115年北一女教甄-數學詳解

 臺北市立第一女子高級中學 115 學年度第一次正式教師甄選

一、填充題 (每格 8 分, 共 72 分)

解答:$$f(x)=a\sin x+b\cos x+c = \sqrt{a^2+b^2} \sin(x+\theta)+c \Rightarrow \cases{\max(f) =\sqrt{a^2+b^2}+c=7\\ \min(f)=-\sqrt{a^2+b^2}+c=3} \\ \Rightarrow \cases{\sqrt{a^2+b^2}=2\\c=5} \Rightarrow \max(f) =f(\pi/3) =2\sin(\pi/3+\theta)+5 \Rightarrow {\pi\over 3}+\theta={\pi\over 2} \Rightarrow \theta={\pi\over 6} \\ \Rightarrow 2\sin(x+ \pi/6)+5= 2({\sqrt 3\over 2}\sin x+{1\over 2}\cos x) +5=\sqrt 3\sin x+\cos x+5 \Rightarrow (a,b,c)=\bbox[red, 2pt]{(\sqrt 3,1,5)}$$

解答:$$\sqrt{4n^2-10}+5n\lt a_n\lt 7n+10 \Rightarrow \sqrt{4-{10\over n^2}}+5\lt {a_n\over n}\lt 7+{10\over n} \\ \cases{\displaystyle \lim_{n\to \infty}\sqrt{4-{10\over n^2}}+5 =7\\ \displaystyle \lim_{n\to \infty} 7+{10\over n} =7} \Rightarrow \lim_{n\to \infty}{a_n\over n}=7 \quad (夾擠定理) \\ 因此\lim_{n\to \infty} {(a_n+3n)^2 \over na_n+3n^2-8} = \lim_{n\to \infty} {((a_n/n)+3)^2 \over (a_n/n)+3-(8/n^2)} = {(7+3)^2\over 7+3} =\bbox[red, 2pt]{10}$$

解答:$$全部的放法（不考慮黑格限制）:5! = 120 種。\\至少有 1 枚棋子放在黑格:\\\qquad 從 5 個黑格選 1 個放棋子，剩下的 4 行 4 列隨意放 ,有 \binom{5}{1} \times 4! = 5 \times 24 = 120 種。\\至少有 2 枚棋子放在黑格:\\\qquad 從 5 個黑格選 2 個放棋子，剩下的 3 行 3 列隨意放 有 \binom{5}{2} \times 3! = 10 \times 6 = 60 種。\\至少有 3 枚棋子放在黑格:\\\qquad 從 5 個黑格選 3 個放棋子，剩下的 2 行 2 列隨意放 有 \binom{5}{3} \times 2! = 10 \times 2 = 20 種。\\至少有 4 枚棋子放在黑格:\\\qquad 從 5 個黑格選 4 個放棋子，剩下的 1 行 1 列隨意放 有 \binom{5}{4} \times 1! = 5 \times 1 = 5 種。\\ 5枚棋子全在黑格:只有1種\\ 排容原理: 120-120+60-20+5-1= \bbox[red, 2pt]{44}$$

解答:$$z=x+ki \Rightarrow z^6=64i =64 \left( \cos{\pi\over 2}+\sin{\pi\over 2} \right) \\\Rightarrow z_n= \sqrt[6]{64} \left[\cos \left( {\pi/2+2n\pi\over 6} \right)+ i\sin \left( {\pi/2+2n\pi\over 6} \right)  \right] \\\qquad = 2 \left[\cos \left( {(4n+1)\pi\over 12} \right)+ i\sin \left( {(4n+1)\pi\over 12} \right)  \right], n=0,1,2,3,4,5 \\ \Rightarrow k=2 \sin \left( {(4n+1)\pi\over 12} \right) \gt 0 \Rightarrow \cases{n=0 \Rightarrow k_0= 2\sin 15^\circ= (\sqrt 6-\sqrt 2)/2 \\ n=1 \Rightarrow k_1=2\sin 75^\circ =(\sqrt 6+\sqrt 2)/2 \\ n=2 \Rightarrow k_2=2 \sin 135^\circ =\sqrt 2 \\ n=3,4,5 \Rightarrow k\lt 0} \\ \Rightarrow k_0+k_1+k_2 = \bbox[red, 2pt]{\sqrt 6+\sqrt 2}$$

解答:$$\cases{a_n^5 =5b_n^5 \Rightarrow \displaystyle {a_n\over b_n} = \sqrt[5]5 \\b_k=b_3\cdot r^{k-3}=b_3\cdot \displaystyle {1\over 5^{k-3}}} \Rightarrow {a_3\over b_k} = {a_3\over b_3\cdot {1\over 5^{k-3}}} ={a_3\over b_3}\cdot 5^{k-3} =\sqrt[5] 5 \cdot 5^{k-3} =5^{k-14/5} \\\Rightarrow \log{a_3\over b_k}=(k-{14\over 5})\log 5 \Rightarrow \sum_{k=1}^{10}\log{a_3\over b_k} = \sum_{k=1}^{10}(k-{14\over 5})\log 5 =(55-28)\log 5=\bbox[red, 2pt]{27\log 5}$$

解答:

$$假設\angle CAD=\theta \Rightarrow \angle BAD=2\theta \Rightarrow \cos \theta={3^2+4^2-2^2 \over 2\cdot 3\cdot 4}={7\over 8} \\ \cases{\displaystyle {\triangle ABD\over \triangle ADC}={\overline{BD} \over CD} ={\overline{BD} \over 2} \\ \displaystyle {\triangle ABD\over \triangle ADC} = {\overline{AB}\cdot \overline{AD}\sin 2\theta \over \overline{AD} \cdot \overline{AC}\sin \theta}  ={2\overline{AB} \cos\theta\over \overline{AC}}={\overline{AB}\cos \theta\over 2} ={\overline{AB}\cdot (7/8) \over 2} ={7\over 16}\overline{AB}} \\ \Rightarrow {\overline{BD} \over 2}={7\over 16}\overline{AB} \Rightarrow \overline{AB}={8\over 7}\overline{BD} \\ \cos \angle ADC={3^2+2^2-4^2 \over 2\cdot 3\cdot 2}=-{1\over 4} \Rightarrow \cos \angle ADB={1\over 4} ={{\overline{BD}^2+3^2-\overline{AB}^2 \over 2\cdot 3\cdot \overline{BD}}} \\= {{\overline{BD}^2+9-{64\over 49}\overline{BD}^2 \over 6\cdot \overline{BD}}} \Rightarrow {10\over 49} \overline{BD}^2+ \overline{BD} -6=0 \Rightarrow 10\overline{BD}^2 +49\overline{BD} -294=0 \\ \Rightarrow (2\overline{BD}-7)(5\overline{BD}+42) =0 \Rightarrow \overline{BD}= \bbox[red, 2pt]{7\over 2}$$

解答:

$$假設\cases{A(0,20)\\ B(26,0) \\C(0,0)} \Rightarrow 圓\Gamma_1 \cases{圓心O_1= \overline{AB}中點=(13,10) \\ 圓半徑R=\overline{AB}/2= \sqrt{269}} \\ 圓\Gamma_2與x軸及y軸均相切 \Rightarrow \cases{圓心O_2=(r,r)\\ 圓半徑r},又\Gamma_1與\Gamma_2 相切\Rightarrow \overline{O_1O_2} =R-r \\ \Rightarrow (r-13)^2+(r-10)^2= (\sqrt{269}-r)^2 \Rightarrow r^2-46r+2\sqrt{269}r=0 \Rightarrow r(r-46+2\sqrt{269}=0) \\ \Rightarrow r= \bbox[red, 2pt]{46-2\sqrt{269}}$$

解答:$$假設\cases{\vec a= \overrightarrow{OA} \\\vec b= \overrightarrow{OB} \\\vec c= \overrightarrow{OC} } \Rightarrow \cases{\vec a\cdot \vec b=3\cdot 4\cdot (\sqrt 3/2)=6\sqrt 3\\ \vec b\cdot \vec c= 4\cdot 5\cdot (\sqrt 3/2)=10\sqrt 3\\ \vec c\cdot \vec a=5\cdot 3\cdot (\sqrt 2/2)= 15\sqrt 3/2} \\ 假設平面六面體的體積為V \Rightarrow V^2= \begin{vmatrix} \vec a\cdot \vec a& \vec a\cdot \vec b& \vec a\cdot \vec c \\ \vec b\cdot \vec a& \vec b\cdot \vec b& \vec b\cdot \vec c \\\vec c\cdot \vec a& \vec c\cdot \vec b& \vec c\cdot \vec c \\\end{vmatrix} = \begin{vmatrix} 9& 6\sqrt 3& 15\sqrt 3/2\\ 6\sqrt 3& 16& 10\sqrt 3\\ 15\sqrt 3/2& 10\sqrt 3& 25 \end{vmatrix} \\=900(3\sqrt 3-5)   \Rightarrow V=30\sqrt{3\sqrt 3-5} \Rightarrow 四面體體積={1\over 6}V = \bbox[red, 2pt]{5\sqrt{3\sqrt 3-5}}$$

解答:$$(2-a_{n+1})(4+a_n)=8 \Rightarrow 2a_n-4a_{n+1}-a_{n+1}a_n=0 \Rightarrow {2\over a_{n+1}} -{4\over a_n}-1=0 \\\Rightarrow {1\over a_{n+1}} =2\cdot {1\over a_n}+{1\over 2} \Rightarrow b_{n+1}=2b_n+{1\over 2} , 其中b_n={1\over a_n} \Rightarrow b_{n+1}+{1\over 2}=2 \left( b_n+{1\over 2} \right) \\ \Rightarrow \langle b_n+{1\over 2}\rangle 為一公比為2的等比數列 \Rightarrow b_1+{1\over 2}={1\over a_1}+{1\over 2}={5\over 2} \Rightarrow b_n+{1\over 2}={5\over 2}\cdot 2^{n-1} \\ \Rightarrow {1\over a_n}=b_n ={5\over 2}\cdot 2^{n-1}-{1\over 2 } \Rightarrow \sum_{k=1}^{n} {1\over a_k} =\sum_{k=1}^{n} \left( {5\over 2}\cdot 2^{k-1}-{1\over 2 } \right) ={5\over 2}\sum_{k=1}^{n} 2^{k-1}-\sum_{k=1}^{n}{1\over 2} \\ ={5\over 2}(2^n-1)-{n\over 2}= \bbox[red, 2pt]{{1\over 2}(5\cdot 2^n-n-5)}$$

二、計算證明題 (共 75 分, 需寫出計算過程, 否則不予計分)

解答:$$\textbf{(1) }最主要的錯誤在於等號兩邊皆為非負數才可以這樣計算 ,而2x-1可能為負值,不適用此方法 \\\textbf{(2) } 方法一: 兩邊平方求解；\\方法二:分段討論, 依|x-1|, |2x-5|正負值拆成x\le 1, 1\le x\lt 5/2, x\ge 5/2 分段計算\\方法三:取\cases{y=f(x)=|x-1|\\ y=g(x)= |2x-5|}, 畫出兩圖形,找出f(x)\ge g(x)的解$$

解答:$$\textbf{高一方法:} \cases{A(1,-1) \\B(3,2) \\ P\in \Gamma:y={1\over 2}x^2+2x+3} \Rightarrow \cases{L=\overleftrightarrow{AB}: 3x-2y=5\\ P(t,{1\over 2}t^2+2t+3)} \Rightarrow d(P,L)= {|-t^2-t-11|\over \sqrt{13}} \\\qquad \Rightarrow t=-{1\over 2}時, d(P,L)有最小值{43\over 4\sqrt{13}} \Rightarrow \triangle ABC ={1\over 2}\cdot \overline{AB}\cdot {43\over 4\sqrt{13}}={43\over 8} \\\textbf{高二方法:}距離最短的點 P，必然是與\overline{AB} 平行的拋物線切線之切點 \\ 平行\overline{AB}的直線L_p:3x-2y+k=0 \Rightarrow y={3\over 2}x+{k\over 2}代入\Gamma \Rightarrow x^2+x+(6-k)=0 \\ \Rightarrow 判別式:1-4(6-k)=0 \Rightarrow k={23\over 4} \Rightarrow d(L,L_p)={43\over 4\sqrt{13}} \Rightarrow \triangle ABC={43\over 8} \\\textbf{高三方法:}y={1\over 2}x^2+2x+3 \Rightarrow y'=x+2 ={3\over 2} \Rightarrow x=-{1\over 2} \Rightarrow P(-{1\over 2},{17\over 8}) \\ \Rightarrow \triangle ABC={1\over 2} \begin{vmatrix} 1&-1& 1\\ 3& 2& 1\\ -{1\over 2}& {17\over 8}& 1 \end{vmatrix} ={43\over 8}$$

解答:$$\textbf{(1) } 小綠的方法僅適用在平面上\\\textbf{(2) }P\in L \Rightarrow  P=(1+2t,t,3+2t) \Rightarrow f(t)= \overline{PA}+ \overline{PB} = \sqrt{9t^2+104}+ \sqrt{9t^2+54t+497} \\=\sqrt{(3t-0)^2+(\sqrt{104})^2}+ \sqrt{(3t+9)^2+ (\sqrt{416})^2} \\ 取u=3t \Rightarrow f(t)= g(u)= \sqrt{u^2+(\sqrt{104})^2} + \sqrt{(u+9)^2 +(\sqrt{416})^2}\\ \Rightarrow g(u) = \overline{QM}+ \overline{QN}, 其中\cases{Q(u,0) \\M(0,\sqrt{104}) \\N(-9,-\sqrt{416})} \Rightarrow M,Q,N在一直線上有最小值=\overline{MN} = \bbox[red, 2pt]{3\sqrt{113}} \\ \Rightarrow {0-\sqrt{104}\over u-0} ={-\sqrt{416}-\sqrt{104}\over -9-0} \Rightarrow u=-3 \Rightarrow t=-1\Rightarrow \bbox[red, 2pt]{P(-1,-1,1)}$$

解答:$$y=f(x)=k-x^2與x軸的交點在x=\sqrt k \quad(在區間[0,2]內) \\\Rightarrow 面積A(k)= \int_0^{\sqrt k}(k-x^2)\,dx + \int_{\sqrt k}^2 (x^2-k)\,dx ={2\over 3}{k\sqrt k} + \left( {8\over 3}-2k+{2\over 3}k\sqrt k \right) \\ \Rightarrow A(k)={4\over 3}k\sqrt k-2k+{8\over 3} \Rightarrow A'(k)=2\sqrt k-2=0 \Rightarrow k=1 \Rightarrow a=1 \Rightarrow b=A(1)=2 \\ \Rightarrow (a,b)= \bbox[red, 2pt]{(1,2)}$$

解答:$$\textbf{(1) }令\cases{b_k\in \{0,1\} \\ a_k=2b_k-1} \Rightarrow \cases{b_k=1 \Rightarrow a_k=1\\ b_k=0\Rightarrow a_k=-1} \Rightarrow 2026=S_{11} =\sum_{k=1}^{11}(2b_k-1)2^k = \sum_{k=1}^{11}b_k2^{k+1} -\sum_{k=1}^{11}2^k \\=\sum_{k =1}^{11}b_k2^{k+1} -4094 \Rightarrow \sum_{k= 1}^{11}b_k2^{k +1}= 2026 +4094 =6120 \Rightarrow \sum_{k=1}^{11}b_k2^{k-1}=1530\\ 1530的二進位表示:01011111101 \Rightarrow \cases{b_1=b_3=b_{10}=0\\ b_k=1, k\ne 1,3,10} \Rightarrow \bbox[red, 2pt]{\cases{a_1=a_3=a_{10}=-1\\ a_k=1,k\ne 1,3,10}} \\\textbf{(2) }簡單來說，二進的表示式(b_n)是唯一的，因此a_k也是唯一的$$

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解題僅供參考，其他教甄試題及詳解