台中市立台中第一高級中等學校 109 學年度
第1次教師甄選-數學科試題
解：
$$\cases{|z|=1\\ |z-1.45|=1.05} \Rightarrow \cases{O_1(0,0),r_1=1\\ O_2(1.45,0), r_2=1.05} \Rightarrow \overline{O_1O_2} =1.45，\\假設兩圓相交於A、B兩點，且\overline{AB}與\overline{O_1O_2}相交於C，如上圖；\\ \cos \angle AO_1O_2 = {r_1^2+\overline{O_1O_2}^2-r_2^2 \over 2r_1\overline{O_1O_2}} ={1+1.45^2-1.05^2 \over 2\times 1.45} ={1\over 1.45} \\ \Rightarrow z的實部=\overline{O_1C} = \overline{O_1A}\cos \angle AO_1O_2 = 1\cdot {1\over 1.45} =\bbox[red,2pt]{20\over 29}$$
$$P(x)=x^5-x^2+1 = (x-\alpha_1)(x-\alpha_2)(x-\alpha_4)(x-\alpha_4)(x-\alpha_5) \\ \Rightarrow \cases{P(i)=i^5-i^2+1=2+i=-(\alpha_1-i)(\alpha_2-i)(\alpha_3-i)(\alpha_4-i)(\alpha_5-i) \\ P(-i)=(-i)^5-(-i)^2+1=2-i= -(\alpha_1+i) (\alpha_2+i)(\alpha_3+i)(\alpha_4+i)(\alpha_5+i)}\\ Q(x)=x^2+1=(x+i)(x-i) \Rightarrow Q(\alpha_1) \cdot Q(\alpha_2)\cdot Q(\alpha_3)\cdot Q(\alpha_4)\cdot Q(\alpha_5) \\ = P(i)P(-i) =(2+i)(2-i)= \bbox[red,2pt]{5}$$
解：
$${8^x +27^x \over 12^x+18^x} = {1+({27\over 8})^x \over ({12\over 8})^x +({18\over 8})^x} = {1+({3\over 2})^{3x} \over ({3\over 2})^{x} +({3\over 2})^{2x}} \equiv{1+u^3 \over u+u^2 } ={7\over 6}，其中u=({3\over 2})^{x}\\ \Rightarrow 6u^3-7u^2-7u+6=0 \Rightarrow (u+1)(6u^2-13u+6)=0 \Rightarrow (u+1)(3u-2)(2u-3)=0\\ \Rightarrow \cases{u=-1(不合)\\ u=3/2 \\ u=2/3} \Rightarrow \bbox[red,2pt]{\cases{x=1\\ x=-1}}$$
解：
$$令g(x)=xf(x)-4040 =ax^4+bx^3+cx^2+dx-4040 \\由於2f(2)=3f(3)=4f(4)=4040 \Rightarrow g(2)=g(3)=g(4)=0\\ \Rightarrow g(x)=(x-2)(x-3)(x-4)(ax+k) \Rightarrow g(0)=-24k = -4040 \Rightarrow k={4040\over 24}\\ 又\cases{g(1)=-6(a+k)=f(1)-4040 \cdots(1)\\ g(5)= 6(5a+k) =5f(5)-4040 \cdots(2)} \Rightarrow 5\times(1)+(2) \Rightarrow -24k=5(f(1)+f(5))-24240 \\ \Rightarrow f(1)+f(5)={24240-24k \over 5} ={24240-4040\over 5} = \bbox[red,2pt]{4040}$$
5. 當桌球比賽比分為 10:10 時， 稱為 deuce，此後必須由兩人輪流各發一球， 直到其中一名球員比對手多勝 2 分時比賽結束。 依過去經驗知道， 甲乙兩人比賽桌球， 當甲發球時，甲得分機率為 3/5；當乙發球時，乙得分機率為 3/5。
今甲乙兩人比賽，目前比分恰為 10:10， 接著輪到甲發球，假設各次得分為獨立事件，則從 deuce 發生後開始計算發球次數， 到比賽結束時，兩人發球總次數的期望值為______次。
$$deuce後再deuce，也就是不分勝負，即甲勝後乙勝，或乙勝後甲勝\\\qquad，其機率為:{3\over 5}\times {3\over 5} + {2\over 5}\times {2\over 5}= {13\over 25};\\能分出勝負的情形：甲連勝2次，或乙連勝2次，其機率為:{3\over 5}\times {2\over 5} +{2\over 5}\times {3\over 5} ={12\over 25}\\ 因此E=2\times {12\over 25}+ (E+2)\times {13\over 25} \Rightarrow {12\over 25}E= 2 \Rightarrow E=\bbox[red, 2pt]{25\over 6}$$
解：
$$令\cases{\overline{PA}=a \\\overline{PB}=b \\\overline{PC}=c } \Rightarrow \cases{\triangle PAB={1\over 2}ab\sin \angle APB \\ \triangle PBC={1\over 2}bc\sin \angle BPC \\\triangle PCA={1\over 2}ac\sin \angle APC } \Rightarrow \cases{{\sqrt 3\over 2}={1\over 2}ab\sin 60^\circ= {\sqrt 3\over 4}ab \\ 2={1\over 2}bc\sin 60^\circ ={\sqrt 3\over 4}bc \\1={1\over 2}ac\sin 60^\circ ={\sqrt 3\over 4}ac} \\ \Rightarrow \cases{ab=2 \\ bc={8\over \sqrt 3} \\ ac={4\over \sqrt 3}},三式相乘\Rightarrow a^2b^2c^2 ={64\over 3} \Rightarrow abc={8\over \sqrt 3}(a,b,c均為正值) \Rightarrow \cases{a=1\\ b=2\\c ={4\over \sqrt 3}}\\ 因此我們可以在\overline{PB}上找一點B'及\overline{PC}上找一點C',使得\overline{PB'}= \overline{PC'}= \overline{PA}=1,\\也就是說PAB'C'為一正四面體,其高(h)為\sqrt{2\over 3}\times 邊長=\sqrt{2\over 3}\\ 由於h=A至平面PB'C'的距離= A至平面PBC的距離,\\因此四面體PABC體積={1\over 3}\times h\times \triangle PBC ={1\over 3}\times \sqrt{2\over 3}\times 2= \bbox[red,2pt]{{2\over 9}\sqrt 6}$$
解：
$$\cases{\sigma(X)=6\\ \sigma(Y)=4} \Rightarrow \sigma(X+Y) =\sqrt{\sigma^2(X) +2Cov(X,Y)+ \sigma^2(Y)} \Rightarrow 2\sqrt{23} =\sqrt{36+2Cov(X,Y)+16} \\ \Rightarrow Cov(X,Y)=20 \Rightarrow 迴歸直線斜率\;m={Cov(X,Y)\over \sigma^2(X)} ={20\over 36} ={5\over 9} \\\Rightarrow 迴歸直線: y-\mu(Y)=m (x-\mu(X)) \Rightarrow y-16 = {5\over 9}(x-54) \Rightarrow \bbox[red,2pt]{y={5\over 9}x-14}$$
解：
$$ 第一個a有16個位置可填，第2個a只能填剩下的9個位置，因此有16\times 9=144種填法\\，但兩個a長的一樣，所以有144\div 2=72種填法；同理，2個b也有72種填法；\\若不考慮重複問題，兩個a兩個b有72^2種填法；若兩個a與兩個b剛好填在相同位置，則有72種；\\ 若一個a與一個b剛好填在相同位置，另一個a與另一個b不再同位置，則有16\times 9\times 8=1152種；\\綜合以上，符合要求的填法共有72^2-72-1152= \bbox[red,2pt]{3960}種$$
解：
$$本題\bbox[red,2pt]{不予計分}$$
解：
$$令\triangle ABC外接圓半徑=R，即\cases{\overline{OA}=R \\\overline{OB}=R \\\overline{OC}=R };\\由題意知:\overrightarrow{AO} = \overrightarrow{AB}+ 2\overrightarrow{AC} \Rightarrow 2\overrightarrow{AC} =\overrightarrow{AO} - \overrightarrow{AB} = \overrightarrow{BO} \Rightarrow \overrightarrow{AC}={1\over 2}\overrightarrow{BO} \\\Rightarrow \cases{\overline{AC}=R/2\\ \overline{AC} \parallel \overline{BO}};因此令\angle ACB=\alpha \Rightarrow \cases{\angle CBO=\alpha (內錯角)\\ \angle AOB=2\alpha (對同弧的圓心角是圓周角的2倍)} \\ \Rightarrow \angle CAO= \angle AOB=2\alpha \Rightarrow \cos \angle BOC = \cos (180^\circ-2\alpha) =-\cos 2\alpha =-\cos \angle CAO \\ =-{R^2+R^2/4-R^2 \over 2\times R\times (R/2)} =-{1\over 4} ={R^2+R^2-\overline{BC}^2 \over 2R^2} \Rightarrow \overline{BC}=\sqrt{5\over 2}R \Rightarrow {\overline{BC} \over \sin\angle BAC}=2R \\ \Rightarrow \sin \angle BAC= \bbox[red,2pt]{\sqrt{10}\over 4}$$
$$向量性質:\cases{\vec a\times (\vec b\times \vec c)=(\vec a\cdot \vec c)\vec b- (\vec a\cdot \vec b)\vec c \\ (\vec a\times \vec b)\cdot \vec c=\vec a \cdot (\vec b\times \vec c) \\ (\vec a\times \vec b)\cdot b =(\vec a\times \vec b)\cdot a=0} \\因此(\vec a\times \vec b)\cdot ((\vec b\times \vec c)\times (\vec c\times \vec a)) =(\vec a\times \vec b)\cdot (((\vec b\times \vec c)\cdot \vec a)\vec c- ((\vec b\times \vec c)\cdot \vec c)\vec a) =(\vec a\times \vec b)\cdot ((\vec b\times \vec c)\cdot \vec a)\vec c \\ =((\vec b\times \vec c)\cdot \vec a) ((\vec a\times \vec b)\cdot \vec c) =(\vec a\cdot (\vec b\times \vec c))(\vec a\cdot (\vec b\times \vec c)) =7 \Rightarrow \vec a\cdot (\vec b\times \vec c)=\pm \sqrt 7\\ (3\vec a+\vec b+\vec c), (\vec a-\vec b+2\vec c), (\vec b+\vec c)三向量張開的平行六面體體積=|(3\vec a+\vec b+\vec c)\cdot ((\vec a-\vec b+2\vec c)\times (\vec b+\vec c))| \\ = \begin{Vmatrix} 3& 1& 1 \\ 1 & -1 & 2 \\ 0 & 1 & 1\end{Vmatrix} \times |\vec a\cdot (\vec b\times \vec c)| =|-9|\times |\pm \sqrt 7| =\bbox[red,2pt]{9\sqrt 7}$$
$$B=(I+A)^{-1}(I-A) \Rightarrow (I+A)B=I-A \Rightarrow B+AB=I-A \\\Rightarrow I+A+B+AB= 2I \Rightarrow (I+A)(I+B)=2I \Rightarrow {1\over 2}(I+A)(I+B)=I \\ \Rightarrow (I+B)^{-1}={1\over 2}(I+A) ={1\over 2}\begin{bmatrix}2 & 0 & 0 \\ -2 & 4 & 0 \\ 0 & -4 & 6 \end{bmatrix} =\bbox[red,2pt]{\begin{bmatrix}1 & 0 & 0 \\ -1 & 2 & 0 \\ 0 & -2 & 3 \end{bmatrix}}$$
$$\cases{a_0=1 \cdots(1)\\ a_{2k+1}=a_k\cdots(2) \\ a_{2k+2}=a_k+a_{k+1}\cdots(3)},k\in N\cup \{0\}\\ 由(2)知: 1=a_0=a_1=a_3=a_7=\cdots = a_{2^{n-1}-1} =a_{2^{n}-1}\cdots(4)\\ S(n)= \sum_{k=0}^{2^n-1} a_k= (a_0+ a_2+ a_4+ \cdots + a_{2^n-2}) +(a_1+a_3 +\cdots + a_{2^n-1})=偶次項+奇次項\\ 偶次項= (a_0+ (a_0+a_1) +(a_1+a_2) + (a_2+a_3)+\cdots +(a_{2^{n-1}-2} +a_{2^{n-1}-1}) \\ \qquad \;= 2(a_0+a_1+\cdots +a_{2^{n-1}-2})+a_{2^{n-1}-1}\\ 奇次項=(a_0+a_1 +a_2 +\cdots + a_{2^{n-1}-1}) \\ \Rightarrow S(n)=3(a_0+a_1+\cdots + a_{2^{n-1}-2})+ 2a_{2^{n-1}-1} =3(a_0+a_1+\cdots + a_{2^{n-1}-1}) -a_{2^{n-1}-1} \\ \qquad \quad= 3(a_0+a_1+\cdots + a_{2^{n-1}-1}) -1(由(4)知a_{2^{n-1}-1}=1) \\ \qquad \quad=3S(n-1)-1 \\ \Rightarrow \sum_{k=0}^{63}a_k =\sum_{k=0}^{2^8-1}a_k = S(8) =3S(7)-1 =3^2S(6)-3-1=\cdots \\= 3^5S(1)-3^4-3^3-3^2-3-1 =3^5(a_0+a_1)-(1+3+\cdots+3^4) =243\times 2-121\\ =\bbox[red,2pt]{365}$$
解：
$$y=2x-x^2=0 \Rightarrow x(x-2)=0 \Rightarrow y=2x-x^2與x軸交於(0,0)及(2,0)\\ \Rightarrow S_1+S_2= \int_0^2 2x-x^2 \;dx= \left. \left[ x^2-{1\over 3}x^3\right]\right|_0^2 =4-{8\over 3}={4\over 3};\\由於S_1:S_2=1:7 \Rightarrow S_1={1\over 8}(S_1+S_2)= {1\over 8}\times {4\over 3}= {1\over 6}\\ y=2x-x^2=kx \Rightarrow x^2+(k-2)x=0 \Rightarrow x=0,2-k \Rightarrow S_1= \int_0^{2-k} 2x-x^2-kx\;dx \\ =\left. \left[x^2-{1\over 3}x^3 -{k\over 2}x^2\right] \right|_0^{2-k} = (1-{k\over 2})(2-k)^2 -{1\over 3}(2-k)^3 ={1\over 6}(2-k)^3 ={1\over 6} \Rightarrow k=1\\ \Rightarrow \Gamma_1繞y軸旋轉體積=\pi\int_0^1y^2\;dy -\pi\int_0^1 (1-\sqrt{1-y})^2\;dy ={1\over 3}\pi-\pi\int_0^1 (2-y-2\sqrt{1-y})\;dy \\ ={1\over 3}\pi-\pi \left.\left[2y-{1\over 2}y^2+{4\over 3}(1-y)^{3/2} \right] \right|_0^1 ={1\over 3}\pi-{1\over 6}\pi =\bbox[red,2pt]{\pi \over 6}$$
解：
解：$${\vec u\over |\vec u|}為單位向量 \Rightarrow {\overrightarrow{PA} \over |\overrightarrow{PA}|} +{\overrightarrow{PB} \over |\overrightarrow{PB}|} +{\overrightarrow{PC} \over |\overrightarrow{PC}|} =\vec 0 \Rightarrow \angle APB=\angle BPC=\angle APC = {360^\circ \over 3} =120^\circ;\\ 令\cases{ \overline{PA}=a \\ \overline{PB}=b \\ \overline{PC}=c } \Rightarrow \cases{\cos \angle APB= (\overline{PA}^2 +\overline{PB}^2 -\overline{AB}^2 )/(2\overline{PA}\cdot \overline{PB} )\\ \cos \angle BPC= (\overline{PB}^2 +\overline{PC}^2 -\overline{BC}^2 )/(2\overline{PB}\cdot \overline{PC} )\\ \cos \angle APC= (\overline{PA}^2 +\overline{PC}^2 -\overline{AC}^2 )/(2\overline{PA}\cdot \overline{PC} )\\ } \\ \Rightarrow \cases{-1/2= (a^2+b^2-1)/2ab \\ -1/2=(b^2+c^2 -4)/2bc \\ -1/2 = (a^2+c^2-3)/2ac} \Rightarrow \cases{a^2+b^2 +ab=1\cdots(1) \\ b^2+c^2 +bc=4 \cdots(2)\\ c^2+a^2+ca = 3\cdots(3)}\\ \Rightarrow (1)+(2)+(3) \Rightarrow 2(a^2+b^2+c^2) +ab+bc+ca =8 \cdots(4)\\ \triangle ABC面積= \triangle PAB+\triangle PBC+\triangle PCA = {1\over 2}ab\sin \angle PAB + {1\over 2}bc\sin \angle PBC + {1\over 2}ac\sin \angle PAC \\ ={1\over 2}(ab+bc+ca)\sin 120^\circ ={\sqrt 3\over 4}(ab+bc+ca) = {1\over 2}\overline{AB}\times \overline{AC}= {\sqrt 3\over 2} \Rightarrow ab+bc+ca=2 \\ 將ab+bc+ca=2代入(4) \Rightarrow a^2+b^2+c^2 =3 \Rightarrow (a+b+c)^2 = a^2+b^2+c^2 +2(ab+ bc+ca) \\ =3+4=7 \Rightarrow a+b+c=\sqrt 7 \\ 又\cases{(1)+(2) \Rightarrow a^2+2b^2+c^2+ab+bc=5 \\ (2)+(3)\Rightarrow a^2+b^2+2c^2 +bc+ca = 7 \\ (1)+(3) \Rightarrow 2a^2+b^2+c^2+ab+ac =4} \Rightarrow \cases{b^2+ab+bc=5-3=2 \\ c^2+bc+ca =7-3=4 \\ a^2+ab+ac =4-3=1} \\ \Rightarrow \cases{b(a+b+c)=2 \\ c(a+b+c)=4 \\ a(a+b+c)=1} \Rightarrow \cases{a=1/\sqrt 7 \\ b=2/\sqrt 7 \\ c=4/\sqrt 7} \Rightarrow (\overline{PA},\overline{PB},\overline{PC}) = \bbox[red,2pt]{\left( {1\over \sqrt 7}, {2\over \sqrt 7}, {4\over \sqrt 7}\right)}$$
解：
$$假設p(n):第n次按鈕出現紅球的機率 \Rightarrow \cases{p(n-1)=第n-1次按鈕出現紅球的機率 \\ 1-p(n-1)=第n-1次按鈕出現白球的機率}\\ \Rightarrow p(n)={1\over 3}p(n-1)+{3\over 5}(1-p(n-1)) = {3\over 5}-{4\over 15}p(n-1)\\ =({-4\over 15})^{n-1}p(1)+{3\over 5}( 1+{-4\over 15}+({-4\over 15})^2+\cdots+({-4\over 15})^{n-2}) \\ =({-4\over 15})^{n-1}\times{1\over 2}+{3\over 5}( {1-({-4\over15})^{n-1} \over 1-{-4\over15}}) =({-4\over 15})^{n-1}\times{1\over 2}+{9\over 19}-{9\over 19}\times ({-4\over15})^{n-1} \\ =\bbox[red,2pt]{{1\over 38}({-4\over15})^{n-1}+ {9\over 19}}$$
解：$$利用Lagrange 乘數來求解，令\cases{f(\alpha,\beta)=\cos \alpha\\ g(\alpha,\beta)=\cos(\alpha+\beta)- \cos \alpha-\cos \beta} \Rightarrow \cases{{\partial f\over \partial \alpha} =\lambda {\partial g\over \partial \alpha} \\{\partial f\over \partial \beta} =\lambda {\partial g\over \partial \beta} \\ g=0} \\ \Rightarrow \cases{-\sin \alpha=\lambda(-\sin(\alpha+\beta) +\sin \alpha) \cdots(1)\\ 0=\lambda(-\sin(\alpha +\beta)+\sin \beta) \cdots(2) \\ \cos(\alpha+\beta)= \cos \alpha+\cos \beta \cdots(3)};\\由(2)知: \lambda(-\sin(\alpha +\beta)+\sin \beta)=0 \Rightarrow \sin \beta= \sin(\alpha+\beta)(\because \lambda\ne 0) \Rightarrow \cases{\alpha=2\pi \\ \alpha=\pi-2\beta} 代入(3) \\ \Rightarrow \cases{\cos \beta= 1+\cos \beta(不合)\\ \cos(\pi-\beta)=-\cos 2\beta+\cos \beta} \Rightarrow -\cos \beta=-(2\cos^2 \beta-1)+\cos \beta \\ \Rightarrow 2\cos^2\beta-2\cos \beta -1=0 \Rightarrow \cos\beta =\cases{(1-\sqrt 3)/2\\ (1+\sqrt 3)/2 (不合,其值> 1)} \\ \Rightarrow \cos \alpha=\cos(\pi-2\beta) = -\cos 2\beta =-2\cos^2\beta+1 =-2\times {2-\sqrt 3\over 2}-1 =\bbox[red,2pt]{\sqrt 3-1}$$
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