105年竹北高中教甄-數學詳解

 國立竹北高中105學年度第1學期第1次教師甄選

壹： 填充題

解答：

$$\cases{z_1=a\\ z_2=-ai\\ z_3=-a\\ z_4=ai},a\in \mathbb{R} \Rightarrow {z_2-z_1\over z_3-z_1} ={-ai-a\over -2a} = \bbox[red, 2pt]{{1\over 2}+ {1\over 2}i}$$

解答：$${1\over p}+{1\over 3q}=12 \Rightarrow {3\over p}+{1\over q}=36 \Rightarrow {1\over p}+ {1\over p}+ {1\over p}+{1\over q} \ge 4\sqrt[4]{{1\over p^3q}} \\ \Rightarrow 36\ge 4\sqrt[4]{{1\over p^3q}} \Rightarrow 9^4 \ge {1\over p^3q} \\ 因此3\log_{1/3} p+\log_{1/3}q = -3\log_3p-\log_3 q =-\log_3(p^3q) =\log_3{1\over p^3q} \le \log_3 9^4 =8\\ \Rightarrow 3\log_{1/3} p+\log_{1/3}q 的最大值為\bbox[red,2pt]{8}$$

解答：

$$\sqrt{(x-1)^2+ (2^x-4)^2} +\sqrt{(x-1)^2+2^x} =\overline{PA}+\overline{PB}，其中\cases{A(1,4)\\ B(1,0)\\ P在y=2^x上}\\ 由於A、B在曲線y=2^x的兩側，因此\overline{PA}+\overline{PB}的最小值=\overline{AB}=\bbox[red, 2pt]{4}$$

解答：

$$依題意\overline{PQ}=\sqrt 3，其中\cases{P在\overline{BC}上\\ Q在\overline{OA}上}；由於\triangle ABC為正\triangle 且\overline{OA}=\overline{OB}=\overline{OC}，所以P為\overline{BC}中點；\\直角\triangle PQB:\overline{BQ}=\sqrt{2^2+3}=\sqrt 7；又\cases{直角\triangle BQA:\overline{QA}=\sqrt{4^2-7}=3\\ 直角\triangle BQO:\overline{QO}=\sqrt{a^2-7}}\\ 因此\overline{OA}=a = \overline{QA}+\overline{QO} =3+\sqrt{a^2-7} \Rightarrow (a-3)^2=a^2-7\Rightarrow 6a=16 \Rightarrow a=\bbox[red, 2pt]{8\over 3}$$

解答：$$f(x)=(x-1)^{20}= (x^2+1)Q(x)+px+r \Rightarrow f(i)=(i-1)^{20}=pi+r\\ \Rightarrow \cases{r=f(i)的實部\\ p=f(i)的虛部} \quad，而f(i)=((i-1)^2)^{10} =(-2i)^{10} =(-2)^{10}i^{10}=1024\cdot (-1)=-1024\\ \Rightarrow (p,r)=\bbox[red, 2pt]{(0,-1024)}$$

解答：

$$\overline{AH}\bot \overline{BC} \Rightarrow \overrightarrow{AH} \cdot \overrightarrow{BC}=0 \Rightarrow (-{1\over 2}\overrightarrow{AB} +{3\over 2}\overrightarrow{AC}) \cdot (-\overrightarrow{AB}+ \overrightarrow{AC})=0 \\ \Rightarrow {1\over 2}|\overrightarrow{AB}|^2 -2\overrightarrow{AB}\cdot \overrightarrow{AC}+{3\over 2} |\overrightarrow{AC}|^2=0 \Rightarrow {25\over 2} -2\overrightarrow{AB}\cdot \overrightarrow{AC}+{27\over 2} =0 \Rightarrow \overrightarrow{AB}\cdot \overrightarrow{AC} =13\\ 又\overrightarrow{AB}\cdot \overrightarrow{AC} =|\overrightarrow{AB}| |\overrightarrow{AC}| \cos \angle A \Rightarrow \cos  \angle A={13\over 5\cdot 3} ={13\over 15} \Rightarrow \sin \angle A= {2\sqrt{14}\over 15}；\\餘弦定理: \cos \angle A={\overline{AB}^2 +\overline{AC}^2-\overline{BC}^2 \over 2\times \overline{AB} \times \overline{AC}} \Rightarrow {13\over 15}={25+9-\overline{BC}^2 \over 30} \Rightarrow \overline{BC}=2\sqrt 2\\ 正弦定理: {\overline{BC} \over \sin \angle A}=2R \Rightarrow R={2\sqrt 2\over 2\sqrt{14}/15} \cdot {1\over 2}= {15\over 2\sqrt 7} \Rightarrow 外接圓面積=R^2\pi =\bbox[red, 2pt]{{225\over 28}\pi}$$

解答：$$\alpha +\beta+\gamma =\pi \Rightarrow {\alpha\over 2} +{\beta \over 2}+{\gamma \over 2}={\pi \over 2} \\\Rightarrow  \cot {\alpha\over 2} =\cot({\pi\over 2}-{\beta\over 2} -{\gamma\over 2}) =\tan({\beta\over 2} +{\gamma\over 2}) ={\tan(\beta/2) +\tan(\gamma/2) \over 1-\tan(\beta/2) \tan(\gamma/2)} \\= {1/\cot(\beta/2) +1/\cot(\gamma/2) \over 1-1/(\cot(\beta/2) \cot(\gamma/2))}  ={\cot{\beta\over 2} +\cot{\gamma\over 2} \over \cot{\beta\over 2}\cot{\gamma\over 2}-1}\\ \Rightarrow \cot {\alpha\over 2}\cot {\beta\over 2}\cot {\gamma\over 2} -\cot {\alpha\over 2} =\cot{\beta\over 2} +\cot{\gamma\over 2} \Rightarrow \cot {\alpha\over 2}\cot {\beta\over 2}\cot {\gamma\over 2}=\cot {\alpha\over 2} +\cot{\beta\over 2} +\cot{\gamma\over 2} \\ =3\cot{\beta \over 2} (\because\cot {\alpha\over 2} +\cot {\gamma\over 2} =2\cot {\beta\over 2}) \Rightarrow \cot {\alpha\over 2}\cot {\gamma\over 2} = \bbox[red,2pt]{3}$$

解答：

$$假設F_1(\sqrt 5,0)對稱L的對稱點為F_1'(5,5-\sqrt 5)，則\overline{F_2F_1'}與L的交點P即為橢圓的切點；\\也就是\overline{F_2F_1'} = \overline{PF_2}+\overline{PF_1}=2a \Rightarrow \sqrt{(5+\sqrt 5)^2 +(5-\sqrt 5)^2} =2\sqrt{15} \Rightarrow a=\sqrt{15}\\ \Rightarrow b^2=a^2-c^2= 15-5=10 \Rightarrow 橢圓方程式:\bbox[red, 2pt]{{x^2 \over 15}+ {y^2\over 10}=1}$$

解答：$$令a=\sqrt{\log_2 x-1} \Rightarrow \log_2 x=a^2+1 \Rightarrow \log_{1/2}x^3 =-\log_2 x^3 =-3\log_2 x=-3(a^2+1)\\ 因此原式:a+ {1\over 2}\cdot (-3a^2-3)+2 \gt 0 \Rightarrow 3a^2-2a-1 \lt 0 \Rightarrow (3a+1)(a-1)\lt 0\\ \Rightarrow -{1\over 3}\lt a\lt 1  \Rightarrow 0\le \sqrt{\log_2 x-1} \lt 1 \Rightarrow 0\le \log_2 x-1 \lt 1 \Rightarrow 1\le \log_2 x \lt 2\\ \Rightarrow \bbox[red, 2pt]{2\le x\lt 4}$$

解答：

$$\cases{C:y^2=ax\\ L:py=x} \Rightarrow 交點A(ap^2,ap) \\\Rightarrow 所圍區域(上圖著色區域)繞x軸旋轉體積= \pi\int_0^{ap^2} (ax-{1\over p^2}x^2)\;dx \\=\pi \left. \left[ {1\over 2}ax^2- {1\over 3p^2}x^3\right]\right|_0^{ap^2} =\left({1\over 2}a^3p^4-{1\over 3}a^3p^4 \right)\pi =\bbox[red, 2pt]{a^3p^4\pi\over 6}$$

解答：

$$圓x^2+y^2-4x-7y+10=0 \Rightarrow (x-2)^2 +(y-{7\over 2})^2={25\over 4} \Rightarrow \cases{圓心C(2,{7\over 2})\\ 半徑r={5\over 2}} \\ \Rightarrow 圓在第一象限及第二象限，並與y軸交於\cases{P(0,5)\\ Q(0,2)}；\\而直線L:y=m(x-3)必經A(3,0)， 因此L斜率介於\overleftrightarrow{AP} 與\overleftrightarrow{BP}之間，即符合所求；\\由於\cases{\overleftrightarrow{AP} :y=-{5\over 3}x+5\\ \overleftrightarrow{BP}: y=-{2\over 3}x+2} \Rightarrow \bbox[red, 2pt]{-{5\over 3} \lt m\lt -{2\over 3}}$$

解答：

$$圓:(x-5)^2+(y-12)^2=20^2\cases{與x軸交於\cases{P(21,0)\\ Q(-11,0)} \\與y軸交於\cases{R(0,12+5\sqrt{15})\\ S(0,12-5\sqrt{15})} }\\ 取\cases{A在x軸上，且\overline{OQ}=\overline{AP} \\C在y軸上，且\overline{OS}=\overline{CR}}\quad，則矩形OABC面積即為所求\\，也就是24\times 10=\bbox[red,2pt]{240}$$

貳： 非選擇題

解答：$$由於試題未附z-表，只能以z=3來推估P(-3\le X\le 3)=99.7\%\\ 因此誤差e \le 3\% \Rightarrow 3 \cdot \sqrt{0.5(1-0.5)\over n} \le 0.03 \Rightarrow {0.5\over \sqrt n} \le 0.01 \Rightarrow 50\le \sqrt n \Rightarrow n=\bbox[red, 2pt]{2500}\\事實上，P(|X|\le z)=99.7\% \Rightarrow z=2.967，依此來計算，n \ge   2445.3 \Rightarrow n=2446$$

解答：$$x^5-1=0 \Rightarrow (x-1)(1+x+x^2 +x^3+x^4)=0 \\\Rightarrow 令\omega,\omega^2,\omega^3,\omega^4為1+x+x^2 +x^3+x^4=0 之四根，其中\omega = e^{i2\pi/5} \\ \Rightarrow f(x)=1+x+x^2 +x^3+x^4 =(x-\omega)(x-\omega^2) (x-\omega^3)(x-\omega^4) \\ \Rightarrow f(1)=1+1+1+1+1 =(1-\omega)(1-\omega^2)(1-\omega^3)(1-\omega^4) \\\Rightarrow (1-\omega)(1-\omega^2)(1-\omega^3)(1-\omega^4)=5\\令\cases{A(1)\\ B(\omega)\\ C(\omega^2)\\ D(\omega^3)\\ E(\omega^4)},\text{where }\omega^k =e^{i2k\pi/5}\\ \Rightarrow \overline{AB}\times \overline{AC}\times \overline{AD}\times \overline{AE}=|\omega-1| \cdot|\omega^2-1|\cdot |\omega^3-1|\cdot |\omega^4-1| \\ =  |(1-\omega)(1-\omega^2)(1-\omega^3) (1-\omega^4)|= \bbox[red, 2pt]{5}$$

解答：$$f(x)=x^4-4p^3x+12 \Rightarrow f'(x)=4x^3-4p^3=0 \Rightarrow x=p\\ f(x)\gt 0 \Rightarrow f(p) \gt 0 \Rightarrow p^4-4p^4+12 \gt 0 \Rightarrow 3p^4 \lt 12 \Rightarrow p^4 \lt 4\\ \Rightarrow p^2\lt 2 \Rightarrow \bbox[red,2pt]{-\sqrt 2
\lt p\lt \sqrt 2}$$

解答：

(1)$$\begin{array}{} a_{2n}-a_{2n-1} & = & ({1\over 5})^{2n}\\ a_{2n-1}-a_{2n-2} &= & ({1\over 5})^{2n-1}-({1\over 3})^{2n-1} \\ a_{2n-2}-a_{2n-3} & = & ({1\over 5})^{2n-2} \\ a_{2n-3}-a_{2n-4} &= & ({1\over 5})^{2n-3}-({1\over 3})^{2n-3} \\\cdots & &\dots\\ a_2-a_1 & = &({1\over 5})^{2} \\ a_1-a_0& = & {1\over 5}-{1\over 3}\\\hline a_{2n}-a_0&=&({1\over 5}+{1\over 5^2} +\cdots {1\over 5^{2n}})-({1\over 3}+{1\over 3^3}+\cdots +{1\over 3^{2n-1}}) \end{array}\\ \Rightarrow \lim_{n\to\infty} a_{2n}={1/5\over 1-1/5}-{1/3\over 1-1/9}={1\over 4}-{3\over 8} =\bbox[red, 2pt]{-{1\over 8}}$$(2)$$\begin{array}{} a_{2n+2}-a_{2n+1} & = & ({1\over 5})^{2n+2}\\a_{2n+1}-a_{2n} & = & ({1\over 5})^{2n+1}-({1\over 3})^{2n+1}   \\\hline a_{2n+2}-a_{2n} & = & {6\over 5}({1\over 5})^{2n+1}-({1\over 3})^{2n+1}   \end{array}\\ \Rightarrow  a_{2n+2}-a_{2n}={{6\over 5}\cdot 3^{2n+1}-5^{2n+1}\over 15^{2n+1}}\\ 當n=0時，a_2-a_0=({18\over 5}-5)/15 \lt 0，成立；\\假設n=k時亦成立，即a_{2k+2}-a_{2k}\lt 0 \Rightarrow {{6\over 5}\cdot 3^{2k+1}-5^{2k+1}\over 15^{2k+1}} \lt 0 \Rightarrow {6\over 5}\cdot 3^{2k+1}\lt 5^{2k+1}\\當n=k+1時，a_{2k+4}-a_{2k+2}= {{6\over 5}\cdot 3^{2k+3}-5^{2k+3}\over 15^{2k+3}}\\ 由於{6\over 5}\cdot 3^{2k+1}\lt 5^{2k+1} \Rightarrow  {6\over 5}\cdot 3^{2k+3}\lt 3^2\cdot 5^{2k+1}\lt 5^2\cdot 5^{2k+1}=5^{2k+3}\\ \Rightarrow {6\over 5}\cdot 3^{2k+3}-5^{2k+3}\lt 0 \Rightarrow a_{2k+4}-a_{2k+2}\lt 0，即n=k+1時亦成立\\由歸納法知:a_{2n+2}-a_{2n} \lt 0，\bbox[red, 2pt]{故得證}\\因此a_{2n} \lt a_{2n-2} \lt \cdots \lt a_2 \lt a_0=0 \Rightarrow a_{2n}\lt 0；\\ 再由\lim_{n\to \infty}a_{2n}=-{1\over 8}，可得-{1\over 8}\le a_n \lt 0,n\in \mathbb{N}$$

解答：

(1)$$C^n_k = C^{n-1}_k + C^{n-1}_{k-1} = \left( C^{n-2}_k +C^{n-2}_{k-1}\right) +\left(C^{n-2}_{k-1} +C^{n-2}_{k-2} \right) = C^{n-2}_{k} + 2C^{n-2}_{k-1} +C^{n-2}_{k-2}，\bbox[red, 2pt]{故得證}$$(2)$$從n個人中挑出k個的方法可區分成三種情況:\\ 甲乙都沒被選中:C^{n-2}_{k}\\ 甲乙其中一人被選中:C^2_1C^{n-2}_{k-1}=2C^{n-2}_{k-1}\\ 甲乙都被選中:C^{n-2}_{k-2}\\ 因此n個人選k個的方法數C^n_k =C^{n-2}_{k} + 2C^{n-2}_{k-1} +C^{n-2}_{k-2}$$

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