113年中山女中教甄-數學詳解

 臺北市立中山女子高級中學 113 學年度第 1 次教師甄選

數學科題目(測驗題型部分)

解答: $$f(n)=\sqrt[3]{(n-1)^2} +\sqrt[3]{(n+1)(n-1)} +\sqrt[3]{(n+1)^2} \\ \Rightarrow {1\over f(n)} ={1\over \sqrt[3]{(n-1)^2} +\sqrt[3]{(n+1)(n-1)} +\sqrt[3]{(n+1)^2}} \\\qquad ={\sqrt[3]{(n+1) } -\sqrt[3]{(n-1)}\over (\sqrt[3]{(n-1)^2} +\sqrt[3]{(n+1)(n-1)} +\sqrt[3]{(n+1)^2}) (\sqrt[3]{(n+1) } -\sqrt[3]{(n-1)})} \\\qquad ={\sqrt[3]{(n+1) } -\sqrt[3]{(n-1)} \over (n+1)-(n-1)} ={1\over 2}(\sqrt[3]{(n+1) } -\sqrt[3]{(n-1)}) \\ \Rightarrow \sum_{k=1}^{500} {1\over f(2k-1)} ={1\over 2} \sum_{k=1}^{500} \left(\sqrt[3]{2k}-\sqrt{2k-2} \right) \\={1\over 2}(\sqrt[3]2-\sqrt[3] 0 +\sqrt[3]4-\sqrt[3]2 + \cdots + \sqrt[3]{1000}-\sqrt[3]{998}) ={1\over 2} \cdot \sqrt[3]{1000}= \bbox[red, 2pt]5$$

解答: $$\cases{n=5\\ k=5}代公式(k-1)^n+(-1)^n(k-1)= 4^5-4=1024-4= \bbox[red, 2pt]{1020}$$

解答: $$已知\begin{cases}P(S\to S)= 0 \\ P(S\to T)=1/3,& S\ne T \end{cases}, 其中S,T\in \{A,B,C,D \}\\ 因此轉換矩陣A=\begin{bmatrix}0 & 1/3& 1/3& 1/3 \\1/3 & 0 &1/3 &1/3 \\1/3& 1/3 & 0   &1/3 \\ 1/3 & 1/3 &1/3 &0 \\\end{bmatrix} \\ =\begin{bmatrix}1 & -1 & -1 & -1 \\1 & 1 & 0 & 0 \\1 & 0 & 1 & 0 \\1 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}1 & 0 & 0 & 0 \\0 & \frac{-1}{3} & 0 & 0 \\0 & 0 & \frac{-1}{3} & 0 \\0 & 0 & 0 & \frac{-1}{3} \end{bmatrix} \begin{bmatrix}\frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\\frac{-1}{4} & \frac{3}{4} & \frac{-1}{4} & \frac{-1}{4} \\\frac{-1}{4} & \frac{-1}{4} & \frac{3}{4} & \frac{-1}{4} \\ \frac{-1}{4} & \frac{-1}{4} & \frac{-1}{4} & \frac{3}{4} \end{bmatrix} \\ \Rightarrow A^n= \begin{bmatrix}1 & -1 & -1 & -1 \\1 & 1 & 0 & 0 \\1 & 0 & 1 & 0 \\1 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}1 & 0 & 0 & 0 \\0 & (\frac{-1}{3})^n & 0 & 0 \\0 & 0 & (\frac{-1}{3})^n & 0 \\0 & 0 & 0 & (\frac{-1}{3})^n \end{bmatrix} \begin{bmatrix} \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\\frac{-1}{4} & \frac{3}{4} & \frac{-1}{4} & \frac{-1}{4} \\\frac{-1}{4} & \frac{-1}{4} & \frac{3}{4} & \frac{-1}{4} \\ \frac{-1}{4} & \frac{-1}{4} & \frac{-1}{4} & \frac{3}{4} \end{bmatrix} \\欲求A^n\begin{bmatrix}1 \\0\\0\\0 \end{bmatrix}= \begin{bmatrix}a \\b\\c\\d \end{bmatrix}的a \Rightarrow a={3\cdot(-1)^n+3^n\over 4\cdot 3^n} =\bbox[red, 2pt]{{1\over 4}+{3\over 4}\cdot \left( -{1\over 3}\right)^n}$$

解答:

$$S_{\triangle PAB} ={1\over 2}\cdot \overline{AB}\cdot \overline{BC}= 625\\ \triangle PAB周長=100+50=150 \Rightarrow S_{\triangle PAB}=\sqrt{75(75-a)(75-50)(25+a)} =625\\ \Rightarrow 3a^2-300a+625=0 \Rightarrow a=50-{50\over \sqrt 6} \Rightarrow \sin \angle APD={25\over a} \\ \Rightarrow \cos \angle APD={\sqrt{a^2-625} \over a} \Rightarrow \overline{PD}= a\cos \angle APD= \sqrt{a^2-625}=\bbox[red, 2pt]{50\sqrt 6-75\over 3}$$

解答:

$$A={1\over 5}\begin{bmatrix}3 & -4 \\4 & 3 \end{bmatrix} =\begin{bmatrix}{3\over 5} & -{4\over 5} \\{4 \over 5}& {3\over 5} \end{bmatrix} =\begin{bmatrix}\cos\theta & -\sin \theta \\\sin \theta & \cos \theta\end{bmatrix} 為一旋轉矩陣\Rightarrow \cases{\sin \theta=4/5\\ \cos \theta=3/5}\\\Rightarrow  \sin \angle P_1OP_2 = \sin 2\theta= 2\sin \theta\cos \theta= {24\over 25}\\ 假設\overline{OP_1}=a, 則\triangle P_{ 1 }P_{ 2 }P_{ 3 }=\triangle OP_{ 1 }P_{ 2 }+\triangle OP_{ 2 }P_{ 3 }-\triangle OP_{ 1 }P_{ 3 }\\ =\frac { 1 }{ 2 } a^{ 2 }\sin { \theta  } +\frac { 1 }{ 2 } a^{ 2 }\sin { \theta  } -\frac { 1 }{ 2 } a^{ 2 }\sin { 2\theta  }   =a^{ 2 }\times \frac { 4 }{ 5 } -a^{ 2 }\times \frac { 3 }{ 5 } \times \frac { 4 }{ 5 } = {\frac { 8 }{ 25 } a^{ 2 }} \\ 現在P_1在y={x^2\over 10}-10上 \Rightarrow P_1(m,{m^2\over 10}-10) \Rightarrow \overline{OP_1}^2={1\over 100}m^4-m^2+100 \\ =\frac { 1 }{ 100 } { \left( m^{ 2 }-50 \right)  }^{ 2 }+75 \Rightarrow \triangle P_{ 1 }P_{ 2 }P_{ 3 }=\frac { 8 }{ 25 } \times \left[ \frac { 1 }{ 100 } { \left( m^{ 2 }-50 \right)  }^{ 2 }+75 \right] \\=\frac { 3 }{ 2500 } { \left( m^{ 2 }-50 \right)  }^{ 2 }+24 \Rightarrow 最小值為\bbox[red,2pt]{24}$$

解答: $$假設三角形三邊長分別為x-1,x,x+1,則面積為\sqrt{{3x\over 2}(1+{x\over 2}) ({x\over 2})({x\over 2}-1)} =3 \\ \Rightarrow {3x\over 2}\cdot {x+2\over 2} \cdot {x\over 2}\cdot {x-2\over 2}=9 \Rightarrow x^4-4x^2-48=0 \Rightarrow x^2= 2+2 \sqrt{13} \\ \Rightarrow x=\sqrt{2+2\sqrt{13}} \Rightarrow 最大邊長= \bbox[red, 2pt]{\sqrt{2+2\sqrt{13}}+1}$$

解答: $$f(x+y)=2f(x)f(y) \Rightarrow f(0)=2f(0)f(0) \Rightarrow f(0)={1\over 2} \\ f'(x)=\lim_{h\to 0}{f(x+h)-f(x)\over h} =\lim_{h\to 0}{2f(x)f(h)-f(x)\over h} =\lim_{h\to 0}{2f(x)f(h)-2f(x)\cdot {1\over 2}\over h} \\\quad = \lim_{h\to 0}{2f(x)f(h)-2f(x)\cdot f(0)\over h} =2f(x)\lim_{h\to 0}{f(h)-f(0)\over h} =2f(x)f'(0)=4f(x) \\ \Rightarrow f'(x)=4f(x) \Rightarrow f''(x)=4f'(x)=16f(x) \Rightarrow {f''(x)\over f(x)}= \bbox[red, 2pt]{16}$$

解答: $${1\over 20}\times {999^{1000}\over 1000^{999}} ={1000\over 20}\times \left( {999\over 1000}\right)^{1000} =50\times\left( 1-{1\over 1000}\right)^{1000} \\ 由於\lim_{x\to \infty}(1-{1\over x})^x={1\over e} \Rightarrow  \left[ 50\times\left( 1-{1\over 1000}\right)^{1000} \right] \approx\left[ {50\over e} \right] \approx[18.39] =\bbox[red, 2pt]{18}$$

解答: $$\cases{A(4,0,0)\\ B(0,4,0)\\ C(0,0,4)} \Rightarrow M=(B+C)/2=(0,2,2) \Rightarrow \overline{CM} =\overline{CB}=2\sqrt 2=\overline{BC'} \\ \Rightarrow  \overrightarrow{AM}=(-4,2,2) \Rightarrow 過A且法向量為 \overrightarrow{AM}的平面E_1: -2x+y+z=4\\ \Rightarrow C'\in E_1且\overline{BC'}=2\sqrt 2 \Rightarrow C'(\sqrt 2,3+\sqrt 2,1+\sqrt 2)\\ 又E_2=\triangle ABC: x+y+z=4 \Rightarrow H({\sqrt 2\over 3},3+{\sqrt 2\over 3},1+{\sqrt 2\over 3}) \Rightarrow 平面HC'B:  \bbox[red, 2pt]{2x-y-z+4=0}$$

解答: $$\tan 2\theta={2\tan\theta\over 1-\tan^2\theta} \Rightarrow \tan 3\theta={\tan 2\theta+\tan \theta\over 1-\tan\theta\tan 2\theta} ={3 \tan \theta-\tan^3\theta \over 1-3\tan^2 \theta} \\ \Rightarrow \tan 5\theta= {\tan 3\theta+ \tan 2\theta\over 1-\tan 2\theta\tan 3\theta} ={\tan^5\theta -10\tan^3\theta +5\tan \theta\over 5\tan^4 \theta-10\tan^2 \theta+1} \\ \Rightarrow \tan 10\theta={2\tan 5\theta\over 1-\tan^2 5\theta} ={10\tan^9x -120\tan^7 x+252\tan^5 x-120\tan^3 x+10\tan x\over -\tan^{10}x+45\tan^8 x-210\tan^6x +210\tan^4 x-45\tan^2 x+1} \\ \Rightarrow \cases{|a_1|+|a_3| +|a_5|+|a_7| +|a_9|=512\\ |a_0|+|a_2|+|a_4| +|a_6|+|a_8| +|a_{10}|=512} \Rightarrow \sum_{k=0}^{10} |a_k|=\bbox[red, 2pt]{1024}$$

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