國立羅東高中 115 學年度第一次教師甄選
一、填充題 (共 16 題，每題 5 分，合計 80 分)
解答:$$\cases{abc=1\\ ab+bc+ca-3abc=0} \Rightarrow ab+bc+ca-3=0 \Rightarrow ab+bc+ca=3 \\ \Rightarrow {1\over 1-bc}+ {1\over 1-ca}+{1\over 1-ab} = {1\over 1-1/a}+ {1\over 1-1/b}+{1\over 1-1/c} ={a\over a-1} +{b\over b-1}+{c\over c-1} \\=1+{1\over a-1} +1+{1\over b-1} +1+{1\over c-1} =3+ \left( {1\over a-1}+{1\over b-1}+{1\over c-1} \right) \\=3+ {(b-1)(c-1)+ (a-1)(c-1)+ (a-1)(b-1)\over (a-1)(b-1)(c-1)} =3+(-2)= \bbox[red, 2pt]1$$
解答:$$101^{15} =(10^2+1)^{15} =(10^2)^0+ C^{15}_1(10^2) +C^{15}_2 (10^2)^2 +C^{15}_3 (10^2)^3 +C^{15}_4 (10^2)^4 + \cdots \\\qquad =1+ C^{15}_1 10^2 +C^{15}_2 10^4 +C^{15}_3 10^6 +C^{15}_4 10^8 + \cdots \\ 一百萬=10^6 \Rightarrow 只需計算前四項，10^8之後皆超過百萬 \\ \Rightarrow 1+1500+ 1,050,000+455,000,000 =456,051,501 \Rightarrow 百萬位數字是\bbox[red, 2pt]6$$
解答:
$$y=x^3-4x \Rightarrow y'=3x^2-4 \Rightarrow y'(1)=-1 \Rightarrow 切線方程式: y=-(x-1)-3 \Rightarrow x+y+2=0 \\ \Rightarrow x^3-4x=-x-2 \Rightarrow x^3-3x+2=0 \Rightarrow (x-1)^2(x+2)=0 \Rightarrow 交點\cases{P(1,-3) \\Q(-2,0)} \\ \Rightarrow 所圍面積=\int_{-2}^1 (x^3-3x+2)\,dx = \left. \left[ {1\over 4}x^4-{3\over 2}x^2+2x \right] \right|_{-2}^1 ={3\over 4}+6= \bbox[red, 2pt]{27\over 4}$$
解答:$$\cases{2c=\overline{F_1 F_2} =\sqrt{2^2+2^2}=2\sqrt 2\\ 2a= |\overline{PF_1}-\overline{PF_2}|=3-1=2} \Rightarrow \cases{a=1\\c=\sqrt 2} \Rightarrow b=1 \Rightarrow a=b \Rightarrow 等軸雙曲線\\ \Rightarrow 兩條漸近線會互相垂直，且與貫軸的夾角為45^\circ\\ 現在貫軸 \overleftrightarrow{F_1F_2} 的斜率={2-0\over 2-0}=1 \Rightarrow 漸近線為垂直線或水平線，\\且通過中心點C={1\over 2}(F_1+F_2)=(1,1) \Rightarrow 漸近線為x=1及y=1 \\ \Rightarrow 雙曲線方程式:(x-1)(y-1)=k 通過P(2+{1\over \sqrt 2}, 2-{1\over \sqrt 2}) \Rightarrow k={1\over 2} \\\Rightarrow (x-1)(y-1)={1\over 2} \Rightarrow \bbox[red, 2pt]{2xy-2x-2y+1=0}$$
解答:$$\cases{a_6=27\\ a_{10} =47} \Rightarrow \cases{a_1+ 5d=27\\a_1+9d=47} \Rightarrow \cases{a_1=2\\ d=5} \Rightarrow a_n=2+5(n-1)=5n-3 \\ \Rightarrow \sum_{n=1}^{10}a_n =\sum_{n=1}^{10}(5n-3) = 5\cdot 55-3\cdot 10=245 \\a_n=\log b_n \Rightarrow b_n= 10^{a_n} \Rightarrow b_1\cdot b_2\cdots b_{10} =10^{a_1+ a_2+ \cdots +a_{10}} =\bbox[red, 2pt]{10^{245}}$$
解答:$$f(x)=(x-2)^3+3(x-2)^2-2(x-2)-6 \Rightarrow f'(x)=3(x-2)^2+6(x-2)-2 \\ \Rightarrow \cases{f(-2)= -14\\f'(-2)= 22} \Rightarrow y=22(x+2)-14 \Rightarrow \bbox[red, 2pt]{y=22x+30}$$
解答:$$y=f(x)=x^2-x+{1\over 4} =(x-{1\over 2})^2 \Rightarrow f(x_1)=f(x_2)=a \Rightarrow (x_1-{1\over 2})^2 =(x_2-{1\over 2})^2 =a \\ x_2\lt {1\over 2}\lt x_1 \Rightarrow x_1-{1\over 2}\gt 0 \Rightarrow \log_a \left|x_1^2-{1\over 4} \right| -\log_a \left|x_1+{1\over 2} \right| +2\log_a \left|x_2-{1\over 2} \right| \\=\log_a (x_1^2-{1\over 4}) -\log_a (x_1+{1\over 2}) +\log_a(x_2-{1\over 2})^2 =\log_a {(x_1-{1\over 2})(x_1+{1\over 2}) \over x_1+{1\over 2}} +\log_a a \\=\log_a (x_1-{1\over 2})+1 ={1\over 2}+1= \bbox[red, 2pt]{3\over 2}$$
解答:$$擲骰子一次有6種可能，擲四次有6^4=1296種可能;而擲骰子兩次可決定一頂點座標，\\共有6\times 6=36個不同頂點，依直線y=x與36個點的相對位置，可區分成: \\ \textbf{在直線上(集合D): }即符合x=y，有(1,1),(2,2), \dots,(6,6)，共6個點\\ \textbf{在直線左上方(集合U):}即符合x\lt y，共有(36-6)/2=15個點\\\textbf{在直線右下方(集合L:}即符合x\gt y，共有(36-6)/2=15個點\\符合題意的A,B兩點:\\ \textbf{Case I: }\cases{A\in U\\ B\in L}或\cases{A\in L\\ B\in U}，各有15\times 15=225種，合計450\\ \textbf{Case II: }\cases{A\in D\\ B\in U} 或\cases{A\in D\\ B\in L}或\cases{B\in D\\ A\in U} 或\cases{B\in D\\ A\in L}，各有6\times 15=90，合計360種 \\ \textbf{Case III: }\cases{A\in D\\ B\in D}, 由於A\ne B, 因此共有6\times 5=30種 \\ \Rightarrow 機率={450+360+30\over 1296} = \bbox[red, 2pt]{35\over 54}$$
解答:
$$取\cases{A(0,0) \\B(4,0)\\C(0,4)\\ P(p,0)} \Rightarrow 重心G=(A+B+C)/3 =(4/3,4/3) \\ 取\cases{P,P_1對稱於y軸\\ P,P_2對稱於 \overleftrightarrow{BC}} \Rightarrow \cases{P_1=(-p,0) \\P_2=(4,4-p)} \Rightarrow P_1,P_2均在 \overleftrightarrow{QR}上，重心G也在\overleftrightarrow{QR}上 \\ \Rightarrow P_1,G,P_2在一直線上 \Rightarrow {4/3-0\over 4/3-(-p)}={(4-p)-4/3\over 4-4/3} \Rightarrow 12p-9p^2=0 \Rightarrow 3p(4-3p)=0 \\ \Rightarrow p= {4\over 3} \Rightarrow \overline{AP} =\bbox[red, 2pt]{4\over 3}$$
解答:$$\begin{vmatrix} a^2+1& ab& ac\\ ab & b^2+1& bc\\ ac& bc& c^2+1 \end{vmatrix}= (a^2+1) \begin{vmatrix} b^2+1& bc\\ bc&c^2+1 \end{vmatrix} -ab \begin{vmatrix} ab& bc\\ac&c^2+1 \end{vmatrix} +ac \begin{vmatrix} ab&b^2+1\\ ac& bc \end{vmatrix}\\=(a^2+1)(b^2+c^2+1)-a^2b^2-a^2c^2 =a^2+b^2+c^2+1=15 \Rightarrow a^2+b^2+c^2=14 \\ 柯西不等式: (a^2+b^2+c^2)(1^2+2^2+3^2) \ge(a+2b+3c)^2 \Rightarrow 14\cdot 14 \ge (a+2b+3c)^2\\ \Rightarrow a+2b+ 3c \le \bbox[red, 2pt]{14}$$
解答:$$本題試題疑義回覆\bbox[cyan,2pt]{送分}$$
解答:$$(1-i)^n=a_n+ib_n \Rightarrow a_{n+1}+ ib_{n+1} =(1-i)^{n+1} =(a_n+ib_n)(1-i) =(a_n+b_n)+i(b_n-a_n) \\ \Rightarrow \cases{a_{n+1} =a_n+b_n\\ b_{n+1}=b_n-a_n} \Rightarrow \begin{bmatrix}a_{n+1} \\ b_{n+1} \end{bmatrix} = \begin{bmatrix}1& 1\\-1&1 \end{bmatrix} \begin{bmatrix}a_n\\ b_n \end{bmatrix} \Rightarrow A= \begin{bmatrix}1& 1\\-1&1 \end{bmatrix} \Rightarrow \det(A)=2 \\ \cases{P'(1,2) \\ Q'(4,6)\\ R'(3,8)} \Rightarrow \triangle P'Q'R' ={1\over 2} \begin{Vmatrix} 1& 2& 1\\4& 6& 1\\3& 8& 1\end{Vmatrix} =5 \Rightarrow \triangle P'Q'R'= \det(A) \cdot \triangle PQR \\ \Rightarrow \triangle PQR ={\triangle P'Q'R'\over \det(A)} = \bbox[red, 2pt]{5\over 2}$$
解答:$$\bbox[cyan,2pt]{題目有誤}，應該是「袋中有編號1,2,...,n號的球各1顆共n顆...」\\n顆球任取2顆有{n\choose 2}={n(n-1)\over 2}種取法\\假設取出的兩球編號為a,b，且a\lt b \Rightarrow X=b-a=1,2,\dots, n-1\\ X=k可能的情形: (a,b)=(1,1+k),(2,2+k),\dots,(n-k,n)，有n-k種組合\\ \Rightarrow P(X=k) ={n-k\over C^n_2} \Rightarrow E(X) =\sum_{k=1}^{n-1} k\cdot P(X=k) =\sum_{k=1}^{n-1} k\cdot{n-k\over C^n_2}={1\over C^n_2}\sum_{k=1}^{n-1} (nk-k^2) \\= {2\over n(n-1)} \left[n\cdot {n(n-1)\over 2} -{(n-1)n(2n-1)\over 6} \right] ={n+1\over 3} \lt 12 \Rightarrow n\lt 35\\ \Rightarrow n= \bbox[red, 2pt]{34}$$
解答:$$\cases{橢圓\Gamma_1: {x^2\over m}+y^2=1 \Rightarrow \cases{a_1^2=m\\ b_1^2=1} \Rightarrow c^2= m-1\\雙曲線\Gamma_2: {x^2\over n}-{y^2 \over 3}=1 \Rightarrow \cases{a_2^2=n\\ b_2^2=3} \Rightarrow c^2=a_2^2+b_2^2=n+3} \Rightarrow m-1=n+3 \Rightarrow m-n=4 \\ 假設\cases{r_1=\overline{PF_1} \\ r_2=\overline{PF_2}} \Rightarrow \cases{r_1+r_2=2a_1= 2\sqrt m\\ |r_1-r_2|=2a_2=2\sqrt n} \Rightarrow \cases{(r_1+r_2)^2 =4m\\ (r_1-r_2)^2=4n}\\ \Rightarrow \cases{(r_1+r_2)^2-(r_1-r_2)^2 =4r_1r_2=4(m-n) =4\cdot 4=16 \Rightarrow r_1r_2=4 \\ (r_1+r_2)^2+(r_2-r_2)^2 =2(r_1^2 +r_2^2) =4(m+n) \Rightarrow r_1^2+r_2^2 = 2(m+n)} \\ 假設 \angle F_1PF_2=\theta \Rightarrow \overline{F_1F_2}^2= r_1^2+r_2^2-2r_1r_2\cos \theta \Rightarrow 4c^2=4(m-1)=2(m+n)-8\cos \theta \\ \Rightarrow 4-8\cos \theta=2(m-n)=2\cdot 4 =8\Rightarrow \cos \theta=-{1\over 2} \Rightarrow \theta={2\pi\over 3} \Rightarrow \tan \theta= \bbox[red, 2pt]{-\sqrt 3}$$
解答:$$f(x)=x^3+3ax^2+3bx+c \Rightarrow f'(x) =3x^2+6ax+3b \Rightarrow f''(x)=6x+6a \\ x=2有極值\Rightarrow f'(2)=0 \Rightarrow 12+12a+3b=0 \Rightarrow 4a+b=-4 \cdots(1) \\ 在x=1處切線斜率=-3 \Rightarrow f'(1)=-3 \Rightarrow 3+6a+3b=-3 \Rightarrow 2a+b=-2 \cdots(2) \\ 由式(1)及式(2)可得\cases{a=-1\\ b=0} \Rightarrow \cases{f(x)=x^3-3x^2+c\\ f'(x)=3x^2-6x \\f''(x)=6x-6} \Rightarrow f'(x)=0 \Rightarrow 3x(x-2)=0 \Rightarrow x=0,2 \\ \Rightarrow \cases{f''(0)=-6 \lt 0\\ f''(2)=6\gt 0} \Rightarrow \cases{f(0)=c為極大值\\ f(2)=-4+c為極小值} \Rightarrow 極大值-極小值= \bbox[red, 2pt]4$$
解答:$$\textbf{Case I: }x\ge {1\over 2}\Rightarrow \log_2(2x-1+5x-2) =\log_2(7x-3)\le 3 \Rightarrow x\le 11/7 \Rightarrow {1\over 2}\le x\le {11\over 7}\\ \textbf{Case II: }{2\over 5}\le x\le {1\over 2} \Rightarrow \log_2(1-2x+5x-2)= \log_2(3x-1)\le 3 \Rightarrow x\le 3 \Rightarrow {2\over 5}\le x\le {1\over 2}\\ \textbf{Case III: }x\le {2\over 5} \Rightarrow \log_2(1-2x+2-5x)= \log_2(3-7x)\le 3 \Rightarrow x\ge -{5\over 7} \Rightarrow -{5\over 7}\le x\le {2\over 5}\\ \Rightarrow -{5\over 7}\le x\le {11\over 7} \Rightarrow \cases{a=-5/7\\ b=11/7} \Rightarrow 2a-b= \bbox[red, 2pt]{-3}$$
二、計算證明題(共 2 題，每題 10 分，合計 20 分)
解答:$$假設\vec a與\vec b的夾角為\theta \Rightarrow \vec a在\vec b投影長度為|\vec a|\cos \theta, 又\vec b的單位向量為{\vec b\over |\vec b|} \\ \Rightarrow 正射影向量\vec p=|\vec a|\cos \theta\cdot {\vec b\over |\vec b|} \\由於\vec a\cdot \vec b=|\vec a||\vec b|\cos \theta \Rightarrow |\vec a|\cos \theta={\vec a\cdot \vec b\over |\vec b|} \Rightarrow \vec p= {\vec a\cdot \vec b\over |\vec b|}\cdot {\vec b\over |\vec b|} = \left( {\vec a\cdot \vec b\over |\vec b|^2} \right)\vec b, \bbox[red, 2pt]{故得證}$$
解答:
$$P,Q對稱於直線L:x-y=k \Rightarrow \overleftrightarrow{PQ}斜率為-1 \Rightarrow \overleftrightarrow{PQ} :x+y=c \\ 將y=c-x代入\Gamma \Rightarrow (x-1)^2=8(c-x+1) \Rightarrow x^2+6x-8c-7=0有相異實根\\ \Rightarrow 判別式:\Delta =36-4(-8c-7) \gt 0 \Rightarrow c\gt -2 \\ 假設\cases{P(x_1,y_1) \\Q(x_2,y_2)} \Rightarrow x_1,x_2 為x^2+6x-8c-7=0的兩根\Rightarrow x_1+x_2=-6 \Rightarrow M(x,y)=\overline{PQ}中點 \\ \Rightarrow x={x_1 +x_2\over 2} =-3 \Rightarrow M(-3,y) \in \overleftrightarrow{PQ} \Rightarrow -3+y=c \Rightarrow y=c+3 \Rightarrow M(-3,c+3)\\ 同時M(-3,c+3) \in L \Rightarrow -3-(c+3)=k \Rightarrow c=-k-6 \gt -2 \Rightarrow \bbox[red, 2pt]{k\lt -4}$$
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解題僅供參考，其他教甄試題及詳解 
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