臺北市立永春高級中學 115 學年度第一次正式教師甄選
一、填充題（每格 7 分，共 70 分）
解答:$$S= \sum_{n=1}^\infty {2n\over 3^n} ={2\over 3}+{4\over 3^2}+{ 6\over 3^3} +{8\over 3^4} +\cdots \Rightarrow {1\over 3}S={2\over 3^2} +{4\over 3^3}+{6\over 3^4}+{8\over 3^5}+ \cdots \\ \Rightarrow S-{1\over 3}S= {2\over 3}+{2\over 3^2}+ {2\over 3^3}+{ 2\over 3^4}+ \cdots \Rightarrow {2\over 3}S= 2 \left( {1/3 \over 1-1/3} \right) =1 \Rightarrow S= \bbox[red, 2pt]{3\over 2}$$
解答:$$2^x=5^k \Rightarrow x\log_5 2=k \Rightarrow x={k\over \log_5 2} \Rightarrow f(2^x) =6\cdot {k\over \log_5 2} \cdot \log_5 2+3=6k+3\\ \Rightarrow f(5)+f(5^2)+\cdots+f(5^8) = \sum_{k=1}^8\left( 6k+3 \right) =6\cdot {9\cdot 8\over 2}+3\cdot 8= \bbox[red, 2pt]{240}$$
解答:$$\det(A-\lambda I)=(\lambda-1)^3 \Rightarrow 取f(x)=x^{10} =(x-1)^3 p(x)+ax^2+bx+c \\ \Rightarrow f'(x)=10x^9= 3(x-1)^2p(x)+(x-1)^3p'(x)+2ax+b\\ \Rightarrow f''(x)=90x^8=6(x-1)p(x)+ 6(x-1)^2p'(x) + (x-1)^3p''(x)+2a \\ \Rightarrow \cases{f(1)=1=a+b+c\\ f'(1)=10=2a+b\\ f''(1)=90=2a} \Rightarrow \cases{a=45\\ b=-80\\ c=36} \Rightarrow f(x)=x^{10} =(x-1)^3p(x)+45x^2-80x+36 \\ \Rightarrow A^{10}=45A^2-80A+36I= 45 \begin{bmatrix}1 & 2 & 1 \\0 & 1 & 2 \\0 & 0 & 1 \end{bmatrix} -80 \begin{bmatrix}1& 1&0\\0& 1& 1\\0&0& 1 \end{bmatrix}+ 36 \begin{bmatrix}1&0&0\\0& 1&0\\0&0& 1 \end{bmatrix} = \begin{bmatrix}1 & 10 & 45 \\0 & 1 & 10 \\0 & 0 & 1 \end{bmatrix} \\ \Rightarrow 1+10+45+1+10+1= \bbox[red, 2pt]{68}$$
解答:$$假設\cases{P(x,y,2x-y) \\A(0,0,-1) \\B(3,1,5)} \Rightarrow f(x,y) =\overline{PA}+ \overline{PB} \Rightarrow f(x)的最小值= \overline{AB} =\sqrt{9+1+36} =\bbox[red, 2pt]{\sqrt{46}} \\ z=2x-y \Rightarrow 平面E:2x-y-z=0 \Rightarrow P\in E為一動點，固定點A,B剛好在E的兩側，\\ 因此\overline{PA}+ \overline{PB} 的最小值=\overline{AB}$$
解答:$$事件A:紅球比黑球早出現，也就是第一顆紅球在第一顆黑球前面\\ \qquad 袋中有紅球2顆、黑球4顆\Rightarrow P(A)={2\over 2+4}={1\over 3} \\事件B:紅球比白球早出現 \Rightarrow P(B)={2\over 2+3}={2\over 5} \\ 事件A\cap B:紅球比黑球早也比白球早出現\Rightarrow P(A\cap B) ={2\over 2+4+3} ={2\over 9} \\ \Rightarrow P(紅球最後出現)=1-P(A)-P(B)+P(A\cap B)=1-{1\over 3}-{2\over 5}+{2\over 9} = \bbox[red, 2pt]{22\over 45}$$
解答:$$|z_1-(6+8i)| =3代表圓\Gamma_1:\cases{圓心O_1(6,8)\\ 圓半徑r_1= 3} \\ |(1+i)z_2-2i|= \left| (1+i) \left( z_2-{2i\over 1+i} \right)\right| = |1+i|\cdot \left| z_2- {2i\over 1+i} \right| =\sqrt 2 |z_2-(1+i)|=\sqrt 2 \\ \Rightarrow |z_2-(1+i)|=1 代表圓\Gamma_2: \cases{圓心O_2(1,1)\\ 圓半徑r_2=1}\\ \Rightarrow |z_1-z_2|最小值代表\Gamma_1上的點到\Gamma_2上的點最短距離，由於\overline{O_1O_2} =\sqrt{5^2+7^2}=\sqrt{74} \gt r_1+r_2 \\ \Rightarrow 兩圓不相交\Rightarrow |z_1-z_2|最小值=\overline{O_1O_2} -r_1-r_2= \bbox[red, 2pt]{\sqrt{74}-4}$$
解答:
$$\overrightarrow{AC} = \alpha \overrightarrow{AB} +\beta \overrightarrow{AD} \Rightarrow \overrightarrow{AC} \times \overrightarrow{AD} =(\alpha \overrightarrow{AB} +\beta \overrightarrow{AD}) \times \overrightarrow{AD} =\alpha \overrightarrow{AB} \times \overrightarrow{AD}+0 \\ \Rightarrow \alpha= {\overrightarrow{AC} \times \overrightarrow{AD} \over \overrightarrow{AB} \times \overrightarrow{AD}} = {\triangle ACD \over \triangle ABD}, 同理可得 \beta= {\overrightarrow{AC} \times \overrightarrow{AB} \over \overrightarrow{AB} \times \overrightarrow{AD}} = {\triangle ABC \over \triangle ABD} \\\Rightarrow \cases{ \alpha={\triangle ACD \over \triangle ABD} ={2\cdot 6\cdot \sin D \over 3\cdot 6 \cdot \sin A} ={2\over 3}\cdot {\sin D\over \sin A}\\ \beta={\triangle ABC \over \triangle ABD}= {2\cdot 3\cdot \sin B\over 3\cdot 6\cdot \sin A} ={1\over 3}\cdot {\sin D\over \sin A}} \; (\angle B與 \angle D互補 \Rightarrow \sin B=\sin D) \\ 現在只要能算出{\sin D\over \sin A}就能求出\alpha與\beta\\ 正弦定理: \cases{{\overline{AC} \over \sin D } =2R\\ {\overline{BD} \over \sin A}=2R} \Rightarrow {\sin D\over \sin A} ={\overline{AC} \over \overline{AD}} = {\overline{AB} \cdot \overline{AD}+\overline{BC}\cdot \overline{CD} \over \overline{AB}\cdot \overline{BC}+ \overline{AD}\cdot \overline{CD}}(第二托勒密定理) ={11\over 9} \\ \Rightarrow \cases{\alpha=(2/3)\cdot (11/9)= 22/27 \\ \beta= (1/3)\cdot (11/9)=11/27} \Rightarrow (\alpha, \beta) = \bbox[red, 2pt]{\left( {22\over 27} ,{11\over 27}\right)}$$
解答:$$假設四根為\cases{a-3D\\ a-D\\ a+D\\ a+3D},其中d=2D \Rightarrow 四根之和=4a =8 \Rightarrow a=2 \Rightarrow 四根\cases{2-3D\\ 2-D\\ 2+D\\ 2+3D} \\ \Rightarrow x^4-8x^3+14^2+kx +m= (x-(2-3D))(x-(2-D)) (x-(2+D)) (x-(2+3D)) \\=((x-2)^2-9D^2)((x-2)^2-D^2) =(x-2)^4-10D^2(x-2)^2+9D^4 \\=x^4-8x^3+(24-10D^2)x^2+ (-32+40D^2)x +16-40D^2+9D^4 \\ \Rightarrow \cases{24-10D^2=14 \\ k=-32+40D^2\\ m=16-40D^2+9D^4} \Rightarrow D^2=1 \Rightarrow \cases{D=1 \Rightarrow d=2\\k=8\\ m=-15} \Rightarrow (d,k,m)= \bbox[red, 2pt]{(2,8,-15)}$$
解答:$$依題意:甲贏乙且乙贏丙就代表甲贏丙，不可以丙贏甲形成一個迴圈\\ 因此\cases{第1名有9勝\\ 第2名有8勝\\ 第3名有7勝\\ \cdots\\ 第10名有0勝 } \Rightarrow 9^2+8^2+\cdots+1^2+0^2= \bbox[red, 2pt]{285}$$
二、計算證明題（每題 10 分，共 30 分）
解答:$$簡單而言，f(a)=f(b) \Rightarrow f(a^2+1) =f(b^2+1) \Rightarrow a=b \Rightarrow f 為1-1函數\\ 又f是實係數多項式，因此f必為一次式 (均值定理) \Rightarrow f(x)=ax+b \\又f(0)=0 \Rightarrow f(x)=ax \Rightarrow \cases{左式:f(x^2+1)=a(x^2+1)=ax^2+a \\ 右式:(f(x))^2+1=a^2x^2+1} \Rightarrow a=1\\ \Rightarrow \bbox[red, 2pt]{f(x)=x}$$
解答:
$$假設\cases{\overline{BC}=a \\\overline{AC}=b\\ \overline{AB}=c \\D\in \overline{BC} \\E\in \overline{CA} \\ F\in \overline{AB} \\ 內切圓圓心I} \Rightarrow \cases{s=(a+b+c)/2 \\ 內切圓半徑=\overline{ID} =\overline{IE} =\overline{IF}=r } \\ \cases{\angle FIE=180^\circ-\angle A \\ \angle DIF=180^\circ-\angle B\\ \angle DIE=180^\circ-\angle C} \Rightarrow \cases{\triangle FIE= {1\over 2} \overline{IE}\cdot \overline{IF} \sin \angle FIE = {1\over 2}r^2 \sin(180^\circ-\angle A) ={1\over 2}r^2 \sin \angle A\\ \triangle DIF ={1\over 2}r^2 \sin \angle B\\ \triangle DIE ={1\over 2}r^2 \sin \angle C} \\ \Rightarrow T={1\over 2}r^2 (\sin \angle A+ \sin \angle B+ \sin \angle C)= {1\over 2}r^2 \left( {a\over 2R} +{b\over 2R} +{c\over 2R}\right) ={r^2\over 4R} (2s)={r^2s\over 2R} \\={r(rs) \over 2R} ={rS\over 2R} \Rightarrow {T\over S} ={r\over 2R}\; \bbox[red, 2pt]{故得證}$$
解答:$$假設 \cases{A= L\cap \Gamma_1=(a,e^{-a})\\ B=L\cap \Gamma_2 =(b,-e^{b+1})} \Rightarrow \overline{AB}斜率= {-e^{b+1}-e^{-a} \over b-a}; 又 \cases{\Gamma_1:y=e^{-x} \Rightarrow y'=-e^{-x} \\ \Gamma_2:y=-e^{x+1} \Rightarrow y'=-e^{x+1}} \\ \Rightarrow \cases{A點切線斜率: -e^{-a} \\ B點切線斜率:-e^{b+1}} \Rightarrow -e^{-a}=-e^{b+1} \Rightarrow -a=b+1 \Rightarrow b=-a-1 \cdots(1) \\ 將b+1=-a 代入 \overline{AB}斜率 ={-e^{-a}-e^{-a} \over b-a} =-e^{-a} \Rightarrow {2\over b-a}=1 \Rightarrow b=a+2\cdots(2) \\ 由(1)及(2)可得-a-1=a+2 \Rightarrow a=-{3\over 2} \Rightarrow b={1\over 2} \Rightarrow A=(-{3\over 2},e^{3/2}) \Rightarrow 斜率=-e^{3/2} \\ \textbf{(a) }L: \bbox[red, 2pt]{y-e^{3/2}=-{e^{3/2}(x+{3\over 2})} } \\ \textbf{(b) }A= \bbox[red, 2pt]{\Gamma_1切點=(-{3\over 2},e^{3/2})} , B= \bbox[red, 2pt]{\Gamma_2 切點= ({1\over 2},-e^{3/2})}$$
解題僅供參考，其他教甄試題及詳解
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