114年北一女教甄-數學詳解

臺北市立第一女子高級中學 114 學年度第一次正式教師甄選

一、填充題(每格 7 分，共 56 分)

解答:$$\cases{\displaystyle \sum_{k=1}^{44} [\log_{45} k] =44\times 0=0\\\displaystyle \sum_{k=45}^{2024} [\log_{45} k] =(2024-25+1)\cdot 1 =1980\\ \displaystyle \sum_{k=2025}^{n} [\log_{45} k] =(n-2025+1)\cdot 2=2n-4048} \Rightarrow 1980+2n-4048=2000 \Rightarrow n= \bbox[red, 2pt]{2034}$$

解答:$$B=A逆時針旋轉60^\circ \Rightarrow \begin{bmatrix}b\\ 25 \end{bmatrix} = \begin{bmatrix}1/2& -\sqrt 3/2\\ \sqrt 3/2& 1/2 \end{bmatrix} \begin{bmatrix}a\\ 20 \end{bmatrix} = \begin{bmatrix}a/2-10\sqrt 3\\ \sqrt 3 a/2+10 \end{bmatrix} \\ \Rightarrow \cases{a/2-10\sqrt 3=b\\ \sqrt 3a/2+10=25} \Rightarrow \cases{a=10\sqrt 3\\ b=-5\sqrt 3} \Rightarrow ab= \bbox[red, 2pt]{-150}$$

解答:$$\cases{a\in S_A=\{2,3,4,5,6,7\} \\ b\in S_B =\{2,3,4,5,6,7,8,9\}} \Rightarrow (a,b)共有6\times 8=48種組合\\ \log_a b= n \in \mathbb N \Rightarrow \cases{n=1\Rightarrow a=b \Rightarrow (a,b) =(2,2),(3,3), \dots,(7,7),共6組\\ n=2 \Rightarrow b=a^2 \Rightarrow (2,4),(3,9),共2組\\ n=3 \Rightarrow b=a^3 \Rightarrow (a,b)=(2,8), 只有1組} \\ \Rightarrow 合計6+2+1=9組 \Rightarrow 機率p={9\over 48} ={3\over 16}\\ 首次滿足條件所需的投擲次數 X符合幾何分佈的定義 \Rightarrow E(X)=1/p= \bbox[red, 2pt]{16\over 3}$$

解答:

$$\overline{AD} \bot 平面PQR \Rightarrow \overrightarrow{AD} \bot \overrightarrow {RP} \Rightarrow \overrightarrow{AD} \cdot \overrightarrow {RP} =0 \Rightarrow \overrightarrow{AD} \cdot (\overrightarrow {AP}- \overrightarrow{AR}) =0 \Rightarrow \overrightarrow{AD} \cdot \overrightarrow{AP} =\overrightarrow{AD} \cdot \overrightarrow{AR} \\ \Rightarrow \overline{AD}\cdot \overline{AP} \cos 60^\circ =\overline{AD}\cdot \overline{AR} \cos 0^\circ (正四面體任兩稜夾角60^\circ) \Rightarrow \overline{AP}=4\\ 同理, \overrightarrow{AD} \cdot \overrightarrow{RQ} =0 \Rightarrow \overrightarrow{AD} \cdot \overrightarrow{AQ} =\overrightarrow{AD}\cdot \overrightarrow{AR} \Rightarrow 5\cdot \overline{AQ}\cdot {1\over 2}=5\cdot 2 \Rightarrow \overline{AQ}=4 \\ \Rightarrow {四面體A-PQR體積\over 四面體A-BCD體積} = {\overline{AP}\times \overline{AQ}\times \overline{AR} \over \overline{AB}\times \overline{AC}\times \overline{AD}} ={4\cdot 4\cdot 2\over 5\cdot 5\cdot 5}={32\over 125} \\ \Rightarrow 五面體PQR-BCD占四面體體積的1-{32\over 125} ={93\over 125} \\ \Rightarrow 五面體PQR-BCD體積={93\over 125} \cdot {1\over 6\sqrt 2}\cdot 5^3= \bbox[red, 2pt]{31\sqrt 2\over 4}$$

解答:$$\lim_{n\to \infty} \sum_{k=1}^n {n\over 2n^2+3kn+k^2} =\lim_{n\to \infty} \sum_{k=1}^n {1/n\over 2+3(k/n)+(k/n)^2} =\int_0^1 {1\over 2+3x+x^2}\,dx \\= \int_0^1 {1\over (x+2)(x+1)}\,dx=\int_0^1 \left( {1\over x+1}-{1\over x+2}  \right)\,dx = \left. \left[ \ln (x+1)-\ln(x+2) \right] \right|_0^1 \\=(\ln 2-\ln 3)-(\ln 1-\ln 2) = \bbox[red, 2pt]{2\ln 2-\ln 3}$$

解答:$$假設\cases{S_0:兩個寶箱都關閉\\ S_1:一個寶箱開啟, 一個關閉\\ S_2:兩個寶箱都開啟} \Rightarrow \cases{P(S_0\to S_0) =(2/3)\cdot (2/3)=4/9\\ P(S_0\to S_1) =2\cdot (2/3)\cdot (1/3)=4/9 \\ P(S_0\to S_2)= (1/3)\cdot (1/3)=1/9\\ P(S_1\to S_1) =2/3\\ P(S_1\to S_2)=1/3} \\假設\cases{E_1= S_1\to S_2天數的期望值 \\ 欲求的E=S_0\to S_2 天數的期望值}\Rightarrow \cases{E_1= P(S_1\to S_1)E_1+1 \\E=P(S_0\to S_0)E+ P(S_0\to S_1)E_1+1} \\\Rightarrow \cases{E_1=(2/3)E_1+1 \\ E=(4/9)E+(4/9)E_1+1} \Rightarrow E_1=3 \Rightarrow E= \bbox[red, 2pt]{21\over 5}$$

 解答:

$$假設\overleftrightarrow{BC}與\overline{DE}交於F, 則ABFD為平行四邊形\Rightarrow \cases{\overline{DF}  =\overline{AB}=4 \\ \overline{BF}= \overline{AD}=8}\\ 圓冪性質:\overline{DF}^2= \overline{FC}\cdot \overline{FB} \Rightarrow 4^2=\overline{FC}\cdot 8 \Rightarrow \overline{CF}=2 \Rightarrow \overline{BC} =6 \\對同弧\stackrel{\Large \frown}{CD} 的圓切角\angle CDE 與圓周角\angle CAD相等,即\angle CDE=\angle CAD=\beta \\ 又\overline{AD} \parallel \overline{BC} \Rightarrow \angle ACB =\angle CAD=\beta \Rightarrow \stackrel{\Large \frown}{CD}=\stackrel{\Large \frown}{AB} \Rightarrow \overline{CD}=\overline{AB}=4 \\ 由於\cases{\angle ACB=\angle CDE=\beta \\\angle BAC=\angle AED =\alpha ( \because\overline{DE} \parallel \overline{AB})} \Rightarrow \triangle ABC \sim \triangle ECD (AAA) \\ \Rightarrow {\overline{AB} \over \overline{BC}} ={\overline{CE} \over \overline{CD}} \Rightarrow {4\over 6} ={\overline{CE} \over 4} \Rightarrow \overline{CE}= \bbox[red, 2pt]{8\over 3}$$

解答:$$z^2-2z+5=0 \Rightarrow z=1\pm 2i \Rightarrow \cases{A(1,2) \\ B(1,-2)} \\ z^2-2az+1=0 \Rightarrow 判別式:4a^2-4\ge 0 (\because a\in \mathbb N) \Rightarrow 有相異二實根 z=a\pm \sqrt{a^2-1}\\ \Rightarrow \cases{C(a+\sqrt{a^2-1},0) \\D(a-\sqrt{a^2-1},0)} \Rightarrow \overline{CD}為圓直徑 \Rightarrow \cases{圓心O(a,0) \\ 半徑r=\sqrt{a^2-1}} \\ \Rightarrow \overline{OA}=r \Rightarrow (a-1)^2+2^2=a^2-1 \Rightarrow a=\bbox[red, 2pt]3$$

二、計算證明題(共 74 分，須寫出計算過程，否則不予計分) 

解答:$$小綠推論: A^2=I^2 \Rightarrow (A-I)(A+I)=0 \Rightarrow A=I 或-I\\ \qquad 兩矩陣相乘為0不代表矩陣為0, 例: A= \begin{bmatrix}1&0\\ 0&-1 \end{bmatrix} \Rightarrow A^2=I, 但A\ne I, A\ne -I \\ 小青推論的錯誤與小綠一樣,以為B^2-B= B(B-I)=0 就推論B=0或I\\ \quad , 例如: A= \begin{bmatrix} 1&0\\0&0 \end{bmatrix}滿足A^2=I,但A\ne I, A\ne 0\\ 結論: 矩陣相乘:AB=0不能推論A=0或B=0$$

解答:$$小綠在平面的推論是正確的, 但在空間的推論是錯誤的!\; b_4=15 \ne 16\\ b_n= {n\choose 0} +{n\choose 1} +{n\choose 2} +{n\choose 3} ={n^3+5n+6\over 6}$$

解答:$$每個橫列有3個正方形，不是塗黑就是塗白，共有2^3=8種塗法:\cases{0黑3白:1種塗法\\ 1黒2白:3種塗法\\ 2黑1白:3種塗法\\ 3黑0白:1種塗法}\\牆面有6列, 因此8種塗法需選出6種，但需符合黑,白各9塊. 若全選有12黑,需扣除3黑的情況\\ 狀況A:全選再扣除「0黑3白」及「3黑0白」,這兩種各自只有1種塗法\\狀況B:全選再扣除「1黑2白」及「2黑1白」,有C^3_1\cdot C^3_1=9種選法\\ 狀況A+狀況B共有10種塗法,而六種不同樣式的橫列有6!=720種排列，\\總共有10\times 720= \bbox[red, 2pt]{7200}種塗法$$

解答:$$隨機變數X:首次抽中大獎的次數 \Rightarrow X\sim Geo(p=0.1) \Rightarrow P(X=k)=(1-p)^{k-1}p\\ 定義\cases{H_0: 中獎機率p=10\% \\H_1: 中獎機率p\lt 10\%}, \alpha=0.1 \Rightarrow 欲求臨界值C,使得 P(X\ge C\mid H_0) \le 0.1\\ X\sim Geo(p) \Rightarrow P(X\ge C)=(1-p)^{C-1} \le 0.1 \Rightarrow 0.9^{C-1}\le 0.1 \Rightarrow C-1 \ge 22(查試題附表) \\ \Rightarrow C\ge 23 \Rightarrow \bbox[red, 2pt]{拒絕域為\{X \mid X\ge 23\},X為首次抽中大獎的次數} \\ 小綠的X值=22 \not \ge 23 \Rightarrow \bbox[red, 2pt]{不能拒絕}$$

解答:$$拋物線y=ax^2+bx+c與直線y=mx+n所截面積={|a|\over 6}(x_2-x_1)^3\\ L:y=mx 代入y=-x^2+2x+3 \Rightarrow x^2+(m-2)x-3=0 \Rightarrow \cases{兩根之和:x_1+x_2=2-m\\ 兩根之積:x_1x_2=-3} \\ \Rightarrow (x_2-x_1)^2= (2-m)^2+12 \Rightarrow 所截面積S={1\over 6} \left( (2-m)^2+12  \right)^{3/2} \\同理, y=mx 代入y=-x^2+6x+7 \Rightarrow x^2+(m-6)x-7=0 \Rightarrow \cases{兩根之和:x_1+x_2=6-m\\ 兩根之積:x_1x_2=-7}  \\\Rightarrow (x_2-x_1)^2=(6-m)^2+28 \Rightarrow 所截面積S={1\over 6} \left( (6-m)^2+28  \right)^{3/2} \\ \Rightarrow (2-m)^2+12 =(6-m)^2+28 \Rightarrow \bbox[red, 2pt]{m=6} \Rightarrow S={1\over 6}(16+12)^{3/2} \Rightarrow \bbox[red, 2pt]{S={28\over 3}\sqrt 7}$$

解答:$$漸近線\cases{y=0\\ 24x-7y=0 } \Rightarrow 雙曲線\Gamma: y(24x-7y)=k經過(1,2) \Rightarrow k=2(24-14)=20 \\ \Rightarrow 24xy-7y^2=20 \Rightarrow [x\; y] \begin{bmatrix}0 & 12\\12& -7 \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix} =20 \Rightarrow A= \begin{bmatrix}0 & 12\\12& -7 \end{bmatrix} \Rightarrow \det(A-\lambda I)=0 \\ \Rightarrow (\lambda+16)(\lambda-9)=0 \Rightarrow \lambda=-16,9 \Rightarrow \Gamma經旋轉後\Gamma': -16x'^2+9y'^2=20 \\ \Rightarrow -{x'^2\over 5/4}+{y'^2\over 20/9}=1 \Rightarrow a^2={20\over 9} \Rightarrow a={2\sqrt 5\over 3} \Rightarrow 2a= \bbox[red, 2pt]{4\sqrt 5\over 3}$$

========================= END ============================

解題僅供參考，其他教甄試題及詳解