113年桃園高中教甄-數學詳解

 桃園高中113學年度第1次教師甄選初試

一、是非題:每題2分，共10分

解答: $$\cases{f(x)=(0.07)^x\\ g(x)=\log_{0.07} x} \Rightarrow f,g互為反函數 \Rightarrow f=g的交點位於直線y=x上\\\Rightarrow y=f(x)與直線y=x只有一個交點，不可能三個,故\bbox[red, 2pt]\times$$

解答: $$依柴比雪夫不等式:P(8\le X\le 12)=P(|X-10|\le 2)\ge 1-{1\over 2^2} =75\%,故\bbox[red, 2pt]\bigcirc$$

解答: $$若y=rx為一水平線，則y=rx不是垂直線，故\bbox[red, 2pt]\times$$

解答:

$$有可能，如上圖,故\bbox[red, 2pt]\bigcirc$$

解答: $$\lim_{x\to 0^+} x\ln x= \lim_{x\to 0^+} {\ln x\over 1/x} = \lim_{x\to 0^+} {(\ln x)'\over (1/x)'} = \lim_{x\to 0^+} {1/ x\over -1/x^2} = \lim_{x\to 0^+} -x=0 \\ \Rightarrow  \lim_{x\to 0^+}x^x =e^{\lim_{x\to 0^+}(x\ln x)} =e^0=1 存在,故\bbox[red, 2pt]\times$$

二、填充A:每格5分，共45分

解答:

$$\Gamma:xy=k \lt 0\Rightarrow 對稱軸為直線y=x \Rightarrow 切線\overline{PA}斜率為\tan 75^\circ\\ \tan 75^\circ= \tan (45^\circ+30^\circ)={1+1/\sqrt 3\over 1-1/\sqrt 3} =2+\sqrt 3 \Rightarrow 切線\overleftrightarrow {PA}方程式y=(2+\sqrt 3)(x-2)+2 \\ \Rightarrow x\left( (2+\sqrt 3)(x-2)+2 \right)=k \Rightarrow (2+\sqrt 3)x^2+(-2-2\sqrt 3)x-k=0 \\ \Rightarrow 判別式:(2+2\sqrt 3)^2+4k(2+\sqrt 3)=0 \Rightarrow k= \bbox[red, 2pt]{-2}$$

解答: $$Q=\begin{bmatrix}0 & 0& 1 \\a & -b& 0\\ b& a & 0 \end{bmatrix} \Rightarrow Q^{-1} = \begin{bmatrix}0 & {a\over a^2+b^2}& {b\over a^2+b^2} \\0 & -{b\over a^2+b^2}& {a\over a^2+b^2}\\ 1& 0 & 0 \end{bmatrix} =\begin{bmatrix}0&a & b \\0& -b & a\\1& 0 & 0 \end{bmatrix} \\ \Rightarrow B=Q^{-1}AQ =\begin{bmatrix} 0&a & b \\0& -b & a\\1& 0 & 0 \end{bmatrix} \begin{bmatrix}-1& 0 & 0 \\0& 1+b^2 & -ab\\0& -ab & 1+a^2 \end{bmatrix} \begin{bmatrix}0 & 0& 1 \\a & -b& 0\\ b& a & 0 \end{bmatrix} \\=\begin{bmatrix}0 & a& b \\0 & -2b& 2a\\ -1& 0 & 0 \end{bmatrix} \begin{bmatrix}0 & 0& 1 \\a & -b& 0\\ b& a & 0 \end{bmatrix} = \begin{bmatrix}a^2+b^2 & 0& 0 \\0 & 2a^2+2b^2& 0\\ 0& 0 & -1 \end{bmatrix} =\bbox[red, 2pt]{\begin{bmatrix} 1 & 0& 0 \\0 & 2& 0\\ 0& 0 & -1 \end{bmatrix}}$$

解答: $${x\over 2021}= {y\over 2024}={z\over 2027} \Rightarrow \cases{x=2021k\\ y=2024k\\ z=2027k} \Rightarrow x+y+z=6072k=2024 \Rightarrow k={1\over 3}\\ \Rightarrow \cases{x=2021/3\\ y=2024/3\\ z=2027/3} \Rightarrow \cases{x=y-1\\ z=y+1} \\\Rightarrow x^3+y^3+z^3-3xyz =(x+y+z)(x^2+y^2+z^2-xy-yz-xz) \\=3y\cdot {1\over 2}((x-y)^2+(y-z)^2+(z-x)^2) ={3\over 2}\cdot {2024\over 3}(1+1+4)= \bbox[red, 2pt]{6072}$$

解答:

$$假設\cases{\overline{AB}=a =\sin48^\circ\\ \overline{MA}= \overline{MB} =b =\sin 42^\circ},則取\cases{A(0,a)\\ B(0,0) \\C(2b,0)\\ D(2b,a)} \Rightarrow E(\cos 12^\circ,-\sin 12^\circ) \\ \Rightarrow N=(E+C)/2 = (b+{1\over 2}\cos 12^\circ,-{1\over 2}\sin 12^\circ) \Rightarrow \overline{DN}^2=(b-{1\over 2}\cos12^\circ)^2+(a+{1\over 2}\sin12^\circ)^2 \\=b^2-b\cos12^\circ+{1\over 4}\cos^2 12^\circ +a^2+ a\sin12^\circ +{1\over 4}\sin^2 12^\circ =a\sin 12^\circ-b\cos 12^\circ+ {5\over 4} \\=\sin 48^\circ\sin 12^\circ-\cos 48^\circ \cos12^\circ+{5\over 4}=-\cos(48^\circ+12^\circ)+ {5\over 4} = -{1\over 2}+{5\over 4}={3\over 4} \Rightarrow \overline{DN}= \bbox[red, 2pt]{\sqrt 3\over 2}$$

解答:

$$假設\cases{O為原點\\ P(z^2)\\ A(z^2+25)\\ B(z^2-25)},由於Arg(z^2-25)-Arg(z^2+25)={\pi\over 2} \Rightarrow \angle BOA ={\pi\over 2} \\\Rightarrow P是直角\triangle OAB外接圓圓心  \Rightarrow \overline{PO}= \overline{PA} =\overline{PB} =25\\ \Rightarrow z^2=25e^{i(2\theta )} \Rightarrow z= \pm 5e^{i\theta } =\pm5(\cos \theta+i\sin \theta) =\pm 5({7\over 25}+i{24\over 25}) = \bbox[red, 2pt]{\pm({7\over 5}+i{24\over 5})}$$

解答:

$$令P為\overline{CD}中點，則\cases{\overline{AP} =\sqrt{\overline{AP}^2-\overline{CP}^2} =\sqrt{5^2-3^2} =4 \\ \overline{BP}=\sqrt{\overline{BC}^2-\overline{CP^2}} =\sqrt{6^2-3^2} =3\sqrt 3} \\\Rightarrow \cos \angle APB ={4^2+(3\sqrt 3)^2-5^2 \over 2\cdot 4\cdot 3\sqrt 3}= {\sqrt 3\over 4}  \Rightarrow \sin \angle APB= {\sqrt{13} \over 4} \\\Rightarrow S_{\triangle ABP} ={1\over 2}\cdot \overline{AP}\cdot \overline{BP}\sin \angle APB= {3\sqrt{39} \over 2}= {1\over 2}\cdot \overline{AP}\cdot d(B,\overline{AP}) \\\Rightarrow d(B,\overline{AP}) ={3\sqrt{39}\over 4} \Rightarrow 六面體體積=3倍四面體體積= S_{\triangle ACD} \cdot d(B,\overline{AP})\\ ={1\over 2}\cdot 6\cdot 4\cdot {3\sqrt{39}\over 4} =\bbox[red, 2pt]{9\sqrt{39}}$$

解答: $$E(x)=0.4\times 0.5\times 10+(0.6+0.4\times 0.5)(E(x)+5) \Rightarrow E(x)=\bbox[red, 2pt]{30}$$

解答: $$x^2=8-x^2 \Rightarrow x=\pm 2\\若正\triangle 邊長為a,則面積={\sqrt 3\over 4}a^2, 因此\Gamma體積=\int_{-2}^2 {\sqrt 3\over 4}(x^2-(8-x^2))^2\,dx ={\sqrt 3\over 2} \int_0^2(2x^2-8)^2\,dx \\={\sqrt 3\over 2}\int_0^2 (4x^4-32x^2+64)\,dx ={\sqrt 3\over 2} \left.\left[ {4\over 5}x^5-{32\over 3}x^3+64x \right] \right|_0^2 ={\sqrt 3\over 2}\cdot {1024\over 15} =\bbox[red, 2pt]{512\sqrt 3\over 15}$$

解答: $$(10+1)C^{10+2}_9-2C^{10+1}_9 =11C^{12_9}-2C^{11}_9=\bbox[red, 2pt]{2310},\\參考資料\text{ 2008 AIME II } \href{https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_12}{第12題}$$

三、填充B:每格7分，共35分

解答: $$假設\cases{狀態1=S1:今天吃團膳\\ 狀態2=S2:昨天吃團膳,今天不吃團膳\\ 狀態3=S3:昨天、今天都不吃團膳 } \Rightarrow \cases{P(S1\to S2) =1\\ P(S2\to S1)=1/2\\ P(S2\to S3)=1/2\\ P(S3\to S1)=1}\\ \Rightarrow 轉換矩陣A=\begin{bmatrix}0 & 1/2 & 1 \\1& 0 &0    \\0 & 1/2 &0\end{bmatrix} \\ 長期而言\begin{bmatrix}0 & 1/2 & 1 \\1& 0 &0    \\0 & 1/2 &0\end{bmatrix}\begin{bmatrix} p1\\p2   \\p3\end{bmatrix} =\begin{bmatrix} p1\\p2   \\p3\end{bmatrix} \Rightarrow \begin{bmatrix}-1 & 1/2 & 1 \\1& -1 &0    \\0 & 1/2 & -1\end{bmatrix}\begin{bmatrix} p1\\p2   \\p3\end{bmatrix}=0 且p1+p2+ p3=1 \\ \Rightarrow \cases{p1=p2=0.4\\ p3=0.2} \Rightarrow 吃團膳機率p1=\bbox[red, 2pt]{0.4}$$

解答: $$y=x+k代入圓\Rightarrow (x-1)^2+(x+k+2)^2=23 \Rightarrow 2x^2+(2k+2)x+k^2+4k-18=0\\ \Rightarrow \cases{兩根之和=-(k+1)\\ 兩根之積=(k^2+4k-18)/2}\\ A,B皆在y=x+k上\Rightarrow 假設\cases{A(a,a+k)\\ B(b,b+k)} \Rightarrow \overrightarrow{OA}\cdot \overrightarrow{OB}=ab+(a+k)(b+k)=0 \\ \Rightarrow k^2+(a+b)k+2ab=0 \Rightarrow k^2-(k+1)k+(k^2+4k-18)=0, 其中\cases{a+b=兩根之和\\ ab=兩根之積} \\ \Rightarrow k^2+3k-18=0 \Rightarrow (k+6)(k-3)=0 \Rightarrow k=\bbox[red, 2pt]{3,-6}$$

解答: $$f(x)=x^{130}-1=(x^4-x^3+2x^2-x+1)Q(x)+Ax^3+Bx^2+Cx+D \\ \qquad =(x^2+1)(x^2-x+1)Q(x)+Ax^3+Bx^2+Cx+D \\ \Rightarrow f(i)=i^{130}-1=-2=-Ai-B+Ci+D \Rightarrow \cases{-B+D=-2 \cdots(1)\\ -A+C=-2 \cdots(2)} \\ \Rightarrow f(\omega={1+\sqrt 3i\over 2})=\omega^{130}-1=-\omega -1=-A+B(\omega-1)+C\omega+D =(B+C)\omega-A-B+D \\\Rightarrow \cases{B+C=-1 \cdots(3)\\ -A-B+D=-1  \cdots(4)},由(1-4)可得\cases{A=-1\\ B=0\\ C=-1\\ D=-2} \Rightarrow 餘式=\bbox[red, 2pt]{-x^3-x-2}$$

解答: $$\begin{array}{l}n& x& x個數\\\hline 0&-200,0,200& 3\\ 1&-201, -199,-1,1,199,201& 6\\ 2& -202,-200,-198,-2,0,2,198,200,202& 9 \\ & \cdots \\ 99& -299,-297,-295,\dots,295,297,299 & 300 \\ 100 & -300,-298,-296, \dots,296, 298,300 & 301 \\\hline\end{array}\\ 當n=0\to 99, f_n=0解的個數比f_{n-1}=0 解的個數多3個; \\且n=99時,相鄰各解差距為1,因此當n=100時,只能增加一個解,因此個數為\bbox[red, 2pt]{301}$$

解答:

$$假設正方形中心點O為原點，由正六邊形邊長為1,因此\overline{NN'}=\sqrt 3 \Rightarrow N({1\over 2},\sqrt 3-{1\over 2})\\ \overleftrightarrow{NP}的斜率為-\sqrt 3 \Rightarrow \overleftrightarrow{NP}:y=-\sqrt 3(x-{1\over 2}) +\sqrt 3-{1\over 2} \Rightarrow A=\overleftrightarrow{NP}\cap  \{直線:y=x\}\\ =\left({5\over 2}-\sqrt 3,{5\over 2}-\sqrt 3 \right)  \Rightarrow \overline{AB} =5-2\sqrt 3 \\\Rightarrow 梯形ABMN面積=(2\sqrt 3-3)\left( 1+5-2\sqrt 3\over 2\right) =9\sqrt 3-15\\ 欲求之面積=4個梯形ABMN面積+1個邊長為\overline{AB}的正方形\\=4(9\sqrt 3-15)+(5-2\sqrt 3)^2=16\sqrt 3-23 \Rightarrow \cases{m=16\\ n=3\\ p=-23} \Rightarrow m+n+p= \bbox[red, 2pt]{-4}\\參考資料:\text{ 2022 AMC 12B }\href{https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_25}{第25題}$$

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