臺北市立建國高級中學 114 學年度第 1 次正式教師甄選
一、填充題（每題 6 分，共計 72 分）
解答:
$$假設\cases{\overline{AB}=c\\ \overline{AC} =b}, \; \tan A={4\over 3} \Rightarrow \cases{\sin A=4/5\\ \cos A=3/5} \Rightarrow \cases{2\triangle ABD=3\sqrt 2b\sin 45^\circ =3b\\ 2\triangle ACD=3\sqrt 2c \sin (A-45^\circ) ={3\over 5}c \\ 2\triangle ABC=bc \sin A={4\over 5}bc} \\ \Rightarrow {4\over 5}bc= 3b+{3\over 5}c \ge 2\sqrt{3b\cdot {3c\over 5}} ={6\over \sqrt 5} \sqrt{bc} \Rightarrow {16\over 25}(bc)^2\ge {36\over 5}bc \Rightarrow bc \ge {45\over 4} \\ \Rightarrow \triangle ABC={2\over 5}bc \ge \bbox[red, 2pt]{9\over 2}$$
解答:
$$\cos \angle AOB ={13+7-10\over 2\sqrt{91}} ={5\over \sqrt{91}}\\ C=\overline{AB}中點\Rightarrow \overrightarrow{OC} ={1\over 2} (\overrightarrow{OA}+ \overrightarrow{OB}), 又D= \overline{OC}中點\Rightarrow \overrightarrow{OD} ={1\over 2} \overrightarrow{OC} ={1\over 4} (\overrightarrow{OA}+\overrightarrow{OB}) \\={1\over 4} \left( {1\over m} \overrightarrow{OP} +{1\over n} \overrightarrow{OQ}\right) \Rightarrow {1\over m}+{1\over n}=4 \cdots(1)\\ \overrightarrow{OP} \bot \overrightarrow{PQ} \Rightarrow \overrightarrow{OP} \cdot ( \overrightarrow{PO}+ \overrightarrow{OQ}) =0 \Rightarrow m \overrightarrow{OA} \cdot ( -m\overrightarrow{OA}+ n\overrightarrow{OB}) =0 \Rightarrow 7m^2=mn \overrightarrow{OA}\cdot \overrightarrow{OB} \\ \Rightarrow 7m^2 =mn \overline{OB}\cdot \overline{OA} \cos \angle AOB=5mn \Rightarrow m={5\over 7}n \cdots(2) \\ 由(1)及(2) 可得(m,n)= \bbox[red, 2pt]{\left( {3\over 7}, {3\over 5} \right)}$$
解答:$$\cases{A(0,0,0) \\B({\sqrt 2\over 2},0, {\sqrt 2\over 2}) \\C(0, {\sqrt 2\over 2},{\sqrt 2\over 2}) \\D({\sqrt 2\over 2},{\sqrt 2\over 2},0)} \Rightarrow E={2B+C\over 3} = \left( {\sqrt 2\over 3},{\sqrt 2\over 6},{\sqrt 2\over 2} \right) \Rightarrow \cases{L_1= \overleftrightarrow{AE}: {x\over 2}= {y\over 1}={z\over 3} \\L_2= \overleftrightarrow{CD}: \cases{x+z =\sqrt 2/2\\y=\sqrt 2/2}} \\ \Rightarrow L_1與L_2歪斜 \Rightarrow 平面E:x-5y +z=0包含L_1且與L_2平行\Rightarrow d(E,L_2)= d(L_1, L_2)= \bbox[red, 2pt]{2\sqrt 6\over 9}$$
解答:$$y={1\over x}代入橢圓 \Rightarrow {x^2\over a^2} +{1\over b^2x^2}=1 \Rightarrow b^2x^4-a^2b^2x^2+a^2 =0 \\ \Rightarrow 判別式: a^4b^4-4a^2b^2=0 \Rightarrow ab=2 \Rightarrow b={2\over a} \Rightarrow x^2={a^2b^2 \over 2b^2} ={a^2\over 2} \Rightarrow P({a\over \sqrt 2}, {\sqrt 2\over a}) \\ \Rightarrow S=\triangle PF_1F_2 ={1\over 2} \overline{F_1F_2}\cdot {\sqrt 2\over a} ={c\over a}\sqrt 2={\sqrt{a^2-b^2}\over a} \sqrt 2 =\sqrt{1-(b/a)^2} \cdot \sqrt 2 \\ \Rightarrow \lim_{a\to \infty}S= \bbox[red, 2pt]{\sqrt 2}$$
解答:$$考慮點數和不為10的情形:\\\textbf{Case I }完全沒有5 \Rightarrow 剩下\{1,2,3,4,6\}\Rightarrow \cases{只有\{1,2,3,6\}(沒有4) \Rightarrow 排列數4^5 \\只有\{1,2,3,4\}(沒有6) \Rightarrow 排列數4^5 \\只有\{1,2,3\}(沒有4,也沒有6) \Rightarrow 排列數3^5} \\ \qquad \Rightarrow 小計2\times 4^5-3^5=1805 \\\textbf{Case II }只有一個5\Rightarrow 剩下\{1,2,3,4,6\}選4個\Rightarrow \cases{只有\{1,2,3,4\} \Rightarrow 排列數4^4 \\只有\{1,2,3,6\} \Rightarrow 排列數4^4 \\只有\{1,2,3\} \Rightarrow 排列數3^4 } \\ \qquad \Rightarrow 小計C^5_1 (2\times4^4-3^4) =2155\\ 因此和不為10的情形共有1805+2155=3960 \Rightarrow 至少有一對和為10的情形: 6^5-3960=3816 \\ \Rightarrow 機率為{3816\over 6^5=7776}= \bbox[red, 2pt]{53\over 108}$$
解答:$$(A+I) \begin{bmatrix}1\\1 \end{bmatrix} =A \begin{bmatrix}1\\ 1 \end{bmatrix} + \begin{bmatrix}1\\1 \end{bmatrix} = \begin{bmatrix}0\\ -2 \end{bmatrix}+ \begin{bmatrix} 1\\ 1 \end{bmatrix}= \begin{bmatrix}1\\ -1 \end{bmatrix} \\ \Rightarrow (A+I) \begin{bmatrix}1\\1 \end{bmatrix} =\begin{bmatrix}1\\ -1 \end{bmatrix} \Rightarrow (A+I)^2 \begin{bmatrix}1\\ 1 \end{bmatrix}=(A+I) \begin{bmatrix}1\\ -1 \end{bmatrix} =0 = \begin{bmatrix}0\\ 0 \end{bmatrix} \\ \Rightarrow \cases{(A+I) \begin{bmatrix}1\\ 1 \end{bmatrix} = \begin{bmatrix}0\\-2 \end{bmatrix} \\[1ex](A+I) \begin{bmatrix}1\\ -1 \end{bmatrix} = \begin{bmatrix}0\\ 0 \end{bmatrix}} \Rightarrow (A +I)\begin{bmatrix}1& 1\\ 1& -1 \end{bmatrix} = \begin{bmatrix}1&0\\ -1 &0\end{bmatrix} \Rightarrow A+I= \begin{bmatrix}1&0\\ -1 &0\end{bmatrix} \begin{bmatrix}1& 1\\ 1& -1 \end{bmatrix}^{-1} \\ \Rightarrow A+I= \begin{bmatrix}1&0\\ -1 &0\end{bmatrix} \begin{bmatrix}1/2& 1/2\\ 1/2& -1/2 \end{bmatrix} = \begin{bmatrix}1/2& 1/2\\ -1/2& -1/2 \end{bmatrix} \\ A^{100} =((A+I)-I)^{100} = \sum_{n=0}^{100} {100\choose n}(A+I)^n(-I)^{100-n} =I+100(A+I)(-I) \\= \begin{bmatrix}1& 0\\0& 1 \end{bmatrix} + \begin{bmatrix}50& 50\\ -50&-50 \end{bmatrix} \begin{bmatrix}-1&0\\ 0&-1 \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix}-49& -50\\50& 51 \end{bmatrix}}$$
解答:$$f(\theta) ={7\cos 2\theta\over 3+\sin \theta} = {7(1-2\sin^2\theta)\over 3+\sin \theta} ={8-2\sin^2\theta\over 3+\sin \theta} \\ t=\sin \theta \Rightarrow g(t)={8-2t^2\over 3+t}, t\in [-1,1] \Rightarrow g'(t)={-2t^2-12t-8\over (3+t)^2} =0 \Rightarrow t^2+6t+4=0 \\ \Rightarrow t=-3+ \sqrt 5 \; (-3-\sqrt 5\lt -1) \Rightarrow \cases{g(-3+\sqrt 5)=12-4\sqrt 5 \\g(-1) =3\\g(1) =3/2} \Rightarrow (M,m) = \bbox[red, 2pt]{(12-4\sqrt 5, {3\over 2})}$$
解答:$$令t=2^x \Rightarrow f(t)=t^2-bt+c=0, t\gt 0; 由於 \cases{x_1\gt 0 \Rightarrow t\gt 1\\ -1\lt x_2\lt 0 \Rightarrow 1/2\lt t\lt 1} \Rightarrow {1\over 2} \lt t_1\lt 1\lt t_2 \\ y=f(t)圖形為凹向上 \Rightarrow \cases{f(1)\lt 0 \Rightarrow b\gt c+1\\ f(1/2)\gt 0 \Rightarrow 2b\lt 4c+1} \Rightarrow c+1 \lt b\lt 2c+{1\over 2}\\ c\lt 10 \Rightarrow \cases{c=9 \Rightarrow 10\lt b\lt 18.5 \Rightarrow b=11,12,\dots,18 \Rightarrow 共8組\\ c=8 \Rightarrow 9\lt b\lt 16.5 \Rightarrow b=10,11,\dots,16 \Rightarrow 共7組 \\ \cdots\\ c=1 \Rightarrow 2\lt b\lt 2.5 \Rightarrow 共0組} \Rightarrow 合計:1+2+\cdots+8= \bbox[red, 2pt]{36} $$
解答:$$假設|x|=|y|=|z|=k \in \mathbb R \Rightarrow xyz= \sqrt 3+\sqrt 5i \Rightarrow |xyz|= \sqrt 8=|x||y||z| =k^3 \Rightarrow k=\sqrt 2 \\ \Rightarrow \cases{x \bar x =k^2=2 \\ y\bar y=k^2 =2\\ z\bar z=k^2 =2 } \Rightarrow \cases{\bar x=2/x\\ \bar y =2/y\\ \bar z=2/z} \Rightarrow \bar x+\bar y+\bar z= 2 \left( {1\over x} +{1\over y} +{1\over z} \right) =2\cdot {xy+ yz+zx \over xyz} \\ \Rightarrow xy+ yz+zx={1\over 2} (\overline{x+y+z})xyz = {1\over 2}\cdot \left( -{\sqrt 3\over 2}+\sqrt 5i \right) (\sqrt 3+\sqrt 5i) = {1\over 4}(-13+\sqrt{15}i) \\ \Rightarrow (x+y+z)^2 = \left( -{\sqrt 3\over 2}-\sqrt 5i \right)^2 = -{17\over 4}+\sqrt{15}i \Rightarrow x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx) \\=-{17\over 4}+\sqrt{15}i-{1\over 2}(-13+\sqrt{15}i) ={9\over 4}+{\sqrt{15}\over 2}i \\ \Rightarrow x_1x_2+ y_1y_2+ z_1z_2= {1\over 2}Im(x^2+y^2+z^2) ={1\over 2} \cdot {\sqrt{15}\over 2} = \bbox[red, 2pt]{\sqrt{15}\over 4}$$
解答:$$K_n=|1-a_1|+ |a_1-a_2|+ \cdots+ |a_{n-1}-a_n|+ |a_n-6| \ge |1-6|=5 \\ \Rightarrow K_n=5(最小值)的條件是1\le a_1\le a_2\le \cdots\le a_n \le 6 \\ \Rightarrow L_n= K_n+|a_4-4|\ge 5+0=5(最小值), 此時1\le a_1\le a_2\le \cdots\le a_n \le 6且a_4=4 \\ \Rightarrow 1\le a_1\le a_2\le a_3\le 4\le a_5\le\cdots\le a_n\le 6\\ \Rightarrow \cases{1\le a_1\le a_2\le a_3\le 4\Rightarrow 方法數=H^4_3=20 \\4\le a_5\le \cdots\le a_n\le 6 \Rightarrow 方法數=H^3_{n-4} =C^{n-2}_2 =(n-2)(n-3)/2} \\ \Rightarrow 滿足兩條件的方法數:20\times {(n-2)(n-3) \over 2} =10(n-2)(n-3) \Rightarrow 機率= \bbox[red, 2pt]{10(n-2)(n-3) \over 6^n} $$
解答:$$假設25筆資料中有m筆資料,y_1, y_2, \dots, y_m,為非負數,另外n筆資料, z_1,z_2, \dots, z_n,為皆為負數25 \\ 假設\sum_{k=1}^{m}y_k= A \Rightarrow \sum_{k=1}^{25}x_k=A+ \sum_{k=1}^{n} z_k=0 \Rightarrow \sum_{k=1}^{n}z_k=-A \Rightarrow \sum_{k=1}^{25}|x_k|=A+|-A|=2A\\ 柯西不等式: (y_1^2+y_2^2+ \cdots+y_m^2) (1^2+1^2 +\cdots +1^2) \ge (y_1+y_2+\cdots+ y_m)^2 \\ \Rightarrow A^2= \left( \sum_{k=1}^{m} y_k\right)^2 \le m \sum_{k=1}^{m}y_k^2\Rightarrow \sum_{k=1}^{m}y_k^2\ge {A^2\over m} \\ 同理, \left( \sum_{k=1}^{n}z_k \right)^2 \le n \sum_{k=1}^n z_k^2 \Rightarrow (-A)^2\le n \sum_{k=1}^n z_k^2 \Rightarrow \sum_{k=1}^n z_k^2 \ge {A^2\over n} \Rightarrow \sum_{k=1}^{25} x_k^2 =1 \ge {A^2\over m}+{A^2\over n} \\ \Rightarrow 1\ge A^2\cdot {m+n\over mn} =A^2\cdot {25\over mn} \Rightarrow A^2 \le {mn\over 25} \Rightarrow A\le {\sqrt{mn} \over 5} \le {\sqrt{13\cdot 12} \over 5} ={2\sqrt{39} \over 5} \\ \Rightarrow \sum_{k=1}^{25}|x_k|=2A\le \bbox[red, 2pt]{4\sqrt{39} \over 5}$$
二、計算題（各題配分列於題後，全部 2 大題，共計 28 分）
解答:$$\textbf{(1) }左式:a_{n+2}-a_{n+1} ={5\over 8}-{1\over 2}(a_{n+1})^2 - \left( {5\over 8}-{1\over 2}(a_{n})^2 \right) = {1\over 2}(a_{n})^2-{1\over 2}(a_{n+1})^2 \\ \qquad 右式:-b_n(a_{n+1}-a_n)=-{1\over 2}(a_{n+1}+a_n)(a_{n+1}-a_n) =-{1\over 2} \left( (a_{n+1})^2-(a_n)^2 \right) \\ \qquad = {1\over 2}(a_{n})^2-{1\over 2}(a_{n+1})^2 \\ \qquad \Rightarrow 左式=右式, \bbox[red, 2pt]{故得證} \\\textbf{(2) }用歸納法證明:\\ \qquad n=1 \Rightarrow a_1={1\over 4} \lt {5\over 8} \Rightarrow 原式成立\\ \qquad 假設n=k時,原式也成立, 即0\lt a_k \lt {5\over 8} \\ \qquad n=k+1 \Rightarrow a_{k+1}={5\over 8}-{1\over 2}a_k^2\\ \qquad \cases{a_k^2 \gt 0 \Rightarrow -{1\over 2}a_k^2 \lt 0 \Rightarrow a_{k+1}={5\over 8}-{1\over 2}a_k^2\lt {5\over 8} \\ a_k\lt {5\over 8} \Rightarrow a_k^2 \lt {25\over 64 } \Rightarrow -{1\over 2}a_k^2 \gt -{25\over 128} \Rightarrow {5\over 8}-{1\over 2}a_k^2 \gt {5\over 8}-{25\over 128} ={55\over 128}\gt 0} \\ \qquad \Rightarrow 0\lt a_{k+1}\lt {5\over 8} \Rightarrow 依數學歸納法可得0\lt a_n\lt {5\over 8},\bbox[red, 2pt]{故得證} \\\textbf{(3) }|a_{n+1}-a_n|= \left| \left( {5\over 8}-{1\over 2}a_n^2 \right)-a_n\right| = \left| -{1\over 2}\left( a_n^2+2a_n-{5\over 4} \right)\right| ={1\over 2} \left| a_n^2+2a_n-{5\over 4}\right| \\ \qquad ={1\over 2} \left| \left( a_n-{1\over 2} \right) \left( a_n+{5\over 2} \right) \right| ={1\over 2} \left| a_n+{5\over 2}\right| \cdot \left| a_n-{1\over 2}\right| \gt {1\over 2}\cdot {5\over 2} \cdot \left| a_n-{1\over 2}\right|\ge \left| a_n-{1\over 2}\right| \\\qquad \Rightarrow |a_{n+1}-a_n| \ge \left| a_n-{1\over 2}\right|, \bbox[red, 2pt]{故得證} \\\textbf{(4) } \left|a_{n+1}-{1\over 2} \right| =\left| \left( {5\over 8}-{1\over a}a_n^2 \right)-{1\over 2} \right| = \left|{1\over 2} \left( {1\over 4}-a_n^2 \right) \right| ={1\over 2} \left| a_n+{1\over 2}\right| \left| a_n-{1\over 2}\right| \\\qquad \lt {1\over 2}\cdot {9\over 8}\left| a_n-{1\over 2}\right| ={9\over 16}\left| a_n-{1\over 2}\right| \; (0\lt a_n\lt {5\over 8} \Rightarrow {1\over 2}\lt a_n+{1\over 2}\lt {9\over 8}) \\ \qquad \Rightarrow \left|a_{n+1}-{1\over 2} \right|\lt {9\over 16} \left|a_{n}-{1\over 2} \right| \lt ({9\over 16})^2 \left|a_{n-1}-{1\over 2} \right| \lt \cdots \lt ({9\over 16})^n \left|a_{1 }-{1\over 2} \right| \\ \qquad \Rightarrow \lim_{n\to \infty} \left| a_n-{1\over 2}\right| =0 \Rightarrow \bbox[red, 2pt]{\lim_{n\to \infty} a_n={1\over 2} 且 \langle a_n \rangle 收斂}$$
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解題僅供參考，其他教甄試題及詳解
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