112年高雄市高中教甄聯招-數學詳解

高雄市 112 學年度市立高級中等學校教師甄選

一、 計算證明題 (1 至 12 題每題 6 分， 13 至 16 題每題 7 分， 共 100 分)

解答： $$\cases{A(5,-1,2) \\B(-5,-1,-6)} \Rightarrow \overline{AB} 中點C=(0,-1,-2),欲求過C且垂直平面E的直線L\\ L的方向向量\vec u=E的法向量=(1,2,3) \Rightarrow L:{x\over 1}={y+1\over 2}={z+2\over 3} \\ 欲求之P=L\cap E; 若P\in L,則P(t,2t-1,3t-2), t\in \mathbb R;\\ 又P\in E \Rightarrow t+2(2t-1)+3(3t-2)-6=0  \Rightarrow t=1 \Rightarrow \bbox[red,2pt]{P(1,1,1)}$$

解答： $$\alpha,\beta,\gamma 為x^3-x-3=0之三根 \Rightarrow \cases{\alpha+\beta+\gamma=0\\ \alpha \beta+ \beta\gamma +\gamma\alpha=-1 \\ \alpha\beta \gamma=3}\\ {\alpha-1\over \alpha+1}+ {\beta -1\over \beta +1}+{\gamma-1\over \gamma+1} =1-{2\over \alpha+1} +1-{2\over \beta+1}+1-{2\over \gamma +1} =3-2\left( {1\over \alpha+1} +{1\over \beta+1} +{1\over \gamma+1}\right) \\= 3-2\left( { (\alpha+1)(\beta+1) +(\beta+1)(\gamma+1) +(\gamma+1)(\alpha+ 1)\over (\alpha+1) (\beta+1) (\gamma+1)}\right) \\=3-2\cdot {\alpha\beta+ \beta\gamma+ \gamma\alpha+ 2(\alpha+\beta+\gamma)+3 \over \alpha\beta\gamma +\alpha\beta+ \beta\gamma+ \gamma\alpha+1} =3-2\cdot {-1+0+3\over 3-1+1} =3-{4\over 3}=\bbox[red,2pt]{5\over 3}$$

解答：

$$\cases{|x-1|\le 2\\ |x+y|\le 1}所圍區域頂點坐標\cases{A(3,-4) \\B(3,-2)\\ C(-1,2)\\ D(-1,0)}\\ 令f(x,y)=x-2y \Rightarrow \cases{f(A)=11\\ f(B)=7\\ f(C)=-5\\ f(D)=-1} \Rightarrow 最大值為\bbox[red,2pt]{11}$$

解答： $$\left(\sqrt{10+\sqrt a}-\sqrt{10-\sqrt a}\right)^2 =20-2\sqrt{100-a} \Rightarrow \sqrt{10+\sqrt a}-\sqrt{10-\sqrt a}= \sqrt 2\cdot \sqrt{10-\sqrt{100-a}}\\ 令L={\sqrt{10+\sqrt 1}+ \sqrt{10+\sqrt 2}+ \cdots +\sqrt{10+\sqrt {99}}\over \sqrt{10-\sqrt 1}+ \sqrt{10-\sqrt 2}+ \cdots +\sqrt{10-\sqrt {99}}} \\\Rightarrow L-1= {\left(\sqrt{10+\sqrt 1} -\sqrt{10-\sqrt 1}\right) +\left(\sqrt{10+\sqrt 2}-\sqrt{10-\sqrt 2} \right)+   \cdots +\left(\sqrt{10+\sqrt {99}} -\sqrt{10-\sqrt {99}}\right) \over \sqrt{10-\sqrt 1}+ \sqrt{10-\sqrt 2}+ \cdots +\sqrt{10-\sqrt {99}}} \\={\sqrt 2\left( \sqrt{10-\sqrt{100-1}}\right) +\sqrt 2\left(\sqrt{10-\sqrt{100-2}} \right)+   \cdots + \sqrt 2\left( \sqrt{10-\sqrt{100-99}}\right)\over \sqrt{10-\sqrt 1}+ \sqrt{10-\sqrt 2}+ \cdots +\sqrt{10-\sqrt {99}}} \\={\sqrt 2\left( \sqrt{10-\sqrt{99}} +\sqrt{10-\sqrt{98} }+   \cdots +   \sqrt{10-\sqrt{1}}\right)\over \sqrt{10-\sqrt 1}+ \sqrt{10-\sqrt 2}+ \cdots +\sqrt{10-\sqrt {99}}} =\sqrt 2\\ \Rightarrow L=\bbox[red, 2pt]{\sqrt 2+1}$$

解答： $$等比數列:a_1=a,a_2=ar,a_3=ar^2 \Rightarrow r={(ar+2a)r\over ar+2a} ={a_3+2a_2\over a_2+2a_1} \\ \Rightarrow r={-4a+4\sqrt 2+ \log_8 2023 +2(2a-2\sqrt 2+ \log_4 2023)\over 2a-2\sqrt 2+ \log_4 2023+ 2(-a+\sqrt 2+\log_2 2023)} \\ ={\log_8 2023+ 2\log_4 2023\over \log_4 2023+ 2\log_2 2023} ={{1\over 3} \log_2 2023 +\log_2 2023\over {1\over 2}\log_2 2023 +2\log_2 2023} ={{4\over 3}\over {5\over 2}} =\bbox[red,2pt]{8\over 15}$$

解答： $$\sin(37^\circ) =\sin(30^\circ+7^\circ) =\sin 30^\circ\cos 7^\circ+ \sin 7^\circ \cos 30^\circ={1\over 2} \cos 7^\circ+{\sqrt 3\over 2} \sin 7^\circ\\ 原式: \sin^2 37^\circ +\cos^2 7^\circ-\sin 37^\circ\times \cos 7^\circ \\=\left( {1\over 2} \cos 7^\circ+{\sqrt 3\over 2} \sin 7^\circ\right)^2  +\cos^2 7^\circ-\left( {1\over 2} \cos 7^\circ+{\sqrt 3\over 2} \sin 7^\circ\right) \cos 7^\circ \\={1\over 4}\cos ^2 7^\circ+ {\sqrt 3\over 2} \sin 7^\circ \cos  7^\circ+{3\over 4} \sin^2 7^\circ +\cos^2  7^\circ-{1\over 2}\cos^2  7^\circ-{\sqrt 3\over 2}\sin  7^\circ\cos  7^\circ \\={3\over 4}\cos ^2  7^\circ+{3\over 4}\sin ^2  7^\circ= \bbox[red, 2pt]{3\over 4}$$

解答： $$取g(x)=f(x)-x,則1,2,3為g(x)=0的三根 \Rightarrow g(x)=(x-1)(x-2)(x-3)(x-k)\\ \Rightarrow f(x)=g(x)+x=(x-1)(x-2)(x-3)(x-k)+x \\ 由於f(x)=0的四根之積=d=1\cdot 2\cdot 3\cdot k=6k \Rightarrow k={d\over 6}\\ \Rightarrow f(x)=(x-1)(x-2)(x-3)(x-{d\over 6})+x\\ 因此\cases{f(0)= d\\ f(4)= 3\cdot 2\cdot 1\cdot (4-{d\over 6})+4=28-d} \Rightarrow {1\over 4}(f(0)+f(4))={1\over 4}\cdot 28=\bbox[red, 2pt]7$$

解答： $$原式=\lim_{n\to \infty}\sum_{k=1}^n{1\over 4n^2}\sqrt{4n^2-k^2} =\lim_{n\to \infty}\sum_{k=1}^n{1\over 4n }\sqrt{4-({k\over n})^2} =\int_0^1 {1\over 4}\sqrt{4-x^2}\,dx\\ 令x=2\sin \theta \Rightarrow dx=2\cos \theta\,d\theta,則原式=\int_0^{\pi/6}{1\over 4}\sqrt{4\cos^2\theta}\cdot 2\cos\theta\,d\theta \\=\int_0^{\pi/6} \cos^2\theta\,d\theta=\int_0^{\pi/6}{1\over 2}(\cos 2\theta+1)\,d\theta =\left.\left[{1\over 4}\sin 2\theta+{1\over 2}\theta \right] \right|_0^{\pi/6} =\bbox[red,2pt]{{\sqrt 3\over 8}+ {\pi \over 12}}$$

解答：

解答： $$\cases{|z_1|=2\\ |z_2|=3\\ |z_1-z_2|=\sqrt 5} \Rightarrow \cases{\overline{OA}=2\\ \overline{OB}= 3\\ \overline{AB}=\sqrt 5} \Rightarrow \overline{OB}^2 =\overline{OA}^2+ \overline{AB}^2 \Rightarrow \angle OAB=90^\circ\\ \Rightarrow \cos \angle AOB=\cos \theta={2^2+3^2-5\over 2\cdot 2\cdot 3} ={2\over 3}\\ 令\cases{A'(iz_1)\\ A''(-iz_1)} \Rightarrow \cases{\overline{A'B}=|z_2-iz_1| \\ \overline{A''B}=|z_2+iz_1| } \Rightarrow \cases{\cos (90^\circ-\theta)=(2^2+3^2-\overline{A'B}^2)/2\cdot 2\cdot 3\\ \cos (90^\circ+\theta)=(2^2+3^2-\overline{A''B}^2)/2\cdot 2\cdot 3 } \\ \Rightarrow \cases{\overline{A'B}^2=13-4\sqrt 5\\ \overline{A''B}^2 =13+4\sqrt 5} \Rightarrow |z_1^2+z_2^2| =|(z_2-iz_1) (z_2+iz_1)| =\sqrt{(13-4\sqrt 5)(13+4\sqrt 5)}\\ =\bbox[red, 2pt]{\sqrt{89}}$$

解答：

$$先求過Ａ(1,8)的圓切線，假設圓切點Ｐ(a,\sqrt{16-a^2}) \Rightarrow \overrightarrow{PA} \cdot \overrightarrow{OP}=0 \\ \Rightarrow (1-a,8-\sqrt{16-a^2}) \cdot(a, \sqrt{16-a^2})=0 \Rightarrow 65a^2-32a-768=0\\ \Rightarrow (13a-48)(5a+16)=0 \Rightarrow a={48\over 13},-{16\over 5} \Rightarrow 兩切點\cases{P(-16/5, 12/5)\\ Q(48/13,20/13)}\\ 又x^2+y^2=16 \Rightarrow 2x+2yy'=0 \Rightarrow y'=-{x\over y} \Rightarrow 兩切線斜率\cases{m_1=4/3\\ m_2=-12/5} \\ \Rightarrow 兩切線\cases{L_1= \overleftrightarrow{AP} : 4x-3y+20=0 \\ L_2 =\overleftrightarrow{AQ}: 12x+5y=52} \Rightarrow \cases{B=L_1\cap (y=-4) =(-8,-4)\\ C=L_2 \cap (y=-4)= (6,-4)}\\ \cases{過Ａ垂直x軸的垂線L_3:x=1\\ 與L_2垂直的垂線L_4斜率為5/12 \Rightarrow L_4: 5x-12y=8} \Rightarrow H=L_3\cap L_4 =\bbox[red,2pt]{(1,-{1\over 4})}$$

解答： $$A^2+4AB=0 \Rightarrow A(A+4B)=0 \Rightarrow A=-4B \Rightarrow A^{-1}=-{1\over 4}B^{-1} \\ \Rightarrow A^{-1}+B^{-1}={3\over 4}B^{-1}= \begin{bmatrix} 3 & 6\\ 9 & 12\end{bmatrix} \Rightarrow B^{-1}=\begin{bmatrix} 4 & 8\\12 & 16\end{bmatrix} \Rightarrow B=\begin{bmatrix} -1/2 & 1/4\\ 3/8 & -1/8\end{bmatrix} =\begin{bmatrix} a & b\\ c & d\end{bmatrix} \\ \Rightarrow a=\bbox[red,2pt]{-{1\over 2}}$$

解答： $$令\overrightarrow{PD} =x\overrightarrow{PA}+ y\overline{PB} , 由於x+y=1 \Rightarrow D\in \overline{BC}\\ 因此\overrightarrow{PD} +z\overrightarrow{PC}=0, 若\cases{z=0 \Rightarrow P=D\\ z=1 \Rightarrow P為\overline{CD}中點\\ z=2 \Rightarrow \overline{PC}: \overline{CD}=1:3\\ z=3 \Rightarrow \overline{PC}: \overline{CD} = 1:4} \\ \Rightarrow {S\over \triangle ABC}={1-1/16}=\bbox[red,2pt]{15\over 16}$$

解答： $$5A,4B,3C串連數至少是3，因此只要考慮增加一個串連的情形：\\ A拆成2個串連：\cases{A+4A:4B與3C插入兩個A串連之間，共有12個\\ 2A+3A:4B與3C插入兩個A串連之間，共有12個}\Rightarrow 合計24種\\ B拆成2個串連:\cases{B+3B:5A與3C插入兩個B串連之間，共有12個\\ 2B+2B: 5A與3C插入兩個B串連之間，共有6個}，合計18個\\ C拆成1C+2C:5A與4B插入兩個B串連之間，共有12個\\ 以上共有24＋18＋12＝\bbox[red,2pt]{54}種$$

解答： $$由於a^{527}+b^{527} =(a+b)(a^{526}-a^{525}b+ a^{524} b^2-\cdots-ab^{526}+ b^{527}) \Rightarrow a+b可以整除a^{527}+b^{527}\\ 而\sum_{k=1}^{2022}k^{527} =(1^{527}+2022^{527})+(2^{527}+2021^{527}) +\cdots+ (1011^{527}+ 1012^{527})\\ 上式每個括號內都是1+2022=2023的倍數,因此2023可以整除\sum_{k=1}^{2022}k^{527},\bbox[red,2pt]{故得證}$$

解答： $$\cases{ax+by+cz=0\\ bx+cy+az=0\\ cx+ay+bz=0}至少有一解(0,0,0)\\ 依題意有無限多解 \Rightarrow {a\over b}={b\over c}={c\over a}=k \Rightarrow \cases{a=bk\\ b=ck\\ c=ak},k\ne 0 \\ \Rightarrow a=bk=ck^2=ak^3 \Rightarrow k=1 \Rightarrow a=b=c, \bbox[red, 2pt]{故得證}$$

解答： $$\cases{{4^n+3^n\over 8^n+3^n} \le {4^n+4^n\over 8^n }={2\cdot 4^n\over 8^n} =2({1\over 2})^n \\ {4^n+3^n\over 8^n+3^n} \ge{4^n \over 8^n+8^n} ={1 \over 2}\cdot {4^n\over 8^n}} \Rightarrow {1 \over 2}\cdot {4^n\over 8^n}\le {4^n+3^n\over 8^n+3^n} \le 2({1\over 2})^n \\ \Rightarrow \left( {1\over 2} \right)^{1/n}\cdot {1\over 2} \le \left({4^n+3^n\over 8^n+3^n} \right)^{1/n} \le 2^{1/n}\cdot {1\over 2} \\ \Rightarrow \lim_{n\to \infty}\left( {1\over 2} \right)^{1/n}\cdot {1\over 2} \le \lim_{n\to \infty} \left({4^n+ 3^n\over 8^n+3^n} \right)^{1/n} \le \lim_{n\to \infty}2^{1/n}\cdot {1\over 2} \\ \Rightarrow {1\over 2}\le \lim_{n\to \infty} \left({4^n+ 3^n\over 8^n+3^n} \right)^{1/n} \le {1\over 2} \Rightarrow \lim_{n\to \infty} \left({4^n+ 3^n\over 8^n+3^n} \right)^{1/n} = \bbox[red, 2pt]{1\over 2}$$

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