107年高雄市市立高中聯合教甄-數學科詳解

高雄市 107 學年度市立高級中等學校
聯合教師甄選數學科試題

計算題：一律詳列過程； 1~12 題每題 7 分， 13~14 題每題 8 分

解：

$$\begin{array}{|c|c|c|c|c|} \hline  \bbox[black,2pt]{\square} &   & \bbox[black,2pt]{\square} &   &\bbox[black,2pt]{\square} \\\hline  & \bbox[black,2pt]{\square} &  & \bbox[black,2pt]{\square} &  \\\hline \bbox[black,2pt]{\square} &   & \bbox[black,2pt]{\square} &   &\bbox[black,2pt]{\square} \\\hline   & \bbox[black,2pt]{\square} &   & \bbox[black,2pt]{\square} &  \\\hline \bbox[black,2pt]{\square} &   & \bbox[black,2pt]{\square} &   &\bbox[black,2pt]{\square} \\\hline\end{array}=\begin{array}{|c|c|c|c|c|} \hline  a_1 & a_2 & a_3 & a_4 &a_5 \\\hline a_6 & a_7 & a_8 & a_9 &a_{10} \\\hline a_{11} & a_{12} & a_{13} & a_{14} &a_{15} \\\hline a_{16} & a_{17} & a_{18} & a_{19} &a_{20} \\\hline a_{21} & a_{22} & a_{23} & a_{24} &a_{25} \\\hline\end{array} \\ 符合要求的選取:\\首項在第1列: (a_1,\{a_8,a_{10}, a_{12}, a_{14},a_{18}, a_{20}, a_{22},a{24}\})\to 8個\\ \qquad \qquad (a_2,-),(a_3,-),(a_4,-),(a_5,-),也是各8個，共5\times 8=40個 \\首項在第2列:(a_6,\{a_{13}, a_{15}, a_{17},a_{19}, a_{23}, a_{25}\})\to 6個\\ \qquad \qquad (a_7,\{a_{14}, a_{16}, a_{18},a_{20}, a_{24}\})\to 5個\\\qquad \qquad (a_8,-),(a_{10},-)各6個,(a_9,-)5個，共3\times 6+2\times 5=28個\\首項在第3列:(a_{11},-),(a_{12},-),(a_{13},-),(a_{14},-),(a_{15},-)各4個，共4\times 5=20個\\ 首項在第4列: (a_{16},-),(a_{18},-),(a_{20},-)各2個，(a_{17},a_{24}),(a_{19},a_{22})，共3\times 2+2=8個\\總共有40+28+20+8 = 96個符合要求，機率為{96\over C^{25}_2} ={96\over 25\times 12} =\bbox[red, 2pt]{8\over 25}$$

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解：

$$f(x)=x^{107}-2018x^{106}+g(x),g(x)為105次實係數多項式\\ \Rightarrow f(x^2+x+1)= (x^2+x+1)^{107}- 2018(x^2+x+1)^{106}+ g(x^2+x+1)\\ \Rightarrow f(x^2+x+1)為214次多項式且f(x^2+x+1)=0所有根的和即為x^{213}的係數再乘上(-1);\\而x^{213}只出現在(x^2+x+1)^{107}內，其係數為107 \Rightarrow f(x^2+x+1)=0所有根的和=\bbox[red, 2pt]{-107}$$

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解：

$$S=\{2^1,2^2,\dots,2^{201}\} = \{a_1,a_2,\dots,a_{201}\},取a_ia_ja_k成等比數列,1\le i,j,k \le 201\\\begin{array} {ll} 數列& 數量\\\hline a_1a_2a_3,a_2a_3a_4,\dots,a_{199}a_{200}a_{201} & 199 \\ a_1a_3 a_5, a_2a_4a_6, \dots, a_{197}a_{199}a_{201} & 197 \\ a_1a_4 a_7,a_2a_5a_8,\dots,a_{195}a_{198}a_{201} & 195 \\ \cdots & \cdots\\ a_1a_{101}a_{201} & 1\\\hline\end{array} \\ \Rightarrow 共有199+197+\cdots + 1=\sum_{k=1}^{100}(2k-1) =10100-100 =\bbox[red, 2pt]{10000}$$

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解：

$$R=\int_0^2 x^3\;dx = \left. \left[ {1\over 4}x^4 \right]\right|_0^2 =4;\\U_n={2\over n}\left( ({2\over n})^3 +({4\over n})^3 +({6\over n})^3 +\cdots +({2n\over n})^3 \right) =({2\over n})^4(1^3+2^3+\cdots +n^3) \\ =({2\over n})^4 \times \left({n(n+1)\over 2} \right)^2 =4(1+{2\over n}+ {1\over n^2});\\ \Rightarrow |U_n-R| < {1\over 100} \Rightarrow {8n+4 \over n^2} < {1\over 100} \Rightarrow n^2-800n-400 > 0 \\令f(n)=n^2-800n-400 \Rightarrow \cases{f(800)=-400\lt 0\\ f(801)=401 \gt 0}\Rightarrow n \ge \bbox[red, 2pt]{801}$$

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解：

$$g(x)=\sqrt{x+a}-b \Rightarrow \cases{g(1)=0\\ g'(1)=1/6} \Rightarrow \cases{\sqrt{a+1}=b\\ {1\over 2\sqrt{a+1}}={1\over 6}}  \Rightarrow (a,b)=\bbox[red, 2pt]{(8,3)}$$

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解：

$$\lim_{n\to \infty}{1\over \sqrt n}({1\over \sqrt n+\sqrt{n+2}} +{1\over \sqrt {n+1} +\sqrt{n+3}} +{1\over \sqrt {n+2}+\sqrt{n+4}} + \cdots +{1\over \sqrt {2n-1}+\sqrt{2n+1}} )\\ = \lim_{n\to \infty}{1\over 2\sqrt n}(\sqrt{n+2} -\sqrt{n} +\sqrt {n+3} -\sqrt{n+1} +  \sqrt {n+4} -\sqrt{n+2} + \cdots +\sqrt {2n+1}-\sqrt{2n-1} )\\ = \lim_{n\to \infty}{1\over 2\sqrt n}(\sqrt{2n+1} +\sqrt{2n}-\sqrt n-\sqrt{n+1}) \\ =\lim_{n\to \infty}{1\over 2}\left(\sqrt{2+{1\over n}} +\sqrt 2-1-\sqrt{1+{1\over n}} \right) = {1\over 2}(\sqrt 2+\sqrt 2-1-1)\\ =\bbox[red,2pt]{\sqrt 2-1}$$

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解：

$$令\cases{A(a,0),a\ge 0\\ B(0,b)\\C(x,y) \\P(-3,0)} \Rightarrow \cases{\overrightarrow{PB}=(3,b)\\ \overrightarrow{BA}=(a,-b) \\ \overrightarrow{BC}=(x,y-b)\\ \overrightarrow{CA}= (a-x,-y)} \Rightarrow \cases{\overrightarrow{BA} \cdot \overrightarrow{BA}=3a-b^2=0 \\ 2\overrightarrow{BC}+ 3\overrightarrow{CA} =(3a-x,-y-2b)=(0,0)} \\ \Rightarrow \cases{ a=x/3\\ b=-y/2} \Rightarrow 3a-b^2 = x-(-y/2)^2 = x-y^2/4=0 \Rightarrow \bbox[red, 2pt]{y^2=4x}$$

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解：

$$\cases{L_1:{x-4\over 1}= {y+3\over -1},z=0\\ L_2:{x-2\over 1}={y-2\over 1},z=1} \Rightarrow \cases{\vec u_1=(1,-1,0)\\ \vec u_2=(1,1,0)}  \Rightarrow \vec n=\vec u_1\times \vec u_2 =(0,0,2)\\令\cases{在L_1上任取一點P(4,-3,0)\\ 在L_2上作取一點Q(2,2,1)} \Rightarrow \overrightarrow {PQ}=(-2,5,1) \\ \Rightarrow 兩歪斜線的距離d={|\overrightarrow{PQ} \cdot \vec n|\over |\vec n|}={2\over 2}=1\Rightarrow 稜邊長=\sqrt{2}d= \bbox[red, 2pt]{\sqrt 2}$$

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解：

$$圓C:x^2+y^2-2x+3y=0 \Rightarrow (x-1)^2 + (y+{3\over 2})^2= {13\over 4} \Rightarrow \cases{圓心O(1,-{3\over 2})\\ 半徑r={\sqrt{13}\over 2}} \\ 令L_1\bot L\Rightarrow L_1: 5x+4y=k，又L_1經過O，因此5-6=-1= k \Rightarrow L_1: 5x+4y+1=0 \\ \Rightarrow P在L_1上可表示成P(a,{-1-5a\over 4}),又\text{dist}(O,L) = \overline{OD} = {4+{15\over 2}-18\over \sqrt{5^2+4^2}} ={13\over 2\sqrt{41}};\\ 由於\cases{\angle AOP= \angle AOD\\ \angle OAP=90^\circ= \angle ODA} \Rightarrow \triangle OAP \sim \triangle ODA \Rightarrow \overline{OA}^2 = \overline{OD}\times \overline{OP}\\ \Rightarrow {13\over 4}= {13\over 2\sqrt{41}}\times \overline{OP} \Rightarrow \overline{OP}= {\sqrt{41} \over 2 } =\sqrt{(a-1)^2 + (-{3\over 2}+{5a+1 \over 4})^2} \\{41\over 4} ={41\over 16}(a-1)^2\Rightarrow a=3 \Rightarrow \bbox[red, 2pt]{P(3,-4)}$$

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解：

$$\angle A_1OA_{12}=360^\circ \div 12=30^\circ \Rightarrow \cases{\overline{QA_{12}}= {1\over 2}r =1\\ \overline{OQ}= {\sqrt 3\over 2}r =\sqrt 3} \\\Rightarrow \cases{矩形A_2A_{12}A8A_6面積= \overline{A_2A_{12}}\times \overline{A_{12}A_8}  =2 \overline{QA_{12}} \times 2 \overline{OQ}=4\sqrt 3; \\正方形ABCD面積= \overline{A_2A_{12}}^2=4};\\ \Rightarrow 陰影面積=2\times 矩形A_2A_{12}A8A_6面積-正方形ABCD面積 =2\times 4\sqrt 3-4= \bbox[red, 2pt]{8\sqrt 3-4}$$

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解：

$$f(x)= \sqrt{2^{2x}+\left(x-1\right)^2}+\sqrt{\left(2^x-4\right)^2+\left(x-1\right)^2}\\ \Rightarrow f'(x)=\dfrac{\ln\left(2\right){\cdot}2^{2x+1}+2\left(x-1\right)}{2\sqrt{2^{2x}+\left(x-1\right)^2}}+\dfrac{\ln\left(2\right)\left(2^x-4\right){\cdot}2^{x+1}+2\left(x-1\right)}{2\sqrt{\left(2^x-4\right)^2+\left(x-1\right)^2}}\\ f'(x)=0 \Rightarrow x=1 \Rightarrow f(1)=2+2= \bbox[red, 2pt]{4}$$

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解：

$${x^2 \over 144}-{y^2\over 25}=1 \Rightarrow \cases{a=12\\ b=5 } \Rightarrow c=13;\\現在\cases{ \overline{PF_1}:\overline{PF_2} =1:3 \\\overline{PF_2}-\overline{PF_1}= 2a=24} \Rightarrow \cases{\overline{PF_1}=12 \\ \overline{PF_2}=36} \\\Rightarrow \cos \angle F_1PF_2 = {\overline{PF_1}^2 +\overline{PF_2}^2-\overline{F_2F_2}^2 \over 2\times \overline{PF_1}\times \overline{PF_2}} ={12^2+36^2-26^2 \over 2\times 12\times 36} =\bbox[red, 2pt]{191\over 216}$$

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解：

$$正弦定理: {\overline{S_1S_2} \over \sin \angle S_1BS_2} = {\overline{BS_1} \over \sin \angle BS_2S_1} \Rightarrow {r \over \sin(180^\circ-\varphi-\theta)}  = {\overline{BS_1} \over \sin \theta} \Rightarrow \overline{BS_1}= {r\sin \theta \over \sin (\varphi+\theta)} \\ \Rightarrow \overline{AB}= \overline{BS_1}\tan \angle AS_1B ={r\sin \theta \tan \phi\over \sin (\varphi+\theta)}\Rightarrow \overline{AB}= {r\sin \theta \tan \phi\over \sin (\varphi+\theta)}，\bbox[red, 2pt]{故得證}$$

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解：

$$\lim_{\theta \to 0}{\tan \theta-\sin \theta \over \theta^3} =\lim_{\theta \to 0}{(\tan \theta-\sin \theta)' \over (\theta^3)'} =\lim_{\theta \to 0}{{1\over \cos^2 \theta}-\cos \theta  \over 3\theta^2} =\lim_{\theta \to 0}{\left({1\over \cos^2 \theta}-\cos \theta\right)'  \over (3\theta^2)'} \\=\lim_{\theta \to 0}{{2\sin \theta \over \cos^3 \theta}+\sin \theta  \over 6\theta} =\lim_{\theta \to 0}{\left({2\sin \theta \over \cos^3 \theta}+\sin \theta\right)'  \over (6\theta) '}=\lim_{\theta \to 0}{{2   \over \cos^2 \theta} +{6\sin^2 \theta   \over \cos^4 \theta}+\cos \theta  \over 6} = {2+0+1\over 6} ={1\over 2}\\，\bbox[red, 2pt]{故得證}$$

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