112年新竹六家高中教甄-數學詳解

新竹縣立六家高中 112 學年度 第 1 次教師甄選

解答: $$z_1^2=z_2^2 = -2-2\sqrt 3i = 4e^{4\pi i/3}, 取\cases{z_1=2e^{2\pi i/3}\\ z_2=-2e^{2\pi i /3}}\\ 令\cases{A(z_1)\\ B(z_2)\\ C(z)\\ O=\overline{AB}中點=原點},由|z-z_1|=|z-z_2|\Rightarrow \overline{AC}=\overline{BC} \Rightarrow C在\overline{AB} 的中垂線上 \\ \Rightarrow \overline{AC}^2 = \overline{OA}^2+\overline{OC}^2 \Rightarrow 4^2 =2^2+|z|^2 \Rightarrow |z|=\bbox[red, 2pt]{2\sqrt 3}$$

解答: $$a_n= {n\over n^2-1} ={n\over (n-1)(n+1)} \Rightarrow a_1\cdot a_2\cdot a_3\cdots a_k = 1\cdot {2\over 1\cdot 3}\cdot {3\over 2\cdot 4}\cdot{4\over 3\cdot 5}\cdots {k\over (k-1)(k+1)} \\={2\over (k-1)!(k+1)} ={2k\over (k+1)! } =2\left( {1\over k!}-{1\over (k+1)!}\right) \\ \Rightarrow \sum_{k=1}^n (a_1\cdot a_2\cdots a_k) =2\sum_{k=1}^n\left( {1\over k!}-{1\over (k+1)!}\right)=2(1-{1\over (n+1)!})= \bbox[red, 2pt]{2-{2\over (n+1)!}}$$

解答: $$f(x)=\int_{x^2}^{x^3} (t^2+t-3)^3\,dt \Rightarrow f'(x)=(x^6+x^3-3)^3(3x^2)-(x^4+x^2-3)^3(2x) \\ \Rightarrow f'(-1)=(-3)^3\cdot 3-(-2)\cdot (-1) =-81-2=\bbox[red ,2pt]{-83}$$

解答: $$令\cases{\vec u=2\vec a+\vec b\\ \vec v=\vec a-4\vec b} \Rightarrow \cases{\vec b=(\vec u-2\vec v)/9\\ \vec a=(4\vec u+\vec v)/ 9} \Rightarrow 3\vec a+5\vec b={1\over 9}(17\vec u- 7\vec v) \\ \Rightarrow \cases{|3\vec a+5\vec b| \le {1\over 9}(17|\vec u|+7|\vec v|)={1\over 9}(34+7) ={41\over 9}\\ |3\vec a+5\vec b| \ge {1\over 9}(17|\vec u|-7|\vec v|)={1\over 9}(34-7)=3} \Rightarrow (M,m)=\bbox[red, 2pt]{({41\over 9},3)}$$

解答: $$f(x)={x^2+x+16\over x} \Rightarrow f'(x)=1-{16\over x^2} \Rightarrow f''(x)={32\over x^3}\\ 若f'(x)=0 \Rightarrow x^2=16 \Rightarrow x=\pm 4 \Rightarrow f''(4) \gt 0 \Rightarrow f(4)=9 為極小值\Rightarrow m\ge 4\\ f(x)=11 \Rightarrow x+1+{16\over x}=11 \Rightarrow x+{16\over x}=10 \Rightarrow \cases{x=2\\ x=8}\\ 因此f(x)\in [9,11] 且 x \in [2,8] \Rightarrow \bbox[red, 2pt]{4\le m\le 8}$$

解答: $$f(a,b)=(a-41b-33)^2 +(a-42b-34)^2+(a-40b-35)^2+ (a-39b-32)^2\\ \qquad +(a-38b-31)^2\\ \Rightarrow f_a=0 \Rightarrow 5a-200b-165=0 \Rightarrow a=40b+33 代回原式 \\ \Rightarrow f(b)=(-b)^2+(-2b-1)^2 +(-2)^2 +(b+1)^2+(2b+2)^2 =10b^2+14b+10 \\ \Rightarrow f'(b)=20b+14=0 有極值,此時b=-{7\over 10} \Rightarrow  a=-{7\over 10} \cdot 40+33=5 \\ \Rightarrow (a,b)=\bbox[red, 2pt]{(5,-{7\over 10})},但公布的答案是\bbox[blue,2pt]{(5,{7\over 10})}$$

解答: $$A:0^\circ N,15^\circ E \Rightarrow A(R\sin 90^\circ \cos 15^\circ,R\sin 90^\circ \sin 15^\circ, R\cos 90^\circ) =(5(\sqrt 6+\sqrt 2) ,5(\sqrt 6-\sqrt 2),0)\\ B:45^\circ S,120^\circ W \Rightarrow B(R\sin 135^\circ\cos (-120^\circ), R\sin 135^\circ\sin (-120^\circ), R\cos 135^\circ) =(-5\sqrt 2,-5\sqrt 6,-10\sqrt 2) \\ \Rightarrow \overline{AB}= \sqrt{(5\sqrt 6+10\sqrt 2)^2 +(10\sqrt 6-5\sqrt 2)^2+(10\sqrt 2)^2}=20\sqrt 3 \\ \Rightarrow \angle AOB=120^\circ ={2\pi\over 3} \Rightarrow \stackrel{\frown}{AB} =20\cdot {2\pi \over 3} =\bbox[red, 2pt]{40\pi\over 3}$$

解答: $$f(x)=x^3-3x+1 \Rightarrow f'(x)=3x^2-3 \Rightarrow \cases{f(a_0)=3\\ f'(a_0)=9} \\ \Rightarrow 在(a_0,f(a_0))的切線L: y=f'(a_0)(x-a_0)+f(a_0) \Rightarrow y=9(x-2)+3 \\ \Rightarrow L與x軸交於 9(x-2)+3=0 \Rightarrow x={5\over 3} \Rightarrow a_1=\bbox[red, 2pt]{5\over 3}$$

解答: $$取\triangle BCD在x-y平面上,原點即為重心,即\cases{B(0,2\sqrt 3,0)\\ C(-3,-\sqrt 3,0)\\ D(3,-\sqrt 3,0)\\ A(0,0,a)},又\overline{AB}=5\Rightarrow a=\pm \sqrt{13} \\ 取A(0,0,\sqrt{13}) \Rightarrow \cases{L_1=\overleftrightarrow{AB} =(0, -2\sqrt 3 t,\sqrt{13}t+\sqrt{13}),t\in \mathbb R\\ L_2= \overleftrightarrow{CD}=(s+3,-\sqrt 3,0),s\in \mathbb R } \\ \Rightarrow d(L_1,L_2)= \sqrt{(s+3)^2 +(-2\sqrt 3t+\sqrt 3)^2 +(\sqrt{13}t+\sqrt{13})^2}\\ 當\cases{s=-3\\ t=7/25}時,d(L_1,L_2)有最小值\bbox[red,2pt]{3\sqrt{39}\over 5}$$

解答: $$一路領先的次數=C^{11}_6-C^{11}_7=462-330=132 \Rightarrow 機率={132\over 462}=\bbox[red, 2pt]{2\over 7}\\ \href{https://chu246.blogspot.com/2021/11/blog-post.html}{公式來源}$$

解答: $$a_{n+2}=2a_{n+1}+a_n \Rightarrow a_n-2a_{n-1}-a_{n-2}=0 \Rightarrow a_n=C_1(1+\sqrt 2)^n +C_2(1-\sqrt 2)^n\\ 初始值\cases{a_1=1\\ a_2=2} \Rightarrow \cases{C_1(1+\sqrt 2) +C_2(1-\sqrt 2) =1\\ C_1(1+\sqrt 2)^2 +C_2(1-\sqrt 2)=2} \Rightarrow \cases{C_1=1/2\sqrt 2\\ C_2=-1/2\sqrt 2} \\ \Rightarrow a_n={1\over 2\sqrt 2}(1+\sqrt 2)^n -{1\over 2\sqrt 2}(1-\sqrt 2)^n  \triangleq C(\alpha^n-\beta^n),其中\cases{\alpha=1+\sqrt 2\\ \beta=1-\sqrt 2\\ C=1/2\sqrt 2} \\因此 \cases{\alpha+\beta =2\\ \alpha\beta=-1 } \Rightarrow \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta =6\\ 現在\cases{a_{k+1}^2 = C^2(\alpha^{2k+2} +\beta^{2k+2}-2\alpha^{k+1}\beta^{k+1}) \\ a_ka_{k+2}= C^2(\alpha^{2k+2}+ \beta^{2k+2}-\alpha^k\beta^{k+2}- \alpha^{k+2}\beta^k)} \\ \Rightarrow a_{k+1}^2-a_ka_{k+2}=C^2(\alpha^k\beta^{k+2}+ \alpha^{k+2}\beta^k -2\alpha^{k+1}\beta^{k+1}) =C^2((\alpha\beta)^k (\alpha^2+\beta^2)-2(\alpha\beta)^{k+1})\\ ={1\over 8}(6(-1)^k-2(-1)^{k+1}) ={1\over 8}(6(-1)^k+2(-1)^{k}) =(-1)^k\\ \Rightarrow \sum_{k=1}^{2023}(-1)^k =(-1+1)+ (-1+1)+\cdots +(-1+1)-1=\bbox[red, 2pt]{-1}$$

解答: $$f(x)=|2-\log_3 x| \Rightarrow f(9)=0為最小值 \Rightarrow \cases{f(x)遞減,x\lt 3\\ f(x)遞增, x\gt 3} \Rightarrow \cases{a\lt b\lt 9\lt c\\ 2-\log_3 c \lt 0}\\ 因此f(b)=f(c) \Rightarrow 2-\log_3 b= \log_3 c-2 \Rightarrow \log_3 bc=4 \Rightarrow bc=3^4\\ 又2-\log_3 a= 2(2-\log_3 b) \Rightarrow \log_3 {b^2\over a}=2 \Rightarrow b^2=9a \Rightarrow {b\over ac}={b^2\over abc}={9\over bc} ={9\over 3^4}=\bbox[red, 2pt]{1\over 9}$$

解答: $$令\cases{a_n:第n個人拿到白球的方法數\\ b_n:第n個人拿到不是白球的方法數}\\ n=1 \Rightarrow \cases{a_1=1:第1個人只能拿到白球\\ b_1=4:第1個人可能拿到紅黃藍黑四色之一} \\n=2\Rightarrow \cases{a_2=5:第2個拿到白球,第1個人可以拿5色任一球\\ b_2=8:第2個人拿到四色之一,第1個人是白色或同色,共4\cdot 2 =8}\\ 依此類推,可得\cases{a_n=a_{n-1}+ b_{n-1}\\ b_n=4a_{n-1}+ b_{n-1}} \Rightarrow \begin{bmatrix}a_n \\ b_n \end{bmatrix}=\begin{bmatrix}1 & 1\\ 4 & 1 \end{bmatrix}\begin{bmatrix}a_{n-1}\\ b_{n-1} \end{bmatrix} \\ \Rightarrow \begin{bmatrix}a_6 \\ b_6 \end{bmatrix}=\begin{bmatrix}1 & 1\\ 4 & 1 \end{bmatrix}\begin{bmatrix}a_{5}\\ b_{5} \end{bmatrix} =\begin{bmatrix}1 & 1\\ 4 & 1 \end{bmatrix}^5\begin{bmatrix}a_{1}\\ b_{1} \end{bmatrix} =\begin{bmatrix}121 & 61\\ 244 & 121 \end{bmatrix} \begin{bmatrix}1\\ 4 \end{bmatrix} = \begin{bmatrix}365\\ 728 \end{bmatrix} \\ \Rightarrow 六個人的總方法數=365+728= \bbox[red, 2pt]{1093}$$

解答: $$D是\overline{BC}中點 \Rightarrow \overrightarrow{AD}={1\over 2}\overrightarrow{AB}+{ 1\over 2} \overrightarrow{AC} ={1\over 2x}\overrightarrow{AM}+{ 1\over 2y} \overrightarrow{AN} \\ \Rightarrow {5\over 3}\overrightarrow{AG}={1\over 2x}\overrightarrow{AM}+{ 1\over 2y} \overrightarrow{AN}\Rightarrow \overrightarrow{AG}={3\over 10x}\overrightarrow{AM}+{ 3\over 10y} \overrightarrow{AN} \\ M,G,N在一直線上\Rightarrow {3\over 10x}+{3\over 10y}=1 \Rightarrow {1\over x}+{1\over y}={10\over 3},並滿足0\lt x,y\le 1\\ 柯西不等式:({1\over x^2}+({2\over y})^2) (1^2+({1\over 2})^2)\ge ({1\over x}+{1\over y})^2,此時 {1/x \over 1}={2/y\over 1/2} \Rightarrow {1\over x}={4\over y}\\ \Rightarrow {1\over x}+{1\over y}={5\over y}={10\over 3}\Rightarrow y={15\over 10}\gt 1違反y\le 1\\ y最大只能取1 \Rightarrow {1\over x}={7\over 3} \Rightarrow {1\over x^2}+{4\over y^2}={49\over 9}+4=\bbox[red, 2pt]{85\over 9}$$

解答: $$\lim_{n\to \infty}{\ln\left((1+{1\over n})(1+{2\over n})\cdots(1+{n\over n}) \right) \over n} =\lim_{n\to \infty} \sum_{k=1}^n{1\over n}\ln(1+{k\over n}) =\int_0^1 \ln(1+x)\,dx \\ =\left. \left[(x+1)\ln(x+1)-x \right]\right|_0^1 = \bbox[red, 2pt]{2\ln 2-1}$$

解答: $$y=f(x)=x^3+kx^2-1 \Rightarrow f'(x)=3x^2+2kx\\ 假設切點P(a,f(a)) \Rightarrow 切線斜率=f'(a) \Rightarrow 切線方程式 y=f'(a)(x-a)+f(a)\\ 切線通過(0,0)\Rightarrow 0=f'(a)(0-a)+f(a) \Rightarrow 0=(3a^2+2ka)(-a)+a^3+ka^2-1 \\ \Rightarrow 2a^3+ka^2+1=0\\ 令g(a)=2a^3+ka^2+1 \Rightarrow g'(a)=6a^2+2ka=0 \Rightarrow 2a(3a+k)=0 \Rightarrow a=0,-k/3\\ g(a)=0 有三相異實根\Rightarrow g(0)\times g(-k/3)\lt 0 \Rightarrow 1\cdot (k^3+27)\lt 0 \Rightarrow \bbox[red, 2pt]{k\lt -3}$$

解答:

$$令\overline{BF_2}=k \Rightarrow \overline{AF_1}=3k,由雙曲線定義:\cases{\overline{AF_1}-\overline{AF_2}=2a \\ \overline{BF_1}-\overline{BF_2}=2a}\\由於\angle AF_2F_1+ \angle BF_2F_1 =180^\circ \Rightarrow \cos \angle AF_2F_1 =-\cos \angle BF_2F_1\\ \Rightarrow {5a^2+9k^2-(2a+3k)^2 \over 6\sqrt 5ak}=-{k^2+5a^2-(2a+k)^2 \over 2\sqrt 5ak } \Rightarrow {a^2-12ak\over 3} ={-a^2+4ak  \over 1} \\ \Rightarrow {a-12k\over 3\sqrt 5} ={a^2-4ak-2k^2 \over 2a+k} \Rightarrow a^2=6ak \Rightarrow a=6k \Rightarrow \cases{\overline{AF_1}= 15k \\ \overline{BF_1}= 13k}\\ 取s=(\overline{AF_1}+ \overline{BF_1}+ \overline{AB})\div 2= 16k \Rightarrow \triangle AF_1B面積= \sqrt{16k\cdot k \cdot 3k \cdot 12k} =24k^2={32\over 3} \\   \Rightarrow k={2\over 3}  \Rightarrow s={32\over 3} \Rightarrow 內切圓半徑={\triangle AF_1B\over s} ={32/3\over 32/3}=\bbox[red, 2pt]1$$

解答: $$\mathbf{(1)}\tan a_{n+1}\cdot \cos a_n=1 \Rightarrow \tan a_{n+1}= \sec a_n \Rightarrow \tan^2 a_n= \sec^2 a_{n-1} =1+\tan^2 a_{n-1} \\ \Rightarrow \tan^2 a_n=1+1+\cdots +(1+\tan^2 a_1) =(n-1)+ \tan^2 {\pi\over 6} =n-1+{1\over 3} =\bbox[red, 2pt]{3n-2\over 3} \\\mathbf{(2)} \tan a_{n+1}\cdot \cos a_n= {\sin a_{n+1} \over \cos a_{n+1}} \cdot \cos a_n=1 \Rightarrow \sin a_{n+1}={\cos a_{n+1} \over \cos a_n} \\ \Rightarrow \prod_{k=1}^{m} \sin a_k = \sin a_1 \cdot {\cos a_2\over \cos a_1}\cdot {\cos a_3\over \cos a_2} \cdot {\cos a_4\over \cos a_3} \cdots {\cos a_m\over \cos a_{m-1}} ={\sin a_1\over \cos a_1}\cdot \cos a_m ={1\over 10} \\ \Rightarrow \sec a_m =10\tan a_1={10\over \sqrt 3 } \Rightarrow \sec^2 a_m =1+ \tan^2 a_m ={100\over 3} \\ \Rightarrow 1+{3m-2\over 3}={100\over 3} \Rightarrow m= \bbox[red, 2pt]{33}$$

解答: $$\cases{a_n= 3a_{n-1}+ 5b_{n-1} \cdots(1)\\ b_n= a_{n-1}+7b_{n-2}\cdots(2)} \Rightarrow \cases{7a_n= 21 a_{n-1}+ 35b_{n-1}\\ 5b_n= 5a_{n-1}+35b_{n-2}},\;兩式相減 \Rightarrow 7a_n-5b_n= 16a_{n-1} \\ \Rightarrow b_n={7\over 5}a_n-{16\over 5}a_{n-1} \Rightarrow b_{n-1}={7\over 5}a_{n-1}-{16\over 5}a_{n-2} 代入(1) \Rightarrow a_n= 10a_{n-1}-16a_{n-2} \\ \Rightarrow a_n-10a_{n-1} +16a_{n-2}=0 \Rightarrow \alpha^2-10\alpha+16=0 \Rightarrow (\alpha-8)(\alpha-2)=0 \\ \Rightarrow \alpha=8,2 \Rightarrow a_n =C_18^n +C_22^n \\ 由(1) 可得 a_1=3a_0+5b_0 =6+5=11 \Rightarrow \cases{a_1=8C_1+2C_2=11\\ a_0=C_1+C_2 = 2} \Rightarrow \cases{C_1=7/6\\ C_2=5/6} \\ \Rightarrow a_n= \bbox[red, 2pt]{{7\over 6}8^n +{5\over 6}2^n,n\ge 1}$$

解答: $$\mathbf{(1)}\;p(X=n)= pq^{n-1} \Rightarrow E(X)=\sum_{n=1}^\infty npq^{n-1} =p{d\over dq} \sum_{n=0}^\infty q^n =p{d\over dq} \left({1\over 1-q}\right) =p\cdot {1\over (1-q)^2} \\ =p\cdot {1\over p^2}=\bbox[red, 2pt]{1\over p} \\\mathbf{(2)}\; E(X(X-1)) =\sum_{n=1}^\infty n(n-1)pq^{n-1} =pq\sum_{n=1}^\infty n(n-1)q^{n-2} =pq\cdot {d^2\over dq^2}\sum_{n=0}^\infty q^n \\=pq \cdot {d^2\over dq^2}\left({1\over 1-q} \right)=pq \cdot {d\over dq} {1\over (1-q)^2} =pq \cdot {2\over (1-q)^3} ={2(1-p)\over p^2}\\ \Rightarrow E(X^2)=E(X(X-1))+E(X)={2(1-p) \over p^2}+{1\over p } ={2-p\over p^2} \\ \Rightarrow Var(X)=E(X^2)-(E(X))^2 ={2-p\over p^2} -{1\over p^2}= \bbox[red,2pt]{1-p\over p^2}$$

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