111年中科實中教甄-數學詳解

國立中科實驗高級中學111學年度第1次教師教師甄選

一、填充題(共16題，每題5分)

解答：$$f(x)={25^x\over 25^x+5}    \Rightarrow f(1-x)= \cfrac{25^{1-x}}{25^{1-x}+5} =\cfrac{{25\over 25^x}}{{25+ 5\cdot 25^x\over 25^x}} =\cfrac{25}{25+5\cdot 25^x} =\cfrac{5}{25^x+5}\\ \Rightarrow f(x)+f(1-x)= 1 \Rightarrow f\left({1\over 221}\right) +f\left({2\over 221}\right) + \cdots +f\left({220\over 221}\right) +f\left({221\over 221}\right) \\ =\left( f\left( {1\over 221}\right)+ f\left({220 \over 221} \right) \right) + \left( f\left( {2\over 221} \right)+ f\left({219 \over 221} \right)\right) +\cdots +\left( f\left({110\over 221}\right)+ f\left({111 \over 221} \right)\right) +f(1)\\ =(1+1+\cdots+ 1)+ f(1) =110+ {25\over 30} =\bbox[red,2pt]{110{5\over 6}}$$

解答：$$\sqrt{12+3\sqrt{12}} =\sqrt{12+6\sqrt{3}} =\sqrt{12+ 2\sqrt{27}} = \sqrt 9+\sqrt 3 =3+\sqrt 3 \Rightarrow \cases{a=4\\ b=\sqrt 3-1} \\ \Rightarrow {2a-3\over a-b-1}+{5\over a+b+1} ={5\over 4-\sqrt 3}+{5\over 4+\sqrt 3} =5\cdot {8\over 16-3} =\bbox[red,2pt]{40\over 13}$$

解答：$$取\cases{u=\ln x\\ dv =dx} \Rightarrow \cases{du ={1\over x}dx\\ v=x} \Rightarrow \int \log_{10} x\,dx= {1\over \ln 10} \int \ln x\,dx ={1\over \ln 10}\left( x\ln x- \int 1\,dx\right) \\ = \bbox[red,2pt]{{1\over \ln 10}\left( x\ln x- x\right) +C, C為常數}$$

解答：$$\begin{array} {c|ccccccccccccc} n & 1 & 2 & 3 & 4 & 5 & 6& 7 & 8 & 9 & 10 & 11 & 12& 13\\\hline 7^n \mod 13 & 7 & 10 & 5 & 9 & 11 & 12 & 6 & 3 & 8 & 4 & 2 & 1& 7\end{array}\\ \Rightarrow 循環數為12 \Rightarrow 121 \mod 12= 1 \Rightarrow 7^{121} \mod 13 = 7^1 \mod 13=\bbox[red,2pt]{7}$$

解答：$$x^2+2022^2 = y^2+2021^2 \Rightarrow 2022^2-2021^2=y^2-x^2 \Rightarrow (y+x)(y-x)= 4043 = 311\times 13\\ \Rightarrow \cases{y+x= 311\\ y-x=13} \Rightarrow \cases{ x=(311-13)\div 2 = 149\\ y=(311+ 13)\div 2 = 162} \Rightarrow (x,y) = \bbox[red, 2pt]{(149,162)}$$

解答：$$y=|x-5|-|x-1|+|x+2|-{1\over 2}x-{5\over 2}\\ \Rightarrow y=\begin{cases} x-5-x+1+x+2-{1\over 2}x-{5\over 2} ={1\over 2}x -{9\over 2} & \text{if } 5\le x\le 7\\5-x-x+1+x+2-{1\over 2}x-{5\over 2} = -{3\over 2}x+ {11\over 2} & \text{if } 1\le x\le 5\\ 5-x-1+x +x+2 -{1\over 2}x-{5\over 2} = {1\over 2}x+ {7\over 2} & \text{if }-2\le x\le 1\\ 5-x-1+x-x-2  -{1\over 2}x-{5\over 2} =-{3\over 2}x-{1\over 2} & \text{if }-3\le x\le -2\end{cases} \\ \Rightarrow \min(y) =\begin{cases}-2 & \text{if } 1\le x\le 7 \\ {5\over 2} & \text{if }-3\le x\le 1\\ \end{cases} \Rightarrow b=\bbox[red, 2pt]{-2}$$

解答：

$$x^2+y^2=|x|+|y| \Rightarrow \begin{cases} (x-1/2)^2+(y-1/2)^2 =1/2 & \text {if }x,y\ge 0 \\  (x+1/2)^2+(y-1/2)^2 =1/2 & \text {if }x\le 0 ,y\ge 0 \\ (x-1/2)^2+(y+1/2)^2 =1/2 & \text {if }x\ge 0,y\le 0 \\ (x+1/2)^2+(y+1/2)^2 =1/2 & \text {if }x,y\le 0 \end{cases}\\ 所圍面積= 邊長為\sqrt 2的正方形ABCD及半徑為1/\sqrt 2的四個半圓=(\sqrt 2)^2+ 2\times ({1\over \sqrt 2})^2\pi \\ =\bbox[red, 2pt]{2 +\pi}$$

解答：$$\cases{L_1=\overleftrightarrow{AB}: 2x-y+2=0\\ L_2= \overleftrightarrow{AC}: 2x+y-6=0} \Rightarrow A= L_1\cap L_2 = (1,4); 又 B\in L_1  \Rightarrow  B(t,2t+2) t\in \mathbb{R};\\由於\overline{OB} = \overline{OA} = 5\sqrt 2 \Rightarrow  t^2 +(2t+5)^2 = 50 \Rightarrow (t+5)(t-1)=0 \Rightarrow t=-5(t=1\Rightarrow B=A)\\ \Rightarrow B(-5,-8) \Rightarrow \overline{AB}= \sqrt{36+144} = \bbox[red,2pt]{6\sqrt 5} =\overline{EF}$$

解答：$$z={-3+ 3\sqrt 3i\over 10} \Rightarrow \cases{|z|= \sqrt{{9\over 100}+ {27\over 100}} ={6\over 10}={3\over 5} \\|z-1|= \sqrt{{169\over 100}+ {27\over 100}} ={14\over 10}= {7\over 5}} \\ \sum_{k=n}^{\infty} |z^{k+1}-z^k| =
\sum_{k=n}^{\infty} |z^k(z-1)| = \sum_{k=n}^{\infty} |z^k||z-1| = {7\over 5}\sum_{k=n}^{\infty} ({3\over 5})^k = {7\over 5}\cdot ({3\over 5})^n\left({1\over 1-3/5} \right) \\ ={7\over 2}\cdot ({3\over 5})^n\lt 10^{-20} \Rightarrow \log 7-\log 2 + n(\log 3-\log 5) =0.8451-0.301+n(0.4771-(1-0.301))\\ =0.5441-0.2219n \lt -20 \Rightarrow 20.5441 \lt 0.2219n \Rightarrow n\gt 92.58 \Rightarrow n=\bbox[red, 2pt]{93}$$

解答：$$f(x)= \begin{vmatrix} x+1 & 2 & x+3 \\ 3 & x+2 & x+1\\ x+2 & x+3 & 1\end{vmatrix} =-2x^3-6x^2+4x+12 = -(x+3)(2x^2-4)\\ \Rightarrow f(x-5)=0 \Rightarrow -(x-2)(2(x-5)^2-4)=0 \Rightarrow 有理根\alpha = 2\\ f(x^2-1)=-2x^6+10x^2+4=0  正實根\beta \Rightarrow \beta^2=1+\sqrt 2\\ 令四角錐高度h \Rightarrow \beta^2= h^2 + (\sqrt 2 \alpha/2)^2 \Rightarrow h^2=\sqrt 2-1\\ 取底面四方形ABCD及頂點E坐標: \cases{A(0,0,0)\\ B(\alpha,0,0)=(2,0,0)\\ C(\alpha,\alpha,0)= (2,2,0)\\ D(0,\alpha,0)=(0,2,2)\\ E(\alpha/2,\alpha/2,h) =(1,1,h)}  \Rightarrow \cases{\overrightarrow{AB}=(2,0,0)\\ \overrightarrow {AE}=(1,1,h)\\ \overrightarrow{AD}=(0,2,0)} \\ \Rightarrow \cases{\vec u= \overrightarrow{AB}\times \overrightarrow{AE} =(0,-2h,2)\\ \vec v= \overrightarrow{AE}\times \overrightarrow{AD} = (2h,0,-2)} \Rightarrow \cos\theta ={\vec u\cdot \vec v\over |\vec u||\vec v|} ={-4\over 4h^2+4} ={-1\over h^2+1}={-1\over \sqrt 2} =\bbox[red,2pt]{-{\sqrt 2\over 2}}$$

解答：$$p^{1\over 6}x = q^{-1\over 4}x^2 \Rightarrow x(p^{1\over 6}-q^{-1\over 4}x)=0 \Rightarrow x=0, p^{1\over 6}q^{1\over 4} \Rightarrow A(p,q)= \int_0^{p^{1/6}\;q^{1/4}} p^{1\over 6}x- q^{-1\over 4}x^2\,dx\\ =\left. \left[ {1\over 2}p^{1\over 6}x^2- {1\over 3} q^{-1\over 4}x^3 \right] \right|_0^{p^{1/6}\;q^{1/4}} =({1\over 2}-{1\over 3})p^{1\over 2}q^{1\over 2}={1\over 6}\sqrt{pq} \\\Rightarrow L=\lim_{n\to \infty} {1\over n^2}\sum_{k=1}^n A(k,k+1) =\lim_{n\to \infty} {1\over n^2}\sum_{k=1}^n {1\over 6}\sqrt{k(k+1)}  \\ \Rightarrow \lim_{n\to \infty} {1\over n^2}\sum_{k=1}^n {1\over 6}k\lt L\lt \lim_{n\to \infty} {1\over n^2}\sum_{k=1}^n {1\over 6}(k+1) \Rightarrow {1\over 12}\lt L\lt {1\over 12} (夾擠定理)\\ \Rightarrow L=\bbox[red, 2pt]{1\over 12}$$

解答：$$X為幾何分佈，即X\sim \text{Geometric}(p=2/10)\Rightarrow p(X=x) = (1-p)^{x-1}p \Rightarrow EX= 1/p =\bbox[red,2pt]{5}$$

解答：$$由題意:\cases{X:A卷成績\\ Y:B卷成績\\ X+Y:總成績} \Rightarrow \cases{\mu(X)= 50\\ \sigma(X)= 15\\ \mu(Y)= 25\\ \sigma(Y) = 10 \\ \sigma(X+Y)= 23} \Rightarrow \cases{Var(X)=15^2\\ Var(Y)= 10^2\\ \mu(X+Y)= (50n+25n)/(n+n)=75/2\\ Var(X+Y)= 23^2} \\ \Rightarrow Cov(X,Y)= (Var(X+Y)-Var(X)-Var(Y))\div 2= (23^2-15^2-10^2)\div 2 = 102\\ 相關係數r={Cov(X,Y)\over \sigma(X)\sigma(Y)} ={102\over 15\times 10} = \bbox[red, 2pt]{0.68}$$

解答：$$利用排容原理:{7^6-3\times 6^6+3\times 5^6 -4^6\over 3!} ={20460 \over 6} = \bbox[red,2pt]{3410}\\ 公布的答案是\bbox[blue,2pt]{3395}$$

解答：$$f(x)=  x^4-4x^3+5x-\left(\int_3^x (5t^3-10t^2+3t+ 1)dt \right) +12 \Rightarrow f(3)=81-108+15-(0)+12=0\\ \Rightarrow f'(x) = 4x^3-12x^2+5-(5x^3-10x^2+3x+1) =-x^3-2x^2-3x+4\\ \Rightarrow f'(3)= -27-18-9+4=-50\\ 因此我們有\cases{f(3)=0\\ f'(3)= -50}，因此\lim_{h\to 0} {f(3+4h) \over 5h} =\lim_{h\to 0} {f(3+4h)-f(3) \over 5h} =\lim_{h\to 0} {4\over 5} \cdot {f(3+4h)-f(3) \over 4h} \\ ={4\over 5}\lim_{k\to 0} {f(3+k)-f(3)\over k}(取k=4h) ={4\over 5}f'(3)= {4\over 5}(-50)= \bbox[red, 2pt]{-40}$$

解答：$$\cases{A(-1,1,3)\\ B(1,2,3)\\ C(2,0,4) \\ D(1,1,1)} \Rightarrow \overleftrightarrow{CD}: {x-2 \over -1}= {y \over 1}={z-4\over -3}; 若P\in \overleftrightarrow{CD}，則P(-t+2,t,-3t+4),t\in \mathbb{R}\\ \Rightarrow \cases{\vec u= \overrightarrow{PA} =(t-3, 1-t,3t-1) \\ \vec v=\overrightarrow{PB} =(t-1,2-t,3t-1) } \Rightarrow \triangle PAB ={1\over 2}\sqrt{|\vec u|^2|\vec v|^2 -(\vec u\cdot \vec v)^2}\\ 取f(t)= |\vec u|^2|\vec v|^2 -(\vec u\cdot \vec v)^2 =(11t^2-14t+11)(11t^2 -12t+6)-(11t^2-13t+6)^2 \\=54t^2-60t+30 \Rightarrow 當t=5/9有最小值，此時P=\bbox[red,2pt]{({13\over 9},{5\over 9},{7\over 3})}$$

二、計算證明題(共2題，每題10分)

解答：$$y= \arctan x \Rightarrow \color{blue}{\tan y= x} \Rightarrow {d\over dx} \tan y= {d\over dx}x \Rightarrow y'\sec^2 y = 1 \\\Rightarrow {d\over dx}\arctan x = y'={1\over \sec^2 y} ={1\over 1+\tan^2 y} ={1\over 1+x^2} \Rightarrow {d\over dx}\arctan x = {1\over 1+x^2},\bbox[red,2pt]{Q.E.D}$$

解答：$$(1)f(2)=-9 \lt 0 \Rightarrow y=f(x)圖形為凹向上且f(x)=a(x-2)^2-9，因此f(x)=0 \Rightarrow x=2\pm {3\over \sqrt a} \\ \Rightarrow \int_2^{2+3/\sqrt a} 9-a(x-2)^2\,dx = 36/2 \Rightarrow \left.\left[ 9x-{1\over 3}a(x-2)^3 \right]\right|_2^{2+3/\sqrt a} =18 \Rightarrow {18\over \sqrt a}=18\Rightarrow a=1\\ \Rightarrow f(x)=(x-2)^2-9 = \bbox[red, 2pt]{x^2-4x-5}\\(2)由(1)知: \cases{\beta=2+3=5 \\ \alpha=2-3=-1}，而g(x)=-x^3+dx^2+ex \Rightarrow g'(x)=-3x^2 +2dx+e \\ \Rightarrow \cases{g'(\alpha)= g'(-1)= -3-2d+e=0\\ g'(\beta)=g'(5) =-75+ 10d+e =0} \Rightarrow \cases{d=6\\ e=15} \Rightarrow g(x)=-x^3 +6x^2+15x\\ \Rightarrow R=\int_{-1}^0 g^2(x)\pi \,dx = \pi\int_{-1}^0 x^6-12x^5+6x^4 +180x^3+225x^2\,dx\\ =\pi \left.\left[{1\over 7}x^7-2x^6+{6\over 5}x^5+45x^4+75x^3 \right]\right|_{-1}^0 =\pi(0-(-32-{47\over 35})) =\bbox[red, 2pt]{{1167\over 35}\pi}$$

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解題僅供參考，其他教甄試題及詳解