112年新竹中學教甄-數學詳解

112年新竹中學數學科代理教師甄選

一、填充題(共60分，每題6分)

解答：$$假設滿足條件且長度為n的數量為f(n)\\ n=3:只有101一種 \Rightarrow f(3)=1\\ n=4:有兩種1011,1101 \Rightarrow f(4)=2\\ n=5:也是兩種10101,11011 \Rightarrow f(5)=2\\只要在合乎條件的字串中後面加上01或011讓長度加長且合乎條件\\ 因此f(n)=f(n-2)+f(n-3),n\ge 6 \Rightarrow \\ \begin{array}{c|ccccccccccccccc}n & 6 & 7 & 8 & 9 & 10 & 11 &  12 & 13 & 14 & 15 & 16 \\\hline f(n) & 3& 4& 5& 7 & 9 & 12& 16& 21& 28& 37 & 49\end{array} \Rightarrow f(16)=\bbox[red, 2pt]{49}\\ \href{https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_25}{參考資料}$$

解答：$$甲乙相鄰且甲不靠窗: \\\fbox{乙}\fbox{甲}\qquad \fbox{Ｄ}\fbox{Ｅ}\fbox{Ｆ}\fbox{Ｇ} \qquad\fbox{Ｈ}\fbox{Ｋ}\cdots(1)\\ \fbox{Ａ}\fbox{Ｃ}\qquad \fbox{甲}\fbox{乙}\fbox{Ｆ}\fbox{Ｇ} \qquad\fbox{Ｈ}\fbox{Ｋ} \cdots(2)\\ \fbox{Ａ}\fbox{Ｃ}\qquad \fbox{乙}\fbox{甲}\fbox{Ｆ} \fbox{Ｇ} \qquad\fbox{Ｈ}\fbox{Ｋ} \cdots(3)\\ \fbox{Ａ}\fbox{Ｃ}\qquad \fbox{Ｄ}\fbox{甲}\fbox{乙} \fbox{Ｇ} \qquad\fbox{Ｈ}\fbox{Ｋ} \cdots(4)\\ \fbox{Ａ}\fbox{Ｃ}\qquad \fbox{Ｄ}\fbox{乙}\fbox{甲} \fbox{Ｇ} \qquad\fbox{Ｈ}\fbox{Ｋ}\cdots(5)\\ \fbox{Ａ}\fbox{Ｃ}\qquad \fbox{Ｄ}\fbox{Ｅ}\fbox{甲} \fbox{乙} \qquad\fbox{Ｈ}\fbox{Ｋ}\cdots(6)\\ \fbox{Ａ}\fbox{Ｃ}\qquad \fbox{Ｄ}\fbox{Ｅ}\fbox{乙} \fbox{甲} \qquad\fbox{Ｈ}\fbox{Ｋ}\cdots(7)\\ \fbox{Ａ}\fbox{Ｃ}\qquad \fbox{Ｄ}\fbox{Ｅ}\fbox{Ｆ} \fbox{Ｇ} \qquad\fbox{甲}\fbox{乙}\cdots(8)\\ 接下來計算(1)-(6)丙丁不相鄰且皆不靠窗\\(1)與(8)各有14種，(2,3,6,7)各有10種，(4)及(5)各有12種，合計92種\\ 最後將剩下的四人任排有4!=24種，一共92\times 24= \bbox[red,2pt]{2208}種$$

解答：$$第１次檢查有心臟病的比率=有心臟病被正確診斷+無心臟病被誤診\\=0.001\times 0.99+0.999\times 0.02\\{第１次檢查有心臟病且第２次檢查也有心臟病而真的有心臟病\over 第１次檢查有心臟病且第２次檢查也有心臟病} \\ ={0.001\times 0.99\times 0.99 \over 0.001\times 0.99\times 0.99+ 0.999\times 0.02\times 0.02} = \bbox[red,2pt]{363\over 511}$$

解答：

$$假設半圓心C，小圓圓心O，如上圖\\小圓圓周角\angle QPR=60^\circ \Rightarrow 圓心角\angle QOR=120^\circ \Rightarrow {\overline{QR} \over \sin \angle QOR}={3\sqrt 3\over \sin 120^\circ} =6=2r \\ 小圓半徑r=3\Rightarrow  \overline{OR}=\overline{OQ} =3\Rightarrow \triangle COR \cong \triangle COQ (SSS) \Rightarrow \angle COQ=\angle COR=120^\circ \\ \Rightarrow \cos \angle COQ=-{1\over 2}={\overline{OC}^2+3^2-7^2\over 2\cdot \overline{CO}\cdot 3} \Rightarrow \overline{OC}=5 \Rightarrow \overline{CP}=4 \\ 令\angle COP=\theta \Rightarrow \cases{\sin \theta=4/5\\ \cos \theta=3/5} 且\cases{\angle POR=120^\circ-\theta\\ \angle POQ=120^\circ+\theta} \\\Rightarrow \triangle POR+\triangle POQ={1\over 2}\cdot 3^2(\sin(120^\circ-\theta)+ \sin(120^\circ+\theta)) ={9\over 2}\sqrt 3\cos \theta={27\over 10}\sqrt 3\\ 又\triangle OPR={1\over 2}\cdot 3^2\sin 120^\circ= {9\over 4}\sqrt 3 \Rightarrow  \triangle PQR={27\over 10}\sqrt 3+{9\over 4}\sqrt 3= \bbox[red, 2pt]{{99\over 20}\sqrt 3}$$

解答：$$假設\omega =\cos(2\pi/7)+ i\sin(2\pi/7)=e^{2\pi i/7} \Rightarrow \omega^6=e^{12\pi i/7}=e^{-2\pi i/7} \Rightarrow \cos(2\pi/7) =(\omega+\omega^6)/2\\ 同理，\cos(4\pi/7)=(\omega^2+ \omega^5)/2, \cos(6\pi/7) =(\omega^3+\omega^4)/2\\ 因此x^3+ax^2+ bx+c=0的三根為\cases{\alpha= {\omega +\omega^6\over 2} \\\beta={\omega^2 +\omega^5\over 2} \\ \gamma= {\omega^3 +\omega^4\over 2}}  \Rightarrow \cases{-a= \alpha+ \beta+ \gamma \\ b= \alpha\beta +\beta \gamma +\gamma\alpha\\ -c= \alpha\beta\gamma } \\ \alpha+\beta+\gamma = {1\over 2}(\omega+\omega^2+\cdots +\omega^6) ={1\over 2}(-1)=-{1\over 2} \Rightarrow a={1\over 2} \\ \alpha\beta +\beta\gamma +\gamma\alpha ={1\over 4}\cdot 2(\omega+ \omega^2+\cdots +\omega^6) =-{1\over 2} \Rightarrow b=-{1\over 2} \\ \alpha\beta \gamma = {1\over 8}(2+\omega+\omega^2+\cdot s+\omega^6) ={1\over 8} \Rightarrow c=-{1\over 8} \\ 因此abc= \bbox[red, 2pt]{1\over 32}$$

解答：$$z_1,z_2,z_3,z_4為P(z)=0的四根 \Rightarrow \cases{z_1+z_2+ z_3+z_4=-4 \\ z_1(z_2+ z_3+ z_4)+ z_2(z_3+z_4)+ z_3z_4  =3\\ z_1z_2z_3 +z_2z_3z_4+ z_3z_4z_1+ z_1z_2z_4= -2 \\ z_1z_2z_3z_4 = 1} \\又\cases{f(z_1)= 4i\bar z_1 \\f(z_2)= 4i\bar z_2 \\f(z_3)= 4i\bar z_3 \\f(z_4)= 4i\bar z_4 } 為Q(z)=0的四根\Rightarrow \cases{B= (4i)^2\times 3=-48\\ D=(4i)^4\times 1=256} \Rightarrow B+D=256-48= \bbox[red,2pt]{208}$$

解答：

$$等腰梯形對角線等長\Rightarrow \overline{BD}= \overline{AC}=2+1+3=6\\ 直角\triangle BDD'(見上圖): \overline{BD}^2= \overline{BD'}^2+\overline{DD'}^2 =(\overline{BY}+ \overline{DX})^2+ \overline{XY}^2 \Rightarrow 6^2=(\overline{BY}+ \overline{DX})^2+ 1^2\\ \Rightarrow \overline{BY}+ \overline{DX}= \sqrt{35} \Rightarrow 梯形面積= \triangle ABC+\triangle ADC ={1 \over 2} \cdot \overline{AC}(\overline{BY}+ \overline{DX})= {1\over 2}\cdot 6\cdot \sqrt{35} \\=\bbox[red, 2pt]{3\sqrt{35}}$$

解答：$$L: y=\tan \theta \cdot x \Rightarrow 過P(a,b)且與L垂直的直線L':y=-{1\over \tan \theta}(x-a)+b \\ \Rightarrow P'=L\cap L' =\left({a+b\tan \theta \over 1+\tan^2\theta}, {a\tan\theta+ b\tan^2\theta \over 1+\tan^2 \theta} \right) =\begin{bmatrix}{1\over 1+\tan^2\theta} & {\tan\theta \over 1+\tan^2\theta}\\ {\tan \theta \over 1+\tan^2\theta} & {\tan^2\theta \over 1+\tan^2\theta} \end{bmatrix} \begin{bmatrix}a  \\b \end{bmatrix}\\ \Rightarrow A=\begin{bmatrix}{1\over 1+\tan^2\theta} & {\tan\theta \over 1+\tan^2\theta}\\ {\tan \theta \over 1+\tan^2\theta} & {\tan^2\theta \over 1+\tan^2\theta} \end{bmatrix} \Rightarrow A^2=A \Rightarrow A^{112}=A= \bbox[red,2pt]{\begin{bmatrix}{1\over 1+\tan^2\theta} & {\tan\theta \over 1+\tan^2\theta}\\ {\tan \theta \over 1+\tan^2\theta} & {\tan^2\theta \over 1+\tan^2\theta} \end{bmatrix}}$$

解答：

$$\cos C={4^2+6^2-28\over 2\cdot 4\cdot 6}={1\over 2} \Rightarrow \angle C=60^\circ，因此令\cases{B(0,0)\\ C(4,0)} \Rightarrow A(1,3\sqrt 3)\\ 假設\overline{AC}上的垂足為Q，則\overline{BQ} =2\sqrt 3 \Rightarrow Q(3,\sqrt 3) \Rightarrow \cases{L_1= \overleftrightarrow{BQ} :3y=\sqrt 3 x \\ L_2=\overleftrightarrow{CP}: x+\sqrt 3y=4}\\ \Rightarrow P=L_1 \cap L_2 =(2,2\sqrt 3/3) \Rightarrow \cases{\overrightarrow{AP} =(1,-7\sqrt 3/3) \\ \overrightarrow{AB} =(-1,-3\sqrt 3) \\ \overrightarrow{AC} =(3,-3\sqrt 3)} \Rightarrow (\alpha,\beta) =\bbox[red, 2pt]{({1\over 3},{4\over 9})}$$

解答：$$本題相當於求迴歸直線:y=ax+b,樣本點為(x,y) =(4,5), (6,5), (8,7),(10,9),(12,9)\\ \begin{array}{} X & Y & X^2 & XY \\\hline 4 & 5 & 16 & 20\\ 6 & 5 & 36& 30\\ 8 & 7 & 64& 56\\ 10 & 9 & 100 & 90\\ 12 & 9 & 144& 108 \\\hdashline 40 & 35 & 360 & 304\end{array} \Rightarrow \cases{\bar x=40/5=8\\ \bar y=35/5=7 } \\ \Rightarrow a={n\sum (xy) -(\sum x)(\sum y) \over n \sum x^2-(\sum x)^2} ={5\times 304-40\times 35\over 5\times 360-40^2}={3\over 5} \\ 又迴歸直線必經(\bar x,\bar y) \Rightarrow y={3\over 5}(x-8)+7 \Rightarrow y={3\over 5}x+{11\over 5} \Rightarrow (a,b)=\bbox[red, 2pt]{({3\over 5},{11\over 5})}$$

解答：$$\bbox[pink,2pt]{敘述}：\\若整係數多項式f(x)=a_nx^2 +a_{n-1}x^{n-1}+ \cdots + a_1x+a_0有一次因式px-q，\\其中p,q為整數且互質，則p\mid a_n且q\mid a_0\\ \bbox[pink,2pt]{證明}:f(x)有一次因式px-q \Rightarrow f(q/p)=0 \\ \Rightarrow a_n\left({q\over p} \right)^n +a_{n-1}\left({q\over p} \right)^{n-1} +\cdots+ a_1\left({q\over p} \right)+a_0=0 \\ \Rightarrow a_n q^n +a_{n-1}pq^{n-1} + \cdots +a_1p^{n-1}q+ a_0p^n=0 \\ \Rightarrow \cases{p\left(a_{n-1}q^{n-1} + \cdots +a_1p^{n-2}q+ a_0p^{n-1} \right)= -a_n q^n \Rightarrow p \mid -a_nq^n \\q \left( a_nq^{n-1}+ a_{n-1}pq^{n-2}+ \cdots+a_1 p^{n-1} \right) =-a_0p^n \Rightarrow q \mid -a_0p^n}\\由於p,q互質，因此 \cases{p\mid a_n \\ q\mid a_0}，故得證。$$

解答：$$ a_n=2a_{n-1}+n =2(2a_{n-2}+ (n-1))+n =2^2a_{n-2}+2(n-1)+n \\=2^2(2a_{n-3}+(n-2)) +2(n-1)+n =2^3a_{n-3}+ 2^2(n-2)+ 2(n-1)+n \\= 2^{n-1}a_1+ 2^{n-2}\cdot 2+ 2^{n-3}\cdot 3+ \cdots+ 2^2(n-2)+ 2(n-1)+n \\=2^{n-1}\cdot 1+ 2^{n-2}\cdot 2+ 2^{n-3}\cdot 3+ \cdots+ 2^2(n-2)+ 2(n-1)+ 2^0\cdot n \\ \Rightarrow 2a_n=2^n+2^{n-1}\cdot 2+ 2^{n-2}\cdot 3+ \cdots +2^3(n-2)+ 2^2(n-1)+ 2\cdot n\\ \Rightarrow 2a_n-a_n=a_n=2^n+2^{n-1}+2^{n-2}+ \cdots +2^2+2-n =\bbox[red, 2pt]{2^{n+1}-n-2}$$

解答：$$\mathbf{(1)}\; 假設三狀態，分別為S_1\cases{甲:1白1紅\\ 乙:1白2紅}，S_2\cases{甲:2紅\\ 乙:2白1紅}，S_3\cases{甲:2白\\ 乙:3紅}\\ P(S_1\to S_1)=P(甲抽白球至乙，乙抽白球至甲)+P(甲抽紅球至乙，乙抽紅球至甲)={5\over 8}\\ P(S_1\to S_2)=P(甲抽白球至乙，乙抽紅球至甲)＝{1\over 4} \\ P(S_1\to S_3)=P(甲抽紅球至乙，乙抽白球至甲)　＝{1\over 8}\\ 同理P(S_2\to S_1)={1\over 2}, P(S_2\to S_2)={1\over 2};P(S_3\to S_3)= {1\over 4}, P(S_3\to S_1)= {3\over 4}\\因此狀態轉換矩陣A=\begin{bmatrix} 5/8 & 1/2 & 3/4 \\ 1/4 & 1/2 & 0\\ 1/8 & 0 & 1/4\end{bmatrix} \Rightarrow A^2 \begin{bmatrix} 1\\ 0 \\0 \end{bmatrix} = \left[\begin{matrix}\frac{39}{64} & \frac{9}{16} & \frac{21}{32} \\\frac{9}{32} & \frac{3}{8} & \frac{3}{16} \\\frac{7}{64} & \frac{1}{16} & \frac{5}{32} \end{matrix}\right]\begin{bmatrix} 1\\ 0 \\0 \end{bmatrix} =\begin{bmatrix} 1\\ 0 \\0 \end{bmatrix} =\begin{bmatrix}\frac{39}{64} \\\frac{9}{32} \\ \frac{7}{64}\end{bmatrix}\\ \Rightarrow 2紅球在甲箱的機率(S_2) =\bbox[red, 2pt]{9\over 32} \\ \mathbf{(2)}\; A=\left[\begin{matrix} -2 & 1 & 6 \\ 1 & -2 & 3 \\1 & 1 & 1 \end{matrix}\right] \left[\begin{matrix} 0 & 0 & 0 \\0 & \frac{3}{8} & 0 \\0 & 0 & 1\end{matrix}\right]\left[\begin{matrix}\frac{-1}{6} & \frac{1}{6} & \frac{1}{2} \\\frac{1}{15} & \frac{-4}{15} & \frac{2}{5} \\\frac{1}{10} & \frac{1}{10} & \frac{1}{10}\end{matrix}\right]  \\\Rightarrow A^\infty =\left[\begin{matrix} -2 & 1 & 6 \\ 1 & -2 & 3 \\1 & 1 & 1 \end{matrix}\right] \left[\begin{matrix} 0 & 0 & 0 \\0 & \color{blue}0 & 0 \\0 & 0 & 1\end{matrix}\right]\left[\begin{matrix}\frac{-1}{6} & \frac{1}{6} & \frac{1}{2} \\\frac{1}{15} & \frac{-4}{15} & \frac{2}{5} \\\frac{1}{10} & \frac{1}{10} & \frac{1}{10}\end{matrix}\right] = \left[\begin{matrix} \frac{3}{5} & \frac{3}{5} & \frac{3}{5} \\\frac{3}{10} & \frac{3}{10} & \frac{3}{10} \\\frac{1}{10} & \frac{1}{10} & \frac{1}{10}\end{matrix} \right] \\ \Rightarrow A^{\infty} \begin{bmatrix} 1\\ 0\\ 0\end{bmatrix}= \left[\begin{matrix}\frac{3}{5} \\\frac{3}{10} \\\frac{1}{10}\end{matrix}\right] \Rightarrow P(S_2)= \bbox[red, 2pt]{3\over 10}$$

解答：$$\mathbf{(1)}\;\cases{L垂直平面Ｎ\\ \overleftrightarrow {FJ}垂直平面N} \Rightarrow L\parallel \overleftrightarrow {FJ}再加上\angle ACF =\angle CAJ=90^\circ \Rightarrow ACFJ為矩形\\ 同理ABEJ也是矩形，因此\cases{\overline{AC}=\overline{FJ} \\\overline{IE} =\overline{AB}= \overline{AC}/2} \Rightarrow {\overline{IE} \over  \overline{FJ}}={1\over 2} ={\overline{DI} \over \overline{DJ}} \Rightarrow I為\overline{DJ}中點\\ 三角形中線定: \overline{AJ}^2 +\overline{AD}^2= 2(\overline{AI}^2+ \overline{IJ}^2) \Rightarrow 100=2(25+ \overline{IJ}^2) \Rightarrow \overline{IJ}= \bbox[red, 2pt]5 \\\mathbf{(2)}\; \overline{IJ}=5 \Rightarrow \overline{DJ}=10 \Rightarrow \overline{DJ}^2 = \overline{AJ}^2 +\overline{AD}^2 \Rightarrow \angle JAD=90^\circ\\ d(L,M)= A至底邊\overline{DJ}的高 = 8\times 6\div 10= \bbox[red,2pt]{24\over 5}$$

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