國立屏東女中 114 年教師甄試數學科試題
一、填充題(每格 5 分，共 85 分)
解答:$$z^3=ai, a\in \mathbb R \Rightarrow |z^3+4-5i|=|4+(a-5)i|=5 \Rightarrow 16+(a-5)^2=25 \Rightarrow (a-5)^2=9 \\ \Rightarrow \cases{a=8\\ a=2} \Rightarrow \cases{z^3=8i\\ z^3=2i} \Rightarrow \cases{z=-2i\\ z=-\sqrt[3]2i} \Rightarrow \cases{|z+i|=|-i|=1\\ |z+i|=|(1-\sqrt[3]2)i|=\sqrt[3]{2}-1} \\ \Rightarrow 最小值為 \bbox[red, 2pt]{\sqrt[3]{2}-1}$$
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$$由題意可知:ABCD為等腰梯形且\overline{CD}=2\overline{AB} \\ 令\overline{AC}與\overline{BD}的交點為P,且\angle APB=\theta \Rightarrow \cos \angle BAC=\cos \angle ABC=\cos {\pi-\theta\over 2} =\sin{\theta\over 2 }\\ =\sqrt{1-\cos \theta\over 2} = \sqrt{1-7/9\over 2} = \bbox[red, 2pt]{1\over 3}$$
解答:
解答:$$\lim_{h\to 0}{f(-1-h)-f(-1) \over 2h} = {1\over 2}\lim_{h\to 0}{f(-1-h)-f(-1) \over h} ={1\over 2}\lim_{k\to 0}{f(-1+k)-f(-1) \over -k} \\= -{1\over 2}f'(-1)=43 \Rightarrow f'(-1)=-86 \Rightarrow \cases{f(-1)=84\\ f'(-1)=-86\\ f''(-1)=0} \\\Rightarrow f(x)=ax^3+3ax^2 +(3a-86)x+(a-2) \\又\int_0^t f(x)dx\le \int_a^b f(x)\,dx, \forall b\gt a\gt -1 \Rightarrow f(0)=0 \Rightarrow a=2 \Rightarrow f(x)=2x^3+6x^2-80x \\ \Rightarrow f(x)=2x(x+8)(x-5) \Rightarrow t=\bbox[red, 2pt]5$$
解答:
解答:$$本題\bbox[cyan,2pt]{送分}$$
解答:$$假設\triangle ACD在平面z=0上，由題意可知:\triangle ABC與\triangle ACD面積已知(固定) \\ 因此在\triangle ABC平面垂直\triangle ACD平面時，體積有最大值\\ 假設\overline{AC}在y軸上 \Rightarrow \overrightarrow{AB}=\sqrt{30}({1\over 2},0,{\sqrt 3\over 2}), \overrightarrow{AD} = 9({\sqrt 3\over 2},{1\over 2},0) \\ \Rightarrow \cos \angle BAD ={\overrightarrow{AB} \cdot \overrightarrow{AD} \over |\overrightarrow{AB} || \overrightarrow{AD}|} ={ \sqrt 3\over 4} \Rightarrow \sin \angle BAD= t=\bbox[red, 2pt]{\sqrt{13}\over 4}$$
$$\textbf{(1) }\begin{bmatrix}{21\over 29} & {20\over 29} \\{20\over 29} & -{21\over 29} \end{bmatrix} =\begin{bmatrix}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{bmatrix} 為一鏡射矩陣 \Rightarrow \cases{\cos 2\theta=21/29\\ \sin 2\theta=20/29} \\ \Rightarrow \tan 2\theta={20\over 21} \Rightarrow \tan \theta={2\over 5} \Rightarrow 直線L: y={2\over 5}x \Rightarrow \Gamma_1\cap \Gamma_2= L\cap \Gamma_1 =P(第一象限交點)\\ \Rightarrow {x^2\over 100}+{(2x/5)^2\over 2}=1 \Rightarrow 9x^2=100 \Rightarrow x={10\over 3} \Rightarrow y={4\over 3} \Rightarrow P= \bbox[red, 2pt]{({10\over 3},{4\over 3})} \\\textbf{(2) }\Gamma_2 \to \Gamma_1 對稱直線為L及與L垂直的直線\Rightarrow k= \bbox[red, 2pt]{{2\over 5}或-{5\over 2}}$$
解答:$$直線L的方向向量\vec \ell=({\sqrt 2\over 2}, -{\sqrt 2\over 2},-\sqrt 3), 母線M的方向向量\vec m=(0,0,-1)\\ 兩直線夾角\theta,則\cos \theta={ \vec \ell \cdot \vec m\over |\vec \ell||\vec m|} ={\sqrt 3\over 2} \Rightarrow \theta=30^\circ 或150^\circ\\ 假設P(x,y,z)在光線邊緣上 \Rightarrow \overrightarrow{SP} =(x-2\sqrt 6,y-3\sqrt 6,z-12)與\vec \ell夾角為30^\circ \Rightarrow \overrightarrow{SP}\cdot \vec \ell =\lvert \overrightarrow{SP}\rvert |\vec \ell| \cos 30^\circ \\ \Rightarrow {\sqrt 2\over 2}(x-2\sqrt 6)-{\sqrt 2\over 2}(y-3\sqrt 6)-\sqrt 3(z-12)=\sqrt 3 \sqrt{(x-2\sqrt 6)^2 +(y-3\sqrt 6)^2 +(z-12)^2} \\ y=0代入上式\Rightarrow 5x^2+2\sqrt 6xz-50\sqrt 6x+12z+318=0 \Rightarrow 判別式:(2\sqrt 6)^2-0=24\gt 0\\ \Rightarrow 圖形為\bbox[red, 2pt]{雙曲線}\\ 對稱軸L:{x-2\sqrt 6\over \sqrt 2/2} ={y-3\sqrt 6\over -\sqrt 2/2} ={z-12\over -\sqrt 3} 在y=0的投影為{x-2\sqrt 6\over \sqrt 2/2} ={z-12\over -\sqrt 3} \\ \Rightarrow x=2\sqrt 6-{z-12\over \sqrt 6} 代入雙曲線方程式可得x=-\sqrt 6\pm 18\sqrt{2\over 7} \Rightarrow z=30\pm36\sqrt{3\over 7}\\ 由圖形可知z=30-36\sqrt{3\over 7},因此頂點坐標為\bbox[red, 2pt]{\left( -\sqrt 6+ 18\sqrt{2\over 7},0,30-36\sqrt{3\over 7}\right)}$$
解答:$$\lim_{h\to 0}{f(-1-h)-f(-1) \over 2h} = {1\over 2}\lim_{h\to 0}{f(-1-h)-f(-1) \over h} ={1\over 2}\lim_{k\to 0}{f(-1+k)-f(-1) \over -k} \\= -{1\over 2}f'(-1)=43 \Rightarrow f'(-1)=-86 \Rightarrow \cases{f(-1)=84\\ f'(-1)=-86\\ f''(-1)=0} \\\Rightarrow f(x)=ax^3+3ax^2 +(3a-86)x+(a-2) \\又\int_0^t f(x)dx\le \int_a^b f(x)\,dx, \forall b\gt a\gt -1 \Rightarrow f(0)=0 \Rightarrow a=2 \Rightarrow f(x)=2x^3+6x^2-80x \\ \Rightarrow f(x)=2x(x+8)(x-5) \Rightarrow t=\bbox[red, 2pt]5$$
$$假設\cases{A(0,a) \\B(0,b)\\C(c,0)\\ D(0,0)} \Rightarrow M={A+C\over 2} =({c\over 2},{a\over 2}) \Rightarrow \cases{\overline{AC}^2= a^2+c^2\\ \overline{BC}^2=b^2+c^2\\ \overline{AB}^2=(a-b)^2} \\ \Rightarrow a^2+c^2=b^2+c^2+{1\over 2}(a-b)^2 \Rightarrow a^2+2ab-3b^2=0 \Rightarrow (a-b)(a+3b)=0 \Rightarrow a=-3b\\ 取\cases{a=6\\ b=-2\\ c=4} \Rightarrow \cases{A(0,6) \\B(0,-2)\\C(4,0)\\ D(0,0)} \Rightarrow M=(2,3) \Rightarrow \overleftrightarrow{BM}:y={5\over 2}x-2 \Rightarrow P=({4\over 5},0) \\ \Rightarrow \cases{\overrightarrow{AP}=({4\over 5},-6) \\\overrightarrow{AB} =(0,-8) \\\overrightarrow{AC} =(4,-6)} \Rightarrow ({4\over 5},-6) =x(0,-8)+y(5,-6) \Rightarrow (x,y) = \bbox[red, 2pt]{({3\over 5}, {1\over 5})}$$
解答:$$共有C^5_2=10場比賽，每隊都2勝2負，因此機率為{1\over 2^{10}}\cdot (5-1)!={24\over 1024} =\bbox[red, 2pt]{3\over 128} \\ 其中(5-1)!代表環狀排列數$$
解答:$$\cases{x+y=z-1\\ xy=z^2-7z+14} \Rightarrow xy=(x+y+1)^2-7(x+y+1)+14 \Rightarrow x^2+xy+y^2-5x-5y+8=0 \\ 令\cases{f(x,y)= x^2+y^2-4x-4y\\ g(x,y)= x^2+xy+y^2-5x-5y+8} \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y\\ g=0 } \Rightarrow \cases{2x-4=\lambda (2x+y-5) \\ 2y-4=\lambda(x+2y-5) \\x^2+xy+y^2-5x-5y+8=0} \\ \Rightarrow \cases{x=2\Rightarrow y=1\\ y=2\Rightarrow x=1\\ x=y \Rightarrow x=y=2,4/3}\Rightarrow (x,y)=(2,1),(1,2),(2,2),(4/3,4/3) \Rightarrow \cases{f(2,1)=-7\\ f(1,2)=-7\\ f(2,2)=-8\\ f(4/3,4/3) =-64/9} \\ \Rightarrow (M,m) = \bbox[red, 2pt]{(-7,-8)}$$
解答:$$x=\sum_{k=2}^{20}{20\choose k}{80\choose k+2} =\sum_{k=2}^{20}{20\choose k}{80\choose 78-k} ={20+80\choose78}-{20\choose 0}{80\choose 78}-{20\choose 1}{80\choose 77} \\={100\choose 22}-{80\choose 2} -20{80\choose 3} \Rightarrow \cases{{100\choose 22} \equiv {2\choose 0}{18\choose 22} \text{ mod }41 \equiv 0 \text{ mod 41} \\ {80\choose 2} \equiv {1\choose 0}{39\choose 2} \text{ mod 41} \equiv 3 \text{ mod }41 \\ 20{80\choose 3} \equiv 740 \text{ mod 41} \equiv 2 \text{ mod 41}} \\ \Rightarrow x \text{ mod 41} \equiv -5 \text{ mod 41} \equiv \bbox[red, 2pt]{36} \text{ mod 41} \\ 註:公式來源\href{https://brilliant.org/wiki/lucas-theorem/}{\text{Lucas's Theorem}}$$
解答:$$假設\cases{\triangle代表1-7的任一數字\\\Box代表1-7及9的任一數字 },符合條件的數字:\\ \cases{\textbf{三個8 }:(\Box888)有8\times 4=32個\\ \textbf{二個8 }:\cases{88(\Box \Box): 8\times7=56個\\ 89(8\triangle):2\times 7=14個\\ 9(88\triangle):7\times 3=21個} \Rightarrow 合計91個\\ \textbf{一個8 }: \cases{89(\triangle \triangle): 7\times 6=42個 \\9(8\triangle \triangle):7\times 6\times 3=126} \Rightarrow 合計168個 \\\textbf{零個8 }:9(\triangle \triangle \triangle): 7\times 6\times 5=210 個} \Rightarrow 合計32+ 91+168+210= \bbox[red, 2pt]{501}$$
解答:$$P=\begin{bmatrix} 0 & 1 & 1 \\1 & 0 & 1 \\1 & 1 & 0 \end{bmatrix} =\begin{bmatrix} -1 & -1 & 1 \\1 & 0 & 1 \\0 & 1 & 1\end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\0 & -1 & 0 \\0 & 0 & 2\end{bmatrix} \begin{bmatrix} \frac{-1}{3} & \frac{2}{3} & \frac{-1}{3} \\\frac{-1}{3} & \frac{-1}{3} & \frac{2}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{bmatrix} \\ \Rightarrow P^n= \begin{bmatrix} -1 & -1 & 1 \\1 & 0 & 1 \\0 & 1 & 1\end{bmatrix} \begin{bmatrix} (-1)^n & 0 & 0 \\0 & (-1)^n & 0 \\0 & 0 & 2^n \end{bmatrix} \begin{bmatrix} \frac{-1}{3} & \frac{2}{3} & \frac{-1}{3} \\\frac{-1}{3} & \frac{-1}{3} & \frac{2}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3}\end{bmatrix} \\= \begin{bmatrix} \frac{2 \left(-1\right)^n +2^n}{3} & \frac{-\left(-1\right)^n+2^n}{3} & \frac{-\left(-1\right)^n+2^n}{3} \\\frac{-\left(-1\right)^n+2^n}{3} & \frac{2 \left(-1\right)^n+2^n}{3} & \frac{-\left(-1 \right)^n +2^n}{3} \\\frac{-\left(-1\right)^n+2^n}{3} & \frac{-\left(-1 \right)^n+2^n}{3} & \frac{2 \left(-1\right)^n+2^n}{3} \end{bmatrix} \Rightarrow \bbox[red, 2pt]{a_n={1\over 3}(2^n-(-1)^n)}$$
解答:$$取a_n=\left({b_{n+1} \over 2b_n} \right)^2 +b_n^2 \Rightarrow a_{n+1} =\left({b_{n+2} \over 2b_{n+1}} \right)^2 +b_{n+1}^2 =\left( {\displaystyle {b_{n+1}^3\over 2b_n^2}-2b_{n+1}b_n^2 \over 2b_{n+1}} \right)^2 +b_{n+1}^2 \\=\left({b_{n+1}^2\over 4b_n^2}-b_n^2 \right)^2+b_{n+1}^2 =\left[ \left( {b_{n+1 }\over 2b_n} \right)^2 +b_n^2\right]^2 =a_n^2 \\ \Rightarrow a_1=\left({b_{2} \over 2b_1} \right)^2+ b_1^2 = \left({96 \over 12} \right)^2+ 6^2 =100=10^2 \Rightarrow a_2=10^4 \Rightarrow a_3=10^8 \Rightarrow a_n=10^{2^n} \\ \Rightarrow \log(\log p)= \log(\log a_{100}) =\log (\log 10^{2^{100}}) =\log 2^{100} =100\log 2 =100\times 0.301=30.1 \Rightarrow \bbox[red, 2pt]{30}$$
解答:$$\cases{\vec a\cdot \vec b=\vec a\cdot \vec c\\ \vec a\times \vec b=\vec a\times \vec c} \Rightarrow \cases{\vec a\cdot(\vec b-\vec c)=0\\ \vec a\times(\vec b-\vec c) =0 }\Rightarrow \vec a\bot (\vec b-\vec c) 且\vec a\parallel (\vec b-\vec c) \Rightarrow \vec b-\vec c=0 \\ \Rightarrow \vec b=\vec c \quad \bbox[red, 2pt]{QED.}$$
解答:$$\textbf{(1) }a_5=\cases{甲甲甲甲甲:1種\\ 甲(乙甲甲甲):4種\\ 甲乙甲(甲乙):2種\\ 甲甲(乙乙甲):3種} \Rightarrow a_5=1+4+2+3= \bbox[red, 2pt]{10} \\\textbf{(2) }由一路領先的\href{https://chu246.blogspot.com/2021/11/blog-post.html}{公式}可知: a_n=1+\sum_{k=1}^{[n/2]} \left(C^n_k-C^n_{{k-1}} \right), 其中k=乙的數量 \\=1+(C^n_1-1)+ (C^n_2-C^n_1) +(C^n_3-C^n_2) + \cdots (C^n_{[n/2]}-C^n_{[n/2]-1}) = \bbox[red, 2pt]{C^n_{[n/2]}}$$
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解題僅供參考，其他教甄試題及詳解
解答:$$共有C^5_2=10場比賽，每隊都2勝2負，因此機率為{1\over 2^{10}}\cdot (5-1)!={24\over 1024} =\bbox[red, 2pt]{3\over 128} \\ 其中(5-1)!代表環狀排列數$$
解答:$$\cases{x+y=z-1\\ xy=z^2-7z+14} \Rightarrow xy=(x+y+1)^2-7(x+y+1)+14 \Rightarrow x^2+xy+y^2-5x-5y+8=0 \\ 令\cases{f(x,y)= x^2+y^2-4x-4y\\ g(x,y)= x^2+xy+y^2-5x-5y+8} \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y\\ g=0 } \Rightarrow \cases{2x-4=\lambda (2x+y-5) \\ 2y-4=\lambda(x+2y-5) \\x^2+xy+y^2-5x-5y+8=0} \\ \Rightarrow \cases{x=2\Rightarrow y=1\\ y=2\Rightarrow x=1\\ x=y \Rightarrow x=y=2,4/3}\Rightarrow (x,y)=(2,1),(1,2),(2,2),(4/3,4/3) \Rightarrow \cases{f(2,1)=-7\\ f(1,2)=-7\\ f(2,2)=-8\\ f(4/3,4/3) =-64/9} \\ \Rightarrow (M,m) = \bbox[red, 2pt]{(-7,-8)}$$
解答:$$假設\cases{\triangle代表1-7的任一數字\\\Box代表1-7及9的任一數字 },符合條件的數字:\\ \cases{\textbf{三個8 }:(\Box888)有8\times 4=32個\\ \textbf{二個8 }:\cases{88(\Box \Box): 8\times7=56個\\ 89(8\triangle):2\times 7=14個\\ 9(88\triangle):7\times 3=21個} \Rightarrow 合計91個\\ \textbf{一個8 }: \cases{89(\triangle \triangle): 7\times 6=42個 \\9(8\triangle \triangle):7\times 6\times 3=126} \Rightarrow 合計168個 \\\textbf{零個8 }:9(\triangle \triangle \triangle): 7\times 6\times 5=210 個} \Rightarrow 合計32+ 91+168+210= \bbox[red, 2pt]{501}$$
解答:$$P=\begin{bmatrix} 0 & 1 & 1 \\1 & 0 & 1 \\1 & 1 & 0 \end{bmatrix} =\begin{bmatrix} -1 & -1 & 1 \\1 & 0 & 1 \\0 & 1 & 1\end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\0 & -1 & 0 \\0 & 0 & 2\end{bmatrix} \begin{bmatrix} \frac{-1}{3} & \frac{2}{3} & \frac{-1}{3} \\\frac{-1}{3} & \frac{-1}{3} & \frac{2}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{bmatrix} \\ \Rightarrow P^n= \begin{bmatrix} -1 & -1 & 1 \\1 & 0 & 1 \\0 & 1 & 1\end{bmatrix} \begin{bmatrix} (-1)^n & 0 & 0 \\0 & (-1)^n & 0 \\0 & 0 & 2^n \end{bmatrix} \begin{bmatrix} \frac{-1}{3} & \frac{2}{3} & \frac{-1}{3} \\\frac{-1}{3} & \frac{-1}{3} & \frac{2}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3}\end{bmatrix} \\= \begin{bmatrix} \frac{2 \left(-1\right)^n +2^n}{3} & \frac{-\left(-1\right)^n+2^n}{3} & \frac{-\left(-1\right)^n+2^n}{3} \\\frac{-\left(-1\right)^n+2^n}{3} & \frac{2 \left(-1\right)^n+2^n}{3} & \frac{-\left(-1 \right)^n +2^n}{3} \\\frac{-\left(-1\right)^n+2^n}{3} & \frac{-\left(-1 \right)^n+2^n}{3} & \frac{2 \left(-1\right)^n+2^n}{3} \end{bmatrix} \Rightarrow \bbox[red, 2pt]{a_n={1\over 3}(2^n-(-1)^n)}$$
二、計算證明題(共 15 分)
解題僅供參考，其他教甄試題及詳解
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