114年金門高中教甄-數學詳解

 國立金門高中114 學年度第一次教師甄選

一、 填充題(每格 5 分，共 70 分) 

解答:$$\log_2 10^{100}= {\log 10^{100} \over \log 2} ={100\over 0.301} \approx 332.2 \Rightarrow \bbox[red, 2pt]{333}位數$$

解答:$$完全平方數的因數個數是奇數，其它為偶數。一開始是關閉，欲求打開的門數量，\\相當求1-100因數是奇數有幾個?完全平方數有1,4,9,\dots,100，共10個;\\ 又6號沒來，所以6的倍數有\lfloor {100\over 6}\rfloor=16個，其因數個數也是奇數。\\ 其中需扣除36(出現兩次)，因此有10+16-2=\bbox[red, 2pt]{24}扇門是打開的$$

解答:$${\triangle A_{n+1} B_{n+1}C_{n+1} \over \triangle A_nB_n C_n} ={1\over 7} \href{http://www.mathland.idv.tw/fun/area.htm}{參考資料} \\ \Rightarrow \sum_{n=1}^\infty a_n =1+{1\over 7}+{1\over 7^2}+ \cdots ={1\over 1-1/7} = \bbox[red, 2pt]{7\over 6}$$

解答:$$2n^2+3n-44= (n-4)(2n+11 )=3P^2\;又 3P^2的因數為1,3,P,3P,P^2,3P^2 \\假設\cases{A=n-4\\ B=2n+11=2A+19}\\ \textbf{Case I }\cases{A=1\\ B=3P^2} \Rightarrow B=2A+19=21 =3P^2 \Rightarrow P=\sqrt 7不合\\ \textbf{Case II }\cases{A=3\\ B=P^2} \Rightarrow B=2A+19 =25=P^2\Rightarrow P=5 \Rightarrow n=7 \\\textbf{Case III }\cases{A=P\\ B=3P} \Rightarrow 3P=2P+19 \Rightarrow P=19 \Rightarrow A=19 \Rightarrow n=23  \\ 因此可得n= \bbox[red, 2pt]{7,23}$$

解答:$$f(x)=x^{114}+2025 = (x-1)^2 p(x)+ ax+b \Rightarrow f'(x)=114x^{113} =2(x-1)p(x)+(x-1)^2p'(x)+a \\ \Rightarrow \cases{f(1) =2026=a+b\\ f'(1) =114=a} \Rightarrow b=1912 \Rightarrow 餘式為\bbox[red, 2pt]{114x+1912}$$

解答:$$\overline{AP}^2 +\overline{BP}^2 +\overline{CP}^2 =3(x-1)^2+(y+1)^2+  (y-1)^2+  y^2+(z-2)^2 +z^2 +(z-4)^2 \\ \Rightarrow \cases{f(x,y,z) =3(x-1)^2+(y+1)^2+  (y-1)^2+  y^2+(z-2)^2 +z^2 +(z-4)^2 \\g(x,y,z)=x +y+ z} \\ \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y\\ f_z= \lambda g_z \\ g=0} \Rightarrow \cases{6(x-1)= \lambda\\ 6y=\lambda\\ 6(z-2) =\lambda\\ x+y+z=0} \Rightarrow \cases{x-1=y\\ y=z-2} \Rightarrow x+y+z=y+1+y+y+2=0 \Rightarrow y=-1\\ \Rightarrow \cases{x=0\\ y=-1\\ z=1} \Rightarrow f(0,-1,1)=3+0+4+1+1+1+9= \bbox[red, 2pt]{19}$$

解答:$$\cases{f(\alpha)=2\\ f(\beta)=1} \Rightarrow \cases{3\sin \alpha+4\cos \alpha=2 \\ 3\sin \beta+4\cos \beta=1} \Rightarrow \cases{\sin (\alpha+\theta)=2/5\\ \sin (\beta+ \theta)=1/5} \Rightarrow \cases{ \cos(\alpha +\theta)=\sqrt{21}/5\\ \cos(\beta+\theta )=2\sqrt 6/5} \\ \Rightarrow \cos(\alpha-\beta) =\cos(\alpha+\theta-(\beta+\theta)) = \cos (\alpha+\theta)\cos(\beta+\theta)+ \sin(\alpha+\theta) \sin(\beta+ \theta) \\={\sqrt{21} \over 5}\cdot {2\sqrt 6\over 5}+{2\over 5}\cdot {1\over 5} ={2+6\sqrt{14}\over 25} \Rightarrow \sin(\alpha-\beta) ={\sqrt{117-8\sqrt{126}}\over 25} =\bbox[red, 2pt]{{4\sqrt 6}-\sqrt{21} \over 25}$$

解答:$$假設\cases{第1區:1-4\\ 第2區:5-8\\ 第3區:9-12},依題意\cases{第1區只有B或C\\ 第2區只有A或C\\ 第3區只有A或B} \\ 假設第1區有k個B \Rightarrow 第1區有4-k個C \Rightarrow \cases{第2區有k個C\\ 第2區有個4-k個A} \Rightarrow \cases{第3區有k個A\\ 第3區有4-k個B} \\ \Rightarrow 排法共有\sum_{k=0}^4 {4\choose k}^3 =1+4^3+6^3+4^3+1= \bbox[red, 2pt]{346}$$

解答:$$a=1!\times 2!\times 3!\times \cdots\times 10!=10 \times 9^2\times 8^3\times \cdots\times2^9\times 1^{10}= 2^{38} \times 3^{17} \times 5^7 \times 7^4 \\\Rightarrow 完全平方數有 \overbrace{2^0, 2^2,\dots, 2^{38}}^{20個}, \overbrace{3^0,3^2,\dots, 3^{16}}^{9個}, \overbrace{5^0,5^2,\dots,5^6}^{4個}, \overbrace{7^0,7^2,7^4}^{3個}\\ \Rightarrow 共20\times 9\times 4\times 3= \bbox[red, 2pt]{2160} 個$$

解答:

$$\textbf{(1)} \overline{BD}^2= \overline{AD}^2 +\overline{AB}^2 = \overline{DQ}^2+ \overline{BQ}^2 \Rightarrow 36+64=25+\overline{BQ}^2 \Rightarrow \overline{BQ} =5\sqrt 3 \\假設\cases{\overline{QR}=a\\ \overline{CR}=b \\ \angle QDC=\alpha\\ \angle DRQ= \beta} \Rightarrow \cases{\overline{DR}=8-b\\ \overline{BR}=5\sqrt 3-a \\ \angle CBQ=\alpha\\ \angle BRC=\beta} \Rightarrow \triangle DQR \sim \triangle BCR (AA) \Rightarrow {\overline{DQ} \over \overline{QR}} ={\overline{BC} \over \overline{CR}} \\ \Rightarrow {5\over a} ={6\over b} \Rightarrow a={5b\over 6} \cdots(1) \\ 直角\triangle DQR: \overline{DR}^2= \overline{DQ}^2+ \overline{QR}^2 \Rightarrow (8-b)^2=25+a^2 =25+ \left( {5b\over 6}\right)^2 \\ \Rightarrow 11b^2-576b+39\cdot 36=0 \Rightarrow b={288-150\sqrt 3\over 11} \Rightarrow \sin \beta={\overline{DQ} \over \overline{DR}} ={5\over 8-b} ={4+3\sqrt 3\over 10} \\ \Rightarrow \triangle ABQ ={1\over 2}\overline{AB} \cdot \overline{BQ} \sin \beta={1\over 2}\cdot 8\cdot 5\sqrt 3 \cdot {4+3\sqrt 3\over 10} = \bbox[red, 2pt]{18+8\sqrt 3} \\\textbf{(2)}\sin \beta={4+3\sqrt 3\over 10} \Rightarrow \cos \beta=  {4\sqrt 3-3\over 10}={\overline{AB}^2+ \overline{BQ}^2 -\overline{AQ}^2 \over 2\cdot \overline{AB} \cdot \overline{BQ}} ={139-\overline{AQ}^2 \over 80\sqrt 3} \\\quad \Rightarrow \overline{AQ}^2 = \bbox[red, 2pt]{43+24\sqrt 3}$$

解答:

$$假設\cases{A(-5,8) \\B(0,y) \\C(x,0) \\D(6,-4)} \Rightarrow \sqrt{x^2+y^2} +\sqrt{4^2+(x-6)^2}+\sqrt{5^2+(y-8)^2} =\overline{BC}+\overline{CD}+ \overline{AB} \\ \Rightarrow 最小值發生在A,B,C,D在一直線上，此時最小值=\overline{AD} = \bbox[red, 2pt]{\sqrt{265}}$$

解答:$$10^n= 2^n\times 5^n \Rightarrow a_n = \left( 1+2+2^2+ \cdots +2^n \right) \left( 1+5+5^2 +\cdots +5^n\right) = \left(2^{n+1}-1 \right) \left({5^{n+1}-1\over 4} \right) \\ \Rightarrow \lim_{n \to \infty}{\left(2^{n+1}-1 \right) \left({5^{n+1}-1\over 4} \right) \over 10^{n+1}} ={1\over 4 } \lim_{n\to \infty} \left(1-{1\over 2^{n+1}} \right) \left(1-{1\over 5^{n+1}} \right) =\bbox[red, 2pt]{1\over 4}$$

解答:$$\cases{\displaystyle {1\over a}-{1\over 2} ={2-a\over 2a} ={b+c\over 2a} \ge {1\over a}\sqrt{bc} \\\displaystyle {1\over b}-{1\over 2} ={2-b\over 2b} ={a+c\over 2b} \ge {1\over b}\sqrt{ac} \\\displaystyle {1\over c}-{1\over 2} ={2-c\over 2c} ={a+b\over 2c} \ge {1\over c}\sqrt{ab}} \Rightarrow \left({1\over a}-{1\over 2} \right) \left( {1\over b}-{1\over 2}\right)\left( {1\over c}-{1\over 2}\right) \ge {1\over abc} \sqrt{a^2b^2 c^2} = \bbox[red, 2pt]1$$

二、 計算與說明題 (每題 10 分，共 30 分) 

解答:$$黃金比例\phi ={1+\sqrt 5\over 2}(註: \phi-1={1\over \phi}), 此題黃金矩形的長L=\phi、寬為W=1 \\第1段圓弧\ell_1: 第一個正方形的邊長為W, 即圓半徑r_1=W \Rightarrow \ell_1= {1\over 4} \times 2\pi r_1={\pi W\over 2} \\第2段圓弧\ell_2: 邊長變為L-W =\phi W-W=(\phi -1)W={W\over \phi} =r_2 \Rightarrow \ell_2={\pi\over 2}\cdot {W\over \phi} \\ 第3段圓弧\ell_3: r_3={r_2 \over \phi} ={W\over \phi^2} \Rightarrow \ell_3={\pi\over 2}\cdot {W\over \phi^2} \\第n段圓弧\ell_n={\pi\over 2}\cdot {W\over \phi^{n-1}} \\因此\sum_{k=1}^\infty \ell_n ={\pi W\over 2} \left( 1+{1\over \phi}+{1\over \phi^2}+ \cdots+ {1\over \phi^{n-1}}+ \cdots\right) ={\pi W\over 2} \cdot{\phi \over \phi-1} ={\pi W\over 2}\cdot \phi^2 \\={\pi \over 2} \left( {1+\sqrt 5\over 2}\right)^2 = \bbox[red, 2pt]{(3+\sqrt 5)\pi\over 4}$$

解答:$$\textbf{(1) }\cases{|z_1|=2\\ |z_2|=3\\ |z_2-z_1|=\sqrt 5} \Rightarrow \cases{\overline{OA}=2\\ \overline{OB}=3\\ \overline{AB}=\sqrt 5} \Rightarrow \cos \angle AOB ={2^2+3^2-(\sqrt 5)^2 \over 2\cdot 2\cdot 3} = \bbox[red, 2pt]{2\over 3} \\ \textbf{(2) }\cases{|z_1+z_2|^2 =(z_1+z_2)(\bar z_1+\bar z_2) =|z_1|^2+|z_2|^2+ z_1\bar z_2+ \bar z_1 z_2 \\ |z_1-z_2|^2 =(z_1-z_2)(\bar z_1-\bar z_2) =|z_1|^2+|z_2|^2 -(z_1\bar z_2+\bar z_1z_2)} \\ \quad \Rightarrow |z_1+z_2|^2+|z_1-z_2|^2 =2 \left( |z_1|^2+|z_2|^2 \right) \Rightarrow |z_1+z_2|^2+5 =2 \left( 4+9 \right) \Rightarrow |z_1+z_2|= \bbox[red, 2pt]{\sqrt{21}} \\\textbf{(3) } {z_1\over z_2} ={|z_1|\over |z_2|} \left( \cos \angle AOB +i\sin \angle AOB\right) ={2\over 3} \left( {2\over 3} \pm i{\sqrt 5\over 3}\right) = \bbox[red, 2pt]{{4\over 9} \pm {2\sqrt 5\over 9}i}$$

解答:

$$\textbf{(1) }  y=x^2+3 \Rightarrow y'=2x \\\quad 假設切點P(t,t^2+3) \Rightarrow 切線斜率y'(t)=2t \Rightarrow \overline{AP} 斜率={t^2+3\over t-1} =2t \Rightarrow t^2-2t-3=0\\\quad \Rightarrow (t-3)(t+1) =0 \Rightarrow \cases{t=3 \Rightarrow 斜率y'(3)=6 \Rightarrow 切線方程式: y=6(x-1) \\ t=-1 \Rightarrow 斜率y'(-1)=-2 \Rightarrow 切線方程式: y=-2(x-1)} \\ \quad \Rightarrow 切線方程式: \bbox[red, 2pt]{y=6x-6, y=-2x+2} \\\textbf{(2) }  兩切點分別為\cases{P(-1,y(-1))=(-1,4) \\Q(3,y(t)) =(3,12)} \\ \quad \Rightarrow 所圍面積= \int_{-1}^1 \left( x^2+3-(-2x+2)\right)\,dx + \int_{-1}^3 \left( x^2+3-(6x-6)\right) \,dx =2+{2\over 3}+ 2+{2\over 3}\\\qquad = \bbox[red, 2pt]{16\over 3}$$

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