 }$ [Figure]
(4) If variables $x , y$ satisfy the constraint conditions $\left\{ \begin{array} { c } x + y \geq - 1 \\ 2 x - y \leq 1 \\ y \leq 1 \end{array} \right.$, then the minimum value of $z = 3 x - y$ is
(A) $- 7$
(B) $- 1$
(C) $1$
(D) $2$
(5) 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 $(0,1)$ (6) Given that the expansion of $\left( \sqrt { \mathrm { x } } - \frac { \mathrm { a } } { \sqrt { \mathrm { x } } } \right) ^ { 5 }$ contains a term with $\mathrm { x } ^ { \frac { 3 } { 2 } }$ whose coefficient is 30, then $\mathrm { a } =$
(A) $\sqrt { 3 }$
(B) $- \sqrt { 3 }$
(C) $6$
(D) $- 6$ (7) In the square shown in Figure 2, 10000 points are randomly thrown. The estimated number of points falling in the shaded region (where curve C is the density curve of the normal distribution $N ( 0,1 )$) is
[Figure]
Figure 2
(A) $2386$
(B) $2718$
(C) $3413$
(D) $4772$
Attachment: If $X \sim N \left( \mu , \sigma ^ { 2 } \right)$, then
$$\begin{aligned} & P ( \mu - \sigma < x \leq \mu + \sigma ) = 0.6826 \\ & P ( \mu - 2 \sigma < x \leq \mu + 2 \sigma ) = 0.9544 \end{aligned}$$
(8) Points $\mathrm { A } , \mathrm { B } , \mathrm { C }$ move on the circle $\mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } = 1$, and $\mathrm { AB } \perp \mathrm { BC }$. If point P has coordinates $( 2,0 )$, then the maximum value of $| \overrightarrow { P A } + \overrightarrow { P B } + \overrightarrow { P C } |$ is
A. $6$
B. $7$
C. $8$
D. $9$ (9) The graph of the function $f ( x ) = \sin 2 x$ is shifted to the right by $\varphi \left( 0 < \varphi < \frac { \pi } { 2 } \right)$ units to obtain the graph of function $g ( x )$. If for $x _ { 1 }$ and $x _ { 2 }$ satisfying $\left| f \left( x _ { 1 } \right) - g \left( x _ { 2 } \right) \right| = 2$, we have $\left| x _ { 1 } - x _ { 2 } \right| _ { \min } = \frac { \pi } { 3 }$, then $\varphi =$
A. $\frac { 5 \pi } { 12 }$
B. $\frac { \pi } { 3 }$
C. $\frac { \pi } { 4 }$
D. $\frac { \pi } { 6 }$

Given a vector $\vec { v } = ( - 2,3 )$ and two points $A$ and $B$ on the coordinate plane, where the $x$-coordinate and $y$-coordinate of point $A$, and the $x$-coordinate and $y$-coordinate of point $B$ all lie in the interval $[ 0,1 ]$, what is the maximum value of $| \vec { v } + \overrightarrow { A B } |$?
(1) $\sqrt { 13 }$
(2) $\sqrt { 17 }$
(3) $3 \sqrt { 2 }$
(4) 5
(5) $\sqrt { 2 } + \sqrt { 13 }$