108年關西高中教甄-數學詳解

國立關西高級中學108學年度第1次教師甄選數學科試題

解：

$$\cases{\Gamma_1:y\le x^2/4 \Rightarrow x^2\ge 4y\\ \Gamma_2:|x-1|+|y-1|=k} \Rightarrow \cases{\Gamma_1:拋物線的外(下)側\\ \Gamma_2:中心(1,1)的正方形} \\ \Rightarrow k要最小相當於找(x,y)\in \Gamma_1使得正方形最小 \Rightarrow (x=a,y=a^2/4) \Rightarrow k=|a-1|+|a^2/4-1|最小 \\\Rightarrow a=1 \Rightarrow (p,q)=(x,y)=(a,a^2/4)=\bbox[red,2pt]{(1,1/4)}$$

解： $$\cases{z_{n+1}= (1+\sqrt 3i)z_n-\sqrt 3i \cdots(1)\\ z_{n}= (1+\sqrt 3i)z_{n-1}-\sqrt 3i\cdots(2)} \Rightarrow (1)-(2) \Rightarrow z_{n+1}-z_n= (1+\sqrt 3i)(z_n-z_{n-1}) \\ \Rightarrow z_{n+1}-z_n= (1+\sqrt 3i)^2(z_{n-1}-z_{n-2}) =\cdots = (1+\sqrt 3i)^{n-1}(z_2-z_1)  \\= (1+\sqrt 3i)^{n-1}(3(1+\sqrt 3i)-\sqrt 3i-3) = (1+\sqrt 3i)^{n-1}(2\sqrt 3i)\\ \Rightarrow |z_{n+1}-z_n| = |(1+\sqrt 3i)^{n-1}(2\sqrt 3i)| =|(1+\sqrt 3i)^{n-1}||2\sqrt 3i| =2^{n-1}\times 2\sqrt 3= 2^n\sqrt 3 \\ \Rightarrow \sum_{n=1}^{\infty} {1\over |z_{n+1}-z_n|} =\sum_{n=1}^{\infty} {1\over 2^n\sqrt 3} = {1\over \sqrt 3}\times {1/2\over 1-1/2} =\bbox[red,2pt]{1\over \sqrt 3}$$

解： $$EY=2(C^n_0p^0(1-p)^n +C^n_2p^2(1-p)^{n-2} +\cdots) -(C^n_1p^1(1-p)^{n-1} +C^n_3p^2(1-p)^{n-3} +\cdots)\\ =({1\over 2}+{3\over 2})(C^n_0p^0(1-p)^n +C^n_2p^2(1-p)^{n-2} +\cdots)\\\qquad +({1\over 2}-{3\over 2})(C^n_1p^1(1-p)^{n-1} +C^n_3p^2(1-p)^{n-3} +\cdots)\\ ={1\over 2}\sum_{k=0}^nC^n_kp^k(1-p)^{n-k} +{3\over 2}\sum_{k=0}^nC^n_k(-1)^kp^k(1-p)^{n-k}\\ ={1\over 2}(p+(1-p))^n +{3\over 2}(-p+(1-p))^n = \bbox[red,2pt]{{1\over 2}+{3\over 2}(1-2p)^n}$$

解： $$\cases{y^2+xy-7x=0\\ x,y\in N} \Rightarrow y^2= x(7-y) > 0 \Rightarrow y=1,2,\dots,6\\ \Rightarrow y=6 \Rightarrow 36=x(7-6) \Rightarrow x=36 \Rightarrow (x,y)=\bbox[red,2pt]{(36,6)}\\ 其它y=1,2,\dots,5代入均不符x,y\in N$$

5. 將正整數1,2,3,4,5,6,7任意分成兩組，使得每組至少有一個數。則第一組數的和與第二組數的和相等的機率為________。

解：

符合要求的分組方式(1,6,7)+(2,3,4,5),(3,4,7)+(1,2,5,6),(2,5,7)+(1,3,4,6), (3,5,6)+(1,2,4,7)，共四種；七個數字分兩組有$C^7_1+ C^7_2+C^7_3$=7+21+ 35=63種分法；因此機率為$\bbox[red,2pt]{4\over 63}$

6. 已知點$A(3,1),B({5\over 3},2)$是平行四邊形$ABCD$的頂點，已知$ABCD$四個頂點都落在函數$f(x)=\log_2 {ax-b\over x-1}$的圖形上，試求平行四邊形$ABCD$的面積=___

解： $$\cases{f(x)=\log_2 {ax-b\over x-1}\\ A(3,1)\\ B({5\over 3},2)\\ A,B,C,D皆在y=f(x)上} \Rightarrow \cases{f(3)=1\\ f({5\over 3})=2} \Rightarrow \cases{\log_2 {3a-b\over 2}=1 \\ \log_2{5a/3-b \over 2/3}=2} \Rightarrow \cases{3a-b= 4\\ 5a-3b=8} \\ \Rightarrow \cases{a=1\\ b=-1} \Rightarrow f(x)= \log_2{x+1\over x-1} \Rightarrow f(-x)=\log_2{-x+1\over -x-1}=\log_2{x-1\over x+1} =-\log_2{x+1\over x-1}=-f(x)\\ \Rightarrow f(-x)=-f(x) \Rightarrow y=f(x)對稱原點 \Rightarrow \cases{A(3,1)的對點A'(-3,-1)\\ B({5\over 3},2)的對稱點B'(-{5\over 3},-2)} \\ \Rightarrow ABA'B即為落在y=f(x)上的平行四邊形的頂點\\ \Rightarrow \cases{\overrightarrow{AB} =(-{4\over 3},1)\\ \overrightarrow{AB'} =(-{14\over 3},-3)} \Rightarrow ABA'B面積=\sqrt{|\overrightarrow{AB}|^2|\overrightarrow{AB'}|^2-(\overrightarrow{AB} \cdot \overrightarrow{AB'})^2} \\=\sqrt{{25\over 9}\times {277\over 9} -({29\over 9})^2} = \sqrt{676\over 9} =\bbox[red,2pt]{26\over 3}$$

解： $$假設拋物線\Gamma方程式: y^2=4cx 與直線L:4x+y-20=0的交點P;\\P在L上\Rightarrow P(t,20-4t)也在\Gamma上\Rightarrow (20-4t)^2=4ct \Rightarrow 4t^2-(40+c)t+100=0 \\  \Rightarrow \cases{t_1=(40+c+ \sqrt{c^2+80c})/8 \\ t_2= (40+c-\sqrt{c^2+ 80c})/ 8} \Rightarrow \cases{B(t_1,20-4t_1) \\C(t_2,20-4t_1)\\ A(x_a,y_a)} \Rightarrow 重心F(c,0)=(A+B+C)\div 3 \\ \Rightarrow \cases{c=(t_1+t_2+x_a)\div 3 =(10+c/4+x_a)\div 3\\ 0=(20-4t_1+20-4t_2+y_a)\div 3 =(y_a-c)\div 3} \Rightarrow \cases{x_a=11c/4-10 \\y_a=c}\\ A(x_a,y_a)在\Gamma上\Rightarrow y_a^2=4cx_a \Rightarrow c^2=4c({11\over 4}c-10) =11c^2-40c \Rightarrow 10c^2-40c=0 \Rightarrow c=4\\ \Rightarrow \Gamma: \bbox[red,2pt]{y^2=16x}$$

解： $$\cases{F_1(-1,0)\\ F_2(1,0)} \Rightarrow c=1 \Rightarrow {x^2\over a^2}+{y^2 \over b^2}=1 ;\\將\cases{直線y=x-\sqrt 3\\ a^2=b^2+1}代入橢圓方程式\Rightarrow {x^2\over b^2+1}+{(x-\sqrt 3)^2 \over b^2}=1 \\ \Rightarrow (2b^2+1)x^2-2\sqrt 3(b^2+1)x+3(b^2+1)-b^2(b^2+1)=0\\ \Rightarrow 判別式為0 \Rightarrow 12(b^2+1)^2-4(2b^2+1)(3(b^2+1)-b^2(b^2+1))=0 \\ \Rightarrow 8b^6-8b^2=0 \Rightarrow 8b^2(b^4-1)=0 \Rightarrow b=1 \Rightarrow a^2=b^2+1=2\\ \Rightarrow 橢圓方程式:\bbox[red,2pt]{{x^2\over 2}+{y^2\over 1}=1}$$

解： $$假設四虛根分別為\cases{z_1=a'+b'i\\\bar z_1=a'-b'i\\ z_2=c'+d'i\\ \bar z_2 =c'-d'i},其中a',b',c',d'\in R;\\\Rightarrow x^4+ax^3 +bx^2+cx+d = (x-z_1)(x-\bar z_1)(x-z_2)(x-\bar z_2)\\   =(x^2-(z_1+\bar z_1)x+z_1\bar z_1) (x^2-(z_2+\bar z_2)x+z_2\bar z_2)\Rightarrow b= z_1\bar z_1+z_2\bar z_2 +(z_1+\bar z_1) (z_2+\bar z_2)\\ = (a'+b'i)(a'-b'i)+(c'+d'i)(c'-d'i)+(2a')\times (2c') =a'^2+b'^2+c'^2+d'^2+4a'c'\\ 由題意知\cases{z_1z_2=(a'c'-b'd')+( a'd'+b'c')i= 13+i \\ \bar z_1 +\bar z_2= (a'+c')-(b'+d')i=3+4i} \Rightarrow \cases{a'c'-b'd' =13\\ a'd'+b'c'=1\\ a'+c'=3\\ b'+d'= -4} \\ \Rightarrow (a'+c')^2+(b'+d')^2 = 3^2+(-4)^2 \Rightarrow a'^2+b'^2+c'^2+d'^2+ 2a'c'+2b'd'=25 \\ \Rightarrow a'^2+b'^2+c'^2+d'^2+2a'c'+2(a'c'-13)=25 \Rightarrow a'^2+b'^2+ c'^2+d'^2+4a'c'=25+26\\=\bbox[red,2pt]{51}$$

解： $$\cases{\Gamma_1: y=1/e^x \Rightarrow y'=-1/e^x\\ \Gamma_2:y=-e^{x+1} \Rightarrow y'=-e^{x+1}}，假設\cases{L與\Gamma_1的切點為A(a,1/e^a) \\ L與\Gamma_2的切點為B(b,-e^{b+1})}  \\ \Rightarrow \cases{切點A的斜率:y'(a)=-1/e^a \cdots(1) \\ 切點B的斜率:y'(b)=-e^{b+1} \cdots(2) \\\overline{AB}的斜率:{-e^{b+1}-e^{-a} \over b-a} \cdots(3)} \Rightarrow \cases{(1)=(2) \Rightarrow -e^{-a}=-e^{b+1} \Rightarrow -a=b+1\cdots(4)\\ (1)=(3) \Rightarrow {-e^{b+1}-e^{-a} \over b-a}=-e^{-a}\cdots(5)} \\(4)代入 (5) \Rightarrow 斜率= {-2e^{-a}\over -2a-1}=-e^a \Rightarrow -2a-1=2 \Rightarrow a=-3/2 \\\Rightarrow 切線L方程式為過A且斜率為-e^{-3/2}  \Rightarrow L: \bbox[red,2pt]{y=-e^{-3/2}(x+{3\over 2})+e^{3/2}}$$

解：

(1) $$f_n(x)=\begin{cases}n-n^2|x|&|x|\le {1\over n}\\ 0& |x| \ge {1\over n} \end{cases} = \begin{cases}n-n^2x & 0\le x \le {1\over n}\\ n+n^2x & -{1\over n} \le x \le 0\\ 0 & 其它 \end{cases} \\ \Rightarrow I_n =\int_{-1}^1 f_n(x)\cos x\;dx = \int_{-1/n}^0 (n+n^2x)\cos x\;dx +\int_{0}^{1/n} (n-n^2x)\cos x\;dx \\ =\left .\left[ n\sin x +n^2x\sin x+n^2\cos x\right] \right|_{-1/n}^0 + \left .\left[ n\sin x-n^2x\sin x-n^2\cos x \right] \right|_{0}^{1/n} \\ = (n^2-n^2\cos {1\over n})+(-n^2\cos{1\over n}+n^2) = \bbox[red,2pt]{2n^2(1-\cos{1\over n})}$$ (2) $$\lim_{n\to\infty} I_n = \lim_{n\to\infty} 2n^2(1-\cos{1\over n})=2\lim_{n\to\infty}{1-\cos(1/n) \over 1/n^2} =2\lim_{n\to\infty}{(-{1\over n^2})\sin(1/n) \over -2/n^3}\\ =2\lim_{n\to\infty}{\sin(1/n) \over 2/n} = 2\times {1\over 2}= \bbox[red,2pt]{1} (\because \lim_{n\to \infty} {\sin(1/n)\over 1/n} =\lim_{x\to 0} {\sin(x)\over x} =1)$$

解：

(1) $$兩圖形\cases{L: y=k(x-a)+a^3-3a \\y=f(x)=x^3-3x}的交點 \Rightarrow k(x-a)+a^3-3a=x^3-3x \\ \Rightarrow  k(x-a)+3(x-a)=x^3-a^3 = (x-a)(x^2+ax+a^2) \\\Rightarrow (x-a)(x^2+ax+a^2-3-k)=0 有三相異實根 \\\Rightarrow x^2+ax+a^2-3-k=0的判別式>0 且a不是其中一根\Rightarrow \cases{a^2-4(a^2-3-k) > 0 \\ a^2+a\cdot a+a^2-3-k \ne 0}\\ \Rightarrow \bbox[red,2pt]{\cases{k > {3\over 4}a^2-3 \\ k\ne 3a^3-3}}$$

$$a=\sqrt 3 \Rightarrow P(\sqrt 3,0);並假設\cases{Q(x_1,y_1)\\ R(x_2,y_2)}，由於斜率k<0 \Rightarrow \cases{x_1> x_2\\ y_2 > y_1}\\ 直線L過P且斜率為k \Rightarrow L: y=k(x-\sqrt 3) \Rightarrow P,Q,R為 x^3-3x= k(x-\sqrt 3)之相異三根\\ 因此x^3-3x=x( x+\sqrt 3)(x-\sqrt 3)= k(x-\sqrt 3) \Rightarrow x(x+\sqrt 3)=k \Rightarrow x^2+\sqrt 3x-k=0\\ \Rightarrow \cases{x_1+x_2=-\sqrt 3\\ x_1x_2= -k} \Rightarrow (x_1-x_2)^2 =(x_1+x_2)^2-4x_1x_2 =3+4k \Rightarrow x_1 -x_2 = \sqrt{4k+3}\\ \Rightarrow x_2-x_1 = -\sqrt{4k+3} \Rightarrow y_2-y_1= k(x_2-\sqrt 3)-k(x_1-\sqrt 3)=k(x_2-x_1) = -k \sqrt{4k+3} \\ = \sqrt{4k^3+3k^2} \le \sqrt{1\over 4} ={1\over 2} (g(k)=4k^3+3k^2 \Rightarrow g'(k)=12k^2+6k \Rightarrow g(-1/2)=1/4為極大值)\\ \Rightarrow S(k)=\triangle OQR = \triangle OPR-\triangle OPQ = {1\over 2}\sqrt 3(y_2 - y_1) \le {\sqrt 3\over 2}\times {1\over 2}={\sqrt 3\over 4} \\ \Rightarrow \bbox[red,2pt]{S(k) \le {\sqrt 3\over 4}}$$

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