23. (Total: 18 points; Part 1: 4 points; Part 2: 6 points; Part 3: 8 points) Given that the general terms of sequences $\{a_n\}$ and $\{b_n\}$ are $a_n = 3n + 6, b_n = 2n + 7$ $(n \in \mathbb{N}^*)$ respectively. The elements in the set $\{x \mid x = a_n, n \in \mathbb{N}^*\} \cup \{x \mid x = b_n, n \in \mathbb{N}^*\}$ are arranged in increasing order to form a sequence $c_1, c_2, c_3, \cdots, c_n, \cdots$ (1) Find the three smallest numbers that are both terms of sequence $\{a_n\}$ and terms of sequence $\{b_n\}$; (2) Among the terms $c_1, c_2, c_3, \cdots, c_{40}$, how many are not terms of sequence $\{b_n\}$? Please explain your reasoning; (3) Find the sum of the first $4n$ terms $S_{4n}$ $(n \in \mathbb{N}^*)$ of sequence $\{c_n\}$. 2011 Shanghai College Entrance Examination Mathematics (Liberal Arts) Answers 1. $\{x \mid x < 1\}$; 2. $-2$; 3. $-\frac{3}{2}$; 4. $\sqrt{5}$; 5. $x + 2y - 11 = 0$; 6. $x < 0$ or $x > 1$; 7. $3\pi$; 8. $\sqrt{6}$; 9. $\frac{5}{2}$; 10. 2; 11. 6; 12. $\frac{15}{2}$; 13. 0.985; 14. $[-2, 7]$. 15. A; 16. D; 17. A; 18. B. 19. Solution: $(z_1 - 2)(1 + i) = 1 - i \Rightarrow z_1 = 2 - i$ Let $z_2 = a + 2i, a \in \mathbb{R}$, then $z_1 z_2 = (2 - i)(a + 2i) = (2a + 2) + (4 - a)i$. Since $z_1 z_2 \in \mathbb{R}$, we have $4 - a = 0$, so $z_2 = 4 + 2i$. 20. Solution: (1) Connect $BD, AB_1, B_1D_1, AD_1$. Since $BD \parallel B_1D_1$ and $AB_1 = AD_1$, the angle between skew lines $BD$ and $AB_1$ is $\angle AB_1D_1$. Let $\angle AB_1D_1 = \theta$. $\cos \theta = \frac{AB_1^2 + B_1D_1^2 - AD_1^2}{2AB_1 \times B_1D_1} = \frac{\sqrt{10}}{10}$. Therefore, the angle between skew lines $BD$ and $AB_1$ is $\arccos \frac{\sqrt{10}}{10}$. (2) Connect $AC, CB_1, CD_1$. The volume of the tetrahedron is [Figure] $V = V_{ABCD - A_1B_1C_1D_1} - 4 \times V_{C - B_1C_1D_1} = 2 - 4 \times \frac{1}{3} = \frac{2}{3}$. 21. Solution: (1) When $a > 0, b > 0$, for any $x_1, x_2 \in \mathbb{R}, x_1 < x_2$, we have $f(x_1) - f(x_2) = a(2^{x_1} - 2^{x_2}) + b(3^{x_1} - 3^{x_2})$. Since $2^{x_1} < 2^{x_2}, a > 0 \Rightarrow a(2^{x_1} - 2^{x_2}) < 0$, and $3^{x_1} < 3^{x_2}, b > 0 \Rightarrow b(3^{x_1} - 3^{x_2}) < 0$, we have $f(x_1) - f(x_2) < 0$, so function $f(x)$ is increasing on $\mathbb{R}$. When $a < 0, b < 0$, by similar reasoning, function $f(x)$ is decreasing on $\mathbb{R}$. (2) $f(x+1) - f(x) = a \cdot 2^x + 2b \cdot 3^x > 0$. When $a < 0, b > 0$, we have $\left(\frac{3}{2}\right)^x > -\frac{a}{2b}$, so $x > \log_{1.5}\left(-\frac{a}{2b}\right)$. When $a > 0, b < 0$, we have $\left(\frac{3}{2}\right)^x < -\frac{a}{2b}$, so $x < \log_{1.5}\left(-\frac{a}{2b}\right)$. 22. Solution: (1) $m = 2$, the ellipse equation is $\frac{x^2}{4} + y^2 = 1$, and $
In the rectangular coordinate system $xOy$, the parametric equation of curve $C _ { 1 }$ is $\left\{ \begin{array} { l } x = 2 \cos \alpha \\ y = 2 + 2 \sin \alpha \end{array} \right.$ ($\alpha$ is the parameter). $M$ is a moving point on $C _ { 1 }$. Point $P$ satisfies $\overrightarrow { OP } = 2 \overrightarrow { OM }$. The locus of point $P$ is curve $C _ { 2 }$.
23. (Total: 18 points; Part 1: 4 points; Part 2: 6 points; Part 3: 8 points)\\
Given that the general terms of sequences $\{a_n\}$ and $\{b_n\}$ are $a_n = 3n + 6, b_n = 2n + 7$ $(n \in \mathbb{N}^*)$ respectively. The elements in the set $\{x \mid x = a_n, n \in \mathbb{N}^*\} \cup \{x \mid x = b_n, n \in \mathbb{N}^*\}$ are arranged in increasing order to form a sequence $c_1, c_2, c_3, \cdots, c_n, \cdots$\\
(1) Find the three smallest numbers that are both terms of sequence $\{a_n\}$ and terms of sequence $\{b_n\}$;\\
(2) Among the terms $c_1, c_2, c_3, \cdots, c_{40}$, how many are not terms of sequence $\{b_n\}$? Please explain your reasoning;\\
(3) Find the sum of the first $4n$ terms $S_{4n}$ $(n \in \mathbb{N}^*)$ of sequence $\{c_n\}$.
2011 Shanghai College Entrance Examination Mathematics (Liberal Arts) Answers\\
1. $\{x \mid x < 1\}$;\\
2. $-2$;\\
3. $-\frac{3}{2}$;\\
4. $\sqrt{5}$;\\
5. $x + 2y - 11 = 0$;\\
6. $x < 0$ or $x > 1$;\\
7. $3\pi$;\\
8. $\sqrt{6}$; 9. $\frac{5}{2}$;\\
10. 2;\\
11. 6;\\
12. $\frac{15}{2}$;\\
13. 0.985;\\
14. $[-2, 7]$.
15. A; 16. D; 17. A; 18. B.\\
19. Solution: $(z_1 - 2)(1 + i) = 1 - i \Rightarrow z_1 = 2 - i$
Let $z_2 = a + 2i, a \in \mathbb{R}$, then $z_1 z_2 = (2 - i)(a + 2i) = (2a + 2) + (4 - a)i$.\\
Since $z_1 z_2 \in \mathbb{R}$, we have $4 - a = 0$, so $z_2 = 4 + 2i$.
20. Solution: (1) Connect $BD, AB_1, B_1D_1, AD_1$. Since $BD \parallel B_1D_1$ and $AB_1 = AD_1$,\\
the angle between skew lines $BD$ and $AB_1$ is $\angle AB_1D_1$. Let $\angle AB_1D_1 = \theta$.\\
$\cos \theta = \frac{AB_1^2 + B_1D_1^2 - AD_1^2}{2AB_1 \times B_1D_1} = \frac{\sqrt{10}}{10}$.\\
Therefore, the angle between skew lines $BD$ and $AB_1$ is $\arccos \frac{\sqrt{10}}{10}$.\\
(2) Connect $AC, CB_1, CD_1$. The volume of the tetrahedron is\\
\includegraphics[max width=\textwidth, alt={}, center]{10c3299f-3e7b-4015-b66c-39d236113438-4_508_435_1249_1361}\\
$V = V_{ABCD - A_1B_1C_1D_1} - 4 \times V_{C - B_1C_1D_1} = 2 - 4 \times \frac{1}{3} = \frac{2}{3}$.\\
21. Solution: (1) When $a > 0, b > 0$, for any $x_1, x_2 \in \mathbb{R}, x_1 < x_2$, we have $f(x_1) - f(x_2) = a(2^{x_1} - 2^{x_2}) + b(3^{x_1} - 3^{x_2})$.\\
Since $2^{x_1} < 2^{x_2}, a > 0 \Rightarrow a(2^{x_1} - 2^{x_2}) < 0$, and $3^{x_1} < 3^{x_2}, b > 0 \Rightarrow b(3^{x_1} - 3^{x_2}) < 0$,\\
we have $f(x_1) - f(x_2) < 0$, so function $f(x)$ is increasing on $\mathbb{R}$.
When $a < 0, b < 0$, by similar reasoning, function $f(x)$ is decreasing on $\mathbb{R}$.\\
(2) $f(x+1) - f(x) = a \cdot 2^x + 2b \cdot 3^x > 0$.\\
When $a < 0, b > 0$, we have $\left(\frac{3}{2}\right)^x > -\frac{a}{2b}$, so $x > \log_{1.5}\left(-\frac{a}{2b}\right)$.\\
When $a > 0, b < 0$, we have $\left(\frac{3}{2}\right)^x < -\frac{a}{2b}$, so $x < \log_{1.5}\left(-\frac{a}{2b}\right)$.
22. Solution: (1) $m = 2$, the ellipse equation is $\frac{x^2}{4} + y^2 = 1$, and $