國立蘭陽女中 114 學年度第二次教師甄試
一、填充題：(每格 5 分，共 70 分)
解答:$$\cases{出現奇數的機率:4/6=2/3\\ 出現偶數的機率=1/3} \Rightarrow P_1={1\over 3} \Rightarrow P_n={1\over 3}P_{n-1}+{2\over 3}(1-P_{n-1}) ={2\over 3}-{1\over 3}P_{n-1} \\ \Rightarrow \bbox[red, 2pt]{\cases{P_1= {1\over 3}\\ P_n={2\over 3}-{1\over 3}P_{n-1},n\ge 2}}$$
解答:$$8個相同的球排成三列,且每列至少一球,有H^3_{8-3} =C^7_5=21排法\\ 四種色球有{8!\over 2!2!3!} =1680種排法，因此共有1680\times 21= \bbox[red, 2pt]{35280}排法$$
解答:
$$直線: y={1\over \sqrt 3}x \Rightarrow 直線與x軸夾角30^\circ \Rightarrow 假設\cases{\overline{OA}=a\\ \overline{OB}=b} \Rightarrow \cases{A(\sqrt 3a/2, a/2) \\B(b,0) \\ \triangle OAB面積=ab/4} \\ \Rightarrow \overline{AB}=4 \Rightarrow ({\sqrt 3a\over 2}-b)^2+{a^2\over 4} =16 \\利用\text{ Lagrange's 算子求極值:} \cases{f(a,b)=ab/4\\ g(a,b)= ({\sqrt 3a\over 2}-b)^2+{a^2\over 4} -16 } \Rightarrow \cases{b/4= \lambda(2a-\sqrt 3b) \\ a/4=\lambda(2b-\sqrt 3a)} \\ \Rightarrow a=b 代入g(a,a)=0 \Rightarrow a^2={16\over 2-\sqrt 3} \Rightarrow {1\over 4}ab={1\over 4}a^2={4\over 2-\sqrt 3} = \bbox[red, 2pt]{8+4\sqrt 3}$$
解答:$$3^5= 243 \equiv 43 \pmod {100} \Rightarrow 3^{10} \equiv 43^2 =1849 \equiv 49 \pmod{100} \\ \Rightarrow 3^{20} \equiv 49^2=2401 \equiv 01\pmod{100} \Rightarrow 3^{200} \equiv 01 \pmod{100} \\ \Rightarrow 3^{200}末兩位數是01 \Rightarrow 3^{20}-2025 \equiv 1-25=-24 \pmod{100} \\\Rightarrow 3^{20}-2025的末兩位數=100-24=76 \Rightarrow 十位數字為\bbox[red, 2pt]7$$
解答:$$A= \begin{bmatrix}1& -\sqrt 3\\ \sqrt 3&1 \end{bmatrix} =2 \begin{bmatrix}\cos \pi/3 & -\sin \pi/3\\ \sin\pi/3& \cos \pi/3 \end{bmatrix} \Rightarrow A^3=8 \begin{bmatrix} -1& 0\\ 0& -1\end{bmatrix} =-8I \\ \Rightarrow A+A^4+A^7+\cdots +A^{3n+1} =A(I+A^3+A^6+ \cdots+A^{3n}) =A(I-8I+64+\cdots +(-8)^nI) \\=A(1-8+8^2+\cdots +(-8)^n)I = A\cdot {1-(-8)^{n+1}\over 9}\cdot I= {1-(-8)^{n+1}\over 9}A \Rightarrow c=\bbox[red, 2pt]{ {1-(-8)^{n+1}\over 9}}$$
解答:$$假設\cases{\overline{TD}=h\\ \overline{CD}=a} \Rightarrow \cases{\tan \theta=h/(210+a) \\\tan 2\theta=h/(60+a)\\ \tan 3\theta =h/a} \Rightarrow \cases{\tan 2\theta= 2\tan \theta/(1-\tan^2\theta) \\ \tan 3\theta=(\tan\theta+\tan 2\theta)/(1-\tan \theta\tan 2\theta)} \\ \Rightarrow \cases{h^2= (a+210)^2-2(a+60)(a+210) \\h^2= (a+60)(a+210)-a(a+60)-a(a+210)} \Rightarrow a={105\over 2} \Rightarrow h = \bbox[red, 2pt]{75\sqrt 7\over 2}$$
解答:$$\vec n=(1,2,3) \times(2,1,-1) =(-5,7,-3) \Rightarrow \vec n\bot (1,-1,a) \Rightarrow \vec n\cdot (1,-1,a)=0 \\ \Rightarrow -5-7-3a=0 \Rightarrow a= \bbox[red, 2pt]{-4}$$
解答:$$$$
解答:$${x^2\over 25}+{y^2\over 18}=1 \Rightarrow \cases{a=5\\ b=3\sqrt 2} \Rightarrow c=\sqrt 7 \Rightarrow \overline{F_1F_1} =2c=2\sqrt 7\\ 假設\cases{\overline{AF_1}=x\\ \overline{AF_2}=y} \Rightarrow x+y=2a=10 \Rightarrow \cos \angle F_1AF_2={1\over 2}={x^2+y^2-(2\sqrt 7)^2\over 2xy} \\={(x+y)^2-2xy-28 \over 2xy}= {100-2xy-28 \over 2xy} \Rightarrow xy=24 \Rightarrow S_{\triangle F_1AF_2} ={1\over 2}xy \sin 60^\circ = \bbox[red, 2pt]{6\sqrt 3}$$
解答:$$z^7-1=0 的七個根為\omega_0, \omega_1,\dots, \omega_6 \Rightarrow (z^7-1) =(z-\omega_0)(z-\omega_1) \cdots (z-\omega_6) \\ \Rightarrow z^7-1=(z-1)(1+z+z^2+\cdots z^6) =(z-1)(z-\omega_1) \cdots (z-\omega_6) \\ \Rightarrow f(z)=1+z+z^2+ \cdots+z^6 =(z-\omega_1)(z-\omega_2) \cdots (z-\omega_6) \\ \Rightarrow \overline{A_0A_1} \cdot \overline{A_0A_2}\cdots \overline{A_0A_6} =|1-\omega_1||1-\omega_2| \cdots |1-\omega_6| =|(1-\omega_1)(1-\omega_2) \cdots (1-\omega_6)| \\=|f(1)|= \bbox[red, 2pt]7$$
解答:$${x-y+3 \over x+y+2} =k \Rightarrow 直線L:(k-1)x+(k+1)y+2k-3=0 \\圓\Gamma:(x+1)^2+(y-1)^2=1 \Rightarrow \Gamma \cap L\ne \varnothing \Rightarrow d(圓心(-1,1),L)\le 1 \Rightarrow {|2k-1|\over \sqrt{(k-1)^2 +(k+1)^2}} \le 1 \\ \Rightarrow (2k-1)^2 \le (k-1)^2+(k+1)^2 \Rightarrow 2k^2-4k-1\le 0 \Rightarrow {2-\sqrt 6\over 2}\le k\le {2+\sqrt 6\over 2} \\ \Rightarrow (M,m)= \bbox[red, 2pt]{\left({2+\sqrt 6\over 2},{2- \sqrt 6\over 2} \right)}$$
解答:$$\begin{array}{}X& Y& XY &X^2\\\hline 4&6&24& 16\\ 6&6& 36& 36\\ 8&8& 64&64\\ 10&10& 100& 100\\ 12& 10& 120& 144\end{array} \Rightarrow \cases{\sum X=40\\ \sum Y=40\\ \sum(XY)=344\\ \sum X^2=360} \Rightarrow \cases{\bar X=40/5=8\\ \bar Y=40/5=8} \\ 斜率m={n\sum XY-\sum X\sum Y\over n\sum X^2-(\sum X)^2} ={5\cdot 344-40\cdot 40\over 5\cdot 360-40^2} ={3\over 5} \Rightarrow y={3\over 5}x+b通過(\bar X,\bar Y) \\ \Rightarrow 8={3\over 5}\cdot 8+b \Rightarrow b= {16\over 5} \Rightarrow 迴歸直線: \bbox[red, 2pt]{y={3\over 5}x+{16\over 5}}$$
解答:$$x^2+y^2+z^2+2x-2y+4z-19=0 \Rightarrow (x+1)^2+(y-1)^2+ (z+2)^2=5^2 \Rightarrow \cases{球心O(-1,1,-2)\\ 球半徑R=5} \\ \Rightarrow d(O,E)= {|-12|\over \sqrt{1+4+4}} =4 \Rightarrow 圓C半徑r 滿足R^2=4^2+r^2 \Rightarrow r=3 \Rightarrow 圓C面積=9\pi\\ \cases{平面E法向量\vec u=(1,-2,2) \\ xy平面法向量\vec v=(0,0,1)} \Rightarrow 兩平面夾角\theta滿足 \cos \theta={\vec u\cdot \vec v\over |\vec u||\vec v|} ={2\over 3} \\ \Rightarrow 投影面積=9\pi \cdot {2\over 3}= \bbox[red, 2pt]{6\pi}$$
解答:$$\cases{A(1,3)\in L\\ B(3,7) \in L} \Rightarrow \cases{A(1,3) \xrightarrow{平移(0,1) )}(1,2) \xrightarrow{旋轉45度} (-\sqrt 2/2,3\sqrt 2/2) \xrightarrow{平移(0,-1)}A'(-\sqrt 2/2,(2+ 3\sqrt 2)/2)\\B(3,7) \xrightarrow{平移(0,1) )}(3,6) \xrightarrow{旋轉45度} (-3\sqrt 2/2, 9\sqrt 2/2) \xrightarrow{平移(0,-1)} B'(-3\sqrt 2/2,(2+9\sqrt 2)/2)} \\ \Rightarrow \overleftrightarrow{A'B'}: y=-3(x+{\sqrt 2\over 2})+{2+3\sqrt 2\over 2} \Rightarrow \bbox[red, 2pt]{3x+y-1=0}$$
解答:$$\left({\log x\over -\log 3}-1 \right) \left( {\log x\over -\log 4} +2\right) \left( {\log x\over -\log 5}-3 \right) \gt 0 \Rightarrow \left(\log x+\log 3 \right)\left( \log x-2\log 4\right) \left(\log x+3\log 5 \right)\lt 0 \\ \Rightarrow \cases{-\log 3\lt \log x\lt 2\log 4\\ \log x\lt -3\log 5} \Rightarrow \cases{{1\over 3}\lt x\lt 16\\ x\lt{1\over 5^3} \\ x\gt 0} \Rightarrow \bbox[red, 2pt]{{1\over 3}\lt x\lt 16 或0\lt x\lt {1\over 125}}$$
二、計算題：共 45 分，請在答案卷第二頁開始作答，並標明題號。
解答:$$\textbf{(1) }\cases{\triangle ABC面積={1\over 2}\overline{BC}\cdot \overline{AH} =\displaystyle {1\over 2}\cdot \sqrt 3\cdot {1\over 2} ={\sqrt 3\over 4}\\ \triangle ABC面積={1\over 2}\overline{AB}\times \overline{AC} \sin\angle BAC =\displaystyle {\sqrt 3\over 4} \overline{AB}\times \overline{AC}} \\\quad \Rightarrow {\sqrt 3\over 4}={\sqrt 3\over 4}\overline{AB}\times \overline{AC} \Rightarrow \overline{AB}\times \overline{AC}=\bbox[red, 2pt]1 \\\textbf{(2)} \cos 60^\circ={\overline{AB}^2+ \overline{AC}^2-3\over 2\overline{AB} \cdot \overline{AC}} \Rightarrow {1\over 2}={(\overline{AB}+ \overline{AC})^2-2\overline{AB} \cdot \overline{AC}-3\over 2\overline{AB} \cdot \overline{AC}} ={(\overline{AB}+ \overline{AC})^2-5 \over 2} \\\quad \Rightarrow (\overline{AB}+ \overline{AC})^2=6 \Rightarrow \overline{AB}+ \overline{AC} =\bbox[red, 2pt]{\sqrt 6} \\\textbf{(3)} \cases{\overline{AB} \times \overline{AC} =1\\ \overline{AB} + \overline{AC} = \sqrt 6} \Rightarrow \cases{\overline{AB} =\displaystyle {\sqrt 6-\sqrt 2\over 2} \\ \overline{AC} =\displaystyle {\sqrt 6+\sqrt 2\over 2}} \\ \overline{AD}為角平分線 \Rightarrow {\overline{BD} \over \overline{DC}} ={\overline{AB} \over \overline{AC}} ={\sqrt 6-\sqrt 2 \over \sqrt 6+\sqrt 2} =2-\sqrt 3 \Rightarrow \cases{\overline{BD}={2-\sqrt 3\over 3-\sqrt 3}\cdot \sqrt 3\\ \overline{CD} ={\sqrt 3\over 3-\sqrt 3}} \\ 直角\triangle AHC: \overline{AC}^2=\overline{AH}^2+\overline{CH}^2 \Rightarrow \overline{CH}={2+\sqrt 3\over 2} \Rightarrow \overline{DH} =\overline{CH}-\overline{CD}={1\over 2} \\ \Rightarrow \angle ADH為等腰直角三角形\Rightarrow \overline{AD}= \bbox[red, 2pt]{\sqrt 2\over 2}$$
解答:$$本題\bbox[cyan,2pt]{送分}$$
解答:$$取f(t)= \sum_{k=1}^n (a_kt+b_k )^2 = \sum_{k=1}^n (a_kt+b_k )^2 = \left(\sum_{k=1}^na_k^2 \right) t^2+ \left(\sum_{k=1}^n 2a_kb_k \right) t+ \sum_{k= 1}^nb_k^2 \ge 0\\ \Rightarrow 判別式:\left(\sum_{k=1}^n 2a_kb_k \right)^2-4\left(\sum_{k=1}^na_k^2 \right)\left(\sum_{k=1}^n b_k^2 \right) \le 0 \Rightarrow \left(\sum_{k=1}^n a_kb_k \right)^2\le \left(\sum_{k=1}^na_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)\\ 等號成立時 \Rightarrow f(t)= \sum_{k=1}^n (a_kt+b_k )^2 =0 \Rightarrow a_kt+b_k=0 \Rightarrow {b_k\over a_k}=-t, k=1,2,\dots,n \\ \Rightarrow \exists \lambda \in \mathbb R, \ni b_k=\lambda a_k, k=1,2,\dots,n \;\bbox[red, 2pt]{故得證}$$
解題僅供參考，其他教甄試題及詳解
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