彰化女中113 學年度教師甄選數學科試題卷
一、填充題(每5 分，共90 分)
解答:$$\sqrt{(x+5)^2 +(y+4)^2+25} +\sqrt{(x-3)^2+ (y-5)^2+49} =\overline{PA} +\overline{PB},其中\cases{P(x,y,0) \\A(-5,-4,-5) \\B(3,5,7)} \\ \Rightarrow \overline{PA} +\overline{PB} \le \overline{AB} =\sqrt{8^2+9^2+12^2} =\sqrt{289} =\bbox[red, 2pt]{17}$$
解答:$$\cases{三點共線有16種\\ 四點共線有4種\\五點共線有12種} \Rightarrow 可形成三角形的機率={ C^{25}_3-16C^3_3- 4C^4_3-12C^5_3 \over C^{25}_3}\\ ={2300-16-16-120 \over 2300} ={ 2148\over 2300}= \bbox[red, 2pt]{537\over 575}$$
解答:$$f(x)=x+{2\over x}+{64\over 9}\left( {x\over x^2-x+2}\right) \\\Rightarrow f'(x)=1-{2\over x^2}+{64\over 9}\left( {1\over x^2-x+2} -{x(2x-1)\over (x^2-x+2)^2}\right) \\={x^2(x^2-x+2)^2 -2(x^2-x+2)^2\over x^2(x^2-x+2)^2} +{64\over 9}\left( {x^2-x+2-2x^2+x \over (x^2-x+2)^2}\right) \\={(x^2-2)(x^2-x+2)^2 \over x^2(x^2-x+2)^2} -{64\over 9} \cdot {x^2-2\over (x^2-x+2)^2} \\={(x^2-2)(x^2-x+2)^2 \over x^2(x^2-x+2)^2} -{64\over 9} \cdot {x^2(x^2-2) \over x^2(x^2-x+2)^2} \\= {x^2-2\over x^2(x^2-x+2)^2} \left((x^2-x+2)^2-({8x\over 3})^2 \right) \\= {x^2-2\over x^2(x^2-x+2)^2}(x^2-{11\over 3}x+2)(x^2+{5\over 3}x+2)\\ f(x)最小值出現在x^2-{11\over 3}x+2＝0\Rightarrow 兩根之和= \bbox[red, 2pt]{11\over 3}$$
解答:$$\lim_{x\to 1}{x^2f(1)-f(x^2) \over x-1} =\lim_{x\to 1}{(x^2f(1)-f(x^2))' \over (x-1)'} =\lim_{x\to 1}{2xf(1)-2xf'(x^2) \over 1} \\=4-10=\bbox[red, 2pt]{-6}$$
解答:$$S_n=\int_0^1 (x^n-x^{n+1})\,dx =\left. \left[ {1\over n+1}x^{n+1}-{1\over n+2}x^{n+2}\right] \right|_0^1 ={1\over n+1}-{1\over n+2} \\ \Rightarrow \sum_{n=1}^\infty S_n=\left({1\over 2}-{1\over 3} \right)+ \left({1\over 3}-{1\over 4} \right)+ \cdots= \bbox[red, 2pt]{1\over 2}$$
$$\left| z(-1+\sqrt 3 i)\right| =|z|\cdot |-1+\sqrt 3 i| =|z|\cdot 2=4 \Rightarrow |z|=2 \Rightarrow \Gamma:x^2+y^2=4 \\ 取\cases{A(0,4)\\ C(0,1)} ,則\Gamma 為滿足\overline{PA}=2\overline{PC}的阿波羅尼斯圓，其中P(z) \in \Gamma \\ 欲求{1\over 2}|z-4i|+|z-3|= {1\over 2} \overline{PA} +\overline{PB} =\overline{PC}+\overline{PB}\; (其中B=(3,0)) \\因此當P、B、C在一直線時，其值最小，即\overline{PC}+\overline{PB} ＝\overline{BC}= \bbox[red, 2pt]{\sqrt{10}}$$
解答:
解答:
$$A,H,D在一直線上\Rightarrow \overrightarrow{AH} ={1\over 3} \overrightarrow{AB}+{1\over 4} \overrightarrow{AC} ={1\over 3} \overrightarrow{AB}+{2\over 3} \overrightarrow{AD} \Rightarrow \overline{AD}:\overline{AC}=3:8\\ \Rightarrow 取\cases{A(0,0) \\ D(3k,0)\\ C(8k,0)\\ H(3k,t)\\ B(3k,s)} \Rightarrow (3k,t)={1\over 3}(3k,s)+{1\over 4}(8k,0) \Rightarrow s=3t \Rightarrow B(3k,3t)\\ \Rightarrow \overrightarrow{AH} \cdot \overrightarrow{BC} =0 \Rightarrow (3k,t)\cdot (5k,-3t)=15k^2=3t^2 \Rightarrow t=\sqrt{5}k \\ \Rightarrow \cos \angle BAC = {\overrightarrow{AB} \cdot \overrightarrow{AC} \over |\overrightarrow{AB} || \overrightarrow{AC}|} ={24k^2 \over \sqrt{9k^2+9t^2} \cdot 8k} ={24k^2 \over \sqrt{54k^2} \cdot 8k} ={1\over \sqrt 6} =\bbox[red, 2pt]{\sqrt 6\over 6}$$
$$\frac{\text{d} }{\text{d}x} \left(\int_1^x f(t)\,dt \right) =\frac{\text{d} }{\text{d}x} \left( {1\over 3}x^3+ax+b \right) \Rightarrow f(x)=x^2+a\\ \Rightarrow f(x),g(x)皆為對稱y軸的偶函數\Rightarrow \overline{AB}為一水平線且\overline{AB}=2 \\ \Rightarrow \cases{A(-1,g(-1)) \Rightarrow A(-1,5/2)\\ B(1,g(1)) \Rightarrow B(1,5/2)} \\ \cases{\int_1^1 f(t)\,dt=0 = {1\over 3}+a+b \Rightarrow b=-{1\over 3}-a \\f(1)=g(1)={5\over 2}=1+a \Rightarrow a={3\over 2}} \Rightarrow (a,b)=({3\over 2},-{11\over 6}) \\ \Rightarrow a-b={3\over 2}+{11\over 6}={20\over 6}= \bbox[red, 2pt]{10\over 3}$$
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解答:$$\cases{X\sim N(60,5^2) \\ Y\sim N(70,6^2)} \Rightarrow X+Y \sim N(60+70,9^2) =N(130,9^2) \\ \Rightarrow Var(X+Y)=9^2=81 =Var(X)+Var(Y)+2 \text{Cov}(X,Y) =25 +36+ 2\text{Cov}(X,Y) \\ \Rightarrow \text{Cov}(X,Y)=10 \Rightarrow 迴歸直線斜率={\text{Cov}(X,Y)\over Var(X) } ={10\over 25}={2\over 5}\\ 迴歸直線經過(\bar X,\bar Y) =(60,70) \Rightarrow \bbox[red, 2pt]{y-50={2\over 5}(x-60)}$$
解答:$$0.216=0.6^3 \Rightarrow 0.6^3\le x\le 1 \Rightarrow 0\le \log_{0.6} x\le 3 \\取 t=\log_{0.6} x \Rightarrow x=0.6^t \Rightarrow f(x)=g(t)= (0.6^t)^{(t-2)^3} =0.6^{t(t-2)^3} \\ h(t)=t(t-2)^3 \Rightarrow h'(t)=(t-2)^3+3t(t-2)^2 =(t-2)^2(4t-2) =0 \Rightarrow t=2,{1\over 2} \\ \Rightarrow \cases{g(2)=1\\ g(1/2)=0.6^{-27/16} \gt 1} \Rightarrow 最大值=\bbox[red, 2pt]{0.6^{-27/16}}$$
解答:$$|z+3|+|z-3|=10為一橢圓,其中\cases{a=5\\ c=3 \\ 左焦點F_1(-3,0)\\ 右焦點F_2(3,0)}\\ |z_1+3|,|z_2+3|, |z_3+3|代表三點至左焦點距離,也就是Re(z_1),Re(z_2),Re(z_3)成等差\\因此 Re(z_1+z_3) =2Re(z_2) =2\times {5\over 4} =\bbox[red,2pt]{5\over 2}$$
解答:
$$\cases{y=-\sqrt{4-x^2} 圖形為x^2+y^2=4的下半圓 \\ y=-x^2+4圖形為一凹向下拋物線，其頂點為(0,4)} \Rightarrow 所圍區域如上圖\\ 直線3x+y=k的斜率為-3 \\ \cases{y=-\sqrt{4-x^2}\\ y=-x^2+4} \Rightarrow \cases{y'={x\over \sqrt{4-x^2}}=-3 \Rightarrow x=-3\sqrt 2/\sqrt 5 \Rightarrow y=-\sqrt{2\over 5}\\y'=-2x =-3 \Rightarrow x=3/2 \Rightarrow y=7/4} \\ \Rightarrow 切點為\cases{（-3\sqrt{2\over 5},-\sqrt{2\over 5})\\ ({3\over 2},{7\over 4})} \Rightarrow 3x+y=\cases{-10\sqrt{2\over 5}=-2\sqrt{10}\\{25\over 4}} \\ \Rightarrow \bbox[red, 2pt]{-2\sqrt{10}\le k\le {25\over 4}}$$
解答:$$x={3\over 2x+1} \Rightarrow 2x^2+x-3=0 \Rightarrow (2x+3)(x-1)=0 \Rightarrow x=1,-{3\over 2} \\ 取b_n={a_n-1\over a_n+3/2} =\frac{{3\over 2a_{n-1}+1}-1}{ {3\over 2a_{n-1}+1}+{3\over 2}} ={2-2a_{n-1}\over 3a_{n-1} +{9\over 2}} =-{2\over 3} \cdot {a_{n-1}-1 \over a_{n-1}+{3\over 2}} =-{2\over 3} b_{n-1} \\ a_1=2 \Rightarrow b_1={2-1\over 2+3/2} ={2\over 7} \Rightarrow b_n={2\over 7} \cdot (-{2\over 3})^{n-1}\\ b_n={a_n-1\over a_n+3/2} \Rightarrow a_n={5\over 2(1-b_n)}-{3\over 2}=1+{{10\over 7}(-2/3)^{n-1} \over 2-{4\over 7}(-2/3)^{n-1}} \\ \Rightarrow \lim_{n\to \infty} (a_n-1)(-{3\over 2})^n =\lim_{n\to \infty} {{10\over 7}(-3/2) \over 2-{4\over 7}(-2/3)^{n-1}} =\lim_{n\to \infty} {-{30\over 14} \over 2 } = \bbox[red, 2pt]{15\over 14}$$
解答:$$((1+x)+x^2(1+x))^6 = (1+x^2)^6 (1+x)^6 \\ \Rightarrow \cases{(1+x^2)^6=1+6x^2+ 15x^4+ 20x^6+ 15x^8+ 6x^{10} + x^{12} \\ (1+x)^6= 1+6x+15x^2 +20x^3+ 15x^4+ 6x^5+ x^6} \\ \Rightarrow x^5係數=1\cdot 6+6\cdot 20+ 15\cdot 6 =\bbox[red, 2pt]{216}$$
$$跟第七題一樣,假設\cases{x=x'\\ y=3y'/5} \Rightarrow {x'^2\over 25}+{(3y'/5-3)^2\over 9}=1 \Rightarrow x'^2+(y'-5)^2=25為一圓\\ 當P'(0,15)有最小外切三角形(正\triangle ),其面積為75\sqrt 3\\ \left| {\partial(x,y) \over \partial(x',y')} \right|= \begin{Vmatrix} 1 & 0\\ 0 & 3/5\end{Vmatrix} ={3\over 5} \Rightarrow \cases{m={3\over 5}\times 15=9\\ M={3\over 5} \times 75\sqrt 3=45\sqrt 3} \Rightarrow (m,M) =\bbox[red, 2pt]{(9,45\sqrt 3)}$$
解答:$$\textbf{(1)}\; A=\begin{bmatrix}\frac{1}{10} & \frac{7}{10} \\\frac{9}{10} & \frac{3}{10}\end{bmatrix} = \begin{bmatrix}-1 & \frac{7}{9} \\1 & 1\end{bmatrix} \begin{bmatrix}\frac{-3}{5} & 0 \\0 & 1\end{bmatrix} \begin{bmatrix}\frac{-9}{16} & \frac{7}{16} \\\frac{9}{16} & \frac{9}{16}\end{bmatrix} \\\Rightarrow A^n= \begin{bmatrix}-1 & \frac{7}{9} \\1 & 1\end{bmatrix} \begin{bmatrix} (\frac{-3}{5})^n & 0 \\0 & 1^n \end{bmatrix} \begin{bmatrix}\frac{-9}{16} & \frac{7}{16} \\\frac{9}{16} & \frac{9}{16} \end{bmatrix}= \begin{bmatrix} {9\over 16}(-{3\over 5})^n +{7\over 16}& -{7\over 16}({-3\over 5})^n+{7\over 16}\\ -{9\over 16}({-3\over 5})^n+{9\over 16}& {7\over 16}({-3\over 5})^n+{9\over 16}\end{bmatrix}\\ \alpha_nA+\beta_nI= \begin{bmatrix} {1\over 10}\alpha_n & {7\over 10}\alpha_n\\ {9\over 10}\alpha_n & {3\over 10}\alpha_n \end{bmatrix} +\begin{bmatrix}\beta_n & 0 \\0 & \beta_n \end{bmatrix} = \begin{bmatrix} {1\over 10}\alpha_n +\beta_n& {7\over 10}\alpha_n\\ {9\over 10}\alpha_n & {3\over 10}\alpha_n +\beta_n \end{bmatrix} \\ \Rightarrow \cases{\alpha_n =-{5\over 8}(-{3\over 5})^n+{5\over 8} \\\beta_n = {5\over 8}(-{3\over 5})^n+ {3\over 8}} \Rightarrow \alpha_n+ \beta_n=1. \;\bbox[red, 2pt]{QED}\\ \textbf{(2)}\; \alpha_n =-{5\over 8}(-{3\over 5})^n+{5\over 8} \Rightarrow -0.6\alpha_n+1={3\over 8}(-{3\over 5})^n-{3\over 8}+1 \\\qquad ={3\over 8} \cdot (-{5\over 3})(-{3\over 5})^{n+1}+{5\over 8}=-{5\over 8}(-{3\over 5})^{n+1}+{5\over 8} =\alpha_{n+1}. \bbox[red, 2pt]{QED.}\\ \lim_{n\to \infty} \alpha_n=\lim_{n\to \infty} \left( -{5\over 8}(-{3\over 5})^n+{5\over 8}\right) =\bbox[red, 2pt]{5\over 8}$$
解答:$$假設\omega=\cos{2\pi\over n}+i\sin {2\pi\over n},則x^{n-1}+x^{n-2}+\cdots +1=0的n-1個根為\omega^k,k=1,2\dots,n-1且\omega^n=1 \\令f(x)=x^{n-1}+x^{n-2}+\cdots +1= (x-\omega^1)(x-\omega^2)\cdots (x-\omega^{n-1}) \\\Rightarrow f(1)=n=(1-\omega^1) (1-\omega^2) \cdots (1-\omega^{n-1}) \\ \Rightarrow 1-\omega^k= 1-(\cos{2 k\pi\over n}+i\sin {2k\pi\over n}) =2\sin^2{k\pi\over n}-2\sin{k\pi\over n} \cos{k\pi \over n}i =2\sin{k\pi\over n} \left(\sin{k\pi\over n}-\cos{k\pi \over n}i \right) \\ \Rightarrow |1-\omega^k|=2\sin{k\pi \over n} \Rightarrow n=|1-\omega^1||1-\omega^2|\cdots |1-\omega^{n-1}| =2\sin{\pi\over n} \cdot 2\sin{2\pi\over n} \cdots \sin{(n-1)\pi\over n} \\ \Rightarrow \sin{\pi\over n}\cdot \sin{2\pi\over n} \cdots \sin{(n-1)\pi \over n}={n\over 2^{n-1}}. \bbox[red, 2pt]{ \text{QED.}}$$
解答:
$$\triangle ABC為邊長30的正三角形，因此假設\cases{A(0,0,0)\\ B(30,0,0)\\ C(15,15\sqrt 3,0)}\\ \Rightarrow \cases{E=\overline{AB}的中點 =(15,0,0)\\G為\triangle ABC的重心=(15,5\sqrt 3,0) } \Rightarrow \cases{\overline{DE} =\sqrt{\overline{AD}^2-\overline{AE}^2} =15\sqrt{15} \\ \overline{EG}={1\over 3}\overline{CE} =5\sqrt 3} \\ \Rightarrow \overline{DG} =\sqrt{\overline{DE}^2 -\overline{EG}^2} =10\sqrt{33} \Rightarrow D(15,5\sqrt 3,10\sqrt{33}) \\ \overline{GG'}= \overline{EG} \tan \theta =5\sqrt 3\cdot {\sqrt{11} \over 5} =\sqrt{33} \Rightarrow G'=(15,5\sqrt 3,\sqrt{33})\\ \Rightarrow 平面E=\triangle G'AB: -\sqrt{11}y+5z=0 \Rightarrow C'= \overleftrightarrow{CD} \cap E =(15,{25\over 2}\sqrt{3},{5\over 2}\sqrt{33}) \\ \Rightarrow \overline{CC'}=\sqrt{({5\over 2}\sqrt 3)^2+ ({5\over 2}\sqrt{33})^2} = \bbox[red, 2pt]{15}$$
二、計算證明題(共10 分)
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解題僅供參考，其他歷年試題及詳解
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