114年台南一中教甄-數學詳解

國立台南第一高級中學114學年度教師甄選

一、填充題 (第1~3題，每題5分；第4~8題，每題6分；第9~11題，每題7分)

解答:$$點數和為4的情形:\{1,3\},\{2,2\}, \{3,1\}, 有3種 \Rightarrow 機率p={3\over 6\times 6} ={1\over 12} \\ 二項式分布B(N=500,p=1/12) \Rightarrow \cases{p_n= {500\choose n}p^n(1-p)^{500-n} \\ p_{n+1} ={500\choose n+1}p^{n+1} (1-p)^{499-n}} \\ \Rightarrow {500\choose n}p^n(1-p)^{500-n} \lt{500\choose n+1}p^{n+1} (1-p)^{499-n} \Rightarrow {500\choose n}(1-p) \lt {500\choose n+1}p \\ \Rightarrow {1-p\over 500-n} \lt {p\over n+1} \Rightarrow {11/12\over 500-n} \lt {1/12\over n+1} \Rightarrow n\lt {489\over 12}=40.75 \Rightarrow n之最大值為\bbox[red, 2pt]{40}$$

解答:$$假設\cases{a= |\overrightarrow{BC}| \\ b= |\overrightarrow{CA}| \\ c=|\overrightarrow{AB}|} \Rightarrow \overrightarrow{AB}+\overrightarrow{BC} = \overrightarrow{AC} \Rightarrow |\overrightarrow{AB}+\overrightarrow{BC}|^2 = |\overrightarrow{AC}|^2 \\ \Rightarrow (\overrightarrow{AB}+\overrightarrow{BC}) \cdot (\overrightarrow{AB}+\overrightarrow{BC}) =(\overrightarrow{CA})(\overrightarrow{CA}) \Rightarrow c^2+a^2+2(-5)=b^2 \Rightarrow c^2+a^2-b^2=10 \cdots(1) \\ 同理可得\cases{a^2+b^2-c^2=12 \cdots(2) \\ b^2+c^2-a^2=14 \cdots(3)} \Rightarrow (1)+(2)+(3) =a^2+b^2+c^2=36 \Rightarrow \cases{a^2=11\\ b^2=13\\ c^2=12} \\ \Rightarrow \triangle ABC={1\over 2} \sqrt{|\overrightarrow{AB}|^2 |\overrightarrow{BC}|^2 -(\overrightarrow{AB} \cdot \overrightarrow{BC})^2} ={1\over 2} \sqrt{12\cdot 11-(-5)^2} = \bbox[red, 2pt]{\sqrt{107}\over 2}$$

解答:$$假設\cases{X_m:第m次取球的編號\\ M:遊戲結束時取出的總球數} \Rightarrow 遊戲結束的條件\cases{X_m\lt X_{m-1}, 2\le m\le 10\\ m=10} \\\textbf{Case I }m=10 \Rightarrow X_1\lt X_2\lt \cdots\lt X_9 \Rightarrow 9球任排中, 只有一種符合此條件 \Rightarrow P(M=10)={1\over 9!} \\ \textbf{Case II }2\le m\le 9 \Rightarrow X_1\lt X_2\lt \cdots\lt X_{m-1}且X_m\lt X_{m-1} \\ \qquad \Rightarrow 在m!種排列中，有(m-1)種符合此條件\Rightarrow P(M=m)={m-1\over m!} \\ 因此期望值=(10\times 10!)\times {1\over 9!}+ \sum_{m=2}^9 \left( (10\times m!)\times {m-1\over m!} \right) =100+360= \bbox[red, 2pt]{460}$$

解答:$$\cases{1\lt \alpha\lt 2\\ 5\lt \beta\lt 6} \Rightarrow \cases{ 6\lt \alpha+\beta\lt 8 \\5\lt \alpha \beta \lt 12} \Rightarrow \cases{6\lt b/a\lt 8 \\ 5\lt 3c/a\lt 12} \Rightarrow \cases{(a,b,c)=(1,7,2) \\ (a,b,c)=(1,7,3)} \\ \Rightarrow \cases{x^2-7x+6=0 \Rightarrow \cases{\alpha=1\\\beta=6} \Rightarrow 不符\cases{1\lt \alpha\lt 2\\ 5\lt \beta\lt 6}\\ x^2-7x+9=0 \Rightarrow \cases{\alpha=(7-\sqrt{13})/2 \approx 1.7\\ \beta=(7+\sqrt{13})/2 \approx5.3}} \Rightarrow (a,b,c)= \bbox[red, 2pt]{(1,7,3)}$$

解答:$$t={x^2+y^2+1\over x+2y+2} \Rightarrow t(x+2y+2)=x^2+y^2+1 \Rightarrow  \left( x-{t\over 2} \right)^2+(y-t)^2={5\over 4}t^2+2t-1 \ge 0 \\ \Rightarrow 5t^2+8t-4\ge 0 \Rightarrow (5t-2)(t+2)\ge 0 \Rightarrow \bbox[red, 2pt]{t\ge {5\over 2} 或t\le -2}$$

解答:$$a_{n+1}a_n-{n\over n+1}a_{n+1}-{n+1\over n+2}a_n -{n+4\over n+2}=0 \\ \Rightarrow \left( a_{n+1}-{n+1\over n+2} \right) \left( a_n-{n\over n+1} \right) ={n+4\over n+2} +{n\over n+2} =2\\ 取b_n=a_n-{n\over n+1} \Rightarrow b_{n+1}\cdot b_n=2 \Rightarrow b_{n+1}={2\over b_n} = {2\over 2/b_{n-1}} =b_{n-1} \Rightarrow 循環數=2 \\ \Rightarrow b_1=a_1-{1\over 1+1}=500-{1\over 2}={999\over 2} \Rightarrow b_{998} ={2\over b_1} ={4\over 999} =a_{998}-{998\over 999} \\ \Rightarrow a_{998} ={1002\over 999} =\bbox[red, 2pt]{334\over 333}$$

解答:$$\cases{\overrightarrow{PA} =-\overrightarrow{AP} \\ \overrightarrow{PB} =\overrightarrow{AB}- \overrightarrow{AP} \\\overrightarrow{PC} =\overrightarrow{AC }-\overrightarrow{AP}} \Rightarrow -\overrightarrow{AP}+ 2 \left( \overrightarrow{AB}-\overrightarrow{AP} \right) +3 \left( \overrightarrow{AC}-\overrightarrow{AP} \right) =k\overrightarrow{AB} \\ \Rightarrow \overrightarrow{AP} ={2-k\over 6} \overrightarrow{AB}+ {1\over 2} \overrightarrow{AC} \Rightarrow \cases{2-k\gt 0\\ {2-k\over 6}+ {1\over 2}\lt 1} \Rightarrow \bbox[red, 2pt]{-1\lt k\lt 2}$$

解答:$$P\in \Gamma:y=x^2 \Rightarrow P(t,t^2) \Rightarrow P'= \begin{bmatrix}2&0\\ a& 1 \end{bmatrix} \begin{bmatrix}t\\ t^2 \end{bmatrix} = \begin{bmatrix}2t\\ at+t^2 \end{bmatrix} \in \Gamma' \Rightarrow \Gamma':y={1\over 4}x^2+ {a\over 2}x\\ y=x^2 代入\Gamma' \Rightarrow x^2={1\over 4}x^2+ {a\over 2}x \Rightarrow 3x^2-2ax=0 \Rightarrow x(3x-2a)=0 \Rightarrow \cases{x=0 \Rightarrow y=0\\ x=2a/3 \Rightarrow y=4a^2/9} \\ \Rightarrow \Gamma與\Gamma'的交點為\cases{A(0,0) \\B(2a/3,4a^2/9)} \Rightarrow 所圍面積= \left| \int_0^{2a/3} \left( {3\over 4}x^2-{a\over 2}x \right)\,dx \right| =\bbox[red, 2pt]{{1\over 27}a^3}$$

解答:$$假設稜長s \Rightarrow 正四面體體積= {\sqrt 2\over 12}s^3 =72 \Rightarrow s=6\sqrt 2 =\overline{AB} =\sqrt{3(a-b)^2} \Rightarrow a-b=2\sqrt 6 \cdots(1)\\ \cases{A,B皆在直線x=y=z上\\ C,D皆在xy平面上} \Rightarrow \cases{\overline{AB}中點M((a+b)/2,(a+b)/2,(a+b)/2)\\ \overline{CD}的中點N(k,k,0) } \\ \Rightarrow \overline{AB}的垂直平分面: (x-{a+b\over 2})+ (x-{a+b\over 2})+ (x-{a+b\over 2})=0 \Rightarrow x+y+z={3\over 2}(a+b) \\ \Rightarrow N(k,k,0)在此垂直平分面\Rightarrow k+k+0={3\over 2}(a+b) \Rightarrow k={3\over 4}(a+b) \Rightarrow N({3\over 4}(a+b),{3\over 4}(a+b),0) \\ \Rightarrow \overline{AN}={\sqrt 3\over 2}s \Rightarrow \overline{AN}^2 =(a-k)^2+(a-k)^2+a^2 =2(a-{3\over 4}(a+b))^2+a^2={3\over 4}s^2\\ \Rightarrow {(a-3b)^2\over 8}+a^2 =54 \cdots(2) \\ 由(1)及(2)可得\cases{a=\bbox[red, 2pt]{3\sqrt 6} \\b=\sqrt 6}$$

解答:$$\cases{P(2x,2y,2x-y) \\A(2,4,-9) \\B(-2,-6,-11)} \Rightarrow f(x,y)=\overline{PA} +\overline{PB} \\P(2x,2y,2x-y) \in 平面E:2x-y-2z=0, 取g(x,y)=2x-y-2z \Rightarrow \cases{g(A)=18\gt 0\\ g(B)=24 \gt 0} \\ \Rightarrow A, B在E的同一側, 因此取A'=A對於E的對稱點 \Rightarrow A'=(-6,8,-1) \Rightarrow \overline{A'B}= \bbox[red, 2pt]{2\sqrt{78}}$$

解答:

$$\omega+ \bar \omega=2 Re(\omega)=-6\sqrt 2 \Rightarrow Re(\omega)=-3\sqrt 2 \Rightarrow \omega=-3\sqrt 2+i y_\omega \\ \Rightarrow \omega'={1+i\over \sqrt 2} \omega = \left( {1\over \sqrt 2}+{i\over \sqrt 2} \right)(-3\sqrt 2+iy_\omega) = \left( -3-{y_\omega\over \sqrt 2} \right) +i \left( -3+{y_\omega\over \sqrt 2} \right) \\ \Rightarrow \omega'的軌跡就是一條直線L: x+y=-6\\ \left|{z-i\over z-1} \right| =k \Rightarrow |z-i|^2=k^2|z-1|^2 \Rightarrow x^2+(y-1)^2=k^2 [(x-1)^2+y^2]\\ \Rightarrow x^2+y^2-{2k^2\over k^2-1}x+{2\over k^2-1}y+1=0 \Rightarrow \left( x-{k^2\over k^2-1} \right)^2+ \left( y+{1\over k^2-1} \right)^2= \left( {\sqrt 2 k\over k^2-1} \right)^2\\ \Rightarrow 圖形為一圓, 且k越小半徑越大 \\ \Rightarrow |z-\omega'|的最小值=圓到直線L的最短距離 \Rightarrow 取k=2 \Rightarrow \cases{圓心O(4/3,-1/3)\\ 半徑r=2\sqrt 2/3} \\ \Rightarrow 最小值=d(O,L)-r={7\over \sqrt 2} -{2\sqrt 2\over 3} = \bbox[red, 2pt]{17\sqrt 2\over 6}$$

二、計算題 (須寫出過程方給分)

解答:$$\left| \log {|\sin 2\theta| \over 2} -2\log |\cos \theta|\right| =\left| \log {|\sin 2\theta| \over 2} -\log \cos^2 \theta\right| = \left| \log {|\sin 2\theta| \over 2 \cos^2 \theta}  \right| \\=\left| \log {2|\sin \theta||\cos \theta| \over 2 \cos^2 \theta}  \right| =\left| \log {|\sin \theta|  \over  |\cos \theta|}  \right| =|\log |\tan \theta|| =5-\tan \theta\\ 取 x=\tan \theta \Rightarrow |\log |x||=5-x \Rightarrow 欲求兩圖形\cases{y=|\log |x||\\ y=5-x} 的交點數\\ x\gt 0 \Rightarrow \cases{y_1=\log x\\ y=5-x}有兩交點, 分別在區間(0,1)及(1,5) \\x\lt 0 \Rightarrow 取u=-x \Rightarrow \cases{y=|\log u|\\ y=5+u}有一交點u\in (0,1)\Rightarrow x\in (-1,0) \\ 因此|\log|x||=5-x有三交點, 由於\tan \theta 在區間(0,2\pi)有兩個週期, 因此共有3\times 2=\bbox[red, 2pt]6個交點$$

解答:$$\textbf{(1) }假設\omega 為x^2+x+1=0之一根 \Rightarrow \omega^2+\omega+1=0 \Rightarrow \omega^3=1 \\\qquad \Rightarrow (\omega^2+ \omega+1)f(\omega) =0=2\omega^{302} +\omega^{100} +\omega^{61}+a =2\omega^2+\omega+\omega+a \Rightarrow 2(\omega^2+\omega)+a=0 \\\qquad \Rightarrow 2(-1)+a=0 \Rightarrow a= \bbox[red, 2pt]2 \\\textbf{(2) }f(x)=(x^2-x+1)p(x)+Ax+B \Rightarrow (x^2+x+1)f(x)= (x^2+x+1) [(x^2-x+1)p(x)+Ax+B] \\ \Rightarrow g(x)=2x^{302}+x^{100}+x^{61}+2= (x^2+x+1) [(x^2-x+1)p(x)+Ax+B] \\\qquad 假設z為x^2-x+1=0的一根\Rightarrow z^3=-1 \Rightarrow g(z)=2z^2-z+z+2=(z^2+z+1)(Az+B) \\ \Rightarrow 2z^2+2=(2z)(Az+B) \Rightarrow 2(z-1)+2= 2Az^2+2Bz=2A(z-1)+2Bz =(2A+2B)z-2A \\\qquad \Rightarrow 2z=(2A+2B)z-2A \Rightarrow \cases{A=0\\ B=1} \Rightarrow 餘式: \bbox[red, 2pt]1 \\\textbf{(3) }f(x)=(x^2+x+1)Q(x)+Cx+D \\\qquad\Rightarrow (x^2+x+1) f(x) = h(x)=2x^{302}+x^{100}+x^{61}+2=(x^2+x+1)^2Q(x)+(x^2+x+1)(Cx+D) \\\qquad \Rightarrow h'(x)=604x^{301} +100x^{99}+ 61x^{60} =(x^2+x+1)R(x)+ (2x+1)(Cx+D) \\ \qquad \Rightarrow h'(\omega) =604\omega+100+61= (2\omega+1)(C\omega+D) \Rightarrow 604\omega+ 161=(-C+2D)\omega-2C+D\\ \qquad \Rightarrow \cases{-C+2D=604\\ -2C+D=161} \Rightarrow \cases{C=94\\ D=349} \Rightarrow 餘式: \bbox[red, 2pt]{94x+349}$$

解答:

$$\triangle ABC三邊長為3-4-5 \Rightarrow \angle ACB=90^\circ, 假設\angle BCR=\theta \Rightarrow \cases{\angle RBC=120^\circ-\theta\\ \angle ACP= 90^\circ-\theta\\ \angle CAP= 30^\circ+\theta}\\ 正弦定理: \cases{\triangle RBC: \displaystyle {\overline{RC} \over \sin  (120^\circ-\theta)} ={4\over \sin 60^\circ } \\ \triangle ACP: \displaystyle {\overline{CP} \over \sin (30^\circ+\theta)} ={3\over \sin 60^\circ}} \Rightarrow \cases{\overline{RC}=8\sin(120^\circ-\theta)/\sqrt 3\\ \overline{CP} =6\sin(30^\circ+\theta)/\sqrt 3}\\ \Rightarrow 正\triangle PQR邊長L=\overline{RP}= \overline{RC}+ \overline{CP} ={1\over \sqrt 3} ((4\sqrt 3+3)\cos \theta+ (3\sqrt 3+4)\sin \theta) \\ \Rightarrow L的最大值 L_{max}={1\over \sqrt 3} \sqrt{(4\sqrt 3+3)^2 +(3\sqrt 3+4)^2} ={1\over \sqrt 3} \sqrt{100+48\sqrt 3} \\ \Rightarrow \triangle PQR最大面積={\sqrt 3\over 4}L_{max}^2= {\sqrt 3\over 4}\cdot {1\over 3}(100+48\sqrt 3) =\bbox[red, 2pt]{{25\over 3}\sqrt 3+12}$$

 

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解題僅供參考，其他教甄試題及詳解