114年明倫高中教甄-數學詳解

 臺北市立明倫高級中學 114 學年度第 1 次正式教師甄選

一、填充題（每格5分，共70分，請寫在答案卷上，答案須化為最簡分數或最簡根式）

解答:$$f(x)=x^3-3x+1 \Rightarrow f'(x)=3x^2-3 \Rightarrow a_2=a_1-{f(a_1)\over f'(a_1)}=-{1\over -3} ={1\over 3} \\ \Rightarrow a_3= a_2-{f(a_2) \over f'(a_2)} ={1\over 3}-{1\over -72} = \bbox[red, 2pt]{25\over 72}$$

解答:$$z={1\over i^{2025}}-{2\over i^{2024}} +{3\over i^{2023}}-{4\over i^{2022}}+\cdots -{2024\over i^2}+{2025 \over i} \\={1\over i}-2-{3\over i}+ 4+{5\over i}-6-{7\over i}+8+\cdots +2024+{2025\over i} \\= {1\over i}\left(1+ 5+\cdots+ 2025 \right)-(2+6+ \cdots +2022)-{1\over  i}(3+7 + \cdots+2023)+(4+8+\cdots+2024) \\={1\over i}\cdot {2026\cdot 507\over 2}-{2024\cdot 506\over 2}-{1\over i}\cdot {2026\cdot 506\over 2}+ {2028\cdot 506\over 2} ={1013\over i}+1012 \\=1012-1013i \Rightarrow a-b= \bbox[red, 2pt]{2025}$$

解答:$$P(X=n)=0.4^{n-1}\cdot 0.6 \Rightarrow \cases{P(X=1)=0.6 \\P(X=2)= 0.24\\ P(X=3) =0.096 \\P(X=4) =0.0384 \\P(X=5) =0.01536 \\P(X=6) =0.006144} \Rightarrow P(X\le 6) =0.995904\\ H_0:命中=0.6, \alpha=0.01 \Rightarrow 雙尾檢定\Rightarrow 拒絕H_0的機率要超過0.995，取n=\bbox[red, 2pt]6 $$

解答:

$$R為一半圓，半徑為2,面積為2\pi \Rightarrow R之質心P(0,{8+6\pi\over 3\pi})\\ \Rightarrow P繞原點一圈距離為2\pi \cdot {8+6\pi\over 3\pi}  \Rightarrow 旋轉體積= 2\pi \cdot {8+6\pi\over 3\pi} \cdot 2\pi=\bbox[red, 2pt]{{32\pi\over 3} +8\pi^2}$$

解答:$$S=10\lim_{n\to \infty}\left(1+{0.5\over n} \right)^n =10\sqrt e \approx 10\times 1.65 \Rightarrow 整數部份= \bbox[red, 2pt]{16}$$

解答:$$\Gamma:{x^2\over 4}+{y^2\over 2}=1 \Rightarrow \cases{a=2\\ b=\sqrt 2} \\ 直線L:x-\sqrt 3 y=1順時針旋轉30^\circ \Rightarrow L':y=-{1\over 2} \Rightarrow \max(d(\Gamma',L)) =\max(d(\Gamma, L'))\\ =b+{1\over 2}=\sqrt 2+{1\over 2} = \bbox[red, 2pt]{1+2\sqrt 2\over 2}$$

解答:$$假設等腰直角三角形的底邊長為a, 則三角形面積為{a^2\over 4} \\ \Rightarrow 帳篷體積=2\int_0^6 {a^2\over 4}\,da = \bbox[red, 2pt]{36}$$

解答:$$假設\cases{A(a,0,0) \\B(0,b,0)\\ C(0,0,c) \\ D(ax,by,cz)} 且\cases{abc=60\\ a\triangle OAB=ab/2\\ a\triangle OAC=ac/2\\ a\triangle OBC=bc/2}  \Rightarrow \cases{OABD體積=abcz/6=10z=12 \\ OACD體積=abcy/6=10y=15\\ OBCD體積=abcx/6=10x=18} \\ \Rightarrow \cases{z=12/10\\ y=15/10\\ x=18/10} \Rightarrow (x,y,z) = \bbox[red, 2pt]{\left({{9\over 5}},{3\over 2},{6\over 5} \right)}$$

解答:$$\lim_{n\to \infty} {1\over 3n-1}\left(f({x^2\over n}) + f({2x^2\over n}) + \cdots +f({nx^2\over n}) \right) =\lim_{n\to \infty} {1\over 3n-1} \sum_{k=1}^n f({kx^2\over n}) \\=\lim_{n\to \infty} {n\over 3n-1} \sum_{k=1}^n {1\over n}f({kx^2\over n}) ={1\over 3}\int_0^1 f(x^2 t)\,dt ={\sqrt[3]{7+x} \over x} \Rightarrow \int_0^1 x^2f(x^2t)\,dt =3x \sqrt[3]{7+x} \\ 取u=x^2t \Rightarrow du=x^2dt \Rightarrow \int_0^{x^2}f(u)\,du =3x\sqrt[3]{7+x} \Rightarrow {d\over dx}\left( \int_0^{x^2}f(u)\,du\right) ={d\over dx}\left( 3x\sqrt[3]{7+x} \right) \\ \Rightarrow 2xf(x^2)=3\sqrt[3]{7+x}+ x(7+x)^{-2/3} \Rightarrow 2f(1)=3\cdot 2+{1\over 4} \Rightarrow f(1)=3+{1\over 8} = \bbox[red, 2pt]{25\over 8}$$

解答:$$令\cases{a=X_1-X_2\\ b=X_3-X_4\\ c=X_5-X_6} \Rightarrow 0\le a,b,c,\le 5\\ \Rightarrow \begin{array}{l} a& b& c& 數量\\\hline0& 5& 5& 6\times 1\times 1=6 \\\hdashline 1& 4& 5& 5\times 2\times 1=10\\ & 5& 4& 5\times 1\times 2=10\\\hdashline 2& 3& 5& 4\times 3\times 1=12\\ & 4& 4& 4\times 2\times 2=16\\ & 5& 3& 4\times 1\times 3=12\\\hdashline 3& 2& 5& 3\times 4\times 1=12\\ & 3& 4& 3\times 3\times 2=18\\ & 4& 3& 3\times 2\times 3=18 \\& 5& 2& 3\times 1\times 4=12\\\hdashline 4& 1& 5& 2\times 5\times 1=10 \\& 2& 4& 2\times 4\times 2=16\\ & 3& 3& 2\times 3 \times 3=18 \\ & 4& 2& 2\times 2\times 4=16 \\ & 5& 1 & 2\times 1\times 5=10\\\hdashline 5& 0& 5 & 1\times 6\times 1=6\\ & 1& 4& 1\times 5\times 2=10 \\ & 2& 3& 1\times 4\times 3=12\\ & 3& 2& 1\times 3\times 4=12\\ & 4& 1& 1\times 2\times 5=10\\ & 5& 0 & 1\times 1\times 6=6 \\\hline \end{array} \Rightarrow 合計252 \Rightarrow 機率={252\over 6^6} = \bbox[red, 2pt]{7\over 1296}$$

解答:$$\cases{\vec a=(2,1,k) \\ \vec b=(1,2,-2) \\\vec c=(-2,2,1)} \Rightarrow \vec u=\vec a+t\vec b+s\vec c=(2+t-2s,1+2t+2s,k-2t+s) \\ \Rightarrow |\vec u|^2=f(s,t)=(t-2s+2)^2+(2t+2s+1)^2+(-2t+s+k)^2 \\=9t^2+ 9s^2+ (8-4k)t +(2k-4)s +k^2+5 \Rightarrow \cases{f_t= 18t+8-4k =0\\ f_s =18s+2k-4=0} \Rightarrow \cases{t=(2k-4)/9\\ s=(2-k)/9} \\ \Rightarrow f({2-k\over 9},{2k-4\over 9}) =\left({4k+10\over 9} \right)^2 + \left({2k+5\over 9} \right)^2 +\left({4k +10\over 9}  \right)^2 ={(2k+5)^2\over 9} =5^2 \\ \Rightarrow \cases{2k+5=15\\ 2k+5=-15} \Rightarrow \cases{k=5\\ k=-10} \Rightarrow 5-10= \bbox[red, 2pt]{-5}$$

解答:$$$$

解答:$$取\cases{a=\overline{A_0A_1} =\sqrt{5-\sqrt 5\over 2} \\ b=\overline{A_0A_2} =\sqrt{5+\sqrt 5\over 2}} \Rightarrow ab=\sqrt 5\Rightarrow P=\prod_{j=0}^4 \prod_{k=j+1}^4 \overline{A_jA_k} \\= \overline{A_0A_1} \cdot \overline{A_0A_2} \cdot \overline{A_0A_3} \cdot\overline{A_0A_4} \cdot \overline{A_1A_2} \cdot \overline{A_1A_3} \cdot \overline{A_1A_4} \cdot \overline{A_2A_3} \cdot \overline{A_2A_4} \cdot \overline{A_3A_4} \\ =a\cdot b\cdot b\cdot a\cdot a\cdot b\cdot b\cdot a\cdot b\cdot a=a^5b^5 =(\sqrt 5)^5 \approx 2.24^5 \approx 56 \Rightarrow (a,b)= \bbox[red, 2pt]{(2,5)}$$

解答:$$f\cdot g=x^2g \Rightarrow f'\cdot g'=f'g+fg' \Rightarrow 2xg'(x)=2xg(x)+x^2g'(x) \\ \Rightarrow (x^2-2x)g'(x)+2xg(x)=0 \Rightarrow g'(x)+{2\over x-2}g(x)=0 一階微分方程 \\ \Rightarrow 積分因子I(x)= e^{\int{2\over x-2}dx} =(x-2)^2 \Rightarrow (x-2)^2g'(x)+2(x-2)g(x)=0 \\ \Rightarrow \left( (x-2)^2 g(x)\right)'=0 \Rightarrow (x-2)^2g(x)=c_1 \Rightarrow g(x)={c_1\over (x-2)^2} \Rightarrow g(3)=c_1=1 \\ \Rightarrow g(x)={1\over (x-2)^2} \Rightarrow g(4)= \bbox[red, 2pt]{1\over 4}$$

二、 計算與證明題（ 共 30 分， 請將詳細解題過程寫在答案卷上， 不完整者酌予計分；請使用黑色或藍色原子筆作答。 ）

解答:$$\textbf{(1) }E(X)=\sum_{x=1}^\infty xq^{x-1}p =p\sum_{x=1}^\infty xq^{x-1} =p{d\over dq}\sum_{x=0}^\infty q^x = p{d\over dq} \left({1\over 1-q} \right) =p\cdot {1\over (1-q)^2} \\=p\cdot {1\over p^2} ={1\over p} \Rightarrow E(X)={1\over p}\quad \bbox[red, 2pt]{QED}$$

解答:$$\textbf{(1) } a_2={10\over 13}\cdot 0+{24\over 13}\sqrt{4-0^2} ={48\over 13} \Rightarrow a_3= {10\over 13} \cdot {48\over 13}+{24\over 13} \sqrt{4^2-({48\over 13})^2} \\= {10\over 13} \cdot {48\over 13}+{24\over 13}\cdot {20\over 13}= \bbox[red, 2pt]{960\over 169}$$

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