115年麗山高中教甄-數學詳解

 臺北市立麗山高級中學 115 學年度第一次正式教師甄選

一、填充題（ 14 題，每題 5 分）

解答:

$$\cases{\log \alpha^3+3\alpha-6=0 \Rightarrow 3\log \alpha=6-3\alpha \Rightarrow \log \alpha=2-\alpha \\ 10^{\beta+1} +10\beta-20=0 \Rightarrow 10\cdot 10^\beta+10\beta-20=0 \Rightarrow10^\beta =2-\beta} \\ \Rightarrow \cases{A( \alpha,\log\alpha) 是y=\log x與直線L:y=2-x的交點 \\B(\beta,10^\beta)是y=10^x與直線L:y= 2-x的交點} \Rightarrow \cases{A\in L \Rightarrow A(\alpha,2-\alpha) \\B\in L \Rightarrow B(\beta,2-\beta)} \\ 由於y=10^x與y=\log x互為反函數，兩圖形對稱y=x;且直線L也對稱y=x \\ 因此\cases{\alpha=2-\beta\\ \beta=2-\alpha} \Rightarrow \alpha+\beta= \bbox[red, 2pt]2$$

解答:

$$圓C:(x+1)^2+(y-2)^2=5 \Rightarrow \cases{圓心M(-1,2)\\ 圓半徑r=\sqrt 5} \Rightarrow \cases{\overline{AM}=\sqrt{4^2+3^2}=5 \\ \overrightarrow{AM}=(-4,3)} \\\Rightarrow \overline{AP} =\sqrt{\overline{AM}^2-r^2} = \sqrt{25-5}=2\sqrt 5 \\ \overrightarrow{AQ} =\overrightarrow{AM}+ \overrightarrow{MQ} \Rightarrow \overrightarrow{AP} \cdot \overrightarrow{AQ} = \overrightarrow{AP} \cdot (\overrightarrow{AM}+ \overrightarrow{MQ} ) =\overrightarrow{AP} \cdot  \overrightarrow{AM} +\overrightarrow{AP} \cdot \overrightarrow{MQ}  \\ =\overrightarrow{AP} \cdot  (\overrightarrow{AP} + \overrightarrow{PM}) +\overline{AP} \cdot \overline{MQ} \cos \theta  =\overline{AP}^2+0+\overline{AP} \cdot r \cos \theta= 20+10\cos \theta\\ \Rightarrow 最大值=20+10= \bbox[red,2pt]{30}$$

解答:$$\Gamma: \cases{2x+3y=1\\ 3x-2y=-5} \Rightarrow \cases{x=-1\\ y=1} \Rightarrow P(-1,1,t_1)\in \Gamma \\ \Lambda:\{(x,y,z) \mid x=4,y=k\} \Rightarrow Q(4,k,t_2) \in \Lambda \\ M=\overline{PQ} 中點 \Rightarrow M={P+Q\over 2} = \left( {3\over 2},{k+1\over 2},{t_1+t_2\over 2} \right) \Rightarrow \cases{x_0=3/2\\ 7=(k+1)/2 \Rightarrow k=13} \\ \Rightarrow Q(4,13,t_2)\Rightarrow m=\sqrt{(4-(-1))^2+(13-1)^2} = 13 \Rightarrow x_0+k+m={3\over 2}+13+13= \bbox[red, 2pt]{55\over 2}$$

解答:$$報數(a_n)依序為1,1,2,(3),5,8,13,(21),34,55,89,(144),\dots \\ \Rightarrow 循環數為4,即a_n=4, n=4k, k\in \mathbb N \Rightarrow 30=7\times 4+2 \Rightarrow 共出現\bbox[red, 2pt]7個3的倍數$$

解答:$$(f(x))^2+(g(x))^2=0 \Rightarrow f(x)=g(x)=0 \\ \cases{f(x)=x^3+4x^2+x-6 =(x-1)(x+2)(x+3)=0 \Rightarrow x=1,-2,-3 \\g(x)=2x^3+(k-2)x^2+kx-2k = (x-1)(2x^2+kx+2k) =0 \Rightarrow x=1,(-k\pm \sqrt{k^2-16k})/4} \\ 取\alpha=1為共同實根，另一實根\beta 可能是-2或-3 \\ 若\beta=-2 \Rightarrow 2(-2)^2-2k+2k=8\ne 0 \Rightarrow -2不是g(x)=0的根\\ 若\beta=-3 \Rightarrow 2(-3)^2-3k+2k=0 \Rightarrow k=18 \Rightarrow \cases{\alpha=1\\ \beta=-3\\k=18} \Rightarrow k+\alpha+\beta= \bbox[red, 2pt]{16}$$

解答:$$\cases{E_1: x+2y+2z=7\\ E_2:x+2y+2z=16} \Rightarrow \overline{BC}=d(E_1,E_2) ={9\over \sqrt{1+4+4}} = 3 \\ \overrightarrow{AD} =\overrightarrow{AB}+ \overrightarrow{BC}+ \overrightarrow{CD} \Rightarrow \overline{AD}^2= (\overrightarrow{AB}+ \overrightarrow{BC}+ \overrightarrow{CD}) \cdot (\overrightarrow{AB}+ \overrightarrow{BC}+ \overrightarrow{CD}) \\=\overline{AB}^2 +\overline{BC}^2+ \overline{CD}^2 +2(\overrightarrow{AB}\cdot \overrightarrow{BC}+ \overrightarrow{BC} \cdot \overrightarrow{CD}+ \overrightarrow{CD} \cdot \overrightarrow{AB}) \\\Rightarrow 5^2=2^2+3^2+3^2+2(0+0+\overrightarrow{CD} \cdot \overrightarrow{AB}) \Rightarrow \overrightarrow{CD} \cdot \overrightarrow{AB} ={3\over 2} \\ \cases{L_1的方向向量= \overrightarrow{AB} \\ L_2 的方向向量= \overrightarrow{CD}} \Rightarrow \overrightarrow{AB} \cdot \overrightarrow{CD} = \overline{AB}\cdot \overline{CD} \cos \theta \Rightarrow {3\over 2}=2\cdot 3 \cos \theta \Rightarrow \cos \theta={1\over 4} \\ \Rightarrow \sin \theta=\sqrt{1-{1\over 16}} = \bbox[red, 2pt]{\sqrt{15}\over 4}$$

解答:$$(2x^2-4)^2= (3x^2-2x-12)(x^2+2x+4) \Rightarrow x^4-4x^3-12x^2+32x+64=0 \\ \Rightarrow (x+2)^2(x-4)^2=0 \Rightarrow \bbox[red, 2pt]{x= 4,-2}$$

解答:

$$底部圓面積為\pi \Rightarrow 圓半徑r=1 \Rightarrow \overline{OA}^2=(\sqrt{143})^2 +r^2 =12^2 \Rightarrow \overline{OA}=12 \Rightarrow \overline{OB}=12/2=6 \\ 將直圓錐展開後成為一個扇形，扇形圓心為O, 扇形半徑R=\overline{OA}=12 \Rightarrow 圓心角\theta \Rightarrow R\theta=圓周長 \\ \Rightarrow 12\theta=2\pi \Rightarrow \theta={\pi\over 6} =30^\circ \\螞蟻爬兩圈，因此將兩個相同的扇形並列，成為圓心角60^\circ的大扇形。\\假設大扇形OAA_2, B_2={O+A_2 \over 2} \Rightarrow 爬兩圈最短離=\overline{AB_2} \\ \triangle OAB_2: \cos \angle AOB_2 ={\overline{OA}^2+ \overline{OB_2}^2 -\overline{AB_2}^2 \over 2\cdot \overline{OA} \cdot \overline{OB_2}} \Rightarrow {1\over 2} ={144+36-\overline{AB_2}^2\over 144} \\ \Rightarrow \overline{AB_2}^2= 108 \Rightarrow \overline{AB_2}= \bbox[red, 2pt]{6\sqrt 3}$$

解答:$$L_1: x-1={y-3\over 2}={z-4\over 2} \Rightarrow \cases{L_1的方向向量\vec u_1=(1,2,2) \\ P(1,3,4) \in L_1} \\ L_2:\cases{x+y=4\\ y+2z=2} \Rightarrow \cases{L_2的方向向量\vec u_2=(1,1,0)\times(0,1,2)=(2,-2,1) \\ Q(2,2,0) \in L_2} \\ \Rightarrow \vec u_1\cdot \vec u_2=2-4+2=0 \Rightarrow L_1\bot L_2 \\ \Rightarrow 兩歪斜線距離 d= {(\vec u_1\times \vec u_2) \cdot \overrightarrow{PQ} \over |\vec u_1\times \vec u_2|}={(6,3,6)\cdot (1,-1,-4) \over |(6,3,6)|} =3 \Rightarrow 體積=3^3=\bbox[red, 2pt]{27}$$

解答:

$$假設正七邊形的七個頂點為z^7=1的七個根, 其中z_0=1, z_k= \cos {2k\pi\over 7}+i\sin {2k\pi \over 7},k=1-6\\ 即A(z_0), B(z_1), \dots,G(z_6) \Rightarrow \cases{\overline{AB}=|z_1-1| \\ \overline{AC}=|z_2-1| \\ \cdots\\ \overline{AG} =|z_6-1} \\ z^7-1=(z-1)(z^6+z^5+\cdots+z+1) \Rightarrow f(z)=z^6+z^5+\cdots +z+1=(z-z_1)(z-z_2) \cdots(z-z_6) \\ \Rightarrow f(1)=7=(1-z_1)(1-z_2)\cdots (1-z_6) \Rightarrow |1-z_1|\cdot |1-z_2| \cdots|1-z_6|=7 \\ \Rightarrow \overline{AB} \times \overline{AC}\times \overline{AD} \times \overline{AE}\times \overline{AF} \times \overline{AG}=7\\由於\cases{\overline{AB} =\overline{AG} \\ \overline{AC} =\overline{AF} \\ \overline{AD} =\overline{AE}} \Rightarrow \left( \overline{AB} \times \overline{AC}\times \overline{AD} \right)^2=7 \Rightarrow \overline{AB} \times \overline{AC}\times \overline{AD} = \bbox[red, 2pt]{\sqrt 7}$$

解答:

$$假設P=\overline{AC}\cap \overline{BD} \Rightarrow \angle APB=90^\circ \Rightarrow \stackrel{\Large \frown}{AB}+\stackrel{\Large \frown}{CD} =180^\circ\\ 假設\cases{圓半徑R\\ \angle AOB=2\alpha\\ \angle COD=2\beta } \Rightarrow 2\alpha+2\beta=180^\circ \Rightarrow \alpha+\beta= 90^\circ \\ 正弦定理: {\overline{AB} \over \sin \alpha} ={\overline{CD} \over \sin \beta =\cos \alpha} =2R \Rightarrow \overline{AB}^2+ \overline{CD}^2=4R^2(\sin^2\alpha+\cos^2\alpha) \\ \Rightarrow 2^2+3^2=4R^2 \Rightarrow R^2={13\over 4} \Rightarrow 圓面積=R^2\pi= \bbox[red, 2pt]{13\pi\over 4}$$

解答:$$等軸雙曲線x^2-y^2=a^2 上任一點P(x_0,y_0) \Rightarrow \cases{\overline{PF_1} =|ex_0+a|\\ \overline{PF_2} =|ex_0-a|}, 其中e=離心率={c\over a} =\sqrt 2 \\ \Rightarrow \overline{PF_1}\cdot \overline{PF_2}=|e^2x_0^2-a^2|=|2x_0^2-(x_0^2-y_0^2)|=|x_0^2+y_0^2| =\overline{PO}^2= \bbox[red, 2pt]{81}$$

解答:$$f(x)代表走到x號房的走法數 \Rightarrow \cases{A\to 1只有1種走法\Rightarrow f(1)=1\\ A\to 2:A到1到2,或A到2,共2種\Rightarrow f(2)=2} \\ \Rightarrow f(3)=f(1)+ f(2)=3 \Rightarrow f(4)=f(2)+f(3)=5 \Rightarrow f(5)=f(3)+f(4)=8 \\ \Rightarrow f(6)=f(5)+f(4)=13 \Rightarrow f(7) =f(5)+ f(6)=21 \Rightarrow f(8)=f(7)+f(6)=34\\ \Rightarrow f(9)=f(8)+f(7)=55 \Rightarrow f(10)=f(9)+f(8)= \bbox[red, 2pt]{89}$$

解答:$$a_{n} =a_{n-1}+2^{n-1} =a_{n-2}+2^{n-2} +2^{n-1} = \cdots=a_1+ 2^1+2^2+\cdots+2^{n-1} \\ =1+2^1+2^2+\cdots+2^{n-1} =2^n-1 \Rightarrow a_n=2^n-1 \\ \Rightarrow {1\over a_n\times a_{n+1}} ={1\over (2^n-1)(2^{n+1}-1)} =2^n \left( {1\over 2^n-1}-{1\over 2^{n+1}-1} \right) \\ \Rightarrow {1\over a_1\times a_2}+{2\over a_2\times a_3}+ \cdots+{2^9\over a_{10} \times a_{11}} ={1\over 2}\left[ \left( {1\over a_1}-{1\over a_2} \right) +\left( {1\over a_2}-{1\over a_3} \right) + \cdots+ \left( {1\over a_{10}}-{1\over a_{11}} \right) \right] \\= {1\over 2} \left( {1\over a_1}-{1\over a_{11}} \right)= {1\over 2} \left( 1-{1\over 2^{11}-1} \right) ={1\over 2}\cdot {2046\over 2047} =\bbox[red, 2pt]{1023\over 2047}$$

二、 計算證明題（ 3 題，每題 10 分）

解答:$$假設S= \sum_{i=1}^n a_i, \\任2個砝碼相加，代表每一個砝碼會與另外的n-1個砝碼相加\\ 因此k=(n-1)a_1+ (n-1)a_2+\cdots+(n-1)a_n =(n-1)S \\任取 3 個代表每一個特定的砝碼會與剩下的 「n-1 個砝碼中任取 2 個」搭配，\\所以每個砝碼出現的次數為{n-1\choose 2}={(n-1)(n-2)\over 2} \Rightarrow 2k={(n-1)(n-2)\over 2}S \\ \Rightarrow 2(n-1)S ={(n-1)(n-2)\over 2}S \Rightarrow n=6 \Rightarrow k=5S \\ 柯西不等式: (a_1^2+a_2^2+ \cdots+a_n^2)(1^2+\cdots +1^2) \ge (a_1+\cdots+a_n)^2 \Rightarrow 6(a_1^2+a_2^2+ \cdots+a_n^2)\ge S^2 \\ \Rightarrow (a_1^2+a_2^2+ \cdots+a_n^2) \ge {S^2\over 6} \Rightarrow {S^2\over 6}=2400 \Rightarrow S=120 \Rightarrow k=5S= \bbox[red, 2pt]{600}$$

解答:$$\textbf{(1) }s={1\over 2}(4+5+6) ={15\over 2} \Rightarrow \triangle ABC={1\over 2} \sqrt{s(s-4)(s-5)(s-6)} ={15\over 4} \sqrt 7 \\ \Rightarrow {15\over 4} \sqrt 7 ={1\over 2}\cdot \overline{BC} \cdot \overline{AH} ={1\over 2}\cdot 6\cdot \overline{AH} \Rightarrow \overline{AH}= \bbox[red, 2pt]{5\sqrt 7\over 4}$$

$$\textbf{(2) }設\cases{H'為H對於\overline{AB}的對稱點\\ H''為H對於\overline{AC}的對稱點} \Rightarrow \cases{P=\overline{H'H''} \cap \overline{AB} \\ Q= \overline{H'H''} \cap \overline{AC}} \\ \Rightarrow \triangle PQH周長= \overline{PH} +\overline{HQ} +\overline{QP} =\overline{PH'} +\overline{H''Q} +\overline{QP} = \overline{H'H''} \\ 假設h=\overline{AH} ={5\sqrt 7\over 4} \Rightarrow \overline{AH'} =\overline{AH''} =h 且\angle H'AH'' =2\angle A \\ \cos \angle A={4^2+5^2-6^2 \over 2\cdot 4\cdot 5} = {1\over 8} \Rightarrow \cos 2\angle A=2\cos^2 \angle A-1= -{31\over 32} ={2h^2-\overline{H'H''}^2\over 2h^2} \\ \Rightarrow \overline{H'H''}^2= {63\over 16}h^2 \Rightarrow \overline{H'H''} ={3\sqrt 7\over 4}\cdot {5\sqrt 7\over 4} = \bbox[red, 2pt]{105\over 16}$$

解答:$$\textbf{(1) } \sum_{i=1}^n (x_i-a)^2 =\sum_{i=1}^n (x_i-\mu+\mu-a)^2 = \sum_{i=1}^n (x_i-\mu)^2+ 2(\mu-a) \sum_{i=1}^n(x_i-\mu) +(\mu-a)^2 \sum_{i=1}^n 1 \\= \sum_{i=1}^n (x_i-\mu)^2+ 2(\mu-a) \cdot 0 +(\mu-a)^2 \cdot n = \sum_{i=1}^n (x_i-\mu)^2+  n(\mu-a)^2 \ge \sum_{i=1}^n (x_i-\mu)^2 \\ \Rightarrow  \sum_{i=1}^n (x_i-a)^2 \ge  \sum_{i=1}^n (x_i-\mu)^2, \bbox[red, 2pt]{故得證} \\ \textbf{(2) } 假設 \cases{M=   \text{max}_{i\le i\le n} x_i \\m=   \text{min}_{i\le i\le n} x_i  } \Rightarrow R=M-m \\ 取a={M+m\over 2},代入第(1)小題可得 \sum_{i=1}^n (x_i-\mu)^2 \le \sum_{i=1}^n \left( x_i-{M+m\over 2} \right)^2\\ 由於 \left| x_i-{M+m\over 2}\right| \le {M-m\over 2} ={R\over 2} \Rightarrow \left( x_i-{M+m\over 2} \right)^2 \le {R^2\over 4} \\ 因此\sum_{i=1}^n (x_i-\mu)^2 \le \sum_{i=1}^n \left( x_i-{M+m\over 2} \right)^2 \le \sum_{i=1}^n {R^2\over 4} ={nR^2\over 4} \\ \Rightarrow {1\over n} \sum_{i=1}^n (x_i-\mu)^2 \le {R^2\over 4} \Rightarrow \sigma^2 \le {R^2\over 4} \Rightarrow \sigma \le {R\over 2}\;\bbox[red, 2pt]{故得證}$$

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解題僅供參考，其他教甄試題及詳解