112年全國教甄聯招-數學詳解

教育部受託辦理112學年度公立高級中等學校教師甄選

第一部分：選擇題（ 共40分）
一、單選題（ 每題3分， 共24分）

解答：$$\begin{vmatrix}2020 &2021  &2022  \\2021 & 2022 & 2023 \\2022 & 2023 & 2025\end{vmatrix} \underrightarrow{-R_1+R_2,-R_1+R_3} \begin{vmatrix}2020 &2021  &2022  \\1 & 1 & 1 \\2 & 2 & 3\end{vmatrix} \\ \underrightarrow{-2R_2+R_3}\begin{vmatrix}2020 &2021  &2022  \\1 & 1 & 1 \\0 & 0 & 1 \end{vmatrix} =2020-2021=-1，故選\bbox[red, 2pt]{(A)}$$

解答：$${n^3+103\over n+11} =n^2-11n+121-{1228\over n+11} \Rightarrow 取1228=n+11 \Rightarrow n=1217，故選\bbox[red, 2pt]{(A)}$$

解答：$$x^2+xy +y^2=3 \Rightarrow 2x+y+xy'+2yy'=0 \Rightarrow y'=-{2x+y\over x+2y} \\ \Rightarrow y'(1,1)=-1 \Rightarrow 過(1,1)且斜率為-1的直線方程式:y=-(x-1)+1\\ \Rightarrow x+y=2，故選\bbox[red, 2pt]{(B)}$$

解答：$$\sum_{k=0}^\infty {1\over 2^k}\cos(120^\circ k) \\=\sum_{k=0}^\infty {1\over 2^{3k}}\cos(120^\circ \cdot 3k)  +{1\over 2^{3k+1}}\cos(120^\circ \cdot (3k+1))+ {1\over 2^{3k+2}}\cos(120^\circ \cdot (3k+2)) \\={1\over 1-{1\over 8}}-{{1\over 4}\over 1-{1\over 8}}-{{1\over 8} \over 1-{1\over 8}} ={{8\over 7}} -{2\over 7} -{1\over 7} ={5\over 7}，故選\bbox[red, 2pt]{(B)}$$

解答：$$原式S={1\over 3^9}\sum_{k=2}^{10} k(k-1)C^{10}_k 2^{k-2}\\ 令f(x)=(1+x)^{10} =\sum_{k=0}^{10}C^{10}_k x^k \Rightarrow f'(x)=10(1+x)^9 =\sum_{k=1}^{10}kC^{10}_k x^{k-1} \\ \Rightarrow f''(x)= 90(1+x)^8 =\sum_{k=2}^{10}k (k-1)C^{10}_k x^{k-2} \Rightarrow S={f''(2)\over 3^9} ={90\cdot 3^8\over 3^9} =30，故選\bbox[red, 2pt]{(A)}$$

解答：$$\omega ={-1+\sqrt 3i\over 2} =\cos{2\pi \over 3}+i\sin {2\pi\over 3} =e^{2\pi i/3} \Rightarrow \omega^3=1 \Rightarrow (\omega-1)(\omega^2+ \omega+1)=0\\ \Rightarrow \cases{\omega^3=1 \\ \omega^2+\omega +1=0 } \\ 因此(\omega^5+ \omega^4+5) (-3\omega^2+2\omega+ 2)(3\omega^2-\omega+3) \\=(\omega^2+\omega +5)(-3(-\omega-1)+2\omega+2)(3(-\omega-1)-\omega+3) \\ =4(5\omega+5)(-4\omega)= 4(5(-\omega^2))(-4\omega)=80\omega^3 =80，故選\bbox[red, 2pt]{(B)}$$

解答：$$對於每種壽司\cases{可能被拿手一盤: 有C^3_1=3種分配方式 \\ 可能被拿走二盤:有C^3_2=3種分配方式}共有六種可能\\ \Rightarrow 10種壽司有6^{10}種可能，故選\bbox[red, 2pt]{(B)}$$

解答：$$\log 2^n\ge 12 \Rightarrow n\ge {12\over \log 2}={12\over 0.301} \approx 39.8 \Rightarrow n=40，故選\bbox[red, 2pt]{(A)}$$

二、複選題（每題 4 分，共 16 分，全對才給分）

解答：

$$狀態1\to狀態1: P(A\;\underrightarrow{2W}\;B\; \underrightarrow{2W}\; A)+P(A\; \underrightarrow{1R1W} \;B\; \underrightarrow{1R1W}\;A) ={1\over 3}+{4\over 9}={7\over 9} \\狀態1\to狀態2: P(A\;\underrightarrow{1R1W}\;B\; \underrightarrow{2W}\; A) ={2\over 3}\cdot {1\over 3}={2\over 9} \\ 狀態轉變圖如上\\(A)\times: 第一回合後仍在狀態一的機率={7\over 9} \\(B) \bigcirc: 第一回合後在狀態二的機率={2\over 9} \\(C)\times: P(S_1\to S_1\to S_1)+ P(S_1\to S_2\to S_1) =({7\over 9})^2 +{2\over 9}\cdot {2\over 3}\gt {20\over 81} \\(D)\bigcirc: P(S_1\to S_1\to S_1\to S_1)+ P(S_1\to S_1\to S_2\to S_1) + P(S_1\to S_2 \to S_2\to S_1) \\\qquad +P(S_1\to S_2\to S_1\to S_1) =({7\over 9})^3+{7\over 9} \cdot {2\over 9} \cdot {2\over 3}+{2\over 9}\cdot {1\over 3}\cdot {2\over 3}+ {2\over 9}\cdot {2\over 3}\cdot {7\over 9} ={547\over 729} \\，故選\bbox[red, 2pt]{(BD)}$$

解答：$$f(x)= (x-\alpha)(x-\beta)(x-\gamma)(x-\delta) =(x^2-(\alpha+\beta)x+\alpha \beta) (x^2-(\gamma+\delta)x +\gamma\delta) \\=(x^2-(3-2i)x+4-i)(x^2-(3+2i)x+4+i) =x^4-6x^3+21x^2 +28x+17 \\(A)\bigcirc: a=-6\\ (B)\times : b=21\\ (C)\bigcirc: c=-28\\ (D)\bigcirc: d=17\\，故選\bbox[red, 2pt]{(ACD)}$$

解答：$$(A)\bigcirc: {x+150\over 3}\lt x \Rightarrow x\gt 75 \Rightarrow 原始分數大於75調整後分數會變低 \\ (B)\times: 相關係數都是1 \\(C)\times: \sum z_i^2 \ne 1\\(D)\bigcirc: 作弊同學分數剛好是全班平均分數,因此移除該同學分數不改變全班平均,\\\qquad 因此甲案平均仍高於乙案\\，故選\bbox[red, 2pt]{(AD)}$$

解答：$$令\cases{\vec a= \overrightarrow{OA} \\ \vec b=\overrightarrow{OB} \\ \vec c=\overrightarrow{OC} \\\vec d=\overrightarrow{OD}}\\(A) \bigcirc: O,A,D共線\Rightarrow {\vec d\over \vec a} ={k\vec a+(2-k)\vec b\over \vec a} =常數\Rightarrow k=2 \\ (B)\bigcirc: \overrightarrow{CD} =-\vec c+\vec d=-\vec a-\vec b+k\vec a+(2-k)\vec b =-(1-k)\vec a+(1-k)\vec b \\ =(1-k)(-\vec a+\vec b)=(1-k)\overrightarrow{AB} \Rightarrow \overrightarrow{CD} \parallel \overrightarrow{AB} \\(C) \bigcirc: \cases{\overrightarrow{AB}=(b_1-a_1,b_2-a_2) \\ \overrightarrow{AD} =((k-1)a_1+(2-k)b_1, (k-1)a_2+(2-k)b_2)} \\\Rightarrow \triangle ABD面積=\begin{vmatrix}b_1-a_1 &b_2-a_2   \\(k-1)a_1+(2-k)b_1 & (k-1)a_2+(2-k)b_2)\end{vmatrix} =|a_2b_1-a_1b_2|與k無關\\ (D)\times: \cases{\overrightarrow{AC} =\overrightarrow{OB}=(b_1,b_2) \\ \overrightarrow{AD} =((k-1)a_1+(2-k)b_1, (k-1)a_2+(2-k)b_2)} \\ \Rightarrow \triangle ACD =\begin{vmatrix}b_1  &b_2  \\(k-1)a_1+(2-k)b_1 & (k-1)a_2+(2-k)b_2)\end{vmatrix}=|(k-1)(a_2b_1-a_1b_2)|與k有關\\，故選\bbox[red, 2pt]{(ABC)}$$

解答：$$鏡射矩陣\begin{bmatrix}\cos 2\theta &  \sin 2\theta\\ \sin 2\theta & -\cos 2\theta \end{bmatrix},其中\cases{L_1:x+y=0 \Rightarrow \theta=135^\circ\\ L_2: x=\sqrt 3y \Rightarrow \theta=30^\circ} \\ \Rightarrow \cases{鏡射矩陣A=\begin{bmatrix}0 &  -1\\ -1 & 0 \end{bmatrix} \\[1ex]鏡射矩陣B= \begin{bmatrix}1/2 &  \sqrt 3/2\\ \sqrt 3/2 & -1/2 \end{bmatrix}} \Rightarrow Q=AP=\begin{bmatrix} -1 \\ -1 \end{bmatrix} \Rightarrow R=BQ =\begin{bmatrix} (-1-\sqrt 3)/2 \\ (1-\sqrt 3)/2 \end{bmatrix} \\ \Rightarrow R=\bbox[red,2pt]{({-1-\sqrt 3\over 2},{1-\sqrt 3\over 2})}$$

解答：$$A=\begin{bmatrix}1 & a \\-1 &-2 \end{bmatrix} \Rightarrow A^3=\begin{bmatrix}1 & -a^2+3a \\a-3 &3a-8 \end{bmatrix} =I \Rightarrow a=3 \\ \Rightarrow A=\begin{bmatrix}1 & 3 \\-1 &-2 \end{bmatrix} \Rightarrow A^2=\begin{bmatrix}-2 & -3 \\1 &1 \end{bmatrix} \Rightarrow A^2+A+I=0 \\ 令B=A+ 2A^2 +\cdots +20A^{20} \Rightarrow AB= A^2+ 2A^3+\cdots +20A^{21} \\ \Rightarrow B-AB= A+A^2 +\cdots +A^{20}-20A^{21} =A^{19}+A^{20}-20A^{21} \\ \Rightarrow (I-A)B=A+A^2-20I= -21I\\ \Rightarrow B=-21I(I-A)^{-1} =\begin{bmatrix}-21 & 0 \\ 0& -21 \end{bmatrix}\begin{bmatrix}0 & -3 \\ 1& -3 \end{bmatrix} =\begin{bmatrix}-21 & 0 \\ 0& -21 \end{bmatrix}\begin{bmatrix}1 & 1 \\ -1/3& 0 \end{bmatrix} =\bbox[red, 2pt]{\begin{bmatrix} -21 & -21\\ 7 & 0\end{bmatrix}}$$

解答：$$X\sim b(n=10,p={1\over 2} ) \Rightarrow \cases{\mu=E(X)=np=5\\ \sigma =\sqrt{np(1-p)} =\sqrt{10}/2 \approx 1.58} \\ \Rightarrow P(\mu-\sigma\le X\le \mu +\sigma) =P(5-1.58\le X\le 5+1.58) =P(3.42\le X\le 6.58)\\ =P(X=4)+ P(X=5)+P(X=6) ={1\over 2^{10}}(C^{10}_4 +C^{10}_5 +C^{10}_6) ={672\over 1024} =\bbox[red, 2pt]{21 \over 32}$$

解答：$$6\times 24^x-2\times 27^x+ 3\times 16^x-18^x = 2\cdot 3\cdot 3^x\cdot 2^{3x} -2\cdot 3^{3x} +3\cdot 2^{4x}-2^x\cdot 3^{2x} \\=2^{3x+1}\cdot 3^{x+1}-2\cdot 3^{3x} +3\cdot 2^{4x}-2^x\cdot 3^{2x}=(2^{3x+1}\cdot 3^{x+1}+3\cdot 2^{4x}) -(2\cdot 3^{3x}+2^x\cdot 3^{2x}) \\ =6\cdot 2^{3x}(3^x +2^{x-1})-2\cdot 3^{2x}(3^x+2^{x-1}) =(6\cdot 2^{3x}-2\cdot 3^{2x})(3^x+ 2^{x-1})=0\\ \Rightarrow 6\cdot 2^{3x}  =2\cdot 3^{2x} \Rightarrow \log 6+3x\log 2=\log 2+2x\log 3 \Rightarrow \log 3=(2\log 3-3\log 2)x \\ \Rightarrow x={\log 3\over 2\log3- 3\log 2} \Rightarrow {2x-1\over 3x}={3\log 2\over 2\log 3-3\log 2} \times {2\log 3-3\log 2\over 3\log 3} \\ ={\log 2\over \log 3}=\bbox[red, 2pt]{\log_3 2} $$

解答：$$a_n= \sum_{k=1}^{n}{n\over n^2+k^2} = \sum_{k=1}^{n}{1/n\over 1+(k/n)^2}  \Rightarrow \lim_{n \to \infty} a_n = \int_0^1 {1\over 1+x^2}\,dx =\left. \left[ \tan^{-1}x \right] \right|_0^1 =\bbox[red, 2pt]{\pi\over 4}$$

解答：$$令f(k)={x\over k+4^2} +{y\over k+5^2} +{z\over k+6^2} -1,依題意:f(1^2)=f(2^2)=f(3^2)=0 \\ 因此(k-1)(k-4) (k-9) \\ \qquad =(k+4^2)(k+5^2)(k+6^2)-x(k+5^2)(k+6^2)-y(k+4^2)(k+6^2)-z(k+4^2)(k+5^2) \\ \Rightarrow \cases{左式k^2係數=-1-4-9=-14\\ 右式k^2係數=4^2+5^2+6^2-(x+y+z)} \Rightarrow x+y+z=77+14=\bbox[red, 2pt]{91}$$

解答：$$x^2f(x)=(3x-2)p(x)+r \Rightarrow x^3f(x) =(3x-2)xp(x)+rx =(3x-2)xp(x)+{r\over 3}(3x-2)+{2r\over 3} \\ \Rightarrow {2r\over 3}=6 \Rightarrow r=\bbox[red, 2pt]9$$

解答：$$Q=a^2-2a =(a-1)^2-1為有理數\Rightarrow a=1+ \sqrt q,其q為有理數，但\sqrt q為無理數\\ 又{P\over Q} ={a^3+3a^2-16a+6\over a^2-2a } =a+5+{-6a+6\over a^2-2a}=6+ \sqrt q+{ - 6\sqrt q\over q-1} 為有理數\\ \Rightarrow q-1=6 \Rightarrow q=7 \Rightarrow a=\bbox[red, 2pt]{1+\sqrt 7}$$

解答：$$\cases{A(2,2,1)\\ B(1,0,2)\\ C(0,1,2)} \Rightarrow \cases{\overrightarrow{AB} =(-1,-2,1)\\ \overrightarrow{AC} =(-2,-1,1)} \Rightarrow \vec n=\overrightarrow{AB} \times \overrightarrow{AC} =(-1,-1, -3) \\ \Rightarrow E=\triangle ABC:x+y+3z =7 \Rightarrow \cases{E\cap 直線(2,0,s) =\bbox[red,2pt]{(2,0,5/3)}\\ E\cap 直線(0,2,t)=\bbox[red,2pt]{(0,2,5/3)}} $$

解答：

$$令\overline{MF} =\overline{FC}=a \Rightarrow 在直角\triangle BMF中，\overline{MF}^2 = \overline{MB}^2 + \overline{BF}^2 \Rightarrow a^2=120^2+(288-a)^2 \Rightarrow a=169\\ 令\overline{DE}= \overline{EG} =b \Rightarrow \overline{AE}=288-b \Rightarrow \cases{直角\triangle MAE: \overline{ME}^2 = \overline{MA}^2 + \overline{AE}^2 =120^2+(288-b)^2 \\ 直角\triangle EDC: \overline{EC}^2 = \overline{ED}^2 + \overline{DC}^2 =b^2+240^2}\\ 由於\overline{EM} =\overline{EC} \Rightarrow 120^2+(288-b)^2 =b^2+240^2 \Rightarrow b=69 \\ 直角\triangle FHE: \overline{EF}^2= \overline{EH}^2 + \overline{HF}^2 =240^2+ (a-b)^2=240^2+ (169-69)^2=260^2 \Rightarrow \overline{EF} =\bbox[red, 2pt]{260}$$

解答：$$\cases{u=\ln x\\ dv =xdx} \Rightarrow \cases{du={dx\over x}\\ v={1\over 2}x^2} \Rightarrow \int x\ln x\,dx ={1\over 2}x^2\ln x -{1\over 2}\int xdx ={1\over 2}x^2 \ln x-{1\over 4}x^2+C\\ \Rightarrow \int_1^e x\ln x\,dx =\left. \left[{1\over 2}x^2 \ln x-{1\over 4}x^2 \right]\right|_1^e ={1\over 2}e^2-{1\over 4}e^2+{1\over 4} =\bbox[red, 2pt]{{1\over 4}e^2+{1\over 4}}$$

解答：

$$假設\overline{AB} \bot \overline{BO} 且\overline{OB}\bot \overline{BC}，則欲求之|\cos \theta | = |\cos \angle ABC|\\ 在直角\triangle OAB中，假設\overline{AB}=1再加上\angle AOB=30^\circ \Rightarrow \cases{\overline{OA}=2\\ \overline{OB}=\sqrt 3}\\ 同理，在直角\triangle OBC中，\angle BOC=45^\circ \Rightarrow \overline{BC}=\overline{OB}=\sqrt 3,且\overline{OC}=\sqrt 6\\ 在\triangle OAC中，\cos \angle AOC =\cos 60^\circ = {1\over 2} ={2^2+(\sqrt 6)^2 -\overline{AC}^2 \over 2\cdot 2\cdot \sqrt 6} \Rightarrow \overline{AC}^2 =10-2\sqrt 6\\ 在\triangle ABC中，\cos \angle ABC={1^2+(\sqrt 3)^2-(10-2\sqrt 6) \over 2\cdot 1\cdot \sqrt 3} = {2\sqrt 6-6\over 2\sqrt 3}=\sqrt 2-\sqrt 3\\ \Rightarrow |\cos \theta| =|\sqrt 2-\sqrt 3|= \bbox[red, 2pt]{\sqrt 3-\sqrt 2}$$

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