113年台北陽明高中教甄-數學詳解

113 年臺北市立陽明高級中學 數學科 第 1 次教師甄選 

解答： $$計算每一個括號內的分數，1,{1\over 2}, {1\over 3}, \dots, {1\over 99},{1\over 100}的數量,剛好{1\over n}出現n^2次\\總和 =\sum_{n=1}^{100} \left( n^2\cdot {1\over n} \right) =\sum_{n=1}^{100} n = {100\cdot 101\over 2} =\bbox[red, 2pt]{5050}$$

解答：

解答： $${a+b+c+d+e +f+ 8+11+12+17\over 10} ={a+b+c+d+e +f \over 6} \Rightarrow a+b+c+d+e +f=72\\ {a^2 +b^2+c^2+d^2+e^2 +f^2+ 8^2+11^2+12^2+17^2\over 10} ={a^2 +b^2+c^2+d^2+e^2 +f^2 \over 6} \\ \Rightarrow a^2 +b^2+c^2+d^2+e^2 +f^2=927 \\ \Rightarrow 變異數={927\over 6}-\left( 72\over 6 \right)^2 = \bbox[red, 2pt]{21\over 2}$$

解答： $$考慮n除以2,2^2,2^3,2^4商數為奇數的情形：\\n是奇數 \Rightarrow f(n)=n \Rightarrow \sum_{n=1}^{25} (2n-1)=625\\ n=2^k \Rightarrow f(n)=1 \Rightarrow \sum_{k=1}^5 f(2^k) =5 \\ n=2k(k是奇數) \Rightarrow f(n)=k \Rightarrow f(6)+f(10)+(14)+ f(18)+ f(22)+ f(26)+ f(30) \\\qquad +f(34)+ f(38)+ f(42)+ f(46) +f(50) \\\qquad = \sum_{n=1}^{12} (2n+1) =168\\ n=2^2k(k是奇數) \Rightarrow f(n)=k \Rightarrow f(12)+ f(20)+   f(28)+ f(36)+f(44) \\\qquad =3+5+ 7+9+11= 35\\ n=2^3k (k是奇數) \Rightarrow f(n)=k \Rightarrow f(24)+ f(40)=3+5=8\\n=2^4k(k是奇數) \Rightarrow f(n)=k \Rightarrow f(48)=3\\ 總和= 625+5+168+35+8 +3=\bbox[red, 2pt]{844}$$

解答： $$\log_{2x-5}(x^3-7x^2+11x+4)=2 \Rightarrow (2x-5)^2=x^3-7x^2+11x+4 \\ \Rightarrow x^3-11x^2+31x-21=0 \Rightarrow (x-1)(x-3)(x-7)=0 \\ \Rightarrow \cases{x=1\\ x=3 \\ x=7} \Rightarrow 底數2x-5=\cases{-3\lt 0 不合\\ 1不合\\ 9} \Rightarrow x=\bbox[red, 2pt] 7$$

解答： $$\cases{f(x,y)=18x+10y^2\\ g(x,y)=2x^2+5y^2-7x} \Rightarrow \cases{f_x=\lambda g_x\\ f_y= \lambda g_y\\ g=0} \Rightarrow \cases{18=\lambda(4x-7)  \cdots(1)\\ 20y=\lambda(10y) \cdots(2)} \\ (2) \Rightarrow \cases{\lambda=2 \Rightarrow x=4 \Rightarrow y無實數解\\ y=0 \Rightarrow 2x^2-7x=0 \Rightarrow x=0,7/2} \Rightarrow \cases{f(0,0)=0\\ f(7/2,0)=63} \Rightarrow 最大值\bbox[red, 2pt]{63}$$

解答： $$\textbf{Case I }甲抽中紅\to 乙抽中紅 \;機率={4\over 15}\times {3\over 14}= {2\over 35} \\\textbf{Case II }甲抽中紅\to 乙抽中白\to 乙抽中紅\;機率={4\over 15}\times {1\over 14}\times {3\over 13} ={2\over 455}\\ \textbf{Case III }甲抽中白\to 甲抽中紅 \to 乙抽中紅\;機率={1\over 15} \times {4\over 14} \times {3\over 13} ={2\over 455}\\ 機率總和={2\over 35}+ {2\over 455}+ {2\over 455}= \bbox[red, 2pt]{6\over 91}$$

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解答：

$$9\times 7+8\times 6=63+48=\bbox[red, 2pt]{111}$$

解答： $$假設10x^3-39x^2+29x-6=0的三根為\alpha, \beta,\gamma \Rightarrow \cases{\alpha+ \beta+ \gamma=39/10\\ \alpha\beta +\beta\gamma + \gamma \alpha =29/10\\ \alpha \beta\gamma =6/10} \\ \Rightarrow (\alpha+2) (\beta+2)(\gamma+2 )=\alpha \beta \gamma+2(\alpha \beta +\beta \gamma+ \gamma \alpha)+4(\alpha+\beta+ \gamma)+8 \\={6\over 10}+2\cdot {29\over 10}+ 4\cdot {39\over 10}+8= \bbox[red, 2pt]{30}$$

解答：

$$八個頂點任取四點，再扣除四點同平面的情形=C^8_4-12=70-12= \bbox[red, 2pt]{58}\\ 四點共平面的情形：除了六面體的六個面外，還有斜平面如上圖，因此共有12個$$

解答：

$$假設正八面體稜長為4，各頂點坐標\cases{A(-2,-2,0)\\ B(2,-2, 0) \\ C(2,2, 0) \\ D(-2,2,0)\\ P(0,0,2\sqrt 2)\\ Q(0,0,-2\sqrt 2)}\\ \Rightarrow 六面體頂點\cases{E=(A+B+P)/3 = (0,-4/3,2\sqrt 2/3 )\\ F=(A+B+Q)/3 =(0,-4/3,-2\sqrt 2/3) } \Rightarrow \overline{EF}={4\sqrt 2\over 3}\\ \Rightarrow {六面體體積 \over 八面體體積} =\cfrac{\left({4\sqrt 2\over 3} \right)^3}{{\sqrt 2\over 3}4^3} =\bbox[red, 2pt]{2\over 9}$$

解答：

$$假設三角形三邊長為x,y, \ell-x-y, 則\cases{0\le x\\ 0\le y\\ 0\le\ell-x-y} \Rightarrow \cases{0\le x\\ 0\le y\\ 0\le x+y \le \ell}\\ \cases{兩邊之和大於第三邊\\ 兩邊之差小於第三邊}  \Rightarrow \cases{x+y \ge \ell/2 \\ 0\le x\le \ell/2\\ 0\le y\le \ell/2}\\ 構成三角形條件的機率={\triangle OAB\over \triangle CDE} ={\ell^2/2 \over \ell^2/8} =\bbox[red, 2pt]{1\over 4}$$

解答： $$\cases{|b-c|\cos(A/2)=5\\ (b+c)\sin(A/2)=10} \Rightarrow (b-c)^2\cos^2(A/2)+ (b+c)^2 \sin^2(A/2)=125 \\ \Rightarrow (b^2+c^2)\cos^2(A/2)+ (b^2+c^2)\sin^2(A/2)-2bc(\cos^2(A/2)-\sin^2(A/2))=125 \\ \Rightarrow b^2+c^2 -2bc\cos A=a^2 =125 \Rightarrow a= \bbox[red, 2pt]{5\sqrt 2}$$

解答： $$\bbox[cyan,2pt]{本題送分}$$

解答： $$A=\begin{bmatrix}-6 & -1 & 2\\3 & -1 & 1\\-7 & -1 & 2 \end{bmatrix} \Rightarrow [A\mid I] =\left[ \begin{array}{rrr|rrr}-6 & -1 & 2 & 1 & 0 & 0\\3 & -1 & 1 & 0 & 1 & 0\\-7 & -1 & 2 & 0 & 0 & 1\end{array} \right] \xrightarrow{(-1/6)R_1\to R_1} \\\left[ \begin{array}{rrr|rrr} 1 & \frac{1}{6} & - \frac{1}{3} & - \frac{1}{6} & 0 & 0\\3 & -1 & 1 & 0 & 1 & 0\\-7 & -1 & 2 & 0 & 0 & 1\end{array} \right]  \xrightarrow{R_2-3R_1\to R_2, R_3+7R_1 \to R_3}\left[ \begin{array}{rrr|rrr}1 & \frac{1}{6} & - \frac{1}{3} & - \frac{1}{6} & 0 & 0\\0 & - \frac{3}{2} & 2 & \frac{1}{2} & 1 & 0\\0 & \frac{1}{6} & - \frac{1}{3} & - \frac{7}{6} & 0 & 1\end{array} \right]\\ \xrightarrow{(-2/3)R_2 \to R_2} \left[ \begin{array}{rrr|rrr}1 & \frac{1}{6} & - \frac{1}{3} & - \frac{1}{6} & 0 & 0\\0 & 1 & - \frac{4}{3} & - \frac{1}{3} & - \frac{2}{3} & 0\\0 & \frac{1}{6} & - \frac{1}{3} & - \frac{7}{6} & 0 & 1\end{array} \right] \xrightarrow{R_1-(1/6)R_2\to R_1,R_3-(1/6)R_2\to R_3} \\ \left[ \begin{array}{rrr|rrr}1 & 0 & - \frac{1}{9} & - \frac{1}{9} & \frac{1}{9} & 0\\0 & 1 & - \frac{4}{3} & - \frac{1}{3} & - \frac{2}{3} & 0\\0 & 0 & - \frac{1}{9} & - \frac{10}{9} & \frac{1}{9} & 1\end{array} \right]  \xrightarrow{-9R_3\to R_3} \left[ \begin{array}{rrr|rrr}1 & 0 & - \frac{1}{9} & - \frac{1}{9} & \frac{1}{9} & 0\\0 & 1 & - \frac{4}{3} & - \frac{1}{3} & - \frac{2}{3} & 0\\0 & 0 & 1 & 10 & -1 & -9\end{array} \right] \\ \xrightarrow{R_1+(1/9)R_3\to R_1, R_2+(4/3)R_3 \to R_2} \left[ \begin{array}{rrr|rrr}1 &0& 0 & 1 & 0 & -1\\0 & 1 & 0 & 13 & -2 & -12\\0 &0 & 1 & 10 & -1 & -9\end{array} \right] \\ \Rightarrow A^{-1}= \bbox[red, 2pt]{\left[ \begin{matrix}1 & 0 & -1\\13 & -2 & -12\\10 & -1 & -9\end{matrix} \right]}$$

解答： $$4!\times 4!=\bbox[red, 2pt]{576}$$

解答： $$\cos(\cos^{-1} x)= \cos\left( \cos^{-1} {1\over 2} +\cos^{-1} {1\over 7}\right) \Rightarrow x={1\over 2}\cdot {1\over 7}- \sin\left( \cos^{-1} {1\over 2}\right) \sin\left( \cos^{-1} {1\over 7}\right) \\ \Rightarrow x={1\over 14}-{\sqrt 3\over 2} \cdot {4\sqrt 3\over 7} = \bbox[red, 2pt]{-{11\over 14}}$$

解答： $$y={1+x\over 1-x}={2\over 1-x}-1 \Rightarrow y'={2\over (1-x)^2}\\ f(y)=x \Rightarrow f'(y)y'=1 \Rightarrow f'(y)={1\over y'}={(1-x)^2\over 2} \\ y=3 \Rightarrow {2\over 1-x}-1=3 \Rightarrow x={1\over 2} \Rightarrow f'(y=3)={(1-(1/2))^2 \over 2} = \bbox[red, 2pt]{1\over 8}$$

解答： $$f(x)= \int_1^x t(t^2-4)\,dt \Rightarrow f'(x)=x(x^2-4)  \Rightarrow f''(x)=3x^2-4\\ f'(x)=0 \Rightarrow x=0,\pm 2 \Rightarrow \cases{f''(0)=-4 \lt 0\\ f''(\pm 2)=8 \gt 0} \Rightarrow x=0時有極大值\\ f(0)=\int_1^0 t(t^2-4)\,dt ={7\over 4} \Rightarrow (a,b)=\bbox[red, 2pt]{(0,{7\over 4})}$$

解答： $$\bbox[cyan,2pt]{學校提供}$$

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