21. Solution: The possible values of $X$ are $-1, 0, 1$.
$$\begin{aligned} & P(X = -1) = (1-\alpha)\beta, \\ & P(X = 0) = \alpha\beta + (1-\alpha)(1-\beta), \\ & P(X = 1) = \alpha(1-\beta), \end{aligned}$$
Therefore the distribution table of $X$ is

$X$	$-1$	$0$	$1$
$P$	$(1-\alpha)\beta$	$\alpha\beta + (1-\alpha)(1-\beta)$	$\alpha(1-\beta)$

(2) (i) From (1) we have $a = 0.4$, $b = 0.5$, $c = 0.1$. Therefore $p_i = 0.4p_{i-1} + 0.5p_i + 0.1p_{i+1}$, so $0.1(p_{i+1} - p_i) = 0.4(p_i - p_{i-1})$, that is $p_{i+1} - p_i = 4(p_i - p_{i-1})$. Since $p_1 - p_0 = p_1 \neq 0$, the sequence $\{p_{i+1} - p_i\}$ $(i = 0, 1, 2, \cdots, 7)$ is a geometric sequence with common ratio 4 and first term $p_1$.
(ii) From (i) we obtain $p_8 = p_8 - p_7 + p_7 - p_6 + \cdots + p_1 - p_0 + p_0 = (p_8 - p_7) + (p_7 - p_6) + \cdots + (p_1 - p_0) = \frac{4^8 - 1}{3}p_1$. Since $p_8 = 1$, we have $p_1 = \frac{3}{4^8 - 1}$, therefore $p_4 = (p_4 - p_3) + (p_3 - p_2) + (p_2 - p_1) + (p_1 - p_0) = \frac{4^4 - 1}{3}p_1 = \frac{1}{257}$. $p_4$ represents the probability of ultimately concluding that drug A is more effective. From the calculation result, we can see that when drug A has a cure rate of 0.5 and drug B has a cure rate of 0.8, the probability of concluding that drug A is more effective is $p_4 = \frac{1}{257} \approx 0.0039$. The probability of reaching an incorrect conclusion is very small, indicating that this experimental scheme is reasonable.

21. (12 points) Given the function $f ( x ) = ( x

22. Solution: (1) From the problem conditions, $|a| = 1$. Thus the parametric equation of $l$ is $\left\{\begin{array}{l} x = 4t + 1 \\ y = 3t - 1 \end{array}\right.$ ($t$ is the parameter). The parametric equation of circle $C$ is $\left\{\begin{array}{l} x = 1 + \cos\theta \\ y = -2 + \sin\theta \end{array}\right.$ ($\theta$ is the parameter). & \hline \end{tabular}
Eliminating parameter $t$, the ordinary equation of $l$ is $3x - 4y - 7 = 0$. ..... 3 marks
Eliminating parameter $\theta$, the ordinary equation of $C$ is $(x-1)^2 + (y+2)^2 = 1$. ..... 5 marks
(2) The equation of $l'$ is $y = \frac{3}{4}(x + m) - \frac{7}{4}$, i.e., $3x - 4y + 3m - 7 = 0$. ..... 6 marks
Since circle $C$ has only one point at distance $\mathbf{1}$ from $l'$, and the radius of circle $C$ is $\mathbf{1}$, the distance from $C(1, -2)$ to $l'$ is 2. ..... 8 marks
That is, $\frac{|3 + 8 + 3m - 7|}{5} = 2$. Solving, we get $m = 2$ ($m = -\frac{14}{3} < 0$ is rejected). ..... 10 marks

22. Solution: (1) From the given conditions, the polar coordinate equations of the circles containing arcs $AB$, $BC$, $CD$ are $\rho = 2\cos\theta$, $\rho = 2\sin\theta$, $\rho = -2\cos\theta$ respectively.
Thus the polar coordinate equation of $M_1$ is $\rho = 2\cos\theta \left(0 \leq \theta \leq \frac{\pi}{4}\right)$, the polar coordinate equation of $M_2$ is $\rho = 2\sin\theta \left(\frac{\pi}{4} \leq \theta \leq \frac{3\pi}{4}\right)$,
the polar coordinate equation of $M_3$ is $\rho = -2\cos\theta \left(\frac{3\pi}{4} \leq \theta \leq \pi\right)$.
(2) Let $P(\rho, \theta)$. From the given conditions and (1), we have
If $0 \leq \theta \leq \frac{\pi}{4}$, then $2\cos\theta = \sqrt{3}$, solving gives $\theta = \frac{\pi}{6}$;
If $\frac{\pi}{4} \leq \theta \leq \frac{3\pi}{4}$, then $2\sin\theta = \sqrt{3}$, solving gives $\theta = \frac{\pi}{3}$ or $\theta = \frac{2\pi}{3}$;
If $\frac{3\pi}{4} \leq \theta \leq \pi$, then $-2\cos\theta = \sqrt{3}$, solving gives $\theta = \frac{5\pi}{6}$.
In summary, the polar coordinates of $P$ are $\left(\sqrt{3}, \frac{\pi}{6}\right)$ or $\left(\sqrt{3}, \frac{\pi}{3}\right)$ or $\left(\sqrt{3}, \frac{2\pi}{3}\right)$ or $\left(\sqrt{3}, \frac{5\pi}{6}\right)$.

23. Solution: (1) When $a = 1$, $f(x) = \left\{\begin{array}{l} 5 - 2x, & x \leq 1 \\ 3, & 1 < x < 4 \\ 2x - 5, & x \geq 4 \end{array}\right.$ ..... 3 marks
Thus the solution set of the inequality $f(x) < x$ is $(3, 5)$. ..... 5 marks
(2) $f(x) = |x - a| + |x - 4| \geq |(x - a) - (x - 4)| = |a - 4|$. ..... 6 marks
$\therefore |a - 4| \geq \frac{4}{a} - 1 = \frac{4 - a}{a}$. ..... 7 marks
When $a < 0$ or $a \geq 4$, the inequality clearly holds. ..... 8 marks
When $0 < a < 4$, $\frac{1}{a} \leq 1$, then $1 \leq a < 4$. ..... 9 marks
Thus the range of $a$ is $(-\infty, 0) \cup [1, +\infty)$. ..... 10 marks
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23. Solution: (1) When $a = 1$, $f ( x ) = | x - 1 | x + | x - 2 | ( x - 1 )$ .
When $x < 1$, $f ( x ) = - 2 ( x - 1 ) ^ { 2 } < 0$ ; when $x \geq 1$, $f ( x ) \geq 0$ .
Therefore, the solution set of the inequality $f ( x ) < 0$ is $( - \infty , 1 )$ .
(2) Since $f ( a ) = 0$, we have $a \geq 1$ .
When $a \geq 1 , x \in ( - \infty , 1 )$, $f ( x ) = ( a - x ) x + ( 2 - x ) ( x - a ) = 2 ( a - x ) ( x - 1 ) < 0$ .
Therefore, the range of values for $a$ is $[ 1 , + \infty )$ .

23. Solution: (1) Since $[(x-1) + (y+1) + (z+1)]^2$
$= (x-1)^2 + (y+1)^2 + (z+1)^2 + 2[(x-1)(y+1) + (y+1)(z+1) + (z+1)(x-1)]$
$\leq 3[(x-1)^2 + (y+1)^2 + (z+1)^2]$,
from the given condition we have $(x-1)^2 + (y+1)^2 + (z+1)^2 \geq \frac{4}{3}$,
with equality if and only if $x = \frac{5}{3}$, $y = -\frac{1}{3}$, $z = -\frac{1}{3}$.
Therefore, the minimum value of $(x-1)^2 + (y+1)^2 + (z+1)^2$ is $\frac{4}{3}$.
(2) Since
$[(x-2) + (y-1) + (z-a)]^2$
$= (x-2)^2 + (y-1)^2 + (z-a)^2 + 2[(x-2)(y-1) + (y-1)(z-a) + (z-a)(x-2)]$
$\leq 3[(x-2)^2 + (y-1)^2 + (z-a)^2]$,
from the given condition $(x-2)^2 + (y-1)^2 + (z-a)^2 \geq \frac{(2+a)^2}{3}$,
with equality if and only if $x = \frac{4-a}{3}$, $y = \frac{1-a}{3}$, $z = \frac{2a-2}{3}$.
Therefore, the minimum value of $(x-2)^2 + (y-1)^2 + (z-a)^2$ is $\frac{(2+a)^2}{3}$.
From the given condition, $\frac{(2+a)^2}{3} \geq \frac{1}{3}$, solving gives $a \leq -3$ or $a \geq -1$.

The figure on the right shows the three views of a polyhedron. One endpoint of a certain edge of this polyhedron corresponds to point $M$ in the front view and point $N$ in the top view. This endpoint corresponds to which point in the side view?
A．$E$
B．$F$
C．$G$
D．$H$

The figure on the right shows the three-view drawing of a certain solid. The surface area of this solid is
A. $6 + 4 \sqrt { 2 }$
B. $4 + 4 \sqrt { 2 }$
C. $6 + 2 \sqrt { 3 }$
D. $4 + 2 \sqrt { 3 }$

Executing the flowchart on the right, the output value of $n =$
A. 17
B. 19
C. 21
D. 23

The figure on the right shows the three-view drawing of a certain geometric solid. The surface area of this solid is
A. $6 + 4 \sqrt { 2 }$
B. $4 + 4 \sqrt { 2 }$
C. $6 + 2 \sqrt { 3 }$
D. $4 + 2 \sqrt { 3 }$

It is known that $\triangle A B C$ is an equilateral triangle with area $\frac { 9 \sqrt { 3 } } { 4 }$ , and all its vertices lie on the surface of sphere $O$. If the surface area of sphere $O$ is $16 \pi$ , then the distance from $O$ to plane $A B C$ is
A．$\sqrt { 3 }$
B．$\frac { 3 } { 2 }$
C． 1
D．$\frac { \sqrt { 3 } } { 2 }$

Let $A , B , C$ be three points on the surface of sphere $O$, and $\odot O _ { 1 }$ be the circumcircle of $\triangle A B C$. If the area of $\odot O _ { 1 }$ is $4 \pi$ and $A B = B C = A C = O O _ { 1 }$ , then the surface area of sphere $O$ is
A. $64 \pi$
B. $48 \pi$
C. $36 \pi$
D. $32 \pi$

A cone has a base radius of 1 and slant height of 3. The volume of the largest sphere that can be inscribed in this cone is $\_\_\_\_$ .

1. B

Solution: Note that $4 \notin A , 3 \in A \cap B$, so the answer is $B$.

6. In a cube, the midpoints of the three edges passing through vertex $A$ are $E, F, G$ respectively. After cutting off the triangular pyramid $A-EFG$ from the cube, the front view of the resulting polyhedron is shown in the figure on the right. Then the corresponding left view is
A. [Figure]
B. [Figure]
C. [Figure]
D. [Figure]
[Figure]

7. In a cube, the midpoints of three edges passing through vertex $A$ are $E , F , G$ respectively. After cutting off the triangular pyramid $A - E F G$ from the cube, the front view of the orthogonal projection of the resulting polyhedron is shown in the figure on the right. Then the corresponding side view is ( [Figure]
A. [Figure]
B. [Figure]
C. [Figure]
D. [Figure]

11. Let $A, B, C$ be three points on the surface of a sphere $O$ with radius 1, and $AC \perp BC$, $AC = BC = 1$. Then the volume of the triangular pyramid $O-ABC$ is
A. $\frac{\sqrt{2}}{12}$
B. $\frac{\sqrt{3}}{12}$
C. $\frac{\sqrt{2}}{4}$
D. $\frac{\sqrt{3}}{4}$

14. A cone has a base radius of 6 and volume $30 \pi$. Then the lateral surface area of the cone is $\_\_\_\_$ .

As shown in the figure, the orthographic projection of a polyhedron is drawn on grid paper, where each small square has side length 1. Then the volume of the polyhedron is:
A. 8
B. 12
C. 16
D. 20

The BeiDou-3 Global Navigation Satellite System is an important achievement of China's space program. In satellite navigation systems, geostationary satellites orbit in the plane of Earth's equator at an orbital altitude of 36,000 km (orbital altitude is the distance from the satellite to Earth's surface). Treating Earth as a sphere with center $O$ and radius $r = 6400$ km, the latitude of a point $A$ on its surface is defined as the angle that $OA$ makes with the equatorial plane. The maximum latitude at which a geostationary satellite can be directly observed from Earth's surface is $a$. The surface area covered by the satellite signal on Earth's surface is $S = 2 \pi r ^ { 2 } ( 1 - \cos a )$ (in $\mathrm { km } ^ { 2 }$). Then $S$ accounts for approximately what percentage of Earth's surface area?
A. $26 \%$
B. $34 \%$
C. $42 \%$
D. $50 \%$

A right square frustum has upper and lower base edges of lengths 2 and 4 respectively, and lateral edge length 2. Its volume is
A. $20 + 12 \sqrt { 3 }$
B. $28 \sqrt { 2 }$
C. $\frac { 56 } { 3 }$
D. $\frac { 28 \sqrt { 2 } } { 3 }$

Execute the program shown in the flowchart, the output $n =$
A. $3$
B. $4$
C. $5$
D. $6$

Execute the program flowchart on the right, the output $n =$
A. 3
B. 4
C. 5
D. 6

Two cones A and B have equal slant heights. The sum of the central angles of their lateral surface developments is $2 \pi$. Let their lateral surface areas be $S _ { \text{A} }$ and $S _ { \text{B} }$, and their volumes be $V _ { \text{A} }$ and $V _ { \text{B} }$. If $\frac { S _ { \text{A} } } { S _ { \text{B} } } = 2$, then $\frac { V _ { \text{A} } } { V _ { \text{B} } } =$
A. $\sqrt { 5 }$
B. $2 \sqrt { 2 }$
C. $\sqrt { 10 }$
D. $\frac { 5 \sqrt { 10 } } { 4 }$