113年內湖高工教甄-數學詳解

 臺北市立內湖高級工業職業學校 113 學年度正式教師甄選

科別： 數學科 範圍： 數學科專業知能 考試時間： 100 分鐘

壹、 填充題 (每題 7 分， 共 70 分；請在答案卷相對應的題號欄中寫下答案。 )

解答: $$x^5+ 2x^2+x+3=x(x^4+x-1) +x^2+2x+3 \\\Rightarrow (x^5+ 2x^2+x+3)^2 =p(x)(x^4+x-1)+ (x^2+2x+3)^2\\  而(x^2+2x+3)^2 =(x^4+x-1)+ 4x^3+ 10x^2+11x+10 \Rightarrow 餘式為 \bbox[red, 2pt]{4x^3+ 10x^2+11x+10}$$

解答: $$令\cases{\alpha= \sqrt[3] {n+1} \\\beta =\sqrt[3]{n-1}} \Rightarrow \alpha-\beta ={\alpha^3-\beta^3\over \alpha^2+\alpha \beta+ \beta^2} ={2\over \sqrt[3] {(n+1)^2} +\sqrt[3]{n^2-1} +\sqrt[3]{(n-1)^2}} ={2\over a_n} \\ \Rightarrow {1\over a_n} ={1\over 2}(\alpha-\beta)  ={1\over 2}(\sqrt[3]{n+1}-\sqrt[3]{n-1})  \\\Rightarrow {1\over a_1}+{1\over a_3}+{1\over a_5}+ \cdots+ {1\over a_{4095}} ={1\over 2}\left( (\sqrt[3]2 -0 )+ (\sqrt[3]4 -\sqrt[3]2) +(\sqrt[3] 6-\sqrt[3]{4}) +\cdots +(\sqrt[3]{4096} -\sqrt[3]{4094})\right) \\={1\over 2}\sqrt[3]{4096} ={1\over 2} 2^{12/3} ={1\over 2}\cdot 16= \bbox[red, 2pt]8$$

解答: $$\overline{AD}^2 =|\overrightarrow{AD}|^2 =|\overrightarrow{AB} +\overrightarrow{BC} +\overrightarrow{CD}|^2 = |\overrightarrow{AB}|^2+ |\overrightarrow{BC}|^2 +|\overrightarrow{CD}|^2 +2 (\overrightarrow{AB} \cdot \overrightarrow{BC} +\overrightarrow{BC} \cdot \overrightarrow{CD} +\overrightarrow{CD} \cdot \overrightarrow{AB}) \\=3^2+2^2+1^2 +2(-\overrightarrow{BA} \cdot \overrightarrow{BC} -\overrightarrow{CB} \cdot \overrightarrow{CD} +\overrightarrow{CD} \cdot \overrightarrow{AB}) \\=14+2(-\cos 120^\circ \overline{BA}\cdot \overline{BC}- \cos120^\circ \overline{BC}\cdot \overline{CD} +\cos 60^\circ \overline{CD}\cdot \overline{AB}) \\=14+2({1\over 2}\cdot 3\cdot 2+ {1\over 2}\cdot 2\cdot 1+{1\over 2}\cdot 1\cdot 3) =14+(6+2+3) =25 \Rightarrow \overline{AD}=\sqrt{25}= \bbox[red, 2pt]5$$

解答: $$兩人比賽有三種情形，兩得分總和為6，平均每局得2分\\假設高三有n位學生，高二與高三學生比賽C^{n+3}_2次，得分總和為2C^{n+3}_2\\ 因此2C^{n+3}_2-20=np \Rightarrow n^2+(5-p)n-14=0 \Rightarrow p={n^2-14 \over n}+5 \\ \Rightarrow p=n+5-{14\over n}  \Rightarrow {14\over n} \in \mathbb N \Rightarrow n=1,2,7,14 \Rightarrow  n=14 \Rightarrow p=\bbox[red, 2pt]{18}$$

解答: $$(a-3)^2 +(b-3)^2 +(c-3)^2 +(d-3)^2 =4 \Rightarrow (a-3,b-3,c-3, d-3) =\cases{(\pm 2,0,0,0)\\ (\pm 1,\pm 1,\pm 1,\pm 1)} \\ \Rightarrow (a,b,c,d)=\cases{(5,3,3,3) \Rightarrow 排列數4\\ (1,3,3,3) \Rightarrow 排列數4\\ (4,4,4,4) \Rightarrow 排列數1 \\ (2,4,4,4) \Rightarrow 排列數4\\ (2,2,4,4) \Rightarrow 排列數6\\(2,2,2,4) \Rightarrow 排列數4\\(2,2,2,2) \Rightarrow 排列數1} \Rightarrow 總排列數24 \Rightarrow 機率={24\over 6^4} = \bbox[red, 2pt]{1\over 54}$$

解答: $$t=10^{675} \Rightarrow \left \lfloor {10^{2025} \over 10^{675} +2025} \right \rfloor= \left \lfloor{t^3 \over t+2025} \right \rfloor= \left \lfloor t^2-2025t+2025^2-{2025^3\over t+2025} \right \rfloor \\ = \left \lfloor  2025^2-{2025^3\over t+2025} \right \rfloor \text{ mod }1000 = \bbox[red, 2pt]{624} (\because 0\lt {2025^3\over t+2025}\lt 1)$$

解答: $$\Gamma: \begin{bmatrix} x'\\ y'\end{bmatrix}= \begin{bmatrix} \cos 45^\circ& -\sin 45^\circ \\ \sin 45^\circ & \cos 45^\circ\end{bmatrix}\begin{bmatrix} x\\y\end{bmatrix} =\begin{bmatrix} {\sqrt 2\over 2}(x-y) \\ {\sqrt 2\over 2}(x+y)\end{bmatrix} \Rightarrow \cases{x=(x'+y')/\sqrt 2\\ y=(y'-x')/\sqrt 2} \\ \Rightarrow {({x'+y'\over \sqrt 2})^2\over 5}+ {({y'-x'\over \sqrt 2})^2 \over 10}=1 \Rightarrow 3x'^2+2x'y'+ 3y'-20=0 \Rightarrow \bbox[red, 2pt]{3x^2+2xy+3y^2-20=0}$$

解答:

$$\cases{-1\le x\le 0:A=\pi \int_{-1}^0 (1-x^2)^2\,dx ={8\over 15}\pi\\ 0\le x\le 1:B= \pi\int_0^1(-1-x)^2\,dx = {7\over 3}\pi \\1\le x\le 2: C=\pi \int_1^2 \left((-1-x)^2-(1-x^2)^2 \right)\,dx ={19\over 5}\pi } \Rightarrow A+B+C={100\over 15}\pi =\bbox[red, 2pt]{{20\over 3}\pi}$$

解答: $$假設x^7-1=0的七個根為1,x_1,x_2,\dots,x_6, \\則x_1,x_2, \dots, x_6為f(x)=x^6+x^5+\cdots +x+1=0的六根 \\ \Rightarrow f(x)=(x-x_1)(x-x_2)\cdots (x-x_6)\\ \Rightarrow f(1)=7=(1-x_1)(1-x_2) \cdots (1-x_6) =\overline{AB} \cdot \overline{AC}\cdots\overline{AG}\\ 又\overline{AB}^2+\overline{AC}^2+ \cdots+ \overline{AG}^2\\ =(\overrightarrow{AO} + \overrightarrow{OB}) \cdot (\overrightarrow{AO} + \overrightarrow{OB}) + (\overrightarrow{AO} + \overrightarrow{OC}) \cdot (\overrightarrow{AO} + \overrightarrow{OC}) + \cdots +(\overrightarrow{AO} + \overrightarrow{OG}) \cdot (\overrightarrow{AO} + \overrightarrow{OG})  \\= (2+2\overrightarrow{AO} \cdot \overrightarrow{OB})+ (2+2\overrightarrow{AO} \cdot \overrightarrow{OC}) +\cdots +(2+2\overrightarrow{AO} \cdot \overrightarrow{OG}) \\=12+ 2\overrightarrow{AO} \cdot (\overrightarrow{OB} +\overrightarrow{OC} + \cdots+ \overrightarrow{OG}) =12 + 2\overrightarrow{AO} \cdot (\overrightarrow{OA}+\overrightarrow{OB} +\overrightarrow{OC} + \cdots+ \overrightarrow{OG} -\overrightarrow{OA}) \\=12+2\overrightarrow{AO} \cdot \overrightarrow{AO}  =14 \\因此(\overline{AB}\cdot \overline{AC} \cdots \overline{AG}) (\overline{AB}^2+ \cdots +\overline{AG}^2)= 7\times 14= \bbox[red, 2pt]{98}$$

解答:

$$取\cases{O(0,0,0) \\P(0,0,2) \\Q(5,0,0) \\A(5,5,0) \\B(5,-5,0) \\C(-5,-5,0)} \Rightarrow \cases{\overrightarrow{PA} =(5,5,-2) \\\overrightarrow{PB} =(5, -5,-2) \\\overrightarrow{PC} =(-5, -5,-2) } \Rightarrow \cases{\vec u= \overrightarrow{PA}\times \overrightarrow{PB} =(-20,0, -50)\\ \vec v= \overrightarrow{PB} \times \overrightarrow{PC} = (0,20,-50)} \\ \Rightarrow -\cos \alpha =-{\vec u\cdot \vec v \over |\vec u|| \vec v|} =-{2500\over 2900} = \bbox[red, 2pt]{-{25\over 29}}$$

貳、 計算證明題(每題 10 分， 共 30 分；請在答案卷相對應的題號中作答，並請寫下完整的計算或證明過程，否則不予計分。 )

解答:

$$y^2=2x \Rightarrow 焦點F({1\over 2}, 0) \Rightarrow L: y=m(x-{1\over 2}) 代入拋物線方程式 \Rightarrow m^2(x-{1\over 2})^2=2x \\ \Rightarrow m^2x^2-(m^2+ 2)x+{1\over 4}m^2 =0 \Rightarrow 兩交點\cases{A(\alpha, \sqrt{2\alpha}) \\ B(\beta,-\sqrt{2\beta})}\Rightarrow  \alpha \beta={1\over4} \\ \Rightarrow \overrightarrow{OA} \cdot \overrightarrow{OB} =(\alpha,\sqrt{2\alpha}) \cdot (\beta,-\sqrt{2\beta}) =\alpha\beta-2\sqrt{\alpha\beta} ={1\over 4}-2\cdot {1\over 2} =\bbox[red,2pt]{-{3\over 4}}$$

解答: $$f(x)={1\over x^2} \Rightarrow f'(x)=-{2\over x^3} \\ 假設P(a,f(a)) =(a,{1\over a^2}),則L:y=-{2\over a^3} (x-a)+{1\over a^2} \Rightarrow \cases{A({3\over 2}a,0) \\B(0,{3\over a^2})} \Rightarrow \overline{AB} =\sqrt{ {9\over 4}a^2+{9\over a^4}} \\ 取g(a) ={9\over 4}a^2+{9\over a^4} \Rightarrow g'(a)=0 \Rightarrow a=\sqrt 2 \Rightarrow \bbox[red, 2pt]{\cases{\overline{AB} ={3\sqrt 3\over 2} \\P(\sqrt 2,{1\over 2})}}$$

解答: $$\sum_{k=1}^{90} 2k\sin 2k^\circ=2(1\cdot \sin 2^\circ +2\sin 4^\circ +\cdots +44\sin 88^\circ+ 45\sin 90^\circ+ 46\sin 92^\circ+\cdots +89\sin 178^\circ+ 90\sin 180^\circ) \\=2(1\cdot \sin 2^\circ +2\sin 4^\circ+ \cdots +44\sin 88^\circ +45\sin 90^\circ+ 46 \sin 88^\circ+\cdots +88 \sin 4^\circ+89\sin 2^\circ+ 0) \\= 2(90\sin 2^\circ +90\sin 4^\circ +\cdots +90\sin 88^\circ)+ 2\cdot 45\sin 90^\circ \\=180(\sin 2^\circ+ \sin 4^\circ+ \cdots +\sin 88^\circ) +90 \\={90\over \sin 1^\circ}(2\sin 2^\circ \sin 1^\circ +2\sin 4^\circ \sin 1^\circ +\cdots +2\sin 88^\circ \sin 1^\circ)+90 \\={90\over \sin 1^\circ}( \cos 1^\circ-\cos 3^\circ +\cos 3-\cos 5^\circ +\cdots +\cos 87^\circ-\cos 89^\circ)+90 \\={90\over \sin 1^\circ} (\cos 1^\circ-\cos 89^\circ)+ 90 ={90\over \sin 1^\circ} (\cos 1^\circ-\sin 1^\circ)+ 90 =90\cot 1^\circ \\ \Rightarrow \sum_{k=1}^{90} 2k\sin 2k^\circ的平均值={90\cot 1^\circ \over 90} =\cot 1^\circ. \bbox[red, 2pt]{故得證}$$

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解題僅供參考，教甄歷年試題及詳解