106年台中一中教甄-數學詳解

臺中市立臺中第一高級中等學校106學年度第 1 次教師甄選

壹、填充題A部分

解：$$\lim_{n\to \infty}a_n = \lim_{n\to \infty}\sum_{k=1}^n{2\over n}\left[\left( 2+{2k-2\over n}\right)^2+1\right] =\int_0^2 \left[ (2+x)^2+1 \right]dx \\ =\left. \left[ {1\over 3}x^2 +2x^2+5x\right] \right|_0^2 ={4\over 3}+18= \bbox[red,2pt]{62 \over 3}$$

解：$$f_2=f(f_1)=f(ax+b)=a(ax+b)+b=a^2x+ab+b \Rightarrow f_3=f(f_2)=a^3x+a^2b+ab+b\\ \Rightarrow f_7=a^7+a^6b+a^5b+\cdots +ab+b= 128x+381 \Rightarrow \cases{a^7 = 128\\ b(a^7-1)=381} \Rightarrow \cases{a=2 \\b=3}\\ \Rightarrow a+b= \bbox[red,2pt]{5}$$

解：$$f(x)=x^3-x \Rightarrow f'(x)=3x^2-1;令P(p,p^3-p)為切點，其切線斜率為3p^2-1 \\ \Rightarrow 切線L:y=(3p^2-1)(x-p)+p^3-p，又L過(2,a) \Rightarrow a=(3p^2-1)(2-p)+p^3-p\\ \Rightarrow g(p)=2p^3-6p^2+2+a=0;\\ g'(p)=0 \Rightarrow 6p^2-12p=0 \Rightarrow 6p(p-2)=0 \Rightarrow p=0,2;\\有三相異切線\Rightarrow g(p)=0 有相異三實根 \Rightarrow g(p)=0有兩極值且異號，即g(0)g(2) < 0 \\ \Rightarrow (2+a)(16-24+2+a)< 0 \Rightarrow (a-6)(a+2) < 0 \Rightarrow \bbox[red,2pt]{-2< a < 6}$$

解：$$A、B在y^2=4x上，則令\cases{A(a^2/4,a)\\ B(b^2/4,b)};又\overline{OA} \bot \overline{OB}，即\overrightarrow{OA} \cdot \overrightarrow{OB}=0 \Rightarrow {a^2b^2\over 16}+ab=0 \\\Rightarrow ab=-16\\ \triangle OAB面積= {1\over 2}\overline{OA}\cdot \overline{OB} = {1\over 2}\sqrt{{a^4\over 16}+a^2} \cdot \sqrt{{b^4\over 16}+b^2} ={1\over 32}\sqrt{ (a^4+16a^2)(b^4+16b^2)} \\ ={1\over 32} \sqrt{(ab)^4 +16a^2b^2(a^2+b^2) +16^2(ab)^2} =2\sqrt{32+a^2+b^2} =2\sqrt{32+(a+b)^2-2ab} \\ =2\sqrt{64+(a+b)^2} \ge 2\sqrt{64} =\bbox[red,2pt]{16}$$

解：$$\cases{z_1=-3i\\ z_2=3i} \Rightarrow \cases{Arg(z+3i)=Arg(z-z_1) \\ Arg(z-3i)=Arg(z-z_2)} \Rightarrow \angle z_1zz_2= 310^\circ -40^\circ =270^\circ \\ \Rightarrow \triangle z_1zz_2為一直角三角形\Rightarrow \cases{原點O為\triangle z_1zz_2外接圓圓心 \\ 外接圓半徑r=\overline{z_1z_2}\div 2=3} \Rightarrow |z|=\overline{Oz}=r= \bbox[red,2pt]{3}$$

解：

$$在\overline{AB}上找一點E、在\overline{CD}上找一點F，使得\overline{EF}=2，則\triangle CDE 面積= \overline{CD}\times \overline{EF} \div 2= \sqrt 3\\ \overline{AB}在平面CDE法向量的投影長為\overline{AB}\sin {\pi\over 3}={\sqrt 3\over 2}\\ \Rightarrow 四面體A-BCD體積=\sqrt 3\times {\sqrt 3\over 2}\times {1\over 3}= \bbox[red,2pt]{1\over 2}$$

解：$$x^4-x^3+x-1= x^3(x-1)+(x-1)= (x^3+1)(x-1)=(x+1)(x-1)(x^2-x+1)\\ 令f(x)=x^{17}+4x^3-3x+1=(x+1)(x-1)(x^2-x+1)P(x) + ax^3+bx^2 +cx+d\\ \Rightarrow \cases{f(1)=3=a+b+c+d \\ f(-1)=-1=-a+b-c+d } \Rightarrow \cases{a+c=2\\ b+d=1}\cdots(1);\\ x^3+1 為f(x)之因式，將x^3=-1代入f(x) \Rightarrow x^2\cdot (x^3)^5+4x^3-3x+1=ax^3+bx^2 +cx+d\\ \Rightarrow -x^2-3x-3 =bx^2+cx+d-a \Rightarrow \cases{b=-1\\ c=-3\\ d-a=-3}代入(1) \Rightarrow \cases{a=5\\ b=-1\\ c=-3\\ d=2} \\ \Rightarrow 餘式為\bbox[red,2pt]{5x^3-x^2-3x+2}$$

解：$$\overline{AD} =\overline{DE} =\overline{EB} \Rightarrow \triangle ACD面積=\triangle DCE面積=\triangle ECB面積 =\triangle ABC\div 3={1\over 2} \\ \Rightarrow {1\over 2}\overline{CA}\cdot \overline{CD}\sin \alpha ={1\over 2}\overline{CD}\cdot \overline{CE}\sin \beta ={1\over 2}\overline{CE}\cdot \overline{CB}\sin \gamma = {1\over 2}\\ \Rightarrow \overline{CD}\sin \alpha = \overline{CD}\cdot \overline{CE}\sin \beta = 3\overline{CE} \sin \gamma =1 \\ \Rightarrow {\sin \beta \over \sin \alpha \sin \gamma} ={1/( \overline{CD}\cdot \overline{CE}) \over 1/\overline{CD} \cdot 1/3\overline{CE}} =\bbox[red,2pt]{3}$$

解：$$x^3-1=0 \Rightarrow (x-1)(x^2+x+1)=0 \Rightarrow x=1,\omega,\omega^2\\ f(x)=(1+x+x^2)^{1000} =\sum_{k=0}^{2000}a_kx^k\\ \Rightarrow f(1)=3^{1000}=a_0+a_1 +a_2+\cdots+a_{2000}\\ \Rightarrow f(\omega)=0=a_0+a_1\omega+a_2\omega^2+a_3+a_4\omega +a_5\omega^2+\cdots +a_{2000}\omega^{2}\\ \Rightarrow f(\omega^2) =0=a_0+a_1\omega^2+a_2\omega+a_3+ a_4\omega^2+ a_5\omega+ \cdots+ a_{2000}\omega \\ \Rightarrow f(0)+f(\omega)+f(\omega^2) = 3a_0+a_1(1+\omega+\omega^2) +a_2(1 +\omega^2+\omega) + 3a_3+\cdots+ a_{2000}(1+\omega^2+\omega)\\ =3(a_0 +a_3+ a_6+\cdots + a_{1998}) \Rightarrow 3^{1000}=3\sum_{k=0}^{666}a_{3k} \Rightarrow \sum_{k=0}^{666}a_{3k}=3^{999}\\ \Rightarrow \log \left( \sum_{k=0}^{666}a_{3k}\right) =\log 3^{999} =999\times \log 3= 999\times 0.4771=476.6229 \\ \Rightarrow \sum_{k=0}^{666}a_{3k}之值為476+1=\bbox[red,2pt]{477}位數$$

解：$$\cases{n=1 \Rightarrow f(x)=a_1x,其中\;{1\over 2}\lt x\le 1 \\ n=2 \Rightarrow f(x)=a_2x^2,其中\;{1\over 3}\lt x \le {1\over 2} \\ n=3 \Rightarrow f(x)=a_3x^3,其中\;{1\over 4}\lt x \le {1\over 3} \\ \cdots \\n=n \Rightarrow f(x)=a_nx^n,其中\;{1\over n+1}\lt x \le {1\over n} };\\f(x)是連續的 \Rightarrow \cases{f({1\over 2})=a_1{1\over 2}=a_2({1\over 2})^2 \Rightarrow a_2=2a_1 \\ f({1\over 3})=a_2({1\over 3})^2=a_3({1\over 3})^3 \Rightarrow a_3=3a_2 \\\cdots \\ f({1\over n})=a_{n-1}({1\over n})^{n-1} =a_n({1\over n})^n \Rightarrow a_n=na_{n-1}} \\ \Rightarrow a_n=na_{n-1}=n(n-1)a_{n-2} =\cdots = n(n-1)\cdots 2a_1= \bbox[red,2pt]{n! }$$

解：$$n位數各位數分別為b_nb_{n-1}\dots b_2b_1，其中\;1\le b_n\le 9，0\le b_{n-1},b_{n-2},\dots,b_1 \le 9，\\若該數字為吉祥數字，則b_n+b_{n-1}+\cdots +b_1=7，因此共有H^n_{7-1=6} 個；\\因此\cases{H^1_6=C^6_6=1\\ H^2_6=C^7_6=7\\ H^3_6= C^{8}_6=28} \Rightarrow 1+7+28=36；\\1開頭的四位數中，吉祥數字有H^3_6=C^8_6=28個，而2005是四位數中第1個吉祥數字，\\因此n=36+28+1=65 \Rightarrow 5n=325；又\cases{H^4_6=C^9_6=84 \\H^5_6= C^{10}_6= 210} \\\Rightarrow \sum_{k=1}^5H^k_6 =330 > 325 >120= \sum_{k=1}^4H^k_6  \\ \Rightarrow a_{325}是五位數;由於\cases{1開頭的五位數中，吉祥數字有H^4_6=84個\\ 2開頭的五位數中，吉祥數字有H^4_5=56個\\  3開頭的五位數中，吉祥數字有H^4_4=35個 \\ 4開頭的五位數中，吉祥數字有H^4_3=20個 \\ 5開頭的五位數中，吉祥數字有H^4_2=10個} \\ \Rightarrow 1-4開頭的五位數中，有84+56+35+20+10=205，再加1-4位數的吉祥數字120個\\，剛好是325個;因此a_{325}=5開頭的五位數中最大的吉祥數字，即\bbox[red,2pt]{52000}$$

解：

    (1)$$n=1，點數總和可以是3,4,5,6;n=2，點數總和可以是5-12;n=3，點數總和可以是9-18;\\ n=4，點數總和可以是17-24;n=5，點數總和至少是33，\\但擲骰子5次總和最多是5\times 6=30 < 33，因此最多能過\bbox[red,2pt]{4}關；$$(2)$$\begin{array}{ccc} n & 點數總和 & 機率\\\hline 1 & 3-6 & 4/6\\\hdashline 2 & 5 & 4/36\\ & 6 & 5/36 & \\ & 7 & 6/36\\ & 8 & 5/36 \\ & 9 & 4/36 \\ & 10 & 3/36 \\ & 11 & 2/36 \\ & 12 & 1/36 \\\hdashline 3 & 9& 25/216 \\ & 10 & 27/216 \\ & 11 & 27/216 \\ & 12 & 25/216 \\ & 13 & 21/216 \\ & 14 & 15/216 \\ & 15 & 10/216 \\ & 16 & 6/216 \\ & 17 & 3/216 \\ & 18 & 1/216\\\hline \end{array}\\ 連過3關的機率:p(n=1) \cdot p(n=2) \cdot p(n=3) = {4\over 6} \cdot {30 \over 36} \cdot {160\over 216} = \bbox[red,2pt]{100\over 243}$$

解：$$\cases{\{b_n\}為等比\Rightarrow b_n=b_1r^{n-1} \\\{ a_n\}為等差\Rightarrow a_n=a_1+(n-1)d};\\ 又\lim_{n\to \infty}\sum_{k=1}^nb_k ={b_1\over 1-r}=\sqrt 2+1 \Rightarrow b_1=(\sqrt 2+1)(1-r),其中0< r< 1;\\\cases{ \cases{a_1=a_2-d\\ a_3=a_2+d} \\\cases{b_1=b_2/r \\ b_3=b_2r} }\Rightarrow \cases{b_1=a_1^2\\ b_2=a_2^2 \\ b_3=a_3^2} \Rightarrow \cases{{b_2\over r}=(a_2-d)^2\\ b_2=a_2^2 \\ b_2r=(a_2+d)^2} \Rightarrow (a_2-d)^2(a_2+d)^2 =a_2^4 \\ \Rightarrow a_2^4 = (a_2^2-d^2)^2 = a_2^4-2a_2^2d^2+d^4 \Rightarrow 2a_2^2d^2 =d^4 \Rightarrow 2a_2^2 =d^2\\ 若\cases{d=\sqrt 2a_2 \\ d=-\sqrt 2a_2}，利用  b_3=b_2r=\color{blue}{a_2^2r =(a_2+d)^2} \Rightarrow \cases{a_2^2r =(a_2+\sqrt 2a_2)^2 \Rightarrow r=(\sqrt 2+1)^2 > 1 \\ a_2^2r =(a_2-\sqrt 2a_2)^2 \Rightarrow r=(1-\sqrt 2)^2 <1}\\ 由於0< r < 1，因此r= (1-\sqrt 2)^2= 3-2\sqrt 2 \Rightarrow b_1=(\sqrt 2+1)(1-r)=(\sqrt 2+1)(2\sqrt 2-2)=2\\ 現在d=-\sqrt 2a_2 \Rightarrow a_1=a_2-d=(1+\sqrt 2)a_2，再加上題意a_1 < a_2 \Rightarrow \cases{a_1 < 0 \\ d > 0}\\ 因此b_1=a_1^2 \Rightarrow 2=a_1^2 \Rightarrow a_1=-\sqrt 2=(1+\sqrt 2)a_2 \Rightarrow a_2={-\sqrt 2\over \sqrt 2+1} =-2+\sqrt 2 \\ \Rightarrow d= a_2-a_1=-2+\sqrt 2-(-\sqrt 2)=-2+2\sqrt 2\Rightarrow \bbox[red,2pt]{\cases{a_1=-\sqrt 2\\ d=-2+2\sqrt 2}}$$

證：$$0\le x\le 1 \Rightarrow 0 \le 1-x \Rightarrow 0 \le x^n(1-x) \Rightarrow 0\le x^n-x^{n-1}\cdots (1)\\ 又 {1\over n+1} =\cfrac{\overbrace{{x\over n}\cdots {x\over n}}^{n個}+(1-x)}{n+1} \ge \sqrt[n+1]{({x\over n})^n(1-x)} \Rightarrow \frac{1}{(n+1)^{n+1}} \ge ({x\over n})^n(1-x)\\ \Rightarrow \frac{n^n}{(n+1)^{n+1}} \ge x^n(1-x) =x^n-x^{n+1} \Rightarrow x^n-x^{n+1} \le  \frac{n^n}{(n+1)^{n+1}}\cdots(2)\\ 由(1)及(2)可得 0\le x^n-x^{n+1} \le  \frac{n^n}{(n+1)^{n+1}}，\bbox[red, 2pt]{故得證}$$