朱式幸福: 113年西松高中教甄-數學詳解

臺北市立西松高級中學 113 學年度高中部第 1 次正式教師甄選

【數學科（ IB）】初試試題
一、填充題：（配分如後標有(*)： 4 分；未標示： 6 分。共 58 分。）

解答：$$\textbf{(1)}\; \cases{\Gamma_1:(x-1)^2=4(y-1) \\ \Gamma_2: (y+2)^2=-4(x+2)} \Rightarrow \cases{\Gamma_1的頂點為A(1,1),圖形凹向上，對稱軸為垂直線\\ \quad \Gamma_2的頂點為B(-2,-2),凹向左，對稱軸為水平線} \\ \Rightarrow 兩圖形對稱直線L:x+y=k，且L線經過A,B中點C=(A+B)/2 =(-1/2,-1/2) \\ \quad \Rightarrow \bbox[red, 2pt]{L: x+y=-1} \\ \textbf{(2)}\; \cases{\Gamma_1: 2x-2-4y'=0 \Rightarrow y'= {1\over 2}x-{1\over 2} \\ \Gamma_2:2yy'+4+4y'=0 \Rightarrow y' =-{2\over y+2}} \Rightarrow \cases{\Gamma_1: y'=-1 \Rightarrow x=-1 \Rightarrow y=2 \\ \Gamma_2: y'=-1 \Rightarrow y=0 \Rightarrow x=-3 } \\\quad \Rightarrow \cases{P(-1,2)\\ Q(-3,0) } \Rightarrow \overline{PQ}= \bbox[red, 2pt]{2\sqrt 2}$$

解答：$$\textbf{(1)}\; 三個馬綁在一起，與二個是形成四個空間；再將三個小插任這四個相鄰空間中，\\\quad 有C^4_3種插入法，而三個綁在一起的馬與二個星有三種排法，因此共有C^4_3\times 3=\bbox[red, 2pt]{12}種排法 \\\textbf{(2)}\; A:全部任排數=8!/(3!3!2!)=560\\
B:馬相鄰=全-馬不相鄰=560-5!/(3!2!)C(6,3)=560-200=360\\
C:小相鄰=馬相鄰=360\\
D:是相鄰=全-是不相鄰=560-6!/(3!3!)C(7,2)=560-420=140\\
E:馬相鄰且小相鄰=6!/2-5!/2-5!/2+4!/2=360-60-60+12=252\\
F:馬相鄰且是相鄰=6!/3!-5!/3!=100\\
G:小相鄰且是相鄰=馬相鄰且是相鄰=100\\
H:馬相鄰且小相鄰且是相鄰 =5!-4!-4!+3!=78\\欲求之排法=A-(B+C+D)+(E+F+G)-H\\\qquad =560-360-360-140+252+100+100-78 =\bbox[red, 2pt]{74}$$

解答：$$\textbf{(1)}\; \cases{\alpha=2+\sqrt 5\\ \beta=2-\sqrt 5} \Rightarrow \cases{\alpha+\beta=4\\ \alpha\beta=-1} \Rightarrow \cases{\alpha^{3}+ \beta^{3} =(\alpha+\beta)^3-3\alpha\beta (\alpha+ \beta) =4^4+3\cdot 4=268 \\ \alpha^2+\beta^2 =(\alpha+\beta)^2 -2\alpha\beta =4^2+2=18}\\ \Rightarrow \alpha^5+\beta^5= (\alpha^3+\beta^3)(\alpha^2+ \beta^2) -\alpha^2\beta^2(\alpha +\beta) = 268 \cdot 18 -4=4820 \\ \Rightarrow \alpha^{10}+ \beta^{10} =(\alpha^{5}+ \beta^{5})^2 -2\alpha^{5} \beta^{5} = 4820^2+2\\ \quad a=\alpha^{20} +\beta^{20} =(\alpha^{10}+ \beta^{10})^2-2\alpha^{10} \beta^{10} = (4820^2+2)^2-2個位數字=2^2-2=\bbox[red, 2pt]2 \\\textbf{(2)}\; a=(4820^2+2)^2-2末兩位數字=02, 由於 0\lt (2-\sqrt 5)^{20} \lt 1, \\因此b的末兩位數字為02-01=\bbox[red, 2pt]{01}$$

解答：$$旋轉矩陣A=\begin{bmatrix} \cos{\pi\over 3} & -\sin {\pi\over 3}\\ \sin {\pi\over 3} & \cos {\pi\over 3}\end{bmatrix} =\begin{bmatrix} 1/2 & -\sqrt 3/2\\ \sqrt 3/2 & 1/2\end{bmatrix}\\ 假設m= \tan \theta \Rightarrow 鏡射矩陣B= \begin{bmatrix} \cos 2\theta & \sin 2\theta\\ \sin 2\theta & -\cos2\theta\end{bmatrix} \\ \Rightarrow BA= \begin{bmatrix} \cos(2\theta-\pi/3) & \sin (2\theta-\pi/3)\\ \sin (2\theta-\pi/3) & -\cos(2\theta-\pi/3)\end{bmatrix} =\begin{bmatrix}\cos 2\alpha & \sin 2\alpha\\ \sin 2\alpha& -\cos 2\alpha \end{bmatrix} \Rightarrow \alpha =\theta-{\pi\over 6} \\ \Rightarrow m'=\tan \alpha=\tan (\theta-{\pi\over 6}) ={\tan \theta-\tan{\pi\over 6} \over 1+\tan \theta\cdot \tan{\pi\over 6}} ={m-{1\over \sqrt 3}\over 1+ {m\over \sqrt 3}} =\bbox[red, 2pt]{\sqrt 3m-1\over m+\sqrt 3}$$

解答：$$a_n=(-1)^n{n^2+n+1 \over 3^n} \Rightarrow a_{n-1}=(-1)^{n-1}{n^2-n+1\over 3^{n-1}} \Rightarrow {1\over 3}a_{n-1}= (-1)^{n-1}{n^2-n+1\over 3^{n}} \\ \Rightarrow a_n+{1\over 3}a_{n-1} =(-1)^n{2n\over 3^n} \\ S=\sum_{n=1}^\infty a_n \Rightarrow S+{1\over 3}S={4\over 3}S =-1+\sum_{n=2}^\infty (-1)^n{ 2n\over 3^n} \Rightarrow T= {4\over 3}S+1=\sum_{n=2}^\infty (-1)^n{ 2n\over 3^n} \\ \Rightarrow T+{1\over 3}T ={4\over 3}T ={4\over 9}+ 2\sum_{n=3}^\infty (-1)^n {1\over 3^n} ={4\over 9}+2 \cdot (-{1\over 36}) ={7\over 18} \\ \Rightarrow T={21\over 72} \Rightarrow S=(T-1)\times{3\over 4} = \bbox[red, 2pt]{-{17\over 32}}$$

解答：

$$假設\cases{\overline{BD}=\overline{AC}=a \\ \overline{AD}=b\\ \angle BAD=\theta},則\angle ADC=180^\circ-60^\circ-40^\circ=80^\circ \Rightarrow \angle B=80^\circ-\theta \\ 正弦定理:\cases{\triangle ADC:\displaystyle {a\over \sin 80^\circ} ={b\over \sin 40^\circ} \\ \triangle ABD: \displaystyle {a\over \sin \theta} = {b\over \sin (80^\circ-\theta)}} \Rightarrow a={b\sin 80^\circ\over \sin 40^\circ} ={b\sin \theta\over \sin(80^\circ-\theta)}\\ \Rightarrow 2\cos 40^\circ={\sin \theta\over \sin(80^\circ-\theta)} \Rightarrow \sin \theta=2\cos 40^\circ \sin(80^\circ-\theta) =\sin(120^\circ-\theta) + \sin(40^\circ-\theta) \\\Rightarrow \sin\theta+\sin(\theta-120^\circ) =\sin(40^\circ-\theta) \Rightarrow 2\sin(\theta-60^\circ) \cos 60^\circ=\sin(40^\circ-\theta)\\ \Rightarrow \sin(\theta-60^\circ) =\sin(40^\circ-\theta) \Rightarrow \theta=50^\circ = \bbox[red, 2pt]{5\pi\over 18}$$

解答：

$$假設\overline{AC}與\overline{BD}交於P,並令\overrightarrow{AP} =t\overrightarrow{AC} ={3t\over 2}\overrightarrow{AB} +{5t\over 2}\overrightarrow{AD} \Rightarrow{3t\over 2}+{5t\over 2}=1 \Rightarrow t={1\over 4}\\ \Rightarrow \overrightarrow{AP} ={3\over 8} \overrightarrow{AB} +{5\over 8}\overrightarrow{AD} \Rightarrow \cases{\overline{DP}=3k\\ \overline{PB}= 5k}\\ 同理, \overrightarrow{BD} =\overrightarrow{BA} +\overrightarrow{AD} =\overrightarrow{BA}+{2\over 5}(\overrightarrow{AC} -{3\over 2}\overrightarrow{AB}) ={8\over 5} \overrightarrow{BA} +{2\over 5}\overrightarrow{AC} ={6\over 5}\overrightarrow{BA} +{2\over 5}\overrightarrow{BC} \\ 假設\overrightarrow{BP} =t\overrightarrow{BD} ={6t\over 5}\overrightarrow{BA} +{2t\over 5}\overrightarrow{BC} \Rightarrow t={5\over 8} \Rightarrow \overrightarrow{BP} ={3\over 4} \overrightarrow{BA} + {1\over 4} \overrightarrow{BC} \Rightarrow \cases{\overline{AP}=m\\ \overline{PC}=3m}\\ 圓冪性質: \overline{AP}\cdot \overline{PC}= \overline{BP} \cdot \overline{PD} \Rightarrow 3m^2=15k^2 \Rightarrow {k\over m}={1\over \sqrt 5}\\ 正弦定理: {\overline{BD} \over \sin \angle DAB} =2R = {\overline{AC} \over \sin \angle ABC} \Rightarrow {\sin \angle DAB \over \sin \angle ABC} = {\overline{AC}\over \overline{BD}} ={8k\over 4m} ={2\over \sqrt 5} \\ \Rightarrow \sin \angle DAB: \sin \angle ABC= \bbox[red, 2pt]{2: \sqrt 5}$$

解答：$$I=\int_{-1}^1 {x^2\over 1+2^x} \,dx =\int_{-1}^1{ u^2\over 1+2^{-u}}du \;(取u=-x) =\int_{-1}^1 {u^2 2^u\over 1+2^u}du =\int_{-1}^1 {x^2 2^x\over 1+2^x}dx \\ \Rightarrow I+I= \int_{-1}^1 {x^2(2^x+1)\over 1+2^x}\,dx \Rightarrow 2I= \int_{-1}^1 x^2\,dx =2\int_0^1 x^2\,dx ={2\over 3} \Rightarrow I= \bbox[red, 2pt]{1\over 3}$$

二、計算證明題：（共 42 分）

解答：

$$\textbf{(1)}\; {\overline{BE} \over \overline{EC}} ={1\over 2} \Rightarrow \overrightarrow{AE}={1\over 3}\overrightarrow{AC}+{2\over 3} \overrightarrow{AB} \Rightarrow {1\over 3} \cdot 3\overrightarrow{AD}+{2\over 3} \overrightarrow{AB} = \overrightarrow{AD}+{2\over 3} \overrightarrow{AB} \\ 假設\overrightarrow{AP} = t\overrightarrow{AE} = t\overrightarrow{AD}+ {2t\over 3} \overrightarrow{AB} \Rightarrow t+{2t\over 3} =1 \Rightarrow t={3\over 5} \Rightarrow \overrightarrow{AP}= {3\over 5} \overrightarrow{AE} \\ \Rightarrow \overrightarrow{PE}={2\over 3} \overrightarrow{AE} \Rightarrow \overline{AP}: \overline{PE}= \bbox[red, 2pt]{3:2} \\\textbf{(2)}\;{\overline{AD} \over \overline{DC}}={1\over 2} \Rightarrow \overrightarrow{BD}={2\over 3} \overrightarrow{BA}+ {1\over 3} \overrightarrow{BC} \\ 假設\cases{\overrightarrow{BP} =t\overrightarrow{BD}={2t\over 3} \overrightarrow{BA}+ {t\over 3} \overrightarrow{BC}={2t\over 3} \overrightarrow{BA}+ t \overrightarrow{BE}\\ \overrightarrow{BQ} =k \overrightarrow{BD} ={2k\over 3} \overrightarrow{BA}+ {k\over 3} \overrightarrow{BC} ={2k\over 3} \overrightarrow{BA}+ {k\over 2} \overrightarrow{BF}} \Rightarrow \cases{ {2t\over 3}+ t=1\\ {2k\over 3}+{k\over 2}=1} \\ \Rightarrow \cases{t=3/5\\ k=6/7} \Rightarrow \overline{BP} :\overline{PQ} :\overline{QD} :{3\over 5}:{6\over 7}-{3\over 5}:1-{6\over 7}=\bbox[red, 2pt]{21:9:5} \\\textbf{(3)}\; \cases{\triangle BPE={1\over 2}\overline{PB}\cdot \overline{PE} \sin \angle BPE\\ \triangle APQ={1\over 2} \overline{PA} \cdot \overline{PQ} \sin \angle APQ} \Rightarrow {\triangle PBE \over \triangle APQ}={\overline{PB}\cdot \overline{PE} \over \overline{PA}\cdot \overline{PQ}}={2\over 3}\cdot {21\over 9}={14\over 9} \\ \qquad \Rightarrow \triangle PBE : \triangle APQ=\bbox[red, 2pt]{14:9}$$

解答：$$\textbf{(1)}\;y={2\over x+1}+3 \Rightarrow \cases{貫軸方程式：y=x+4\\垂直漸近線：x=-1\\ 水平漸近線：y=3} \Rightarrow 頂點坐標\cases{A(\sqrt 2-1,\sqrt 2+3)\\ B(-\sqrt 2-1, -\sqrt 2+3) }\\\Rightarrow \cases{中心點O(-1,3）\\ a=b=2} \Rightarrow c=2\sqrt 2 \Rightarrow 第1象限焦點坐標O+(2,2)=\bbox[red, 2pt]{(1,5)} $$