112年台中市國中教甄聯招-數學詳解

112 學年度臺中市政府教育局受託辦理本市立國民中學
(含本市立高級中等學校附設國中部)教師甄選

選擇題（共 40 題，每題 2.5分，共 100 分）

解答： $$\begin{cases}x\ge 1 \Rightarrow  & x+3-(x+1)-(x-1)=0 \Rightarrow x=3\\-3\le x\le 1 \Rightarrow  & x+3-(x+1)-(1-x)=0 \Rightarrow x=-1\\ x\le -3 \Rightarrow & -x-3-(x+1)-1+x=0 \Rightarrow x=-5\end{cases} \\ \Rightarrow 3-1-5=-3，故選\bbox[red, 2pt]{(A)}，公布的答案是\bbox[cyan,2pt]{(B)}$$

解答： $$f(x)=x^{-3} \Rightarrow f(0)不存在，故選\bbox[red, 2pt]{(D)}$$

解答： $$\lim_{n\to \infty}{\sin(1/n)\over (1/n)} =\lim_{n\to \infty}{\sin(1/n)\over (1/n)} =\lim_{n\to \infty}{-(1/n^2)\cos (1/n)\over -1/n^2}=\lim_{n\to \infty}\cos{1\over n}=1\\ 由極限比較審斂法(LCT)得知:\sum_{n=1}^\infty \sin{1\over n}與\sum_{n=1}^\infty {1\over n}同斂散，\\ 而\sum_{n=1}^\infty {1\over n}發散，因此\sum_{n=1}^\infty \sin{1\over n}發散，故選\bbox[red, 2pt]{(D)}$$

解答： $$假設A^{-1}=\begin{bmatrix}a & b \\c & d \end{bmatrix},依題意  \cases{A\begin{bmatrix}1 \\3 \end{bmatrix}= \begin{bmatrix}1 \\0 \end{bmatrix}\\ A\begin{bmatrix}4 \\2 \end{bmatrix}= \begin{bmatrix}0 \\2 \end{bmatrix}} \Rightarrow \cases{ \begin{bmatrix}1 \\3 \end{bmatrix}=A^{-1} \begin{bmatrix}1 \\0 \end{bmatrix} =\begin{bmatrix}a \\c \end{bmatrix}\\  \begin{bmatrix}4 \\2 \end{bmatrix}=A^{-1} \begin{bmatrix}0 \\2 \end{bmatrix}=\begin{bmatrix}2b \\2d \end{bmatrix}}\\ \Rightarrow \cases{a=1\\b=2\\ c=3\\ d=1} \Rightarrow A^{-1}=\begin{bmatrix}1 & 2 \\3 & 1 \end{bmatrix}，故選\bbox[red, 2pt]{(D)}$$

解答： $$H的階數必須是3的倍數，故選\bbox[red, 2pt]{(C)}$$

解答： $$假設\cases{A:戴戒指\\ B:戴項鍊}，依題意\cases{1-P(A\cup B)=0.6 \Rightarrow P(A\cup B)=0.4\\ P(A)=0.2\\ P(B)=0.3}\\ \Rightarrow P(A\cap B)=0.2+0.3-0.4=0.1，故選\bbox[red, 2pt]{(A)}$$

解答： $$16^2=256，故選\bbox[red, 2pt]{(D)}$$

解答： $$E[(X+1)^2]=E[X^2+2X+1]=E[X^2]+2E[X]+E[1]=E[X^2]+2+1=E[X^2]+3=8\\ \Rightarrow E[X^2]=5 \Rightarrow Var(X)=E[X^2]-(E[X])^2= 5-1=4 \Rightarrow Var(X)=4\\ 因此Var(1-3x)=9Var(X)=36，故選\bbox[red, 2pt]{(D)}$$

解答： $$f(x)={1\over x(x+1)(x+2)}={1\over 2}\left({1\over x(x+1)}-{1\over (x+1)(x+2)}\right)\\ \Rightarrow f(1)+f(2)+\cdots+f(10)={1\over 2}\left({1\over 2}-{1\over 11\times 12} \right) ={65\over 264}={a\over b} \Rightarrow a+b=329，故選\bbox[red, 2pt]{(D)}$$

解答：

$$取\cases{B(0,0)\\ A(0,4)\\ C(3,0)} \Rightarrow\cases{D(0,3)\\ E(3/4,3)} \Rightarrow \cases{L_1=\overleftrightarrow{CD}: x+y=3\\ L_2=\overleftrightarrow{BE}:y=4x} \Rightarrow F=L_1\cap L_2=(3/5,12/5) \\ \Rightarrow \cases{\overline{CD}=3\sqrt 2\\ \overline{DF}={3\over 5}\sqrt 2} \Rightarrow k={3\over 3/5}=5，故選\bbox[red, 2pt]{(C)}$$

解答：

$$將y=x-1代入y^2=2x+6 \Rightarrow x^2-2x+1=2x+6 \Rightarrow x^2-4x-5=0 \\ \Rightarrow (x-5)(x+1) =0 \Rightarrow \cases{x=-1 \Rightarrow y=-2\\ x=5 \Rightarrow y=4} \Rightarrow 交點\cases{A (-1,-2)\\B(5,4)}\\ 又\cases{y=x-1 \\ y^2=2x+6} \Rightarrow \cases{x=y+1 \\ x=(y^2-6)/2} \Rightarrow \int_{-2}^4 (y+1)-(y^2-6)/2\,dy = \int_{-2}^4 -{1\over 2}y^2+y+4\,dy\\，故選\bbox[red, 2pt]{(A)}$$

解答： $$圖1:\int_0^1 e^{\sqrt x}\,dx = \left. \left[ 2e^{\sqrt x}(\sqrt x-1)\right]\right|_0^1=2\\ 圖2: \int_0^1 xe^x\,dx = \left. \left[ e^x(x-1)\right]\right|_0^1=1\\ 圖3: \int_0^{\pi/2} e^{\sin x}\sin(2x)\,dx = \left. \left[ 2e^{\sin x}(\sin(x)-1) \right]\right|_0^{\pi/2} = 2\\ \Rightarrow 圖1面積=圖3面積，故選\bbox[red, 2pt]{(B)}$$

解答： $$f=\cos(xyz)-x^2y^2-z \Rightarrow \cases{f_x=-yz\sin(xyz)-2xy^2\\ f_y=-xz\sin(xyz)-2x^2y\\ f_z=-xy\sin(xyz)-1} \Rightarrow \nabla f(1,-1,0) =(-2,2,-1) \\ \Rightarrow 切平面:-2(x-1)+2(y+1)-z=0 \Rightarrow z=-2x+2y+4，故選\bbox[red, 2pt]{(B)}$$

解答： $$，故選\bbox[red, 2pt]{()}$$

解答： $$f\left({1+x\over 1-x} \right) =f\left({2\over 1-x}-1 \right) =x \Rightarrow f'\left({2\over 1-x}-1 \right)\cdot {2\over (1-x)^2} =1\\ x={1\over 3}\Rightarrow f'(2)\cdot {2\over 4/9}=1 \Rightarrow f'(2)={2\over 9}，故選\bbox[red, 2pt]{(C)}$$

解答： $$f'(x)\ge 6 \Rightarrow 斜率\ge 6 \Rightarrow f(4) \ge f(1)+(4-1)\times 6=30，故選\bbox[red, 2pt]{(A)}$$

解答： $$\begin{array}{} x & y &數量\\\hline 1& 1-11& 11\\ 2 & 1-11& 11 \\ 3 & 1-10& 10\\ 4& 1-9&9\\ 5 &1-7& 7\\\hline  \end{array} \Rightarrow 共有11+11+10+9+7=48組解，故選\bbox[red, 2pt]{(B)}$$

解答： $$所有偶數\cases{\square \square \square2 \\\square \square \square4 \\\square \square \square6 }\quad 各有5\times 4\times 3=60個，共60\times 3=180個\\ 需扣除3在百位數\cases{\square 3 \square2 \\\square 3 \square4 \\\square 3 \square6 }\quad 各有4\times 3=12個,共有12\times 3=36個\\ 因此合乎要求的有180-36=144個偶數，故選\bbox[red, 2pt]{(D)}$$

解答： $$，故選\bbox[red, 2pt]{()}$$

解答： $$假設\cases{紅球有a個\\ 白球有20-a個} \Rightarrow \cases{取到1紅1白的機率=a(20-a)/C^{20}_2\\ 取到2紅球的機率=C^a_2/C^{20}_2} \\ \Rightarrow 取到紅球個數的期望值={a(20-a)\over C^{20}_2}+2 {C^a_2\over C^{20}_2} ={19a\over 190} ={a\over 10} ={4\over 5} \Rightarrow a=8，故選\bbox[red, 2pt]{(A)}$$

解答： $$(x^2-2kx +k^2)+ (y^2-2ky+k^2)=k^2-4k+2a \Rightarrow (x-k)^2+(y-k)^2 = (k-2)^2+2a-4\\ \Rightarrow 圓半徑^2 =(k-2)^2+2a-4 \gt 0 \Rightarrow 2a-4\gt 0 \Rightarrow a\gt 2，故選\bbox[red, 2pt]{(D)}$$

解答： $$第6,7,8,9次進球合計24+14+12+22=72球\\ 假設\cases{前5次投籃共進a球\\ 第10次投籃投進b球} \Rightarrow \cases{前5次平均進球數=a/5\\ 前9次平均進球數=(a+72)/9\\ 投10次平均進球數=(a+b+72)/10} \Rightarrow \cases{(a+72)/9 \gt a/5\\ (a+b+72)/10 \gt 18} \\ \Rightarrow \cases{a\lt 90\\ a+b\gt 108} \Rightarrow a最大值為89 \Rightarrow b \gt 108-89=19 \Rightarrow b=20，故選\bbox[red, 2pt]{(C)}$$

解答： $$A與B都不發生的機率={4^3\over 6^3}={64 \over 216} \Rightarrow 至少發生A或B的機率=1-{64\over 216} ={152\over 216} ={19\over 27} \\ \Rightarrow a+b=19+27=46，故選\bbox[red, 2pt]{(D)}$$

解答：

$$取\cases{E(0,0)\\ B(0,4)\\ C(0,-4)\\ D(-8,0)} \Rightarrow F=(B+D)\div 2=(-4,2) \Rightarrow A=3F-2E=(-12,6)\\ \Rightarrow L=\overleftrightarrow{AC}:5x+6y+24=0 \Rightarrow k={\triangle ABC\over \triangle ACD} ={d(B,L)\over d(D,L)} ={48\over 16} =3，故選\bbox[red, 2pt]{(B)}$$

解答： $$f(x)=\cos^2(2x)+ 6\sin^2x+3 = \cos^2(2x)+6\cdot {1-\cos 2x\over 2}+3 = \cos^2(2x) -3\cos (2x)+6\\ =(\cos(2x)-{3\over 2})^2+{15\over 4} \Rightarrow \cases{最大值b=f(\pi/2)=10\\ 最小值a=f(0)=4} \Rightarrow 2b-a=20-4=16，故選\bbox[red, 2pt]{(C)}$$

解答： $$\lim_{n\to \infty} \sum_{k=1}^n {1\over \sqrt{k^2+nk+n^2}}= \lim_{n\to \infty} \sum_{k=1}^n {1/n\over \sqrt{(k/n)^2+(k/n)+1}}= \int_0^1 {1\over \sqrt{x^2+x+1}}\,dx \\=\left. \left[\ln \left| {2\over \sqrt 3}x+ {1\over \sqrt 3}+ {2\over \sqrt 3}\sqrt{(x^2+x+1)}\right| \right] \right|_0^1 =\ln(2+\sqrt 3)-\ln \sqrt 3 \\ =\ln{2+\sqrt 3\over \sqrt 3}=\ln \alpha \Rightarrow \alpha={2\over \sqrt 3}+1 =1+{2\sqrt 3\over 3}，故選\bbox[red, 2pt]{(B)}$$

解答： $$R=\{(x,y)\mid 0\le y\le \sqrt{2x-x^2}\} 為圓(x-1)^2+y^2=1的上半部\\取\cases{x=r\cos \theta\\ y=r\sin \theta}，由於y\le \sqrt{2x-x^2} \Rightarrow x^2+y^2\le 2x \Rightarrow r^2\le 2r\cos \theta \Rightarrow r\le 2\cos \theta\\ 因此\iint_R 9\sqrt{x^2+y^2}\,dA = \int_0^2 \int_0^\sqrt{2x-x^2} 9\sqrt{x^2+y^2}\,dydx = \int_0^{\pi/2} \int_0^{2\cos \theta} 9r^2\,drd\theta \\= \int_0^{\pi/2 } 24\cos^3\theta\,d\theta = \left.\left[ 24\sin x-8\sin^3 x \right]\right|_0^{\pi/2} =16，故選\bbox[red, 2pt]{(D)}$$

解答： $$A=\begin{bmatrix}1 & -1 & 0 \\-1 & 3 & -1 \\0 & -1 &2\end{bmatrix}\Rightarrow  \det(A-\lambda I) = \lambda^3- 6\lambda^2+9\lambda-3 =0\Rightarrow \cases{\alpha_1 +\alpha_2 +\alpha_3=6 \\ \alpha_1 \alpha_2 +\alpha_2\alpha_3 +\alpha_3\alpha_1=9\\\alpha_1 \alpha_2 \alpha_3=3} \\ \Rightarrow (\alpha_1 +\alpha_2 +\alpha_3)^2= \alpha_1^2 +\alpha_2^2 +\alpha_3^2+ 2(\alpha_1 \alpha_2 +\alpha_2\alpha_3 +\alpha_3\alpha_1) \Rightarrow 36=\alpha_1^2 +\alpha_2^2 +\alpha_3^2+18 \\ \Rightarrow \alpha_1^2 +\alpha_2^2 +\alpha_3^2=36-18=18 \\又 \alpha_1^3 +\alpha_2^3 +\alpha_3^3-3\alpha_1 \alpha_2 \alpha_3 =(\alpha_1 +\alpha_2 +\alpha_3)(\alpha_1^2 +\alpha_2^2 +\alpha_3^2-(\alpha_1 \alpha_2 +\alpha_2\alpha_3 +\alpha_3\alpha_1)) \\ \Rightarrow \alpha_1^3 +\alpha_2^3 +\alpha_3^3-9=6(18-9) \Rightarrow \alpha_1^3 +\alpha_2^3 +\alpha_3^3=54+9=63，故選\bbox[red, 2pt]{(C)}$$

解答： $${x+y/2+y/2+z/3+z/3+z/3\over 6}\ge \sqrt[6]{xy^2z^3\over 36}\\ xy^2z^3有極有值時，x={y\over 2}={z\over 3} \Rightarrow \cases{z=3x\\ y=2x} \Rightarrow x+y+ z=6x =2 \Rightarrow \cases{x=1/3 =\alpha\\ y=2/3 =\beta\\ z=1= \gamma} \\ \Rightarrow 3\alpha+\gamma =1+1=2，故選\bbox[red, 2pt]{(A)}$$

解答： $$令\cases{a=\sqrt[3]{\sqrt{17}+3} \\b= \sqrt[3]{\sqrt{17}-3}} \Rightarrow x=a-b \Rightarrow x^3=(a-b)^3 =a^3-b^3 -3ab(a-b) =6-3\sqrt[3]{8}x\\ \Rightarrow x^3=6-6x \Rightarrow x^3+6x+7=6-6x+6x+7=13，故選\bbox[red, 2pt]{(D)}$$

解答： $$\lim_{k\to \infty}\left| {a_{k+1}\over a_k}\right| =\lim_{k\to \infty}\left| {{(k+1)!\over (k+1)^{k+1}}(x-1)^{k+1} \over {k!\over k^k}(x-1)^k}\right| =\lim_{k\to \infty}\left| {({k\over k+1}})^k (x-1)\right| = \left| {{1\over e}}  (x-1)\right|  \lt 1 \\ \Rightarrow 1-e\lt x\lt 1+e \Rightarrow \cases{a=1-e\\ b=1+e} \Rightarrow a+2b=1-e+2+2e=3+e，故選\bbox[red, 2pt]{(D)}$$

解答： $$\sin x =x-{x^3\over 3!}+{x^5\over 5!}-\cdots \Rightarrow \sin{1\over x} ={1 \over x}-{1\over 3! x^3} +{1\over 5! x^5}-\cdots \\ \Rightarrow x\sin{1\over x} =1-{1\over 3!x^2}+ {1\over 5!x^4} -\cdots \cdots(1)\\ 而\lim_{x\to \infty} \sqrt{x^2-x}-x =\lim_{x\to \infty} {-x\over \sqrt{x^2-x}+x} =\lim_{x\to \infty} {-1\over \sqrt{1- 1/x}+1} =-{1\over 2}\cdots(2) \\ 由(1)及(2)可得: \lim_{x\to \infty} x\left(\sqrt{x^2-x}-x \right) \sin({1\over x}) =-{1\over 2}，故選\bbox[red, 2pt]{(A)}$$

解答： $$f(t)\in P_2(\mathbb R) \Rightarrow f(t)=at^2+bt +c, a,b,c\in \mathbb R \Rightarrow f'(x)=2ax+b\\ \Rightarrow T(f)(x) =3(2ax+b)+ \int_0^x (2at^2+ 2bt +2c) \,dt ={2\over 3}ax^3+ bx^2 +(6a+2c)x+3b\\ \text{If }T(f)(x) =0, \text{ then we have }a=b=c=0 \Rightarrow Null(T)=\{0\} \Rightarrow \text{nullity}(T)=0，故選\bbox[red, 2pt]{(A)}$$

解答： $$\det(A-\lambda I)=0 \Rightarrow -\lambda^3+8\lambda^2-4\lambda-48=0 \Rightarrow -(\lambda+2)(\lambda-4) (\lambda-6)=0\\ \Rightarrow 特徵值=-2,4,6，故選\bbox[red, 2pt]{(B)}$$

解答： $$取\cases{p=a-3\\ q=b-2\\ r=c-1}  \Rightarrow a+b+c\le 15 \Rightarrow p+q+r\le 9 \Rightarrow 共有H^4_9=220組整數解，故選\bbox[red, 2pt]{(Ｂ)}$$

解答： $$\cases{\cases{x+y+1=0\\ x-y+3=0} \Rightarrow (x,y)=(-2,1)\\ \cases{x+2y+4=0\\ x-2y+12=0} \Rightarrow (x,y)=(-8,2)} \Rightarrow r=4,s=2 \Rightarrow r+s=6 ，故選\bbox[red, 2pt]{(B)}$$

解答： $$A=\begin{bmatrix}4 & 4 \\-1 & -1 \end{bmatrix} =\begin{bmatrix}-1 & -4\\1 & 1 \end{bmatrix} \begin{bmatrix}0 & 0 \\0 & 3 \end{bmatrix} \begin{bmatrix}1/3 & 4/3 \\-1/3 & -1/3 \end{bmatrix} \equiv PDP^{-1} \\ \Rightarrow A^n=PD^nP^{-1} \Rightarrow A+ A^2+\cdots +A^n =P(D+D^2+\cdots +D^n)P^{-1} \\=P\begin{bmatrix}0 & 0 \\0 & 3+3^2+\cdots +3^n \end{bmatrix} P^{-1} =P\begin{bmatrix}0 & 0 \\0 & (3^{n+1}-3)/2 \end{bmatrix} P^{-1} =\begin{bmatrix} 2(3^n-1) & 2(3^n-1) \\{1\over 2}(1-3^n) & {1\over 2}(1-3^n) \end{bmatrix} \\ \Rightarrow b+c={1\over 2}(1-3^n)+ {1\over 2}(1-3^n)=1-3^n，故選\bbox[red, 2pt]{(B)}$$

解答： $$f(x,y)=x^3-y^3+6x^2+ 2x+y+2 =(x^3+ 3x^2 +2x)+3x^2 -(y^3-y)+2 \\=x(x^2+3x+2)+ 3x^2-y(y^2-1)+2= x(x+1)(x+2)+3x^2-(y-1)y(y+1)+2 \\ 連續三整數之積必為3的倍數 \Rightarrow \cases{x(x+1)(x+2)=3k\\ 3x^2=3m\\ (y-1)y(y+1)= 3n} \\ \Rightarrow f(x,y)=3(k+m-n)+2 \ne 0, \forall k,m,n\in \mathbb Z，故選\bbox[red, 2pt]{(A)}$$

解答： $$\lim_{n\to \infty}\left[2n^2 \left({r\over n}-\sum_{k=1}^r {1\over n+k}\right) \right] =\lim_{n\to \infty}\left[2nr -\sum_{k=1}^r {2n^2\over n+k} \right]\\ =\lim_{n\to \infty}\left[2nr -\sum_{k=1}^r \left((2n-2k)+{2k^2\over n+k} \right) \right] =\lim_{n\to \infty}\left[2nr - \left(2nr-r(r+1)+\sum_{k=1}^r{2k^2\over n+k} \right) \right] \\=\lim_{n\to \infty}\left[r(r+1)-\sum_{k=1}^r{2k^2\over n+k}  \right] =r(r+1)-\lim_{n\to \infty}\sum_{k=1}^r{2k^2\over n+k}  =r(r+1)，故選\bbox[red, 2pt]{(D)}$$

解答： $${b\over a}={|b^2+c^2-a^2|\over bc} ={b^2+c^2-a^2\over bc} (c\gt b\gt a \Rightarrow c^2+b^2-\gt a^2 )\\ \Rightarrow {b\over 2a}={b^2+c^2-a^2\over 2bc} \Rightarrow {\sin B\over 2\sin A}=\cos  A \Rightarrow \sin B=\sin(2A) \Rightarrow \angle B=2\angle A\\ 同理,{c\over b}={|c^2+a^2-b^2| \over ca} \Rightarrow \sin C= \sin(2B) \Rightarrow C=2\angle B \\ 因此\angle A+ \angle B+ \angle C=\angle A+2\angle A+4\angle A=7\angle A=\pi \Rightarrow \angle A={\pi \over 7}，故選\bbox[red, 2pt]{(A)}$$

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